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14500	https://www.tutoringhour.com/worksheets/integers/opposite-value/	Search Functionality Update! To optimize your search experience, please refresh the page. Windows: Press Ctrl + F5 Mac: Use Command + Shift+ R or Command + Option + R Mobile: Tap and hold the refresh icon, then select "Hard Refresh" or "Reload Without Cache" for an instant upgrade! See More Sign In Child Login Sign Up Math Worksheets By Grade KG 1 2 3 4 5 6 7 8 By Topic Number Sense Measurement Statistics Geometry Pre Algebra Algebra ELA Worksheets By Grade KG 1 2 3 4 5 6 7 8 By Topic Grammar Vocabulary Figurative Language Phonics Reading Comprehension Reading and Writing Science Worksheets Social Studies Worksheets Interactive Worksheets MATH KG 1 2 3 4 5 6 7 8 ELA KG 1 2 3 4 5 6 7 8 Browse By Grade MATH KG 1 2 3 4 5 6 7 8 ELA KG 1 2 3 4 5 6 7 8 Log In Child Login Math Worksheets ELA Worksheets Science Worksheets Social Studies Worksheets Interactive Worksheets Browse By Grade Become a Member Math Worksheets By Grade KG 1 2 3 4 5 6 7 8 By Topic Number Sense Measurement Statistics Geometry Pre Algebra Algebra ELA Worksheets By Grade KG 1 2 3 4 5 6 7 8 By Topic Grammar Vocabulary Figurative Language Phonics Reading Comprehension Reading and Writing Interactive Worksheets Math KG 1 2 3 4 5 6 7 8 ELA KG 1 2 3 4 5 6 7 8 Browse By Grade Math KG 1 2 3 4 5 6 7 8 ELA KG 1 2 3 4 5 6 7 8 Math Worksheets Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Charts Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Numbers Prime and Composite Numbers Roman Numerals Ordinal Numbers Patterns Rounding Fractions Decimals Integers Measurement Big vs Small Long vs Short Tall vs Short Heavy vs Light Full, Half-Full, or Empty Time Calendar Money Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature Statistics and Data Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Mean Mean, Median, Mode, Range Mean Absolute Deviation Stem and Leaf Plot Box and Whisker Plot Factorials Permutations Combinations Geometry Positions 2D Shapes 3D Shapes Lines, Rays, and Line Segments Points, Lines, and Planes Angles Symmetry Transformation Area Perimeter Circles Ordered Pairs Slope Midpoint Formula Distance Formula Parallel and Perpendicular Lines Surface Area Volume Pythagorean Theorem Pre-Algebra Factors GCF LCM Fractions Integers Decimals Significant Figures Percent Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Scientific Notation Logarithms Absolute Value Algebra Translating Algebraic Phrases Simplifying Algebraic Expressions Evaluating Algebraic Expressions Equations Systems of Equations Slope of a Line Equation of a Line Quadratic Equations Functions Polynomials Inequalities Determinants Arithmetic Sequence Arithmetic Series Geometric Sequence Complex Numbers Trigonometry ELA Worksheets Science Worksheets Social Studies Worksheets Support Opposite Value of Integers Worksheets Math > Pre-Algebra > Integers > Opposite Value Gather momentum with our printable worksheets on opposite values of integers! Each pdf contains three sections that will ensure students know opposite integers like the back of their hand! First, they write down the opposite values of positive and negative integers. Next, they mark the given integer as well as its opposite value on number lines. Finally, they find the opposite values of expressions such as +(−5) and −(−2); the opposite values are 5 and −2 in this case. Remind them that adding a pair of opposite integers will give the sum as the zero, and let them carefully verify the solutions with the included answer key. We recommend these worksheets for 6th grade and 7th grade students. Learn to Solve Grab the Worksheet Grab the Worksheet Grab the Worksheet Related Printable Worksheets ▶ Absolute Value and Opposite Integers ▶ Integers on a Number Line ▶ Additive Inverse Tutoringhour What We Offer Math Worksheets ELA Worksheets Science Worksheets Social Studies Worksheets Interactive Worksheets Lessons Games Membership Benefits How to Use Interactive Worksheets How to Use Printable Worksheets FAQ Printing Help What's New Testimonial About Us Privacy Policy Refund Policy Terms Contact Us Copyright © 2025 - Tutoringhour You must be a member to unlock this feature! Sign up now for only $29.95/year — that's just 8 cents a day! Printable Worksheets 20,000+ Worksheets Across All Subjects Access to Answer Key Add Worksheets to "My Collections" Create Custom Workbooks Interactive Practice 2000+ Math and ELA Worksheets Preview and Assign Worksheets Create Groups and Add Children Track Progress
14501	https://physics.stackexchange.com/questions/199622/how-to-determine-the-direction-of-induced-current-flow	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How to determine the direction of induced current flow? Ask Question Asked Modified 7 years, 1 month ago Viewed 280k times 6 $\begingroup$ There are three ways of inducing current in a loop/coil of wire as shown in my book. We can have a magnet approach a coil of wire, or a wire approaching a magnet. Both can be understood in the same way. On the other hand, we can also change the magnetic flux by pushing a loop of wire into a magnetic field. This confuses me. Two vertical lines are cutting across the field. But since they are connected, the induced current, I conjecture, would cancel each other. I saw in a YouTube video that to determine the direction in such situations as 2, one curl the fingers of their right hand along the wire, with the thumb pointing in the direction of the field. So the curled fingers are in the direction of the current. Basically the directions of current indicated by the thumb is always opposite the direction of the changing field. magnetic-fields Share Improve this question edited Aug 25, 2018 at 5:59 knzhou 108k2424 gold badges306306 silver badges511511 bronze badges asked Aug 10, 2015 at 16:31 most venerable sirmost venerable sir 1,3811414 gold badges3232 silver badges5353 bronze badges $\endgroup$ Add a comment \| 4 Answers 4 Reset to default 7 $\begingroup$ The rule is called Lenz's Law. You already appear to know how to determine the direction of the magnetic field due to a current in a loop, which is part of the answer. What Lenz's Law tells us is that the direction of the induced current in the loop is such that it "opposes the change in the flux". Here's a picture I grabbed from to illustrate this. The B-field is pointing up and increasing. So we say there is an "induced B field" opposing this increase (so it points down). The current in the loop is such that it would create this induced field according to the usual right-hand rule for B-fields due to loops. So imagine that the B-field was decreasing instead. So if B is up the change in the B-field is down. Thus the induced B-field would point up and the current would be opposite to what is in the diagram. Here is another diagram from the same website showing some other cases. Practice working through the reasoning of Lenz's Law to verify that you see why the induced current is the in the direction indicated for each case: A word of warning about this concept. The change in the B-field does not really induce an "induced B-field". What it really does is create a circular E-field, and it is this which drives the current in the loop. The "induced B-field" is a fictional construct which makes it easier to figure out the direction of the current. Share Improve this answer answered Aug 12, 2015 at 20:06 gleedadswellgleedadswell 1,26366 silver badges99 bronze badges $\endgroup$ 4 $\begingroup$ Thanks for the warning. It's doesn't really bother me at this level. $\endgroup$ most venerable sir – most venerable sir 2015-08-14 23:01:31 +00:00 Commented Aug 14, 2015 at 23:01 $\begingroup$ Is there a trick for searching up good sources, or did you spend a lot of time searching? $\endgroup$ most venerable sir – most venerable sir 2015-08-14 23:03:14 +00:00 Commented Aug 14, 2015 at 23:03 $\begingroup$ The method I learned from the video was actually new to me. The method is basically curling the four fingers in the direction of the current with the thumb pointing against the direction of incoming B-field. But my method is actually 1. Outstretching the four fingers in the direction of B-field (the trick is to do this at a place where the B-field lines are perpendicular to the segment of the loop) 2. Palm facing in the direction of the force (so say if the magnet is coming towards the loop, it can also be thought of as the loop coming towards the magnet, so palm will be facing magnet).... $\endgroup$ most venerable sir – most venerable sir 2015-08-14 23:11:34 +00:00 Commented Aug 14, 2015 at 23:11 $\begingroup$ (Continued)...3. Then finally the thumb will be in the direction of the wrong current flow (according to Lenz's Law). So the correction current direction will the opposite. But Does my method always work though? (I started using it because it resembles other methods) $\endgroup$ most venerable sir – most venerable sir 2015-08-14 23:12:49 +00:00 Commented Aug 14, 2015 at 23:12 Add a comment \| 2 $\begingroup$ I think the main idea that you are missing is that the induced current is caused by the change in the magnetic "flux" (which is not the same as the magnetic field). Without resorting to any calculus (which is needed for the real definition of flux) the flux is like the magnetic field times the cross-sectional area that it "goes through" inside your coil/loop. So there are two key ideas here to be able to understand all cases: It is the flux, not the field, that matters. It is the rate of change of flux, not the absolute size of the flux, that matters. So, in situation 1 the flux is changing because the field strength inside the loop is changing. For the magnet approaching the coil the strength of the field inside the coil is increasing, so this makes the flux also decrease. For the magnet moving away from the coil the strength of the field inside the coil is decreasing, so the flux is also decreasing. This increasing/decreasing difference is why the induced current direction depends on which way you move the magnet. Notice that if you hold the magnet stationary no current is induced (remember, it is the change in flux that matters). Now for situation 2 maybe it is easier to think about having a coil whose diameter you can change. If you increase the diameter (holding the nearby magnet stationary) then the field strength is the same but the flux increases because the area that the field "goes through" has increased. So this would have the same effect as increasing the field, such as by moving the magnet closer. Similarly, decreasing the coil diameter will have the same effect as moving the magnet farther away. So finally, looking at the scanned picture for your situation 2 we have a coil moving from a place where there is no field to a place where there is a field. So the flux increases as the loop moves into this region and a current will be induced. How would this be achieved in practice? This would be difficult but it would be easy to move a coil from a place with a weak field to a place with a strong field. Hope that helps! In my opinion induced currents are hands-down the most difficult things to figure out in elementary electromagnetism, so don't be surprised that you're having some trouble wrapping your head around them. Share Improve this answer answered Aug 10, 2015 at 17:20 gleedadswellgleedadswell 1,26366 silver badges99 bronze badges $\endgroup$ 2 $\begingroup$ What is then a good method using hands to determine direction? $\endgroup$ most venerable sir – most venerable sir 2015-08-12 02:09:20 +00:00 Commented Aug 12, 2015 at 2:09 $\begingroup$ I answer this in the post below... $\endgroup$ gleedadswell – gleedadswell 2015-08-12 20:06:51 +00:00 Commented Aug 12, 2015 at 20:06 Add a comment \| 2 $\begingroup$ The Maxwell equation says $$\operatorname{curl} \vec E = - {\partial \vec B \over \partial t},$$where $\vec B$ is the magnetic field and $\vec E$ is the induced electric field.That may appear, to an elementary student, to be a bunch of gobbledygook, but it means "the electric field curls clockwise around a change in magnetic field;" normally the + orientation is counterclockwise by the right-hand rule; the negative sign makes it clockwise. Okay, so we need to first define "magnetic field." One popular way to see the lines of a magnetic field directly to look at the effect of the magnets on iron filings, which will naturally show you some "lines" when you bring a magnet nearby; see these images if you've never seen the effect before. The idea of "magnetic field" is basically just that we're going to take these lines and add an idea of "forwards" or "backwards" to them: these lines therefore "come out" of the North pole of a magnet and "go into" the South pole of a magnet. Please reread this paragraph until that convention is firm in your head. You will know it is when you can appreciate that the Earth's North Pole must be the South pole of some big magnet, which is why the North poles of magnets point North: magnets generally want to align with an existing magnetic field, which means the magnetic field at the surface must point Northward, which means it must be going into Earth's North Pole, which makes it a South magnetic pole. Now you need to understand the change in magnetic field. The magnetic field has both a strength (how close the field lines are together) and a direction (the direction the iron filings point, combined with the orientation forwards/backwards defined above). If a change increases the strength of the magnetic field, like when you get closer to a bar magnet, we point the change in the same direction as the magnetic field. But if the magnetic field gets weaker, then the change points opposite. If the direction of the magnetic field is changing, then we have to also include a component which points perpendicular to the original magnetic field, pointing in the direction that it changes. The full description of how to do this is known as "vector calculus" and I can only give a couple of basic guidelines on how to calculate these "changes" without that framework. Now: point your left-hand thumb in the direction of the change of magnetic field: then your fingers curl in the orientation of the induced electric field. It is "clockwise" when you look at your hand thumb-on, or when you look at the change pointing towards you. This means that if a current follows that curling, it goes to a higher voltage; or if it opposes that curling, it goes to a lower voltage. This same "reverse" rule can also be phrased as Lenz's law. This says that induction works like inertia: changing magnetic fields produce electric fields that would cause a current that would oppose the change. As you may know, a wire also produces a magnetic field. The direction of this magnetic field looks like this: point your right thumb in the direction of the current, then your fingers curl in the direction of the induced magnetic field. If you combine your hands together, pointing your left thumb up to indicate an upward change of magnetic field, then putting your right thumb against your left index finger, you'll see that your right fingers curl into your palm, opposite your left thumb. Lenz's law gives some awesome quick intuitions. For example: the basic electrical component known as an inductor is a loop of wire usually wrapped around some ferromagnetic material. By Lenz's law, when you try to change the current that's going through it, it induces a voltage which tries to keep the same current going through it. It's like an inertial term, it fights any change in the electron velocity. So as you try to ramp up the current, you fight its voltage; as you try to ramp down the current the same thing happens. Share Improve this answer answered Aug 12, 2015 at 22:01 CR DrostCR Drost 39.6k33 gold badges4444 silver badges119119 bronze badges $\endgroup$ 1 $\begingroup$ Amazing answer. Some questions. Q1 Is the "component which points perpendicular to the original magnetic field" supposed to be a curve? Q2 what is the significance of saying (last sentence of the second to last paragraph) "if you combine your...opposite your left thumb"? (I can't picture it in my head due to the lack of sufficient description. I thought you were to show me how the magnetic field around a current-carrying wire look, which I think is sufficiently described in the second to last sentence of that paragraph.) $\endgroup$ most venerable sir – most venerable sir 2015-08-15 01:54:26 +00:00 Commented Aug 15, 2015 at 1:54 Add a comment \| 0 $\begingroup$ The Fleming right hand rule say that:Hold the thumb,the forefinger and the centre finger of your right-hand at angle to one another.Adjust your hand in such a way that four finger point in the direction of magnetic field,and thumb point in the direction of motion of conductor. Share Improve this answer answered Sep 22, 2015 at 5:21 PrateekPrateek 1 $\endgroup$ 1 $\begingroup$ True, but Fleming's right hand rule is only useful in some cases. It is useful when you have a single conductor moving through a B-field, which is connected to a circuit such that the rest of the circuit is not moving through a field. So it works for the cases in the original poster's second diagram. But for cases where we have a stationary loop of wire in a changing field Fleming's right hand rule doesn't tell us anything. Lenz's Law is more powerful than Fleming's right hand rule because it can be applied to any situation. $\endgroup$ gleedadswell – gleedadswell 2022-06-24 19:23:02 +00:00 Commented Jun 24, 2022 at 19:23 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions magnetic-fields See similar questions with these tags. 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14502	https://pmc.ncbi.nlm.nih.gov/articles/PMC11691105/	Acquisition of Certifiable Competencies in Undergraduate Medical Curriculum of National Medical Commission: Role of Rapid Cycle Deliberate Practice - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice editorial Int J Appl Basic Med Res . 2024 Nov 1;14(4):211–213. doi: 10.4103/ijabmr.ijabmr_478_24 Search in PMC Search in PubMed View in NLM Catalog Add to search Acquisition of Certifiable Competencies in Undergraduate Medical Curriculum of National Medical Commission: Role of Rapid Cycle Deliberate Practice Rajiv Mahajan Rajiv Mahajan 1 Department of Pharmacology, Adesh Institute of Medical Sciences and Research, Bathinda, Punjab, India Find articles by Rajiv Mahajan 1,✉ Author information Article notes Copyright and License information 1 Department of Pharmacology, Adesh Institute of Medical Sciences and Research, Bathinda, Punjab, India ✉ Address for correspondence: Dr. Rajiv Mahajan, Department of Pharmacology, Adesh Institute of Medical Sciences and Research, Bathinda - 151 101, Punjab, India. E-mail: drrajivmahajan01@gmail.com Received 2024 Oct 6; Revised 2024 Oct 15; Accepted 2024 Oct 17; Issue date 2024 Oct-Dec. Copyright: © 2024 International Journal of Applied and Basic Medical Research This is an open access journal, and articles are distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as appropriate credit is given and the new creations are licensed under the identical terms. PMC Copyright notice PMCID: PMC11691105 PMID: 39749170 On September 12, 2024, the National Medical Commission (NMC) released revised “Guidelines for Competency-Based Medical Education (CBME) Curriculum 2024,” to be implemented from MBBS admission batch 2024-2025. Along with the guidelines, the CBME topics and competencies, i.e., Volume I, II, and III of the “Competency-Based Undergraduate Curriculum for the Indian Medical Graduate 2024” were also released in the same document. Volume I of the curriculum contains competencies of Anatomy, Physiology, Biochemistry, Pathology, Microbiology, Pharmacology, and Forensic Medicine and Toxicology. Volume II of the curriculum contains competencies of Medicine and allied subjects (including community medicine), whereas Volume III contains competencies of Surgery and allied subjects. Except for the subjects of Anatomy, General Surgery, Radiology, and Orthopedics, the curriculum has notified various certifiable competencies for other subjects [Table 1]. The number of times that competency needs to be performed correctly so as to get a certification has also been mentioned as “number required to certify P,” as a last column in the curriculum table. This “number required to certify P” varies from 1 to 5 for different competencies. Table 1. Number of certifiable competencies of different subjects \| Subject \| Total number of competencies \| Number of certifiable competencies \| :---: \| Physiology \| 136 \| 10 \| \| Biochemistry \| 84 \| 10 \| \| Pharmacology \| 92 \| 11 \| \| Pathology \| 182 \| 9 \| \| Microbiology \| 74 \| 7 \| \| Forensic medicine \| 158 \| 6 \| \| Community medicine \| 136 \| 10 \| \| General medicine \| 525 \| 22 \| \| Pediatrics \| 406 \| 15 \| \| Psychiatry \| 17 \| 2 \| \| Dermatology \| 48 \| 2 \| \| Ophthalmology \| 60 \| 1 \| \| Otorhinolaryngology \| 63 \| 6 \| \| Obstetrics and Gynecology \| 141 \| 16 \| \| Anesthesiology \| 52 \| 4 \| Open in a new tab Let’s examine what the guidelines state about acquisition of these certifiable competencies. The guidelines explicitly state that “learners must have completed the required certifiable competencies for that phase of training and completed the logbook appropriate for that phase of training to be eligible for appearing at the final university examination of that subject.” This indicates that the record of number of times the competency/skill is performed correctly so as to complete the desired “number required to certify P,” must be entered in the logbook, and all such sessions will be assessment sessions. So if a competency is required to be correctly performed five times for certification purposes, provision must be provided in the logbooks to record those five encounters. No learner will be able to perform any skill/competency correctly from the word go. Every learner will have to attend some learning sessions along with learning attempts before fully learning to perform the skill/competency correctly. Even after learning, there are chances that learner may fail to perform it correctly on 3 rd or 4 th attempt and need remedial sessions. Provision must be provided in the logbooks to record these learning, remedial, and assessment sessions. Can a learner make so many attempts on the patients, to learn a skill/competency? It is noteworthy that for the bulk of the psychomotor/procedural skill-related competencies, simulation-based learning is the recommended method, given in the curriculum. The suggested teaching–learning method for most of these competencies is by “Demonstration-Observation-Assist-Perform (DOAP)” method. The effectiveness of simulation-based learning, particularly the simulations using part-task trainers, in the acquisition of skills in medical training is well documented in the literature.[2,3] With the introduction of the CBME curriculum from the MBBS admission batch 2019, the effectiveness of the DOAP method for the acquisition of practical and clinical skills has been well established in Indian medical colleges.[4,5] Any simulation-based session is structured into three phases: prebriefing, immersive experience, and debriefing. Prebriefing is concerned with the preparation and a briefing to the learners before the commencement of the experiential learning activity. During the immersive phase, the simulation in actual is performed and learned. This is the phase when the DOAP method will be used for learning the concerned skill. Debriefing is the phase involving reflection on action and feedback. Various models of debriefing have been documented in literature. Let’s evaluate why the “Rapid Cycle Deliberate Practice (RCDP)” method is the “best fit model” of debriefing for learning procedural skill-based certifiable competencies, using simulation-based learning. As discussed above, the acquisition of certifiable competencies will require multiple learning and remedial sessions, repeated at short intervals. Moreover, the learners are not going to practice the learned skill on humans immediately after learning. There will be a long gap between the time of acquisition of skills in simulation environment and its actual practice on humans, thereby leading to knowledge attrition. The undergraduate students are also not mature enough to undergo learner-led debriefing sessions as stipulated in other popular models. All these factors favor the adoption of the RCDP model for debriefing and learning procedural skills in simulation-based learning. Let’s examine the theoretical underpinnings of RCDP. Literature has documented that RCDP sessions typically involve repeated performance of a simulation activity by the learners at short intervals, with facilitator-led microbriefing, where feedback sessions are more frequent compared to traditional postsimulation feedback. The repetitive simulation activity with microbriefing has been documented to minimize knowledge attrition also. As one can decipher, the requirements of learning procedural skills using simulation-based learning are perfectly complemented by the conceptual basis of RCDP. Studies have also demonstrated the better efficacy of RCDP in learning procedural skills using simulation-based learning compared to traditional post-simulation feedback.[11,12] As elaborated above, most of the certifiable competencies of the MBBS curriculum of NMC are to be acquired through simulation-based learning using the DOAP method of skill acquisition, and most of the certifiable competencies are procedural skills. The learners are not going to practice the learned skill on humans immediately. Learners will need multiple learning and remedial sessions, with recording of each session on logbook. So undoubtedly, combining RCDP for debriefing with DOAP for immersive simulation activity is the best bet for acquiring certifiable competencies. Subsequent assessment sessions can be planned using simulations – with or without prebrief and debrief. Direct observation will be the best tool to assess these skills/competencies for certification purposes. The conceptualized framework for learning and assessment of all such procedural skill-based certifiable competencies has been depicted in Figure 1. Figure 1. Open in a new tab A typical framework for learning and assessment of a skill-based certifiable competency of the National Medical Commission undergraduate curriculum Let’s try to understand the framework through an example. In Pharmacology, the competency number PH 1.5 in the revised curriculum is – Describe various routes of drug administration, their advantages and disadvantages and demonstrate administration of, e.g., SC, IV, IM, SL, rectal, spinal, sublingual, intranasal sprays, and inhalers. The skill part has been suggested to be acquired through DOAP method using simulations. If we choose one objective of this competency as – demonstrate administration of intramuscular injection correctly on part-task trainer, the “number required to certify P” has been mentioned as two. This means the student will be required to correctly demonstrate the skill of administrating intramuscular injection two times on part-task trainer (assessment sessions), before he/she can be certified to acquire the skill. He/she will require multiple learning sessions too. Need of remedial sessions may be felt as well. The framework depicted in Figure 1 can be used for arranging learning and remedial sessions using DOAP and RCDP methods. And when the students are supposed to demonstrate this skill on the real patients? The competency number GM 1.27 of General Medicine is – Administer an intramuscular injection with an appropriate communication to the patient. This reflects that the skill learned during Phase 2 of the training is going to be demonstrated on real patients in Final Phase Part I/II. Such a long gap can lead to attrition of the knowledge. So, even after certification of skill in Phase 2, RCDP sessions should be arranged during this gap period. In nutshell, RCDP method is ‘the best fit’ method for the acquisition of skills using part-task trainers in simulation environment, thereby promoting deliberate practice and preventing knowledge attrition, as skills are repeated at short intervals with microbriefings. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Numbers of combinations modulo m, efficiently Ask Question Asked 2 years, 4 months ago Modified2 years, 3 months ago Viewed 3k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. First of all I'm solving a programming problem rather than a math problem now. The question is Anish got an unbiased coin and he tossed it n times and he asked Gourabh to count all the number of possible outcomes with j heads, for all j from 0 to n. Since the number of possible outcome can be huge, he will tell the values modulo m. To be clear, we need to return one integer per value of j. The question is simple, but the problem arises with the time limit, being 1.5 seconds, but with input n as large as 200000. I used math.comb to calculate the values, but it took more than 1.5 seconds to run. So, are there any ways to calculate combinations in a faster way? Edit#1: Sample input: 2 998244353 Sample output: 1 2 1 Edit#2: Here is the code that I've tried: ```python import math n,m=input().split() n = int(n) m = int(m) l = [] for i in range(n+1): l.append(math.comb(n,i)%m) print(l) ``` P.S: Please let me know if this is off topic for this site and suggest a suitable SE site to post this question. Thanks in advance! This question is from an inter college contest which ended 2 months ago. Here is the original problem: (you'll need an account, and first time follow the "Contest Link" here to enter). In case you are not able to view the problem, please find the picture below. python algorithm performance math combinations Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited May 29, 2023 at 23:56 Stefan Pochmann 29k 9 9 gold badges 48 48 silver badges 116 116 bronze badges asked May 29, 2023 at 13:35 AdithyaAdithya 238 4 4 silver badges 12 12 bronze badges 26 9 Since it's modulo m, I strongly suspect that there is some math you can do here to convert it into an easy problem. If so, it is more a math problem than a programming problem. If you think I may be right, consider math.stackexchange.com .Berthur –Berthur 2023-05-29 13:39:21 +00:00 Commented May 29, 2023 at 13:39 2 @Berthur I have posted it on Math.SE, but they suggested to post it here :)Adithya –Adithya 2023-05-29 13:40:04 +00:00 Commented May 29, 2023 at 13:40 1 Is there any constraint on m in the original question? If m is a prime number, then this is definitely a question about using Lucas' theorem. Note that 998244353 is, in fact, a prime number.ken –ken 2023-05-29 14:04:33 +00:00 Commented May 29, 2023 at 14:04 1 It's probably enough to just compute the next value from the previous value instead of recomputing from scratch every time.user2357112 –user2357112 2023-05-29 15:57:20 +00:00 Commented May 29, 2023 at 15:57 1 Think about the mathematical relationship between n choose i and n choose i+1.user2357112 –user2357112 2023-05-29 15:58:14 +00:00 Commented May 29, 2023 at 15:58 \|Show 21 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 12 Save this answer. Show activity on this post. Using the usual multiplicative formula to compute the next number from the previous, but with keeping the numbers small. Let's first look at a naive version for clarity. Naive ```python def naive(n, m): c = 1 yield c for k in range(n): c = c (n-k) // (k+1) yield c % m n, m = map(int, input().split()) print(naive(n, m)) ``` Takes me ~30 seconds with n=200000. Because c grows very large, up to 60204 digits (199991 bits). And calculations with such large numbers are slow. Fast Instead of naively computing those large c and using modulo m only for output, let's keep c small throughout, modulo m. Got accepted on the site, taking ~0.68 seconds. ```python from math import gcd def fast(n, m): c = 1 G = 1 yield c for k in range(n): mul = n - k while (g := gcd(mul, m)) > 1: mul //= g G = g div = k + 1 while (g := gcd(div, m)) > 1: div //= g G //= g c = c mul pow(div, -1, m) % m yield c G % m n, m = map(int, input().split()) print(fast(n, m)) ``` Attempt This Online! Multiplication is fine under modulo. If it were only c = c (n-k), we could just do c = c (n-k) % m. Division doesn't allow that. So instead of dividing by k+1, we multiply with its inverse (k+1)-1 modulo m. The inverse of some number x is the number x-1 so you get x·x-1 = 1. For example, 7-1 modulo 10 is 3. Because multiplying 7 and 3 gives you 21, which is 1 (modulo 10). Next issue: Not all numbers have an inverse modulo m. For example, 6 doesn't have an inverse modulo 10. You can't multiply 6 with any integer and get 1 (modulo 10). Because 6 and 10 have common divisor 2. What we'll do is invert as much of 6 as possible. Extract the common divisor 2, leaving us with 3. That does have an inverse modulo 10 (namely 7). So extract prime factors in the multipliers/divisors common with m into a separate number G. And update c with what remains, modulo m. Then combine c and G for output. Rough times I get for n=200000, m=998244353 (the large prime from the question): naive: 30.0 seconds fast: 1.0 seconds Matt's: 1.0 seconds For n=200000, m=23571113171923: naive: 30.0 seconds fast: 1.2 seconds Matt's: 4.8 seconds I think worst case is a modulus with many primes like m=23571113171923, that maximizes my G. With n=200000, G grows up to 127 bits. Nothing to worry about. My solution/explanation for a similar problem on Leetcode. That had modulus 10 and I hardcoded factors 2 and 5 and counted them instead of multiplying them into a number G like I did here. Maybe I'll revisit it with this general solution... Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited May 30, 2023 at 0:08 answered May 29, 2023 at 17:02 Stefan PochmannStefan Pochmann 29k 9 9 gold badges 48 48 silver badges 116 116 bronze badges 2 Comments Add a comment Simon Goater Simon GoaterOver a year ago How do you know g divides G? 2023-05-29T18:28:15.58Z+00:00 0 Reply Copy link Stefan Pochmann Stefan PochmannOver a year ago @SimonGoater Similar to knowing that the // (k+1) in the naive version has no remainder. We only divide out what we multiply in. In the naive one it's more obvious, as we know the result is an integer. For the optimized one... We still only divide out what we multiplied in. Consider 7C3. That's (765) / (123), but we compute it as 1 7 / 1 6 / 2 5 / 3. At the point where we divide by 3, we have already multiplied by the 7, 6 and 5. Three consecutive numbers, one of them must have factor 3. So when we divide by 3, we just take out what we've already put in by multiplying with 6. 2023-05-29T18:43:01.333Z+00:00 0 Reply Copy link This answer is useful 8 Save this answer. Show activity on this post. Since you need to output values for all j, you should use this generating function: If j >= 1, then C(n,j) = C(n,j-1) (n+1-j) / j Normally, when this sort of question is asked, the modulus m is a prime larger than n. This makes it relatively easy to do all these calculations mod m, because every j will have a multiplicative inverse. In fact, it is so unusual to ask this question with a non-prime modulus, that I bet the codeforces problem description just fails to mention it. I would try it with the prime modulus assumption. If you're using python 3.8 or better, then there is a modular inverse built into the language, and you can do it just like this: python def getBinomialCoefficientsMod(n,m): result = for j in range(1,n+1): result.append(( result[j-1] (n+1-j) pow(j,-1,m) )%m) return result Edit: Well, it turns out that m is not always a large enough prime, and I don't want to leave this answer incomplete, so here's a version that works with composite or small m: ```python def getBinomialCoefficientsMod(n,m): # get the prime factors of the modulus facs=[] fac=2 mleft = m while facfac<=mleft: if m%fac==0: facs.append(fac) while mleft%fac==0: mleft//=fac fac+=1 if mleft>1: facs.append(mleft) result = # factor of the last result that is relatively prime to m rpresult = 1 # powers of the prime factors of m in the last result exponents = len(facs) for j in range(1,n+1): p=1 num = n+1-j den = j # remove factors of the modulus from num and den, # track their exponents, and get their product for i in range(len(facs)): fac = facs[i] while num%fac==0: exponents[i]+=1 num//=fac while den%fac==0: exponents[i]-=1 den//=fac p = ppow(fac,exponents[i],m) rpresult = (rpresult num pow(den,-1,m)) % m result.append(( rpresult p )%m) return result ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited May 29, 2023 at 19:10 answered May 29, 2023 at 16:29 Matt TimmermansMatt Timmermans 60.7k 3 3 gold badges 55 55 silver badges 105 105 bronze badges 6 Comments Add a comment Matt Timmermans Matt TimmermansOver a year ago Perhaps the OP will try it and we shall see. Otherwise you can keep track of the exponents of factors of the modulus, but that's annoying :) 2023-05-29T16:48:57.927Z+00:00 0 Reply Copy link Adithya AdithyaOver a year ago No, m need not be a prime number. The second test case has n=4 and m=6. 2023-05-29T17:25:08.257Z+00:00 0 Reply Copy link Matt Timmermans Matt TimmermansOver a year ago Oh well, I guess I'll make it work for composite m, then... 2023-05-29T19:13:26.54Z+00:00 0 Reply Copy link Stefan Pochmann Stefan PochmannOver a year ago Did you try it at the site? Seems about 2-4 times slower than mine, which got accepted in ~0.7 seconds, so yours could be close. Our results match. 2023-05-29T19:23:14.583Z+00:00 0 Reply Copy link Matt Timmermans Matt TimmermansOver a year ago It could very well be slower, but that probably depends on the modulus. It took 0.6 seconds on my macbook for n,m = 200000,23454. I don't have a codeforces account. 2023-05-29T19:28:22.183Z+00:00 0 Reply Copy link Add a comment\|Show 1 more comment This answer is useful 2 Save this answer. Show activity on this post. I will sugest to use something like Modulo Multiplication. If you have two numbers a and b, and you calculate their product modulo mod, you can take the modulo of each number first and then calculate the product. That is: python (a b) % mod = ((a % mod) (b % mod)) % mod Taking into account that: C(n, r) = n! / (r! (n - r)!) You can probably get some nice reduction of computation passing mod inside a getting a general result. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 23, 2023 at 13:23 answered May 29, 2023 at 16:17 Jose Manuel de FrutosJose Manuel de Frutos 994 1 1 gold badge 10 10 silver badges 21 21 bronze badges 1 Comment Add a comment Petr L. Petr L.Over a year ago Yes, and to divide n! by (r! (n - r)!) in modular arithmetics, you need to compute the modular multiplicative inverse of (r! (n - r)!), and then % mod the answer. Since simple division does not work there. 2024-04-07T20:33:54.9Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. 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14504	https://www.chegg.com/homework-help/questions-and-answers/volume-v-4-3-pi-r-3-spherical-balloon-changes-radius--rate-3-volume-change-respect-radius--q19538352	Solved The volume V = 4/3 pi r^3 of a spherical balloon \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Calculus Calculus questions and answers The volume V = 4/3 pi r^3 of a spherical balloon changes with the radius. a. At what rate (in^3/in) does the volume change with respect to the radius when r = 7 in? b. Using the rate from part a. by approximately how much does the volume Increase when the radius changes from 7 to 7.2 in? a. At what rate (in^3/in) does the volume change with respect to the Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: The volume V = 4/3 pi r^3 of a spherical balloon changes with the radius. a. At what rate (in^3/in) does the volume change with respect to the radius when r = 7 in? b. Using the rate from part a. by approximately how much does the volume Increase when the radius changes from 7 to 7.2 in? a. At what rate (in^3/in) does the volume change with respect to the Show transcribed image text There are 2 steps to solve this one.Solution 100%(3 ratings) Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! To approach the first step of this problem, calculate the derivative of the volume function V=4 3 π r 3 with respect to r. Step 1 Solution : Given that , Volume of spherical balloon : π V=4 3 π r 3 View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question Transcribed image text: The volume V = 4/3 pi r^3 of a spherical balloon changes with the radius. a. At what rate (in^3/in) does the volume change with respect to the radius when r = 7 in? b. Using the rate from part a. by approximately how much does the volume Increase when the radius changes from 7 to 7.2 in? a. At what rate (in^3/in) does the volume change with respect to the radius when r = 7 in? in^3/in (Type an exact answer in terms of pi.) b. Using the rate from part a. by approximately how much does the volume Increase when the radius changes from 7 to 7.2 in? in^3 (Use the answer from part a to find this answer. Round to two decimal places as needed.) Not the question you’re looking for? Post any question and get expert help quickly. 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14505	https://www.onlinemathlearning.com/multiply-divide-word-problems-worksheet.html	Multiply Divide Word Problems Worksheets Related Pages Math Worksheets Lessons for First Grade Free Printable Worksheets Word Problems Worksheets Word Problems (Addition & Subtraction) Word Problems (Multiplication & Division) Multiplication & Division Word Problems Worksheets In these free math worksheets, students learn to solve multiplication & division word problems. There are two sets of multiplication & division word problems worksheets: Word Problems Worksheets (one-step) Word Problems Worksheets (two-step) How to solve multiplication & division word problems? Solving multiplication and division word problems involves breaking down the problem, identifying the key information, and using appropriate mathematical operations to find the solution. Distinguishing between multiplication and division word problems involves understanding the context of the problem and identifying key phrases or clues that indicate which operation is required. Here are some tips to help you differentiate between multiplication and division word problems: Multiplication Word Problems: Example: There are 6 boxes, and each box contains 8 chocolates. How many chocolates in all? Example: There are 4 shelves, and each shelf has 15 books. How many books are there in total? Division Word Problems: Example: There are 24 cookies, and they need to be shared equally among 6 friends. How many cookies does each friend get? Example: A baker has 36 cupcakes, and he wants to pack them into boxes. If each box can hold 6 cupcakes, how many boxes will he need? Keywords: Pay attention to keywords that indicate multiplication or division. Words like “times,” “multiply,” “each,” and “total” often suggest multiplication, while words like “share,” “divide,” and “split” suggest division. Context: Consider the overall context of the problem. If you are dealing with the total amount or combining quantities, it’s likely a multiplication problem. If you are dividing or distributing a quantity, it’s likely a division problem. By carefully reading the word problem and recognizing these clues, you can determine whether the problem requires multiplication or division. Practice with a variety of problems to strengthen your skills in distinguishing between the two operations. Have a look at this video if you need to review how to solve multiplication or division word problems. Multiplication/Division Word Problems Click on the following worksheet to get a printable pdf document. Scroll down the page for more Multiplication/Division Word Problems Worksheets. More Multiplication/Division Word Problems Worksheets Printable (Answers on the second page.) Multiply Divide Word Problems Worksheet #1 (one-step) Multiply Divide Word Problems Worksheet #2 (one-step) Multiply Divide Word Problems Worksheet #3 (two-step) Multiply Divide Word Problems Worksheet #4 (two-step) Online Multiplication Word Problems (at most 3-digit) Multiplication Word Problems (at most 4-digit) 2-step Word Problems (using +, -, ×) Multi-step Word Problems (using +, -, ×) Related Lessons & Worksheets Multiplication Word Problems Multiplication Word Problems (using tape diagrams) More Printable Worksheets Try out our new and fun Fraction Concoction Game. Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. Copyright © 2005, 2025 - OnlineMathLearning.com. Embedded content, if any, are copyrights of their respective owners.
14506	https://sandbox.trascendit-corp.com/ucuenca/ocw/pluginfile.php/79/mod_folder/content/0/Clinical%20manifestations%20and%20diagnosis%20of%20irritable%20bowel%20syndrome%20in%20adults%20-%20UpToDate.pdf?forcedownload=1	Official reprint from UpToDate www.uptodate.com ©2020 UpToDate, Inc. and/or its affiliates. All Rights Reserved. Clinical manifestations and diagnosis of irritable bowel syndrome in adults Author: Arnold Wald, MD Section Editor: Nicholas J Talley, MD, PhD Deputy Editor: Shilpa Grover, MD, MPH, AGAF All topics are updated as new evidence becomes available and our peer review process is complete. Literature review current through: Jun 2020. \| This topic last updated: Aug 29, 2019. INTRODUCTION Irritable bowel syndrome (IBS) is a functional disorder of the gastrointestinal tract characterized by chronic abdominal pain and altered bowel habits. However, only a small percentage of those affected seek medical attention [ 1-5 ]. Approximately 40 percent of individuals who meet diagnostic criteria for IBS do not have a formal diagnosis [ 6]. IBS is associated with increased health care costs and is the second highest cause of work absenteeism [ 7,8 ]. In the United States, IBS accounts for 25 to 50 percent of all referrals to gastroenterologists [ 9]. This topic will review the clinical manifestations and diagnosis of IBS. The pathophysiology and management of IBS are discussed in detail separately. (See "Pathophysiology of irritable bowel syndrome" and "Treatment of irritable bowel syndrome in adults" .) EPIDEMIOLOGY Prevalence — The prevalence of irritable bowel syndrome (IBS) in North America estimated from population-based studies is approximately 10 to 15 percent [ 1,2,10-14 ]. In a meta-analysis that included eight international studies, the pooled prevalence of IBS was estimated to be 11 percent, with wide variation by geographic region [ 15 ]. The prevalence of IBS was 25 percent lower in those aged over 50 years as compared with those who were younger (OR, 0.75; 95% CI, 0.62-0.92) [ 14 ]. The overall prevalence of IBS in women was higher as compared with men (odds ratio 1.67 [95% CI 1.53–1.82]) [ 16 ]. This relative difference reflects an absolute difference in prevalence of approximately 5 percent between the sexes, with a prevalence in women and ® men of 14 and 9 percent, respectively. Women may be more likely to have constipation- predominant IBS as compared with men [ 16 ]. Associated conditions — IBS is associated with other conditions including fibromyalgia, chronic fatigue syndrome (also known as systemic exertion intolerance disease), gastroesophageal reflux disease, functional dyspepsia, non-cardiac chest pain, and psychiatric disorders including major depression, anxiety, and somatization [ 17-21 ]. (See "Pathophysiology of irritable bowel syndrome", section on 'Psychosocial dysfunction' .) CLINICAL MANIFESTATIONS Irritable bowel syndrome (IBS) is characterized by chronic abdominal pain and altered bowel habits [ 17 ]. Chronic abdominal pain — Abdominal pain in IBS is usually described as a cramping sensation with variable intensity and periodic exacerbations. The location and character of the pain can vary widely [ 17,22 ]. The severity of the pain may range from mild to severe. The pain is frequently related to defecation. While in some patients abdominal pain is relieved with defecation, some patients report worsening of pain with defecation [ 23 ]. Emotional stress and meals may exacerbate the pain. Patients with IBS also frequently report abdominal bloating and increased gas production in the form of flatulence or belching. Altered bowel habits — Symptoms of IBS include diarrhea, constipation, alternating diarrhea and constipation, or normal bowel habits alternating with either diarrhea and/or constipation. Diarrhea — Diarrhea is usually characterized as frequent loose stools of small to moderate volume. Bowel movements generally occur during waking hours, most often in the morning or after meals. Most bowel movements are preceded by lower abdominal cramping pain, urgency, and a sensation of incomplete evacuation or tenesmus. Approximately one-half of all patients with IBS complain of mucus discharge with stools [ 24 ]. Large volume diarrhea, bloody stools, nocturnal diarrhea, and greasy stools are not associated with IBS. Constipation — Stools are often hard and may be described as pellet-shaped. Patients may also experience tenesmus even when the rectum is empty. DIAGNOSIS Overview of diagnostic approach — Irritable bowel syndrome (IBS) should be suspected in patients with chronic abdominal pain and altered bowel habits (constipation and/or diarrhea). A clinical diagnosis of IBS requires the fulfillment of symptom-based diagnostic criteria and a limited evaluation to exclude underlying organic disease [ 25,26 ]. (See 'Diagnostic criteria' below and 'Initial evaluation' below.) Diagnostic criteria — In the absence of a biologic disease marker, several symptom-based criteria have been proposed to standardize the diagnosis of IBS. The most widely used among them are the Rome IV criteria. Rome IV criteria for IBS – According to the Rome IV criteria, IBS is defined as recurrent abdominal pain, on average, at least one day per week in the last three months, associated with two or more of the following criteria [ 17,25 ]: ● Related to defecation • Associated with a change in stool frequency • Associated with a change in stool form (appearance) • IBS subtypes – Subtypes of IBS are recognized based on the patient's reported predominant bowel habit on days with abnormal bowel movements. The Bristol stool form scale (BSFS) should be used to record stool consistency [ 27 ]. Subtypes can only confidently be established when the patient is evaluated off medications used to treat bowel habit abnormalities. IBS subtypes are defined for clinical practice as follows: ● IBS with predominant constipation – Patient reports that abnormal bowel movements are usually constipation (type 1 and 2 in the BSFS) • IBS with predominant diarrhea – Patient reports that abnormal bowel movements are usually diarrhea (type 6 and 7 in the BSFS) • IBS with mixed bowel habits – Patient reports that abnormal bowel movements are usually both constipation and diarrhea (more than one-fourth of all the abnormal bowel movements were constipation and more than one-fourth were diarrhea) • IBS unclassified – Patients who meet diagnostic criteria for IBS but cannot be accurately categorized into one of the other three subtypes. • Other criteria – The Manning criteria include relief of pain with bowel movements, looser and more frequent stools with onset of pain, passage of mucus, and a sense of incomplete emptying ( table 1 ) [ 24 ]. There have been conflicting data regarding the predictive ability of the Manning criteria [ 28-30 ]. The Kruis criteria are less frequently used in clinical practice [31,32 ]. A number of studies have assessed the accuracy of the Rome and Manning criteria ● Initial evaluation History and physical examination — The medical history serves to identify clinical manifestations of IBS as well as identify other possible causes of similar symptoms. The BSFS should be used to record stool consistency [ 27 ]. We perform a thorough history with particular attention to the symptoms that are concerning for organic disease. The history should include exposure to a variety of medications that can cause similar symptoms ( table 2 and table 3 ). A subgroup of patients report an acute viral or bacterial gastroenteritis prior to the onset of IBS symptoms. Family history assessment should include the presence of inflammatory bowel disease, colorectal cancer, and celiac disease. The physical examination is usually normal in patients with IBS. However, patients may have mild abdominal tenderness to palpation. In patients with constipation a rectal examination may be useful in identifying dyssynergic defecation [ 37 ]. (See "Etiology and evaluation of chronic constipation in adults", section on 'Physical examination' and 'Additional evaluation based on the presence of alarm features' below.) Laboratory testing — There is no definitive diagnostic laboratory test for IBS. The purpose of laboratory testing is primarily to exclude an alternative diagnosis. A fecal calprotectin above a threshold level of 50 mcg/g had a pooled sensitivity and specificity for IBD of 81 and 87 percent, respectively [ 26 ]. Lactoferrin threshold values above the threshold range of 4.0 to 7.25 mg/g have a sensitivity and specificity for IBD of 79 and 93 percent, respectively. In a meta-analysis, patients with IBS symptoms and a CRP level of ≤0.5 or calprotectin level of ≤40μg/g, there was a ≤1 percent probability of IBD [ 26,38 ]. in a variety of practice settings [ 33-36 ]. As a result, some investigators continue to use the Manning criteria or a combination of both. No symptom-based criteria have ideal accuracy for diagnosing IBS; however, the Manning and Kruis criteria perform at least as well as the Rome I criteria [ 10 ]. In all patients with suspected IBS, we perform a complete blood count. ● In patients with diarrhea, we perform the following [ 26 ]: ● Fecal calprotectin or fecal lactoferrin • Stool testing for giardia (antigen detection or nucleic acid amplification assay) • Serologic testing for celiac disease • C-reactive protein levels, only if fecal calprotectin and fecal lactoferrin cannot be performed •Data to support testing for celiac disease are conflicting. In a meta-analysis of 14 studies that included 4204 individuals, of whom 2278 (54 percent) met diagnostic criteria for IBS, 4 percent of patients had celiac disease [ 39 ]. However, almost all the studies included in this analysis were conducted outside the United States. A prospective multicenter study performed in the United States compared the prevalence of abnormal celiac antibodies and biopsy proven celiac disease in patients with non-constipated IBS to that of healthy controls. Although more than 7 percent of non-constipated IBS patients had celiac disease associated antibodies suggesting gluten sensitivity, the prevalence of biopsy proven celiac disease was similar to controls [ 40 ]. (See "Epidemiology, pathogenesis, and clinical manifestations of celiac disease in adults" and "Diagnosis of celiac disease in adults", section on 'Serologic evaluation' .) The diagnostic role of antibodies to cytolethal distending toxin B (CdtB) and vinculin requires confirmation before they can be used in the evaluation of patients with suspected IBS [ 41-43 ]. One study that evaluated anti-CdtB and anti-vinculin titers in 2375 patients with IBS diarrhea, found that patients anti-CdtB were significantly higher in IBS diarrhea as compared to patients with IBD, healthy controls, celiac disease, and IBS constipation [ 42 ]. The specificity of anti-CdtB for IBS diarrhea was 92 percent but the sensitivity was only 44 percent. Anti-vinculin had a sensitivity and specificity of 33 and 84 percent, respectively. Other tests — In addition, we perform a limited number of studies guided by the clinical presentation. These include the following: Additional evaluation based on the presence of alarm features — The extent of additional testing depends on the presence of alarm features. Although the presence of concerning features may identify patients more likely to have an organic disease, most patients will ultimately have a negative evaluation. Age-appropriate colorectal cancer screening in all patients. ● In IBS patients with constipation, abdominal radiograph to assess for stool accumulation and determine the severity. ● We perform physiologic testing (anorectal manometry and balloon expulsion testing) to rule out dyssynergic defecation in patients with severe constipation that is refractory to management with dietary changes and osmotic laxative therapy. (See "Etiology and evaluation of chronic constipation in adults", section on 'Dyssynergic defecation' and "Etiology and evaluation of chronic constipation in adults", section on 'Motility studies' and "Treatment of irritable bowel syndrome in adults", section on 'Constipation' .) ● Alarm features – Alarm features include [ 10 ]: ●DIFFERENTIAL DIAGNOSIS The differential diagnosis of irritable bowel syndrome (IBS) is broad. In patients with diarrhea- predominant symptoms, other important causes of chronic diarrhea include celiac disease, microscopic colitis, small intestinal bacterial overgrowth, and inflammatory bowel disease. Constipation may be secondary to organic disease, dyssynergic defecation, or slow colonic transit. While some of these alterative diagnoses are excluded during the course of evaluation in patients with suspected IBS, other diagnoses require additional diagnostic testing and need only be performed in selected patients with alarm features. Other causes of chronic diarrhea and constipation are discussed in detail separately. (See 'Initial evaluation' above and "Approach to the adult with chronic diarrhea in resource-rich settings" and "Etiology and Age of onset after age 50 years • Rectal bleeding or melena • Nocturnal diarrhea • Progressive abdominal pain • Unexplained weight loss • Laboratory abnormalities (iron deficiency anemia, elevated C-reactive protein or fecal calprotectin/lactoferrin) • Family history of IBD or colorectal cancer • Patients without alarm features – In patients who meet diagnostic criteria for IBS and have no alarm features, we do not routinely perform any additional testing beyond the initial evaluation. This limited diagnostic approach rules out organic disease in over 95 percent of patients [ 44,45 ]. ● Patients with alarm features – In patients with alarm features, we perform additional evaluation to exclude other causes of similar symptoms [ 46 ]. The diagnostic evaluation is based on the clinical presentation and usually includes endoscopic evaluation in all patients and imaging in selected cases. In patients with diarrhea, we perform colonoscopy to evaluate for the presence of IBD and perform biopsies to exclude microscopic colitis [ 47- 49 ]. We reserve colonic imaging (eg, abdominal computed tomography scan) if there is a clinical suspicion for a structural lesion [ 46 ]. The imaging modality is guided by the clinical presentation. As an example, if pain, bloating, early satiety and constipation are of recent onset in a postmenopausal woman, then we perform a pelvic imaging with an ultrasound and/or abdominal CT scan [ 50 ]. (See "Etiology and evaluation of chronic constipation in adults", section on 'Endoscopy' and "Colorectal cancer: Epidemiology, risk factors, and protective factors", section on 'Incidence' .) ● evaluation of chronic constipation in adults" and 'Additional evaluation based on the presence of alarm features' above.) DISEASE COURSE Most patients with irritable bowel syndrome (IBS) have chronic symptoms that vary in severity over time. In a systematic review that included clinic-based IBS patients with variable long term follow-up (six months to six years), 2 to 5 percent of patients were diagnosed with an alternate gastrointestinal disease. Symptoms remained unchanged or progressed in 30 to 50 percent and 2 to 18 percent, respectively [ 51 ]. An improvement in symptoms was reported in 12 to 38 percent of patients. Patients may also experience a change in IBS subtype over time with the most frequent change being from predominant constipation or diarrhea to mixed bowel habits. SOCIETY GUIDELINE LINKS Links to society and government-sponsored guidelines from selected countries and regions around the world are provided separately. (See "Society guideline links: Irritable bowel syndrome" .) INFORMATION FOR PATIENTS UpToDate offers two types of patient education materials, "The Basics" and "Beyond the Basics." The Basics patient education pieces are written in plain language, at the 5 to 6 grade reading level, and they answer the four or five key questions a patient might have about a given condition. These articles are best for patients who want a general overview and who prefer short, easy-to-read materials. Beyond the Basics patient education pieces are longer, more sophisticated, and more detailed. These articles are written at the 10 to 12 grade reading level and are best for patients who want in-depth information and are comfortable with some medical jargon. Here are the patient education articles that are relevant to this topic. We encourage you to print or e-mail these topics to your patients. (You can also locate patient education articles on a variety of subjects by searching on "patient info" and the keyword(s) of interest.) th th th th Basics topics (see "Patient education: Irritable bowel syndrome (The Basics)" )● Beyond the Basics topics (see "Patient education: Irritable bowel syndrome (Beyond the ●SUMMARY AND RECOMMENDATIONS Basics)" and "Patient education: Chronic diarrhea in adults (Beyond the Basics)" ) Irritable bowel syndrome (IBS) is a functional disorder of the gastrointestinal tract characterized by chronic abdominal pain and altered bowel habits. The estimated prevalence of IBS globally is approximately 11 percent with a higher prevalence in younger individuals and in women. (See 'Epidemiology' above.) ● Abdominal pain in IBS is usually described as a cramping sensation with variable intensity and periodic exacerbations. The pain is frequently related to defecation. While in some patients abdominal pain is relieved with defecation, a substantial portion of patients report worsening of pain with defecation. (See 'Chronic abdominal pain' above.) ● Symptoms of IBS include diarrhea, constipation, alternating diarrhea and constipation, or normal bowel habits alternating with either diarrhea and/or constipation. Diarrhea is usually characterized as frequent loose stools of small to moderate volume. Bowel movements generally occur during waking hours, most often in the morning or after meals. Patients may have constipation with interludes of diarrhea or normal bowel function. (See 'Altered bowel habits' above.) ● IBS should be suspected in patients with chronic abdominal pain and altered bowel habits (constipation and/or diarrhea). A clinical diagnosis of IBS requires the fulfillment of symptom-based diagnostic criteria and a limited evaluation to exclude underlying organic disease. (See 'Overview of diagnostic approach' above.) ● According to the Rome IV criteria, IBS is defined as recurrent abdominal pain, on average, at least one day per week in the last three months, associated with two or more of the following criteria (see 'Diagnostic criteria' above): Related to defecation • Associated with a change in stool frequency • Associated with a change in stool form (appearance) • Initial evaluation in all patients with suspected IBS includes a history and physical examination, and limited testing to evaluate for the presence of alarm features concerning for organic disease. (See 'Initial evaluation' above.) ● In all patients with suspected IBS, we perform a complete blood count and age •Use of UpToDate is subject to the Subscription and License Agreement . REFERENCES Talley NJ, Zinsmeister AR, Van Dyke C, Melton LJ 3rd. Epidemiology of colonic symptoms appropriate colorectal cancer screening. In patients with diarrhea, we perform the following: • Fecal calprotectin or fecal lactoferrin - Stool testing for giardia - Serologic testing for celiac disease - C-reactive protein levels only if fecal calprotectin and fecal lactoferrin cannot be performed - Alarm features concerning for underling organic disease include: ● Age of onset after age 50 years • Rectal bleeding or melena • Nocturnal diarrhea • Progressive abdominal pain • Unexplained weight loss • Laboratory abnormalities (eg, iron deficiency anemia, elevated C-reactive protein or fecal calprotectin/lactoferrin) • Family history of inflammatory bowel disease or colorectal cancer • In patients who meet diagnostic criteria for IBS and have no alarm features, we do not routinely perform any additional testing beyond the initial evaluation. In patients with alarm features, we perform additional evaluation to exclude other causes of similar symptoms. The diagnostic evaluation is based on the clinical presentation and usually includes endoscopic evaluation in all patients and imaging in selected cases. (See 'Additional evaluation based on the presence of alarm features' above.) Most patients with IBS have chronic symptoms that vary in severity over time. Patients may also experience a change in IBS subtype over time with the most frequent change being from predominant constipation or diarrhea to mixed bowel habits. (See 'Disease course' above.) ● and the irritable bowel syndrome. Gastroenterology 1991; 101:927. Drossman DA, Li Z, Andruzzi E, et al. U.S. householder survey of functional gastrointestinal disorders. Prevalence, sociodemography, and health impact. Dig Dis Sci 1993; 38:1569. Jones R, Lydeard S. Irritable bowel syndrome in the general population. BMJ 1992; 304:87. Heaton KW, O'Donnell LJ, Braddon FE, et al. Symptoms of irritable bowel syndrome in a British urban community: consulters and nonconsulters. 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Site of pain from the irritable bowel. Lancet 1980; 2:443. Simren M, Palsson OS, Whitehead WE. Update on Rome IV Criteria for Colorectal Disorders: Implications for Clinical Practice. Curr Gastroenterol Rep 2017; 19:15. Manning AP, Thompson WG, Heaton KW, Morris AF. Towards positive diagnosis of the irritable bowel. Br Med J 1978; 2:653. Mearin F, Lacy BE, Chang L, et al. Bowel Disorders. Gastroenterology 2016. Smalley W, Falck-Ytter C, Carrasco-Labra A, et al. AGA Clinical Practice Guidelines on the Laboratory Evaluation of Functional Diarrhea and Diarrhea-Predominant Irritable Bowel Syndrome in Adults (IBS-D). Gastroenterology 2019; 157:851. Blake MR, Raker JM, Whelan K. Validity and reliability of the Bristol Stool Form Scale in healthy adults and patients with diarrhoea-predominant irritable bowel syndrome. Aliment Pharmacol Ther 2016; 44:693. Talley NJ, Phillips SF, Melton LJ, et al. Diagnostic value of the Manning criteria in irritable bowel syndrome. Gut 1990; 31:77. Smith RC, Greenbaum DS, Vancouver JB, et al. Gender differences in Manning criteria in the irritable bowel syndrome. Gastroenterology 1991; 100:591. Taub E, Cuevas JL, Cook EW 3rd, et al. Irritable bowel syndrome defined by factor analysis. Gender and race comparisons. Dig Dis Sci 1995; 40:2647. Kruis W, Thieme C, Weinzierl M, et al. A diagnostic score for the irritable bowel syndrome. Its value in the exclusion of organic disease. Gastroenterology 1984; 87:1. Frigerio G, Beretta A, Orsenigo G, et al. Irritable bowel syndrome. Still far from a positive diagnosis. Dig Dis Sci 1992; 37:164. Hammer J, Talley NJ. Diagnostic criteria for the irritable bowel syndrome. Am J Med 1999; 107:5S. Fass R, Longstreth GF, Pimentel M, et al. Evidence- and consensus-based practice guidelines for the diagnosis of irritable bowel syndrome. Arch Intern Med 2001; 161:2081. Vanner SJ, Depew WT, Paterson WG, et al. Predictive value of the Rome criteria for diagnosing the irritable bowel syndrome. Am J Gastroenterol 1999; 94:2912. Ford AC, Bercik P, Morgan DG, et al. Validation of the Rome III criteria for the diagnosis of irritable bowel syndrome in secondary care. Gastroenterology 2013; 145:1262. Talley NJ. How to do and interpret a rectal examination in gastroenterology. Am J Gastroenterol 2008; 103:820. Menees SB, Powell C, Kurlander J, et al. A meta-analysis of the utility of C-reactive protein, erythrocyte sedimentation rate, fecal calprotectin, and fecal lactoferrin to exclude inflammatory bowel disease in adults with IBS. Am J Gastroenterol 2015; 110:444. Ford AC, Chey WD, Talley NJ, et al. Yield of diagnostic tests for celiac disease in individuals with symptoms suggestive of irritable bowel syndrome: systematic review and meta-analysis. Arch Intern Med 2009; 169:651. Cash BD, Rubenstein JH, Young PE, et al. The prevalence of celiac disease among patients with nonconstipated irritable bowel syndrome is similar to controls. Gastroenterology 2011; 141:1187. Rezaie A, Park SC, Morales W, et al. Assessment of Anti-vinculin and Anti-cytolethal Distending Toxin B Antibodies in Subtypes of Irritable Bowel Syndrome. Dig Dis Sci 2017; 62:1480. Pimentel M, Morales W, Rezaie A, et al. Development and validation of a biomarker for diarrhea-predominant irritable bowel syndrome in human subjects. PLoS One 2015; 10:e0126438. Schmulson M, Balbuena R, Corona de Law C. Clinical experience with the use of anti- CdtB and anti-vinculin antibodies in patients with diarrhea in Mexico. Rev Gastroenterol Mex 2016; 81:236. Schmulson MW, Chang L. Diagnostic approach to the patient with irritable bowel syndrome. Am J Med 1999; 107:20S. Svendsen JH, Munck LK, Andersen JR. Irritable bowel syndrome--prognosis and diagnostic safety. A 5-year follow-up study. Scand J Gastroenterol 1985; 20:415. O'Connor OJ, McSweeney SE, McWilliams S, et al. Role of radiologic imaging in irritable bowel syndrome: evidence-based review. Radiology 2012; 262:485. Guagnozzi D, Arias Á, Lucendo AJ. Systematic review with meta-analysis: diagnostic overlap of microscopic colitis and functional bowel disorders. Aliment Pharmacol Ther 2016; 43:851. Kamp EJ, Kane JS, Ford AC. Irritable Bowel Syndrome and Microscopic Colitis: A Systematic Review and Meta-analysis. Clin Gastroenterol Hepatol 2016; 14:659. Gudsoorkar VS, Quigley EM. Distinguishing Microscopic Colitis From Irritable Bowel Syndrome. Clin Gastroenterol Hepatol 2016; 14:669. Goff B. Symptoms associated with ovarian cancer. Clin Obstet Gynecol 2012; 55:36. El-Serag HB, Pilgrim P, Schoenfeld P. Systemic review: Natural history of irritable bowel syndrome. Aliment Pharmacol Ther 2004; 19:861. Topic 2630 Version 33.0 GRAPHICS Manning criteria for the diagnosis of irritable bowel syndrome Pain relieved with defecation More frequent stools at the onset of pain Looser stools at the onset of pain Visible abdominal distention Passage of mucus Sensation of incomplete evacuation The likelihood of irritable bowel syndrome is proportional to the number of Manning criteria that are present. Graphic 63034 Version 1.0 Medications associated with diarrhea System targeted by drug Type of agent Examples Cardiovascular Antiarrhythmics Digoxin Procainamide Quinidine Antihypertensives ACE inhibitors Angiotensin II receptor blockers Beta blockers Hydralazine Methyldopa Cholesterol-lowering agents Clofibrate Gemfibrozil Statins Diuretics Acetazolamide Ethacrynic acid Furosemide Central nervous system Antianxiety drugs Alprazolam Meprobamate Antiparkinsonian drugs Levodopa Other agents Anticholinergic agents Fluoxetine Lithium Tacrine Endocrine Oral hypoglycemic agents Metformin Thyroid replacement therapy Synthroid Gastrointestinal Antiulcer/antacid drugs H2RAs Magnesium-containing antacids Misoprostol Proton pump inhibitors Bile acids Chenodeoxycholic acid Ursodeoxycholic acid Laxatives Cathartics Lactulose Sorbitol Treatments for inflammatory bowel disease 5-aminosalycilates (particularly olsalazine) Musculoskeletal Gold salts Auranofin Nonsteroidal antiinflammatory drugs Ibuprofen Mefenamic acid Naproxen Phenylbutazone Treatments for periodic fever syndrome or gout Colchicine Other Antibiotics Amoxicillin Ampicillin Cephalosporins Clindamycin ¶ Neomycin Tetracycline Antineoplastic agents Many Dietary Alcohol Sugar substitutes (eg, sorbitol) Vitamins Magnesium Vitamin C ACE: angiotensin-converting enzyme; H2RA: histamine-2 receptor antagonist. Olmesartan has been associated with sprue-like enteropathy. ¶ Most antibiotics have been associated with diarrhea. Data from: Holt PR. Diarrhea and malabsorption in the elderly. Gastroenterol Clin North Am 2001; 30:427. Ratnaike RN, Jones TE. Mechanisms of drug-induced diarrhoea in the elderly. Drugs Aging 1998; 13:245. Graphic 71449 Version 7.0 Drugs associated with constipation Analgesics Anticholinergics Antihistamines Antispasmodics Antidepressants Antipsychotics Cation-containing agents Iron supplements Aluminum (antacids, sucralfate) Barium Neurally active agents Opiates Antihypertensives Ganglionic blockers Vinca alkaloids Calcium channel blockers 5HT3 antagonists Graphic 62307 Version 2.0 Contributor Disclosures Arnold Wald, MD Nothing to disclose Nicholas J Talley, MD, PhD Employment: University of Newcastle [PVC Global Research]; John Hunter Hospital [Senior Sta ff Specialist in gastroenterology]; Medical Journal of Australia [Editor-In-Chief]. Patent Holder: Biomarkers of irritable bowel syndrome [Irritable bowel syndrome]; Licensing Questionnaires [Mayo Clinic Talley Bowel Disease Questionnaire - Mayo Dysphagia Questionnaire (Nepean Dyspepsia Index)]; Nestec European Patent [Application 127353589; Nanotechnology]; Singapore 'Provisional' Patent [NTU Ref TD/129/17 "Microbiota Modulation Of BDNF Tissue Repair Pathway"]. Grant/Research/Clinical Trial Support: Commonwealth Labs [IBS (diagnostic blood test]); VN National Science Challenge [Research funding for IBS]. Consultant/Advisory Boards: Adelphi Values [Functional dyspepsia (Patient-reported outcome measures)]; Viscera Labs [Bile acid sequestrant]; twoXAR [IBS]; Anatara Lifesciences [IBS, IBD (Dietary supplement)]; Avrio [Gastroenterology (OTC drugs)]; Censa [Diabetic gastroparesis epidemiology]; Sanofi [Probiotic product (Bacillus clausii)]; Pfizer [IBS (OTC drugs in gastroenterology)]; Planet Innovation [Gastroenterology (Gas capsule)]; Takeda [Gastroparesis (TAK906)]; Allakos [Gastric eosinophilic disease (AK002)]; Progenity Inc, San Diego [SIBO (Intestinal capsule device)]; IM HealthScience [IBS, FD (Peppermint oil)]. Other Financial Interest: IM Health Science [Medicine, clinical skills]. Shilpa Grover, MD, MPH, AGAF Nothing to disclose Contributor disclosures are reviewed for conflicts of interest by the editorial group. When found, these are addressed by vetting through a multi-level review process, and through requirements for references to be provided to support the content. Appropriately referenced content is required of all authors and must conform to UpToDate standards of evidence. Conflict of interest policy
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14508	https://tutorial.math.lamar.edu/classes/calcii/crossproduct.aspx	Paul's Online Notes Go To Notes Practice Problems Assignment Problems Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections Dot Product 3-Dimensional Space Introduction Chapters Series & Sequences 3-Dimensional Space Classes Algebra Calculus I Calculus II Calculus III Differential Equations Extras Algebra & Trig Review Common Math Errors Complex Number Primer How To Study Math Cheat Sheets & Tables Misc Contact Me MathJax Help and Configuration Notes Downloads Complete Book Practice Problems Downloads Complete Book - Problems Only Complete Book - Solutions Assignment Problems Downloads Complete Book Other Items Get URL's for Download Items Print Page in Current Form (Default) Show all Solutions/Steps and Print Page Hide all Solutions/Steps and Print Page Paul's Online Notes Home / Calculus II / Vectors / Cross Product Prev. Section Notes Practice Problems Assignment Problems Next Section Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width. Section 11.4 : Cross Product In this final section of this chapter we will look at the cross product of two vectors. We should note that the cross product requires both of the vectors to be three dimensional vectors. Also, before getting into how to compute these we should point out a major difference between dot products and cross products. The result of a dot product is a number and the result of a cross product is a vector! Be careful not to confuse the two. So, let’s start with the two vectors (\vec a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle ) and (\vec b = \left\langle {{b_1},{b_2},{b_3}} \right\rangle ) then the cross product is given by the formula, [\vec a \times \vec b = \left\langle {{a_2}{b_3} - {a_3}{b_2},{a_3}{b_1} - {a_1}{b_3},{a_1}{b_2} - {a_2}{b_1}} \right\rangle ] This is not an easy formula to remember. There are two ways to derive this formula. Both of them use the fact that the cross product is really the determinant of a 3x3 matrix. If you don’t know what that is don’t worry about it. You don’t need to know anything about matrices or determinants to use either of the methods. The notation for the determinant is as follows, [\vec a \times \vec b = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\{{a_1}}&{{a_2}}&{{a_3}}\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right\|] The first row is the standard basis vectors and must appear in the order given here. The second row is the components of (\vec a) and the third row is the components of (\vec b). Now, let’s take a look at the different methods for getting the formula. The first method uses the Method of Cofactors. If you don’t know the method of cofactors that is fine, the result is all that we need. Here is the formula. [\vec a \times \vec b = \left\| {\begin{array}{{20}{c}}{{a_2}}&{{a_3}}\{{b_2}}&{{b_3}}\end{array}} \right\|\vec i - \left\| {\begin{array}{{20}{c}}{{a_1}}&{{a_3}}\{{b_1}}&{{b_3}}\end{array}} \right\|\vec j + \left\| {\begin{array}{{20}{c}}{{a_1}}&{{a_2}}\{{b_1}}&{{b_2}}\end{array}} \right\|\vec k] where, [\left\| {\begin{array}{{20}{c}}a&b\c&d\end{array}} \right\| = ad - bc] This formula is not as difficult to remember as it might at first appear to be. First, the terms alternate in sign and notice that the 2x2 is missing the column below the standard basis vector that multiplies it as well as the row of standard basis vectors. The second method is slightly easier; however, many textbooks don’t cover this method as it will only work on 3x3 determinants. This method says to take the determinant as listed above and then copy the first two columns onto the end as shown below. [\vec a \times \vec b = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\{{a_1}}&{{a_2}}&{{a_3}}\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right\|\,\,\,\,\begin{array}{{20}{c}}{\vec i}&{\vec j}\{{a_1}}&{{a_2}}\{{b_1}}&{{b_2}}\end{array}] We now have three diagonals that move from left to right and three diagonals that move from right to left. We multiply along each diagonal and add those that move from left to right and subtract those that move from right to left. This is best seen in an example. We’ll also use this example to illustrate a fact about cross products. Example 1 If (\vec a = \left\langle {2,1, - 1} \right\rangle ) and (\vec b = \left\langle { - 3,4,1} \right\rangle ) compute each of the following. (\vec a \times \vec b) (\vec b \times \vec a) Show All Solutions Hide All Solutions a (\vec a \times \vec b) Show Solution Here is the computation for this one. [\begin{align}\vec a \times \vec b & = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\2&1&{ - 1}\{ - 3}&4&1\end{array}} \right\|\,\,\,\,\begin{array}{{20}{c}}{\vec i}&{\vec j}\2&1\{ - 3}&4\end{array}\ & = \vec i\left( 1 \right)\left( 1 \right) + \vec j\left( { - 1} \right)\left( { - 3} \right) + \vec k\left( 2 \right)\left( 4 \right) - \vec j\left( 2 \right)\left( 1 \right) - \vec i\left( { - 1} \right)\left( 4 \right) - \vec k\left( 1 \right)\left( { - 3} \right)\ & = 5\vec i + \vec j + 11\vec k\end{align}] b (\vec b \times \vec a) Show Solution And here is the computation for this one. [\begin{align}\vec b \times \vec a & = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\{ - 3}&4&1\2&1&{ - 1}\end{array}} \right\|\,\,\,\,\begin{array}{{20}{c}}{\vec i}&{\vec j}\{ - 3}&4\2&1\end{array}\ & = \vec i\left( 4 \right)\left( { - 1} \right) + \vec j\left( 1 \right)\left( 2 \right) + \vec k\left( { - 3} \right)\left( 1 \right) - \vec j\left( { - 3} \right)\left( { - 1} \right) - \vec i\left( 1 \right)\left( 1 \right) - \vec k\left( 4 \right)\left( 2 \right)\ & = - 5\vec i - \vec j - 11\vec k\end{align}] Notice that switching the order of the vectors in the cross product simply changed all the signs in the result. Note as well that this means that the two cross products will point in exactly opposite directions since they only differ by a sign. We’ll formalize up this fact shortly when we list several facts. There is also a geometric interpretation of the cross product. First we will let (\theta) be the angle between the two vectors (\vec a) and (\vec b)and assume that (0 \le \theta \le \pi ), then we have the following fact, [\begin{equation}\left\\| {\vec a \times \vec b} \right\\| = \left\\| {\vec a} \right\\|\,\,\left\\| {\vec b} \right\\|\,\sin \theta \label{eq:eq1} \end{equation}] and the following figure. There should be a natural question at this point. How did we know that the cross product pointed in the direction that we’ve given it here? First, as this figure implies, the cross product is orthogonal to both of the original vectors. This will always be the case with one exception that we’ll get to in a second. Second, we knew that it pointed in the upward direction (in this case) by the “right hand rule”. This says that if we take our right hand, start at (\vec a) and rotate our fingers towards (\vec b)our thumb will point in the direction of the cross product. Therefore, if we’d sketched in (\vec b \times \vec a) above we would have gotten a vector in the downward direction. Example 2 A plane is defined by any three points that are in the plane. If a plane contains the points (P = \left( {1,0,0} \right)), (Q = \left( {1,1,1} \right)) and (R = \left( {2, - 1,3} \right)) find a vector that is orthogonal to the plane. Show Solution The one way that we know to get an orthogonal vector is to take a cross product. So, if we could find two vectors that we knew were in the plane and took the cross product of these two vectors we know that the cross product would be orthogonal to both the vectors. However, since both the vectors are in the plane the cross product would then also be orthogonal to the plane. So, we need two vectors that are in the plane. This is where the points come into the problem. Since all three points lie in the plane any vector between them must also be in the plane. There are many ways to get two vectors between these points. We will use the following two, [\begin{align}\overrightarrow {PQ} & = \left\langle {1 - 1,1 - 0,1 - 0} \right\rangle = \left\langle {0,1,1} \right\rangle \ \overrightarrow {PR} & = \left\langle {2 - 1, - 1 - 0,3 - 0} \right\rangle = \left\langle {1, - 1,3} \right\rangle \end{align}] The cross product of these two vectors will be orthogonal to the plane. So, let’s find the cross product. [\begin{align}\overrightarrow {PQ} \times \overrightarrow {PR} & = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\0&1&1\1&{ - 1}&3\end{array}} \right\|\,\,\,\,\begin{array}{{20}{c}}{\vec i}&{\vec j}\0&1\1&{ - 1}\end{array}\ & = 4\vec i + \vec j - \vec k\end{align}] So, the vector (4\vec i + \vec j - \vec k) will be orthogonal to the plane containing the three points. Now, let’s address the one time where the cross product will not be orthogonal to the original vectors. If the two vectors, (\vec a) and (\vec b), are parallel then the angle between them is either 0 or 180 degrees. From (\eqref{eq:eq1}) this implies that, [\left\\| {\vec a \times \vec b} \right\\| = 0] From a fact about the magnitude we saw in the first section we know that this implies [\vec a \times \vec b = \vec 0] In other words, it won’t be orthogonal to the original vectors since we have the zero vector. This does give us another test for parallel vectors however. Fact If (\vec a \times \vec b = \vec 0) then (\vec a) and (\vec b) will be parallel vectors. Let’s also formalize up the fact about the cross product being orthogonal to the original vectors. Fact Provided (\vec a \times \vec b \ne \vec 0) then (\vec a \times \vec b) is orthogonal to both (\vec a) and (\vec b). Here are some nice properties about the cross product. Properties If (\vec u), (\vec v) and (\vec w) are vectors and (c) is a number then, [\begin{align} & \vec u \times \vec v = - \vec v \times \vec u & \hspace{0.75in} & \left( {c\vec u} \right) \times \vec v = \vec u \times \left( {c\vec v} \right) = c\left( {\vec u \times \vec v} \right)\ &\vec u \times \left( {\vec v + \vec w} \right) = \vec u \times \vec v + \vec u \times \vec w & \hspace{0.75in} & \vec u\centerdot \left( {\vec v \times \vec w} \right) = \left( {\vec u \times \vec v} \right)\centerdot \vec w\ & \vec u\centerdot \left( {\vec v \times \vec w} \right) = \left\| {\begin{array}{{20}{c}}{{u_1}}&{{u_2}}&{{u_3}}\{{v_1}}&{{v_2}}&{{v_3}}\{{w_1}}&{{w_2}}&{{w_3}}\end{array}} \right\| & & \end{align}] The determinant in the last fact is computed in the same way that the cross product is computed. We will see an example of this computation shortly. There are a couple of geometric applications to the cross product as well. Suppose we have three vectors (\vec a), (\vec b) and (\vec c) and we form the three dimensional figure shown below. The area of the parallelogram (two dimensional front of this object) is given by, [{\rm{Area}} = \left\\| {\vec a \times \vec b} \right\\|] and the volume of the parallelepiped (the whole three dimensional object) is given by, [{\rm{Volume}} = \left\| {\vec a\centerdot \left( {\vec b \times \vec c} \right)} \right\|] Note that the absolute value bars are required since the quantity could be negative and volume isn’t negative. We can use this volume fact to determine if three vectors lie in the same plane or not. If three vectors lie in the same plane then the volume of the parallelepiped will be zero. Example 3 Determine if the three vectors (\vec a = \left\langle {1,4, - 7} \right\rangle ), (\vec b = \left\langle {2, - 1,4} \right\rangle ) and (\vec c = \left\langle {0, - 9,18} \right\rangle ) lie in the same plane or not. Show Solution So, as we noted prior to this example all we need to do is compute the volume of the parallelepiped formed by these three vectors. If the volume is zero they lie in the same plane and if the volume isn’t zero they don’t lie in the same plane. [\begin{align}\vec a\centerdot \left( {\vec b \times \vec c} \right) & = \left\| {\begin{array}{{20}{c}}1&4&{ - 7}\2&{ - 1}&4\0&{ - 9}&{18}\end{array}} \right\|\,\,\,\,\begin{array}{{20}{c}}1&4\2&{ - 1}\0&{ - 9}\end{array}\ & = \left( 1 \right)\left( { - 1} \right)\left( {18} \right) + \left( 4 \right)\left( 4 \right)\left( 0 \right) + \left( { - 7} \right)\left( 2 \right)\left( { - 9} \right) - \ & \hspace{0.75in}\left( 4 \right)\left( 2 \right)\left( {18} \right) - \left( 1 \right)\left( 4 \right)\left( { - 9} \right) - \left( { - 7} \right)\left( { - 1} \right)\left( 0 \right)\ & = - 18 + 126 - 144 + 36\ & = 0\end{align}] So, the volume is zero and so they lie in the same plane. 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14509	https://www.intmath.com/functions-and-graphs/understanding-two-point-form-in-geometry.php	Understanding Two Point Form in Geometry Skip to main content Interactive Mathematics Home Tutoring Features Reviews Pricing FAQ Problem Solver More Lessons Forum Interactives Blog Contact About SearchSearch IntMath Search for: Close Login Create Free Account Search IntMath Search: Close Home Tutoring Features Reviews Pricing FAQ Problem Solver More Lessons Forum Interactives Blog Contact About Login Create Free Account Want Better Math Grades? ✅ Unlimited Solutions ✅ Step-by-Step Answers ✅ Available 24/7 ➕Free Bonuses ($1085 value!) Invalid email address Thank you for booking, we will follow up with available time slots and course plans. Home Introduction to Geometry Introduction to Geometry On this page Introduction to Geometry Functions and Graphs 1. Introduction to Functions 2. Functions from Verbal Statements 3. Rectangular Coordinates 4. The Graph of a Function 4a. Domain and Range of a Function 4b. Domain and Range interactive applet 4c. Comparison calculator BMI - BAI 5. Graphing Using a Computer Algebra System 5a. Online graphing calculator (1): Plot your own graph (JSXGraph) 5b. Online graphing calculator (2): Plot your own graph (SVG) 6. Graphs of Functions Defined by Tables of Data 7. Continuous and Discontinuous Functions 8. Split Functions 9. Even and Odd Functions Related Sections Math Tutoring Need help? Chat with a tutor anytime, 24/7. Chat now Online Math Solver Solve your math problem step by step! Online Math Solver IntMath Forum Get help with your math queries: See Forum Understanding Two Point Form in Geometry The two-point form is a concept for finding the equation of a straight line when given two points on the line. This technique is commonly used in geometry to calculate linear equations. It follows the formula y – y1 = m(x – x1), where (x1,y1) and (x2,y2) are two points on the line, and m is the slope of the line. In this blog post, we will discuss how to use two-point form and how it can be applied in geometry. Calculating Slope with Two-Point Form The first step in using the two-point form is calculating the slope of your line. The slope is defined as "the rate at which one variable changes with respect to another." In other words, it tells you how steep your line is. To calculate slope using the two-point form, you need to find the difference between your x-coordinates divided by the difference between your y-coordinates. For example, if you have two points (3,4) and (6,7), your slope would be calculated as (6-3)/(7-4) = 3/3 = 1. This means that your line has a rise-over run of 1:1 or 45 degrees—a perfectly diagonal line! Finding Equations with Two-Point Form Once you know the slope of your line, you can use the two-point form to find its equation. The basic formula for this calculation is y – y1 = m(x – x1). You'll start by replacing both “m” and “y” with their respective values from before—in our example above “m” was 1 and “y” was 4—and then solving for “x”. This gives us 4 - 4 = 1(x - 3), which simplifies to 0 = x - 3; therefore x = 3. Now that we have both our x and y values, we plug them into our equation to get 4 - 4 = 1(3 - 3), which simplifies to 4 - 4 = 0; therefore y = 4. So our final equation for this example would be y=4x+0! Conclusion The two-point form is an important concept in geometry that can help us find equations for lines when given only two points on said lines. By following a few simple steps—calculating slope with two-point form and then plugging that value into an equation—we can quickly determine what type of straight line we're dealing with and its corresponding equation! With practice and dedication, students should soon become comfortable applying this concept when working with geometry problems involving linear equations. FAQ What is a two-point form? The two-point form is a concept for finding the equation of a straight line when given two points on the line. This technique is commonly used in geometry to calculate linear equations and follows the formula y – y1 = m(x – x1). How do you find the two-point form of a line? To find the two-point form of a line, you need to first calculate the slope by finding the difference between your x-coordinates divided by the difference between your y-coordinates. Then, plug that value into an equation (y – y1 = m(x – x1)) with both “m” and “y” replaced with their respective values to solve for “x.” Finally, plug both the x and y values into the equation to get your final result. How do you use two points? Two points can be used to calculate the slope of a line using the two-point form, as well as its corresponding equation. By finding the difference between your x-coordinates divided by the difference between your y-coordinates, you can determine the slope of the line. You then plug that value into an equation (y – y1 = m(x – x1)) with both “m” and “y” replaced with their respective values to solve for “x.” Finally, plug both the x and y values into the equation to get your final result. What is a point-slope form with two points? The point-slope form with two points is a technique used to find the equation of a straight line when given two points on the line. This technique follows the formula y - y1 = m(x - x1), where “m” is the slope of the line and “y” and “x ” are the y- and x-coordinates of the two points. By finding the difference between your x-coordinates divided by the difference between your y-coordinates, you can determine the slope of the line and then plug those values into an equation to solve for “x”. Finally, plug both the x and y values into the equation to get your final result. Functions and Graphs Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class. 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14510	https://apps.dtic.mil/sti/tr/pdf/ADA377256.pdf	^4f 7A fk J: $4 00 00 < Ü < NATIONAL ADVISORY COMMITTEE FOR AERONAUTICS TECHNICAL NOTE 2881 AERODYNAMIC CHARACTERISTICS OF A TWO-BLADE NACA 10-(3)(062)-045 PROPELLER AND OF A TWO-BLADE NACA 10-(3)(08) -045 PROPELLER By William Solomon Langley Aeronautical Laboratory Langley Field, Va. DISTRIBUTION STATEMENT A Approved for Public Release Distribution Unlimited Reproduced From Best Available Copy Washington January 1953 J 20000508 228 hoo-M-Mit IM NATIONAL ADVISORY COMMITTEE FOR AERONAUTICS TECHNICAL NOTE 2881 AERODYNAMIC CHARACTERISTICS OF A TWO-BLADE NACA 10-(3)(062)-045 PROPELLER AND OF A TWO-BLADE NACA 10-(3)(08)-0^5 PROPELLER1 By William Solomon SUMMARY As part of the investigation to determine the influence of blade- section thickness ratio on propeller performance, tests were made in the Langley l6-foot high-speed tunnel to determine the aerodynamic characteristics of the two-blade NACA 10-(3)(o62)-OV? propeller and of the two-blade NACA 10-(3)(o8)-0^5 propeller. Data were obtained over a blade-angle range from 20° to 55° as measured at the 0.75-radius station, the various constant values of rotational speed used giving a range of advance ratio from 0.5 to 3.8. Maximum efficiencies of the order of 91.5 to 92 percent were obtained for the propellers. The propeller with the thinner airfoil sections over the outboard portion of the blades, the NACA 10-(3)(062)-0^5 pro- peller, had lower losses at high tip speeds, the difference amounting to about 5 percent at a helical tip Mach number of 1.10. INTRODUCTION The aerodynamic characteristics of a series of 10-foot-diameter propellers were investigated in the Langley l6-foot high-speed tunnel in a comprehensive propeller research program. Having high-critical- speed NACA l6-series airfoil sections (ref. 1), these propellers were designed to have Betz minimum induced-energy-loss loading (ref. 2) for a blade angle of V?° at the 0.7 radius when used as a four-blade pro- peller operating at an advance ratio of approximately 2.1. The ultimate purpose of the program was to determine the influence upon propeller Supersedes the recently declassified NACA RM L8E26, "Aerodynamic Characteristics at High Speeds of a Two-Blade NACA 10-(3)(o62)-0^5 Pro- peller and of a Two-Blade NACA 10-(3)(08)-0^5 Propeller" by William Solomon, 19^. NACA TN 2881 performance of propeller design factors and of compressibility; the pro- peller tests reported herein form part of the investigation of the effects of blade-section thickness ratio. SYMBOLS b blade width/ ft C7 blade-section design lift coefficient Cp power coefficient, P/pn3D5 CT thrust coefficient, T/pn2D^ D propeller diameter, ft h blade-section maximum thickness, ft J propeller advance ratio, v/nD M air-stream Mach number • "f7^ M, helical tip Mach number n propeller rotational speed, rps P power absorbed by the propeller, ft-lb/sec r radius to a blade element, ft R propeller tip radius, ft T propeller thrust, lb V airspeed, 'fps ß blade angle, deg ß0 yep blade angle at 0.75 tip radius, deg Cm T) propeller efficiency, — J Cp V NACA TW 2881 • 3 "HJL optimum efficiency (zero profile-drag losses) p mass density of air, slugs/cu ft APPARATUS Propeller dynamometer.- A diagram showing the configuration for these propeller tests with the 2000-horsepower dynamometer is shown in figure 1. The test apparatus and thrust and torque measuring devices are described in detail in reference 3. Propeller blades.- The blades are of NACA 10-(3)(062)-0^5 and mCA 10-(3)(08) -0^5 design which have designation numbers descriptive of the propeller shape, size, and design aerodynamic characteristics. The numerals of the first group give the diameter in feet; the numerals within the first parentheses give the design lift coefficient, in tenths, of the blade section at the 0.7 radius; the numerals within the second parentheses give the thickness ratio at the 0.7 radius; and the last group designates the solidity of one blade of the propeller at the 0.7 radius. Solidity is the ratio of the blade width at any radius to the circumference of the circle traversed by that blade section. Blade- form curves for these propellers are shown in figure 2. As can be seen the blades differ only in thickness ratio and, very slightly, in blade angle; blade-width ratio and design lift coefficient are identical for the two propellers. A photograph of typical propeller blades mounted on the propeller dynamometer in the tunnel is shown as figure 3. Effi- cient airfoil sections extend to the spinner of the two-blade test pro- peller. Spinner fairing plates with 0.062-inch clearance between blade and spinner were provided for each blade angle tested. TEST PROCEDURE Tests of the propellers were made at blade angles varying in 5 increments from 20° to 55° at the 0.75 radius (45-in. radius). Most tests were run at constant rotational speed, the range of advance ratio (J = V/nD) being obtained by changing the tunnel airspeed. A chart of the various rotational speeds giving the blade angles tested at each speed is given as table I. Because of power limitations, tests could not be made at high blade angle and high rotational speed. Therefore, in order to obtain propeller characteristics over a greater range of tip Mach number, tests were run at a blade angle of 45° for maximum dynamometer power and airspeed at or near maximum tunnel airspeed. For these tests the tunnel airspeed was kept constant and the dynamometer rotational speed was varied to obtain the range of advance ratio required. + NACA TN 2881 RESULTS AND DISCUSSION Reduction of data.- The test data, corrected for spinner force and tunnel-wall interference, are presented as the usual propeller charac- teristics, thrust and power coefficient and propeller efficiency plotted as functions of advance ratio. Reference 3 describes the corrections applied. Propeller characteristics as presented herein represent the forces exerted by the propeller from the spinner surface to the blade tips acting in free air. These data are believed to be accurate to within 1 percent. Because the tests were made over a period of time (therefore, under different conditions of tunnel air temperature), small discontinuities occurred in the curves of air-stream Mach number plotted against advance ratio. Also, discrepancies occurred between the values of stream Mach number at the same value of advance ratio for tests run at different times. Therefore, in order to obtain uniformity in the presented data, all air-stream Mach number curves and helical tip Mach number curves have been corrected to those presented for the same advance ratio and rotational speed for the basic propeller of the NACA series, the NACA 10-(3)(08)-03 propeller, in reference 3. Propeller characteristics.- Faired curves of thrust coefficient, power coefficient, and propeller efficiency plotted as functions of advance ratio are presented in figures h to 10 for the two-blade NACA 10-(3)(o62)-045 propeller and in figures 11 to IT for the two-blade NACA 10-(3)(08)-045 propeller. The variation of air-stream Mach number and helical tip Mach number with advance ratio is shown in the figures presenting propeller efficiency. Envelope efficiency.- Envelope curves of propeller efficiency at the various test rotational speeds are given for the two-blade NACA 10-(3)(o62)-045 propeller in figure 18 and for the two-blade NACA 10-(3)(08)-0^5 propeller in figure 19- Also shown in these figures is the optimum efficiency T^ of a two-blade propeller with Betz mini- mum induced-energy-loss loading operating at 1350 rpm. The optimum- efficiency curve was calculated by a method, given in reference 4, which neglects all profile-drag losses for a two-blade propeller operating at the same values of power coefficient as attained in the tests of the NACA propellers. At the higher values of advance ratio, 2.0 to 2.8, the calculated values of optimum efficiency are higher for the NACA 10-(3)(062)-045 propeller than for the NACA 10-(3)(08)-045 pro- peller because the thinner propeller operates at slightly lower values of power coefficient for maximum efficiency than does the thicker Pro- peller. This higher optimum-efficiency value for the NACA 10-(3)(0b2)-045 propeller is in agreement with propeller theory which indicates that at NACA TN 2881 maximum efficiency the blade-section profile-drag losses should approxi- mately equal the induced losses; therefore, the propeller with the lower profile-drag loss, the propeller with thinner blade sections, will oper- ate under lighter load and, consequently, with lower induced loss. Comparison of this optimum-efficiency curve with the envelope effi- ciency curve (1350 rpm) for both propellers indicates induced losses of the order of 5 percent and profile-drag losses of the order of 4 percent at an advance ratio of 2.2, approximately the design advance ratio. The high peak of the envelope efficiency curve (91.5 to 92 percent) is a result of the use of a Betz loading in the design and the maintaining of efficient airfoil sections to the spinner surface. Comparison of NACA 10-( 3) (08) -045 data with NACA 4-(3) (08)-045 data. The envelope efficiency of the two-blade NACA 10-( 3) (0ö)-045 propeller at 1350- rpm is compared in figure 20 with corresponding results from tests of a 4-foot-diameter model of this propeller (ref. 5). Also shown in the figure are the optimum-efficiency curves for the values of power coefficient at maximum efficiency obtained with the respective propel- lers. Since, during the tests of the 4-foot-diameter propeller, the propeller was mounted on the forward portion of a model with a wing spanning the tunnel and since thrust was measured by the tunnel balance system, some of the slipstream rotation was converted into measured thrust. Therefore, the propulsive efficiencies which were obtained con- siderably exceeded the propeller efficiencies obtained with the 10-foot- diameter propeller. Also, for the 4-foot-diameter propeller, the peak values of efficiency occurred at higher values of advance ratio and, consequently, at lower values of power coefficient. Hence, the values of optimum efficiency for the 4-foot-diameter propeller are higher than optimum-efficiency values for the 10-foot-diameter propeller. In an analysis made in England by A. B. Haines shortly after publication of reference 5, comments were presented to point out an apparent discrep- ancy, as follows: envelope efficiency for the model propeller (NACA 4-(3)(o8)-045) approached optimum propeller efficiency with increasing advance ratio and then equaled the optimum at the higher values of advance ratio. This discrepancy exists 'only if the envelope efficiency for the 4-foot-diameter propeller is compared with optimum efficiency calculated from excessively high values of power coefficient. At low advance ratio where small amounts of rotational energy are imparted to the slipstream, the effect of the wing on recovery of the energy in the form of thrust is very small. At higher blade angles and advance ratios, where relatively much larger amounts of rotational energy are imparted to the slipstream, a significant increment of thrust is obtained from the partial recovery of the rotational energy by the wing. Calculations based on the method of reference 6 show that at high advance ratios the wing recovers approximately 50 percent of the slip- stream rotational energy in the form of thrust; this amounts to approxi- mately 2 percent in efficiency. Therefore, a major part of the total NACA TN 2881 difference between the results of the 4-foot and the 10-foot-diameter- propeller tests is accounted for by this one factor. Compressibility effects.- The maximum efficiency of the two-blade NACA 10-(3)(08)-0^5 and the two-blade NACA 10-(3)(o62)-0^5 propellers at a blade angle of ^5° is shown in figure 21 for a range of helical tip Mach number. At the lower tip Mach numbers there is little differ- ence between the maximum efficiency of the propellers. Above a tip Mach number of 0-90, the maximum efficiency of both propellers decreases rapidly. However, the rate of this maximum-efficiency decrease is lower for the propeller with the thinner outboard blade sections. With an increase of helical tip Mach number from 0.92 to 1.10, the loss in max- imum efficiency is 13 percent for the NACA 10-(3)(062)-0^5 propeller as compared with an l8-percent loss for the NACA 10-(3)(08)-0^5 propeller, the propeller with the thicker outboard blade sections. The higher efficiency of the propeller with the thinner outboard blade sections is expected because airfoil section data show that thin sections maintain their lift to higher Mach numbers than thick sections (ref. 1). CONCLUDING REMARKS An investigation was made in the Langley l6-foot high-speed tunnel to determine the aerodynamic characteristics of the two-blade NACA 10-(3)(062)-045 and the two-blade NACA 10-( 3) (08)-0^5 propellers/ differing principally in thickness distribution. Data are presented over a range of advance ratio from 0.5 to 38 and a blade-angle range from 20° to 55 measured at the 0-75 radius. These propellers, as a result of the use of a Betz loading in the design and the maintaining of efficient airfoil sections to the spinner surface, have a maximum efficiency of the order of 91-5 to 92 percent. The propeller with the thinner airfoil sections over the outboard portion of the blades has lower losses in maximum efficiency caused by compressibility, the difference amounting to about 5 percent at a heli- cal tip Mach number of 1.10. Langley Aeronautical Laboratory, - National Advisory Committee for Aeronautics, Langley Field, Va., June 11, 19^8. NACA TN 2881 REFERENCES 1. Stack, John: Tests of Airfoils Designed To Delay the Compressibility- Burble. NACA Rep. 763, 19^3- (Supersedes NACA TN 976.) 2. Hartman, Edwin P., and Feldman, Lewis: Aerodynamic Problems in the Design of Efficient Propellers. NACA WR L-753, 19 te. (Formerly NACA ACR, Aug. 19^2.) 3. Corson, Blake W., Jr., and Maynard, Julian D.: The Langley 2,000-Horsepower Propeller Dynamometer and Tests at High Speed of an NACA 10-(3)(o8)-03 Two-Blade Propeller. NACA TN 2859, 1952. (Supersedes NACA RM L7L29.) h. Crigler, John L., and Talkin, Herbert W.: Charts for Determining Propeller Efficiency. NACA WR L-lM, 19^. (Formerly NACA ACR Lit-129.) 5. Stack, John, Draley, Eugene C, Delano, James B., and Feldman, Lewis: Investigation of the NACA M3)(o8)-03 and NACA k-(3)(08)-0^5 Two- Blade Propellers at Forward Mach Numbers to 0.725 To Determine the Effects of Compressibility and Solidity on Performance. NACA Rep. 999, 1950. (Supersedes NACA ACR»s 4A10 and kBl6.) 6. Betz, Albert: The Theory of Contra-Vanes Applied to the Propeller. NACA TM 909, 1939. 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'■; m ., , . ^ ... . ■. ■ ■' ^e 0 -2 .4 .6 .8 I.O I.2 1.4 I.6 Advcnce ratio, J 0 :.■ •' : , ■ ■ . (' ■ • ; . . ■ ■ CM 1 ■ -] ' ■. ; ■ ■J ■ . ■ . ■IT .._ CD O. 1 -JL. '-~\ ::'■'. 1 ;■ ;.: : ■ '■ : .. ,- «r -■]': ,.■■ .i .;: ' ■, ., -Ö ! . ; K m U ■ _jj ;;. :;: ;,.■ . ■' . ; .. .: . ■ ■ : ■ _y :.. ■. !■ 1 ■ " ■ . ■ . : 0 ;;: ii^ . 1 i ■ : ■ r.. < 1 - i ^ ■ ' ■ -. t Ü ;:: - ■:: .! ; ■■ ■•: - - ' f ::.' '.: :'■ . . . " -\ g : ,; .: ~-^r : ,:;.- , -.. y : , < J. ' i ' CD ■ . -.. ■ I : ' cd :'■ J& . ■ ■ ■;: Ji • £> r ■■ ■ ■ ■ ;• CO O . ■ ..> . :nt::.i C£ ^ ■■ :; ■ ' «M . . . . : : .: •; ■- ■ #^i .; ..-. ;■ ■ O w Ü : ■^ : • :-■: .■ 0 • ■''■ -W ;•■ 1 '■■ . „ ■ ■■ . •r-l ^1 ■ ■' j. ■ '■■ ■' -ii ■'■ ■ : iii. S p ■■.,' . .- ■ CD . ;■■ ' J ,: O cd : ..; .. : a " !' ' j '; . . ■ O i- ■j ... - ... . - .. :■- j '1; , . : ■ :' •;. 1 ■ : ' i ! :. :- t> . i-L. -; :, . 1 . , ■ ■ ■ t .:-•<■ •;^i .1; CD U :: :■]. ^ : -• ".;'! • 0 !;. .-■■•-; _ J : : 1 ..!..... - -.. • .:;;;, U ■1 t 0 J c ) a C > c > > c r CM Q O J-0 'luaioj^aoo isnjijj. 20 NACA TK 2881 a ts CD M u> -3 •rH ■a O Q) O S O O 0 0 U 1 u c 0 is c^ ■0 0 <u CJ < w Ix, do 'juapjjjaoo Jamod NACA TW 2881 21 Yi 'jsqiunu i)oow op <P IPP T) CO T3 "3 .—f H g Q) 0 0 u «fH c W c- •u < '""^ 0) Ü tuO •r-4 to m CD 1—1 .—1 CD o u (X m o 1 CXI CD 00 I O < u t CD T) CO O CD +-> o CO H CO A U CO CD to •r-l J-0 'juapi^aoo jsrugj. NACA TN 288l 23 o CD •rH Ü a CD O Ü fH CD O ft CD •1-1 +-> Ö o O 00 CD fc dQ 'luajoj^aoo -laMOd 2k NACA TW 2881 2 '";a J6 w 1.0 12 1.4 Advance ratio, J 18 -m 2.0 (c) Efficiency. Figure 8.- Concluded. WACA TN 288l 25 .O O 0) o c o i Ö 0 •rH o CD O O at a 8 rH CM TT (I) CD a. CO 13 Ö o •rH 3 rH • ««« •>■■■■■!■>■ •»•■■■■■«» ■•■■•■■■! aa •••■■■• Baa>aaaa■■■■■«•■■•■«■■«■•«■«■■■■■■■] ■ ■Its ■■»!• •■■■•< aaaataaaaata ■■■■)■■■■■■ ■■«■■■■«■■nntni •••■•■••a tKIlMl»>M»«rlini»i«H)HHa«« «■«•■■■■■I ■•••■■■■■•■»■» « i"5HS55aS! "5C?;22S!S CM —- ■ -> 9 u c o 3 •1-1 o 0 3 50 =§ 0 Ü Ü Ü u u 0) & OS 0 ft pi >■••■•••■■••■■•■■•■•• IsisssH :sas::s:ss isHi laaaaaaaaaaaI•••••■••< laaaa Saaaaaaiaa «a•■>:' iHiuMMnniiMii ■ •«••■»■■••■«•••••■'••ui iaaaasaaiiBaiBaaaa-.D8Baig: IHIMIItmnMII «1MMI eg laiaaaKiaMBaaaaaiiaHBivvaaiiaaaaf »Baavia3aa«anHDfl«iaaMaai •<•■••••••••■ .»»-.••» l«l)IHtHMBaiK«HII''M)<llli< laaaaavaa inMMMiri ««»«»#«•>• i ■•«aiMi" «■■■< la ■«•■tHMi »ni«u«a auaanaBwajBtaBBaaiiBa« ■■»■■»«■« «aw»i !iiaBaaBBaaaaaBBaäB«BiaiiBM>naitiaaaa«aiaaaBBaBaaBa[BaaaaaBani )■ B«BtiBBBBaaaaB«a3jiaaic»aa'a'-.sataiaauaaNBitB>BaBBBaBa« ■■>!>! iaca»«aaaatM aM-iMaaaataaaiMiaaiaiafaaitaait ::u ::::::::::::::::i^:::::::3:::r:;:::::::::;:::-::: I:::;::::;.:;:::::::::::::::::::::::::::::::::::::::: CM CO O <0 Q Q CM Q dQ 'juapj^soo jamod NACA TW 2881 27 is .8 n F'O .. 19? -5 ;UJ 1:4 ;3 ".2 M '';; ■l :t - I:' rJ 1 , )"' ■if ;: 1 ,' :■: w — - ■ ^ ■';f!; . ~^_ :.r ■•]■'. Ft' 1 -■ "•:■;'- .>t:n^ \| r \ ■■ i-.-i; tif ru ..g\|,;. •):. tiS it ^\|;T -■-■[■MX ^ -i- :,- ■•.••■^■S r :■ (' ... A: ir :wuii.:.iit ";'>:}::. «:. "'.■;! t ■■■ ■[:■.;■ ■ ■". ;-i "4£, «,.'" :--r __ ;-':\| ' :--::\|-;. ■- ■.''.■ - ■:: i ..!..' (iji k~ . ■■t l"-."^ ■> :.'-:U: : - ■ j. \ ■ i :. '■■ Li :M'- ::;,:, .; ■ ' V . ':'i .' i . ' : r.;."- ■-f-; L;}::-, •/; I • .:. 1 ..,;-i-. ■■ i-;~ ■4! ■pv '•:.• tlT— -..,/. -:f:; •, A ■■ fi\ ■ .. „:J;__J L..„.X ...i.i^rJ';,. -...: -'- > ■ii : Helical tip Mach numbe far k» -' ;;-' 1 -; :'..r.. ;::" :"T~; .1 ■ ■ TiT fH J^ \ /r;.v 'H:^ '■;cs h f -1 i rF' J'' !;...l.... ;'■!.'. '■•■, .fdj .4: V '';ll L^rj .:!: «1 nifi ^ i;S J f. -^ ^l> r4 r^- iä \|j- ':\ -'"P =f°" jii- ; i" ■fi. ;■ ; 1 ;..'. , ' ! Trtr rcrct r.li ~:!5 — ■>— r; -.;; ■ I1' ' -H !'- i-,r--. '.;:-;. '~:ä ..; ,,.. ■;; !'': : Ji V-' ■■■;- '.. 1. '.;:: ~V: ■■ ■'■ TT ■< ■ ^ )•' #t!:: ' 1:.: ':':; :~n ' j:. ;i . .. ' j ' '-'i :..■' ":j .- 1 i "~ ! ■ I ;; 1 .: 1 ■ -' ! i-r-i ■ '. !' i- 1' - 4- ■ ! ' ■ ( ■■. 1 1 .., , r- ~.l_j_i. .1 1 -'.-I .1 ' T. Air-stream W ach number « .... _ ,-L, • ! ':'"! 1"T I -,.-, - N. ! 1 i ! ! i .1 :.:..! i ! i ... „ _ 1 ! : j 1 ■ i :■ '• 1 ■ 1 . !,> x-p\| j ! 1 ' I r' _j_j^ ■■■■LLLL- f t ■:ji i- j . ■ ■.i ■ J 1 ! - -j— i ,;,i i. . -. J .i i ' 1 V 1. 1 i- ■ 1 1 ! "n 7R R = 20^ "1 ? 51° ':' 3iQ [.■.(■; 1 I , — -- :4- !..\| ...1 ................. .,-:-. -j. 3 .2 A .6 .8 1.0 I 2 1 4 I.( L4 1.2 .0 •8 S; -Q e c o o .2 Advance ratio, J (c) Efficiency. Figure 9.- Concluded. 28 NACA TN 2881 o. ü c 'o 03 O Ü 0) $ o Q. C o Ü c Q) 0) o u ■3 1.4 1.6 1.8 2.0 2.2 "Advance ratio, J 2.4 2.6 2.8 (a) Air-stream Mach number, 0.56. Figure 10.- Characteristics of two-blade NACA 10-(3)(062) -045 propeller at high forward speeds, p = 45°. U • f DJC\t NACA TN 2881 29 .22 F a. ü O 03 O O 0> o a. c o ü c «3 °o it a> o u ZJ 2.0 2.2 2.4 Advance ratio, J (b) Air-stream Mach number, 0.60. Figure 10.- Continued. 30 NACA TN 288l 0. o c. a> o o s o. Q. Ü c g> 'o it: « o u CO 3 1.8 2.0 2.2 2.4 2,6 Advance ratio, J (c) Air-stream Mach number, 0.63. Figure 10.- Concluded. NACA TN 2881 31 8 CD O 3 EH "c? In O CD CO CO o CD .—I f—I CD D. O OH LO d 9 •rH I CO o, I o 1-1 < Ü t CD " TJ CO o CO 0 Ö CO TJ CO -t-> Ü CO u cd o CD U •■-I J-O Juapijjaoo jsnjqx 32 NACA TN 2881 CM rd O rO t T) +2 CD CM cJ -3 0 Ö CD Ü 3 0 0 £ 0 0 O Cvi U c CD O O 0 0 > U m ■0 CD T-H < £ TH O f\|> Ut h 10 s> •iH £> pH <: fO dg 'juapu^oo JSMOd KACA TN 2881 )M 33 y\l jsquuni' I\|ODV\\| ^ > r-H r) O Ei 0 0) 0 •l-l O •i-t «M «H w i-i i-H 0 g, pq CT> 00 10 üo <$ to c\j 3^ NACA TN 2881 a 0 ■5 O 0 O O EH a 00 T3 0 CD to O 8 0 r—I 0) a. o a. 00 o, CO o o t 0 T) cd 1 o £ O CO Ü Ö 0 Ü cd cd .0 O 1 CNJ 0 u 3, Fn -i-O 'juapyjaoo lön-iqi MCA TN 288l 35 Si g 5° 5£ _ _ ^ w o dO 'tU8!O.IJJ800 J9M0d 00 to 73 (I) 0) 3 O •1-1 •a (I) 0 O C) Ü k 1 <1) • O ft 0 36 NACA TN 2881 <M y\l 'jsquunu qoD\|/\\| o 00 CD CJ 1 1 ■ :. - .. .■ : , ;... ■ ■ - . . : ' ■'■:• 00 •JrO jlD ■ fO - ro CJ rd O ro GO .to OJ H CD •rH Ü w .-. ■.. ; ■ ■ ■ ■ ... , 1 •' ■ .. : .• !, ., .. •:-r! . _ ■ . ;.: •yj T\ ■ '■'■ '' :; z? . ; ■ .-.- ■' ■ .. 'li' ■ ' ; .' : •■ -r ■ -" . .■ I . .- . , :■ -.. ; . ■' . ■ ■ ■■■ ! . .:'■ ■ ■ ■ : ;-;. ■::■ ■1 ■ . . :; : !.: ' ■ ■■: ,■■' » . . ..■ ■ ,, ■ ■ :. ! . : . ■ . ■ . : ; . ' • ■ :■:' ■ ■ ■ .. : ■. ■ :. ... ■■ ,- [a-J ■ 1 ■ , : ■ ■ ,■:; : : : :.. . ij , -!;= s-i- -, . •• ■ ■ ■1- "I : : ■ ■ . . : . .. ; n-:.. ■■ : ■pi G1 ' :. . ■ L. _i ±=',J '■•! '■ :. :. ;:: ■ r^" [' ; !4i ft; : .:■ . ■. — , . \ . : — ■■■! \, ! ,; ■:\|. !. .~h 3:1= = ^4- \ ;' ... - j , L V -, j . ;. .'j ; :.- 1- ,, ■[■■< 4 V ■ ": 1 4 ■ / , ■ ■ -.■■ :.: ,A ',.. . :' . ' ■ :' pi ... : ' : ■ 1 ; i- ; '; \ ■ ■ . 1 ■ :: ' - ^ : F ■ \ : ■ : ' ■ ■■: '■ : ' ■■■ -^- ■ '■■ ;, ■'■■ H . ■4, !'»=i ~ :.: r- .. ' . -"- ■" «1- ■ . L- .;" H- : .— .. . ■ ■: :.:; F" ' OJ CO -3 :■-. . .1. 1 , ^ . : .-■ :i I-, ■ ■ ; ;; ■ ■ : • . ;,: ::■ ■ \ , : ': ,:i-i !;:. ; TJ ■ ■i ' . ■ ' , ■ \ ' ' ' : fl) ■•■ : y . , ■ : ' . . . \ \ ".' T? il ; : j.. ■■; . ■■ 1 ■ \ u ■ ■ •■ . ■ : :■ ■■i • ^ ':. — - :,; .. h _ 0 0 - 1 ' "-. ~ - — a ^ - — ■ 4- fe — '- .2 0 Q CO g c 0 > . . "" ■ ■ .: 1 . \ ' .. ; 1 : ■ ■ . . , .. : 1 ■■; ■ : :. ■:•:■ . \ , ;. \ ■ .. ■ : !. CO ■ ■ ■ '. - 4- \ -r— rH -^ i ■ ,; i—; 1__ r: ==> " 71 —. J ■ ii- 4 '■ 1 .' J — < CD •rH : . i ■ .. : ■-.: . ■■ ' ■ ■ -1- -. ■'■J ..:, ■ 1 r~ : , '! . ■ . . 1^ 4 . . - : , : -.. .:: ■... ) \ . ::: ^- ' ^ -.- :.. ■■ : . S4" :■ h ^ "" ■:■ , , . . — ' Bi . <£ ■:. • c 1 ■ T I- ■ A ■ ' :■ . ; . ' : .( . ;. " - rr ■I! t. ; V :■ •' ' r~ . — — —. ■ ; \ — ■=r. j —- —'■ - " S" ^ Q °P iß CJ . 0 .;.-■ ■ «v ■ . \ . M : :: ■ f) ' \ ;. .: \ .: • .... „ ^ -^ — '■ ■r :' a» ,™ ,.;. : '\ -a he r .... b- . :\ r 1 ■-V ■ N .:' ' .' . V -! \ { ... ■ 3 -~ C ■-- __ h- L 0 :\Qa" : s -. : £. > : J 1 .. £ s, : 1 ' ? i J M _\| , • «~- ■: ■ : ' (J O . . .. f Tl • . c » 1 ■ . 5 . :► ! , E 0 1 w : "; . C L { ■ : : ■ a:-i- ■ ■ :; ' . -. : \| ■ . ..; : .... ■ . ;■ j ,,;-;r ■■ \ ■ J..L i il . C > t'ji-J- , '■ : ■• 1 • :. ; ;:; .;\| :1 1. . j . j i ii( . 1 ' ■ ' ■'■! .' ':}: ■ ■: ■! :':r .■'■ 31 :i, i4J- ■ —T ;.:; ■ ; ■V 1' .'.■I :,\| i- ■• , :- -w T ■ ■ M ;;;'. fin v J.\ Ji -:' O & c 0 f I Q I c' k I )U 31 °! < »3 ro c M 0 NACA TN 2881 37 Q. Ü c 0) o gj o o .0 $ o a. c o Ü c » 'o it «0 -5 IB 2.0 22 2.4 Advance rqtiq, J Figure 13.- Characteristics of two-blade NACA 10-(3)(08)-045 propeller. Rotational speed, 1500 rpm; ß0 75R = 45°. 38 NACA TW 2881 " ., , : . -•" ■ I>I , 11 , 1 1 ■ 00 !:' wit:) ;1 :' CO a (D •rH O CD O Ü +J En u~ M ..-J i^i^T ":■ ■1 '::' "■ ~^ : ; : :■ 1 ■-; "L :3 5\F--i^" fH , 4 i! kf 2 T ii-;»; ix;- cvj O O ^iL^^? •■ CD ,%w~ ' ■ ■' '-''el 73 CD ■:'<\ "''& ■ ■ 1! ^! ' J ^. ■ <&'' ' j:;i. ■,' -:.;i ■ jä^Ll ;■:.!.-{ ■-} '■ !■ . ::i':\|iEii w CO - ' ' :"'> w - 0 en :;.'■ ;. ^ .,; ^'■''■Hat- '' • ■ '■■ l<^%~ -f-> q ■ . ' 'Ä' tf 1 ■ ' &'' ' ;'>'i[T: c\i ■ITl"JF ^1 ■ ^7 ■:■ ■■■■'■IT! CD r-H rj^ : ■'■;. :. ■ ; • '-" Ml ■■']-; m CD 9 , ■ • • -■ J ■■' • ■ Ül^ ~ 0 . .'. , ■:■: ■■: -■■•■' ■ ■'■ i >' t^ .iS- " , : f' i ^11:^ LO '; sj7 ■ r .': . 1 ■' . Ui-.-. ;. - ^'' JJcU-d. m X ■ : j,tf ■ . '! :-: '■■ . r^i+ to 1 M -f 1- ■ f ■ — ^ 9 ..■ 1 ■ ■;■ ef ■ \ 1; ;•■! PT ii;; ti' 0" 00 i ' ,'■ : ■ .;] , ... . :\|... 0 . ■ •; .; :■ : i; !; '■■■.;;'. \ F": 77 ^PmU ° CO . ■ ;: :. : i ■;.^ '■■'. \| O' 1 ■ f rT' 'S —' • , . : . ; .; :. ■ ' ■ i :• j ;. jUil §' 0 - < ^ ; ' - ■ : •'2'' . ■ "Iftlfe^ s T-l i":, .■ . .. .: ::■ . : -'"'ii/v ■;■. ;: FT i H" trt s^ Si ■ 2' 0 t -iiitJUirS <M < r ! ■:-: - ': ■ .: •: : j "; H ;■.: 3'Bl "J ~ ■... ■ :■ . ■< : : : ■". > >-< ' :'-ie clit • ■.. ^' ; '.!'••>■• !lu\|- -. CD ■ ■ ■■; :.■ 1 1 ■'■ ■ -"■-.. >'~\ „tpjip-p-'" O T3 CO 3 '■ : : S7- ~~t~'''; L_ . ; . , ^ ;.. !. 1 -, ■:■, , ■'•■'■ vu-4: ä^Ä;' . ■. - ' ■ P ^-^ffiii ■ 0 ■ ■- ' I... . irijtief.-f - . ■ >' ■ ' IT®. °R O . : :-; s-" .■';..;. ;, ;. :rf co : ■ . :."T iH[ £ Ü id :: :■■ -; , , ;. ■■•,: T -I^ Si«, ' : M . .. trtTTT ... ■ ^^01 \ (°- •rH ^1 j.-; ;. ■ ■: I . ;; :■ "TTTT ;■;.:: i ■" fe'M CD -t-> v .- ■: i ■ .TtlTv,^ - O CO ; : : :■' : ;■: ■' '■ ^'i'SErt- fH ; ■ ■: 1 '! ^ CO "Ti-i i t^r 0 ■ ^^S' : ' I I'tt ? i' t. H'T'^ w ■■ t .- _^i. ä. : -.! = -= : M luliifc;" i^igjri CD ;;' r.'iS^ 0 Mi ■ .:• ■:- iJ '■': ■:' ■ y- ^äi^iiOo ' hO £> ^ C J c > a c ) > 10 Q Q Q ° (X( J-0 '\|U3!0!^.9O0 isnjqi WACA TN 2881 39 —: •H T! a (1) CD 3 •H rl <0 •>' Ü la S •« 0) 0 0 O 2 Ü : 0) u 1 c\i c 0 > •0 < 1 ft 1-1 «V d0 %U8!9j^803 JSMOd ko NACA TW 2881 yi 'jaqwnu IJODW 00 to. 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14511	https://www.wikihow.com/Simplify-Algebraic-Fractions	Home Random Browse Articles Quizzes All QuizzesTrending Love Quizzes Personality Quizzes Trivia Quizzes Learn Something New Forums Courses Happiness Hub Play Games This Or That Game Train Your Brain Explore More Support wikiHow About wikiHow Log in / Sign up Terms of Use Categories Education and Communications Studying Mathematics Fractions How to Simplify Algebraic Fractions Download Article Co-authored by David Jia Last Updated: February 24, 2025 References Download Article Simplifying Fractions \| Simplifying Algebraic Fractions \| Tricks for Difficult Problems \| Q&A \| Tips \| Warnings X This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 148,418 times. Algebraic fractions look incredibly difficult at first, and can seem daunting to tackle for the untrained student. With a mixture of variables, numbers, and even exponents, it is hard to know where to begin. Luckily however, the same rules needed to simplify regular fractions, like 15/25, still apply to algebraic fractions. Steps Method 1 Method 1 of 3: Simplifying Fractions Download Article 1 Know the vocabulary for algebraic fractions. The following terms will be used throughout the examples, and are common in problems involving algebraic fractions: X Research source Numerator: The top part of a fraction (ie. (x+5)/(2x+3)). Denominator: The bottom part of the fraction (ie. (x+5)/(2x+3)). Common Denominator: This is a number that you can divide out of both the top and bottom of a fraction. X Expert Source David Jia Academic Tutor Expert Interview For example, in the fraction 3/9, the common denominator is 3, since both numbers can be divided by 3. Factor: One number that multiples to make another. For example, the factors of 15 are 1, 3, 5, and 15. The factors of 4 are 1, 2, and 4. Simplified Equation: This involves removing all common factors and grouping similar variables together (5x + x = 6x) until you have the most basic form of a fraction, equation, or problem. If you cannot do anything more to the fraction, it is simplified. 2. 2 Review how to solve simple fractions. These are the exact same steps you will take to solve algebraic fractions. X Research source Take the example, 15/35. In order to simplify a fraction, we need to find a common denominator. In this case, both numbers can be divided by five, so you can remove the 5 from the fraction: 15 → 5 3 35 → 5 7 Now you can cross out like terms. In this case you can cross out the two fives, leaving your simplified answer, 3/7. Advertisement 3. 3 Remove factors from algebraic expressions just like normal numbers. X Research source In the previous example, you could easily remove the 5 from 15, and the same principle applies to more complex expressions like, 15x – 5. Find a factor that both numbers have in common. Here, the answer is 5, since you can divide both 15x and -5 by the number five. Like before, remove the common factor and multiply it by what is “left.” 15x – 5 = 5 (3x – 1) To check your work, simply multiply the five back into the new expression – you will end up with the same numbers you started with. 4. 4 Know you can remove complex terms just like simple ones. The same principle used in common fractions works for algebraic ones as well. This is the easiest way to simplify fractions while you work. X Research source Take the fraction: (x+2)(x-3) (x+2)(x+10) Notice how the term (x+2) is common in both the numerator (top) and denominator (bottom). As such, you can remove it to simplify the algebraic fraction, just like you removed the 5 from 15/35: (x+2)(x-3) → (x-3) (x+2)(x+10) → (x+10) This leaves us with our final answer: (x-3)/(x+10) Advertisement Method 2 Method 2 of 3: Simplifying Algebraic Fractions Download Article 1 Find a common factor in the numerator, or top part of the fraction. The first thing to do when simplifying an algebraic fraction is to simplify each part of the fraction. Start with the top part, factoring out as many numbers as you can. X Research source For an example, this section will use the problem: 9x-3 15x+6 Start with the numerator: 9x – 3. There is a common factor to both 9x and -3: 3. Factor out the 3 like you would any other number, leaving you with 3 (3x-1). This is your new numerator: 3(3x-1) 15x+6 2. 2 Find a common factor in the denominator. X Research source Continuing the example from above, isolate the denominator, 15x+6. Again, look for a number that can divide into both parts. Here you can again factor out a 3, leaving you with 3 (5x +2). Write in your new denominator: 3(3x-1) 3(5x+2) 3. 3 Remove like terms. This is the stage where you really get to simplify the fraction. Take any terms that are in both the numerator and the denominator and remove them. In this case, we can remove the 3 from both the top and the bottom. X Research source 3(3x-1) → (3x-1) 3(5x+2) → (5x+2) 4. 4 Know when the equation is fully simplified. A fraction is simplified when there are no more common factors in the top or the bottom. X Research source Remember that you cannot remove factors from inside the parenthesis – in the example problem you cannot factor out the x from 3x and 5x, since the full terms are actually (3x -1) and (5x + 2). Thus, the example is fully simplified, making the final answer: (3x-1) (5x+2) 5. 5 Try a practice problem. The best way to learn is to keep trying to simplify algebraic fractions. The answers are underneath the problems. 4(x+2)(x-13) (4x+8) Answer: (x=13) 2x2-x 5x Answer:(2x-1)/5 Advertisement Method 3 Method 3 of 3: Tricks for Difficult Problems Download Article 1 ”Reverse” parts of the fraction by factoring out negative numbers. For example, let’s say we have the equation: 3(x-4) 5(4-x) Notice how (x-4) and (4-x) are ‘’almost’’ identical, but you can’t cross them out because they are reversed. However, (x - 4) can be written as -1 \ (4 - x) in the same way that you rewrite (4 + 2x) as 2 \ (2 + x). This is called “factoring out the negative.” -1 3(4-x) 5(4-x) Now we can easily remove the two identical (4-x): -1 3(4-x) 5(4-x) Leaving us with our final answer -3/5 2. 2 Recognize the difference of two squares when working. The difference of two squares is simply one squared number subtracted by another, like the expression (a2 - b2). The difference of perfect squares always simplifies into two parts, adding and subtracting the square roots. In every case, you can simplify the difference of perfect square as follows: a2 - b2 = (a+b)(a-b) This can be incredibly helpful when trying to find like terms in algebraic fractions. Example: x2 - 25 = (x+5)(x-5) 3. 3 Simplify any polynomial expressions. Polynomials are complex algebraic expressions with more than two terms, like x2 + 4x + 3. Luckily, many polynomials can be simplified using polynomial factorization. The previous expression, for example, can be rewritten as (x+3)(x+1). 4. 4 Remember that variables can be factored out as well. This is especially helpful in expressions with exponents, like x4 + x2. You can remove the largest exponent as a factor. In this case, x4 + x2 = x2(x2 + 1). Advertisement Community Q&A Search Add New Question Question Why does it work exactly like this all the time? Donagan Top Answerer Because the properties of and operations with numbers never change. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 6 Helpful 9 Question When factoring algebraic fractions, how do I know which one will have a plus or minus sign? For example, 12x^2 + 43x+ 35 / 6x^2 +5x - 21? Donagan Top Answerer This is known as "solving by inspection." In other words, you just look at ("inspect") the expression you're factoring, and choose plus and minus signs so that the factors, when multiplied together, result in the original expression. In the example cited, 12x² + 43x +35 factors into (3x + 7) and (4x + 5). 6x² + 5x - 21 factors into (3x + 7) and (2x - 3). In each case the signs were chosen so that multiplying the factors together would result in the original expression. In this case there is a factor of (3x + 7) in both the numerator and denominator, so that they cancel each other. The final, simplified expression is (4x + 5) / (2x - 3). Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 6 Helpful 6 Question What if there's a question like this: (1–y²) / (–y² – y + 2)? Donagan Top Answerer First simplify a little by factoring (-1) from both numerator and denominator: -(y²-1) / -(y²+y-2). The (-1)'s will cancel each other, and you're left with (y²-1) / (y²+y-2). Factor numerator and denominator: [(y+1)(y-1)] / [(y+2)(y-1)]. The two (y-1)'s cancel each other, and the final fraction is (y+1) / (y+2). Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 4 Helpful 6 See more answers Ask a Question 200 characters left Include your email address to get a message when this question is answered. Submit Advertisement Tips Check your work when factoring by multiplying the factor back into the equation -- you will get the same number you started with. Thanks Helpful 9 Not Helpful 3 Always factor out the largest numbers you can to simplify your equation fully. Thanks Helpful 6 Not Helpful 3 Submit a Tip All tip submissions are carefully reviewed before being published Name Please provide your name and last initial Submit Thanks for submitting a tip for review! Advertisement Warnings Forget the laws of indices, and you're dead meat. Therefore, stuff them into your brain at all costs. Thanks Helpful 8 Not Helpful 4 Advertisement You Might Also Like How to Add and Multiply FractionsHow to Change Mixed Numbers to Improper Fractions How Do I Measure ¾ Cup?Easy-to-Follow Methods for Adding FractionsMultiplying Fractions with Whole NumbersHow to Find the Least Common DenominatorHow to Write a Fraction on a Computer (PC, Mac, or Word)How to Divide Mixed NumbersHow to Solve Fraction Questions in MathHow to Convert Repeating Decimals to FractionsHow to Multiply FractionsHow to Understand FractionsHow to Convert Improper Fractions Into Mixed NumbersHow to Divide Fractions by Fractions: 4 Easy Steps Advertisement References ↑ ↑ David Jia. Academic Tutor. Expert Interview ↑ ↑ ↑ ↑ ↑ ↑ ↑ About This Article Co-authored by: David Jia Math Tutor This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 148,418 times. 67 votes - 54% Co-authors: 15 Updated: February 24, 2025 Views: 148,418 Categories: Fractions Article SummaryX To simplify algebraic fractions, start by factoring out as many numbers as you can for the numerator, which is the top part of the fraction. Next, find a common factor in the denominator, which is the bottom part of the fraction, by looking for a number that can divide into both parts. Then, take any terms that are in both the numerator and the denominator and remove them. When there are no more common factors in the top or the bottom, the fraction is simplified! For more tips on factoring, read on! Did this summary help you? In other languages German French Indonesian Spanish Italian Russian Dutch Print Send fan mail to authors Thanks to all authors for creating a page that has been read 148,418 times. Reader Success Stories Anonymous Aug 2, 2017 "It has fantastic ideas. Thank you wikiHow." Share your story Did this article help you? Advertisement Cookies make wikiHow better. By continuing to use our site, you agree to our cookie policy. Co-authored by: David Jia Math Tutor 67 votes - 54% Click a star to vote % of people told us that this article helped them. 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14512	https://faculty.gardner.wwu.edu/gorizia12.pdf	THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 R. J. GARDNER Abstract. In 1978, Osserman wrote a rather comprehensive survey on the isoperimetric in-equality. The Brunn-Minkowski inequality can be proved in a page, yet quickly yields the classical isoperimetric inequality for important classes of subsets of Rn, and deserves to be better known. We present a guide that explains the relationship between the Brunn-Minkowski inequality and other inequalities in geometry and analysis, and some of its recent applications. 1. Introduction About a century ago, not long after the ﬁrst complete proof of the classical isoperimetric inequality was found, Minkowski proved the following inequality: V ((1 −λ)K + λL)1/n ≥(1 −λ)V (K)1/n + λV (L)1/n. (1) Here K and L are convex bodies (compact convex sets with nonempty interiors) in Rn, 0 < λ < 1, V denotes volume, and + denotes vector or Minkowski sum. The inequality (1) had been proved for n = 3 earlier by Brunn, and now it is known as the Brunn-Minkowski inequality. It is a sharp inequality, equality holding if and only if K and L are homothetic. The Brunn-Minkowski inequality was inspired by issues around the isoperimetric problem, and was for a long time considered to belong to geometry, where its signiﬁcance is widely recognized. It implies, but is much stronger than, the intuitively clear fact that the function that gives the volumes of parallel hyperplane sections of a convex body is unimodal. It can be proved on a single page (see Section 6), yet it quickly yields the classical isoperimetric inequality (21) for convex bodies and other important classes of sets. The fundamental geometric content of the Brunn-Minkowski inequality makes it a cornerstone of the Brunn-Minkowski theory, a beautiful and powerful apparatus for conquering all sorts of problems involving metric quantities such as volume, surface area, and mean width. By the mid-twentieth century, however, when Lusternik, Hadwiger and Ohmann, and Henstock and Macbeath had established a satisfactory generalization of (1) and its equality conditions to Lebesgue measurable sets, the inequality had begun its move into the realm of analysis. The last twenty years have seen the Brunn-Minkowski inequality consolidate its role as an analytical tool, 1991 Mathematics Subject Classiﬁcation. 26D15, 52A40. Key words and phrases. Brunn-Minkowski inequality, Minkowski’s ﬁrst inequality, Pr´ ekopa-Leindler inequality, Young’s inequality, Brascamp-Lieb inequality, Barthe’s inequality, isoperimetric inequality, Sobolev inequality, en-tropy power inequality, covariogram, Anderson’s theorem, concave function, concave measure, convex body, mixed volume. Supported in part by NSF Grant DMS 9802388. 1 2 R. J. GARDNER and a compelling picture (see Figure 1) has emerged of its relations to other analytical inequali-ties. In an integral version of the Brunn-Minkowski inequality often called the Pr´ ekopa-Leindler inequality (12), a reverse form of H¨ older’s inequality, the geometry seems to have evaporated. Largely through the eﬀorts of Brascamp and Lieb, this can be viewed as a special case of a sharp reverse form (32) of Young’s inequality for convolution norms. A remarkable sharp inequality (36) proved by Barthe, closely related to (32), takes us up to the present time. The modern viewpoint entails an interaction between analysis and convex geometry so potent that whole conferences and books are devoted to “analytical convex geometry” or “convex geometric analysis.” The main development of this paper includes historical remarks and several detailed proofs that amplify the previous paragraph and show that even the latest developments are accessible to graduate students. Several applications are also discussed at some length. Extensions of the Pr´ ekopa-Leindler inequality can be used to obtain concavity properties of probability measures generated by densities of well-known distributions. Such results are related to Anderson’s theorem on multivariate unimodality, an application of the Brunn-Minkowski inequality that in turn is useful in statistics. The entropy power inequality (48) of information theory has a form similar to that of the Brunn-Minkowski inequality. To some extent this is explained by Lieb’s proof that the entropy power inequality is a special case of a sharp form of Young’s inequality (31). This is given in detail along with some brief comments on the role of Fisher information and applications to physics. We come full circle with consequences of the later inequalities in convex geometry. Ball started these rolling with his elegant application of the Brascamp-Lieb inequality (35) to the volume of central sections of the cube and to a reverse isoperimetric inequality (45). The whole story extends far beyond Figure 1 and the previous paragraph. The ﬁnal Section 19 is a survey of the many other extensions, analogues, variants, and applications of the Brunn-Minkowski inequality. Essentially the strongest inequality for compact convex sets in the direction of the Brunn-Minkowski inequality is the Aleksandrov-Fenchel inequality (51). Here there is a remarkable link with algebraic geometry: Khovanskii and Teissier independently discovered that the Aleksandrov-Fenchel inequality can be deduced from the Hodge index theorem. Analogues and variants of the Brunn-Minkowski inequality include Borell’s inequality (57) for capacity, employed in the recent solution of the Minkowski problem for capacity; Milman’s reverse Brunn-Minkowski inequality (64), which features prominently in the local theory of Banach spaces; a discrete Brunn-Minkowski inequality (65) due to the author and Gronchi, closely related to a rich area of discrete mathematics, combinatorics, and graph theory concerning discrete isoperimetric inequalities; and inequalities (67), (68) originating in Busemann’s theorem, motivated by his theory of area in Finsler spaces and used in Minkowski geometry and geometric tomography. Around the corner from the Brunn-Minkowski inequality lies a slew of related aﬃne isoperimetric inequalities, such as the Petty projection inequality (62) and Zhang’s aﬃne Sobolev inequality (63), much more powerful than the isoperimetric inequality and the classical Sobolev inequality (24), respectively. There are versions of the Brunn-Minkowski inequality in the sphere, hyperbolic space, Minkowski spacetime, and Gauss space, and there is a Riemannian version of the Pr´ ekopa-Leindler inequality, obtained very recently by Cordero-Erausquin, McCann, and Schmuckenschl¨ ager. Finally, pointers are given to other applications of the Brunn-Minkowski inequality. Worthy of special mention here is the derivation of logarithmic Sobolev inequalities from the Pr´ ekopa-Leindler inequality by Bobkov and Ledoux, and work of Brascamp and Lieb, Borell, McCann, and others on diﬀusion equations. Measure-preserving convex gradients and transportation of mass, utlilized by McCann THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 3 in applications to shapes of crystals and interacting gases, were also employed by Barthe in the proof of his inequality. The reader might share a sense of mystery and excitement. In a sea of mathematics, the Brunn-Minkowski inequality appears like an octopus, tentacles reaching far and wide, its shape and color changing as it roams from one area to the next. It is quite clear that research opportunities abound. For example, what is the relationship between the Aleksandrov-Fenchel inequality and Barthe’s inequality? Do even stronger inequalities await discovery in the region above Figure 1? Are there any hidden links between the various inequalities in Section 19? Perhaps, as more connections and relations are discovered, an underlying comprehensive theory will surface, one in which the classical Brunn-Minkowski theory represents just one particularly attractive piece of coral in a whole reef. Within geometry, the work of Lutwak and others in developing the dual Brunn-Minkowski and Lp-Brunn-Minkowski theories (see Section 19) strongly suggests that this might well be the case. An early version of the paper was written to accompany a series of lectures given at the 1999 Workshop on Measure Theory and Real Analysis in Gorizia, Italy. I am very grateful to Franck Barthe, Apostolos Giannopoulos, Paolo Gronchi, Peter Gruber, Daniel Hug, Elliott Lieb, Robert McCann, Rolf Schneider, B´ ela Uhrin, and Gaoyong Zhang for their extensive comments on previous versions of this paper, as well as to many others who provided information and references. 2. A first step An old saying has it that even a journey of a thousand miles must begin with a single step. Ours will be the following easy result (see Section 3 for deﬁnitions and notation). Theorem 2.1. (Brunn-Minkowski inequality in R.) Let 0 < λ < 1 and let X and Y be nonempty bounded measurable sets in R such that (1 −λ)X + λY is also measurable. Then V1 ((1 −λ)X + λY ) ≥(1 −λ)V1(X) + λV1(Y ). (2) Proof. Suppose that X and Y are compact sets. It is straightforward to prove that X + Y is also compact. Since the measures do not change, we can translate X and Y so that X ∩Y = {o}, X ⊂{x : x ≤0}, and Y ⊂{x : x ≥0}. Then X + Y ⊃X ∪Y , so V1(X + Y ) ≥V1(X ∪Y ) = V1(X) + V1(Y ). If we replace X by (1 −λ)X and Y by λY , we obtain (2) for compact X and Y . The general case follows easily by approximation from within by compact sets. □ Simple though it is, Theorem 2.1 already raises two important matters. Firstly, observe that it was enough to prove the theorem when the factors (1 −λ) and λ are omitted. This is due to the positive homogeneity (of degree 1) of Lebesgue measure in R: V1(rX) = rV1(X) for r ≥0. In fact, this property allows these factors to be replaced by arbitrary nonnegative real numbers. For reasons that will become clear, it will be convenient for most of the paper to incorporate the factors (1 −λ) and λ. Secondly, the set (1 −λ)X + λY may not be measurable, even when X and Y are measurable. We discuss this point in more detail in Section 9. The assumption in Theorem 2.1 and its n-dimensional forms, Theorem 5.1 and Corollary 5.3 below, that the sets are bounded is easily removed and is retained simply for convenience. 4 R. J. GARDNER Prekopa-Leindler (12) Holder (11) General Brunn-Minkowski (14) Aleksandrov-Fenchel (51) Barthe (36) Brascamp-Lieb (35) Reverse Young (32) Young (31) Brunn-Minkowski for C1 domains Sobolev for C1 functions (24) Brunn-Minkowski for convex bodies (1) Minkowski’s first for convex bodies (20) Isoperimetric for C1 domains Isoperimetric for convex bodies (21) Entropy power (48) . ' . Figure 1. Relations between inequalities labeled as in the text. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 5 3. A few preliminaries We denote the origin, unit sphere, and closed unit ball in n-dimensional Euclidean space Rn by o, Sn−1, and B, respectively. The Euclidean scalar product of x and y will be written x · y, and ∥x∥denotes the Euclidean norm of x. If u ∈Sn−1, then u⊥is the hyperplane containing o and orthogonal to u. Lebesgue k-dimensional measure Vk in Rn, k = 1, . . . , n, can be identiﬁed with k-dimensional Hausdorﬀmeasure in Rn. Then spherical Lebesgue measure in Sn−1 can be identiﬁed with Vn−1 in Sn−1. In this paper dx will denote integration with respect to Vk for the appropriate k and integration over Sn−1 with respect to Vn−1 will be denoted by du. The term “measurable” applied to a set in Rn will mean Vn-measurable unless stated otherwise. If X is a compact set in Rn with nonempty interior, we often write V (X) = Vn(X) for its volume. We shall do this in particular when X is a convex body, a compact convex set with nonempty interior. We also write κn = V (B). In geometry, it is customary to use the term volume, more generally, to mean the k-dimensional Lebesgue measure of a k-dimensional compact body X (equal to the closure of its relative interior), i.e. to write V (X) = Vk(X) in this case. Let X and Y be sets in Rn. We deﬁne their vector or Minkowski sum by X + Y = {x + y : x ∈X, y ∈Y }. If r ∈R, let rX = {rx : x ∈X}. If r > 0, then rX is the dilatation of X with factor r, and if r < 0, it is the reﬂection of this dilatation in the origin. If 0 < λ < 1, the set (1 −λ)X + λY is called a convex combination of X and Y . Minkowski’s deﬁnition of the surface area S(M) of a suitable set M in Rn is S(M) = lim ε→0+ Vn(M + εB) −Vn(M) ε . (3) In this paper we will use this deﬁnition when M is a convex body or a compact domain with piecewise C1 boundary. A function f on Rn is concave on a convex set C if f ((1 −λ)x + λy) ≥(1 −λ)f(x) + λf(y), for all x, y ∈C and 0 < λ < 1, and a function f is convex if −f is concave. A nonnegative function f is log concave if log f is concave. Since the latter condition is equivalent to f ((1 −λ)x + λy) ≥f(x)1−λf(y)λ, the arithmetic-geometric mean inequality implies that each concave function is log concave. If f is a nonnegative measurable function on Rn and t ≥0, the level set L(f, t) is deﬁned by L(f, t) = {x : f(x) ≥t}. (4) By Fubini’s theorem, Z Rn f(x) dx = Z Rn Z f(x) 0 1 dt dx = Z ∞ 0 Z L(f,t) 1 dx dt = Z ∞ 0 Vn (L(f, t)) dt. (5) 6 R. J. GARDNER If E is a set, 1E denotes the characteristic function of E. The formula f(x) = Z ∞ 0 1L(f,t)(x) dt (6) follows easily from f(x) = R f(x) 0 dt. In [91, Theorem 1.13], equation (6) is called the layer cake representation of f. 4. The Pr´ ekopa-Leindler inequality Theorem 4.1. (Pr´ ekopa-Leindler inequality in R.) Let 0 < λ < 1 and let f, g, and h be nonnegative integrable functions on R satisfying h ((1 −λ)x + λy) ≥f(x)1−λg(y)λ, (7) for all x, y ∈R. Then Z R h(x) dx ≥ µZ R f(x) dx ¶1−λ µZ R g(x) dx ¶λ . Two proofs of this fundamental result will be presented after a comment about the strange-looking assumption (7) that ensures h is not too small. Fix a z ∈R and choose 0 < λ < 1 and any x, y ∈R such that z = (1 −λ)x + λy. Then the value of h at z must be at least the weighted geometric mean (it is the geometric mean if λ = 1/2) of the values of f at x and g at y. Note also that the logarithm of (7) yields the equivalent condition log h ((1 −λ)x + λy) ≥(1 −λ) log f(x) + λ log g(y). If f = g = h, we would have log f ((1 −λ)x + λy) ≥(1 −λ) log f(x) + λ log f(y), which just says that f is log concave. Of course, the previous theorem does not say anything when f = g = h. First proof. We can assume without loss of generality that f and g are bounded with sup x∈R f(x) = sup x∈R g(x) = 1. If t ≥0, f(x) ≥t, and g(y) ≥t, then by (7), h ((1 −λ)x + λy) ≥t. With the notation (4) for level sets, L(h, t) ⊃(1 −λ)L(f, t) + λL(g, t), THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 7 for 0 ≤t < 1. The sets on the right-hand side are nonempty, so by (5), the Brunn-Minkowski inequality (2) in R, and the arithmetic-geometric mean inequality, we obtain Z R h(x) dx ≥ Z 1 0 V1 (L(h, t)) dt ≥ Z 1 0 V1 ((1 −λ)L(f, t) + λL(g, t)) dt ≥ (1 −λ) Z 1 0 V1 (L(f, t)) dt + λ Z ∞ 0 V1 (L(g, t)) dt = (1 −λ) Z R f(x) dx + λ Z R g(x) dx ≥ µZ R f(x) dx ¶1−λ µZ R g(x) dx ¶λ . □ Second proof. We can assume without loss of generality that Z R f(x) dx = F > 0 and Z R g(x) dx = G > 0. Deﬁne u, v : [0, 1] →R such that u(t) and v(t) are the smallest numbers satisfying 1 F Z u(t) −∞ f(x) dx = 1 G Z v(t) −∞ g(x) dx = t. (8) Then u and v may be discontinuous, but they are strictly increasing functions and so are diﬀer-entiable almost everywhere. Let w(t) = (1 −λ)u(t) + λv(t). Take the derivative of (8) with respect to t to obtain f (u(t)) u′(t) F = g (v(t)) v′(t) G = 1. Using this and the arithmetic-geometric mean inequality, we obtain (when f (u(t)) ̸= 0 and g (u(t)) ̸= 0) w′(t) = (1 −λ)u′(t) + λv′(t) ≥ u′(t)1−λv′(t)λ = µ F f (u(t)) ¶1−λ µ G g (v(t)) ¶λ . Therefore Z R h(x) dx ≥ Z 1 0 h (w(t)) w′(t) dt ≥ Z 1 0 f (u(t))1−λ g (v(t))λ µ F f (u(t)) ¶1−λ µ G g (v(t)) ¶λ dt = F 1−λGλ. 8 R. J. GARDNER □ There are two basic ingredients in the second proof of Theorem 4.1: the introduction in (8) of the volume parameter t, and use of the arithmetic-geometric mean inequality in estimating w′(t). The same ingredients appear in the ﬁrst proof, though the parametrization is somewhat disguised in the use of the level sets. Theorem 4.2. (Pr´ ekopa-Leindler inequality in Rn.) Let 0 < λ < 1 and let f, g, and h be nonnegative integrable functions on Rn satisfying h ((1 −λ)x + λy) ≥f(x)1−λg(y)λ, (9) for all x, y ∈Rn. Then Z Rn h(x) dx ≥ µZ Rn f(x) dx ¶1−λ µZ Rn g(x) dx ¶λ . Proof. The proof is by induction on n. It is true for n = 1, by Theorem 4.1. Suppose that it is true for all natural numbers less than n. For each s ∈R, deﬁne a nonnegative function hs on Rn−1 by hs(z) = h(z, s) for z ∈Rn−1, and deﬁne fs and gs analogously. Let x, y ∈Rn−1, let a, b ∈R, and let c = (1 −λ)a + λb. Then hc ((1 −λ)x + λy) = h ((1 −λ)x + λy, (1 −λ)a + λb) = h ((1 −λ)(x, a) + λ(y, b)) ≥ f(x, a)1−λg(y, b)λ = fa(x)1−λgb(y)λ. By the inductive hypothesis, Z Rn−1 hc(x) dx ≥ µZ Rn−1 fa(x) dx ¶1−λ µZ Rn−1 gb(x) dx ¶λ . Let H(c) = Z Rn−1 hc(x) dx, F(a) = Z Rn−1 fa(x) dx, and G(b) = Z Rn−1 gb(x) dx. Then H(c) = H ((1 −λ)a + λb) ≥F(a)1−λG(b)λ. So, by Fubini’s theorem and Theorem 4.1, Z Rn h(x) dx = Z R Z Rn−1 hc(z) dz dc = Z R H(c) dc ≥ µZ R F(a) da ¶1−λ µZ R G(b) db ¶λ = µZ Rn f(x) dx ¶1−λ µZ Rn g(x) dx ¶λ . □ THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 9 Suppose that fi ∈Lpi(Rn), pi ≥1, i = 1, . . . , m are nonnegative functions, where 1 p1 + · · · + 1 pm = 1. (10) H¨ older’s inequality in Rn states that Z Rn m Y i=1 fi(x) dx ≤ m Y i=1 ∥fi∥pi = m Y i=1 µZ Rn fi(x)pi dx ¶1/pi . (11) Let 0 < λ < 1. If m = 2, 1/p1 = 1 −λ, 1/p2 = λ, and we let f = fp1 1 and g = fp2 2 , we get Z Rn f(x)1−λg(x)λ dx ≤ µZ Rn f(x) dx ¶1−λ µZ Rn g(x) dx ¶λ . The Pr´ ekopa-Leindler inequality in Rn can be written in the form Z Rn sup{f(x)1−λg(y)λ : (1 −λ)x + λy = z} dz ≥ µZ Rn f(x) dx ¶1−λ µZ Rn g(x) dx ¶λ , (12) because we can use the supremum for h in (9). A straightforward generalization is Z Rn sup ( m Y i=1 fi(xi) : m X i=1 xi pi = z ) dz ≥ m Y i=1 ∥fi∥pi, (13) where pi ≥1 for each i and (10) holds. So we see that the Pr´ ekopa-Leindler inequality is a reverse form of H¨ older’s inequality and that some condition such as (7) is therefore necessary for it to hold. Notice that the upper Lebesgue integral is used on the left in (12) and (13). This is because the integrands there are generally not measurable. We shall return to this point in Section 9. 5. The Brunn-Minkowski inequality In this section the Brunn-Minkowski inequality is derived from the Pr´ ekopa-Leindler inequality. A diﬀerent and self-contained short proof can be found in Section 6. Theorem 5.1. (General Brunn-Minkowski inequality in Rn, ﬁrst form.) Let 0 < λ < 1 and let X and Y be bounded measurable sets in Rn such that (1 −λ)X + λY is also measurable. Then Vn ((1 −λ)X + λY ) ≥Vn(X)1−λVn(Y )λ. (14) Theorem 5.2. The Pr´ ekopa-Leindler inequality in Rn implies the general Brunn-Minkowski in-equality in Rn. Proof. Let h = 1(1−λ)X+λY , f = 1X, and g = 1Y . If x, y ∈Rn, then f(x)1−λg(y)λ > 0 (and in fact equals 1) if and only if x ∈X and y ∈Y . The latter implies (1−λ)x+λy ∈(1−λ)X +λY , which is true if and only if h ((1 −λ)x + λy) = 1. Therefore (9) holds. We conclude by Theorem 4.2 that Vn ((1 −λ)X + λY ) = Z Rn 1(1−λ)X+λY (x) dx ≥ µZ Rn 1X(x) dx ¶1−λ µZ Rn 1Y (x) dx ¶λ = Vn(X)1−λVn(Y )λ. 10 R. J. GARDNER □ Corollary 5.3. (General Brunn-Minkowski inequality in Rn, standard form.) Let 0 < λ < 1 and let X and Y be nonempty bounded measurable sets in Rn such that (1 −λ)X + λY is also measurable. Then Vn ((1 −λ)X + λY )1/n ≥(1 −λ)Vn(X)1/n + λVn(Y )1/n. (15) Proof. Let λ′ = Vn(Y )1/n Vn(X)1/n + Vn(Y )1/n and let X′ = Vn(X)−1/nX and Y ′ = Vn(Y )−1/nY . Then 1 −λ′ = Vn(X)1/n Vn(X)1/n + Vn(Y )1/n and Vn(X′) = Vn(Y ′) = 1, by the positive homogeneity (of degree n) of Lebesgue measure in Rn (Vn(rA) = rnVn(A) for r ≥0). Therefore (14), applied to X′, Y ′, and λ′, yields Vn ¡ (1 −λ′)X′ + λ′Y ′¢ ≥1. But Vn ¡ (1 −λ′)X′ + λ′Y ′¢ = Vn µ X + Y Vn(X)1/n + Vn(Y )1/n ¶ = Vn(X + Y ) ¡ Vn(X)1/n + Vn(Y )1/n¢n . This gives Vn(X + Y )1/n ≥Vn(X)1/n + Vn(Y )1/n. To obtain (15), just replace X and Y by (1 −λ)X and λY , respectively. □ Remark 5.4. Using the homogeneity of volume, it follows that for all s, t > 0, Vn (sX + tY )1/n ≥sVn(X)1/n + tVn(Y )1/n. (16) Note the advantages of the ﬁrst form (14) of the general Brunn-Minkowski inequality. One need not assume that X and Y are nonempty, and the inequality is independent of the dimension n. The two forms are equivalent, however; to get from the standard to the ﬁrst form, just use Jensen’s inequality for means (see (28) below with p = 0 and q = 1/n). 6. History, alternative proofs, and equality conditions For detailed remarks and references concerning the early history of the Brunn-Minkowski in-equality for convex bodies, see [134, p. 314]. Brieﬂy, the inequality for convex bodies in R3 was discovered by Brunn around 1887. Minkowski pointed out an error in the proof, which Brunn corrected, and found a diﬀerent proof himself. Both Brunn and Minkowski showed that equality holds if and only if K and L are homothetic (i.e., K and L are equal up to translation and dilata-tion). The proof presented in [134, Section 6.1], due to Kneser and S¨ uss in 1932, is very similar to the proof we gave above of the Pr´ ekopa-Leindler inequality, restricted to characteristic functions of convex bodies; note that the case n = 1 is trivial, and the equality condition vacuous, in this case. This is perhaps the simplest approach for the equality conditions for convex bodies. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 11 Another quite diﬀerent proof, due to Blaschke in 1917, is worth mentioning. This uses Steiner symmetrization. Let K be a convex body in Rn and let u ∈Sn−1. The Steiner symmetral SuK of K in the direction u is the convex body obtained from K by sliding each of its chords parallel to u so that they are bisected by the hyperplane u⊥, and taking the union of the resulting chords. Then V (SuK) = V (K) by Cavalieri’s principle, and it is not hard to show that if K and L are convex bodies in Rn, then Su(K + L) ⊃SuK + SuL. (17) One can also prove that there is a sequence of directions um ∈Sn−1 such that if K is any convex body and Km = SumKm−1, then Km →rKB as m →∞, where rK is the constant such that V (K) = V (rKB). Repeated application of (17) now gives V (K + L)1/n ≥ V (rKB + rLB)1/n = (rR + rL)V (B)1/n = V (rKB)1/n + V (rLB)1/n = V (K)1/n + V (L)1/n. See [53, Chapter 5, Section 5] or [150, pp. 310–314] for more details. The general Brunn-Minkowski inequality and its equality conditions were ﬁrst proved by Lus-ternik . The equality conditions he gave were corrected by Henstock and Macbeath , who basically used the method in the second proof of Theorem 4.1 to derive the inequality. Another method, found by Hadwiger and Ohmann , is so beautiful that we cannot resist reproducing it in full (see also [37, Section 8], [51, Section 6.6], [58, Theorem 3.2.41], or [150, Section 6.5]). The idea is to prove the result ﬁrst for boxes, rectangular parallelepipeds whose sides are parallel to the coordinate hyperplanes. If X and Y are boxes with sides of length xi and yi, respectively, in the ith coordinate directions, then V (X) = n Y i=1 xi, V (Y ) = n Y i=1 yi, and V (X + Y ) = n Y i=1 (xi + yi). Now Ã n Y i=1 xi xi + yi !1/n + Ã n Y i=1 yi xi + yi !1/n ≤1 n n X i=1 xi xi + yi + 1 n n X i=1 yi xi + yi = 1, by the arithmetic-geometric mean inequality. This gives the Brunn-Minkowski inequality for boxes. One then uses a trick sometimes called a Hadwiger-Ohmann cut to obtain the inequality for ﬁnite unions X and Y of boxes, as follows. By translating X, if necessary, we can assume that a coordinate hyperplane, {xn = 0} say, separates two boxes in X. Let X+ (or X−) denote the union of the boxes formed by intersecting the boxes in X with {xn ≥0} (or {xn ≤0}, respectively). Now translate Y so that V (X±) V (X) = V (Y±) V (Y ) , (18) where Y+ and Y−are deﬁned analogously to X+ and X−. Note that X+ + Y+ ⊂{xn ≥0}, X−+ Y−⊂{xn ≤0}, and that the numbers of boxes in X+ ∪Y+ and X−∪Y−are both smaller 12 R. J. GARDNER than the number of boxes in X ∪Y . By induction on the latter number and (18), we have V (X + Y ) ≥ V (X+ + Y+) + V (X−+ Y−) ≥ ³ V (X+)1/n + V (Y+)1/n´n + ³ V (X−)1/n + V (Y−)1/n´n = V (X+) Ã 1 + V (Y )1/n V (X)1/n !n + V (X−) Ã 1 + V (Y )1/n V (X)1/n !n = V (X) Ã 1 + V (Y )1/n V (X)1/n !n = ³ V (X)1/n + V (Y )1/n´n . Now that the inequality is established for ﬁnite unions of boxes, the proof is completed by using them to approximate bounded measurable sets. A careful examination of this proof allows one to conclude that if Vn(X)Vn(Y ) > 0, equality holds only when Vn ((conv X) \ X) = Vn ((conv Y ) \ Y ) = 0, where conv X denotes the convex hull of X. Putting the equality conditions above together, we see that if Vn(X)Vn(Y ) > 0, equality holds in the general Brunn-Minkowski inequality (5.1) or (5.3) if and only if X and Y are homothetic convex bodies from which sets of measure zero have been removed. See [37, Section 8] and [150, Section 6.5] for more details and further comments about the case when X or Y has measure zero. Since H¨ older’s inequality (11) in its discrete form implies the arithmetic-geometric mean in-equality, there is a sense in which H¨ older’s inequality implies the Brunn-Minkowski inequality. The dotted arrow in Figure 1 reﬂects the controversial nature of this implication. Though the Hadwiger-Ohmann proof looks quite diﬀerent from the Henstock-Macbeath ap-proach, it shares the same two basic ingredients mentioned after Theorem 4.1, since the Hadwiger-Ohmann cut (18) is tantamount to a parametrization by volume. The Pr´ ekopa-Leindler inequality was explicitly stated and proved by Pr´ ekopa , and Leindler . (See also the historical remarks after Theorem 10.2, however.) The ﬁrst proof of Theorem 4.1 presented here, which follows that of Brascamp and Lieb [35, Theorem 3.1], is also reproduced in [127, Theorem 1.1]. The parametrization idea in the second proof of Theorem 4.1 goes back to Bonnesen; see and the references given there. The induction in Theorem 4.2 can be avoided and the inequality proved at once in Rn by means of the so-called Knothe map (see [120, p. 186]). Quite complicated equality conditions for the Pr´ ekopa-Leindler inequality in R are given in and , but equality conditions in Rn seem to be unknown. Recently, Borell found a “Brownian motion” proof of the Brunn-Minkowski inequality that depends on a generalization of the Pr´ ekopa-Leindler inequality, too complicated to be stated here, involving diﬀusion equations; see and also Section 19.14. 7. Minkowski’s first inequality and the isoperimetric inequality For some classes of sets such as convex bodies, the Brunn-Minkowski inequality is equivalent to another inequality of Minkowski that immediately yields the isoperimetric inequality. This involves a quantity V1(K, L) depending on two convex bodies K and L in Rn that can be deﬁned THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 13 by nV1(K, L) = lim ε→0+ V (K + εL) −V (K) ε . (19) Note that if L = B, then S(K) = nV1(K, B); it is this relationship that will quickly lead us to the isoperimetric inequality and its equality condition. An even shorter path (see [67, Theorem B.2.1]) yields the inequality but without the equality condition. The quantity V1(K, L) is a special mixed volume, and its existence requires just a little of the theory of mixed volumes to establish; see [150, Section 6.4]. In fact, Minkowski showed that if K1, . . . , Km are compact convex sets in Rn, and t1, . . . , tm ≥0, the volume V (P{tiKi : i = 1, . . . , m}) is a polynomial of degree n in the variables t1, . . . , tm. The coeﬃcient V (Kj1, . . . , Kjn) of tj1 · · · tjn in this polynomial is called a mixed volume. Then V1(K, L) = V (K, n −1; L), where the notation means that K appears (n −1) times and L appears once. See [67, Appendix A] for a gentle introduction to mixed volumes. Theorem 7.1. (Minkowski’s ﬁrst inequality for convex bodies in Rn.) Let K and L be convex bodies in Rn. Then V1(K, L) ≥V (K)(n−1)/nV (L)1/n, (20) with equality if and only if K and L are homothetic. Minkowski’s ﬁrst inequality plays a role in the solution of Shephard’s problem: If the projection of a centrally symmetric (i.e., −K is a translate of K) convex body onto any given hyperplane is always smaller in volume than that of another such body, is its volume also smaller? The answer is no in general in three or more dimensions; see [67, Chapter 4] and [99, p. 255]. Theorem 7.2. The Brunn-Minkowski inequality for convex bodies in Rn (and its equality condi-tion) implies Minkowski’s ﬁrst inequality for convex bodies in Rn (and its equality condition). Proof. Substituting ε = t/(1 −t) in (19) and using the homogeneity of volume, we obtain nV1(K, L) = lim ε→0+ V (K + εL) −V (K) ε = lim t→0+ V ((1 −t)K + tL) −(1 −t)nV (K) t(1 −t)n−1 = lim t→0+ V ((1 −t)K + tL) −V (K) t + lim t→0+ (1 −(1 −t)n))V (K) t = lim t→0+ V ((1 −t)K + tL) −V (K) t + nV (K). Using this new expression for V1(K, L) (see [107, p. 7]) and letting f(t) = V ((1 −t)K + tL)1/n, for 0 ≤t ≤1, we see that f′(0) = V1(K, L) −V (K) V (K)(n−1)/n . Therefore (20) is equivalent to f′(0) ≥f(1) −f(0). Since the Brunn-Minkowski inequality says that f is concave, Minkowski’s ﬁrst inequality follows. 14 R. J. GARDNER Suppose that equality holds in (20). Then f′(0) = f(1) −f(0). Since f is concave, we have f(t) −f(0) t = f(1) −f(0) for 0 < t ≤1, and this is just equality in the Brunn-Minkowski inequality. The equality condition for (20) follows immediately. □ The following corollary is obtained by taking L = B in Theorem 7.1. Corollary 7.3. (Isoperimetric inequality for convex bodies in Rn.) Let K be a convex body in Rn. Then µV (K) V (B) ¶1/n ≤ µS(K) S(B) ¶1/(n−1) , (21) with equality if and only if K is a ball. It can be shown (see ) that if M is a compact domain in Rn with piecewise C1 boundary and L is a convex body in Rn, the quantity V1(M, L) deﬁned by (19) with K replaced by M exists. From the Brunn-Minkowski inequality for compact domains in Rn with piecewise C1 boundary and the above argument, one obtains Minkowski’s ﬁrst inequality when the convex body K is replaced by such a domain. Taking L = B, this immediately gives the isoperimetric inequality for compact domains in Rn with piecewise C1 boundary. Essentially the most general class of sets for which the isoperimetric inequality in Rn is known to hold comprises the sets of ﬁnite perimeter; see, for example, the book of Evans and Gariepy [57, p. 190], where the rather technical setting, sometimes called the BV theory, is expounded. It is still possible to base the proof on the Brunn-Minkowski inequality, as Fonseca [62, Theorem 4.2] demonstrates, by ﬁrst obtaining the isoperimetric inequality for suitably smooth sets and then applying various measure-theoretic approximation arguments. In fact, Fonseca’s result is more general (see the material in Section 19.14 on Wulﬀshape of crystals). A strong form of the Brunn-Minkowski inequality is also used by Fonseca and M¨ uller , again in the more general context of Wulﬀshape, to establish the corresponding equality conditions (the same as for (21)). There is (see [134, Theorem 5.1.6] and ) an integral representation for mixed volumes, and in particular, V1(M, L) = 1 n Z ∂M hL(ux) dx, (22) where hL is the support function of L and ux is the outer unit normal vector to ∂M at x. (If we replace hL by an arbitrary function f on Sn−1, then up to a constant, this integral represents the surface energy of a crystal with shape M, where f is the surface tension; see Section 19.14.) When M = K is a suﬃciently smooth convex body, (22) can be written V1(K, L) = 1 n Z Sn−1 hL(u)fK(u) du, (23) where fK is the reciprocal of the Gauss curvature of K at the point on ∂K where the outer unit normal is u; for general convex bodies, fK(u) du must be replaced by dS(K, u), where S(K, ·) is the surface area measure of K. Minkowski’s existence theorem gives necessary and suﬃcient THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 15 conditions for a measure µ in Sn−1 to be the surface area measure of some convex body. Now (20) and (23) imply that if S(K, ·) = µ, then K minimizes the functional L → Z Sn−1 hL(u) dµ under the condition that V (L) = 1, and this fact motivates the proof of Minkowski’s existence theorem. See [134, Section 7.1], where pointers can also be found to the vast literature surrounding the so-called Minkowski problem, which deals with existence, uniqueness, regularity, and stability of a closed convex hypersurface whose Gauss curvature is prescribed as a function of its outer normals. 8. The Sobolev inequality Theorem 8.1. (Sobolev inequality.) Let f be a C1 function on Rn with compact support. Then Z Rn ∥∇f(x)∥dx ≥nκ1/n n ∥f∥n/(n−1). (24) The previous inequality is only one of a family, all called Sobolev inequalities. See [91, Chap-ter 8], where it is pointed out that such inequalities bound averages of gradients from below by weighted averages of the function, and can thus be considered as uncertainty principles. Theorem 8.2. The Sobolev inequality is equivalent to the isoperimetric inequality for compact domains with C1 boundaries. Proof. Suppose that the isoperimetric inequality holds, and let f be a C1 function on Rn with compact support. The coarea formula (a sort of curvilinear Fubini theorem; see [57, p. 112]) implies that Z Rn ∥∇f(x)∥dx = Z R Vn−1(f−1{t}) dt = Z ∞ 0 S (L(\|f\|, t)) dt, where L(\|f\|, t) is a level set of \|f\|, as in (4). Applying the the isoperimetric inequality for compact domains with C1 boundaries to these level sets, we obtain Z Rn ∥∇f(x)∥dx ≥nκ1/n n Z ∞ 0 V (L(\|f\|, t))(n−1)/n dt. On the other hand, by (6) and Minkowski’s inequality for integrals (see [77, (6.13.9), p. 148]), we have µZ Rn \|f(x)\|n/(n−1) dx ¶(n−1)/n = ÃZ Rn µZ ∞ 0 1L(\|f\|,t)(x) dt ¶n/(n−1) dx !(n−1)/n ≤ Z ∞ 0 µZ Rn 1L(\|f\|,t)(x)n/(n−1) dx ¶(n−1)/n dt = Z ∞ 0 V (L(\|f\|, t))(n−1)/n dt. 16 R. J. GARDNER Therefore (24) is true. Suppose that (24) holds, let M be a compact domain in Rn with C1 boundary ∂M, and let ε > 0. Deﬁne fε(x) = 1 if x ∈M, fε(x) = 0 if x ̸∈M + εB, and fε(x) = 1 −d(x, M)/ε if x ∈(M + εB) \ M, where d(x, M) is the distance from x to M. Since fε can be approximated by C1 functions on Rn with compact support, we can assume that (24) holds for fε. Note that fε →1M as ε →0. Also, ∥∇fε(x)∥= 1/ε if x ∈(M + εB) \ M and is zero otherwise. Therefore, by (3), S(M) = lim ε→0+ V (M + εB) −V (M) ε = lim ε→0+ Z Rn ∥∇fε(x)∥dx ≥ lim ε→0+ nκ1/n n µZ Rn \|fε(x)\|n/(n−1) dx ¶(n−1)/n = nκ1/n n µZ Rn 1M(x) dx ¶(n−1)/n = nκ1/n n V (M)(n−1)/n, which is just a reorganization of the isoperimetric inequality (21) . □ As for the isoperimetric inequality, there is a more general version of the Sobolev inequality in the BV theory. This is called the Gagliardo-Nirenberg-Sobolev inequality and it is equivalent to the isoperimetric inequality for sets of ﬁnite perimeter; see [57, pp. 138 and 192]. The Pr´ ekopa-Leindler inequality can also be used to obtain logarithmic Sobolev inequalities; see Section 19.14. 9. Measurability in Brunn-Minkowski and Pr´ ekopa-Leindler If X and Y are Borel sets, then (1 −λ)X + λY , being a continuous image of their product, is analytic and hence measurable. (Erd¨ os and Stone proved that this set need not itself be Borel.) However, an old example of Sierpi´ nski shows that the set (1 −λ)X + λY may not be measurable when X and Y are measurable. There are a couple of ways around the measurability problem. One can simply replace the measure on the left of the Brunn-Minkowski inequality by inner Lebesgue measure Vn∗, the supremum of the measures of compact subsets, thus: Vn∗((1 −λ)X + λY )1/n ≥(1 −λ)Vn(X)1/n + λVn(Y )1/n. A better solution is to obtain a slightly improved version of the Pr´ ekopa-Leindler inequality, and then deduce a corresponding improved Brunn-Minkowski inequality, as follows. Recall that the essential supremum of a measurable function f on Rn is deﬁned by ess sup x∈Rn f(x) = inf{t : f(x) ≤t for almost all x ∈Rn}. Brascamp and Lieb proved the following result. (According to Uhrin , the idea of using the essential supremum in connection with our topic occurred independently to S. Dancs.) THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 17 Theorem 9.1. (Pr´ ekopa-Leindler inequality in Rn, essential form.) Let 0 < λ < 1 and let f, g ∈L1(Rn) be nonnegative. Let s(x) = ess sup y f µx −y 1 −λ ¶1−λ g ³y λ ´λ . (25) Then s is measurable and ∥s∥1 ≥∥f∥1−λ 1 ∥g∥λ 1. Proof. First note that s is measurable. Indeed, s(x) = sup φ∈D Z Rn f µx −y 1 −λ ¶1−λ g ³y λ ´λ φ(y) dy, where D is a countable dense subset of the unit ball of L1(Rn). Therefore s is the supremum of a countable family of measurable functions. With the measurability of s in hand, the proof follows that of the usual Pr´ ekopa-Leindler inequality presented in Section 4. □ The essential form of the Pr´ ekopa-Leindler inequality in Rn implies the usual form, Theorem 4.2. To see this, replace x by z and y by λy′ in (25) and then let x = (z −λy′)/(1 −λ) to obtain s(z) = ess sup y′ f µz −λy′ 1 −λ ¶1−λ g(y′)λ = ess sup{f(x)1−λg(y)λ : z = (1 −λ)x + λy}. Now if h is any integrable function satisfying h ((1 −λ)x + λy) ≥f(x)1−λg(y)λ, we must have h ≥s almost everywhere. It follows from Theorem 9.1 that ∥h∥1 ≥∥s∥1 ≥∥f∥1−λ 1 ∥g∥λ 1. The corresponding improvement of the Brunn-Minkowski inequality requires one new concept. Note that the usual Minkowski sum of X and Y can be written X + Y = {z : X ∩(z −Y )} ̸= ∅. Adjust this by deﬁning the essential sum of X and Y by X +e Y = {z : Vn (X ∩(z −Y )) > 0}. While 1X+Y (z) = sup x∈Rn 1X(x)1Y (z −x), it is easy to see that 1X+eY (z) = ess sup x∈Rn 1X(x)1Y (z −x). (26) Theorem 9.2. (General Brunn-Minkowski inequality in Rn, essential form.) Let 0 < λ < 1 and let X and Y be nonempty bounded measurable sets in Rn. Then Vn ((1 −λ)X +e λY )1/n ≥(1 −λ)Vn(X)1/n + λVn(Y )1/n. (27) 18 R. J. GARDNER Proof. In Theorem 9.1, let f = 1(1−λ)X and g = 1λY . Then, by (26), 1(1−λ)X+eλY (z) = ess sup x∈Rn 1(1−λ)X(x)1λY (z −x) = ess sup x∈Rn 1X µ x 1 −λ ¶ 1Y µz −x λ ¶ = ess sup y∈Rn 1X µz −y 1 −λ ¶ 1Y ³y λ ´ = s(z). The inequality Vn ((1 −λ)X +e λY ) ≥Vn(X)1−λVn(Y )λ, and hence (27), now follow exactly as in Section 5. □ A direct proof of the previous theorem is given in [35, Appendix]. Here is a sketch. One ﬁrst shows that X +e Y is measurable (indeed, open). This is proved using the set A∗of density points of a measurable set A, that is, A∗= ½ x ∈Rn : lim ε→0+ Vn (A ∩B(x, ε)) Vn(B(x, ε)) = 1 ¾ , where B(x, ε) is a ball with center at x and radius ε. Then Vn(A△A∗) = 0, where △denotes symmetric diﬀerence, and this implies that X +e Y = X∗+e Y ∗. Now it can be shown that X∗+e Y ∗is open and X∗+e Y ∗= X∗+ Y ∗. The Brunn-Minkowski inequality (15) in Rn then implies (27). 10. p-concave functions and measures If f is a nonnegative integrable function deﬁned on a measurable subset A of Rn, and µ is deﬁned by µ(X) = Z A∩X f(x) dx, for all measurable subsets X of Rn, we say that µ is generated by f and A. The Pr´ ekopa-Leindler inequality implies that if f is log concave and C is an open convex subset of its support, then the measure µ generated by f and C is also log concave. Indeed, if 0 < λ < 1, X and Y are measurable sets, and z = (1 −λ)x + λy, then the log concavity of f implies f(z)1C∩((1−λ)X+λY )(z) ≥(f(x)1C∩X(x))1−λ (f(y)1C∩Y (y))λ , THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 19 so we can apply Theorem 4.2 to obtain µ ((1 −λ)X + λY ) = Z C∩((1−λ)X+λY ) f(z) dz = Z Rn f(z)1C∩((1−λ)X+λY )(z) dz ≥ µZ Rn f(x)1C∩X(x) dx ¶1−λ µZ Rn f(x)1C∩Y (x) dx ¶λ = µZ C∩X f(x) dx ¶1−λ µZ C∩Y f(x) dx ¶λ = µ(X)1−λµ(Y )λ. This observation has been generalized considerably, as follows. If 0 < λ < 1 and p ̸= 0, we deﬁne Mp(a, b, λ) = ((1 −λ)ap + λbp)1/p if ab ̸= 0 and Mp(a, b, λ) = 0 if ab = 0; we also deﬁne M0(a, b, λ) = a1−λbλ, M−∞(a, b, λ) = min{a, b}, and M∞(a, b, λ) = max{a, b}. These quantities and their natural generalizations for more than two numbers are called pth means. The classic text of Hardy, Lit-tlewood, and P´ olya is still the best general reference. (Note, however, the diﬀerent convention here when p > 0 and ab = 0.) Jensen’s inequality for means (see [77, Section 2.9]) implies that if −∞≤p < q ≤∞, then Mp(a, b, λ) ≤Mq(a, b, λ), (28) with equality if and only if a = b or ab = 0. A nonnegative function f on Rn is called p-concave on a convex set C if f ((1 −λ)x + λy) ≥Mp(f(x), f(y), λ), for all x, y ∈C and 0 < λ < 1. Analogously, we say that a ﬁnite (nonnegative) measure µ deﬁned on (Lebesgue) measurable subsets of Rn is p-concave if µ ((1 −λ)X + λY ) ≥Mp(µ(X), µ(Y ), λ), for all measurable sets X and Y in Rn and 0 < λ < 1. Thus 1-concave is just concave in the usual sense and 0-concave is log concave. The term quasiconcave is sometimes used for −∞-concave. Also, if p > 0 (or p < 0), then f is p-concave if and only if fp is concave (or convex, respectively). It follows from Jensen’s inequality (28) that a p-concave function or measure is q-concave for all q ≤p. Probability density functions of some important probability distributions are p-concave for some p. Consider, for example, the multivariate normal distribution on Rn with mean m ∈Rn and n × n positive deﬁnite symmetric covariance matrix A. This has probability density f(x) = c exp µ −(x −m) · A−1(x −m) 2 ¶ , 20 R. J. GARDNER where c = (2π)−n/2(det A)−1/2. Since A is positive deﬁnite, the function (x −m) · A−1(x −m) is convex and so f is log concave. The probability density functions of the Wishart, multivariate β, and Dirichlet distributions are also log concave; see . The argument above then shows that the corresponding probability measures are log concave. Pr´ ekopa explains how a problem from stochastic programming motivates this result. However, Borell noted that the density functions of the multivariate Pareto (the Cauchy distribution is a special case), t, and F distributions are not log concave, but are p-concave for some p < 0. To obtain similar concavity conditions for the corresponding probability measures, a technical lemma is required. Lemma 10.1. Let 0 < λ < 1 and let a, b, c, and d be nonnegative real numbers. If p + q ≥0, then Mp(a, b, λ)Mq(c, d, λ) ≥Ms(ac, bd, λ), where s = pq/(p + q) if p and q are not both zero, and s = 0 if p = q = 0. Proof. A general form of H¨ older’s inequality (see [77, p. 24]) states that when 0 < λ < 1, p1, p2, r > 0 with 1/p1 + 1/p2 = 1, and a, b, c, and d are nonnegative real numbers, then Mr(ac, bd, λ) ≤Mrp1(a, b, λ)Mrp2(c, d, λ), and that the inequality reverses when r < 0. Suppose that p+q > 0. If p, q > 0, we can let r = s, p1 = p/s, and p2 = q/s, and the desired inequality follows immediately. If p < 0, then q > 0 and we let r = p, p1 = s/p, and p2 = −q/p; then replace a, b, c, and d, by ac, bd, 1/c, and 1/d, respectively. The remaining cases follow by continuity. □ The following theorem generalizes the Pr´ ekopa-Leindler inequality in Rn, which is just the case p = 0. The number p/(np+1) is interpreted in the obvious way; it is equal to −∞when p = −1/n and to 1/n when p = ∞. Theorem 10.2. (Borell-Brascamp-Lieb inequality.) Let 0 < λ < 1, let −1/n ≤p ≤∞, and let f, g, and h be nonnegative integrable functions on Rn satisfying h ((1 −λ)x + λy) ≥Mp (f(x), g(y), λ) , for all x, y ∈Rn. Then Z Rn h(x) dx ≥Mp/(np+1) µZ Rn f(x) dx, Z Rn g(x) dx, λ ¶ . Proof. This is very similar to the proof of the Pr´ ekopa-Leindler inequality. To deal with the case n = 1, follow the second proof of Theorem 4.1, deﬁning F, G, u, v, and w as in that theorem. Then, by Lemma 10.1 with q = 1, Z R h(x) dx ≥ Z 1 0 h (w(t)) w′(t) dt ≥ Z 1 0 Mp (f (u(t)) , g (v(t)) , λ) M1 µ F f (u(t)), G g (v(t)), λ ¶ dt ≥ Z 1 0 Mp/(p+1)(F, G, λ) dt = Mp/(p+1)(F, G, λ). The general case follows as in Theorem 4.2 by induction on n. □ THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 21 Theorem 10.2 was proved (in slightly modiﬁed form) for p > 0 by Henstock and Macbeath (when n = 1) and Dinghas . The limiting case p = 0, as we noted above, was also proved by Pr´ ekopa and Leindler, and rediscovered by Brascamp and Lieb . In general form Theorem 10.2 is stated and proved by Brascamp and Lieb [35, Theorem 3.3] and by Borell [28, Theorem 3.1] (but with a much more complicated proof; see also the paper of Rinott ). The proof above may be found in and (see also [49, Theorem 3.15]), but still draws on methods introduced by Henstock, Macbeath, and Dinghas. Das Gupta’s survey contains a very thorough examination and assessment of the various contributions and proofs before 1980. Brascamp and Lieb obtain an “essential” form of Theorem 10.2, as in the case p = 0 (Theorem 9.1 above). Dancs and Uhrin also oﬀer a version of Theorem 10.2 for −∞≤p < −1/n. In calling Theorem 10.2 the Borell-Brascamp-Lieb inequality we are following the authors of (who also generalize it to a Riemannian manifold setting; see Section 19.13) and placing the emphasis on the negative values of p. In fact, the proof of [42, Corollary 1.1] shows that the strongest inequality in this family is that for p = −1/n; that is, Theorem 10.2 for p = −1/n implies Theorem 10.2 for all p > −1/n. This follows from a suitable rescaling of the functions f, g, and h, Lemma 10.1 with q = −p/(np + 1), and the observation that Mp(a, b, λ)−1 = M−p(1/a, 1/b, λ). The approach of Brascamp and Lieb , incidentally, was to observe that Theorem 10.2 also holds for n = 1 and p = −∞(the argument is contained in the ﬁrst proof of Theorem 4.1), and then to derive Theorem 10.2 for n = 1 and p ≥−1 from this and Lemma 10.1. Corollary 10.3. Let −1/n ≤p ≤∞and let f be an integrable function that is p-concave on an open convex set C in Rn contained in its support. Then the measure generated by f and C is p/(np + 1)-concave. Proof. This follows from Theorem 10.2 in exactly the same way as the special case p = 0 follows from the Pr´ ekopa-Leindler inequality (see the beginning of this section). □ The Brunn-Minkowski inequality says that Lebesgue measure in Rn is 1/n-concave, and The-orem 10.2 supplies plenty of measures that are p-concave for −1/n ≤p ≤∞. Borell (see also [49, Theorem 3.17]) proves a sort of converse to Corollary 10.3: Given −∞≤p ≤1/n and a p-concave measure µ with n-dimensional support S, there is a p/(1 −np)-concave function on S that generates µ. Borell also observed that when p > 1/n, no nontrivial p-concave measures exist in Rn, and that any 1/n-concave measure is a multiple of Lebesgue measure; see [49, The-orem 3.14]. Dancs and Uhrin [44, Theorem 3.4] ﬁnd a generalization of Theorem 10.2 in which Lebesgue measure is replaced by a q-concave measure for some −∞≤q ≤1/n. It is convenient to mention here a sharpening of the Brunn-Minkowski theorem proved by Bonnesen in 1929 (see and [134, p. 314]). If X is a bounded measurable set in Rn, the inner section function mX of X is deﬁned by mX(u) = sup t∈R Vn−1 ³ X ∩(u⊥+ tu) ´ , for u ∈Sn−1. (In 1926, Bonnesen asked if this function determines a convex body in Rn, n ≥ 3, up to translation and reﬂection in the origin, a question that remains unanswered; see [67, 22 R. J. GARDNER Problem 8.10].) Bonnesen proved that if 0 < λ < 1 and u ∈Sn−1, then Vn ((1 −λ)X + λY ) ≥M1/(n−1) (mX(u), mY (u), λ) µ (1 −λ) Vn(X) mX(u) + λ Vn(Y ) mY (u) ¶ . (29) Lemma 10.1 with p = 1/(n −1) and q = 1 shows that this is indeed stronger than (15). As Dancs and Uhrin [44, Theorem 3.2] show, an integral version of (29), in a general form similar to Theorem 10.2, can be constructed from the ideas already presented here. At present the most general results in this direction are contained in the papers of Uhrin; see , , and the references given there. In particular, Uhrin states in [147, p. 306] that all previous results of this type are contained in [147, (3.42)]. The latter inequality has as an ingredient a “curvilinear Minkowski addition,” and its proof reintroduces geometrical methods. 11. Convolutions The convolution of measurable functions f and g on Rn is f ∗g(x) = Z Rn f(x −y)g(y) dy. The next two theorems, on concavity of products and sections of functions, are useful in obtaining a result on the concavity of convolutions. Theorem 11.1. Let p1 + p2 ≥0, and let p = p1p2/(p1 + p2) if p1 and p2 are not both zero, and p = 0 if p1 = p2 = 0. For i = 1, 2, let fi be a pi-concave function on a convex set Ci in Rn. Then the function f(x, y) = f1(x)f2(y) is p-concave on C1 × C2. Proof. Suppose that 0 < λ < 1, and let xi ∈C1 and yi ∈C2 for i = 0, 1. By Lemma 10.1, f ((1 −λ)(x0, y0) + λ(x1, y1)) = f1 ((1 −λ)x0 + λx1) f2 ((1 −λ)y0 + λy1) ≥ Mp1 (f1(x0), f1(x1), λ) Mp2 (f2(y0), f2(y1), λ) ≥ Mp (f1(x0)f2(y0), f1(x1)f2(y1), λ) = Mp (f(x0, y0), f(x1, y1), λ) . □ Theorem 11.2. Let p ≥−1/n and let f be an integrable p-concave function on an open convex set C in Rm+n. For each x in the projection C\|Rm of C onto Rm, let C(x) = {y ∈Rn : (x, y) ∈C}. Then F(x) = Z C(x) f(x, y) dy is p/(np + 1)-concave on C\|Rm. Proof. For i = 0, 1, let xi ∈C\|Rm and let gi(y) = f(xi, y) for y ∈C(xi). Suppose that 0 < λ < 1 and that x = (1 −λ)x0 + λx1, and let g(y) = f(x, y) for y ∈C(x). The p-concavity of f implies that g ((1 −λ)y0 + λy1) ≥Mp (g0(y0), g1(y1), λ) whenever yi ∈C(xi), i = 0, 1. Also, C(x) ⊃(1 −λ)C(x0) + λC(x1). THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 23 Then Theorem 10.2 yields Z C(x) g(y) dy ≥Mp/(np+1) ÃZ C(x0) g0(y) dy, Z C(x1) g1(y) dy, λ ! . This shows that F is p/(np + 1)-concave on C\|Rm. □ If we apply the previous theorem with n = 1 and f = 1C when C is the interior of a convex body K in Rm+1, and let p →∞, we see that the function giving volumes of parallel hyperplane sections of K is 1/n-concave. This statement is equivalent to the Brunn-Minkowski inequality for convex bodies. Theorem 11.3. Let p1 + p2 ≥0, and let p = p1p2/(p1 + p2) if p1 and p2 are not both zero, and p = 0 if p1 = p2 = 0. Suppose further that p ≥−1/n. For i = 1, 2, let fi be an integrable pi-concave function on an open convex set Ci in Rn. Then f1∗f2 is p/(np+1)-concave on C1+C2. Proof. By Theorem 11.1, the function f1(x −y)f2(y) is p-concave for (x −y, y) ∈C1 × C2 ⊂R2n, that is, for x ∈C1 + C2. The result follows from Theorem 11.2. □ For extensions to measures and some examples that limit the possibility of weakening the conditions on p1, p2, and p in Theorem 11.3, see [49, Section 3.3], whose general approach we have followed in this section. Theorem 11.2 can be found in and . The early history of Theorem 11.3 (when p = 0, this says that the convolution of two log concave functions is also log concave) is discussed by Das Gupta [47, p. 313]. 12. The covariogram Theorem 12.1. Let K and L be convex bodies in Rn. Then the function gK,L(x) = V (K ∩(L + x))1/n , for x ∈Rn, is concave on its support. Proof. For x, y ∈Rn and 0 < λ < 1, we have K ∩(L + (1 −λ)x + λy) = K ∩((1 −λ)(L + x) + λ(L + y)) ⊃ (1 −λ) (K ∩(L + x)) + λ (K ∩(L + y)) . Using the Brunn-Minkowski inequality (15), we obtain gK,L ((1 −λ)x + λy) ≥ V ((1 −λ) (K ∩(L + x)) + λ (K ∩(L + y)))1/n ≥ (1 −λ)V (K ∩(L + x))1/n + λV (K ∩(L + y))1/n = (1 −λ)gK,L(x) + λgK,L(y), as required. □ As a corollary, we conclude that the covariogram gK of a convex body K in Rn, deﬁned for x ∈Rn by gK(x) = V (K ∩(K + x)) , 24 R. J. GARDNER is 1/n-concave (and hence log concave) on its support, which, it is easy to check, is the diﬀerence body DK = K + (−K) of K. Obviously gK is unchanged when K is translated or replaced by its reﬂection −K in the origin. Note that gK(x) = Z Rn 1K∩(K+x)(y) dy = Z Rn 1K(y)1K+x(y) dy = Z Rn 1K(y)1K(y −x) dy = 1−K ∗1K(x). The name “covariogram” stems from the theory of random sets, where the covariance is deﬁned for x ∈Rn as the probability that both o and x lie in the random set. The covariogram is also useful in mathematical morphology. See [135, Chapter 9]) and [140, Section 6.2]. In 1986, G. Math´ eron (see the references in ) asked if the covariogram determines convex bodies, up to translation and reﬂection in the origin. Remarkably, this question is open even for n = 2! Nagel proved that the answer is aﬃrmative when K and L are convex polygons in the plane. Bianchi has shown that the answer is aﬃrmative for much larger class of planar convex bodies. He has also found pairs of convex polyhedra that represent counterexamples in R4, but these are still reﬂections of each other in a plane. See also [70, Section 6], and the references given in connection with chord-power integrals in [67, p. 267]. 13. Anderson’s theorem Anderson used the Brunn-Minkowski theorem in his work on multivariate unimodality. He began with the following simple observation. If f is a (i) symmetric (f(x) = f(−x)) and (ii) unimodal (f(cx) ≥f(x) for 0 ≤c ≤1) function on R, and I is an interval centered at the origin, then Z I+y f(x) dx is maximized when y = 0. In probability language, if a random variable X has probability density f and Y is an independent random variable, then Prob {X ∈I} ≥Prob {X + Y ∈I}. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 25 To see this, recall that if g is the probability density of Y , then f ∗g is the probability density of X + Y ; see [82, Section 11.5]. So, by Fubini’s theorem, Prob {X + Y ∈I} = Z I Z R f(z −y)g(y) dy dz = Z R Z I f(z −y)g(y) dz dy = Z R Z I−y f(x)g(y) dx dy ≤ Z R Z I f(x)g(y) dx dy = Z I f(x) dx = Prob {X ∈I}. Anderson generalized this, as follows. If f is a nonnegative function on Rn, call f unimodal if the level sets L(f, t) (see (4)) are convex for each t ≥0. Note that every quasiconcave function and hence all p-concave functions are unimodal. Theorem 13.1. (Anderson’s theorem.) Let K be an origin-symmetric (i.e., K = −K) convex body in Rn and let f be a nonnegative, symmetric, and unimodal function integrable on Rn. Then Z K f(x + cy) dx ≥ Z K f(x + y) dx, for 0 ≤c ≤1 and y ∈Rn. Proof. Suppose initially that f(x) = 1L(x), where L is an origin-symmetric convex body in Rn. Then f(x + y) = 1L(x + y) = 1L−y(x) and Z K f(x + y) dx = Z K 1L−y(x) dx = V (K ∩(L −y)) = gK,L(−y) = gK,L(y). Theorem 12.1 implies that gK,L is log concave. Let λ = (1 −c)/2. Since gK,L(cy) = gK,L ((1 −2λ)y) = gK,L ((1 −λ)y + λ(−y)) ≥ gK,L(y)1−λgK,L(−y)λ = gK,L(y)1−λgK,L(y)λ = gK,L(y), 26 R. J. GARDNER the theorem follows. In the general case, L(f, t) is an origin-symmetric convex body, so by (6), Fubini’s theorem, and the special case just proved, Z K f(x + cy) dx = Z K Z ∞ 0 1L(f,t)(x + cy) dt dx = Z ∞ 0 Z K 1L(f,t)(x + cy) dx dt ≥ Z ∞ 0 Z K 1L(f,t)(x + y) dx dt = Z K f(x + y) dx. □ Anderson’s theorem says that the integral of a symmetric unimodal function f over an n-dimensional centrally symmetric convex body K does not decrease when K is translated towards the origin. Since the graph of f forms a hill whose peak is over the origin, this is intuitively clear. However, it is no longer obvious, as it was in the 1-dimensional case! There may be points x ∈K at which the value of f is larger than it is at the corresponding translate of x. As above, we can conclude from Anderson’s theorem that if a random variable X has probability density f on Rn and Y is an independent random variable, then Prob {X ∈K} ≥Prob {X + Y ∈K}, where K is any origin-symmetric convex body in Rn. We noted above that density functions of some well-known probability distributions are p-concave for some p, and hence unimodal. If they are also symmetric, Anderson’s theorem applies. Suppose K is a convex body in Rn, y ∈Rn, p ≥−1/n, and f is an integrable p-concave function on Rn. Corollary 10.3 implies that the measure µ generated by f and Rn is p/(np + 1)-concave on Rn. Let h(y) = µ(K −y) = Z K−y f(x) dx = Z K f(x + y) dx. Since K −(1 −λ)y0 −λy1 = (1 −λ)(K −y0) + λ(K −y1), we have h ((1 −λ)y0 + λy1) = µ (K −(1 −λ)y0 −λy1) = µ ((1 −λ)(K −y0) + λ(K −y1)) ≥ Mp/(np+1) (µ(K −y0), µ(K −y1), λ) = Mp/(np+1) (h(y0), h(y1), λ) . Therefore h is p/(np + 1)-concave on Rn and hence unimodal. In particular, h(cy) is unimodal in c for a ﬁxed y. This shows that Corollary 10.3 and Anderson’s theorem are related. Anderson’s theorem replaces the restriction p ≥−1/n with a much weaker condition, but requires in exchange the symmetry of f and K. Anderson’s theorem has many applications in probability and statistics, where, for example, it can be applied to show that certain statistical tests are unbiased. See , , , and . THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 27 14. Young’s inequality We saw in the previous sections how the Brunn-Minkowski inequality and convolutions come together naturally. The next theorem provides two convolution inequalities with sharp constants, the ﬁrst proved independently by Beckner and Brascamp and Lieb , and the second by Brascamp and Lieb . (See Section 17 for more information.) We shall soon see that the second inequality actually implies the Brunn-Minkowski inequality. Theorem 14.1. Let 0 < p, q, r satisfy 1 p + 1 q = 1 + 1 r, (30) and let f ∈Lp(Rn) and g ∈Lq(Rn) be nonnegative. Then (Young′s inequality) ∥f ∗g∥r ≤Cn∥f∥p∥g∥q, for p, q, r ≥1, (31) and (Reverse Young inequality) ∥f ∗g∥r ≥Cn∥f∥p∥g∥q, for p, q, r ≤1. (32) Here C = CpCq/Cr, where C2 s = \|s\|1/s \|s′\|1/s′ (33) for 1/s + 1/s′ = 1 (that is, s and s′ are H¨ older conjugates). The inequality (31), when expanded, reads as follows: µZ Rn µZ Rn f(x −y)g(y) dy ¶r dx ¶1/r ≤Cn µZ Rn f(x)p dx ¶1/p µZ Rn g(x)q dx ¶1/q . Inequalities (31) and (32) together show that equality holds in both when p = q = r = 1. In fact, since Cp →1 as p →1, when p = q = r = 1 we have C = 1, and substituting u = x −y, v = y in the left-hand side of (31), we obtain Z Rn Z Rn f(u)g(v) dv du ≤ Z Rn f(x) dx Z Rn g(x) dx. But equality holds here and therefore also in (31), and similarly in (32). Theorem 14.2. The limiting case r →0 of the reverse Young inequality is the essential form of the Pr´ ekopa-Leindler inequality in Rn (Theorem 9.1). Proof. Let fm and gm be sequences of bounded measurable functions with compact support converging in L1(Rn) to f and g, respectively, as m →∞and satisfying fm ≤f and gm ≤g. Let sm(x) = ess sup y fm µx −y 1 −λ ¶1−λ gm ³y λ ´λ . (34) Let s(x) be deﬁned by replacing fm by f and gm by g in (34). As in the proof of Theorem 9.1, s and each sm is measurable. Also, ∥s∥1 ≥∥sm∥1, so if ∥sm∥1 ≥∥fm∥1−λ 1 ∥gm∥λ 1 28 R. J. GARDNER for each m we have ∥s∥1 ≥∥f∥1−λ 1 ∥g∥λ 1. Therefore it suﬃces to prove the theorem when f and g are bounded measurable functions with compact support. Assuming this, note that s(x) = limm→∞Sm(x), where Sm(x) = ÃZ Rn f µx −y 1 −λ ¶(1−λ)m g ³y λ ´λm dy !1/(m−1) . (If we replaced the exponent, 1/(m −1) by 1/m, this would follow from the fact that the mth mean tends to the supremum as m →∞; compare [77, p. 143]. But this replacement is irrelevant in the limit.) Note also that ∥s∥1 = limm→∞∥Sm∥1 (we can interchange the limit and integral because the Sm’s are uniformly bounded and have supports lying in some common compact set). Applying the reverse Young inequality to Sm with m > max{(1 −λ)−1, λ−1}, p = 1/((1 −λ)m), q = 1/(λm), and r = 1/(m −1), we obtain ∥Sm∥1 = Z Rn Sm(x) dx = Z Rn ÃZ Rn f µx −y 1 −λ ¶(1−λ)m g ³y λ ´λm dy !1/(m−1) dx ≥ Ã Cn µZ Rn f µx −y 1 −λ ¶ dx ¶(1−λ)m µZ Rn g ³y λ ´ dy ¶λm!1/(m−1) = Cn/(m−1) ((1 −λ)n∥f∥1)(1−λ)m/(m−1) (λn∥g∥1)λm/(m−1) . Therefore ∥s∥1 = lim m→∞∥Sm∥1 ≥ ³ (1 −λ)1−λλλ lim m→∞C1/(m−1)´n ∥f∥1−λ 1 ∥g∥λ 1. It remains only to check that lim m→∞C1/(m−1) = (1 −λ)−(1−λ)λ−λ. □ 15. The Brascamp-Lieb inequality and Barthe’s inequality The inequalities presented in this section approach the most general known in the direction of Young’s inequality and its reverse form, and represent a research frontier that can be expected to move before too long. Each m × n matrix A deﬁnes a linear transformation from Rn to Rm, and this linear map can also be denoted by A. The Euclidean adjoint A∗of A is then an n × m matrix or linear transformation from Rm to Rn satisfying Ax · y = x · A∗y for each y ∈Rm and x ∈Rn. Theorem 15.1. Let ci > 0 and ni ∈N, i = 1, . . . , m, with P i cini = n. Let fi ∈L1(Rni) be nonnegative and let Bi : Rn →Rni be a linear surjection, i = 1, . . . , m. Then THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 29 (Brascamp-Lieb inequality) Z Rn m Y i=1 fi(Bix)ci dx ≤D−1/2 m Y i=1 µZ Rni fi(x) dx ¶ci (35) and (Barthe’s inequality) Z Rn sup ( m Y i=1 fi(zi)ci : x = X i ciB∗ i zi, zi ∈Rni ) dx ≥D1/2 m Y i=1 µZ Rni fi(x) dx ¶ci , (36) where D = inf ½det (Pm i=1 ciB∗ i AiBi) Qm i=1(det Ai)ci : Ai is a positive deﬁnite ni × ni matrix ¾ . (37) For comments on equality conditions and ideas of proof, including a proof of an important special case of (36), see Section 17. We can begin to understand (35) by taking ni = n, Bi = In, the identity map on Rn, replacing fi by f1/ci i , and letting ci = 1/pi, i = 1, . . . , m. Then P i 1/pi = 1 and the log concavity of the determinant of a positive deﬁnite matrix (see, for example, [20, p. 63]) yields D = 1. Therefore Z Rn m Y i=1 fi(x) dx ≤ m Y i=1 ∥fi∥pi, H¨ older’s inequality in Rn. Next, take m = 2, n1 = n2 = n, B1 = B2 = In, c1 = 1 −λ, and c2 = λ in (36). Again we have D = 1, so Z Rn sup n f1(z1)1−λf2(z2)λ : x = (1 −λ)z1 + λz2 o dx ≥ µZ Rn f1(x) dx ¶1−λ µZ Rn f2(x) dx ¶λ , the Pr´ ekopa-Leindler inequality (12) in Rn. Theorem 15.2. (Young’s inequality in Rn, second form.) Let 0 < p, q, r satisfy 1 p + 1 q + 1 r = 2, and let f ∈Lp(Rn), g ∈Lq(Rn), and h ∈Lr(Rn) be nonnegative. Then Z Rn Z Rn f(x)g(x −y)h(y) dy dx ≤C n∥f∥p∥g∥q∥h∥r, (38) where C = CpCqCr is deﬁned using (33). Theorem 15.3. The second form of Young’s inequality in Rn is equivalent to the ﬁrst (31). 30 R. J. GARDNER Proof. Let p, q, r ≥1 satisfy (30). By H¨ older’s inequality (11), sup ½ ∥f ∗g∥r ∥f∥p∥g∥q : f ∈Lp(Rn), g ∈Lq(Rn) ¾ = = sup ½R Rn(f ∗g)(x)h(x) dx ∥f∥p∥g∥q∥h∥r′ : f ∈Lp(Rn), g ∈Lq(Rn), h ∈Lr′(Rn) ¾ = sup ½R Rn R Rn f(x −y)g(y)h(x) dx dy ∥f∥p∥g∥q∥h∥r′ : f ∈Lp(Rn), g ∈Lq(Rn), h ∈Lr′(Rn) ¾ = sup ½R Rn R Rn f(x)g(x −y)h(y) dy dx ∥f∥p∥g∥q∥h∥r : f ∈Lp(Rn), g ∈Lq(Rn), h ∈Lr(Rn) ¾ , where the last equality is obtained by replacing f, g, h, p, q, and r′, by g, h, f, q, r, and p, respectively, so that 1 p + 1 q + 1 r = 2. □ Theorem 15.4. The Brascamp-Lieb inequality (35) implies Young’s inequality in Rn. Proof. In (35), let m = 3, n1 = n2 = n3 = n, and let Bi : R2n →Rn, i = 1, 2, 3 be the linear maps taking (z1, . . . , z2n) to (z1, . . . , zn), (z1−zn+1, . . . , zn−z2n), and (zn+1, . . . , z2n), respectively; then replace fi by f1/ci i , i = 1, 2, 3 and let c1 = 1/p, c2 = 1/q, and c3 = 1/r. In this case D = C−2, where C is as in Theorem 14.1; see [34, Theorem 5]. This gives Z Rn Z Rn f1(x)f2(x −y)f3(y) dy dx ≤C∥f1∥p∥f2∥q∥f3∥r, which is (38). □ As a side remark, we note that there is a version of Young’s inequality in its second form (38), called the weak Young inequality, which only requires that g ∈Lq w(Rn), the weak Lq space. See [91, Section 4.3] for details. This allows one to conclude in particular that under the (slightly weakened) hypotheses of Theorem 15.2, with q = n/λ, Z Rn Z Rn f(x)∥x −y∥−λh(y) dy dx ≤k(n, λ, p)∥f∥p∥h∥r. (39) This was proved in Lieb with a sharp constant k(n, λ, p). The classical form without the sharp constant is called the Hardy-Littlewood-Sobolev inequality. The case λ = n −2 is of particular interest in potential theory, as is explained in [91, Chapter 9]. 16. Back to geometry As Ball remarks, some geometry comes back into view if we replace f(x) in Young’s inequality (38) by f(−x): Z R Z R f(−x1)g(x1 −x2)h(x2) dx2 dx1 ≤C∥f∥p∥g∥q∥h∥r. (40) THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 31 Deﬁne φ : R2 →R3 by φ(x1, x2) = z = (z1, z2, z3), where z1 = −x1, z2 = x1 −x2, and z3 = x2. Then φ(R2) = S, where S is the plane {(z1, z2, z3) : z1 + z2 + z3 = 0} through the origin. Let f = g = h = 1[−1,1] and C0 = [−1, 1]3. By (40), V2(C0 ∩S) = Z S 1C0(z) dz = Z S f(z1)g(z2)h(z3) dz = J(φ)−1 Z R Z R f(−x1)g(x1 −x2)h(x2) dx2 dx1, where J(φ) is the Jacobian of φ. So Young’s inequality might be used to provide upper bounds for volumes of central sections of cubes. In fact, Ball used the following special case of the Brascamp-Lieb inequality to do just this. Suppose that ci > 0 and ui ∈Sn−1, i = 1, . . . , m satisfy x = m X i=1 ci(x · ui)ui, for all x ∈Rn. This says that the ui’s are acting like an orthonormal basis for Rn. The condition is often written m X i=1 ciui ⊗ui = In, (41) where u ⊗u denotes the rank one orthogonal projection onto the span of u, the map that sends x to (x · u)u. Taking traces in (41), we see that m X i=1 ci = n. (42) Theorem 16.1. Let ci > 0 and ui ∈Sn−1, i = 1, . . . , m be such that (41) and hence (42) holds. If fi ∈L1(R) is nonnegative, i = 1, . . . , m, then (Geometric Brascamp-Lieb inequality) Z Rn m Y i=1 fi(x · ui)ci dx ≤ m Y i=1 µZ R fi(x) dx ¶ci (43) and (Geometric Barthe inequality) Z Rn sup ( m Y i=1 fi(zi)ci : x = X i ciziui, zi ∈R ) dx ≥ m Y i=1 µZ R fi(x) dx ¶ci . (44) Proof. Let ni = 1 and for x ∈Rn, let Bix = x · ui, i = 1, . . . , m. Then B∗ i zi = ziui ∈Rn for zi ∈R. The inequalities (35) and (36) become (43) and (44), respectively, because the hypotheses of the theorem and (37) imply that D = 1 (see [17, Proposition 9] for the details). □ 32 R. J. GARDNER Note that the geometric Barthe inequality (44) still implies the Pr´ ekopa-Leindler inequality in R, with the geometric consequences explained above. Ball used (43) to obtain the best-possible upper bound Vk(C0 ∩S) ≤( √ 2)n−k for sections of the cube C0 = [−1, 1]n by k-dimensional subspaces S, 1 ≤k ≤n−1, when 2k ≥n. (For smaller values of k, the best-possible bound is not known except for some special cases; see .) He also showed that (43) provides best-possible upper bounds for the volume ratio vr(K) of a convex body K in Rn, deﬁned by vr(K) = µV (K) V (E) ¶1/n , where E is the ellipsoid of maximal volume contained in K. The ellipsoid E is called the John ellipsoid of K. The following theorem is a reﬁnement of Ball of a theorem proved by Fritz John. Theorem 16.2. The John ellipsoid of a convex body K in Rn is B if and only if B ⊂K and there is an m ≥n, ci > 0 and ui ∈Sn−1 ∩∂K, i = 1, . . . , m such that (41) holds and P i ciui = o. Ball’s argument is as follows. Let K be a convex body in Rn. Since vr(K) is aﬃne invariant, we may assume that the John ellipsoid of K is B. If we can show that V (K) ≤2n, then vr(K) ≤vr(C0), where C0 = [−1, 1]n. Let ci and ui be as in John’s theorem, and note that the points ui are contact points, points where the boundaries of K and B meet. If K is origin-symmetric and ui is a contact point, then so is −ui; therefore K ⊂L, where L = {x ∈Rn : \|x · ui\| ≤1, i = 1, . . . , m} is the closed slab bounded by the hyperplanes {x : x · ui = ±1}. Also, if fi = 1[−1,1], then 1L(x) = m Y i=1 fi(x · ui)ci. By (43) and (42), V (K) ≤V (L) = Z Rn m Y i=1 fi(x · ui)ci dx ≤ m Y i=1 µZ R fi(x) dx ¶ci = m Y i=1 2ci = 2n. This argument shows that vr(K) is maximal for centrally symmetric K when K is a parallelotope. One consequence of this estimate is the following result of Ball (Behrend proved the result for n = 2). Theorem 16.3. (Reverse isoperimetric inequality for centrally symmetric convex bodies in Rn.) Let K be a centrally symmetric convex body in Rn and let C0 = [−1, 1]n. There is an aﬃne THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 33 transformation φ such that µS(φK) S(C0) ¶1/(n−1) ≤ µV (φK) V (C0) ¶1/n . (45) Proof. Choose φ so that the John ellipsoid of φK is B. The above argument shows that V (φK) ≤ 2n. Since B ⊂φK, we have, by (3), S(φK) = lim ε→0+ V (φK + εB) −V (φK) ε ≤ lim ε→0+ V (φK + εφK) −V (φK) ε = V (φK) lim ε→0+ (1 + ε)n −1 ε = nV (φK) = nV (φK)(n−1)/nV (φK)1/n ≤2nV (φK)(n−1)/n. This is equivalent to (45). □ Of course, one cannot expect a reverse isoperimetric inequality without use of an aﬃne trans-formation, since we can ﬁnd convex bodies of any prescribed volume that are very ﬂat and so have large surface area. In , Ball used the same methods to show that for arbitrary convex bodies, the volume ratio is maximal for simplices, and to obtain a corresponding reverse isoperimetric inequality. The fact that the volume ratio is only maximal for parallelotopes (in the centrally symmetric case) or simplices was shown by Barthe as a corollary of his study of the equality conditions in the Brascamp-Lieb inequality. For other results of this type that employ Theorem 16.1, see , , and . Barthe states a multidimensional generalization of Theorem 16.1, also derived from Theorem 15.1, that leads to a multidimensional Brunn-Minkowski-type theorem. 17. More on history, proofs, and equality conditions The classical Young inequality is ∥f ∗g∥r ≤∥f∥p∥g∥q, for p, q, r ≥1, that is, (31) with the better constant Cn there replaced by 1, under the same assumptions. This can be proved in a few lines using H¨ older’s inequality (11); see [91, p. 99]. It was proved by W. H. Young in 1912–13 (see [77, Sections 8.3 and 8.4] and the references given there), and is related to the classical Hausdorﬀ-Young inequality: If 1 ≤p ≤2 and f ∈Lp(Rn), then ∥ˆ f∥p′ ≤∥f∥p, (46) where ˆ f denotes the Fourier transform ˆ f(x) = Z Rn f(y)e2πix·y dy 34 R. J. GARDNER of f, and p and p′ are H¨ older conjugates. This was proved by Hausdorﬀand Young for Fourier series, and extended to integrals by Titchmarsh in 1924. Beckner , improving earlier partial results of Babenko, showed that when 1 ≤p ≤2, ∥ˆ f∥p′ ≤Cn p ∥f∥p, (47) where Cp is given by (33). (Lieb proved that equality holds only for Gaussians.) This improvement on (46) is related to Young’s inequality (31); in fact, the classical Young inequality was motivated by (46). To see the connection, suppose that (47) holds, n = 1, and 1 ≤p, q, r′ ≤2. If p, q, r satisfy (30), then their H¨ older conjugates satisfy 1/p′+1/q′ = 1/r′. Using this and H¨ older’s inequality (11), we obtain ∥f ∗g∥r ≤ Cr′∥ˆ fˆ g∥r′ ≤ Cr′∥ˆ f∥p′∥ˆ g∥q′ ≤ Cr′(Cp∥f∥p)(Cq∥g∥q) = C∥f∥p∥g∥q. A similarly easy argument (see [21, pp.169–70]) shows that Young’s inequality (31) yields (46) when p′ is an even integer. Young’s inequality in the sharp form (31) was proved independently by Beckner and Bras-camp and Lieb . The reverse Young inequality without the sharp constant (that is, with C replaced by 1) is due to Leindler ; the sharp version was obtained by Brascamp and Lieb . The latter also found the connection to the Pr´ ekopa-Leindler inequality, Theorem 14.2, and established the following equality conditions: When n = 1 and p, q ̸= 1, equality holds in (31) or (32) if and only if f and g are Gaussians: f(x) = ae−c\|p′\|(x−α)2, g(x) = be−c\|q′\|(x−β)2, for some a, b, c, α, β with a, b ≥0 and c > 0. The simplest known proof of Young’s inequality and its reverse form, with the above equality conditions, was found by Barthe . The Brascamp-Lieb inequality in the general form (35), with equality conditions, was proved by Lieb . The special case ni = 1 and Bix = x · vi, where x ∈Rn and vi ∈Rn, i = 1, . . . , m is the main result of Brascamp and Lieb . Let A be an n × n positive deﬁnite symmetric matrix, and let GA(x) = exp(−Ax · x), for x ∈Rn. The function GA is called a centered Gaussian. Lieb proved that the supremum of the left-hand side of (35) for functions fi of norm one is the same as the supremum of the left-hand side of (35) for centered Gaussians of norm one; in other words, the constant D can be computed using centered Gaussians. There is also a version of (35) in which a ﬁxed centered Gaussian appears in the integral on the left-hand side and the constant is again determined by taking the functions fi to be Gaussians; see [34, Theorem 6], where an application to statistical mechanics is given, and [90, Theorem 6.2]. Barthe proved (36), giving at the same time a simpler approach to (35) and its equality conditions. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 35 The fact that the constant D in the geometric Brascamp-Lieb inequality (43) becomes 1 was observed by Ball . Inequality (44) was ﬁrst proved by Barthe . As in the general case, equality holds in (43) and (44) for centered Gaussians. The main idea behind Barthe’s approach is the use of a familiar construction from measure theory. Let µ be a ﬁnite Borel measure in Rn and T : Rn →Rn a Borel-measurable map deﬁned µ-almost everywhere. For Borel sets M in Rn, let ν(M) = (Tµ)(M) = µ(T −1(M)). The Borel measure ν = Tµ is sometimes called the push-forward of µ by T, and T is said to push forward or transport the measure µ to ν. Suppose for simplicity that µ and ν are absolutely continuous with respect to Lebesgue measure, so that µ(M) = Z M f(x) dx and ν(M) = Z M g(x) dx for Borel sets M in Rn, and T is a diﬀerentiable bijection. Then f(x) = g(T(x))J(T)(x), where J(T) is the Jacobian of T, and we can talk of T transporting f to g. If µ and ν are measures on R, absolutely continuous with respect to Lebesgue measure and with µ(R) = ν(R), then we can always ﬁnd a T that transports µ to ν, by deﬁning T(t) to be the smallest number such that Z t −∞ f(x) dx = Z T(t) −∞ g(x) dx. Moreover, if f and g are continuous and positive, then T is strictly increasing and C1, and f(x) = g(T(x))T ′(x). In fact, the same parametrization was used in proving the Pr´ ekopa-Leindler inequality in R. To see this, replace the functions f and g in the second proof of Theorem 4.1 with g1 and g2, respectively. If fi = Fi1[0,1], i = 1, 2, then 1 Gi Z Ti(t) −∞ gi(x) dx = Z t −∞ 10,1 dx = t, so the functions u and v in the second proof of Theorem 4.1 are just T1 and T2, respectively. In other words, u and v transport a suitable multiple of the characteristic function of the unit interval to g1 and g2, respectively. Barthe saw that this is all that is needed to prove (35) and (36) simultaneously in the special case ni = 1 and Bix = x · vi, where x ∈Rn and vi ∈Rn, i = 1, . . . , m. To see this, let ci > 0 satisfy P i ci = n and let fi and gi be nonnegative functions in L1(R) with Z R fi(x) dx = Fi and Z R gi(x) dx = Gi, for i = 1, . . . , m. Standard approximation arguments show that there is no loss of generality in assuming fi and gi are positive and continuous. Deﬁne strictly increasing maps Ti as above, so that 1 Fi Z t −∞ fi(x) dx = 1 Gi Z Ti(t) −∞ gi(x) dx 36 R. J. GARDNER and hence fi(x) Fi = gi(Ti(x))T ′ i(x) Gi , for i = 1, . . . , m. For x ∈Rn, let V (x) = m X i=1 ciTi(x · vi)vi, so that dV (x) = m X i=1 ciT ′ i(x · vi)(vi ⊗vi)(dx). Finally, note that if Bix = x·vi for x ∈R, then B∗ i = xvi, so B∗ i Bx = vi ⊗vi(x), and the constant D in (37) becomes D = inf ½det (Pm i=1 ciaivi ⊗vi) Qm i=1 aci i : ai > 0 ¾ . In the following, we can assume that D ̸= 0. Using the expression for D with ai = T ′ i(x · vi), i = 1, . . . , m to provide a lower bound for the Jacobian of the injective map V , we obtain D Ã m Y i=1 µGi Fi ¶ci! Z Rn m Y i=1 fi(x · vi)ci dx = D Z Rn m Y i=1 ¡ gi(Ti(x · vi))T ′ i(x · vi) ¢ci dx ≤ Z Rn m Y i=1 gi(Ti(x · vi))ci det Ã m X i=1 ciT ′ i(x · vi)(vi ⊗vi) ! dx ≤ Z Rn sup ( m Y i=1 gi(zi)ci : V = X i cizivi, zi ∈R ) dV ≤ Z Rn sup ( m Y i=1 gi(zi)ci : x = X i cizivi, zi ∈R ) dx. To see how centered Gaussians play a role in the equality conditions, note that if fi(x) = exp(−aix2), then since P i ci = n, m Y i=1 µZ R fi(x) dx ¶ci = m Y i=1 µZ R e−aix2 dx ¶ci = m Y i=1 a−ci/2 i µZ R e−x2 dx ¶ci = m Y i=1 µ π ai ¶ci/2 = µ πn Qm i=1 aci i ¶1/2 , THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 37 while Z Rn m Y i=1 fi(x · vi)ci dx = Z Rn m Y i=1 ³ e−ai(x·vi)2´ci dx = Z Rn e−( Pm i=1 ciai(x·vi)vi)·x dx = µ πn det (Pm i=1 ciaivi ⊗vi) ¶1/2 . (The last equality follows from Z Rn e−Ax·x dx = µ πn det A ¶1/2 , where A is a positive deﬁnite symmetric n × n matrix.) To summarize, we have shown that in the special case under consideration, the left-hand side of (36) is greater than or equal to the left-hand side of (35), and that equality holds in (35) for centered Gaussians. This is already enough to prove (36). One more computation is needed to prove (35), but we shall omit it, since it needs some (quite basic) tools of geometry, and refer the reader to . If one wants to apply the same sort of argument in the general situation of Theorem 15.1, one needs an answer to the following question: If µ and ν are measures on Rn, absolutely continuous with respect to Lebesgue measure and with µ(Rn) = ν(Rn), can we ﬁnd a T with some suitable monotonicity property that transports µ to ν? It turns out that an ideal answer has recently been found, called the Brenier map: Providing µ vanishes on Borel sets of Rn with Hausdorﬀ dimension n −1, there is a convex map ψ : Rn →R such that if T = ∇ψ, then T transports µ to ν. See for more details and references. It is appropriate to highlight the contribution of McCann, whose 1994 PhD thesis shows the relevance of measure-preserving convex gradients to geometric inequalities and helped attract the attention of the convex geometry community to Brenier’s result. In and , the Brenier map is exploited as a localization technique to derive new global convexity inequalities which imply the Brunn-Minkowski and Pr´ ekopa-Leindler inequalities as special cases. Barthe [15, Section 2.4] also discovered a generalization of Young’s inequality in Rn that con-tains the geometric Brascamp-Lieb and geometric Barthe inequalities as limiting cases. 18. The entropy power inequality and physics Suppose that X is a discrete random variable taking possible values x1, . . . , xm with probabili-ties p1, . . . , pm, respectively, where P i pi = 1. Shannon introduced a measure of the average uncertainty removed by revealing the value of X. This quantity, Hm(p1, . . . , pm) = − m X i=1 pi log pi, is called the entropy of X. It can also be regarded as a measure of the missing information; indeed, the function Hm is concave and achieves its maximum when p1 = · · · = pm = 1/m, that is, when all outcomes are equally likely. The words “uncertainty” and “information” already suggest a 38 R. J. GARDNER connection with physics, and a derivation of the function Hm from a few natural assumptions can be found in textbooks on statistical mechanics; see, for example, [6, Chapter 3]. If X is a random vector in Rn with probability density f, the entropy h1(X) of X is deﬁned analogously: h1(X) = h1(f) = − Z Rn f(x) log f(x) dx. The notation we use is convenient when h1(X) is regarded as a limit as p →1 of the pth R´ enyi entropy hp(X) of X, deﬁned for p > 1 by hp(X) = hp(f) = p 1 −p log ∥f∥p. The entropy of X may not be well deﬁned. However, if f ∈L1(Rn) ∩Lp(Rn) for some p > 1, then h1(X) = h1(f) is well deﬁned, though its value may be +∞. The entropy power N(X) of X is N(X) = 1 2πe exp µ 2 nh1(X) ¶ . Theorem 18.1. (Entropy power inequality.) Let X and Y be independent random vectors in Rn with probability densities in Lp(Rn) for some p > 1. Then N(X + Y ) ≥N(X) + N(Y ). (48) The entropy power inequality was proved by Shannon [136, Theorem 15 and Appendix 6] and applied by him to obtain a lower bound [136, Theorem 18] for the capacity of a channel. (Via the web site at this paper can be downloaded.) Shannon’s proof shows that equality holds in (48) if X and Y are multivariate normal with proportional covariances. In fact equality holds only for such X and Y , as Stam’s diﬀerent proof (simpliﬁed in and ) of (48) shows. The most accessible direct proof of (48) seems to be that of Blachman . We present a derivation from Young’s inequality and the following lemma, due to Lieb . Lemma 18.2. Let f and g be nonnegative functions in Ls(Rn) for some s > 1, such that Z Rn f(x) dx = Z Rn g(x) dx = 1. Then for 0 < λ < 1, h1(f ∗g) −(1 −λ)h1(f) −λh1(g) ≥−n 2 ((1 −λ) log(1 −λ) + λ log λ) . (49) Proof. For r ≥1, let p = p(r) = r (1 −λ) + λr and q = q(r) = r λ + (1 −λ)r. (50) Then p, q ≥1, 1 p + 1 q = 1 + 1 r, THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 39 and p(1) = q(1) = 1. If r < s is close to 1, then p, q < s, and since f, g ∈L1(Rn) ∩Ls(Rn), we have f ∈Lp(Rn) and g ∈Lq(Rn). Let F(r) = ∥f ∗g∥r ∥f∥p∥g∥q and G(r) = Cn, where C is as Theorem 14.1. By Young’s inequality (31), f ∗g ∈L1(Rn)∩Lr(Rn) and F(r) ≤G(r) for r close to 1. As we noted after Theorem 14.1, the equation F(1) = G(1+) holds. Therefore F(r) −F(1) r −1 ≤G(r) −G(1+) r −1 , for r close to 1, which implies that F ′(1+) ≤G′(1+). We can assume that h1(f ∗g) < ∞and therefore that h1(f) < ∞and h1(g) < ∞. Now if φ ∈Lr(Rn), ∥φ∥1 = 1, and h1(φ) < ∞, then d dr∥φ∥r = 1 r∥φ∥1−r d dr Z Rn φ(x)r dx = 1 r∥φ∥1−r Z Rn φ(x)r log φ(x) dx → Z Rn φ(x) log φ(x) dx = −h1(φ) as r →1. Using this and (50), we see that F ′(1+) = −h1(f ∗g) + (1 −λ)h1(f) + λh1(g). A calculation, helped by the fact that p′ = r′/(1 −λ) and q′ = r′/λ, where p′, q′, r′ denote as usual the H¨ older conjugates of p, q, r, respectively, shows that G′(1+) = n 2 ((1 −λ) log(1 −λ) + λ log λ) . Finally, (49) follows from the inequality F ′(1+) ≤G′(1+). □ Corollary 18.3. Young’s inequality (31) implies the entropy power inequality (48). Proof. In (49), put λ = N(Y ) N(X) + N(Y ). Simpliﬁcation of the resulting inequality leads directly to (48). □ Presumably Lieb, via his papers and , ﬁrst saw the connection between the entropy power inequality (48) and the Brunn-Minkowski inequality (15), the former being a limiting case of Young’s inequality (31) as r →1 and the latter a limiting case of the reverse Young inequality (32) as r →0. Later, Costa and Cover speciﬁcally drew attention to the analogy between the two inequalities, apparently unaware of the work of Brascamp and Lieb. Dembo, Cover, and Thomas explore further connections with other inequalities. These include some involving Fisher information and various uncertainty inequalities. Fisher information was employed by Stam in his proof of (48). Named after the statistician R. A. Fisher, Fisher information is claimed in a recent book by Frieden to be at the heart of 40 R. J. GARDNER a unifying principle for all of physics! If X is a random variable with probability density f on R, the Fisher information I(X) of X is I(X) = I(f) = − Z R f(x)(log f(x))′′ dx = Z R f′(x)2 f(x) dx, assuming these integrals exist. The multivariable form of I is a matrix, the natural extension of this deﬁnition. The quantity I is another measure of the “sharpness” of f or the missing information in X; see [64, Section 1.3] for a comparison of I and h1. Stam (see also ) showed that I can be used to obtain the Weyl-Heisenberg uncertainty inequality, and this inspired Frieden’s work. Frieden’s idea is that for any physical system, I represents how much information can possibly be obtained by measurements, while another quantity, J, is the amount of information bound up in the system. Then I −J leads to a Lagrangian, and the corresponding law of physics arises from its minimization, the second derivative usually present in such a law arising from the ﬁrst derivative present in I. Needless to say, Frieden’s claim has stirred some controversy. Some opinions can be found on the web site at and in the Mathematical Reviews review. The many related inequalities involving entropy and Fisher information are also connected to other consequences of Young’s inequality, such as Nelson’s hypercontractive inequality and various logarithmic Sobolev inequalities; see and Section 19.14. The papers , , and provide still more connections between information theory and convex geometry. 19. A survey In the subsections below we attempt an overview of the various known extensions and analogs of the Brunn-Minkowski inequality not covered above. Without being comprehensive, it should alert the reader to the main developments. 19.1. The Aleksandrov-Fenchel inequality. Theorem 19.1. (Aleksandrov-Fenchel inequality.) Let K1, . . . , Kn be compact convex sets in Rn and let 1 ≤i ≤n. Then V (K1, K2, . . . , Kn)i ≥ i Y j=1 V (Kj, i; Ki+1, . . . , Kn) . (51) See [37, p. 143] and [134, (6.8.7)]. The quantities V (K1, K2, . . . , Kn) and V (Kj, i; Ki+1, . . . , Kn) (where the notation means that Kj appears i times) are mixed volumes, like the quantity V1(K, L) we met in Section 7. In fact, if we put i = n in (51) and then let K1 = L and K2 = · · · = Kn = K, we retrieve Minkowski’s ﬁrst inequality (20) for compact convex sets. Therefore the Aleksandrov-Fenchel inequality implies the Brunn-Minkowski inequality for compact convex sets. In fact, there is a more general version of the latter that is equivalent to (51): Theorem 19.2. (Generalized Brunn-Minkowski inequality for compact convex sets.) Let K1, . . . , Kn be compact convex sets in Rn and let 1 ≤i ≤n. For 0 ≤λ ≤1, let f(λ) = V ((1 −λ)K0 + λK1, i; Ki+1, . . . , Kn)1/i . Then f is a concave function on [0, 1]. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 41 See [37, p. 146] and [134, Theorem 6.4.3]. The Brunn-Minkowski inequality for compact convex sets is the case i = n of Theorem 19.2. Readers familiar with the basic properties of mixed volumes can derive (51) by setting i = 2 in Theorem 19.2 and expanding the resulting inequality to extract the constants (1 −λ) and λ. Inequality (51) with i = 2 results, and the general case follows by induction on i. For compact convex sets, (51) is essentially the most powerful extension of the Brunn-Minkowski inequality known. Proofs of Theorems 19.1 and 19.2, discovered by A. D. Aleksandrov and by W. Fenchel and B. Jessen independently around 1937, can be found in [134, Theorems 6.3.1 and 6.4.3]. Equality conditions are not fully settled even today. An analog of the Aleksandrov-Fenchel inequality for mixed discriminants (see [134, Theorem 6.8.1]) was used by G. P. Egorychev in 1981 to solve the van der Waerden conjecture concerning the permanent of a doubly stochastic matrix. See [134, Chapter 6] for a wealth of information and references. Khovanskii, who with Teissier independently discovered that the Aleksandrov-Fenchel inequal-ity can be deduced from the Hodge index theorem, wrote a readable account of this surprising development in [37, Section 27]. The connection originates in the fact (due to D. M. Bernstein) that the number of complex roots of a generic system of n polynomial equations in n variables equals n! times the mixed volume of the corresponding Newton polytopes, P1, P2, . . . Pn, say. (The Newton polytope is the smallest convex polytope in Rn containing each point (m1, . . . , mn) for which czm1 1 · · · zmn n is a term of the polynomial.) The (n −2) of these n polynomial equations corresponding to P3, . . . , Pn deﬁne an algebraic surface in Cn on which the remaining polynomial equations describe two complex curves. The number of intersection points of these two curves is the number of roots of the system of n equations. Roughly speaking, the Hodge index theorem is an inequality involving the number of intersections of two complex curves Γ1, Γ2 in a compact complex algebraic surface and those of each curve with a slightly deformed copy of itself: ⟨Γ1, Γ2⟩2 ≥⟨Γ1, Γ1⟩⟨Γ2, Γ2⟩. Using the above observations, this can be translated into V (P1, P2, P3, . . . , Pn)2 ≥V (P1, P1, P3, . . . , Pn)V (P2, P2, P3, . . . , Pn). The case i = 2 of (19.1) (and hence, by induction, (19.1) itself) can be shown to follow by approximation by polytopes with rational coordinates. See [37, Section 27] for many more details and also and for more recent advances in this direction. Alesker, Dar, and Milman are able to use the Brenier map (see the end of Section 17) to prove some of the inequalities that follow from the Aleksandrov-Fenchel inequality, but the method does not seem to yield a new proof of (51) itself. In contrast to the Brunn-Minkowski inequality, the Aleksandrov-Fenchel inequality and some of its weaker forms, and indeed mixed volumes themselves, have only partially successful extensions to nonconvex sets. See [37, pp. 177–181], [134, p. 343], and . 19.2. Minkowski-concave functions. A real-valued function φ deﬁned on a class of sets in Rn closed under Minkowski addition and dilatation is called Minkowski concave if φ((1 −λ)X + λY ) ≥(1 −λ)φ(X) + λφ(Y ), (52) for 0 < λ < 1 and sets X, Y in the class. For example, the Brunn-Minkowski inequality implies that V 1/n n is Minkowski concave on the class of convex bodies. When Hadwiger published his 42 R. J. GARDNER extraordinary book in 1957, many other Minkowski-concave functions had already been found, and several more have been discovered since. We shall present some of these; all the functions have the required degree of positive homogeneity to allow the coeﬃcients (1 −λ) and λ to be deleted in (52). Other examples can be found in [75, Section 6.4] and in Lutwak’s papers and . Knothe gave a proof of the Brunn-Minkowski inequality for convex bodies, sketched in [134, pp. 312–314], and the following generalization. For each convex body K in Rn, let F(K, x), x ∈K, be a nonnegative real-valued function continuous in K and x. Suppose also that for some m > 0, F(λK + a, λx + a) = λmF(K, x) for all λ > 0 and a ∈Rn, and that log F ((1 −λ)K + λL, (1 −λ)x + λy) ≥(1 −λ) log F(K, x) + λ log F(L, y) whenever x ∈K, y ∈L, and 0 ≤λ ≤1. For each convex body K in Rn, deﬁne G(K) = Z K F(K, x) dx. Then G(K + L)1/(n+m) ≥G(K)1/(n+m) + G(L)1/(n+m), (53) for all convex bodies K and L in Rn and 0 < λ < 1. This is a consequence of the Pr´ ekopa-Leindler inequality. Indeed, taking f = F((1 −λ)K + λL, ·), g = F(K, ·), and h = F(L, ·), Theorem 4.2 implies that G is log concave. The method of Section 5 can then be used to derive the 1/(n+m)-concavity (53) of G from its log concavity. The Brunn-Minkowski inequality for convex bodies is obtained by taking F(K, x) = 1 for x ∈K. Dinghas found further results of this type. Let 0 ≤i ≤n. The mixed volume V (K, n −i; B, i) is denoted by Wi(K), and called the ith quermassintegral of a compact convex set K in Rn. Then W0(K) = Vn(K). It can be shown (see [134, (5.3.27), p. 295]) that if K is a convex body and 1 ≤i ≤n −1, then Wi(K) = κn κn−i Z G(n,n−i) V (K\|S) dS, (54) where dS denotes integration with respect to the usual rotation-invariant probability measure on the Grassmannian G(n, n−i) of (n−i)-dimensional subspaces of Rn. Thus the quermassintegrals are averages of volumes of projections on subspaces. Letting Ki+1 = · · · = Kn = B in Theorem 19.2 yields: Theorem 19.3. (Brunn-Minkowski inequality for quermassintegrals.) Let K and L be convex bodies in Rn and let 0 ≤i ≤n −1. Then Wi(K + L)1/(n−i) ≥Wi(K)1/(n−i) + Wi(L)1/(n−i), (55) with equality for 0 < i < n −1 if and only if K and L are homothetic. See [134, (6.8.10), p. 385]. The special case i = 0 is the usual Brunn-Minkowski inequality for convex bodies. The quermassintegral W1(K) equals the surface area S(K), up to a constant, so the case i = 1 of (55) is a Brunn-Minkowski-type inequality for surface area. When i = n −1, (55) becomes an identity. The equality conditions for 0 < i < n −1 follow from those known for the corresponding special case of Theorem 19.2. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 43 Let K be a convex body in Rn, deﬁne ˆ W0(K) = V (K) and for 1 ≤i ≤n −1, deﬁne ˆ Wi(K) = κn κn−i ÃZ G(n,n−i) V (K\|S)−1 dS !−1 , the ith harmonic quermassintegral of K. Similarly, deﬁne Φ0(K) = V (K) and for 1 ≤i ≤n −1, deﬁne Φi(K) = κn κn−i ÃZ G(n,n−i) V (K\|S)−n dS !−1/n , the ith aﬃne quermassintegral of K. Note the similarity to (54); the ordinary mean has been replaced by the −1- and −n-means, respectively. As its name suggests, Φi(K) is invariant under volume-preserving aﬃne transformations. Hadwiger [75, p. 268] proved the following inequality. Theorem 19.4. (Hadwiger’s inequality for harmonic quermassintegrals.) If K and L are convex bodies in Rn and 0 ≤i ≤n −1, then ˆ Wi(K + L)1/(n−i) ≥ˆ Wi(K)1/(n−i) + ˆ Wi(L)1/(n−i). Lutwak showed that the same inequality holds for aﬃne quermassintegrals. Theorem 19.5. (Lutwak’s inequality for aﬃne quermassintegrals.) If K and L are convex bodies in Rn and 0 ≤i ≤n −1, then Φi(K + L)1/(n−i) ≥Φi(K)1/(n−i) + Φi(L)1/(n−i). (56) Let K be a convex body in Rn, n ≥3. The capacity Cap (K) of K is deﬁned by Cap (K) = inf ½Z Rn ∥∇f∥2 dx : f ∈C∞ c (Rn), f ≥1K ¾ , where C∞ c (Rn) denotes the inﬁnitely diﬀerentiable functions on Rn with compact support. Here we are following Evans and Gariepy [57, p. 147], where Cap (K) = Cap n−2(K) in their notation. Several deﬁnitions are possible; see and [111, pp. 110–116]. The notion of capacity has its roots in electrostatics and is fundamental in potential theory. Note that capacity is an outer measure but is not a Borel measure, though it enjoys some convenient properties listed in [57, p. 151]. Borell proved the following theorem. Theorem 19.6. (Borell’s inequality for capacity.) If K and L are convex bodies in Rn, n ≥3, then Cap (K + L)1/(n−2) ≥Cap (K)1/(n−2) + Cap (L)1/(n−2). (57) Caﬀarelli, Jerison, and Lieb showed that equality holds if and only if K and L are homo-thetic. Jerison employed the inequality and its equality conditions in solving the correspond-ing Minkowski problem (see Section 7). 44 R. J. GARDNER 19.3. Blaschke addition. If K and L are convex bodies in Rn, then there is a convex body K ∔L such that S(K ∔L, ·) = S(K, ·) + S(L, ·), where S(K, ·) denotes the surface area measure of K. This is a consequence of Minkowski’s existence theorem; see [67, Theorem A.3.2] or [134, Section 7.1]. The operation ∔is called Blaschke addition. Theorem 19.7. (Kneser-S¨ uss inequality.) If K and L are convex bodies in Rn, then V (K ∔L)(n−1)/n ≥V (K)(n−1)/n + V (L)(n−1)/n, (58) with equality if and only if K and L are homothetic. See [134, Theorem 7.1.3] for a proof. Using Blaschke addition, a convex body called a mixed body can be deﬁned from (n −1) other convex bodies in Rn. Lutwak [98, Theorem 4.2] exploits this idea, due to Blaschke and Firey, to produce another strengthening of the Brunn-Minkowski inequality for convex bodies. 19.4. The Lp-Brunn-Minkowski theory. For convex bodies K and L in Rn, Minkowski addi-tion can be deﬁned by hK+L(u) = hK(u) + hL(u), for u ∈Sn−1, where hK denotes the support function of K. If p ≥1 and K and L contain the origin in their interiors, a convex body K +p L can be deﬁned by hK+pL(u)p = hK(u)p + hL(u)p, for u ∈Sn−1. The operation +p is called p-Minkowski addition. Firey proved the following inequality. (Both the deﬁnition of p-Minkowski addition and the case i = 0 of Firey’s inequality are extended to nonconvex sets by Lutwak, Yang, and Zhang .) Theorem 19.8. (Firey’s inequality.) If K and L are convex bodies in Rn containing the origin in their interiors, 0 ≤i ≤n −1 and p ≥1, then Wi(K +p L)p/(n−i) ≥Wi(K)p/(n−i) + Wi(L)p/(n−i), (59) with equality when p > 1 if and only if K and L are equivalent by dilatation. The Brunn-Minkowski inequality for quermassintegrals (55) is the case p = 1. Note that translation invariance is lost for p > 1. Firey’s ideas were transformed into a remarkable extension of the Brunn-Minkowski theory by Lutwak , , who also calls it the Brunn-Minkowski-Firey theory. Lutwak found the appropriate p-analog Sp(K, ·), p ≥1, of the surface area measure of a convex body K in Rn containing the origin in its interior. In , Lutwak generalized Firey’s inequality (59). He also generalized Minkowski’s existence theorem, deduced the existence of a convex body K ∔p L for which Sp(K ∔p L, ·) = Sp(K, ·) + Sp(L, ·) (when K and L are origin-symmetric convex bodies), and proved the following result. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 45 Theorem 19.9. (Lutwak’s p-surface area measure inequality.) If K and L are origin-symmetric convex bodies in Rn and n ̸= p ≥1, then V (K ∔p L)(n−p)/n ≥V (K)(n−p)/n + V (L)(n−p)/n, with equality when p > 1 if and only if K and L are equivalent by dilatation. Note that the Kneser-S¨ uss inequality (58) corresponds to p = 1. Lutwak, Yang, and Zhang study the Lp version of the Minkowski problem (see Section 7). A version corresponding to p = 0 is treated by Stancu . 19.5. Random and integral versions. Let X be a random set in Rn, that is, a Borel measurable map from a probability space Ωto the space of nonempty compact sets in Rn with the Hausdorﬀ metric. A random vector X : Ω→Rn is called a selection of X if Prob (X ∈X) = 1. If C is a nonempty compact set in Rn, let ∥C∥= max{∥x∥: x ∈C}. Then the expectation EX of X is deﬁned by EX = {EX : X is a selection of X and E∥X∥< ∞}. It turns out that if E∥X∥< ∞, then EX is a nonempty compact set. Theorem 19.10. (Vitale’s random Brunn-Minkowski inequality.) Let X be a random set in Rn with E∥X∥< ∞. Then Vn(EX)1/n ≥EVn(X)1/n. (60) See (and for a stronger version). By taking X to be a random set that realizes values (nonempty compact sets) K and L with probabilities (1 −λ) and λ, respectively, we see that Theorem 19.10 generalizes the Brunn-Minkowski inequality for compact sets. A version of (60) for intrinsic volumes (weighted quermassintegrals) of random convex bodies, and applications to stationary random hyperplane processes, are given by Mecke and Schwella . Earlier integral forms of the Brunn-Minkowski inequality, using a Riemann approach to pass from a Minkowski sum to a “Minkowski integral,” were formulated by A. Dinghas; see [37, p. 76]. 19.6. Other strong forms of the Brunn-Minkowski inequality for convex sets. McMullen deﬁnes a natural generalization of Minkowski addition of convex sets that he calls ﬁbre addition, and proves a corresponding Brunn-Minkowski inequality. Several strong forms of the Brunn-Minkowski inequality hold in special circumstances, for example, the stability estimates due to V. Diskant, H. Groemer, and R. Schneider referred to in [71, Section 3] and [134, p. 314], and an inequality of Ruzsa . Dar conjectures that if K and L are convex bodies in Rn and m = maxx∈Rn V (K ∩(L+x)), then V (K + L)1/n ≥m1/n + µV (K)V (L) m ¶1/n . (61) He shows that (61) implies the Brunn-Minkowski inequality for convex bodies and proves that it holds in some special cases. 46 R. J. GARDNER 19.7. Related aﬃne inequalities. A wide variety of fascinating inequalities lie (for the present) one step removed from the Brunn-Minkowski inequality. The survey paper of Osserman indicates connections between the isoperimetric inequality and inequalities of Bonnesen, Poincar´ e, and Wirtinger, and since then many other inequalities have been found that lie in a complicated web around the Brunn-Minkowski inequality. Some of these related inequalities are aﬃne inequalities in the sense that they are unchanged under a volume-preserving linear transformation. The Brunn-Minkowski and Pr´ ekopa-Leindler inequalities are clearly aﬃne inequalities. Young’s inequality and its reverse are aﬃne inequalities, since if φ ∈SL(n), we have φ(f ∗g) = (φf) ∗(φg) and ∥φf∥p = ∥f∥p. The Brascamp-Lieb and Barthe inequalities are also aﬃne inequalities. The sharp Hardy-Littlewood-Sobolev inequality (39) is not aﬃne invariant, but it is invariant under conformal transformations; see [91, Theorem 4.5]. The isoperimetric inequality is also not an aﬃne inequality (if it were, the equality for balls would imply that equality also held for ellipsoids), and neither is the Sobolev inequality (24). There is a remarkable aﬃne inequality that is much stronger than the isoperimetric inequality for convex bodies. The Petty projection inequality states that V (K)n−1V (Π∗K) ≤ µ κn κn−1 ¶n , (62) where K is a convex body in Rn, and Π∗K denotes the polar body of the projection body ΠK of K. (The support function of ΠK at u ∈Sn−1 equals V (K\|u⊥).) Equality holds if and only if K is an ellipsoid. See [67, Chapter 9] for background information, a proof, several other related inequalities, and a reverse form due to Zhang. Zhang has also recently found an astounding aﬃne Sobolev inequality, a common generalization of the Sobolev inequality (24) and the Petty projection inequality (62): If f ∈C1(Rn) has compact support, then µZ Sn−1 ∥Duf∥−n 1 du ¶−1/n ≥2κn−1 n1/nκn ∥f∥n/(n−1), (63) where Duf is the directional derivative of f in the direction u. This is only a taste of a banquet of known aﬃne isoperimetric inequalities. Lutwak wrote an excellent survey. For still more recent progress, the reader can do no better than consult the work of Lutwak, Yang, and Zhang, for example, and . 19.8. A restricted Brunn-Minkowski inequality. Let X and Y be measurable sets in Rn, and let E be a measurable subset of X × Y . Deﬁne the restricted Minkowski sum of X and Y by X +E Y = {x + y : (x, y) ∈E}. Theorem 19.11. (Restricted Brunn-Minkowski inequality.) There is a c > 0 such that if X and Y are nonempty measurable subsets of Rn, 0 < t < 1, t ≤ µVn(X) Vn(Y ) ¶1/n ≤1 t , and Vn(E) Vn(X)Vn(Y ) ≥1 −c min{t√n, 1}, then Vn(X +E Y )2/n ≥Vn(X)2/n + Vn(Y )2/n. THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 47 Szarek and Voiculescu proved Theorem 19.11 in the course of establishing an analog of the entropy power inequality in Voiculescu’s free probability theory. (Voiculescu has also found analogs of Fisher information within this noncommutative probability theory with applications to physics.) Barthe also gives a proof via restricted versions of Young’s inequality and the Pr´ ekopa-Leindler inequality. 19.9. Milman’s reverse Brunn-Minkowski inequality. At ﬁrst such an inequality seems impossible, since if K and L are convex bodies in Rn of volume 1, the volume of K + L can be arbitrarily large. As with the reverse isoperimetric inequality (45), however, linear transformations come to the rescue. Theorem 19.12. (Milman’s reverse Brunn-Minkowski inequality.) There is a constant c in-dependent of n such that if K and L are centrally symmetric convex bodies in Rn, there are volume-preserving linear transformations φ and ψ for which V (φK + ψL)1/n ≤c ³ V (φK)1/n + V (ψL)1/n´ . (64) First proved by V. Milman in 1986, this result is important in the local theory of Banach spaces. See [92, Section 4.3] and [127, Chapter 7]. 19.10. Discrete versions. The Cauchy-Davenport theorem, proved by Cauchy in 1813 and re-discovered by Davenport in 1935, states that if p is prime and X and Y are nonempty ﬁnite subsets of Z/pZ, then \|X + Y \| ≥min{p, \|X\| + \|Y \| −1}. Here \|X\| is the cardinality of X. Many generalizations of this result, including Kneser’s extension to Abelian groups, are surveyed in . The lower bound for a vector sum is in the spirit of the Brunn-Minkowski inequality. We now describe a closer analog. Let Y be a ﬁnite subset of Zn with \|Y \| ≥n + 1. For x = (x1, . . . , xn) ∈Zn, let wY (x) = x1 \|Y \| −n + n X i=2 xi. Deﬁne the Y -order on Zn by setting x <Y y if either wY (x) < wY (y) or wY (x) = wY (y) and for some j we have xj > yj and xi = yi for all i < j. For m ∈N, let DY m be the union of the ﬁrst m points in Zn + (the points in Zn with nonnegative coordinates) in the Y -order. The set DY m is called a Y -initial segment. The points of DY \|Y \| are o <Y e1 <Y 2e1 <Y · · · <Y (\|Y \| −n)e1 <Y e2 <Y · · · <Y en, where e1, . . . , en is the standard orthonormal basis for Rn. Note that the convex hull of DY \|Y \| is a simplex. Roughly speaking, Y -initial segments are as close as possible to being the set of points in Zn + that are contained in a dilatate of this simplex. Theorem 19.13. (Brunn-Minkowski inequality for the integer lattice.) Let X and Y be ﬁnite subsets of Zn with dim Y = n. Then \|X + Y \| ≥ ¯ ¯ ¯DY \|X\| + DY \|Y \| ¯ ¯ ¯ . (65) 48 R. J. GARDNER See , and also for a similar result in ﬁnite subgrids of Zn. That (65) is indeed a Brunn-Minkowski-type inequality is clear by comparing V (K + L) ≥V (rKB + rLB), the consequence of (17) given above. Indeed, (65) is proved by means of a discrete version, called compression, of an anti-symmetrization process related to Steiner symmetrization. 19.11. The dual Brunn-Minkowski theory. Let M be a body in Rn containing the origin in its interior and star-shaped with respect to the origin. The radial function of M is deﬁned by ρM(u) = max{c : cu ∈M}, for u ∈Sn−1. Call M a star body if ρM is positive and continuous on Sn−1. Let M and N be star bodies in Rn, let p ̸= 0, and deﬁne a star body M e +pN by ρM e +pN(u)p = ρM(u)p + ρN(u)p. The operation e +p is called p-radial addition. Theorem 19.14. (p-dual Brunn-Minkowski inequality.) If M and N are star bodies in Rn, and 0 < p ≤n, then V (M e +pN)p/n ≤V (M)p/n + V (N)p/n. (66) The reverse inequality holds when p > n or when p < 0. Equality holds when p ̸= n if and only if M and N are equivalent by dilatation. The inequality (66) follows from the polar coordinate formula for volume and Minkowski’s integral inequality (see [77, Section 6.13]). It was found by Firey for convex bodies and p ≤−1. The general inequality forms part of Lutwak’s highly successful dual Brunn-Minkowski theory, in which the intersections of star bodies with subspaces replace the projections of convex bodies onto subspaces in the classical theory; see, for example, . The cases p = 1 and p = n−1 are called the dual Brunn-Minkowski inequality and dual Kneser-S¨ uss inequality, respectively. A renormalized version of the case p = n+1 of (66) was used by Lutwak in his work on centroid bodies (see also [67, Section 9.1]). There is an inequality equivalent to the dual Brunn-Minkowski inequality called the dual Minkowski inequality, the analog of Minkowski’s ﬁrst inequality (20); see [67, p. 373]. This plays a role in the solution of the Busemann-Petty problem (the analog of Shephard’s problem mentioned after Theorem 7.1): If the intersection of an origin-symmetric convex body with any given hyperplane containing the origin is always smaller in volume than that of another such body, is its volume also smaller? The answer is no in general in ﬁve or more dimensions, but yes in less than ﬁve dimensions. See , , , , and . Lutwak also discovered that integrals over Sn−1 of products of radial functions behave like mixed volumes, and called them dual mixed volumes. In the same paper, he showed that a suitable version of H¨ older’s inequality in Sn−1 then becomes a dual form of the Aleksandrov-Fenchel inequality (51), in which mixed volumes are replaced by dual mixed volumes (and the inequality is reversed). Special cases of dual mixed volumes analogous to the quermassintegrals are called dual quermassintegrals, and it can be shown that an expression similar to (54) holds for these; instead of averaging volumes of projections, this involves averaging volumes of intersections THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 49 with subspaces. Dual aﬃne quermassintegrals can also be deﬁned (see [67, p. 332]), but apparently an inequality for these corresponding to (56) is not known. 19.12. Busemann’s theorem. Let S be an (n −2)-dimensional subspace of Rn, let u ∈Sn−1 ∩ S⊥, and let Su denote the closed (n −1)-dimensional half-subspace containing u and with S as boundary. Let u, v ∈Sn−1 ∩S⊥, and let X and Y be subsets of Su and Sv, respectively. If 0 < λ < 1, let u(λ) be the unit vector in the direction (1 −λ)u + λv, and let (1 −λ)X +h λY be the set of points in Su(λ) lying on a line segment with one endpoint in X and the other in Y . We call the operation +h harmonic addition. Theorem 19.15. (Busemann-Barthel-Franz inequality.) In the notation introduced above, let X and Y be compact subsets of Su and Sv, respectively, of positive Vn−1-measure. If 0 < λ < 1, then Vn−1 ((1 −λ)X +h λY ) ∥u(λ)∥ ≥M−1(Vn−1(X), Vn−1(Y ), λ). (67) Though Theorem 19.15 looks strange, it has the following nice geometrical consequence called Busemann’s theorem. If K is a convex body in Rn containing the origin in its interior and S is an (n −2)-dimensional subspace, the curve r = r(θ) in S⊥such that r(θ) is the (n −1)-dimensional volume of the intersection of K with the half-space Sθ forms the boundary of a convex body in S⊥. Proved in this form by H. Busemann in 1949 and motivated by his theory of area in Finsler spaces, it is also important in geometric tomography (see [67, Theorem 8.1.10]). As stated, Theorem 19.15 and precise equality conditions were proved by W. Barthel and G. Franz in 1961; see [67, Note 8.1] for more details and references. Milman and Pajor [119, Theorem 3.9] found a proof of Busemann’s theorem similar to the second proof of Theorem 4.1 given above. Generalizations along the lines of Theorem 10.2 are possible, such as the following (stated and proved in [15, p. 9]). Theorem 19.16. Let 0 < λ < 1, let p > 0, and let f, g, and h be nonnegative integrable functions on [0, ∞) satisfying h (M−p(x, y, λ)) ≥f(x) (1−λ)yp (1−λ)yp+λxp g(y) λxp (1−λ)yp+λxp , (68) for all nonnegative x, y ∈R. Then Z ∞ 0 h(x) dx ≥M−p µZ ∞ 0 f(x) dx, Z ∞ 0 g(x) dx, λ ¶ . The previous inequality is very closely related to one found earlier by Ball . For other associated inequalities, see [70, Theorem 4.1] and [118, Lemma 1]. 19.13. Brunn-Minkowski and Pr´ ekopa-Leindler inequalities in other spaces. Let X be a measurable subset of Rn and let rX be the radius of a ball of the same volume as X. If ε > 0, the Brunn-Minkowski inequality (16) implies that Vn(X + εB) ≥ ³ Vn(X)1/n + εVn(B)1/n´n = ³ Vn(rXB)1/n + εVn(B)1/n´n = Vn(rXB + εB). (69) For any set A, write Aε = A + εB = {x : d(x, A) ≤ε}. (70) 50 R. J. GARDNER Then we can rewrite (69) as Vn(Xε) ≥Vn((rXB)ε). (71) Notice that (71), by virtue of (70), is now free of the addition and involves only a measure and a metric. With the appropriate measure and metric replacing Vn and the Euclidean metric, (71) remains true in the sphere Sn−1 and hyperbolic space, equality holding if and only if X is a ball of radius rX. (Of course, in these spaces, the ball B(x, r) centered at x and with radius r > 0 is the set of all points whose distance from x is at most r. In Sn−1, balls are just spherical caps.) Though in Rn (71) is only a special case of (16), in Sn−1 and hyperbolic space, (71) is called the Brunn-Minkowski inequality. According to Dudley [52, p. 184], (71) was ﬁrst proved in Sn−1 under extra assumptions by P. L´ evy in 1922, with weaker assumptions by E. Schmidt in the 1940’s, and in full generality by Figiel, Lindenstrauss, and Milman in 1977. In hyperbolic space, (71) is due to E. Schmidt. A proof using symmetrization techniques for both Sn−1 and hyperbolic space can be found in [37, Section 9]. Perhaps more signiﬁcant than (71) for recent developments is a surprising result that holds in Sn−1, n ≥3, with the chordal metric. It can be shown that if Vn−1(X)/Vn−1(B) ≥1/2 and 0 < ε < 1, then Vn−1(Xε) Vn−1(B) ≥1 − ³π 8 ´1/2 e−(n−2)ε2/2. (72) Results of the form (72) are called approximate isoperimetric inequalities, and can be derived from the general Brunn-Minkowski inequality in Rn, as in [4, Theorem 2]. In particular, by taking X to be a hemisphere, we see that for large n, almost all the measure is concentrated near the equator! This result, which again goes back to P. L´ evy, is proved in [120, p. 5]. It is an example of the concentration of measure phenomenon that Milman applied in his 1971 proof of Dvoretzky’s theorem, and that with contributions by Talagrand and others has quickly generated an extensive literature surveyed by Ledoux . An excellent, but more selective, introduction is Ball’s elegant and insightful expository article [13, Lecture 8]. Analogous results hold in Gauss space, Rn with the usual metric but with the standard Gauss measure γn in Rn with density dγn(x) = (2π)−n/2e−∥x∥2/2 dx. Indeed, for bounded Lebesgue measurable sets X and Y in Rn for which (1−λ)X+λY is Lebesgue measurable, we have the inequality γn((1 −λ)X + λY ) ≥γn(X)1−λγn(Y )λ (73) corresponding to (14). This follows from the Pr´ ekopa-Leindler inequality (because the density function is log concave); see, for example, . It can also be derived directly from the general Brunn-Minkowski inequality in Rn by means of the “Poincar´ e limit,” a limit of projections of Lebesgue measure in balls of increasing radius; this and an abundance of additional information and references can be found in Ledoux and Talagrand’s book [86, Section 1.1]. To describe some of this work brieﬂy, let Φ(r) = γ1((−∞, r)) for r ∈R. Borell and Sudakov and Tsirel’son independently showed that if X is a measurable subset of Rn and γn(X) = Φ(rX), then γn(Xε) ≥Φ(rX + ε), with equality if X is a half-space. Ehrhard , gave a new proof using THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 51 symmetrization techniques that also yields the following Brunn-Minkowski-type inequality. If K and L are convex bodies in Rn and 0 < λ < 1, then Φ−1 (γn((1 −λ)K + λL)) ≥(1 −λ)Φ−1 (γn(K)) + λΦ−1 (γn(L)) . (74) While (74) is stronger than (73) for convex bodies, it is unknown whether it holds for Borel sets; see and [86, Problem 1]. An approximate isoperimetric inequality similar to (72) also holds in Gauss space; Maurey (see also see [13, Theorem 8.1]) found a simple proof employing the Pr´ ekopa-Leindler inequality. As in Sn−1, there is a concentration of measure in Gauss space, this time in spherical shells of thickness approximately 1 and radius approximately √n. Closely related work on logarithmic Sobolev inequalities is outlined in the next section. Bahn and Ehrlich ﬁnd an inequality that can be interpreted as a reversed form of the Brunn-Minkowski inequality in Minkowski spacetime, that is, Rn+1 with a scalar product of index 1. Cordero-Erausquin utilizes results of R. McCann to prove a version of the Pr´ ekopa-Leindler inequality on the sphere, remarking that a similar version can be obtained for hyperbolic space. These results are generalized in a remarkable paper by Cordero-Erausquin, McCann, and Schmuckenschl¨ ager, who establish a beautiful Riemannian version of Theorem 10.2. 19.14. Further applications. The Brunn-Minkowski inequality has been used in the study of crystals. A crystal in contact with its melt (or a liquid in contact with its vapor) is modeled by a bounded Borel subset M of Rn of ﬁnite surface area and ﬁxed volume. (We shall ignore measure-theoretic subtleties in this description.) The surface energy is given by F(M) = Z ∂M f(ux) dx, where ux is the outer unit normal to M at x and f is a nonnegative function on Sn−1 representing the surface tension, assumed known by experiment or theory. By the Gibbs-Curie principle, the equilibrium shape of such a crystal minimizes this surface energy among all sets of the same volume. This shape is called the Wulﬀshape. For a soapy liquid drop in air, f is a constant (we are neglecting external potentials such as gravity) and the Wulﬀshape is a ball, by the isoperimetric inequality. For crystals, however, f will generally reﬂect certain preferred directions. In 1901, Wulﬀgave a construction of the Wulﬀshape W: W = ∩u∈Sn−1{x ∈Rn : x · u ≤f(u)}; each set in the intersection is a half-space containing the origin with bounding hyperplane orthog-onal to u and containing the point f(u)u at distance f(u) from the origin. The Brunn-Minkowski inequality can be used to prove that, up to translation, W is the unique shape among all with the same volume for which F is minimum; see, for example, [143, Theorem 1.1]. This was done ﬁrst by A. Dinghas in 1943 for convex polygons and polyhedra and then by various people in greater generality. In particular, Busemann solved the problem when f is continuous, and Fonseca and Fonseca and M¨ uller extend the results to include sets M of ﬁnite perimeter in Rn. Good introductions with more details and references are provided by Taylor and McCann . In fact, McCann also proves more general results that incorporate a convex external potential, by a technique developed in his paper on interacting gases. A gas of particles in Rn is modeled by a nonnegative mass density ρ(x) of total integral 1, that is, a probability 52 R. J. GARDNER density on Rn, or, equivalently, by an absolutely continuous probability measure in Rn. To each state corresponds an energy E(ρ) = U(ρ) + G(ρ) 2 = Z Rn A(ρ(x)) dx + 1 2 Z Rn Z Rn V (x −y) dρ(x) dρ(y). Here U represents the internal energy with A a convex function deﬁned in terms of the pressure, and G(ρ)/2 is the potential energy deﬁned by a strictly convex interaction potential V . The problem is that E(ρ) is not generally convex, making it nontrivial to prove the uniqueness of an energy minimizer. McCann gets around this by deﬁning for each pair ρ, ρ′ of probability densities on Rn and 0 < t < 1 an interpolant probability density ρt such that U(ρt) ≤(1 −t)U(ρ) + tU(ρ′) (75) (and similarly for G and hence for E). McCann calls (75) the displacement convexity of U; ρt is not (1 −t)ρ + tρ′, but rather is deﬁned in the natural way by means of the Brenier map that transports ρ to ρ′ (see the last paragraph of Section 17). McCann is also able to recover the Brunn-Minkowski inequality from (75) by taking A(ρ) = −ρ(n−1)/n and ρ and ρ′ to be the densities corresponding to the uniform probability measures on the two sets. Next we turn to applications to diﬀusion equations. Let V be a nonnegative continuous poten-tial deﬁned on a convex domain C in Rn and consider the diﬀusion equation ∂ψ ∂t = 1 2△ψ −V (x)ψ(x, t) (76) with zero Dirichlet boundary condition (i.e., ψ tends to zero as x approaches the boundary of C for each ﬁxed t). Denote by f(t, x, y) the fundamental solution of (76); that is, ψ(t, x) = f(t, x, y) satisﬁes (76) and its boundary condition, and lim t→0+ f(t, x, y) = δ(x −y). For example, if V = 0 and C = Rn, then f(t, x, y) = (2πt)−n/2e−\|x−y\|2/2t. Brascamp and Lieb proved that if V is convex, then f(t, x, y) is log concave on C2. This is an application of the Pr´ ekopa-Leindler inequality, via Theorem 11.3 with p = 0; basically, it is shown that f is given as a pointwise limit of convolutions of log concave functions (Gaussians or exp(−tV (x))). Borell uses a version of Theorem 10.2 to show that the stronger assumption that V is −1/2-concave implies that t log(tnf(t2, x, y)) is concave on R+ ×C2. In a further study, Borell generalizes all of these results (and the Pr´ ekopa-Leindler inequality) by considering potentials V (σ, x) that depend also on a parameter σ. Another rich area of applications surrounds the logarithmic Sobolev inequality proved by Gross : Entγn(f) ≤1 2Iγn(f), (77) THE BRUNN-MINKOWSKI INEQUALITY OCTOBER 25, 2001 53 where f is a suitably smooth nonnegative function on Rn, γn is the Gauss measure deﬁned in the previous subsection, Entγn(f) = Z Rn f log f dγn − µZ Rn f dγn ¶ µZ Rn log f dγn ¶ , and Iγn(f) = Z Rn ∥∇f∥2 f dγn. Note (see Section 18) that Entγn(f) and Iγn(f) are essentially the negative entropy −h1(f) and Fisher information, respectively, of f, deﬁned with respect to Gauss measure. Bobkov and Ledoux derive (77) from the Pr´ ekopa-Leindler inequality (the “Brascamp-Lieb” in the title of refers to a diﬀerent inequality of Brascamp and Lieb proved in ). Cordero-Erausquin proves (77) directly using the transportation of mass idea seen in action above. McCann’s displacement convexity (75) plays an essential role in very recent work involving several of the above topics. Otto observed that various diﬀusion equations can be viewed as gradient ﬂows in the space of probability measures with the Wasserstein metric (formally, at least, an inﬁnite-dimensional Riemannian structure). McCann’s interpolation using the Brenier map gives the geodesics in this space, and Otto uses the displacement convexity to derive rates of convergence to equilibrium. The same ideas are utilized by Otto and Villani , who ﬁnd a new proof of an inequality of Talagrand for the Wasserstein distance between two probability measures in an n-dimensional Riemannian manifold, and show that Talagrand’s inequality is very closely related to the logarithmic Sobolev inequality (77). The interested reader may also consult Ledoux’s survey . The Brunn-Minkowski inequality was used by Firey in an investigation of the shapes of worn stones, related to the p = 0 version of the Lp-Minkowski problem (see Section 19.4). There is a connection here (as well as for the topic of shapes of crystals described above) with an active area concerning curvature-driven ﬂows; see, in particular, Andrews’ solution of a conjecture of Firey in . Borell applies Theorem 11.2 and his Brunn-Minkowski inequality in Gauss space (see the previous subsection) to option pricing, assuming that underlying stock prices are governed by a joint Brownian motion. Kannan, Lov´ asz, and Simonovits obtain some inequal-ities involving log-concave functions by means of a “localization lemma” that reduces certain inequalities involving integrals over convex bodies in Rn to integral inequalities over “inﬁnitesi-mal truncated cones”—line segments with associated linear functions—and hence to inequalities in a single variable. The proof of this localization lemma uses the Brunn-Minkowski inequality; see [93, Lemma 2.5], where an application to the algorithmic computation of volume is discussed. 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Department of Mathematics, Western Washington University, Bellingham, WA 98225-9063 E-mail address: gardner@baker.math.wwu.edu
14513	https://www.khanacademy.org/math/algebra/ck12-algebra-1/v/pythagorean-theorem	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
14514	http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratres.html	\| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Resolvance of Grating Resolvance or "chromatic resolving power" for a device used to separate the wavelengths of light is defined as \| \| \| Examples of resolvance \| The limit of resolution is determined by the Rayleigh criterion as applied to the diffraction maxima, i.e., two wavelengths are just resolved when the maximum of one lies at the first minimum of the other. \| \| \| \| \| \| \| --- --- --- \| \| Since the space between maxima for N slits is broken up into N-2 subsidiary maxima, the distance to the first mimimum is essentially 1/N times the separation of the main maxima. This leads to a resolvance for a grating of \| \| \| \| --- \| \| \| \| \| Show \| \| \| where N is the total number of slits illuminated and m is the order of the diffraction. \| \| \| \| Example \| \| \| Index Grating concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| \| --- \| Examples of Resolvance A standard benchmark for the resolvance of a grating or other spectroscopic instrument is the resolution of the sodium doublet. The two sodium "D-lines" are at 589.00 nm and 589.59 nm. Resolving them corresponds to resolvance Another standard example is the resolution of the hydrogen and deuterium lines, often done with a Fabry-Perot Interferometer. The red lines of hydrogen and deuterium are at 656.3 nm and 656.1 nm, respectively. This requires a resolvance of \| \| \| Fabry-Perot resolution \| \| Index Grating concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| \| \| --- --- \| \| Resolvance of Grating \| \| \| --- \| \| An approximate development of the resolvance expression can be done by using the small angle approximation to the condition for maxima. \| \| This gives the basic ideas, but the assumptions are shaky, so you might want a real derivation. \| Index Grating concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| --- \| \| Resolvance of Grating This approach to the resolvance of a grating has made use of the fact that the phase is a continuous variable which can be represented analytically, and that the differential of this variable is also well-defined. Since the Rayleigh criterion places the peak of one order at the first minimum of the adjacent order, the phase associated with being "just resolved" is determined to be 2π/N. Taking the differential of that phase gives an expression which contains the differential of wavelength dλ which allows the quantity λ /dλ to be evaluated. In practice, the resolvance is stated in the form R=λ /Δλ for applications like the observation of the sodium doublet. We know the wavelength difference to be Δλ = .59 nm, so the resolvance can help us to anticipate whether a particular diffraction grating could resolve that difference. \| Index Grating concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \|
14515	https://ess.unm.edu/resources/esspresentations_and_files/college-algebra-word-problems.pdf	1 College Algebra Practice Word Problems 1. A basketball team played thirty-two games and won three times as many games as it lost. How many games did the team win? 2. The sum of three consecutive integers is forty-two. Find the three numbers. 3. Ninety-six golf balls were picked up at the driving range and placed into two buckets. One bucket has twenty-eight more golf balls than the other bucket. How many golf balls are in each bucket? 4. When the sum of a number and 3 is subtracted from 10 the result is 5. Identify the integer. 5. Find two numbers whose sum is sixty-eight and whose difference is twenty-two. 6. A freight train starts from Los Angeles and heads for Chicago at 40 mph. Two hours later, a passenger train leaves the same station for Chicago traveling 60mph. How long before the passenger train overtakes the freight train? Further practice: 1. There are three consecutive even numbers such that twice the first is 20 more than the second. Find the numbers. 2. Jay’s father is twice as old as Jay. In 20 years, Jay will be two-thirds as old as his father. How old is each now? 3. Chris and Sandra worked as electricians at $14 and $12 per hour, respectively. One month, Sandra worked 10 hours more than Chris. If their total income for the month was $3520, how many hours did each work during the month? 4. Tickets for a baseball game were $2.50 for general admission and $0.50 for kids. If there were six times as many general admission tickets sold as there were kids’ tickets sold, and total receipts were $7750, how many of each type of ticket were sold? 5. The Hailu family is on a cross-country trip traveling with the Garcias. One day they get separated and the Garcias are 20 miles ahead of the Hailus on the same road. If the Garcias average 50 mph and the Hailus travel at 60 mph, how long will it be before the Hailus catch up with the Garcias? 1 College Algebra Practice Word Problems with Solutions 1. A basketball team played thirty-two games and won three times as many games as it lost. How many games did the team win? Solution: Let 𝑥 equal the number of games the team won and let 𝑦 equal the number of games the team lost. Write an equation for their sum: 𝑥+𝑦=32 From the problem, we can say that 𝑥=3𝑦 Substitute this expression into the above equation for 𝑥: (3𝑦)+𝑦=32 4𝑦=32 𝑦=8 The team lost 8 games. Use 𝑦=8 to find the number of games the team won: 𝑥=3(8)=24 Answer: 24 games won 2. The sum of three consecutive integers is forty-two. Find the three numbers. Solution: Let 𝑥,𝑦 and 𝑧 be the three numbers. Write an equation for their sum: 𝑥+𝑦+𝑧=42 Since the numbers are consecutive, 𝑥 is the smallest number 𝑦=𝑥+1 𝑧=𝑥+2 Substitute these expressions into the above equation and solve for 𝑥: 𝑥+(𝑥+1)+(𝑥+2)=42 3𝑥+3=42 3𝑥=39 𝑥=13 Use 𝑥=13 to find 𝑦 and 𝑧: 2 𝑦=13+1=14 𝑧=13+2=15 𝑥+𝑦+𝑧=13+14+15=42 Answer: 13, 14, 15 3. Ninety-six golf balls were picked up at the driving range and placed into two buckets. One bucket has twenty-eight more golf balls than the other bucket. How many golf balls are in each bucket? Solution: Let 𝑥 equal the number of golf balls in the first bucket and let 𝑦 equal the number of golf balls in the second bucket. Write an equation for their sum: 𝑥+𝑦=96 From the problem, we can say that 𝑦=𝑥+28 Substitute this expression into the equation above for 𝑦 and solve for 𝑥: 𝑥+(𝑥+28)=96 2𝑥+28=96 2𝑥=68 𝑥=34 Use 𝑥=34 to find 𝑦: 𝑦=34+28=62 Answer: 34,62 4. When the sum of a number and 3 is subtracted from 10 the result is 5. Identify the integer. Solution: The sum of a number and 3 is: x+3 Write the problem as a mathematical sentence, meaning convert each phrase into the corresponding part of the equation: 10 - (x+3) = 5 Now solve with algebra: 3 10 - x -3 = 5 7 - x = 5 7 - 5 = x 2 = x x = 2 Answer: 2 5. Find two numbers whose sum is sixty-eight and whose difference is twenty-two. Solution: Let 𝑥 equal the first number and let 𝑦 equal the second number. Write an equation for their sum and for their difference: 𝑥+𝑦=68 𝑥−𝑦=22 In the second equation, isolate 𝑥 by adding 𝑦 to both sides: 𝑥−𝑦+𝑦=22+𝑦 𝑥=22+𝑦 Substitute this expression into the first equation for 𝑥 and solve for 𝑦: (22+𝑦)+𝑦=68 22+2𝑦=68 2𝑦=46 𝑦=23 Use 𝑦=23 to find 𝑥: 𝑥=22+23=45 Answer: 23,45 6. A freight train starts from Los Angeles and heads for Chicago at 40 mph. Two hours later, a passenger train leaves the same station for Chicago traveling 60mph. How long before the passenger train overtakes the freight train? Solution: Rate Time Distance Freight 40 4 Passenger 60 Let 𝑥 = time in hours for the passenger train. The passenger train started 2 hours after the freight train, so the freight train took 2 hours longer. You can represent the time for the freight by 𝑥 + 2: Rate Time Distance Freight 40 𝑥 + 2 Passenger 60 𝑥 Rate times time equals distance (r × t = d), so multiply what you have in the rate box times what you have in the time box and put the result in the distance box: Rate Time Distance Freight 40 𝑥 + 2 40(𝑥 + 2) Passenger 60 𝑥 60𝑥 Set these two distances equal for your equation: 40(𝑥 + 2) = 60𝑥 40𝑥 + 80 = 60𝑥 -20𝑥 = -80 Answer: 𝑥 = 4 Check: 40(4 + 2) = 60(4) 240 = 240 5 Further practice: 1. There are three consecutive even numbers such that twice the first is 20 more than the second. Find the numbers. 2. Jay’s father is twice as old as Jay. In 20 years, Jay will be two-thirds as old as his father. How old is each now? 3. Chris and Sandra worked as electricians at $14 and $12 per hour, respectively. One month, Sandra worked 10 hours more than Chris. If their total income for the month was $3520, how many hours did each work during the month? 4. Tickets for a baseball game were $2.50 for general admission and $0.50 for kids. If there were six times as many general admission tickets sold as there were kids’ tickets sold, and total receipts were $7750, how many of each type of ticket were sold? 5. The Hailu family is on a cross-country trip traveling with the Garcias. One day they get separated and the Garcias are 20 miles ahead of the Hailus on the same road. If the Garcias average 50 mph and the Hailus travel at 60 mph, how long will it be before the Hailus catch up with the Garcias? Answers: 1. 22, 24, 26 2. 20, 40 3. 130, 140 4. 6500, 3000 5. 2 hours
14516	https://anatomypubs.onlinelibrary.wiley.com/doi/10.1002/ar.21051	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Volume 293, Issue 2 pp. 320-337 Review Free Access The Lymph Node Revisited: Development, Morphology, Functioning, and Role in Triggering Primary Immune Responses Guy Sainte-Marie, Corresponding Author Guy Sainte-Marie bernard.sainte-marie@dfo-mpo.gc.ca Département de pathologie et biologie cellulaire, Université de Montréal, Montréal, Québec, Canada Tel: 450-449-1319. Fax: 418-775-0740 Faculté de médecine, Département de pathologie et biologie cellulaire, Université de Montréal, C.P. 6128 succursale Centre-ville, Montréal, QC, Canada H3C 3J7Search for more papers by this author Guy Sainte-Marie, Corresponding Author Guy Sainte-Marie bernard.sainte-marie@dfo-mpo.gc.ca Département de pathologie et biologie cellulaire, Université de Montréal, Montréal, Québec, Canada Tel: 450-449-1319. Fax: 418-775-0740 Faculté de médecine, Département de pathologie et biologie cellulaire, Université de Montréal, C.P. 6128 succursale Centre-ville, Montréal, QC, Canada H3C 3J7Search for more papers by this author First published: 25 January 2010 Citations: 60 Abstract Scenarios have been proposed to explain how lymphoid components of a lymph node favor the encounter of a drained antigen with a circulating competent naïve lymphocyte to trigger a primary immune response. However, these scenarios rest on incorrect concepts about the organ. This situation resulted from a loss of interest for studies on in vivo lymphoid organs due to a widespread switch, decades ago, to work on suspended lymphoid cells. However, an in vivo holistic study of the organ continued in our laboratory. The present review synthesizes resulting knowledge on lymph node morphology and global functioning. We show that the opening of an afferent lymphatic vessel into the subcapsular sinus is the focal point from which the related portion of a lymph node—a node compartment—is developed. As to the formation of a compartment's lymphoid components, it is neonatally orchestrated by the dichotomic nature and distribution of antigens in this subcapsular sinus, which determines a dichotomic recruitment of circulating cells and the compartment's architectural complexity. The transport process of an antigen from a given tissue territory into restricted sites of the draining compartment further defines its local morphological features and activities, while providing the possibility to reduce the wandering of a short-lived naïve cell through innumerable target-devoid sites. We also explain that the nodal lymphoid components are not implicated in the triggering of primary responses, but are rather products of such responses. Scenarios for the triggering of primary responses, consistent with real node morphology and functioning, are proposed. Anat Rec, 293:320–337, 2010. © 2010 Wiley-Liss, Inc. In1984, Nossal wrote “A readership consisting of primarily anatomists has every right to question the favorite sport of research workers in cell immunology. This is to take a lymphoid tissue and totally destroy its beautiful and elaborately designed architecture to obtain simple cell suspension of lymphocytes, which are then asked to do more or less all the jobs of the original anatomic masterpiece.” Nonetheless, in vitro work prevailed largely because of the great difficulty of in vivo experimentation on lymphoid tissues. More recently, Gretz et al. (1996) recalled that while in vitro studies yielded notions on the activation of T and B cells at the subcellular level, the efficiency of the immune system in vivo depends on a higher level of orchestration that occurs in lymph nodes, where most primary immune responses are presumed to be triggered. They added that activation requiring the encounter of an antigen (or an antigen-presenting cell: APC) and a rare competent circulating naïve lymphocyte—and how that encounter occurs—depend on morphological features of the lymph node's parenchymal components, so that knowledge of lymph node morphology is essential for understanding its immune function. This opinion was shared by Cyster (1999), Crivaletto et al. (2004), and Sixt et al. (2005) who along with Gretz et al. (1996) proposed scenarios for the triggering of a primary response in the lymph node. However, these scenarios may directly transpose in vitro findings to the in vivo condition, while ignoring the organ's complexity. Moreover, as will be demonstrated below, the lymph node's assumed morphology rests on some inadequate morphological concepts and it is inconsistent across the various scenarios, reflecting confusion in the field. This review describes the basic morphology of a normal lymph node. It explains how some of the more popular misconceptions about lymph node morphology and functioning arose, and why they are inadequate as a basis for elaborating scenarios on the triggering of a primary response. Alternative views are presented that are derived mainly from a synthesis of our long-term, mostly in vivo work on the rat lymph node undergoing various spontaneous immunological states that reveal natural reactions. We explain how drained antigens and the influenced subcapsular sinus play a primordial role in the neonatal emergence of nodal lymphoid components and in orchestrating their activity. We show how antigen dichotomy—the partitioning of antigens into those causing either a cellular or humoral response—can account for the organ's complexity and dichotomic recruitment of circulating lymphocytes. These features facilitate the encounter of lymph-carried and blood-borne related immunogenic elements, and therefore, the triggering of a response. We also consider how the recruitment of circulating cells may have become more efficient through the elimination of time-consuming aspects of a lymphocyte quest for its target. Finally, we elaborate new and morphologically more realistic proposals on the triggering of primary cellular and humoral responses in the lymph node. MORPHOLOGY OF THE LYMPH NODE1 Since the morphological features of the lymph node are essential for understanding its functions, they are summarized here. Though our work focused on the rat, nodal morphology is basically similar among eight examined mammal species including the human and the mouse (Bélisle and Sainte-Marie,1981a). However, interpretation of single lymph node sections can be complicated by: (1) irregular occurrence, but more often in some species and in large than in small lymph nodes, of variably developed connective tissue trabeculae or septae arising from the capsule, which may partition a lymph node to various extents (Yoffey and Courtice,1970); (2) changes in the appearance of structural components with cutting angle because lymph node and compartment architectures are non-uniform; (3) difficulty of clearly outlining the diverse nodal lymphoid components in standard tissue sections; and (4) intrinsic and extensive variability in the number, size, shape, and content of these components within and among compartments, lymph nodes, and individuals (see for example, the extreme variability in lymphocyte content of the subcapsular sinus described in Sainte-Marie et al.,1982; Sainte-Marie and Peng,1990b; Sainte-Marie,2001). These complications, combined with often limited organ samples, contribute to explain why the lymph node has remained poorly understood and why contradictory reports, conclusions, and functional hypotheses abound in the literature. Although random sections of lymph nodes may be variable and puzzling in appearance, their basic architectural organization is simple and consists of a structure called a compartment. The compartment is the portion of a lymph node associated with the opening of each afferent lymphatic vessel, or each of its terminal branches, into the subcapsular sinus (Bélisle and Sainte-Marie,1981b; Aijima et al.,1986; Hoshi et al.,1997). The smallest lymph nodes have a single afferent lymphatic vessel and a single compartment (Bélisle and Sainte-Marie,1981b). In larger organs, the number of compartments equals the number of afferent lymphatics in small mammals or the number of terminal branches of afferent lymphatics in large mammals (Bélisle and Sainte-Marie,1981b; Aijima et al.,1986). Here and there at the margin of each compartment, cortical gaps connect the subcapsular sinus to medullary sinuses (Fig. 1A, B). A compartment also has a subsinus layer, a peripheral and a deep cortex, as well as medullary cords and sinuses (Fig. 1C, D). Compartmentation was demonstrated by the subcutaneous injection of a small dose of tracer (Sainte-Marie et al.,1982). The tracer entered the subcapsular sinus of the single compartment associated with the lymphatic vessel draining the injected tissue territory. The tracer then flowed from the lymphatic opening towards and into the closest cortical gaps (Fig. 1C). Thus, the drained material from a given tissue territory usually influences only those components of the connected compartment (Sainte-Marie et al.,1982). This is contrary to the common view of the lymph node being one functional unit, such that the content of any of its afferent lymphatics is assumed to spread throughout and influence the whole organ. This view arose from observations made after locally injecting a large dose of tracer that filled all sinuses of the draining lymph node (Drinker et al.,1934). The dose was administered to demonstrate the lymph node's great filtration capacity, but it was physiologically unrealistic. Recently, Gretz et al. (1997) recognized the existence of the compartment, but renamed it a “lobule.” Their description of a “lobule” in the rat lymph node is compatible with ours; however, their statement that the mouse lymph node has a single “lobule” disagrees with observations in Bélisle and Sainte-Marie (1981a). Gretz et al. (1997) reported that in larger mammals, the “lobules” are outlined by radial fibrous trabeculae (more correctly, septae). Although septae may indeed be seen in some species (Yoffey and Courtice,1970), they are not a basic feature of lymph nodes with multiple compartments (Sainte-Marie et al.,1982). For example, there are few septae in the rat's multi-compartmented lymph nodes. Peripheral Cortex and Subsinus Layer Until about 1962, the lymphoid population present between the subcapsular sinus and the medulla of a lymph node was termed cortex, which was further resolved into a peripheral and a vaguely demarcated deep cortex. It was then found that lymphocytes comprise B and T cells, and that T cells, responsible for cellular responses, are located in the deep cortex (Oort and Turk,1965). Subsequently, components that had been so far termed peripheral and deep cortex became known as the “cortex” and “paracortex,” respectively. For reasons that will become apparent below, we continued to use the original terms of cortex including a peripheral and a deep cortex. The peripheral cortex underlies the subcapsular sinus entirely. It tends to stretch symmetrically around the lymphatic opening and to thin with increasing distance (Fig. 1A). It contains a superficial row of folliculo-nodules, each formed of a B-cell follicle covering to various extents a nodule (or germinal center) (Sainte-Marie and Sin,1970). Folliculo-nodules are separated by an extrafollicular zone (Fig. 1C, D); they are not discussed further since they are not implicated in primary responses. An elusive subsinus layer is sandwiched between the subcapsular sinus and the peripheral cortex. Its fiber network supports reticular-like cells interspaced by lymphocytes and APCs (Lind,1968). The subsinus layer is about 25 μm thick over the extrafollicular zone but thins over outwardly bulging folliculo-nodules. This layer complements the screening activity of the sinus wall (Sainte-Marie and Peng,1985a). Deep Cortex The “paracortex” was initially defined as a uniform layer of T cells underlying the whole peripheral cortex of a lymph node (Gutman and Weisman,1972). Actually, the deep cortex of a compartment is hemispherical or semi-ovoid, centered on the lymphatic opening (Fig. 1A), and it usually does not extend horizontally as much as the peripheral cortex above (Bélisle and Sainte-Marie,1981c; Aijima et al.,1986; Okada et al.,2002; Willard-Mack,2006). The deep cortex has a center, where T cells of cellular responses are engendered, and a periphery underlying the center (Fig. 1C) (Bélisle and Sainte-Marie,1981d). The periphery is actually an extension of the extrafollicular zone of the overlying peripheral cortex, with the additional feature of lymphatic culs-de-sac, termed deep cortex sinuses, which are continued by medullary sinuses. Hence, the deep cortex is not totally unlike the peripheral cortex and it is, therefore, inappropriate to call it “paracortex,” which etymologically implies a different component situated beside (para) the cortex. Note that because the deep cortex does not form a continuous layer in a lymph node, we previously referred to its discontinuous portions as deep cortex “units” (Bélisle and Sainte-Marie,1990). Considering that each lymph node compartment has a single portion of deep cortex, the word unit is superfluous and is now discarded except when quoting. Representation of the deep cortex and of its relationship to the peripheral cortex in single tissue sections may be puzzling, due to variable development of individual compartments and the cutting plane (Fig. 2A–D). In some lymph nodes, the peripheral cortex of neighboring compartments does not extend beyond their respective deep cortex, and therefore, these deep cortices may appear to form a continuous layer (Bélisle and Sainte-Marie,1981a). In some compartments, to the contrary, the peripheral cortex extends far beyond the limit of the deep cortex, and then, depending on the cutting plane, tissue sections may exhibit only peripheral cortex so that a compartment appears to be devoid of deep cortex (Fig. 2C). Various other configurations may occur (Sainte-Marie et al.,1990). The fact that the deep cortex usually underlies only part of the peripheral cortex of a compartment makes the cortex as a whole non-uniform, contrary to the state of uniformity implicit in the popular concept of “paracortex.” This difference is of paramount importance, for either state implies different patterns for recruitment of circulating cells. A uniform cortex would result in uniform recruitment. Instead, the peripheral cortex and subcapsular sinus are actually divided into domains, one above the deep cortex center (aDCC) and one extending beyond the deep cortex center (bDCC), that recruit differently (Fig. 1C). The aDCC recruits elements involved in both cellular and humoral responses, whereas the bDCC recruits only elements committed to humoral responses (see below). A third domain located above the deep cortex periphery was also distinguished (Sainte-Marie and Peng,1996), but it is not essential to the present discussion. High Endothelial Venules These peculiar vessels recruit blood cells and were called “postcapillary venules” (Pirro,1954; Burwell,1962) before the term HEV was largely adopted (Anderson and Anderson,1975). HEVs were thought to be restricted to, and scattered throughout, the “paracortex” (Anderson and Anderson,1975; Ford and Smith,1982). It was and continues to be claimed that lymphocytes enter lymph nodes via HEVs only and at random (Gowans and Knight,1964; Rouse et al.,1984). How these misconceptions arose will be examined in a later subsection. Actually, HEVs occur in the extrafollicular zone of the peripheral cortex and the continuing deep cortex periphery; they transform into regular venules upon entering the medullary cords (Fig. 1C) (Sainte-Marie and Sin,1970). Exceptionally, an HEV may cross the deep cortex center and directly enter a cord; nonetheless, lymphocytes traveling in the HEV perivascular channels do not mix with lymphocytes in the center as demonstrated in athymic animals (Sainte-Marie et al.,1984). In addition, HEVs are commonly described as being uniform, but they have generally been studied under conditions that do not reveal the great plasticity in the features of their individual endothelial (or HEV) cells. Striking differences may be apparent even in adjacent cells (for a general review on HEVs, see Sainte-Marie and Peng,1996). Hence, HEV cells can vary markedly in shape and size: the lowest are squamous, the highest are narrowed and may exhibit a blastoid aspect (Dabelow,1938; Burwell,1962; Sainte-Marie,1966). Lymphocytes are present in variable numbers in individual interendothelial spaces (Norberg and Rydgren,1978) and in the expandable subendothelial space present between each HEV cell and the basement membrane (Fig. 3A) (Sainte-Marie,1966; Mikata and Niki,1971). Note that in comparison to the rat, the architecture of the lymph node of the commonly investigated mouse is more difficult to interpret because the peripheral cortex and associated HEVs are less developed, and this in turn may explain why the opinion that HEVs are absent from the peripheral cortex is so persistent. Reticular Fiber Network and Pathways of Transcortical Cell Migration Coarse fibers 1–4 μm wide, lined by pleomorphic fibroblast- or reticular-like cells, arise perpendicularly from the capsule but almost only over the extrafollicular zone (Fig. 1D). They run parallel through the subcapsular sinus, subsinus layer, and extrafollicular zone. Between the coarse fibers in the extrafollicular zone, fine fibers knit small communicating alveoles, each hosting a few lymphocytes (Sainte-Marie and Sin,1970). Fiber material outlines one, but more often two or three, roughly concentric perivascular channels that surround each HEV (Sainte-Marie,1966). The first channel, immediately cuffing the HEV endothelium, is narrow and hosts distorted lymphocytes. The second and third channels are dilated to varying degrees and contain more rounded lymphocytes. At the flattened face of the deep cortex center, the parallel fibers and HEVs bend and course into the deep cortex periphery to join medullary cords below. The extrafollicular zone and continuing deep cortex periphery connect both gateways—the subcapsular sinus and HEV network—for entry of lymphocytes into the compartment. Together, they form a continuing pathway of parallel fibers and perivascular channels that orient the migration of incoming lymph-carried or blood-borne cells. This pathway guides a competent cell committed to either a cellular or a humoral response towards the deep cortex center or a medullary cord, respectively, where it yields effector-cells of a response (Fig. 1C). In other respects, it is noteworthy that the pattern of the fiber network differs in the different components of the compartment (Fig. 1D), allowing them to be better delimited with silver impregnation (Sin,1967; Sainte-Marie and Sin,1970). The Concept of “Paracortical Cords” Kelly (1975) suggested that “paracortical cords,” favoring T cell activation, are the basic elements of a “paracortical area.” A “cord” was said to comprise a central HEV with a perivascular cuff of lymphocytes enclosed in a reticulin framework surrounded by an alleged “paracortical sinus.” The area shown in Kelly's supporting Fig. 2 is in fact the deep cortex of a compartment, whose center is the site of formation of effector-cells of cellular responses. However, no “cord” is apparent in the illustrated deep cortex center and this is consistent with the demonstrated absence of HEVs in the center (refer previous section). More recently, Gretz et al. (1997) reviewed lymph node organization and contemplated how a primary cellular response might be triggered in the “paracortex.” Their review was offered as a functionally-oriented summary of their understanding of “paracortex” architecture and presented a hypothesis regarding its microanatomy and the rules governing an orchestrated movement of cells and soluble factors within it. Actually, they adopted Kelly's general concept of “paracortical cords,” while stating as a caveat that they had not been able to identify deep cortex “units,” and therefore, could not integrate them in their hypothesis. However, Fig. 7B in Gretz et al. (1997) shows an accumulation of lymphocytes under peripheral cortex, which is probably a deep cortex “unit” that appears round due to the cutting plane. Intriguingly, in an article published only a year earlier, Gretz et al. (1996) had acknowledged the existence of deep cortex “units.” Moreover, as pointed out above, the existence of deep cortex “units” was confirmed by others through histology (Ajima et al.,1986) and more recently by plastic casting of lymph nodes (Okada et al.,2002). Gretz et al. (1997) modified Kelly's (1975) concept of “paracortical cords.” They correctly recognized the narrow first perivascular channel cuffing an HEV endothelium as described earlier (Sainte-Marie,1966). However, they renamed it “perivenular” channel, renamed the second and third channels “corridors,” and renamed Kelly's “paracortical” sinus a “cortical” sinus and added that such sinuses fill the extrafollicular zone. Figure 4 of Gretz et al. (1997) was presented to support the existence of “cortical” sinuses, but none can be seen. Recently, Sixt et al. (2005) reported that drained dextrans accumulate in “paracortical” sinuses, but none were illustrated and to our knowledge they do not exist. Gretz et al. (1997) said “that the elegant form” of the paracortical cords “dictates, in large part, the marvelous efficiency of secondary lymphoid tissues in facilitating and regulating immune responses.” Ultimately, they asked “If the foregoing description of the paracortical cord is substantially correct, why has its existence been such a well kept secret?” The answer was that its complex geometry probably “precludes simple assessment of precisely how much of the paracortex consists of paracortical cords.” They stated that Kelly's (1975) serendipitous study had revealed that cords are the essential constituent of “paracortex.” It is not clear how this could be concluded since Kelly had not carried out a tridimensional analysis of the presumed cords, a necessary task given their reportedly complex geometry and the acknowledged difficulty of distinguishing them in normal lymph nodes (Kelly,1975; Gretz et al.,1997). Moreover, it is disconcerting that the “paracortical cords” were not mentioned 1 year earlier in a proposal on how “paracortex” morphology facilitates the triggering of a cellular response (Gretz et al.,1996) or a few years later in an article examining APC-lymphocyte interaction (Gretz et al.,2000). In our opinion, the existence of these cords is unsubstantiated. FUNCTIONAL ASPECTS OF THE LYMPH NODE COMPARTMENT Transport of Soluble Immunogenic Factors by the Reticular Fiber Network How do antigens penetrate the compartment's parenchyma from the subcapsular sinus? An antigen causing a cellular response is carried by an APC, one causing a humoral response is transported by fibers. In1963, Moe suggested that the lymph node's reticular fibers transport various soluble substances. Anderson and Anderson (1975) later reported that drained horseradish peroxidase is dispersed in the reticulum and sheaths around HEVs before permeating their interendothelial spaces and entering their lumen. However, we could not detect the tracer in these spaces and lumen in their supporting Fig. 19. Besides, permeation obtained with horseradish peroxidase was suspicious because this tracer alters vascular permeability (Contran and Karnovsky,1967). Injecting instead a foreign albumin, Sainte-Marie and Peng (1986) observed its transport by fibers running through the thickness of the subcapsular sinus, subsinus layer, extrafollicular zone, deep cortex periphery, and adjoining HEV basement membranes; no endothelial permeation was detected. Recently, Gretz et al. (2000) stated that drained fluorophore-labeled low MW dextrans and proteins diffused from peri-HEV fibers in between HEV cells and into an HEV's lumen. However, in the absence of counterstaining with a standard technique, it is questionable whether the small pale structures in their supporting Fig. 3A, B are HEVs. This uncertainty also applies to the only magnified structure presented in their Fig. 3D, which shows no HEV cells—an essential feature for identifying an HEV. In addition, no interendothelial fluorescence is detectable to support permeation as the delivery mode for the astonishingly abundant fluorescent material seen in the presumed HEV lumen. Hence, in our opinion, HEV permeation has not been demonstrated. Still, a fiber-carried soluble factor can reach HEV cells, as confirmed by the presence in fibers and HEV basement membranes of typical metachromatic products spontaneously released by mast cells in the extrafollicular zone (Sainte-Marie and Peng,1990a). Incidentally, on the basis of a conductive activity, the reticular fiber network was renamed “reticular conduit” by Gretz et al. (1996), which is a reductionist term unlike the accepted one. Transport of soluble immunogenic factors is unidirectional. A barrier, present in each capsule-borne parallel fiber at the outer wall of the subcapsular sinus, prevents transported antigens from disseminating in the capsule and perinodal tissue (Sainte-Marie and Peng,1986). Conversely, antigens deposited on the capsule spread in its fibers but not in their sinusal extensions. By contrast, antigens deposited on the thymic capsule are transported by capsule-borne fibers into the thymic parenchyma (Sainte-Marie et al.,1986a). Restricted Lymphocyte Wandering for a Successful Quest To trigger a response, a naïve lymphocyte must contact its target, which may or may not be present in a body, and when present usually occurs in only a small territory. In addition, at best one out of 100,000 circulating lymphocytes may be competent to react with any one of a large mosaic of antigens (Ager,1994). This context, at face value, makes it highly unlikely that a naïve cell will encounter its target. Gretz et al. (1997) realized that this apparent hurdle had been overcome by an evolved process of antigen transport from tissues into secondary lymphoid organs. Instead of a lymphocyte circulating throughout the entire body, they argued, it needs only to circulate through these organs and mainly through lymph nodes, where virtually all primary responses would be triggered. This is especially important for rare circulating naïve cells that survive only a few days if inactivated (Stites et al.,1994). Sainte-Marie and Peng (1996) further concluded that a circulating cell enters only a compartment hosting its target, not excluding the possibility that a cell might enter for an activity other than triggering a response. The quest is further facilitated by restricting the delivery of those antigens causing cellular responses only to the domain of the subcapsular sinus and peripheral cortex above the deep cortex center (see below). This prevents the wasteful wandering of concerned T cells into the peripheral cortex beyond the center, where only humoral responses are dealt with. Moreover, if T cells committed to cellular responses entered the peripheral cortex extending beyond the deep cortex center, obstacles would prevent them from reaching the center. First, T cells would have to migrate around the large folliculo-nodules and HEVs in this domain of peripheral cortex and cross its numerous reticulin walls set perpendicularly to the path of their migration towards the center. Second, such a migration of T cells would be against the flow of numerous cells committed to humoral responses moving towards the medulla, thus hindering both migrations. Third, cell congestion would increase with proximity to the deep cortex center as all incoming T cells moved towards an increasingly smaller volume of peripheral cortex, rendering opposing migrations even more difficult. Clearly, the entry of concerned T cells in the peripheral cortex above the deep cortex center, demonstrated below, provides the most expedient and effective pathway for these cells to reach the deep cortex center. The above constraints on T cell quest/migration have not been perceived so far because of the widely held concepts that cells enter a lymph node only at HEVs and that HEVs occur only in the deep cortex. Consequently, a randomly incoming T cell would directly enter the deep cortex. These concepts have endured, even though a few studies have demonstrated them to be incorrect, because it is difficult to perceive how unrealistic they are, especially when one is unaware of their genesis as is common nowadays. Therefore, it is appropriate to review the origin of these concepts and examine alternative views. Origin of Popular Concepts Around 1960, the role of thymic lymphocytes was revealed by neonatal thymectomy (Miller,1961); this breakthrough eventually allowed great advances in knowledge of immune reactions at the cellular and subcellular levels. But it also initially generated a rush of in vivo experiments, implicating lymph nodes whose morphology and functioning were still poorly understood. This resulted in many premature observations that could not be put into proper morphological context, and therefore, the interpretations or functional concepts that followed were questionable. A striking example is an initial report that lymph nodes of thymectomized mice “showed no proper structures and no plasma cells, often being reduced to a small piece of adipose tissue” (Miller,1962). Reports followed stating instead that the deep cortex of the lymph nodes of thymectomized mice is devoid of lymphocytes, hence, it was called a “thymus-dependent zone” (Parrott et al.,1966) but also a “T cell area” or a “paracortex” (Goldschneider and McGregor,1973), however, these terms still conveyed partly unrealistic concepts. The term “thymus-dependent zone” is a misnomer (Fossum,1990) because the deep cortex does form in athymic animals, except that its center is devoid of T cells while its periphery is still populated by B cells (Sainte-Marie et al.,1984). As for the term “T cell area,” it implies that the deep cortex contains only T cells, which is true only of its center. The inadequacy of the term “paracortex” was explained above. Incidentally, in two of our earlier works (Sainte-Marie et al.,1981,1984), we mistakenly labeled the area of peripheral cortex over the deep cortex center (aDCC) as being thymus-dependent, because of the occurrence of numerous T cells there. We should have stated simply that the extrafollicular zone aDCC is the site of entrance and migration of T cells towards the deep cortex center. In the absence of a thymus, the extrafollicular zone aDCC is still entered by B cells as is the continuing deep cortex periphery. Some of the most enduring misconceptions arose from the results of an intravenous transfer of labeled thoracic duct cells to determine whether lymphocytes re-circulate from blood to lymph (Gowans and Knight,1964). At the organismal level, it was reported that labeled cells appeared to occur in equal concentrations in the various lymph nodes of the host, which was taken to indicate random entry. This interpretation was not supported by quantitative data. Regardless, even if the mix of T and B cells of various antigen-specificities had been more or less equally distributed among the various lymph nodes, this would not prove or disprove either a random or an antigen-specific entry. At the level of individual lymph nodes, the earliest observation at 24 hr after transfer, revealed that labeled cells present in a host's lymph node were mostly located in the deep cortex, where some were associated with HEVs, and that only “trivial” numbers occurred in the subcapsular sinus and peripheral cortex. These observations yielded the concepts that circulating lymphocytes enter lymph nodes only at HEVs and that these venules are restricted to the deep cortex. The concept that cells do not enter a lymph node at its subcapsular sinus was immediately questionable because in the course of a study of the thymus (Sainte-Marie and Leblond,1958) lymphocytes were observed in the subcapsular sinus of parathymic lymph nodes. This discrepancy prompted an analysis of the experimental basis for this and other concepts summarized above, which showed them to be unfounded, largely due to a lack of observations made at short time-intervals within the first 24 hr after the transfer (Sainte-Marie,1975). To resolve the mode of lymphocyte entry into lymph nodes, we performed an intravenous transfer of labeled thoracic duct cells and followed them through high-frequency observations starting soon after injection (Sainte-Marie et al.,1975). Labeled cells were first seen in the subcapsular sinus and peripheral cortex, and later they accumulated in the deep cortex. However, concerns about the mode of lymphocyte entry were cast into shadow by a blinding interest for the process of lymphocyte recirculation. The early questionable concepts continued to be uncritically accepted and became entrenched in the literature despite later divergent findings (Sainte-Marie et al.,1975; Elves,1977), which were overlooked but never rebutted. Moreover, the great difficulty of in vivo work on lymph nodes soon fostered a switch of research efforts to in vitro work using suspended lymphoid cells. As a result, understanding of the lymph node stagnated. Hence, Gowans wrote in1996, that it is still not possible to explain the evolution of any immune response in terms of the highly dynamic structure of lymphoid tissue in vivo and of the migratory pathways of co-operating lymphocytes within it, qualifying this situation a “disgraceful gap in medical knowledge.” His review unfortunately did not mention some studies discussing the route of lymphocyte entry into lymph nodes and intranodal migration pathways (Sainte-Marie and Peng,1980,1983,1990), which did contribute to partly fill this gap (Fig. 2). The entrenchment of the concept that cells enter the lymph node only via HEVs diverted interest away from the subcapsular sinus and subsinus layer. Similarly, reported variability of individual HEV cells and the existence of subendothelial spaces and lymphocytes were overlooked because the concept of random cell entry was accepted. These oversights formed a serious obstacle to understanding lymph node functioning since these elements and features are of primordial importance, as will become apparent in the next three subsections. Entry of Circulating Lymphocytes at the Subcapsular Sinus Recent findings revealed that for several days after birth, lymphocytic cells colonizing the lymphocyte-free lymph node compartment enter it virtually only at the subcapsular sinus (Sainte-Marie,2001). Thereafter, to a variable extent, lymphocytes continue to enter the compartment via this route as evidenced by many facts (Sainte-Marie and Peng,1996), some of which are recalled here. First, is the frequent occurrence of lymphocytes in the afferent lymphatic vessels (as seen with a favorable cutting plane of tissue in Fig. 19 in Sainte-Marie and Sin,1970) and the subcapsular sinus of older animals. Entry by this route is also supported by the presence in this sinus of newly formed marrow B cells (Brahim and Osmond,1973) and of lymphocytes soon after transfusion (Sainte-Marie et al.,1975). At times, moreover, lymphocytes are retained on the wall of the subcapsular sinus while their number in the corresponding area of subsinus layer and outer stratum of the extrafollicular zone is reduced (Sainte-Marie and Peng,1990b). Additionally, in some lymph node compartments of Xid mice almost devoid of B and T cells, occasional B cells occur in areas of the subcapsular sinus and accumulate in corresponding areas of the subsinus layer while the almost HEV-devoid underlying peripheral cortex remains quite lymphocyte-free (Sainte-Marie and Peng,1985b). A contribution of drained cells to the lymphocytic population of the adult lymph node is supported as well by unusual events in some compartments under an athymic state. One is the replacement of normal cells of this population by drained altered lymphocytes (Sainte-Marie and Peng,1990c). A second event is the atrophy of the extrafollicular zone in a compartment whose subcapsular sinus is deprived of drained lymphocytes (Sainte-Marie and Peng,1990d). A third event is the formation of a compartment replica on the outer wall of the subcapsular sinus, which represents a compensatory reaction to the degradation of the original compartment (Sainte-Marie and Peng,1987). Briefly, a replica can arise next to the lymphatic opening as an intracapsular lymphocytic islet with a high level of blast-related cells, which recalls the neonatal situation. In the absence of HEVs in the capsule, the colonizing cells of the islet should come from the sinus, as they do in the neonatal state. Dichotomic Recruitment of Circulating Lymphocytes due to Antigen Dichotomy One factor that facilitates the encounter of related elements for triggering an immune response is the topographically dichotomic recruitment of circulating cells arising from antigen dichotomy, that is, the partitioning of antigens into groups causing either a cellular or a humoral response. After birth, the earliest drained antigens induce the lymphocytic colonization and development of a compartment. As the compartment develops, antigen dichotomy causes the functional division of both the subcapsular sinus and nascent peripheral cortex into two domains, with each domain having a different recruitment pattern (Fig. 1C). The result is a non-uniform cortex. The formation of the two cortex domains is induced by sequential delivery and differential spreading in the subcapsular sinus of antigens provoking cellular and humoral responses, as follows. Although popular concepts imply that colonization happens by way of blood lymphocytes entering randomly at HEVs (Harris and Ford,1963), in fact only T cells are initially implicated (Eikelenboom et al.,1979). Moreover, they enter at the subcapsular sinus, mostly as blast-related elements, settling on a small surface of the sinus wall opposite the lymphatic opening (Sainte-Marie,2001). From there, they gather at a short distance beneath the wall and form a tiny spheric deep cortex center within a day (Fig. 4A–D) (Bélisle and Sainte-Marie,1981e). Days after, the sinus domain above the center begins to also handle late-arriving drained antigens causing humoral responses that induce the emergence of peripheral cortex above the center, which is initially constituted of extrafollicular zone alone. As the latter thickens, it flattens the outer half of the center and extends under it, prefiguring the deep cortex periphery. Hence, the tardy arrival of B cells prevents mixing of T and B cells in the emerging center, whereas the subsequent formation of the deep cortex periphery continues to preserve the integrity of the center's milieu: B cells do not cross the center to reach the underlying medullary cords, traveling instead via the periphery as confirmed by immunofluorescence (van Roojen,1987). Meanwhile, HEV-precursors of the nascent peripheral cortex and deep cortex periphery develop HEV traits and a cell-recruiting activity, thus enhancing colonization. Progressively, the sinus domain above the center becomes somehow saturated with antigens causing humoral responses. As such antigens continue to enter the sinus, they spread beyond the center into a nascent second sinus domain, which will thus deals only with antigens causing humoral responses. Therefore, beneath this second domain one finds only peripheral cortex underlaid by plasmocytic medullary cords. Indeed, antigens retained by a sinus site enter the portion of peripheral cortex directly beneath, stimulating the development of HEV-precursors and inducing them to recruit relevant circulating cells. Thus, antigen dichotomy establishes a topographically dichotomic pattern of antigen presence and influence in the sinus as well as in the nascent peripheral cortex, which results in a corresponding dichotomic pattern of recruitment of circulating cells. A few factors account for this scenario. At birth, antigens causing humoral responses are controlled for some time by maternal antibodies, probably then halting their drainage towards lymph nodes. This explains why APC-carried antigens causing cellular responses and relevant T cells are the first to be drained from tissues and to enter a lymph node compartment. There, APCs contact fibers crossing the subcapsular sinus and settle, with attracted T cells, on the sinus wall opposite to the lymphatic opening. This leads to the emergence beneath of the deep cortex center with a peculiar milieu propitious to the development of effector-cells of cellular responses. By contrast, antigens causing humoral responses are minute, soluble, and possibly more abundant, so they eventually flow farther than APCs, that is, beyond the center, where they promote the expansion of only peripheral cortex beneath the sinus. In axenic animals there is very little such expansion (Bélisle et al.,1982) because the sinus above the deep cortex can cope with an unusually small amount of antigens of both categories. Paradoxically at first glance, such expansion is also very limited in some neighboring compartments of normal animals, and particularly, in their mesenteric lymph nodes. The reason for this peculiarity is a marked development of humoral responses directly in the gut, as revealed by abundant plasmocytes there, which reduces the arrival of antigens causing humoral responses to draining mesenteric compartments. Consequently, the peripheral cortex does not expand much beyond the deep cortices, which are therefore close to one another (Sainte-Marie et al.,1982). This situation is a vestige of the phylogenesis of the immune system: the formation of gut plasmocytes precedes the development of lymph nodes (Good and Finstad,1967). By contrast, the peripheral cortex expands considerably in some compartments at other sites and this is matched underneath by formation of plasmocytic medullary cords (Fig. 2A, C). It thus appears that the extent to which peripheral cortex and plasmocytic cords are developed beyond the deep cortex is proportional to the supply of antigens causing humoral responses to individual compartments. The lymph node is to be viewed as a complement, not a substitute, to ancestral immune tools. A multitude of potential encounter sites of related elements improves the chance of triggering a response. For greatest efficiency of the system, a naïve cell should be activated wherever it encounters its target and essential mediators within a basic territory of the immune system. Such a territory comprises: (1) a portion of body tissue that is drained by (2) a collecting lymphatic vessel, termed an afferent lymphatic at its junction with (3) a compartment of a local lymph node, over which it spreads its content. Within a territory, a continuum of events unfolds that is initiated by the intrusion of an antigen in the drained tissue, leading to the triggering of a response in the tissue itself and/or in the compartment where the lymphatic carries drained immunogenic elements. That activation can also occur in the drained tissue is evidenced postnatally by a late entry of blast-related cells in the subcapsular sinus and their presence in the sinus beyond the deep cortex, with these cells yielding plasmocytes in underlying medullary cords (Sainte-Marie,2001). In the adult, moreover, humoral responses appear to unfold to a greater degree in some tissues rather than in draining compartments, as is the case of the gut where plasmocytes are known to be abundant. Antigen-Specific Recruitment of Circulating Lymphocytes A process that would further favor the encounter of a naïve cell and its target is the antigen-specific recruitment of circulating cells, which provides a rational explanation for a suite of intriguing observations on HEVs in particular (reviewed in Sainte-Marie and Peng,1996). As explained above, antigen-specific recruitment is a corollary of the beneficial directed transport of antigens from a tissue territory into a single lymph node compartment or even limited sites of that compartment. Moreover, specificity is strongly supported, for instance, by the lymphocytic colonization of neonatal compartments (Sainte-Marie,2001). In a random process, colonization by B and T cells would be expected to occur concomitantly, as both cells would occur irrespective of the presence or absence of proper targets in a compartment. Instead, T cells account for nearly all colonization activity during the first postnatal week (Eikelenboom et al.,1979). Moreover, blast-related forms represent 84% of colonizing cells at 16-20 hr after birth, declining to 21% at day 21, mostly due to an accumulation of their maturing progeny-lymphocytes prior to emigration from the lymph node. Such percentages are not compatible with a random entry of circulating cells of which 0.001% at most are competent to respond to any given antigen (Goodman,1994). In other respects, a specific recruitment of lymph-carried cells is consistent with the observation that a drained soluble antigen adheres to the wall of the subcapsular sinus (Sainte-Marie and Peng,1985c). An APC does so as well, and may also become associated with the subsinus layer (Hendricks et al.,1980). There, an APC could release exosomes by long cytoplasmic processes (Murk et al.,2002; Théry et al.,2002) to expose its antigens at the surface of the sinus wall. A relevant lymphocyte, probing this wall, may thus contact its target and be selectively recruited there. As to the specific recruitment of blood-carried cells, it is supported by the non-uniform (i.e., focalized and selective) adherence pattern of suspended live nodal lymphocytes to HEVs in sections of frozen lymph nodes (Sainte-Marie and Peng,1995) as well as by other observations on HEVs and lymph nodes under various states, which can be explained only by specific recruitment (Sainte-Marie and Peng,1996). In addition, the marked variations in the features of individual and even adjacent HEV cells (Fig. 3A) as well as in the numbers of lymphocytes hosted in their subendothelial spaces reflect the individualism of their activity (Fig. 3C). By itself, the existence of subendothelial lymphocytes betrays a specific recruitment: the presence of lymphocytes in a subendothelial space makes little sense unless they are related cells, recruited by a concerned HEV cell to participate in a specific interaction. Actually, these lymphocytes are in contact with abluminal endothelial processes and, moreover, they interact in groups (Soderström,1967; Bailey and Weiss,1975), which implies a common competence. In turn, the interaction of related cells in a given space requires that they group there, that is, that they be recruited specifically. Indeed, the number of circulating cells that can at a given time enter one of a multitude of exiguous subendothelial spaces is quite limited, so that selection is the only process by which rare cells competent to carry out a particular interaction could become associated in a same space. That HEV cells are stimulated individually and react specifically is also indicated by an unusual excess of lymphocytes in the subendothelial space of an isolated HEV cell(s) of a restricted HEV area(s). Such excessive loading betrays recruitment of related lymphocytes by HEV cells influenced by the same material, the recruited cells then unusually being retained in their subendothelial spaces due to an anomaly that prevents them from entering the compartment's parenchyma. This explanation is consistent with the fact that lymphocyte retention is by far most often observed under the state of immunodeficiency (Sainte-Marie et al., 1986b); moreover, the phenomenon mimics excessive lymphocyte retention in the subsinus layer of immunodeficient Xid mice (Sainte-Marie and Peng,1985b). Finally, it is noteworthy that subendothelial lymphocytes have never been included in models of random cell recruitment (see Ulrich and Thorsten,2003; and Fig. 3B), with which they are incompatible. It was suggested that drained cytokines permeating into an HEV lumen can “either in solution or immobilized on the luminal endothelial surface act directly on lymphocytes” of the blood (Gretz et al.,1996), but the authors omitted to consider a comparable role for drained soluble antigenic materials despite having reported that they too permeate the HEV endothelium. Although one may doubt that antigens usually do so, they nonetheless can reach an HEV cell via its basement membrane as seen above. There, the antigen could be immobilized on the luminal face of the membrane and/or the abluminal face of the HEV cell, becoming accessible to blood cells probing its subendothelial space (Fig. 3C). An antigenic material causing a cellular response could similarly infiltrate a subendothelial space through the release of exosomes by cytoplasmic processes of an APC migrating in a perivascular channel. A contact between a probing blood cell and its target would be the decisive signal in the cascade of steps leading to cell recruitment (Sainte-Marie and Peng,1996). DEVELOPMENT OF ENTRANCES FOR CIRCULATING LYMPHOCYTES Additional aspects of the subcapsular sinus and HEVs will be examined, especially their interrelated and comparable mode of development or expansion. The almost concomitant formation of a new sinus area and of fiber-linked HEV areas beneath, both influenced by the same antigenic material, enhances the recruitment of competent cells and the chances of triggering of responses. Moreover, consideration of the mode of expansion of the subcapsular sinus and HEVs can provide clues on their functioning. Subcapsular Sinus Despite rather uniform appearance, the subcapsular sinus is functionally complex. Because it is shallower above the bulging folliculo-nodules, lymph flows more freely over the extrafollicular zone where the cell-lined fibers crossing the sinus are concentrated. This favors the screening of the lymph content there, so that selected drained elements by far mostly enter the extrafollicular zone (Sainte-Marie and Peng,1985c,1986a). Besides, antigen distribution and recruitment of drained cells differ in the two sinus domains, as discussed earlier. Furthermore, recruitment of particular cells can involve only part of a domain: in some compartments, for instance, the sinus contains numerous mast cells entering the extrafollicular zone selectively next to cortical gaps (Sainte-Marie and Peng,1990a). How does the sinus develop or expand after antigen dichotomy has neonatally delimited it into two functional domains? After the existing areas of the sinus domain above the deep cortex become almost or completely saturated with antigens, the entry in the sinus of additional antigenic material causing humoral responses promotes its expansion. Sinus expansion results from the addition of new sinus area at the margin of a compartment to handle2 excesses of such material and is accompanied by the emergence of peripheral cortex and plasmocytic medullary cords beneath. This mode of expansion strongly suggests that the added area selectively recruits drained cells related to the antigens that initiated its emergence, an interpretation supported by a neonatal event. Days after birth, there is a high level of blast-related cells in the subcapsular sinus and the nascent peripheral cortex beneath. Thereafter, the lymphocyte level increases, greatly predominating at about day 13. Nonetheless, blastic forms in the sinus of some compartments are still abundant beyond the deep cortex, where they can yield a trail stretching from a limited sinus area to the plasmocytic medullary cords directly beneath (Sainte-Marie,2001). This betrays a focalized and specific recruitment of these cells, which are probably activated in the drained tissue by the same antigens that contributed to induce the formation of the sinus area where they are selectively recruited. Expansion of the subcapsular sinus further indicates that the capacity of a sinus area to handle lymph-carried elements is limited. This capacity is at least partly constrained by the correlated diameter and length of local fibers crossing the sinus. In the neonate, the foremost antigens causing humoral responses are transported beneath the subcapsular sinus by short extensions of the fine sinus fibers. These extensions are connected to the basement membranes of underlying HEV-precursors. The antigens carried by these extensions stimulate the development of HEV features in the linked areas of the HEV-precursors and the recruitment of relevant blood cells which, in conjunction with drained cells recruited from the sinus, contribute to the emergence of a thin nascent peripheral cortex. With further antigenic stimuli, the peripheral cortex thickens while fibers and HEVs elongate, matching the growing thickness of the peripheral cortex, as these elements evolve in unison (Bélisle and Sainte-Marie,1981e). Fiber diameter simultaneously increases, probably due to the increasing volume of transported drained factors. Fiber growth and peripheral cortex thickening continue until antigenic stimuli peak or some maximal handling capacity of a sinus area is reached. These events apply as well to the sinus fibers of the peripheral cortex above the deep cortex center. There, moreover, expansion of the sinus wall favors more crossings of APCs at spaces between the parallel fibers, thus allowing a matching expansion of the deep cortex center. Clearly, the neglected subcapsular sinus plays a primordial role in the organization and functioning of the compartment, by screening the content of its afferent lymph and offering passage to focally selected elements into the compartment's parenchyma. Moreover, the rarely observed and almost ignored lymphatic opening is the central point from which this architecture is spatially established; this knowledge is essential for understanding the lymph node compartment. High Endothelial Venules An HEV arises from the transformation of squamous endothelial cells of a regular venule under the influence of an antigenic stimulation. Observations made under various conditions indicate that development of HEV cells results from focal and individual cell stimulation, which is manifest both neonatally and after wherever a new HEV area emerges (Sainte-Marie and Peng,1996). Macrophages were proposed to be the main stimulators of HEV formation (Hendricks et al.,1980). However, HEV formation is hindered in Xid animals with abundant macrophages but few lymphocytes (Sainte-Marie and Peng,1985b) and more recent observations point to an essential stimulatory role for drained lymphocytes and, in particular, antigen-presenting T cells (Pichler and Wisscoray,1994). Hence, it appears that development of HEV cells necessitates stimulation involving antigenic material, drained lymphocytes, and mediators: where they meet in a vessel, a squamous cell becomes an HEV cell (Sainte-Marie and Peng,1996). Neonatally, the foremost antigens causing humoral responses induce initial HEV cell development in a small area of regular venules slightly below the subcapsular sinus (Sainte-Marie and Sin,1970), and possibly on one side only of a vessel (Bailey and Weiss,1975). Moreover, the extent of individual cell transformation differs, reflecting uneven individual stimulation. Subsequent antigens can further transform the earliest HEV cells to the extent that they can still be influenced; they also transform squamous cells in a contiguous area of the same vessels. Thus, HEV elongation mimics subcapsular sinus expansion. Similarly, the HEV scenario also implies limits to the number of HEV cells that can be stimulated at a given time by a given material and to the extent to which a given cell can be stimulated, with these limits partly depending on the amount of material handled by the sinus area above. The scenario of HEV elongation is repeated until the associated sinus area has attained a maximal handling capacity, thereby ending elongation of underlying HEVs. However, the HEV does not remain in a steady state after. Since the influence of a stimulation is transient (Hendricks et al.,1980), an HEV cell regresses when stimulation fades and redevelops if re-stimulated. In fact, such dynamics concern all cell populations of a compartment that are involved in ongoing responses. Because these populations have a defensive role, they would atrophy if stimuli waned and were not followed by new ones. Later in life, new antigenic stimuli accentuate the transformation of lesser developed HEV cells (Henry and Beverly,1976), which are either variably understimulated or regressed cells. The stimulus can also form new HEV cells in areas of transition from an HEV to a regular vessel, mainly at the corticomedullary junction (Burwell,1962), where once again the neonatal features betraying individual stimulation of squamous cells can be observed. The fact that a new stimulation adds areas of HEV cells at this junction should increase the recruitment of competent blood cells. This would likely be the case if a new HEV area selectively recruits cells related to the antigens having caused its emergence, as proposed for a sinus area added by the same stimulus. Indeed, the HEV network starts as fine branches fusing into progressively larger ones that form an HEV trunk upon nearing the junction (Sainte-Marie and Sin,1970). Thus, if a competent blood cell has not been recruited prior to reaching the trunk, it may be recruited there in an area of new HEV cells; otherwise it would exit the compartment through the continuing medullary venule in an inactivated state—a waste of time and energy. Hence, the HEV layout matches that of a new sinus area under the influence of the same stimulation. Their combined features and advantages should ensure the triggering of a response. Functional Link Between a Subcapsular Sinus Area and Associate HEV Areas As we have seen above, the handling of antigens causing cellular responses by the subcapsular sinus domain around the lymphatic opening determines the emergence of the underlying rounded deep cortex center where cells of these responses form. Hence, the nature of the antigens handled in a sinus area determines the nature of the responding effector cells formed beneath it. By analogy, since a newly formed sinus area and underlying HEV area(s) are stimulated by the same antigens, it is reasonable to propose that they recruit the same competent cells to yield the same responses. This would represent a further step in the optimization of the immune process, consisting in the transport of a given tissue antigen into a restricted site(s) of the same sinus and peripheral cortex of the draining compartment. Such a layout would increase the likelihood of the encounter of related elements by eliminating their wandering in target-devoid sites of the peripheral cortex. A common stimulation of a sinus and HEV area(s) occurs because of topographical and functional links between them, whose modality will now be examined. When a humoral response is in cause, fibers transport the antigen and ensure a functional link between a new sinus area and the new underlying area(s) of HEV(s). When a cellular response involving an APC-carried antigen is in cause, two functional links are plausible. One is the release of an antigen from exosomes extruded by an APC associated with an area of sinus wall or subsinus layer. The antigen could then be transported by local fibers as above. A second possibility is that the APC migrates in the perivascular channel of an underlying HEV or HEV-precursor. Upon reaching one or more endothelial cells open to stimulation, the APC influences it or them by way of antigenic material released from exosomes excreted by cytoplasmic processes extending through the HEV's basement membrane. Since the influence of an antigen is transient, a functional link produced by a given antigen evolves following scenarios orchestrated by the importance and duration of its presence in the afferent lymph. For instance, when an antigen handled by a given sinus area fades, that area eventually can handle another material proportionally to its regained capability, with the cells of functionally linked HEV areas behaving likewise (Sainte-Marie and Peng,1996). Such a dynamic state causes variability in the antigen-specificity of recruitment at a given sinus and HEV sites, entailing a variably important distribution of sites recruiting the same cells. This makes it difficult to experimentally prove cell recruitment specificity, as does the varying degree of stimulation of a given HEV cell by diverse antigenic materials. Although not strictly comparable, the transient neonatal recruitment of neutrophils by HEVs illustrates a similar variability in recruitment (Sainte-Marie and Guay,1995). After birth, neutrophils are recruited in the initial HEV area(s) of a regular venule. Recruitment gradually declines there and is shifted towards the venule's next developing HEV area(s); this is repeated for about a month until neutrophil recruitment ceases altogether. This betrays a temporal change in the nature of cells recruited at a given HEV site. To summarize, the proposed scenario represents a more efficient mode for lymphocyte searching through the transport of antigens from a tissue territory into only one compartment of the draining lymph node and, in the case of antigens causing cellular responses, into only one domain of its subcapsular sinus and peripheral cortex. One can reasonably infer that evolution also led to a further improvement consisting in the handling of a given antigenic material by a restricted site(s) of this sinus and peripheral cortex, thus limiting the recruitment of relevant cells to these linked sites. Experimental proof will be difficult. Meanwhile, considering that complex and efficient means to ensure the encounter of related elements have evolved in the brain, one may wonder why comparable refinements would not have occurred in the lymph node compartment as well, especially considering that its reticular fiber network seems well suited for that task and that prompt immune responses can be critical to individual survival. ANTECHAMBERS TO THE LYMPH NODE PARENCHYMA At both entrances by which circulating cells can access the lymph node compartment, evolution may have further favored the formation of a confining place for the rapid encounter of an antigen and recruited competent cells, thus optimizing encounter. Such antechambers to the parenchyma of a compartment exist: the subsinus layer and subendothelial spaces. Subsinus Layer Because of the common belief that lymphocytes do not enter the lymph node via the subcapsular sinus, this elusive component receives very little attention. Overall, observations indicate that it complements the screening of drained cells by the sinus, which can explain why some may be retained in this layer, delaying or preventing their passage into the extrafollicular zone. Thus, in some lymph node compartments of Xid mice, B cells accumulate in the subsinus layer while the cortex remains lymphocyte-free (Sainte-Marie and Peng,1985b). This means that an element that would normally allow their entry into the cortex is lacking or deficient. In fact, these B cells fail to undergo blastogenesis and do not switch to IgG production (Mond et al.,1980). In other respects, since drained immunogenic elements can penetrate the subsinus layer, activation can occur there. Subendothelial Spaces The subendothelial spaces also are largely ignored, as it is assumed that the interaction of circulating lymphocytes with an HEV cell happens exclusively at its luminal face. However, this interaction also proceeds at its abluminal face that is exposed to subendothelial lymphocytes, to factors transported by fibers and conceivably to antigens presented by APCs. In a subendothelial space, endothelial processes contact probing blood lymphocytes. There, these cells form groups that may include drained lymphocytes having migrated into a first perivascular channel (Sainte-Marie and Peng,1996). The groups reflect cell interaction (Soderström,1967; Bailey and Weiss,1975), which is likely a prerequisite, allowing passage into the parenchyma of blood cells entering an endothelial space rather than directly entering the first perivascular channel. Consistent with these observations, suspended live nodal lymphocytes bind selectively to individual HEV cells or restricted HEV sites in sections of frozen lymph nodes (Sainte-Marie and Peng,1995). They bind to the plasma membrane at both faces of HEV cells as well as to subendothelial spaces, basement membranes and first perivascular channels, where they likely become attached to lymphocytes or antigenic materials. Moreover, lymphocytes that bind to a same site are interconnected by cytoplasmic bridges. Note that an endothelial space can also serve as a waiting room for asynchronously recruited partner-cells (see below). Subendothelial lymphocytes predominate in HEV trunks, supporting the idea that a subendothelial space is a site of essential cell interaction. HEV cells in trunk areas have often been stimulated by new antigenic material, which suggests that blood cells entering their subendothelial spaces are often naïve and in need of an essential interaction. This idea is also supported by the fact that unusually large accumulations of subendothelial lymphocytes occur primarily in HEV trunks where they exhibit a peculiar distribution (Sainte-Marie and Peng,1996). Indeed, such accumulations can occur under an isolated HEV cell, a few to several cells in a small HEV area, and/or many cells in a larger area and possibly on one side only of an HEV. This distribution pattern is similar to that of the foremost HEV cells arising in a neonatal regular vessel under the influence of the earliest antigenic stimuli (see above). By analogy, therefore, the similar distribution pattern of accumulated subendothelial cells betrays a stimulation of the implicated HEV cells by the same material and their recruitment of the same lymphocytes. Like lymphocyte retention in the subsinus layer of Xid mice, a subendothelial accumulation must imply that an element required for an essential interaction is lacking or inefficient, so that recruited cells are withheld. The B nature of the retained cells is indicated by the predominance of the phenomenon next to medullary cords, where plasmocytes form. The missing element is probably thymic helper-cells and/or factors, especially since the phenomenon occurs mainly in athymic animals (Sainte-Marie and Peng,1996). In other respects, as factors essential to activation can reach a concerned subendothelial space and the narrow adjacent area of the first perivascular channel where lymphocytes also interact (Schoefl,1972), activation can occur in both places. ROLE OF THE LYMPH NODE IN TRIGGERING PRIMARY IMMUNE RESPONSES A “Paracortical” Hypothesis A proposal by Gretz et al. (1996) dealt with features of the “paracortex” that could facilitate the triggering of a primary cellular response within it. The proposal is summarized using their terminology and conception of lymph node organization as follows: APCs, which have processed antigens in a tissue and are lymph-carried into the subcapsular sinus of a local lymph node, migrate from the sinus through the “intrafollicular” zone to the “paracortex” where they cling to the fiber network. Circulating lymphocytes come from the direction opposite to arriving APCs, by transmigrating into the “paracortex” from HEVs; then T cells committed to cellular responses progress along the fiber network studded with APCs. Activation would occur in this meshwork of intertwining fibers of the “paracortex” when a proper APC and naïve T cell meet. However, the fiber network spreading between the subcapsular sinus and the HEVs in Gretz et al.'s supporting Fig. 1 lies next to a folliculo-nodule (“follicle” in their terminology), and therefore, represents a portion of the extrafollicular zone of a compartment's peripheral cortex, not its deep cortex (“paracortex” in their terminology). Hence, if both APCs and transmigrating naïve T cells did indeed move into the deep cortex, which in fact lies below the extrafollicular zone, then both would move in the same direction, not in opposite directions. In a later article, Gretz et al. (1997) repositioned HEVs inside presumptive “paracortical cords” forming the “paracortex” (see above). In this case, a cell leaving an HEV would settle within the “paracortex,” thus not needing to move towards it from any particular direction. Gretz et al. (1996) remarked that the dense fiber network of the structure presumed to be “paracortex” in their Fig. 1 could facilitate contact between a clinging APC and a relevant wandering naïve T cell. We argued earlier that this structure is in fact a portion of the fiber-rich extrafollicular zone. The deep cortex center itself is actually poor in fibers (Bélisle and Sainte-Marie,1981d; Okada et al.,2002). It is therefore unlikely that these scarce fibers play an important role in mediating contact between rare wandering competent T cells and numerous APCs. Moreover, a quest in the deep cortex center would be hindered by congestion arising from the many proliferating or maturing lymphocytes and by abundant randomly-incoming circulating cells. This is particularly true if one considers the minuteness of a naïve cell with respect to the large volume of a deep cortex center to be probed. Moreover, one wonders what mechanism could guide a minute rare cell through this large volume and permit it to encounter a relevant APC that may be present, but most likely is not. Almost as a closing remark, Gretz et al. (1996) recognized that the fiber network is looser in areas of cell proliferation such as the deep cortex “units,” adding that activation in these areas has already progressed enough that rapid access to information transported by fibers is no longer required. Since the fiber-poor center represents a substantial portion of the deep cortex, the question of where the site of activation is localized in Gretz et al.'s (1996) scheme remains wide open. In another respect, Gretz et al. (1996) initially considered that T cells enter a lymph node only at HEVs, in agreement with a popular view, but soon after Gretz et al. (1997) stated that the afferent lymph carries lymphocytes from tissues and upstream lymph nodes. Why then did they not examine the possibility of a triggering of responses by lymph-carried T cells? Probably because one of the authors had postulated that virtually only memory cells exercise tissue surveillance (Adams and Shaw,1994). If so, naïve cells would be absent from the afferent lymph so that drained cells need not be considered in a proposal dealing with primary responses. However, the concept of tissue surveillance based on memory cells alone is questionable, especially since it is assumed to ensure an optimal immune function. Quite to the contrary, such a surveillance mode would compromise the survival of neonates in particular, for memory cells do not exist when an antigen intrudes for the first time. Besides, lymphocytes occur in the gut lamina propria and in the subcapsular sinus of draining lymph nodes soon after birth, indicating that naïve cells assume surveillance during this crucial period (Sainte-Marie,2001). In addition, Gretz et al. (1996) recognized that the afferent lymph of central lymph nodes can contain some efferent lymph from peripheral lymph nodes. On the basis of random entry of cells into lymph nodes, a concept accepted by Gretz et al. (1996), this afferent lymph would contain naïve T cells having already entered and left a peripheral lymph node because it was devoid of their target. Hence, the participation of lymph-carried T cells should have been considered in the elaboration of a proposal on T cell activation. However, such participation implies lymphocyte intranodal migration in the same direction as lymph-carried APCs, not the opposite direction as proposed by Gretz et al. (1996). A Chemokine Hypothesis Cyster (1999) ascribed a central role to chemokines in the intranodal encounter of circulating cells with their target. It was proposed that T and B cells leaving HEVs at random would be oriented towards a T- or a B-area, respectively, by corresponding attracting chemokines produced there. It is hardly possible to conceive how such an orientation process could work in vivo. For instance, to attract into the deep cortex center T cells committed to cellular responses—which are hypothetically recruited by HEVs situated anywhere in the peripheral cortex—chemokines produced in the center would need to diffuse throughout the peripheral cortex. Since the center underlies only a portion of the peripheral cortex, chemokines would have to first diffuse outwardly into this portion and then laterally while forming a gradient potent enough to attract all incoming T cells from as far as a possibly distant compartment margin. These numerous cells would then all have to migrate towards the center, something also inconceivable as explained above. Thereafter, each cell would have to wander throughout the center in search of a relevant APC, a quest hindered by cellular congestion. In addition, an encounter could happen only if a naïve cell had randomly entered not only a rare relevant lymph node hosting its target, but also the node's relevant compartment. This seems unlikely, for the blood vascular network of a compartment represents a small fraction of the network of a whole lymph node which is usually multi-compartmented, and an even quite smaller fraction of the body's whole vascular network. Hence, such a scenario would be grossly inefficient for rare competent short-lived naïve T cells, which with increasing age become even rarer due to thymic involution. In other respects, it must be considered how a naïve T cell, having randomly entered a center devoid of its target, could succeed in exiting it rather than endlessly and vainly wandering in it due to the local elevated concentration and steep gradient of attracting chemokines. If somehow the cell succeeded in exiting, it would have to repeat, over and over again, the same hazardous scenario in any of a great number of compartments, few (if any) of which contain its target. The cell would likely die before such an encounter. Cyster (1999) stressed that a lymph node contains an immensely complex chemokine landscape, its multiple gradients attracting a cell in opposite directions. If so, a resulting erratic migration would jeopardize the benefit provided by a compartment architecture well organized for an efficient directed migration. He added that it will be exciting to see how many “turns” a cell might make as it travels through such multiple gradients. He did not indicate where a cell would turn. In the case of a T cell committed to a cellular response, turns would not be useful outside of a deep cortex center where there is at least a minute probability that its target might be present. In addition, the randomness of such turns would decrease a cell's chance of encountering its target. Turns would furthermore represent an additional waste of time, which would be a detrimental behavior, especially for short-lived cells. All of this does not fit with a condition identified by Cyster (1999), that, for a mature dendritic cell to function in an immunogenic manner, it must interact rapidly with competent T cells. In fact, Cyster's proposal could work only if the compartment morphology was as simple as presented in his Fig. 3, where a “T area” is reduced to a cell islet close to the subcapsular sinus with HEVs conveniently situated next to it. The reality, as we have seen, is far different, yet this does not preclude a role for chemokines. However, this role must be consistent with the orderly design of the compartment, which imposes definite patterns for factor transport and cell migration. This context suggests that a chemokine has a focal or local influence, as illustrated by the recruitment of T cells committed to cellular responses occurring only above the deep cortex center. The proliferation of unrealistic proposals based on a misunderstanding of lymph node morphology and oversight of relevant literature is self-fostering and will continue. This trend is illustrated by a recent proposal by Crivaletto et al. (2004). These authors also emphasized that a proper understanding of lymph node morphology is paramount for proposing relevant hypotheses and theoretical models for immune responses based on in vitro experiments in cell and molecular immunology. However, their conception of lymph node morphology is inadequate in many ways. For instance, in their Fig. 1, the afferent lymphatics are incorrectly shown to be as numerous as folliculo-nodules whereas the extrafollicular zone between these structures is delimited by an “intermediate sinus”—a component that simply does not exist. A Morphologically Realistic Proposal Previous proposals on the triggering of primary immune responses attempted to explain how morphological features of components of a developed lymph node, mainly its HEVs and “paracortex,” allow triggering of a primary response, concluding that “forms dictate function” (Gretz et al.,1997). In reality, antigens (i.e., function) dictate compartment morphology as we have seen above. From the outset, previous proposals are fundamentally unrealistic because the components (i.e., forms) invoked do not exist prior to the occurrence of the first postnatal antigenic stimuli and immune responses: these components are products of responses, not prerequisites. Therefore, to be plausible, a proposal on the triggering of primary responses must above all be consistent with the level of compartment organization at birth, a requirement not considered so far. At birth, only lymphatic sinuses exist (Fig. 4A) and among them the subcapsular sinus is the first to be exposed to the foremost antigens of the afferent lymph. Sinusal cells handling these incoming antigens contribute to the foremost immunological reactions of the lymph node compartment. This neonatal potential for triggering primary responses in the absence of developed nodal components (i.e., independently of form) persists throughout life. Since each primary response is specific, its triggering at all ages requires particular lymphocytes to be recruited by sinusal or HEV cells influenced by a new antigen. These recruiting cells are newly formed and/or existing cells having regained the ability to be influenced by new antigenic material. Hence, existing sinusal and HEV cells of a developed compartment can be implicated in the triggering of a primary response because of a newly acquired recruiting property that is elicited by a new antigen. To formulate a proposal on events associated with the triggering of a response in a developed lymph node compartment, one must consider that the latter is basically constituted of (1) a receptacle for drained immunogenic elements—the subcapsular sinus, (2) distinct entrances for lymph-borne and blood-carried lymphocytes—the subcapsular sinus and HEVs, respectively, (3) an antechamber to the parenchyma for each type of entrance—the subsinus layer and subendothelial spaces, (4) a migration pathway—the extrafollicular zone and the continuing deep cortex periphery—which is connected to the antechambers and guides a recruited competent lymphocyte to the structure where it will yield its progeny, and (5) distinct structures propitious to the production of effector-cells of either cellular or humoral responses—the deep cortex center and medullary cords, respectively. Where does activation occur? Observations indicate that a recruited T or B cell is activated before it enters the structure where it yields its progeny. Neonatally, a large fraction of the cells entering the nascent deep cortex center or medullary cords are blast-related, that is, already activated cells. Thereafter, plasmablasts still congregate in the initial portion of cords, followed deeper by progressively less immature plasmocytes (Sainte-Marie,1964). These facts point to an activation occurring somewhere upstream of the center and cords. The immediate upstream component is the migration pathway. For sake of efficiency, this pathway for cell traffic should not be a place where a rare short-lived naïve cell searches for a related, yet by far most often absent, element; it should instead serve only to guide an already activated cell directly to either the center or a cord. Therefore, activation should ideally happen even more upstream, in components providing a confined milieu favoring the rapid encounter of related elements and where a naïve cell can be exposed to all factors essential to activation. Such components are the subsinus layer and the subendothelial spaces of relevant HEV cells or the adjacent first perivascular channel. This allows the subsequent direct migration of an activated cell towards either the center or a cord, the cell being oriented by fibers and/or perivascular channels of the migration pathway studded with guiding determinants. On its way, the activated cell undergoes blastogenesis, thus explaining the presence of possibly numerous blast-related cells in the extrafollicular zone and deep cortex periphery (Sainte-Marie,1966). If unfinished, blastogenesis is completed in the center or a cord. Cellular Response After having processed an antigen in a tissue and having been carried by the lymph into the subcapsular sinus of the draining compartment, an APC probes the sinus over the extrafollicular zone present above the deep cortex center until it reaches a receptive site where it attaches to or enters the subsinus layer. The APC's attracting chemokines induce drained T cells committed to cellular responses to probe the sinus there. Chemokines are also transported by local fibers towards the center. Somewhere along their course, these fibers join the basement membrane of HEVs overlying the center, so that the transported chemokines attract similar blood T cells to probe the endothelium of these same HEVs, as can do chemokines released by APCs migrating in their perivascular channels. Thus, the recruitment of T cells committed to cellular responses is restricted to the domain of subcapsular sinus and peripheral cortex present above the center, so as to reach the center with minimal delay (Fig. 1C). A lymph-carried naïve T cell encountering a relevant APC associated with the sinus wall or the subsinus layer can be activated in either place, where they are exposed to all essential factors (Fig. 5A, left). This ensures that a mature APC and a competent T cell will interact quickly. The activated T cell then migrates to the center where it engenders its progeny. The APC can likewise travel to the center (Fig. 5A, left), so that, from a given site of the subsinus layer, both cells are guided by the same fibers to the same place in the center where their interaction may continue. An observation suggests that an APC may cling to an element of the loose fiber network of a deep cortex center, from where it could influence related T cells close by. Indeed, in a large center with a reduced population of T cells, the latter concentrate into broadly spaced ribbons (Bélisle and Sainte-Marie,1981d). Conceivably, factors from APCs clinging to distant fibers attract T cells in their proximity explaining the fewer T cells between ribbons. As for competent naïve blood T lymphocytes, they are recruited by relevant HEV cells, that is, cells that are presenting the implicated antigenic material (Fig. 5A, right). Activation can occur in the subendothelial spaces or in the adjacent first perivascular channel since both places may contain all essential factors. The activated T cell (and perhaps a related APC) then migrates to the center. This leads to their aggregation at the same place of the center where related drained cells also gather, all being guided by the same fibers or channels. Finally, it is conceivable that in the center an APC might activate newly formed memory cells, rapidly reinforcing a response by sparing them the quest for their target. Humoral Response In the case of a thymo-independent response, a drained naïve B cell can be recruited at a site of the subcapsular sinus exposing its target. Activation may occur there or next to it in the subsinus layer (Fig. 5B, left). The activated cell is then guided by local fibers to an underlying medullary cord, either indirectly via the deep cortex periphery for a cell recruited above the deep cortex center or directly for a cell recruited beyond it. The activation of a B cell recruited by a relevant HEV cell can occur in the subendothelial space or in the adjacent perivascular channel; the activated B cell then migrates to the underlying medullary cord (Fig. 5B, right). Reaching a cord, the activated cell completes blastogenesis if it is unfinished. Thus, resulting plasmoblasts congregate at the entrances of cords. In the case of a thymo-dependent response, triggering implicates a T helper-cell and a naïve B cell. In the subcapsular sinus, both cells can be recruited over the extrafollicular zone at a site presenting their target and there, they can enter the subsinus layer. When all essential factors are encountered in the sinus or the layer, activation unfolds. The activated B cell then travels to an underlying medullary cord, guided by local parallel fibers (Fig. 5C). If instead either only the B or the T helper-cell is recruited on the sinus wall where the target is encountered, the recruited cell may wait there or in the subsinus layer for the arrival of a partner. Alternatively, it may leave this site in search of a partner, which could have been recruited by an underlying relevant HEV cell (Fig. 5D, E). To do so, the recruited cell moves along local fibers connecting to a perivascular channel until it reaches a relevant HEV cell. If, instead, a partner is present in the subendothelial space or the first perivascular channel, activation can happen when all essential factors are present. The activated B cell then proceeds towards an underlying medullary cord. If a partner is absent where a relevant HEV cell is first encountered, the recruited cell may still adhere to its target, bind to the abluminal face of the relevant HEV cell or a facing site of its basement membrane, and wait for a partner to arrive. Alternatively, in the absence of a partner there, the recruited B cell may eventually move to a relevant HEV cell located deeper, thus possibly reaching the HEV trunk where recruitment of cells concerned in the triggering of a primary response is most intense. In a simpler scenario, a naïve blood B cell and T helper-cell are both recruited by a same relevant HEV cell (Fig. 5F). To conclude, although knowledge of the lymph node compartment is still incomplete, it is possible to elaborate proposals on its global functioning and on basic tissue and cellular events related to the triggering of primary responses there. Nevertheless, these proposals would benefit from improved knowledge, in particular, on the neonatal emergence of the compartment parenchyma, which could provide further clues on the unfolding of primary responses. A full understanding of the triggering of a response will depend on knowledge of the cellular and subcellular events occurring in the subsinus layer and subendothelial spaces, where triggering would commonly take place. Such understanding is a remote goal. Acknowledgements The author is grateful, in alphabetical order, to Dr. C. Bélisle, Mrs. G. Guay, Mr. F.-S. Peng, and Dr. Y.M. Sin for their contributions to data and discussions. The author thanks Dr. É. Gagnon for helpful information, Dr. J. 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Qatarneh, Ion-Christian Kiricuta, Anders Brahme, Ulf Tiede, Bengt K. Lind, The Anatomical Record Part B: The New Anatomist Morphological Studies of Lymphatic Labyrinths in the Rat Mesenteric Lymph Node Limin Jia, Zunjiang Xie, Jinhua Zheng, Li Liu, Yan He, Fu Liu, Yechun He, The Anatomical Record Lymph node – an organ for T‐cell activation and pathogen defense Georg Gasteiger, Marco Ataide, Wolfgang Kastenmüller, Immunological Reviews Cords, channels, corridors and conduits: critical architectural elements facilitating cell interactions in the lymph node cortex J. Elizabeth Gretz, Arthur O. Anderson, Stephen Shaw, Immunological Reviews Metrics 60 Details Copyright © 2010 Wiley-Liss, Inc. Keywords lymph node development lymph node morphology lymphnode functioning lymphocyte quest triggering of primary immune responses rat Publication History 25 January 2010 25 January 2010 18 September 2009 23 December 2008 Close Figure Viewer Previous FigureNext Figure Caption Download PDF Referenced in 1 Wikipedia pages 67 readers on Mendeley See more details
14517	https://www.aimsciences.org/article/doi/10.3934/dcds.2019303	\| \| \| `x^2+y_1+z_12^34` \| AIMS Journals {{journal.titleEn}} {{journal.titleEn}} {{journal.titleEn}} {{journal.titleEn}} 3. AIMS Press Math Journals 4. {{journal.titleEn}} {{journal.titleEn}} 5. Book Series 6. {{book.nameEn}} 1. {{subColumn.name}} AIMS Journals {{journal.titleEn}} {{journal.titleEn}} AIMS Press Math Journals {{journal.titleEn}} {{journal.titleEn}} Book Series {{book.nameEn}} {{subColumn.name}} Advanced Search {{subColumn.name}} Discrete and Continuous Dynamical Systems {{newsColumn.name}} {{subColumn.name}} Home PDF Cite Share This issue Previous Article Next Article Article Contents Article Contents 2019, Volume 39, Issue 12: 7265-7290. Doi: 10.3934/dcds.2019303 This issue Previous Article A new proof of the boundedness results for stable solutions to semilinear elliptic equations Next Article On the optimal map in the $ 2 $-dimensional random matching problem Sharp large time behaviour in $ N $-dimensional Fisher-KPP equations Jean-Michel Roquejoffre1, ,, Luca Rossi2, and Violaine Roussier-Michon3, 1. Institut de Mathématiques de Toulouse; UMR 5219, Université de Toulouse; CNRS, Université Toulouse Ⅲ, 118 route de Narbonne, 31062 Toulouse, France 2. Centre d'Analyse et de Mathématique Sociales; UMR 8557, Paris Sciences et Lettres; CNRS, EHESS, 54 Bv. Raspail, 75006 Paris, France 3. Institut de Mathématiques de Toulouse; UMR 5219, Université de Toulouse; CNRS, INSA Toulouse, 135 av. Rangueil, 31077 Toulouse, France Corresponding author Corresponding author Dedicated to L. Caffarelli, as a sign of friendship, admiration and respect Received: February 2019 Revised: August 2019 Published: September 2019 The first and second authors are supported by the European Union's Seventh Framework Programme (FP/2007-2013) / ERC Grant Agreement n. 321186 - ReaDi - "Reaction-Diffusion Equations, Propagation and Modelling". The third author is supported by the ANR project NONLOCAL ANR-14-CE25-0013. Abstract / Introduction Full Text(HTML) Related Papers Cited by Abstract Abstract We study the large time behaviour of the Fisher-KPP equation $ \partial_t u = \Delta u +u-u^2 $ in spatial dimension $ N $, when the initial datum is compactly supported. We prove the existence of a Lipschitz function $ s^\infty $ of the unit sphere, such that $ u(t, x) $ approaches, as $ t $ goes to infinity, the function $ U_{c_}\bigg(\|x\|-c_t + \frac{N+2}{c_} \mathrm{ln}t + s^\infty\Big(\frac{x}{\|x\|}\Big)\bigg), $ where $ U_{c} $ is the 1D travelling front with minimal speed $ c_ = 2 $. This extends an earlier result of Gärtner. Keywords:+ Fisher-KPP equation, + logarithmc delay, + large time behaviour. Mathematics Subject Classification: Primary: 35K57. \| \| \| Citation: \| \begin{equation} \ \end{equation} Full Text(HTML) Related Papers Cited by References References \| \| \| --- \| \| \| D. G. Aronson and H. F. Weinberger, Multidimensional nonlinear diffusion arising in population genetics, Adv. in Math., 30 (1978), 33-76. doi: 10.1016/0001-8708(78)90130-5. \| \| \| H. Berestycki and F. Hamel, Reaction-diffusion Equations and Propagation Phenomena, Applied Mathematical Sciences, 2014. \| \| \| H. Berestycki, The inluence of advection on the propagation of fronts in reaction-diffusion equations, in: Nonlinear PDE's in Condensed Matter and Reactive Flows, eds. H. Berestycki, Y. Pomeau, NATO Science Series C, Mathematical and Physical Sciences, Kluwer Acad. Publ., Dordrecht, NL, 569 (2002). \| \| \| J. Berestycki, E. Brunet and J. 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Media, 8 (2013), 275-289. doi: 10.3934/nhm.2013.8.275. \| \| \| F. Hamel, J. Nolen, J.-M. Roquejoffre and L. Ryzhik, The logarithmic delay of KPP fronts in a periodic medium, Journal of the European Mathematical Society, 18 (2016), 465-505. doi: 10.4171/JEMS/595. \| \| \| D. Henry, Geometric Theory of Semlinear Parabolic Equations, Lecture Notes in Mathematics, 840. Springer-Verlag, Berlin-New York, 1981. \| \| \| C. K. R. T. Jones, Asymptotic behaviour of a reaction-diffusion equation in higher space dimensions, Rocky Mountain J. Math., 13 (1983), 355-364. doi: 10.1216/RMJ-1983-13-2-355. \| \| \| A. N. Kolmogorov, I. G. Petrovskii and N. S. Piskunov, Étude de l'équation de la diffusion avec croissance de la quantité de matière et son application à un problème biologique, Bull. Univ. État Moscou, Sér. Inter. A, 1 (1937), 1-26. \| \| \| J. Nolen, J.-M. Roquejoffre and L. Ryzhik, Convergence to a single wave in the Fisher-KPP equation, Chinese Ann. Math. Ser. B (special issue in honour of H. Brezis), 38 (2017), 629–646. doi: 10.1007/s11401-017-1087-4. \| \| \| J. Nolen, J.-M. Roquejoffre and L. Ryzhik, Refined long time asymptotics for the Fisher-KPP fronts, Comm. Contemp. Math., 2018. doi: 10.1142/S0219199718500724. \| \| \| J.-M. Roquejoffre and V. Roussier-Michon, Nontrivial large-time behaviour in bistable reaction-diffusion equations, Annali Mat. Pura Appl., 188 (2009), 207-233. doi: 10.1007/s10231-008-0072-7. \| \| \| J.-M. Roquejoffre and V. Roussier-Michon, Nontrivial dynamics beyond the logarithmic shift in two-dimensional Fisher-KPP equations, Nonlinearity, 31 (2018), 3284-3307. doi: 10.1088/1361-6544/aaba3b. \| \| \| L. Rossi, The Freidlin-Gärtner formula for general reaction terms, Adv. Math., 317 (2017), 267-298. doi: 10.1016/j.aim.2017.07.002. \| \| \| L. Rossi, Symmetrization and anti-symmetrization in parabolic equations, Proc. Amer. Math. Soc., 145 (2017), 2527-2537. doi: 10.1090/proc/13391. \| \| \| V. 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Differential Equations, 13 (2001), 323-353. doi: 10.1023/A:1016632124792. \| Access History Access History PDF XML Export Citation SHARE Email to a Friend Article Metrics HTML views(3116) PDF downloads(267) Cited by(0) ###### Access History Other Articles By Authors on this site Jean-Michel Roquejoffre Luca Rossi Violaine Roussier-Michon on Google Scholar Jean-Michel Roquejoffre Luca Rossi Violaine Roussier-Michon Catalog Export File Citation Format DownLoad: Full-Size Img PowerPoint Return Site map Copyright © 2025 American Institute of Mathematical Sciences
14518	http://www.mmrc.iss.ac.cn/~xgao/paper/book-area.pdf	MACHINE PROOFS IN GEOMETRY Automated Production of Readable Proofs for Geometry Theorems Shang-Ching Chou Department of Computer Science The Wichita State University Xiao-Shan Gao Institute of Systems Science Academia Sinica, Beijing Jing-Zhong Zhang Chengdu Institute of Computer Application Academia Sinica, Chengdu World Scientiﬁc 1994 To Wu Wen-Ts¨ un v Foreword Polya named Descartes’ Rules for the Direction of the Mind his favorite book on how to think. In the Rules, Descartes speculates on the possible existence of a powerful, secret method known to the ancients for doing geometry: But my opinion is that these writers then with a sort of low cunning, deplorable indeed, suppressed this knowledge. Possibly they acted just as many inventors are known to have done in the case of their discoveries, i.e., they feared that their method being so easy and simple would become cheapened on being divulged, and they preferred to exhibit in its place certain barren truths, deduc-tively demonstrated with show enough of ingenuity, as the results of their art, in order to win from us our admiration for these achievements, rather than to disclose to us that method itself which would have wholly annulled the admi-ration accorded. Descartes, that master of both method and geometry, would have been amazed to see the publication of this volume by Chou, Gao, and Zhang, who here present a method for proving extremely diﬃcult theorems in geometry, a method so simple and eﬃcient it can be carried out by high school students or computers. In fact, the method has been implemented in a computer program which can prove hundreds of diﬃcult theorems and moreover can produce simple, elegant proofs. In my view, the publication of this book is the single most important event in automated reasoning since Slagle and Moses ﬁrst implemented programs for symbolic integration. All too often, research in automated reasoning is concerned with technical questions of marginal interest to the mathematics community at large. Various strategies for eﬃcient substitution and propositional or equational reasoning have dominated the ﬁeld of auto-mated reasoning, and the mechanical proving of diﬃcult theorems has been a rare event. Mechanical searches for the proofs of diﬃcult theorems are usually guided extensively by the ‘user’. Almost never do we ﬁnd exhibited a computer program that can routinely treat hard problems in any area of mathematics, but in this book we do! The skeptical reader is urged to ﬂip through the 400 diﬃcult theorems in Chapter 6, all mechanically proved, and try his hand. The exceptional simplicity of the mechanically generated proofs presented for many of these theorems illustrate that the authors’ method is not some algorithm of mere “in principle,” proof-theoretic relevance, such as Tarski’s method. No, by applying the simple method formulated by the authors, the reader may also quickly become an expert vii viii Forward at geometry proving. The stunning beauty of these proofs is enough to rivet the reader’s attention into learning the method by heart. The key to the method presented here is a collection of powerful, high level theorems, such as the Co-side and Co-angle Theorems. This method can be contrasted with the earlier Wu method, which also proved astonishingly diﬃcult theorems in geometry, but with low-level, mind-numbing polynomial manipulations involving far too many terms to be carried out by the human hand. Instead, using high level theorems, the Chou-Gao-Zhang method employs such extremely simple strategies as the systematic elimination of points in the order introduced to produce proofs of stunning brevity and beauty. Although theorems are generally regarded as the most important of mathematical results by the mathematics community, it is methods, i.e., constructions and algorithms, that have the most practical signiﬁcance. For example, the reduction of arithmetical computation to a mechanical activity is the root of the computing industry. Whenever an area of mathematics can be advanced from being an unwieldy body of theorems to a uniﬁed method, we can expect serious practical consequences. For example, the reduction of parts of the calculus to tables of integrals and transforms was crucial to the emergence of modern engineering. Although arithmetic calculation and even elementary parts of analysis have reached the point that computers are both faster at them and more trustworthy than people, the impact of the mechanization of geometry has been less palpable. I believe this book will be a milestone in the inevitable endowment of computers with as much geometric as arithmetic prowess. Yet I greet the publication of this volume with a tinge of regret. Students of artiﬁ-cial intelligence and of automated reasoning often suﬀer from having their achievements disregarded by society at large precisely because, as Descartes observed, any simple, in-genious invention once revealed, seems ﬁrst obvious and then negligible. Progress in the automation of mathematics is inherently dependent upon the slow, deep work of ﬁrst rate mathematicians, and yet this fundamentally important line of work receives negligible soci-etal scientiﬁc support in comparison with those who build bigger bombs, longer molecules, or those multi-million-line quagmires called ‘systems’. Is it possible that keeping such a work of genius as this book secret would be the better strategy for increased research sup-port, using the text as the secret basis for generating several papers and research proposals around each of the 400 mechanically proved theorems, never revealing the full power of the method? Robert S. Boyer December, 1993 Preface This book reports a recent major advance in automated theorem proving in geometry which should be of interest to both geometry experts and computer scientists. The authors have developed a method and implemented a computer program which, for the ﬁrst time, produces short, readable, and elegant proofs for hundreds of geometry theorems. Modern computer technology and science make it possible to produce proofs of theo-rems automatically. However, in practice computer theorem proving is a very diﬃcult task. Historically, geometry theorem proving on computers began in earnest in the ﬁfties with the work of Gelerntner, J. R. Hanson, and D. W. Loveland . This work and most of the subsequent work [129, 144, 83] were synthetic, i.e., researchers focused on the automation of the traditional proof method. The main problem of this approach was controlling the search space, or equivalently, guiding the program toward the right deductions. Despite initial success, this approach did not make much progress in proving relatively diﬃcult theorems. On the other hand, in the 1930s, A. Tarski, introduced a quantiﬁer elimination method based on the algebraic approach to prove theorems in elementary geometry. Tarski’s method was later improved and redesigned by A. Seidenberg , G. Collins and others. In particular, Collins’ cylindrical decomposition algorithm is the ﬁrst Tarski type algorithm which has been implemented on a computer. Solutions of several nontrivial problems of elementary geometry and algebra have been obtained using the implementation [45, 114]. A breakthrough in the use of the algebraic method came with the work of Wen-Ts¨ un Wu, who introduced an algebraic method which, for the ﬁrst time, was used to prove hun-dreds of geometry theorems automatically [164, 36]. Many diﬃcult theorems whose tradi-tional proofs need an enormous amount of human intelligence, such as Feuerbach’s theo-rem, Morley’s trisector theorem, etc., can be proved by computer programs based on Wu’s method within seconds. In Chou’s earlier book , there is a collection of 512 geometry theorems proved by a computer program based on Wu’s method. However, if one wishes to look at the proofs produced using Wu’s method, he/she will ﬁnd tedious and formidable computations of polynomials. The polynomials involved in the proofs can contain hundreds of terms with more than a dozen variables. Because of this, producing short, readable proofs remains a prominent challenge to researchers in the ﬁeld of automated theorem proving. Recently, the authors developed a method which can produce short and readable proofs ix x Preface for hundreds of geometric statements in plane and solid geometries [195, 72, 73, 74]. The starting point of this method is the mechanization of the area method, one of the oldest and most eﬀective methods in plane geometry. One of the most important theorems in geometry, the Pythagorean theorem, was ﬁrst proved using the area method. But the area method has generally been considered just some sort of special trick for solving geometry problems. J. Zhang recognized the generality of the method and developed it into a sys-tematic method to solve geometry problems [40, 41, 42]. By a team eﬀort, we have further developed this systematic method into a mechanical one and implemented a prover1 which has been used to produce elegant proofs for hundreds of geometry theorems. This book contains 478 geometry problems solved entirely automatically by our prover, including machine proofs of 280 theorems printed in full. The area method is a combination of the synthetic and algebraic approaches. In the machine proofs, we still use polynomial computation; but we use geometric invariants like areas and Pythagoras diﬀerences instead of coordinates of points as the basic quantities, and the geometric meaning for each step of the proof is clear. Another important feature of the area method is that the machine proofs produced by the method/program are generally very short; the formulas in the proofs usually have only a few terms, and hence are readable by people. The method is complete for constructive geometry theorems, i.e., those statements whose diagrams can be drawn using a ruler and a pair of compasses only. The area is used to deal with geometry relations like incidence and parallelism. Another basic geometry quantity in our method is the Pythagoras diﬀerence, which is used to deal with geometry relations like perpendicular and congruence of line segments. Besides the area and Pythagoras diﬀer-ence, we also use other geometry quantities, such as the full-angle, the volume, the vector, and the complex number. The reason we use more geometry quantities is that for each geometry quantity, there are certain geometry theorems which can be proved easily using this quantity. Another aspect of automated geometry theorem proving relates to the diﬃculty of learn-ing and teaching geometry. About two thousand years ago, an Egyptian king asked Euclid whether there was an easier way to learn geometry. Euclid’s reply was, “There is no royal road to geometry.” Of course, the diﬃculty here was not the basic geometry concepts, such as points, lines, angles, triangles, lengths, areas, etc. The diﬃculty has been with many other fascinating facts (theorems) and how to use logical reasoning to justify (prove) these theorems based on only a few basic facts (axioms) that are so obvious that they can be taken for granted. One can draw dozens of triangles with three medians and ﬁnd the fact that the three medians of a triangle intersect at the same points. However, the empirical observation is only a justiﬁcation of formation of a conjecture whose correctness must be proved by other means – logical reasoning. 1The prover is available via ftp at emcity.cs.twsu.edu: pub/geometry. Preface xi Unlike algebra, in which most problems can be solved according to some systematic method or algorithm, a human proof requires a diﬀerent set of tricks for each geometry theorem, making it diﬃcult for students to get started with a proof. In the two thousand years since the time of the Egyptian king, people have continued to wish for an easy way to learn geometry. In response to this diﬃculty and in working with middle school students, J. Z. Zhang has established a new geometry axiom system based on the notion of area [40, 41, 42]. Using his new system, Zhang has made a great eﬀort in promoting a new reform in high school geometry education in China. The successful implementation of his method has led to its use in Chinese geometry textbooks for teachers’ colleges. In addition, the area method has been used in recent years to train students of Chinese teams for participation in the International Mathematical Olympiads. One of the goals of this book is to make learning and teaching of geometry easy. The machine proofs generated have a shape that a student of mathematics could learn to de-sign with pencil and paper. By reading the machine produced proofs in this book, many readers might be able to use the mechanical method to prove diﬃcult geometry theorems themselves. This book consists of six chapters. The ﬁrst chapter is about the basic concepts of geometry, area and ratio of lengths. Then we introduce the new method, the area method for proving geometry theorems. This chapter can stand alone as a supplement to textbooks of high school or college geometry. We present this chapter at an elementary level with many interesting examples solved by the new method, with the intention of attracting many readers at various levels, from high school students to university professors. It is also our hope that people will be able to prove many diﬃcult geometry theorems using the method introduced in this chapter. Beginning with Chapter 2, we formalize or mechanize the method by describing the method in an algorithmic way. Those who are interested in geometry only will have a clearer idea about this mechanical procedure of proving geometry theorems. Those who wish to write their own computer programs will be able to produce short and readable proofs of diﬃcult geometry theorems. Experts will be able to ﬁnd further extensions, de-velopments and improvements in this new direction. Only when more and more people are participating in projects of this kind will real advances in geometry education be possible. One of our goals in writing this book is to encourage such research and advances. The last chapter, Chapter 6, is a collection of 400 theorems proved by our computer program, including machine proofs of 205 theorems printed in full. Most of this chap-ter was generated mechanically, including machine proofs in T EX typesetting form. This chapter is an integral and important part of the book, because it alone shows the power of our mechanical method and computer program. Even reading the proofs produced by our computer program is enjoyable. xii Preface Funding from the National Science Foundation gave us a unique opportunity to form a very strong research team at Wichita State University and to make this book possible. Also the Chinese National Science Foundation provided further support for Gao and Zhang. The authors wish to thank many people for their support of our research project. Dr. K. Abdali of National Science Foundation constantly encouraged Chou and recommended fur-ther funding to invite the third author, J. Zhang, to the U.S. This turned out to be the crucial step for this exciting new development and publication of this book. The encouragements from R. S. Boyer, L. Wos, W. T. Wu, W. W. Bledsoe, and J S. Moore when they heard about our early work were invaluable. They also thank Frank Hwang for his encouragement. In Wichita, Mary Edgington has also played a key role in promoting the project. The authors also thank Patsy P. Sheﬃeld and L. Yang for their careful reading of the manuscript and helpful suggestions, and thank Tom Wallis, the computer system manager, for provid-ing an excellent computing environment. Chou used a part of this work in his graduate course “Symbolic and Algebraic Com-putation;” the manuscript of this book was also read by many students in his class. In particular, we wish to thank Joe G. Moore for his proof-reading and Nirmala Navaneethan for her suggestions. The support and patience of the authors’ families were also very important for the suc-cess of this project. S. C. Chou X. S. Gao J. Z. Zhang December, 1993 Contents 1 Geometry Preliminaries 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Directed Line Segments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Areas and Signed Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4 The Co-side Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.5 Parallels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.6 The Co-angle Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.7 Pythagoras Diﬀerences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.8 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 1.9 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.10 Full-Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Summary of Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 2 The Area Method 51 2.1 Traditional Proofs Versus Machine Proofs . . . . . . . . . . . . . . . . . . 51 2.2 Signed Areas of Oriented Triangles . . . . . . . . . . . . . . . . . . . . . . 54 2.2.1 The Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 2.2.2 Basic Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . 57 2.3 The Hilbert Intersection Point Statements . . . . . . . . . . . . . . . . . . 59 2.3.1 Description of the Statements . . . . . . . . . . . . . . . . . . . . 60 2.3.2 The Predicate Form . . . . . . . . . . . . . . . . . . . . . . . . . . 63 2.4 The Area Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 2.4.1 Eliminating Points from Areas . . . . . . . . . . . . . . . . . . . . 65 2.4.2 Eliminating Points from Length Ratios . . . . . . . . . . . . . . . 66 2.4.3 Free Points and Area Coordinates . . . . . . . . . . . . . . . . . . 67 2.4.4 Working Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 71 xiii xiv Preface 2.5 More Elimination Techniques . . . . . . . . . . . . . . . . . . . . . . . . . 76 2.5.1 Reﬁned Elimination Techniques . . . . . . . . . . . . . . . . . . . 76 2.5.2 The Two-line Conﬁguration . . . . . . . . . . . . . . . . . . . . . 79 2.6 Area Method and Aﬃne Geometry . . . . . . . . . . . . . . . . . . . . . . 81 2.6.1 Aﬃne Plane Geometry . . . . . . . . . . . . . . . . . . . . . . . . 82 2.6.2 Area Method and Aﬃne Geometry . . . . . . . . . . . . . . . . . . 83 2.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.7.1 Formula Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.7.2 Existence of n3 Conﬁgurations . . . . . . . . . . . . . . . . . . . . 90 2.7.3 Transversals for Polygons . . . . . . . . . . . . . . . . . . . . . . 94 Summary of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 3 Machine Proof in Plane Geometry 103 3.1 The Pythagoras Diﬀerence . . . . . . . . . . . . . . . . . . . . . . . . . . 103 3.1.1 Pythagoras Diﬀerence and Perpendicular . . . . . . . . . . . . . . 103 3.1.2 Pythagoras Diﬀerence and Parallel . . . . . . . . . . . . . . . . . . 105 3.1.3 Pythagoras Diﬀerence and Area . . . . . . . . . . . . . . . . . . . 107 3.2 Constructive Geometry Statements . . . . . . . . . . . . . . . . . . . . . . 109 3.2.1 Linear Constructive Geometry Statements . . . . . . . . . . . . . . 109 3.2.2 A Minimal Set of Constructions . . . . . . . . . . . . . . . . . . . 112 3.2.3 The Predicate Form . . . . . . . . . . . . . . . . . . . . . . . . . . 113 3.3 Machine Proof for Class CL . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.3.1 The Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.3.2 Reﬁned Elimination Techniques . . . . . . . . . . . . . . . . . . . 120 3.4 The Ratio Constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 3.4.1 More Ratio Constructions . . . . . . . . . . . . . . . . . . . . . . 122 3.4.2 Mechanization of Full-Angles . . . . . . . . . . . . . . . . . . . . 128 3.5 Area Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 3.5.1 Area Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . 133 3.5.2 Area Coordinates and Special Points of Triangles . . . . . . . . . . 136 3.6 Trigonometric Functions and Co-Circle Points . . . . . . . . . . . . . . . . 141 3.6.1 The Co-circle Theorems . . . . . . . . . . . . . . . . . . . . . . . 141 3.6.2 Eliminating Co-Circle Points . . . . . . . . . . . . . . . . . . . . . 144 3.7 Machine Proof for Class C . . . . . . . . . . . . . . . . . . . . . . . . . . 148 3.7.1 Eliminating Points from Geometry Quantities . . . . . . . . . . . . 148 Preface xv 3.7.2 Pseudo Divisions and Triangular Forms . . . . . . . . . . . . . . . 150 3.7.3 Machine Proof for Class C . . . . . . . . . . . . . . . . . . . . . . 154 3.8 Geometry Information Bases and Machine Proofs Based on Full-Angles . . 157 3.8.1 Building the Geometry Information Base . . . . . . . . . . . . . . 158 3.8.2 Machine Proof Based on the Geometry Information Base . . . . . . 162 Summary of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 4 Machine Proof in Solid Geometry 171 4.1 The Signed Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 4.1.1 Co-face Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 4.1.2 Volumes and Parallels . . . . . . . . . . . . . . . . . . . . . . . . 175 4.1.3 Volumes and Aﬃne Geometry of Dimension Three . . . . . . . . . 177 4.2 Constructive Geometry Statements . . . . . . . . . . . . . . . . . . . . . . 181 4.2.1 Constructive Geometry Statements . . . . . . . . . . . . . . . . . . 181 4.2.2 Constructive Conﬁgurations . . . . . . . . . . . . . . . . . . . . . 184 4.3 Machine Proof for Class SH . . . . . . . . . . . . . . . . . . . . . . . . . 187 4.3.1 Eliminating Points from Volumes . . . . . . . . . . . . . . . . . . 188 4.3.2 Eliminating Points from Area Ratios . . . . . . . . . . . . . . . . . 190 4.3.3 Eliminating Points from Length Ratios . . . . . . . . . . . . . . . 193 4.3.4 Free Points and Volume Coordinates . . . . . . . . . . . . . . . . . 196 4.3.5 Working Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 198 4.4 Pythagoras Diﬀerences in Space . . . . . . . . . . . . . . . . . . . . . . . 202 4.4.1 Pythagoras Diﬀerence and Perpendicularity . . . . . . . . . . . . . 202 4.4.2 Pythagoras Diﬀerence and Volume . . . . . . . . . . . . . . . . . . 205 4.5 The Volume Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 4.5.1 The Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 4.5.2 Working Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 210 4.6 Volume Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . 215 Summary of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 5 Vectors and Machine Proofs 221 5.1 Metric Vector Spaces of Dimension Three . . . . . . . . . . . . . . . . . . 221 5.1.1 Inner Products and Metric Vector Space . . . . . . . . . . . . . . . 223 5.1.2 Exterior Products in Metric Vector Space . . . . . . . . . . . . . . 226 5.2 The Solid Metric Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 229 5.2.1 Inner Products and Exterior Products . . . . . . . . . . . . . . . . 230 xvi Preface 5.2.2 Constructive Geometry Statements . . . . . . . . . . . . . . . . . . 233 5.3 Machine Proof by Vector Calculation . . . . . . . . . . . . . . . . . . . . . 235 5.3.1 Eliminating Points From Vectors . . . . . . . . . . . . . . . . . . . 235 5.3.2 Eliminating Points from Inner and Exterior Products . . . . . . . . 238 5.3.3 The Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 5.4 Machine Proof in Metric Plane Geometries . . . . . . . . . . . . . . . . . 242 5.4.1 Vector Approach for Euclidean Plane Geometry . . . . . . . . . . . 243 5.4.2 Machine Proof in Minkowskian Plane Geometry . . . . . . . . . . 246 5.5 Machine Proof Using Complex Numbers . . . . . . . . . . . . . . . . . . . 249 Summary of Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 6 A Collection of 400 Mechanically Proved Theorems 259 6.1 Notation Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 6.2 Geometry of Incidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 6.2.1 Menelaus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 263 6.2.2 Ceva’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 6.2.3 The Cross-ratio and Harmonic Points . . . . . . . . . . . . . . . . 270 6.2.4 Pappus’ Theorem and Desargues’ Theorem . . . . . . . . . . . . . 274 6.2.5 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 6.3 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 6.3.1 Medians and Centroids . . . . . . . . . . . . . . . . . . . . . . . . 285 6.3.2 Altitudes and Orthocenters . . . . . . . . . . . . . . . . . . . . . . 297 6.3.3 The Circumcircle . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 6.3.4 The Euler Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 6.3.5 The Nine-Point Circle . . . . . . . . . . . . . . . . . . . . . . . . 328 6.3.6 Incircles and the Excircles . . . . . . . . . . . . . . . . . . . . . . 333 6.3.7 Intercept Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . 347 6.3.8 Equilateral Triangles . . . . . . . . . . . . . . . . . . . . . . . . . 352 6.3.9 Pedal Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 6.3.10 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 6.4 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 6.4.1 General Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . 364 6.4.2 Complete Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . 372 6.4.3 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 6.4.4 Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 Preface xvii 6.4.5 Cyclic Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . 395 6.4.6 Orthodiagonal Quadrilaterals . . . . . . . . . . . . . . . . . . . . . 401 6.4.7 The Butterﬂy Theorems . . . . . . . . . . . . . . . . . . . . . . . 405 6.5 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 6.5.1 Chords, Secants, and Tangents . . . . . . . . . . . . . . . . . . . . 409 6.5.2 Intersectional Circles . . . . . . . . . . . . . . . . . . . . . . . . . 425 6.5.3 The Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 6.5.4 Orthogonal Circles . . . . . . . . . . . . . . . . . . . . . . . . . . 436 6.5.5 The Simson Line . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 6.5.6 The Pascal Conﬁguration . . . . . . . . . . . . . . . . . . . . . . . 445 6.5.7 Cantor’s Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 449 6.6 A Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 Chapter 1 Geometry Preliminaries In this chapter, we will provide geometry background for the rest of this book. The pre-sentation is informal and the prerequisite in geometry is minimal. Anyone with a semester of high school geometry can read and understand this chapter. We will present a new sys-tematic proof method, the area method, which can be used to solve numerous geometry problems at various levels of diﬃculty, ranging from problems in high school textbooks to those in mathematics competitions. Those who are mainly interested in machine proofs may skip this chapter and start from Chapter 2 directly. 1.1 Introduction Geometry, like other sciences, is concerned with the laws in a speciﬁc domain. For ge-ometry this domain is space. However, reasoning plays a much more important role in geometry and in other branches of mathematics than in other sciences. There are two kinds of reasoning: inductive reasoning and deductive reasoning. Necessity and curiosity have at all times caused people to investigate phenomena and to ﬁnd the laws governing the physical universe. Drawing three medians of a triangle, one ﬁnds that the three medians intersect at the same point. By making repeated experiments, one can come to the conclusion that the three medians of any triangle always intersect at the same point. This kind of inductive reasoning, which is fundamental in experimental sciences, is also very important for observing new facts and laws in mathematics. However, in geometry or mathematics, inductive reasoning generally cannot serve as the justiﬁcation of the correctness of the observed fact. To justify the observed fact, we have to give a proof of the fact. Here the terminology “proof” has a distinctive meaning. Based on already proved facts (theorems), we need to use logical or deductive reasoning to derive this new fact about three medians. Once we prove a fact in this way, we call it a 1 2 Chapter 1. Geometry Preliminaries theorem. The truth of the fact is thus beyond doubt. Here we prove a theorem “based on already proved facts.” However, the proof of these “already proved facts” is based on other “ already proved facts.” In the ﬁnal analysis, we have to choose a few basic facts whose correctness is so evident in our everyday life that we can take them for granted without any proofs. For example, the fact that for any two distinct points there is one and only one line passing through them is so evident that we can use it without proof. Such kinds of facts are very basic and are called axioms or postulates. The selection of axioms is by no means evident. For example, Euclid’s ﬁfth postulate, “from any point not on a line, there is one and only one line passing through the point and parallel to the given line,” is a very evident fact for many people. However, for over two thousand years, mathematicians tried to prove this fact using other basic postulates. The failure of these attempts ﬁnally led to the Non-Euclidean geometries and a revolution in mathematics. There is a special branch of mathematics, foundations of geometry, which is exclusively concerned with the related topics. Generally, the traditional proof method in geometry proceeds as follows: ﬁrst we es-tablish (prove) various basic propositions or lemmas, e.g., the theorems of congruence of two triangles. Then we use these basic propositions to prove new theorems. We can en-large the set of basic propositions by including some of these newly proved theorems. In the traditional methods used by Euclid and many other geometers since then, theorems of congruence and similarity of triangles are the very basic tools. Although the method based on congruence and similarity leads to elegant proofs for many geometry theorems, it also has weaknesses: (1) In a diagram of a geometry statement to be proved, rarely do there exist congruent or similar triangles. In order to use the propositions on congruent or similar triangles, one has to construct auxiliary lines. As we know, adding auxiliary lines is one of the most diﬃcult and tricky steps in the proofs of geometry theorems. This also leads to a changeable and uncertain strategy in the eﬀort of ﬁnding a proof. (2) In propositions on congruent or similar triangles, there is asymmetry between the hypothesis and conclusion. For example, in order to prove the congruence of the segments AB and XY, we need to construct (or ﬁnd) △ABC and △XYZ and to prove the congruence of the two triangles. Generally, this in turn requires us to prove the congruence of three pairs of geometry elements (segments or angles). In order to prove one identity, we need to ﬁnd three other identities! As a consequence, it is very diﬃcult to ﬁnd an eﬀective method for solving geometry problems based on congruence and similarity. In this book, we will use the area of triangles as the basic tool for solving geometry problems. The traditional area method is one of the oldest and most eﬀective methods in plane geometry. One of the most important theorems in geometry, the Pythagorean theorem, was ﬁrst proved using the area method. But the area method has generally been considered as a set of special tricks for solving geometry problems. J. Z. Zhang has been 1.2 Directed Line Segments 3 studying the area method since 1975. He has recognized its generality and has developed it into a systematic method for solving geometry problems [40, 41, 42].1 A large number of geometry problems at various levels of diﬃculty, ranging from basic propositions in high school textbooks to those in mathematics competitions, have been provided with elegant proofs using the area method. This chapter is a summary of the basic facts about the area method together with many geometry theorems solved by this method. Another feature of the area method is that the deduction in the method is achieved mainly by algebraic computation. This makes the area method ready for mechanization, which is the main theme of this book. 1.2 Directed Line Segments For most of the book, we are concerned with plane geometry. Thus our starting point is a plane, which we sometimes refer to as the Euclidean plane. The basic geometry objects on a plane are points and lines. We use capital English letters A, B, C, ... to denote points on the Euclidean plane. For two distinct points A and B, there is one and only one line l that passes through points A and B. We use AB or BA to denote this line. In addition, we can give a line one of the two directions and talk about directed lines. Thus directed line AB has the direction from point A to point B, whereas the directed line BA has the opposite direction from point B to A. Two points A and B on a directed line determine a directed line segment whose length AB is positive if the direction from A to B is the same as the direction of the line and negative if the direction from A to B is in the opposite direction. Thus (1.1) AB = −BA and AB = 0 if and only if A = B. Let A, B, P, and Q be four points on the same line such that A , B. Then the ratio of PQ and AB is meaningful; let it be t. We have PQ AB = t, or PQ = tAB. If the two directed line segments AB and PQ have the same direction then t ≥0; if they have opposite directions then t ≤0. If P or Q is not on line AB, then we cannot compare 1This systematic method has been used to train students of Chinese teams for participat-ing in the International Mathematical Olympiad in solving geometry problems. 4 Chapter 1. Geometry Preliminaries or do arithmetic operations between AB and PQ, because the sign of AB depends on the direction of line AB which has nothing to do with the direction of line PQ. If point P is on line AB then AB = AP + PB or (1.2) AP AB + PB AB = 1. We call AP AB and PB AB the position ratios or the position coordinates of point P with respect to AB. It is clear that for two real numbers s and t such that s + t = 1, there is a unique point P on AB which satisﬁes AP AB = t, PB AB = s. In particular, the statement that point O is the midpoint of segment AB means AO AB = OB AB = 1 2. Two distinct points always determine a line. However, three points are generally not on the same line. If they are, we say that the three points are collinear. In fact, the most fasci-nating facts about many elegant geometry theorems discovered over the past two thousand years have been that no matter how you draw a certain geometry ﬁgure, the three partic-ular points in the ﬁgure are always collinear. Connected with each theorem, there is the collinear line named after the mathematician who discovered it, for example, Pappus’ line, Euler’s line, Gauss’ line, Pascal’s line, Simson’s line, etc. A B 1 A 1 B C 1 C P Q S Figure 1-1 Example 1.1 (Pappus’ Theorem) Let points A, B and C be on one line, and A1, B1 and C1 be on another line. Let AB1 meet A1B in P, AC1 meet A1C in Q, and BC1 meet B1C in S . Show that P, Q, and S are collinear. Please draw a few diagrams for this geometric conﬁguration on a piece of paper. Note that the three intersection points P, Q and S are always collinear. If you have our software package, you can see this fact more vividly. First, the program helps you to draw one diagram in a few seconds on the computer screen. Then you can use the mouse to move any of the user-chosen points, e.g., point C1. The diagram is continuously changed on the screen while you can see that the three moving points P, Q, and S are always on the same line. This observation convinces almost everyone that this statement is always true. However, in order to justify the truth of a fact in mathematics, empirical observations are not enough. We need a proof of the truth of the fact using the mathematical reasoning. The proof of Pappus’ theorem (or of many other geometry theorems) is by no means easy, especially to high school students. The main objective of this book is to present a new systematic proof method. We believe that the serious reader can learn this method easily. 1.3 Areas and Signed Areas 5 Once you know this method and work with a few examples, you can prove a diﬃcult theorem in a few minutes. For a proof of Pappus’ theorem, see page 14. Exercises 1.2 1. Let A, B, and C be three collinear points. Show that AB 2 + BC 2 = AC 2 +2AB·CB. (Use (1.1) and (1.2). Do the same for the following exercises.) 2. Let A, B, C, and D be four collinear points. Then AB · CD + AC · DB + AD · BC = 0. 3. Four collinear points A, B, C, and D are called a harmonic sequence if AC BC = −AD BD. Show that four collinear points A, B, C, and D form a harmonic sequence if and only if AB CB + AB DB = 2. 4. Show that four collinear points A, B, C, and D form a harmonic sequence if and only if OC · OD = OA 2 where O is the midpoint of AB. 1.3 Areas and Signed Areas The next geometric object is the triangle. As we know, three non-collinear points A, B, and C form a triangle which is denoted by △ABC. The area of △ABC is denoted by ▽ABC. The reader can safely assume that ▽ABC is 1 2h · BC where h is the altitude of the triangle on the side BC. In our area method, however, we do not consider this a basic fact. Instead, we use other simple facts about the area as basic propositions. If A, B, and C are collinear, △ABC is called degenerate and ▽ABC is deﬁned to be zero. For any four points A, B, C, and P in the same plane, we can form four triangles △ABC, △PAB, △PBC, and △PAC and the areas of the four triangles satisfy seven diﬀerent rela-tions, depending on the relative position of the points A, B, C and P. A B C P Figure 1-2 If point P is inside the triangle ABC as shown in Figure 1-2, we have ▽ABC = ▽PAB + ▽PBC + ▽PCA. 6 Chapter 1. Geometry Preliminaries A B C P A B C P A B C P Figure 1-3 For the three cases shown in Figure 1-3, we have ▽ABC = ▽PAB + ▽PBC −▽PCA If ABCP is convex. ▽ABC = ▽PAB −▽PBC + ▽PCA If ABPC is convex. ▽ABC = −▽PAB + ▽PBC + ▽PCA If APBC is convex. A B C P A B C P A B C P Figure 1-4 For the three cases shown in Figure 1-4, we have ▽ABC = ▽PAB −▽PBC −▽PCA If C is in the interior of △PAB. ▽ABC = −▽PAB + ▽PBC −▽PCA If A is in the interior of △PBC. ▽ABC = −▽PAB −▽PBC + ▽PCA If B is in the interior of △PAC. We see that the above relations among the areas of triangles formed from the four points are very complicated. However, if we introduce the signed area of an oriented triangle, we can greatly simplify the relations of these areas; the seven relations can be reduced to just one equality. A triangle ABC has two orientations: if A–B–C is counterclockwise, triangle ABC has the positive orientation; otherwise triangle ABC has the negative orientation. Thus △ABC, △BCA, and △CAB have the same orientation, whereas △ACB, △CBA, and △BAC have the opposite orientation. The signed area of an oriented triangle ABC, denoted by S ABC, has the same absolute value as ▽ABC and is positive if the orientation of triangle ABC is positive; otherwise S ABC is negative. We thus have S ABC = S BCA = S CAB = −S ACB = −S BAC = −S CBA. Now the seven equations for ▽ABC, ▽PAB, ▽PBC, and ▽PCA can be summarized as 1.3 Areas and Signed Areas 7 just one equation: regardless of the position among points A, B, C, and P, we always have (1.3) S ABC = S PBC + S PCA + S PAB Remark for Advanced Readers. Based on analytic (coordinate) geometry, the proof of (1.3) is more strict and straightforward, and we do not have to discuss each of the seven cases separately. Suppose we have a coordinate system on the plane and the coordinates for all four points. For example, the coordinates of the points A, B and C are (ax, ay), (bx, by), and (cx, cy), respectively. Twice the signed area S ABC of the triangle ABC is the polynomial (ax −bx)(by −cy)−(ay −by)(bx −cx). Then checking the validity of (1.3) is a straightforward computation. Furthermore, we do not even need to use the brute force computation of polynomials to prove (1.3). Since (1.3) is valid when P is inside the triangle ABC and (1.3) is equivalent to a polynomial equation, that equation must be true. Since the truth of the equation is independent of the relative position of the four points, (1.3) is valid in all cases. With this argument in mind, the understanding of many other identities in this book or in geometry is much easier. For many geometry statements or theorems of equality type, the order relation (inside, between, etc.) is irrelevant. By the above argument based on polynomials, if the statement is true in one case, then it is true in all cases regardless of the relative order position of the points involved. There are many geometry theorems in which the order relation is essential. Proving of these theorems is beyond the scope of our current computer program based on the area method. One of the important examples is “A triangle with two equal internal angle bi-sectors is an isosceles triangle.” In this chapter we will prove many such theorems using the area method. However, the proofs are informal in the sense that we use some facts other than the basic propositions of the area method used in our computer program. End of Remark. Similarly, we can also deﬁne oriented quadrilaterals. Given four points A, B, C, and D, we deﬁne an oriented quadrilateral ABCD according to the point orientation A–B–C–D. Thus BCDA, CDAB, and DABC denote the same oriented quadrilateral as ABCD because their point orientation is the same. Four points can form six (4!/4) diﬀerent orientations, hence six diﬀerent oriented quadrilaterals: ABCD, ADCB, ACBD, ADBC, ACDB, and ABDC, as shown in Figure 1-5. A B C D A B C D A B C D Figure 1-5 8 Chapter 1. Geometry Preliminaries Now we can deﬁne the area of an oriented quadrilateral ABCD to be S ABCD = S ABC + S ACD. In order to justify that S ABCD is well-deﬁned, we need to prove that for the above deﬁnition, we have S ABCD = S BCDA = S CDAB = S DABC which is a direct consequence of (1.3). The deﬁnition of the signed area here can be generalized to an oriented n–polygon with any n > 4. Exercises 1.3 1. Prove the following properties of the areas of quadrilaterals. S ABCD = S BCDA = S CDAB = S DABC = −S DCBA = −S CBAD = −S BADC = −S ADCB, S ABCD = S ABC −S ADC = S BCD −S BAD, and S ABBC = S ABCC = S AABC = S ABCA = S ABC. 2. For any ﬁve points A, B, C, P, and Q in the same plane, we have S PAQB+S PBQC = S PAQC. 1.4 The Co-side Theorem The following is the ﬁrst basic proposition of the area method. A B P C Figure 1-6 Proposition 1.4 Let A, B, and C be three dis-tinct collinear points, and P a point not on line AB. Then we have S PBC S PAB = BC AB, or if BC = tAB then S PBC = tS PAB. Proposition 1.4 is obvious if we use the area formulas: ▽PBC = 1 2h·BC, ▽PAB = 1 2h·AB where h is the distance from point P to line AB. Notice that the areas involved here are signed areas of oriented triangles. Two triangles with a common side are said to be a pair of co-side triangles. From Propo-sition 1.4 we can easily infer the co-side theorem, which is the most important proposition of the area method. By using this theorem alone, we can prove many diﬃcult theorems easily. 1.4 The Co-side Theorem 9 A B P Q M N A B P Q M A B P Q M A B P Q M Figure 1-7 Proposition 1.5 (The Co-side Theorem) Let M be the intersection of the lines AB and PQ and Q , M. Then we have S PAB S QAB = PM QM. Figure 1-7 shows four possible cases of the co-side theorem. Here we give two proofs of Proposition 1.5, which are valid for all the four cases. Proof 1. Let N be a point on the line AB such that MN = AB. By Proposition 1.4 S PAB S QAB = S PMN S QMN = PM QM . Proof 2. S PAB S QAB = S PAB S PAM · S PAM S QAM · S QAM S QAB = AB AM · PM QM · AM AB = PM QM. We mentioned early that in the diagrams of geometry theorems, rarely are there similar or congruent triangles. But there are plenty of co-side triangles. For instance, in each of the four diagrams in Figure 1-7 there are 18 pairs of co-side triangles! Propositions 1.4 and 1.5 are two basic propositions. We will expand the set of basic propositions and, based on them, introduce our area proof method. Using the area method to prove theorems, no skillful trick is needed: we need only to follow systematic or pre-scribed steps to reach the completion of a proof. We will use several non-trivial examples to illustrate how to use the basic propositions to prove theorems. Example 1.6 Let △ABC be a triangle and P be any point in the plane (inside or outside of the triangle). Let D be the intersection of lines AP and CB, i.e., D = AP ∩CB. Also let E = BP ∩AC, and F = CP ∩AB. Show that PD AD + PE BE + PF CF = 1. A B C P F E D A B C P F E D A B C P F E D Figure 1-8 The hypotheses of most geometry theorems can be stated in a constructive way: 10 Chapter 1. Geometry Preliminaries beginning with some arbitrarily chosen points, lines, and circles, we introduce new points by taking arbitrary points on or taking intersections of the lines and circles; from the constructed points, new lines and circles can be formed; we now can introduce new points from these new lines and circles, etc; ﬁnally, a ﬁgure is formed consisting of points, lines, and circles. Such kinds of geometry theorems are called theorems of constructive type. Example 1.6 is a theorem of constructive type. Beginning with the arbitrarily chosen (free) points A, B, C and P, we introduce new points D, E, and F by constructing intersec-tions of lines AP and BC, lines BP and AC, and lines CP and AB. Our aim is to eliminate the constructed points from the left-hand side of the conclusion PD AD + PE BE + PF CF in the reverse order of the points in which they are introduced until all points in the expres-sion are free points. Then the expression is equal to or is easily proved to be equal to the right-hand side of the conclusion. Proof. By using the co-side theorem three times we can eliminate points D, E, and F respectively: PD AD = S PBC S ABC , PE BE = S PCA S ABC , PF CF = S PAB S ABC . By (1.3) on page 7, PD AD + PE BE + PF CF = S PBC + S PCA + S PAB S ABC = S ABC S ABC = 1. If not using the signed area, we would have to discuss several (seven) cases when P is inside or outside the triangle ABC. Figure 1-8 shows three possible cases of the example. The use of the signed area makes the proof concise and more strict. The reader will see this advantage throughout this book in other examples. At this point, we need to mention that the non-degenerate conditions which are nec-essary for a geometry statement to be true are not stated explicitly in the example. Here triangle ABC must be non-degenerate, i.e., S ABC , 0, as we generally implicitly assume in geometry textbooks. But this is not enough. We need each of the intersection points D, E and F to be normal, i.e., there is one and only one intersection point. This imposes con-ditions on the point P. For detailed discussion of non-degenerate conditions, see Chapter 2. 1.4 The Co-side Theorem 11 Example 1.7 (Ceva’s Theorem) The same hypotheses (constructions) as Example 1.6. Show that AF FB · BD DC · CE EA = 1. Proof. Our aim is still to eliminate the constructed points F, E and D from the left-hand side of the conclusion. Using the co-side theorem three times, we can eliminate E, F, and D AF FB · BD DC · CE EA = S APC S BCP · S BPA S CAP · S CPB S ABP = 1. The above proof for Ceva’s theorem can be extended immediately to prove Ceva’s the-orem for an arbitrary (2m + 1)-polygon with m ≥1. Example 1.8 (Ceva’s Theorem for a (2m + 1)-polygon) Let a point O and an arbitrary polygon V1...V2m+1 be given. Let Pi be the intersection of line OVi and the side Vi+mVi+m+1. Then 2m+1 Y i=1 Vi+kPi PiVi+k+1 = 1 where the subscripts are understood to be mod 2m + 1. Proof. By the co-side theorem, Vi+kPi PiVi+m+1 = S OViVi+m S OVi+m+1Vi , i = 1, ..., 2m + 1. Multiplying the above equations together and noticing that the (i + m)-th element in the denominator S OV(i+m)+m+1Vi+m = S OViVi+m is just the i-th element in the numerator, we prove the result. A B C D E F A B C D E F Figure 1-9 Example 1.9 (Menelaus’ Theorem) F, D, and E are three points on sides AB, BC, and CA of a triangle ABC respectively. Show that E, F, and D are collinear if and only if AF FB · BD DC · CE EA = −1. 12 Chapter 1. Geometry Preliminaries Proof. A transversal may meet two sides of a triangle and the third side produced, or all three sides produced (Figure 1-9). The following proof is valid for both cases. If E, F, and D are collinear, by the co-side theorem, AF FB = −S AEF S BEF , BD DC = −S BEF S CEF , CE EA = −S CEF S AEF . Then AF FB · BD DC · CE EA = −1. Conversely, let E, F, and D be points on AC, AB, and BC such that AF FB · BD DC · CE EA = −1. Let EF meet BC in H. Then we need only to show D = H. By what we just proved AF FB · BH HC · CE EA = −1. As a consequence BH HC = BD DC, i.e., D = H. We can extend the Menelaus’ theorem to m-sided polygons. Example 1.10 (Menelaus’ Theorem for an m-polygon) Let V1 · · · Vm be an m-polygon. A line XY meets ViVi+1, i = 1, ..., m in Pi. Then m Y i=1 ViPi PiVi+1 = (−1)m. Proof. By the co-side theorem, ViPi PiVi+1 = −S ViXY S Vi+1XY , i = 1, ..., m. Multiplying the m equations together, we obtain the result. We can use the area method to solve many geometry problems other than theorem prov-ing. Example 1.11 Let ABC be a triangle, D and E be two points on the lines AC and AB such that CD = uAD and AE = vBE. Let P be the intersection of BD and CE. Express PD PB in terms of u and v. A B C E D P Figure 1-10 Solution. We need to eliminate the con-structed points P, E, and D successively. By the co-side theorem we have2 PD PB = S DCE S BCE . 2Here we mention triangle DCE. The side DE of the triangle is not in the ﬁgure. Our method will add those kinds of auxiliary lines automatically. 1.4 The Co-side Theorem 13 Now the right-hand side of the above equality is free of the point P. We still need to eliminate the points E and D from it. By the co-side theorem again, we have: S DCE S BCE = S EDC S EAC · S EAC S EBC = DC AC · EA EB = −uAD AD −uAD · vEB EB = uv (u −1). Let us look at the special case when u = v = −1, i.e., D and E are the midpoints of the side AC and AB respectively. Then PD PB = −1 2. The following theorem follows from this fact immediately. Example 1.12 (The Centroid Theorem) The three medians of a triangle meet in a point, and each median is trisected by this point. Example 1.13 The same hypotheses as Example 1.11. Express S PBC S ABC in terms of u and v. Solution. By (1.3) on page 7 and the co-side theorem, we have S ABC S PBC = S APC S PBC + S ABP S PBC + S PBC S PBC = AE EB + AD DC + 1 = −v −1 u + 1 = u −1 −uv u . Therefore, S PBC S ABC = u u−1−uv. R Q P F E D A C B Figure 1-11 Example 1.14 Take three points D, E, and F on sides AC, AB, and BC of the triangle ABC such that CD AD = u, AE BE = v, BF CF = w. Let R = AF ∩BD, P = BD ∩EC, and Q = AF ∩CE. Express S PQR S ABC in terms of u, v, and w. Solution. By (1.3) on page 7 and Example 1.13, S PQR = S ABC −S PBC −S QCA −S RAB = (1 − u u −1 −uv − w w −1 −wu − v v −1 −wv)S ABC = (1 + wuv)2 (1 −u + uv)(1 −v + wv)(1 −w + wu)S ABC. Remark 1.15 14 Chapter 1. Geometry Preliminaries 1. In Example 1.14, if u = v = w = −1 2 then we have S PQR S ABC = 1 7. 2. As a consequence of Example 1.14, we have proved a stronger version of the Ceva theorem: lines AF, BD, and CE are concurrent if and only if S PQR = 0, i.e., if and only if uvw + 1 = 0 or CD AD · AE BE · BF CF = −1. L K B C A D F G Figure 1-12 Example 1.16 3 Let L be the intersection of AB and CD, K the intersection of AD and BC, F the intersection of BD and KL, and G the intersection of AC and KL. Then LF KF = LG GK . Proof 1. By the co-side theorem, we can eliminate point F: LF KF = S LBD S KBD . We need further to eliminate points K and L: S LBD = LB AB S ABD = S BCDS ABD S BCAD , S KBD = KB CB S CBD = S BADS CBD S BACD . Then LF KF = S BCAD S BACD . Similarly, we can prove that LG GK = S BCAD S BACD = LF KF. Proof 2. The following proof is shorter. LF KF = S LBD S KBD = S LBD S KBL S KBL S KBD = DA AK · LC DC = S DAC S AKC S LAC S DAC = S LAC S AKC = LG GK . Example 1.17 (Pappus’ Theorem) Let points A, B and C be on one line, and A1, B1 and C1 be on another line. Let P = AB1 ∩A1B, Q = AC1 ∩A1C, and S = BC1 ∩B1C. Show that P, Q, and S are collinear. 3This theorem is sometimes referred as the basic principle of the projective geometry. It can be used to deﬁne the harmonic sequence geometrically. For more details see page 372. 1.4 The Co-side Theorem 15 A B 1 A 1 B C 1 C P Q S Figure 1-13 Proof. We generally convert a problem of collinearity into a ratio problem in the following way. Let Z1 = PQ ∩BC1 and Z2 = PQ ∩B1C. We need only to prove that Z1 = Z2. This is equivalent to PZ1 QZ1 · QZ2 PZ2 = 1. (1) We can eliminate points Z2 and Z1 from (1) using the co-side theorem. PZ1 QZ1 · QZ2 PZ2 =S PBC1 S QBC1 · S QCB1 S PCB1 . (2) Now we want to eliminate points Q and P from the right-hand side of (2) using the follow-ing identities which can be easily obtained using the co-side theorem. S QBC1 = QC1 AC1 · S ABC1 = S A1CC1 ·S ABC1 S ACC1 A1 S QCB1 = QC A1C · S A1CB1 = S ACC1 ·S A1CB1 S ACC1 A1 S PBC1 = PB A1B · S A1BC1 = S ABB1·S A1BC1 S ABB1A1 S PCB1 = PB1 AB1 · S ACB1 = S A1BB1·S ACB1 S ABB1A1 . Substituting these into (2) and applying Proposition 1.4, we have S PBC1 S QBC1 · S QCB1 S PCB1 = S ABB1 S ACB1 · S A1BC1 S A1BB1 · S ACC1 S ABC1 · S A1CB1 S A1CC1 = AB AC · A1C1 A1B1 · AC AB · A1B1 A1C1 = 1. The following extension of the co-side theorem is also very useful. A B P Q R Figure 1-14 Proposition 1.18 Let R be a point on line PQ. Then for any two points A and B S RAB = PR PQ S QAB + RQ PQ S PAB. 16 Chapter 1. Geometry Preliminaries Proof. By the co-side theorem S RAP S QAP = PR PQ, S RPB S QPB = PR PQ. By (1.3) on page 7, S RAB = S PAB + S RAP + S RPB = S PAB + PR PQ (S APQ −S BPQ) = S PAB + PR PQ (S APB + S ABQ) = (1 −PR PQ )S PAB + PR PQ S QAB = PR PQ S QAB + RQ PQ S PAB. Example 1.19 (0.116, 4, 7) The circumdiameters AP, BQ, CR of a triangle ABC meet the sides BC, CA, AB in the points K, L, M. Show that (KP/AK) + (LQ/BL) + (MR/CM) = 1. A B C O P Q R K L M Figure 1-15 Proof. By the co-side theorem, RM CM =S ABR S ABC , QL BL =−S ACQ S ABC , PK AK =S BCP S ABC . Note that O is the midpoint of AP, BQ, and CR. Us-ing Proposition 1.18, we can eliminate points R, Q, and P. S ABR =2S ABO −S ABC, S ACQ =2S ACO + S ABC, S BCP =2S BCO −S ABC. Thus RM MC + QL LB + PK KA = −(2S BCO + 2S AOC + 2S ABO −3S ABC) S ABC = S ABC S ABC = 1. Remark 1.20 In summary, we see that to prove a geometry theorem using the area method systematically, we generally follow three steps: ﬁrst, formulate the geometry theorem in a constructive way and state the conclusion of the theorem as an expression in areas and ratios of directed line segments; second, based on the basic propositions about areas, try to eliminate the points from the conclusion in the reverse order in which the points are introduced; ﬁnally state whether the conclusion is true or not. Exercises 1.21 1. There are 48 pairs of co-side triangles in each diagram of Figure 1-8. Try to count how many co-side triangles are there in Figure 1-12. 1.5 Parallels 17 2. If A, B, C, and D are on the same line, then for points P and Q such that S PCQD , 0, we have S PAQB/S PCQD = AB/CD. 3. The same hypotheses as Example 1.6. Show that AP AD + BP BE + CP CF = 2. 4. Let ABCD be a quadrilateral and O a point. Let E, F, G, and H be the intersections of lines AO, BO, CO, and DO with the corresponding diagonals BD, AC, BD, and AC of the quadrilateral. Show that AH HC CF FA BE ED DG GB = 1. 5. (Lesening’s Theorem) Continuing from Example 1.17, let L1, L2, L3 be the intersections of lines OP and CC1, lines OQ and BB1, and lines OS and AA1. Show that L1, L2, L3 are collinear. 1.5 Parallels Deﬁnition 1.22 Let AB and CD be two non-degenerate lines. If AB and CD do not have any common point, we say that AB is parallel to CD. We use the notation AB ∥CD to denote the fact that A, B, C, and D satisfy one of the following conditions: (1) AB and CD are parallel; (2) A = B or C = D; or (3) A, B, C and D are on the same line. A parallelogram is an oriented quadrilateral ABCD such that AB ∥CD, BC ∥AD, and no three vertices of it are on the same line. Let ABCD be a parallelogram. It is clear that line AB and line DC have the same direction. Let ABCD be a parallelogram and P, Q two points on CD. We deﬁne PQ AB = PQ DC to be the ratio of two parallel line segments. We have the following very useful and simple result which is also basic to our method. Proposition 1.23 Let A, B, C, and D be four points. AB ∥CD if and only if S ABC = S ABD or S ADBC = 0. If we assume the area for a triangle to be ah/2, then Proposition 1.23 is obvious. However, we can prove it using Proposition 1.4 (see page 57 for details). With Proposition 1.23, we can prove many other theorems easily. A B C D O Figure 1-16 Example 1.24 Let O be the intersection of the two diagonals AC and BD of a parallelogram ABCD. Show that AO = OC, or AO OC = 1. 18 Chapter 1. Geometry Preliminaries The construction of the ﬁgure proceeds as follows. First we have three arbitrarily chosen points A, B and C. Then we take a point D such that AB = DC. Finally, we take the intersection of two lines AC and BD to obtain the point O. Thus we need to eliminate O from the expression AO OC, then eliminate the point D from the new expression. Here is the proof. Proof. AO OC = S ABD S BCD by the co-side theorem = S ABC S BCA (since AB ∥CD and AD ∥BC, S ABD = S ABC, S BCD = S BCA.) = S ABC S ABC = 1. A B X C Y Z Figure 1-17 Example 1.25 Three parallel lines cut two lines at A, B, C, and X, Y, Z respectively. Show that AB CB = XY ZY . This proposition is considered a very basic property of parallel lines in high school geometry. Here we can prove it very elegantly . Proof. By the co-side theorem and Proposition 1.23, we have AB CB = S ABY S CBY = S XBY S ZBY = XY ZY . A B C D P Q M Figure 1-18 Example 1.26 Lines AB and CD are parallel. Let P = AC ∩BD and Q = AD ∩BC. Line PQ intersects line AB at M. Show that M is the midpoint of AB, i.e., AM = MB. Proof 1. By the co-side theorem, we can eliminate M: AM MB = S PAQ S PQB. By the co-side theorem again, we can eliminate Q: S PAQ = AQ AD S PAD = S ABC S ABDC S PAD, S PQB = BQ BC S PCB = S BDA S BDCA S PCB. Now by Proposition 1.23, AM MB = S PAD S PCB = S PCD+S CAD S PCD+S CBD = 1 Proof 2. The following proof is shorter. AM MB = S PAQ S PQB = S PAQ S QAB · S QAB S PQB = PD DB · CA PC = S CPD S CDB · S CDA S CPD = S CDA S CDB = 1. 1.5 Parallels 19 A B R C Q P Figure 1-19 Example 1.27 (Pascalian Axiom4) Let A, B and C be three points on one line, and P, Q, and R be three points on another line. If AQ ∥RB and BP ∥QC then AP ∥RC. Proof. We need to prove S RAP = S CAP. Since AQ ∥RB and BP ∥QC, by Proposition 1.23 we have S RAP = S RAQ + S APQ = S BAQ + S APQ = S BAPQ = S BAP + S BPQ = S BAP + S BPC = S CAP. S A B C A B C 1 1 1 Figure 1-20 Example 1.28 (Desargues’ Axiom) S AA1, S BB1, and SCC1 are three distinct lines. If AB ∥A1B1 and AC ∥A1C1 then BC ∥B1C1. Proof. We need to show S B1BC = S C1BC. Noting that AB ∥A1B1 and AC ∥A1C1, by the co-side theorem we can eliminate B1 and C1 S BCC1=CC1 CS S S BC = AA1 AS S S BC; S BCB1= = BB1 BS S S BC = AA1 AS S S BC. Hence S B1BC = S C1BC. 2 D 2 C 2 B 2 A 1 D 1 C 1 B 1 A D C B A Figure 1-21 Example 1.29 Let A1, B1, C1, and D1 be points on the sides CD, DA, AB, and BC of a parallelogram ABCD such that CA1/CD = DB1/DA = AC1/AB = BD1/BC = 1/3. Show that the area of the quadrilateral formed by the lines AA1 BB1, CC1, and DD1 is one thirteenth of the area of parallelogram ABCD. 4This result is a special case of Pappus’ theorem. It was referred to as Pascal’s theorem by Hilbert in and was used as an axiom in . 20 Chapter 1. Geometry Preliminaries Proof. By the co-side theorem, S ABCD S ABA2 = 2 · S ABA2 + S A2BD + S A2AD S ABA2 = 2(1 + DB1 B1A + S AA1D S ABA1 ) = 2(1 + 1 2 + 2 3) = 13 3 Then S A2B2C2D2 S ABCD = S ABCD −S ABA2 −S BCB2 −S CDC2 −S DAD2 S ABCD = 1 −12 13 = 1 13. P Q A B C D Figure 1-22 Proposition 1.30 Let ABCD be a parallelo-gram, P and Q be two points. Then S APQ + S CPQ = S BPQ + S DPQ or S PAQB = S PDQC. Proof. Let O be the intersection of AC and BD. Since O is the midpoint of AC, by Propo-sition 1.18, S APQ + S CPQ = 2S OPQ. For the same reason, S BPQ + S DPQ = 2S OPQ. We have proved the ﬁrst formula. The second formula is just another form of the ﬁrst one. Exercises 1.31 1. In Example 1.28, let AA1 ∥BB1 ∥CC1 are three parallel lines. If AB ∥A1B1 and AC ∥A1C1 then BC ∥B1C1. 2. Let l be a line passing through the vertex of M of a parallelogram MNPQ and intersect-ing the lines NP, PQ, and NQ at points R, S , and T. Show that 1/MR+1/MS = 1/MT. 3. The diagonals of a parallelogram and those of its inscribed parallelogram are concurrent. 4. Use the same notations as Example 1.29. If CA1/CD = DB1/DA = AC1/AB = BD1/BC = r then S A2B2C2D2 S ABCD = r2 r2−2r+2. 5. Let P, Q be the midpoints of the diagonals of a trapezoid ABCD. Then PQ is half of the diﬀerence of the two parallel sides of ABCD. 6. A line parallel to the base of trapezoid ABCD meets its two sides and two diagonals at H, G, F, and E. Show that EF = GH. 1.6 The Co-angle Theorem In this section, we will discuss more basic theorems, the co-angle theorems, in the area method. However we have not incorporated these theorems into our computer program. 1.6 The Co-angle Theorem 21 An angle is the ﬁgure consisting of a point O and two rays emanating from O. The symbol for angle is ∠. We assume the value of the angle is ≥0 and ≤180◦. In a triangle ABC, we use ∠A, ∠B, and ∠C to represent the three inner angles of the triangle at vertices A, B, and C respectively. Then we have the following formula for the area of a triangle ABC (for details see Section 1.8 below.) (1.4) ▽ABC = 1 2AB · BC sin(∠B) = 1 2AC · CB sin(∠C) = 1 2AB · AC sin(∠A). If ∠ABC = ∠XYZ or ∠ABC + ∠XYZ = 180◦, we say △ABC and △XYZ are co-angle trian-gles. B A C X Z 1 Z Figure 1-23 Proposition 1.32 (The Co-angle Theorem) If ∠ABC = ∠XYZ or ∠ABC + ∠XYZ = 180◦, we have ▽ABC ▽XYZ = AB·BC XY·YZ. Proof 1. This is a consequence of (1.4). Proof 2. The following proof uses the co-side theorem only. Without loss of generality, we assume that B = Y and Z is on line BC (Figure 1-23). If ∠ABC = ∠XYZ, X is on line AB. We have ▽ABC ▽XYZ = ▽ABC ▽ABZ · ▽ABZ ▽XYZ = BC BZ · AB XY = AB · BC XY · YZ . If ∠ABC + ∠XYZ = 180◦, the result can be proved similarly. The co-angle theorem, though obvious, can be used to prove nontrivial geometry theo-rems easily. Example 1.33 In triangle ABC, if ∠B = ∠C then AB = AC. Proof. By the co-angle theorem, 1 = ▽ABC ▽ACB = AB·BC AC·BC = AB AC. Example 1.34 If ABCD is a parallelogram, then AB = CD. Proof. Since ABCD is a parallelogram, we have ∠CAB = ∠ACD and ▽ABC = ▽BCD = ▽ACD. By the co-angle theorem, we have 1 = ▽ABC ▽ACD = AC · AB AC · CD = AB CD. Example 1.35 F is a point on side BC of △ABC such that AF is the bisector of ∠BAC. Then AB AC = FB FC. 22 Chapter 1. Geometry Preliminaries Proof. By the co-side and co-angle theorems, FB FC = ▽FAB ▽FAC = FA · AB FA · AC = AB AC. Example 1.36 In triangles ABC and XYZ, if ∠A = ∠X, ∠B = ∠Y then AB XY = BC YZ = CA ZX. Proof. From the hypotheses, we also have ∠C = ∠Z. By the co-angle theorem, ▽ABC ▽XYZ = AB · AC XY · XZ = AB · BC XY · YZ = AC · CB XZ · ZY . The result follows from the above formula immediately. Two triangles are said to be similar if their corresponding angles are equal. The above example implies that the corresponding sides of two similar triangles are proportional. Example 1.37 In triangle ABC, AB = AC, AB⊥AC, and M is the midpoint of AB. The perpendicular from A to CM meets BC in P. Show that PC = 2PB. A B C M P Figure 1-24 Proof. Note that ∠PAC = ∠AMC and ∠PAB = ∠ACM. By the co-side and co-angle theorems, PC PB = ▽PAC ▽PAB = ▽PAC ▽MAC · ▽MAC ▽PAB = PA · AC MA · MC · AC · MC PA · AB = AC · AC MA · AB = AB · AB MA · AB = AB MA = 2. Example 1.38 AM is the median of triangle ABC. D, E are points on AB, AC such that AD = AE. DE and AM meet in N. Show that DN NE = AC AB. A B C M E D N Figure 1-25 Proof. From BM = MC, we have ▽ABM = ▽ACM. Now by the co-angle theorem, DN NE = ▽ADN ▽ANE = ▽ADN ▽ABM · ▽ACM ▽ANE = AD · AN AB · AM · AC · AM AN · AE = AD · AC AB · AE = AC AB. 1.6 The Co-angle Theorem 23 Example 1.39 Four rays passing through a point O meet two lines sequentially in A, B, C, D and P, Q, R, S . Show that AB·CD AD·BC = PQ·RS PS ·QR. O A P B Q C D R S Figure 1-26 Proof. By the co-side and co-angle theorem AB · CD · PS · QR AD · BC · PQ · RS = AB AD · CD BC · PS PQ · QR RS = ▽OAB ▽OAD · ▽OCD ▽OBC · ▽OPS ▽OPQ · ▽OQS ▽ORS = ▽OAB ▽OPQ · ▽OCD ▽ORS · ▽OPS ▽OAD · ▽OQR ▽OBC = OA · OB · OC · OD · OP · OS · OQ · OR OP · OQ · OR · OS · OA · OD · OB · OC = 1. Example 1.40 AM is the median of triangle ABC. A line meets the rays AB, AC, and AM at P, Q, and N respectively. Show that AM AN = 1 2( AC AQ + AB AP). A B C M P Q N Figure 1-27 Proof. By the co-angle theorem, AP · AQ AB · AC = ▽APQ ▽ABC = ▽APN ▽ABC + ▽AQN ▽ABC = ▽APN 2 ▽ABM + ▽AQN 2 ▽ACM = AP · AN 2AB · AM + AQ · AN 2AC · AM = 1 2(AP AB + AQ AC ) AN AM Multiplying the two sides of the above formula by AM·AB·AC AN·AP·AQ, we obtain the result. B C A M N Figure 1-28 Example 1.41 Take two points M, N on the sides AB, AC of triangle ABC such that ∠MCB = ∠NBC = ∠A/2. Then BM = CN. Proof. Since ∠BMC = ∠A + ∠ACB −∠A/2 and ∠CNB = ∠A + ∠ABC −∠A/2, we have ∠BMC + ∠CNB = ∠A + ∠ACB + ∠ABC = 180◦. Hence in triangles BMC and CNB, we have ∠MCB = ∠NBC, ∠BMC + ∠CNB = 180◦. By the co-angle theorem, MC · BC NB · BC = ▽BMC ▽CNB = BM · MC CN · NB 24 Chapter 1. Geometry Preliminaries i.e., 1 = BM/CN. B C A D Figure 1-29 Example 1.42 (Pythagorean Theorem5) In triangle ABC, let BC = a, CA = b, AB = c, and ∠ACB = 90◦. Show that a2 + b2 = c2. Proof. Let the altitude on side AB be CD = h. Then ▽ACD + ▽BCD = ▽ABC, i.e., ▽ACD ▽ABC + ▽BCD ▽ABC = 1. Since ∠ACD = ∠ABC, ∠BCD = ∠CAB, by the co-angle theorem we have bh/(ac) + ah/(bc) = 1. Also note that ab = 2 ▽ABC = ch, we have h = ab/c. Substituting this into the above equation, we prove the result. Proposition 1.43 (The Co-angle Inequality) If ∠ABC > ∠XYZ and ∠ABC + ∠XYZ < 180◦then ▽ABC ▽XYZ > AB·BC XY·YZ. Proof 1. This is a consequence of (1.4) on page 21 and the property of the sine function. Proof 2. The following proof does not use the trigonometric functions. Draw an isosceles triangle PQR such that QP = PR and ∠QPR = ∠ABC −∠XYZ. Produce QR to S such that ∠RPS = ∠XYZ. Then we have ∠QPS = ∠ABC and ▽QPS > ▽RPS . By the co-angle theorem Q R P S Figure 1-30 ▽ABC ▽XYZ = ▽ABC ▽QPS ▽QPS ▽RPS ▽RPS ▽XYZ = AB · BC QP · PS ▽QPS ▽RPS RP · PS XY · YZ = AB · BC XY · YZ ▽QPS ▽RPS (QP = RP) > AB · BC XY · YZ . Corollary 1.44 1. If ∠ABC > ∠XYZ and ∠ABC + ∠XYZ > 180◦then ▽ABC ▽XYZ < AB·BC XY·YZ. 5In Chinese literature, this theorem is called the Gou-Gu theorem and is attributed to Shang-Gao (1100 B.C.). This celebrated theorem is one of the most important theorems in the whole realm of geometry. There are about 370 proofs for this theorem in . 1.6 The Co-angle Theorem 25 2. (The Converse of the Co-angle Theorem) If ▽ABC ▽XYZ = AB·BC XY·YZ then ∠ABC = ∠XYZ or ∠ABC + ∠XYZ = 180◦. Example 1.45 In triangle ABC, if ∠B > ∠C then AC > AB. Proof. By the co-angle inequality, 1 = ▽ABC ▽ACB > AB·BC AC·BC = AB AC. . Corollary 1.46 1. Among the segments from a point to any point on a line, the perpendicular is the shortest. 2. The area of any quadrilateral is less than or equal to half of the products of its two diagonals. Example 1.47 If AB ≥AC and P is a point between B and C then AB > AP. A B C P Figure 1-31 Proof. Since AB ≥AC, we have ∠ACB ≥∠ABC. Then ∠APB = ∠ACB + ∠CAP > ∠ABC = ∠ABP. By the co-angle inequality, we have AB > AP. Example 1.48 The sum of any two sides of a triangle is larger than the third one. A B C D Figure 1-32 Proof. Produce BC to D such that CD = AC. Then ∠BDA = ∠CDA = ∠CAD = ∠BAD−∠BAC < ∠BAD. In triangle ABD, by Example 1.45 we have AB < BD = BC + CD = BC + AC. Example 1.49 6 Take three points M, K, and L on the three sides AB, BC, and CA of triangle ABC respectively. Show that at least one of ▽AML, ▽BMK, and ▽CKL is less that 1 4▽ABC. Proof. Let AM = r ·AB, BK = s· BC, and CL = t·AC. Then r, s, and t are positive numbers less that 1. Also BM = (1 −r)AB, CK = (1 −s)BC, and AL = (1 −t)AC. By the co-angle theorem, 6This is a problem from the 1966 International Mathematical Olympiad. 26 Chapter 1. Geometry Preliminaries A B C M L K Figure 1-33 ▽AML ▽ABC · ▽BMK ▽ABC · ▽CKL ▽ABC = AM AB · AL AC · BM AB · BK BC · CK BC · CL AC = r · (1 −t) · (1 −r) · s · (1 −s) · t = r(1 −r) · s(1 −s) · t(1 −t) ≤(1/4) · (1/4) · (1/4). As a consequence, one of ▽AML ▽ABC , ▽BMK ▽ABC , and ▽CKL ▽ABC must be less than 1/4. A B C P X Y Z N M Figure 1-34 Example 1.50 (Erd¨ os’ Inequality) P is a point inside the triangle ABC. Let the distances between the point P and the lines BC, CA, and AB be x, y, and z respec-tively. Show that PA + PB + PC ≥2(x + y + z). Proof. Take points N and M on AB and AC such that AM = AB, AN = AC. Then MN = BC and z · AC + y · AB = z · AN + y · AM = 2(▽APN + ▽APM) ≤PA · MN = PA · BC i.e., z · (AC/BC) + y · (AB/BC) ≤PA. Similarly x · (AB/AC) + z · (BC/AC) ≤PB; x · (AC/AB) + y · (BC/AB) ≤PC. Adding the three equations together, we have PA + PB + PC ≥x · (AB AC + AC AB) + y(BC AB + AB BC) + z · (BC AC + AC BC) ≥2(x + y + z). Example 1.51 (The Steiner-Lehmus Theorem7) In triangle ABC, if the bisectors for angles B and C are equal then AB = AC. B C I A E D P Figure 1-35 Proof. Without loss of generality, we assume AB ≥ AC. Then ∠ACB ≥ABC. Let I be the intersection of BD and CE. Then ∠DCI ≥EBI. Take a point P between DI such that ∠PCI = ∠EBI. We need only to show that P = D. In triangles PCI and EBI, ∠PCI = ∠EBI, ∠PIC = ∠EIB. Then ∠CPI = ∠BEI. In triangles PBC and EBC, ∠CPB = ∠BEC, ∠PCB ≥∠EBC. By the co-angle theorem and the co-angle inequality, PC · PB BE · CE = ▽PBC ▽EBC ≥PC · BC BE · BC, 1.7 Pythagoras Diﬀerences 27 i.e., PB CE ≥1. Therefore PB ≥CE = DB. Since P is on DB, we have P = D. Exercises 1.52 1. We mentioned earlier that rarely are there congruent and similar triangles in the diagram of a geometry theorem, but there are many co-side triangles. There are also many co-angle triangles in any ﬁgure. Try to count the co-angle triangles in Figures 1-23, 1-24, and 1-25. This is why the co-angle theorem works well for many geometry theorems. 2. On the two sides AB and AC two squares ABDE and ACFG are erected externally. Show that ▽ABC = ▽AEG. 3. In triangle ABC, the bisector of the external angle (at vertex A) meets BC in D. Show that AB AC = BD CD. 4. In triangle ABC, the bisectors of the inner and external angles (at vertex A) meet BC in D and E. Show that BD CD = BE CE. 1.7 Pythagoras Diﬀerences To solve geometry problems involving perpendiculars and congruence of line segments, we need to introduce a new geometry quantity: the Pythagoras diﬀerence. To do that, we ﬁrst introduce the concept of co-area of triangles. A B Q P C Figure 1-36 On side AB of a triangle ABC, a square ABPQ is erected such that S ABC and S ABPQ have the same sign (Figure 1-36). The co-area CBAC is a real number such that CBAC = ( ▽ACQ if ∠A ≤90◦; −▽ACQ if ∠A > 90◦. Similarly CABC is equal to ▽BPC or −▽BPC according to whether ∠B is acute or obtuse. Generally speaking, CBAC, CABC, and CACB are diﬀerent. But we have CBAC = CCAB, CABC = CCBA, and CACB = CBCA. Proposition 1.53 For triangle ABC, we have CABC + CBAC = AB2/2. Proof. As in Figure 1-36, if both ∠A and ∠B are acute then CABC + CBAC = ▽BPC + ▽ACQ = ▽ABPQ/2 = AB2/2. 7For interesting extensions of this theorem, see . 28 Chapter 1. Geometry Preliminaries If ∠A is obtuse and ∠B is acute then CABC + CBAC = ▽BPC −▽ACQ = ▽ABPQ/2 = AB2/2. If ∠A is acute and ∠B is obtuse then CABC + CBAC = −▽BPC + ▽ACQ = ▽ABPQ/2 = AB2/2. By Proposition 1.53, CABC + CBAC = AB2 2 , CBCA + CABC = BC2 2 , CBAC + CBCA = CA2 2 . From the above equations, we have CABC = (AB2 + BC2 −AC2)/4, CBAC = (AB2 + AC2 −BC2)/4, CACB = (AC2 + BC2 −AB2)/4. Deﬁnition 1.54 We call AB2 + BC2 −AC2 the Pythagoras diﬀerence of triangle ABC with regard to B, and denote it by PABC, i.e., PABC = 4CABC = AB2 + BC2 −AC2. Proposition 1.55 (Pythagorean Theorem) 1. PABC = 0 if and only if ∠ABC = 90◦. 2. PABC > 0 if and only if ∠ABC < 90◦. 3. PABC < 0 if and only if ∠ABC > 90◦. Proof. As shown in Figure 1-36, it is clear that ▽BPC = 0 if and only if ∠ABC = 90◦. The second and third cases come from the deﬁnition of the co-areas directly. Proposition 1.56 (The Pythagoras Diﬀerence Theorem) If ∠ABC , 90◦, we have 1. ∠ABC = ∠XYZ if and only if PABC PXYZ = AB·BC XY·YZ; 2. ∠ABC + ∠XYZ = 180◦if and only if PABC PXYZ = −AB·BC XY·YZ. Proof. This proposition is a consequence of the deﬁnition of Pythagoras diﬀerence and the converse of the co-angle theorem on page 25. 1.7 Pythagoras Diﬀerences 29 Example 1.57 In triangle ABC, let BC = a, CA = b, and AB = c. Express the length m of the median CM in terms of a, b, and c. A B C M Figure 1-37 Solution. Without loss of generality, let us assume that ∠A is not a right angle. Applying Proposition 1.56 to triangles BAC and MAC, we have PMAC PBAC = MA · CA AB · CA = 1 2 (AB = 2MA). Hence PBAC = 2PMAC. i.e., b2 + c2 −a2 = 2(b2 + c2/4 −m2). Now it is clear that m2 = (a2 + b2 −c2/2)/2. Example 1.58 In triangle ABC, let BC = a, CA = b, and AB = c. Express the length h of the altitude CD in terms of a, b, and c. A B C D Figure 1-38 Solution. Without loss of generality, we assume that ∠A is acute. Applying Proposition 1.56 to triangles DAC and BAC, we have PDAC/PBAC = (AD · b)/(b · c) = AD/c. Let x = AD. By the Pythagorean theorem, b2 −h2 = x2. Then PDAC = x2 + b2 −h2 = 2x2. Thus x/c = PDAC/PBAC = 2x2/(b2 + c2 −a2), i.e., x = (b2 + c2 −a2)/(2c). Since x2 = b2 −h2, we obtain the ﬁnal result: h2 = b2 −(b2 + c2 −a2)2/(4c2). Example 1.59 In triangle ABC, let BC = a, CA = b, and AB = c. Express the length of the bisector CF in terms of a, b, and c. A B C F Figure 1-39 Solution. We assume that ∠A , 90◦. Applying Proposition 1.56 to triangles BAC and FAC PFAC/PBAC = (AF · b)/b · c = AF/c. 30 Chapter 1. Geometry Preliminaries By Example 1.36, AF/BF = b/a; hence AF = (bc)/(a + b). Let x = CF. We have AF/c = PFAC/PBAC = (AF2 + b2 −x2)/(b2 + c2 −a2). Substituting AF = bc/(a + b) into the above equation, we obtain x2 = ab((a + b)2 −c2)/(a + b)2. We deﬁne the Pythagoras diﬀerence for an oriented quadrilateral ABCD as follows. PABCD = AB2 −BC2 + CD2 −DA2. It is easy to derive the following properties for the Pythagoras diﬀerences of quadrilaterals. 1. PABCD = PCDAB = PBADC = PDCBA. 2. PABCD = −PBCDA = −PDABC = −PADCB = −PCBAD. 3. PABCD = PBAC −PDAC = PABD −PCBD = PDCA −PBCA = PCDB −PADB. 4. PABBC = PABC, PAABC = −PBAC, PABCC = −PACB, PABCA = PBAC. 5. PABAC = 0, PABCB = 0. 6. PAPBQ + PBPCQ = PAPCQ. (The additivity of the Pythagoras diﬀerence.) Proposition 1.60 If A, B, and C are collinear, we have PABC = 2BA · BC. Proof. Since AC = AB −CB, we have PABC = AB 2 + CB 2 −AC 2 = AB 2 + CB 2 −(AB −CB)2 = 2BA · BC. Deﬁnition 1.61 We use the notation AB⊥CD to denote that four points A, B, C, and D satisfy one of the following conditions: line AB is perpendicular to line CD; A = B; or C = D. The following result gives a criterion for AB⊥CD using the Pythagoras diﬀerence. A B D C M Figure 1-40 Proposition 1.62 AC⊥BD if and only if PABCD = PABD −PCBD = 0. Proof. Let M and N be the feet of the perpendicu-lars from points A and C to line BD. Then by the Pythagorean theorem, 1.7 Pythagoras Diﬀerences 31 PABD = AB 2 + BD 2 −AD 2 = AM 2 + BM 2 + BD 2 −AM 2 −MD 2 = BM 2 + BD 2 −(BD −BM)2 = 2BM · BD. Similarly PCBD = 2BN · BD. We have PABD = PCBD if and only if MB = NB, i.e., M = N which is equivalent to AC⊥BD. Proposition 1.63 Let P and Q be the feet of the perpendiculars from points A and C to BD. Then PABCD = 2QP · BD. Proof. By Propositions 1.62 and 1.60, PABCD = PABD −PCBD = PPBD −PQBD = 2BP · BD −2BQ · BD = 2QP · BD. Proposition 1.64 Let ABCD be a parallelogram. Then for two points P and Q, we have PPAQB = PPDQC. Proof. This is a consequence of Proposition 1.63. Proposition 1.65 Let D be the foot drawn from point P to a line AB (A , B). Then we have AD DB = PPAB PPBA , AD AB = PPAB 2AB 2, DB AB = PPBA 2AB 2. Proof. By Proposition 1.62, PPAB = PDAB = 2AB · AD, PPBA = PDBA = 2BA · BD. The result is clear now. Proposition 1.66 Let AB and PQ be two nonperpendicular lines and Y be the intersection of line PQ and the line passing through A and perpendicular to AB (Figure 1-41). Then PY QY = PPAB PQAB , PY PQ = PPAB PPAQB , QY PQ = PQAB PPAQB . 32 Chapter 1. Geometry Preliminaries A B P Q Y P Q 1 1 Figure 1-41 Proof. We need only to show the ﬁrst equa-tion. Let P1 and Q1 be the orthogonal projec-tions from P and Q to line AB respectively. By Proposition 1.62, PPAB PQAB = PP1AB PQ1AB = AP1·AB AQ1·AB = AP1 AQ1 = PY QY . Example 1.67 (The Orthocenter Theorem) Show that the three altitudes of a triangle are con-current. C B A E F H Figure 1-42 Proof. Let the two altitudes AF and BE of triangle ABC meet in H. We need only to prove CH⊥AB, i.e., PACH = PBCH. By Proposition 1.62, PACH = PACB = PBCA = PBCH. F E D O C B A Figure 1-43 Example 1.68 (Orthocenter-dual) Let ABCO be a quadrilateral. From point O perpendiculars to OA, OB, and OC are drawn which meet BC, CA, and AB in D, E, and F respectively. Show that D, E, and F are collinear. Proof. Let DE and AB meet in Z. We need only to show Z = F, i.e., AF BF = AZ BZ. By Proposition 1.66, we can eliminate F: AF BF = PAOC PBOC. By the co-side theorem, we can eliminate Z: BZ AZ = S BDE S ADE . To eliminate E, by Proposition 1.66 we have S BDE = EC AC S BDA = PCOB PCOAB S BDA, S ADE = EA AC S ACD = PAOB PCOAB S ACD. Then AF BF · BZ AZ = PAOC PBOC · S BDAPCOB PCOAB · PCOAB PAOBS ACD = PAOC PAOB · S BDA S ACD = PAOC PAOB · BD CD = PAOC PAOB · PAOB PAOC = 1. A B C D E F G Figure 1-44 Example 1.69 On the two sides AB and AC of trian-gle ABC, two squares ABDE and ACFG are drawn externally. Show that BG⊥CE. 1.8 Trigonometric Functions 33 Proof. PBCGE = PBCGA + PBAGE (Additivity) = PBCAA + PACGA + PBAAE + PAAGE (Additivity) = −PBAC −PGAE (AG⊥AC, AB⊥AE) = 0 (∠BAC + ∠GAE = 180◦) B C A D J M N Figure 1-45 Example 1.70 In triangle ABC, take a point J on the altitudes AD. Lines BJ and CJ meet AC and AB in N and M respectively. Show that ∠MDA = ∠ADN. Proof. To prove ∠MDA = ∠ADN, by the co-angle theorem and Proposition 1.56 we need only to show PMDA/S MDA = PADN/S ADN. S MDA = AM AB S BDA = S AJC S AJBC S BDA the co-side theorem; S ADN = AN AC S ADC = S ABJ S ABCJ S ADC the co-side theorem; PADN = NC AC PADA = S BCJ S ABCJ PADA Proposition 1.65; PMDA = MB AB PADA = S BCJ S AJBC PADA Proposition 1.65; Then PADN PMDA S MDA S ADN = S BDA S ADC S AJC S ABJ = BD DC DC BD = 1. Exercises 1.71 1. In Example 1.69, show that BG = CE. 2. In Example 1.69, let M be the midpoint of BC. Show that AM⊥EG and EG = 2AM. 3. The sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of the four sides of the given parallelogram. (Use Example 1.57.) 4. In the quadrilateral ABCD, if ∠ABC = ∠CDA = 90◦and P is the intersection of AC and BD then PBAD PBCD = AP CP. 1.8 Trigonometric Functions We have discussed two major geometric quantities: the area and the Pythagoras diﬀer-ence. Area is a well-known concept and has been used since the time of Euclid. On the other hand, the Pythagoras diﬀerence is unfamiliar to most readers. In this section, we will introduce the trigonometric functions and use them to represent areas and Pythagoras diﬀerences. 34 Chapter 1. Geometry Preliminaries A B C Figure 1-46 The sine and cosine functions can be deﬁned in the usual way. Let ABC be a triangle with ∠B = 90◦. Then sin(∠A) = BC AC, cos(∠A) = AB AC. We can also use the area to deﬁne trigonometric functions. Deﬁnition 1.72 sin(∠A) is twice the signed area of a triangle ABC such that AB = AC = 1. It is easy to see that the two deﬁnitions are consistent. The following properties of the trigonometric functions can be derived from the deﬁnition directly. 1. sin(0) = sin(180◦) = 0; sin(90◦) = 1. 2. sin(∠X) = sin(180◦−∠X). 3. sin(∠A) = sin(∠B) if and only if ∠A = ∠B or ∠A + ∠B = 180◦. 4. Let us assume that ∠A + ∠B ≤180◦. Then A < B implies sin(∠A) < sin(∠B), and vise versa. (This is a consequence of Proposition 1.43.) Proposition 1.73 S ABC = 1 2AB · AC · sin(∠A). Proof. Take two points X and Y on AB and AC respectively such that AX = AY = 1. By the co-angle theorem ▽ABC = ▽AXY AB · AC AX · AY = ▽AXY · AB · AC = 1 2AB · AC · sin(∠A). Applying the above proposition to the three angles of triangle ABC respectively, we have ▽ABC = 1 2bc sin(∠A) = 1 2ac sin(∠B) = 1 2ab sin(∠C). As a direct consequence, we have Proposition 1.74 (The Sine Law) In triangle ABC, let BC = a, CA = b, and AB = c. Then sin(∠A)/a = sin(∠B)/b = sin(∠C)/c. The following is a very useful property of trigonometric functions. 1.8 Trigonometric Functions 35 A B P C Figure 1-47 Proposition 1.75 (The Visual Angle Theorem) Emanating from P, there are three rays PA, PB, and PC such that ∠APC = α, ∠CPB = β, and ∠APB = γ = α + β < 180◦. Then A, B, and C are collinear if and only if sin(∠γ)/PC = sin(∠α)/PB + sin(∠β)/PA. Proof. If points A, B, and C are collinear then we have ▽PAB = ▽PAC + ▽PCB, i.e., PA · PB · sin(∠γ) = PA · PC · sin(∠α) + PB · PC · sin(∠β). We obtain the result by dividing PA · PB · PC from both sides of the above formula. Conversely, from the above formula, we have ▽PAB = ▽PAC + ▽PCB which implies ▽ABC = \| ▽PAB −▽PAC −▽PCB\| = 0, i.e., A, B, and C are collinear. Proposition 1.75 has many applications. A B C E Figure 1-48 Example 1.76 In triangle ABC, ∠ACB = 120◦; CE is the bisector of angle C. Show that 1/CE = 1/CA + 1/CB. Proof. By Proposition 1.75, sin(120◦)/CE = sin(60◦)/CA + sin(60◦)/CB. Since sin(120◦) = sin(180◦−120◦) = sin(60◦), we obtain the result by dividing sin(60◦) from both sides of the above equation. C B A D Figure 1-49 Example 1.77 In the right triangle ABC , let BC = a, AC = b, and CD = h be the altitude on the hy-potenuse. Show that 1/h2 = 1/a2 + 1/b2. Proof. Let α = ∠ACD, β = ∠BCD. By Proposition 1.75 sin(α + β)/h = sin(α)/a + sin(β)/b. We have sin(α + β) = sin(∠ACB) = 1, sin(α) = sin(∠CBD) = h/a, sin(β) = sin(∠CAD) = h/b. Substituting these into the above equation, we obtain the result. 36 Chapter 1. Geometry Preliminaries Deﬁnition 1.78 Let ∠A be an angle. The cosine of ∠A is deﬁned as follows. cos(∠A) = ( sin(90◦−∠A) if ∠A ≤90◦; −sin(∠A −90◦) if ∠A > 90◦. Proposition 1.79 (The Cosine Law) In triangle ABC, we have PABC = 2AB · CB · cos(∠B). Proof. In Figure 1-36, if ∠B ≤90◦ CABC = 1 2AB · BC · sin(90◦−∠B) = 1 2AB · BC · cos(∠B). Thus PABC = 4CABC = 2AB · BC · cos(∠B). Other cases can be proved similarly. Example 1.80 Let α and β both be acute angles. Show that sin(α + β) = sin(α) cos(β) + sin(β) cos(α). Proof. In Proposition 1.75, let us assume that PC⊥AB, PC = h, PA = b, and PB = a. Then sin(α + β) = (h/a) sin(α) + (h/b) sin(β) = sin(α) cos(β) + sin(β) cos(α). It is easy to show that the angles in the preceding example are not necessarily acute. In Example 1.80, let α + β = 90◦. We have the well-known formula: sin(α)2 + cos(α)2 = 1. Example 1.81 sin(30◦) = 1 2; sin(45◦) = √ 2/2; sin(60◦) = √ 3/2. Proof. In Proposition 1.80, setting α = β = 30◦, we have sin(60◦) = 2 sin(30◦) cos(30◦). Since cos(30◦) = sin(90◦−30◦) = sin(60◦), we have sin(30◦) = 1 2. Similarly setting α = 30◦, β = 60◦, we have sin(60◦) = √ 3/2; setting α = β = 45◦, we have sin(45◦) = √ 2/2. So far, the angles are always positive. To represent the signed areas using trigonometric functions, we need to introduce the oriented angle, which is also denoted by ∠. Deﬁnition 1.82 The oriented angle ) ∠ABC is a real number such that (1) the absolute value of ) ∠ABC is the same as the ordinary angle ∠ABC, and (2) ) ∠ABC has the same sign with S ABC. 1.8 Trigonometric Functions 37 It is clear that for any angle α, we have −180◦< α ≤180◦. The arithmetic for oriented angles is understood to be mod 360◦and between −180◦and 180◦, Deﬁnition 1.83 We extend the deﬁnition of the sine and cosine functions to the oriented angles as follows. Let ) ∠α be a negative angle. Then sin() ∠α) = −sin(−) ∠α), cos() ∠α) = cos(−) ∠α). With the above deﬁnition, the properties for these two functions proved before are still valid. In particular, we have Proposition 1.84 S ABC = 1 2AB · BC · sin() ∠ABC), PABC = 2AB · BC · cos() ∠ABC). Example 1.85 (The Herron-Qin Formula8) In triangle ABC, 16S 2 ABC = 4AB 2 · CB 2 −P2 ABC. Proof. By Proposition 1.84, sin() ∠ABC) = 2S ABC AB·BC, cos() ∠ABC) = PABC 2AB·BC. Since sin() ∠ABC)2 + cos() ∠ABC)2 = 1, we have 4S 2 ABC AB2 · BC2 + P2 ABC 4AB2 · BC2 = 1. Thus 16S 2 ABC = 4AB 2 · CB 2 −P2 ABC. A B P Q O Y X Figure 1-50 Deﬁnition 1.86 The oriented angle between two directed lines PQ and AB, denoted by ) ∠(PQ, AB), is deﬁned as follows. Take points O, X, and Y such that OYQP and OXBA are parallelograms. Then we deﬁne ) ∠(PQ, AB) = ) ∠XOY. With the above deﬁnition, it is easy to check the following properties for the oriented angles. 1. ) ∠(PQ, AB) = −) ∠(AB, PQ). 2. ) ∠(PQ, AB) = 180◦+ ) ∠(QP, AB) = 180◦+ ) ∠(PQ, BA). 8Qin (1247 A.D.) is an ancient Chinese mathematician who discovered the area formula for triangles in exact form of this proposition . 38 Chapter 1. Geometry Preliminaries 3. ) ∠(PQ, AB) = ) ∠(QP, BA). 4. ) ∠(PQ, AB) + ) ∠(AB, CD) = ) ∠(PQ, CD). We can now represent the signed area and the Pythagoras diﬀerence of quadrilaterals using trigonometric functions. Proposition 1.87 S ABCD = 1 2AC · BD · sin() ∠(AC, BD)). Proof. Take a point X such that AXDB is a parallelogram. Then AX = BD. By Proposition 1.30, S ABCD = S AACX = S XAC = 1 2XA · AC · sin() ∠XAC) = 1 2AC · BD · sin() ∠(AC, BD)). Proposition 1.88 PABCD = 2AC · BD · cos() ∠(AC, DB)). Proof. Take a point X such that CXDB is a parallelogram. Then CX = BD. By Propo-sition 1.64, PABCD = −PCBAD = −PCCAX = PXCA = 2XA · AC · cos() ∠XCA) = 2AC · BD · cos() ∠(AC, DB)). Example 1.89 (The Herron-Qin Formula for Quadrilaterals) For a quadrilateral ABCD, we have 16S 2 ABCD = 4AC 2 · BD 2 −P2 ABCD. Proof. By Propositions 1.87 and 1.88, sin() ∠(AC, BD)) = 2S ABCD AC·BD , cos() ∠(AC, DB)) = PABCD 2AC·BD. Since sin() ∠(AC, BD))2 + cos() ∠(AC, DB))2 = 1, we have 4S 2 ABCD AC2 · BD2 + P2 ABCD 4AC2 · BD2 = 1. Thus 16S 2 ABCD = 4AC 2 · BD 2 −P2 ABCD. Exercises 1.90 1. Use Example 1.80 to prove the following formulas sin(α −β) = sin(α) cos(β) −cos(α) sin(β) cos(α + β) = sin(α) sin(β) −cos(α) cos(β) cos(α −β) = sin(α) sin(β) + cos(α) cos(β) 2. Let tan(α) = sin(α) cos(α). Show that tan(α + β) = tan(α) + tan(β) 1 −tan(α) tan(β), tan(α −β) = tan(α) −tan(β) 1 + tan(α) tan(β). 1.9 Circles 39 3. Prove the following version of the Herron-Qin formula for quadrilaterals. S 2 ABCD = (p −a)(p −b)(p −c)(p −d) −a · b · c · d · cos(∠B + ∠D 2 )2 where a = AB, b = BC, c = CD, d = DA, and P = (a + b + c + d)/2. (Hint. We have S 2 ABCD = (S ABC + S ACD)2. Then use Proposition 1.84.) 4. Continue from the above exercise. If ABCD is convex and cyclic then S 2 ABCD = (p − a)(p −b)(p −c)(p −d). 5. Continue from the above exercise. If ABCD also has an inscribed circle then S 2 ABCD = abcd. 1.9 Circles In this section, we will discuss another important geometric object: the circle. Points on the same circle are called co-circular or cyclic. I J X A B Figure 1-51 O J A B C Figure 1-52 Let J be a ﬁxed reference point on the circle, I the antipodal of J, and X a point on the tangent of the circle at point J such that S IJX > 0 (Figure 1-51). For any point A on the circle, ) ∠A is deﬁned to be the oriented angle ) ∠AJX, i.e., the (oriented) inscribed angle corresponding to the arc JA. Then for two points A and B on the circle, ) ∠B −) ∠A = ) ∠BJA (Figure 1-51). We now deﬁne the oriented chord on the circle. The absolute value of the oriented chord f AB is AB while its sign is the same as the sign of the oriented angle ) ∠BJA, or equivalently the same as the sign of S BJA. In Figure 1-51, since S BJA > 0, we have f AB > 0. The oriented chord f JA is always positive. Proposition 1.91 With the above notations, we have f AB = d sin() ∠BJA) = d sin() ∠B −) ∠A) 40 Chapter 1. Geometry Preliminaries f JA = d sin() ∠AJX) where d is the diameter of the circle. Proof. The signs of the two sides of the equation are equal. Thus we need only to check the absolute values of both sides of the equation. Since ∠BAJ = ∠BIJ or ∠BAJ +∠BIJ = 180◦, we have AB = sin(∠BJA) BJ sin ∠BAJ = sin(∠BJA) BJ sin ∠BIJ = sin(∠BJA) · IJ. Proposition 1.92 Let the diameter of the circumcircle of triangle ABC be d. Then S ABC = f AB · f BC · f AC/(2d). Proof. As shown in Figure 1-52, we have ∠ABC = ∠CJA or ∠ABC = 180◦−∠CJA. By Propositions 1.73 and 1.91, ▽ABC = 1 2AB · BC\| sin(∠ABC)\| = 1 2AB · BC · \| sin(∠CJA)\| = 1 2dAB · BC · CA. We still need to check whether the signs of both sides of the formula are the same. At ﬁrst, it is easy to see that when interchanging two vertices of the triangle, the signs of both sides of the equation will change. Therefore, we need only to check a particular position for the three vertices A, B, and C, e.g., the case shown in Figure 1-52. In this case, we have S ABC ≥0, S CJB ≥0, S BJA ≥0, and S CJA ≥0. Hence f AB ≥0, f BC ≥0, and f AC ≥0. Proposition 1.93 (The Co-circle Theorem) If the circumcircles of triangles ABC and XYZ are the same or equal then S ABC S XYZ = f AB· f BC· f CA f XY·f YZ·f ZX . Proof. This is a direct consequence of Proposition 1.92. Example 1.94 (Ptolemy’s Theorem) Let A, B, C, and D be four points on the same circle. Then f AB · g CD + g AD · f BC = f AC · g BD. Proof. We chose A to be the reference point. By Proposition 1.91, f AB · g CD + g AD · f BC −f AC · g BD = d2(sin() ∠B) sin() ∠D −) ∠C) + sin() ∠D) sin() ∠C −) ∠B) −sin() ∠C) sin() ∠D −) ∠B)) = 0. Example 1.95 (Brahmagupta’s Formula) Let A, B, C, and D be four points on the same circle. Then S 2 ABCD = (p −f AB)(p −f BC)(p −g CD)(p −g AD) where p = 1 2(f AB + f BC + g CD + g AD). 1.9 Circles 41 Proof. By Example 1.89 and Ptolemy’s theorem 16S 2 ABCD = 4 f AC 2 · g BD 2 −P2 ABCD = 4(f AB · g CD + g AD · f BC)2 −P2 ABCD = ((f AB + g CD)2 −(g AD −f BC)2)((g AD + f BC)2 −(f AB −g CD)2) = 16(p −f AB)(p −f BC)(p −g CD)(p −g AD). O A B C D E G F Figure 1-53 Example 1.96 A, B, C, and D are four co-circle points. For any point E on the same circle, lines DE and CE meet AB in F and G. Show that AF BF · BG AG is independent of E. Proof. By the co-side theorem and the co-circle theorem, AF BF · BG AG = S ADE S BDE S BCE S ACE = g AD · g DE · f EA · f BC · g CE · f EB g BD · g DE · f EB · f AC · g CE · f EA = g AD · f BC g BD · f AC which is independent of E. S Q P C B O A A B C Figure 1-54 1 1 1 Example 1.97 (Pascal’s Theorem) Let A, B, C, A1, B1, and C1 be six points on a circle. Let P = AB1 ∩A1B, Q = AC1 ∩A1C, and S = BC1 ∩B1C. Show that P, Q, and S are collinear. Proof. Note that the points P, Q, and R in this example are constructed in the same way as in Example 1.17 on page 14. Let Z1 = PQ ∩BC1 and Z2 = PQ ∩B1C. We need only to show G = PZ1 QZ1 · QZ2 PZ2 = 1. By Example 1.17, we have G = S ABB1S A1BC1S ACC1 S A1CB1 S ACB1 S A1BB1S ABC1 S A1CC1 . Now G = 1 follows from the co-circle theorem immediately. Example 1.98 (The General Butterﬂy Theorem) A, B, C, D, E, and F are six co-circle points. Lines CD and EF meet AB in M and N. Lines CF and DE meet AB in G and H. Show that MG AG · BH NH = BM AN . (Figure 1-55) 42 Chapter 1. Geometry Preliminaries O A B C D E F M N H G Figure 1-55 O A B D F G C E M N Figure 1-56 Proof. By the co-side theorem G = MG AG · BH NH · AN BM = S MCF S ACF S BDE S NDE AN AB · AB BM = S MCF S ACF S BDE S NDE S AFE S AFBE S ADBC S BDC . By the co-side theorem again, S MCF S DCF = MC DC = S ABC S ADBC , S FDE S NDE = FE NE = S AFBE S ABE . Then G = S DCFS ABCS BDES AFE S ACFS ABES FDES BDC . Now G = 1 follows from the co-circle theorem immediately. In Example 1.98, when G = H becomes the midpoint of AB, we obtain the ordinary butterﬂy theorem: Example 1.99 (The Butterﬂy Theorem) C, D, E, and F are four points on circle O (Figure 1-56). G is the intersection of DE and CF. Through G draw a line perpendicular to OG, meeting CD in M and EF in N. Show that G is the midpoint of MN. In the above examples, no Pythagoras diﬀerences occur in the proof. To deal with Pythagoras diﬀerences, we need to develop some new tools. We still assume that there is a ﬁxed reference point J on the circle whose diameter is d. O A B C J O A B C J Figure 1-57 1.9 Circles 43 Proposition 1.100 Let d be the diameter of the circumcircle of triangle ABC. Then PABC = 2f AB · f CB cos() ∠CJA). Proof. By the cosine theorem for oriented angles, \|PABC\| = 2AB · BC\| cos(∠ABC)\|. As shown in Figure 1-57, we have ∠ABC = ∠CJA or ∠ABC = 180◦−∠CJA. Then \|PABC\| = 2AB · BC · \| cos(∠CJA)\|. We still need to check whether the signs of both sides of the equation are equal. At ﬁrst, when interchanging the position of A and C, the signs of both sides of the equation will not change. Therefore, we need only to consider the following two cases: J is on the arc AC or on the arc AB. In the ﬁrst case, we have f AB ≥0, f CB ≤0. Since ) ∠ABC + ) ∠CJA = 180◦, PABC and cos() ∠CJA) always have opposite signs. Therefore the proposition is true in this case. In the second case, we have f AB ≤0, f CB ≤0. By the inscribe angle theorem, ) ∠CJA = ) ∠CBA. Thus PABC and cos() ∠CJA) always have the same sign. Proposition 1.101 (Co-circle Theorem for Pythagoras Diﬀerences) If the circumcircles of trian-gles ABC and XYZ are the same and PXYZ , 0, then PABC PXYZ = f AB · f BC cos() ∠AJC) f XY · f YZ cos() ∠XJZ) . Proof. This is a direct consequence of Proposition 1.100. A B C O D E F G Figure 1-58 Example 1.102 (Simson’s Theorem) Let D be a point on the circumcircle of triangle ABC. From D three perpendiculars are drawn to the three sides BC, AC, and AB of triangle ABC. Let E, F, and G be the three feet respectively. Show that E, F and G are collinear. Proof. By Menelaus’ theorem (Example 1.9 on page 11), we need only to show G = AG GB · BE EC · CF FA = −1. By Propositions 1.65 and 1.101 G = PBADPACDPDBC PABDPDACPDCB = f BA · g AD cos() ∠BJD) f AC · g CD cos() ∠AJD)g DB · f BC cos() ∠DJC) f AB · g BD cos() ∠AJD)g DA · f AC cos() ∠DJC)g DC · f CB cos() ∠DJB) = −1 44 Chapter 1. Geometry Preliminaries Example 1.103 Let V1, · · · , Vm, and P be m+1 co-circle points, Pi the orthogonal projections from P to ViVi+1, i = 1, ..., m. Show that m Y i=1 ViPi PiVi+1 = (−1)m where the subscripts are understood to be mod m. Proof. By Propositions 1.65 and 1.101 ViPi PiVi+1 = PPViVi+1 PPVi+1Vi = g PVi · g ViVi+1 cos() ∠PJVi+1) g PVi+1 · g Vi+1Vi cos() ∠PJVi) , i = 1, ..., m. Multiplying the m equations together, we obtain the result. Exercises 1.104 1. A, B, C, and D are four co-circle points. AB and CD meet in P. Show that f PA · f PB = f PC · g PD. 2. Show that the diameter of the circumcircle of triangle ABC is equal to the product of two sides dividing the altitude on the third side of the given triangle. 3. Prove Example 1.99 directly. 4. In Example 1.102, if D is an arbitrary point. Show that DO2 = AO2(1 −4S EFG S ABC ). 5. Continue from the above exercise. Show that D is on a ﬁxed circle with O as its center if and only if S EFG is a constant. 1.10 Full-Angles To deﬁne the concept of angles, we need the concept of rays. As a consequence, we need to compare the order of points on a line. In algebraic language, this means that we need to use inequalities to describe angles. In this section, we introduce a new kind of angle, the description of which does not require inequalities. Angles of this kind will be used to simplifying the proofs of many geometry theorems. Deﬁnition 1.105 A full-angle consists of an ordered pair of lines l and m and is denoted by ∠[l, m]. Two full-angles ∠[l, m] and ∠[u, v] are equal if there exists a rotation K such that K(l) ∥u and K(m) ∥v. If A, B and C, D are distinct points on l and m respectively, then ∠[l, m] is also denoted by ∠[AB, CD], ∠[BA, CD], ∠[AB, DC], ∠[AB, m], and ∠[l, DC]. For three points A, B, and C, let ∠[ABC] = ∠[AB, BC]. 1.10 Full-Angles 45 Deﬁnition 1.106 If l⊥m, ∠[l, m] is said to be a right full-angle and is denoted by ∠. If l ∥m, ∠[l, m] is said to be a ﬂat full-angle and is denoted by ∠. To give a criterion for the equality of two full-angles, we need to introduce the tangent function for full-angles. Deﬁnition 1.107 The tangent function for the full-angle ∠[PQ, AB] is deﬁned to be tan(∠[PQ, AB]) = sin() ∠(PQ, AB)) cos() ∠(PQ, AB)). We need to check that this deﬁnition is well-deﬁned. That is when interchanging P, Q or A, B, sin() ∠(PQ,AB)) cos() ∠(PQ,AB)) does not change. This comes from the following equations sin() ∠(PQ, AB)) = −sin() ∠(PQ, BA)) = −sin() ∠(QP, AB)), cos() ∠(PQ, AB)) = −cos() ∠(PQ, BA)) = −cos() ∠(QP, AB)). As a consequence, we see that the sine and cosine functions for a full-angle are mean-ingless. Proposition 1.108 ∠[AB, PQ] = ∠[XY, UV] if and only if ) ∠(AB, PQ) = ) ∠(XY, UV) or ) ∠(AB, PQ)− ) ∠(XY, UV) = 180◦. Proof. Without loss of generality, we may assume AB ∥XY. Then ∠[AB, PQ] = ∠[XY, UV] if and only if PQ ∥UV, i.e., if and only if ) ∠(AB, PQ) = ) ∠(XY, UV) or ) ∠(AB, PQ) − ) ∠(XY, UV) = 180◦. Proposition 1.109 tan(∠[AB, PQ]) = 4S APBQ PAQBP . Proof. This is a consequence of Propositions 1.87 and 1.88. Proposition 1.110 ∠[AB, PQ] = ∠[XY, UV] if and only if tan(∠[AB, PQ]) = tan(∠[XY, UV]). Proof. If ∠[AB, PQ] = ∠[XY, UV], by Proposition 1.108 we have ) ∠(AB, PQ) = ) ∠(XY, UV) or ) ∠(AB, PQ) −) ∠(XY, UV) = 180◦. It is clear that tan(∠[AB, PQ]) = tan(∠[XY, UV]) for both cases. If tan(∠[AB, PQ]) = tan(∠[XY, UV]) then we have sin() ∠(AB, PQ)) cos() ∠(AB, PQ)) = sin() ∠(XY, UV)) cos() ∠(XY, UV)). The above equation holds if and only if ) ∠(AB, PQ) = ) ∠(XY, UV) or ) ∠(AB, PQ)−) ∠(XY, UV) = 180◦, i.e., ∠[AB, PQ] = ∠[XY, UV] by Proposition 1.108. 46 Chapter 1. Geometry Preliminaries Example 1.111 Let ABC be a triangle such that AB = AC. Then ∠[ABC] = ∠[BCA]. Con-versely, if ∠[ABC] = ∠[BCA] and A, B, and C are not collinear then AB = AC. Proof. If AB = AC, we have tan(∠[ABC]) = 4S ABC PABC = 4S BCA BC2 = 4S BCA PBCA = tan(∠[BCA]). Then ∠[ABC] = ∠[BCA]. Conversely, if ∠[ABC] = ∠[BCA] and S ABC , 0, by the deﬁnition of the tangent function we have PABC = PBCA, i.e., AB2 = AC2. In the above example, we do not need to say that the corresponding “inner” angles of an isosceles triangles are equal. To describe the “inner” angle, we need inequalities. Example 1.112 Let l, m, and t be three lines. Then l ∥m if and only if ∠[l, t] = ∠[m, t]. Notice that in the above criterion for parallel, we do not need to mention that the angles should be the “corresponding angles,” the exact description of which needs inequalities. Example 1.113 (The Inscribed Angle Theorem) If A, B, C, and D are cyclic points then ∠[AB, BC] = ∠[AD, DC]. If using angle in the usual sense, we need two conditions: ∠ABC = ) ∠ADC or ∠ABC + ∠ADC = 180◦and to distinguish these two cases, we need inequalities. The proofs for the above two examples will be given later. From these examples, we see that the concept of full-angles makes many geometry relations concise. We will see later that this will lead to short proofs for many geometry theorems. Deﬁnition 1.114 Let l, m, u, and v be four lines. Let K be a rotation such that K(l) ∥v. We deﬁne ∠[u, v] + ∠[l, m] = ∠[u, K(m)]. It is easy to check that the addition of full-angles is associative and commutative. The main properties of the full-angles are summarized as follows. Q1 ∠[u, v] = ∠ if and only if u ∥v. Q2 ∠[u, v] = ∠ if and only if u⊥v. Q3 ∠ + ∠ = ∠. Q4 ∠[u, v] + ∠ = ∠[u, v]. Q5 ∠[u, v] + ∠[l, m] = ∠[l, m] + ∠[u, v]. 1.10 Full-Angles 47 Q6 ∠[u, v] + (∠[l, m] + ∠[s, t]) = (∠[u, v] + ∠[l, m]) + ∠[s, t]. Q7 ∠[u, s] + ∠[s, v] = ∠[u, v]. Q8 if ∠[u, v] = ∠ then for any line t we have ∠[u, t] = ∠[v, t]. Conversely, if for a line t we have ∠[u, t] = ∠[v, t] then ∠[u, v] = 0. Q8 is Example 1.112 which can be derived from Q1–Q7 as follows. If ∠[u, v] = 0, we have ∠[u, t] = ∠[u, v] + ∠[v, t] (Q7) = ∠ + ∠[v, t] the hypothesis = ∠[v, t] + ∠ (Q5) = ∠[v, t] (Q4). Conversely, if ∠[u, t] = ∠[v, t] ∠[u, v] = ∠[u, t] + ∠[t, v] (Q7) = ∠[v, t] + ∠[t, v] the hypothesis = ∠[v, v] (Q7) = ∠ (Q1). The following properties of the full-angles are also often used. Q9 If AB = AC, we have ∠[AB, BC] = ∠[BC, AC]. Conversely, if ∠[AB, BC] = ∠[BC, AC] then AB = AC or A, B, and C are collinear. Q10 Points A, B, C, and D are on the same circle or on same line if and only if ∠[AB, BC] = ∠[AD, DC]. Q11 If AB is the diameter of the circumcircle of triangle ABC then ∠[AC, BC] = ∠ Q12 If O is the circumcenter of triangle ABC then ∠[BO, OC] = 2∠[AB, AC]. Q9 is Example 1.111. For the proofs of Q10, Q11, and Q112, see Example 3.52 on page 131. Example 1.115 Two circles O and Q meet in two points A and B. A line passing through A meets circles O and Q in C and E. A line passing through B meets circles O and Q in D and F. Show that CD ∥EF. Figure 1-59 shows ﬁve possible cases for this example. The following proof based on full-angles is valid for all cases. If we do not use full-angles, we must give diﬀerent proofs for diﬀerent ﬁgures. 48 Chapter 1. Geometry Preliminaries A B O Q C D E F A B O Q C D E F A B O Q C D E F A B O Q C D E F A B O Q C D E F Figure 1-59 Proof. ∠[DC, FE] = ∠[DC, DB] + ∠[DB, FE] (by Q7) = ∠[AC, AB] + ∠[FB, FE] (by Q10 and Q8 since A, B, C, D are cyclic and D ∈BF) = ∠[AE, AB] + ∠[AB, AE] (by Q8 and Q10 since E ∈AC and A, B, F, E are cyclic) = ∠[AE, AE] (by Q7) = ∠. (by Q2) A B C E D H G A B C E D H G Figure 1-60 Example 1.116 In triangle ABC, two altitudes AD and BE meet in H. G is the foot of the perpendicular from point H to AB. Show that ∠[DG, GH] = ∠[HG, GE]. Proof. ∠[DG, GH] + ∠[GE, HG] = ∠[DB, BH] + ∠[AE, HA] (Q10;B,D,G,H;A,E,G,H cyclic.) = ∠[BC, BE] + ∠[AC, AD] (Q8;D∈BC;B∈EH;E∈AC;H∈AD.) = ∠[BC, AC] + ∠[AC, BE] + ∠[AC, BC] + ∠[BC, AD] (Q7) = ∠[BC, AC] + ∠ + ∠[AC, BC] + ∠ (AC⊥BE and BA⊥AD) = ∠[BC, BC] + ∠ (Q7,Q3) = ∠. 1.10 Full-Angles 49 As in Figure 1-60, if both ∠BAC and ∠ABC are acute angles, we have ∠DGH = ∠EGH; if one of them is obtuse, then ∠DGH + ∠EGH = 180◦. Thus, if we do not use full-angles, we must give two proofs for the two cases. A B C D L M N N Figure 1-61 Example 1.117 (The Nine Point Circle) Let the midpoints of the sides AB, BC, and CA of △ABC be L, M, and N, and AD the altitude on BC. Show that L, M, N, and D are on the same circle. Proof. We need to show ∠[LM, MD] = ∠[LN, ND] or equivalently, ∠[LM, MD] + ∠[ND, LN] = ∠. ∠[LM, MD] + ∠[ND, LN] = ∠[AC, BC] + ∠[ND, LN] (Q8; LM ∥AC; M, D, B, C are collinear) = ∠[AC, BC] + ∠[ND, BC] + ∠[BC, LN] (Q7) = ∠[AC, BC] + ∠[BC, AC] + ∠ (Q9,Q1; ND = NC; BC ∥LN) = ∠[AC, AC] = ∠. A B C O D E M Figure 1-62 Example 1.118 The circumcenter of triangle ABC is O. AD is the altitude on side BC. Show that ∠OAD = \|∠C −∠B\|. Proof. Let M be the midpoint of BC, MO and AB meet in E. We need only to show that ∠[AD, AO] = ∠[CE, AC]. ∠[AD, AO] + ∠[AC, CE] = ∠[AD, AC] + ∠[AC, AO] + ∠[AC, BC] + ∠[BC, CE] (Q7) = ∠[AD, BC] + ∠[CO, AC] + ∠[BE, BC] (Q7,Q9 (BE = CE)) = ∠ + ∠[CO, MO] + ∠[MO, AC] + ∠[BA, BC] (Q7; E∈BA;AD⊥BC) = ∠ + ∠[AC, BA] + ∠[MO, AC] + ∠[BA, BC] (Q12) = ∠ + ∠[MO, BC] (Q7) = ∠ + ∠ = ∠. Summary of Chapter 1 50 Chapter 1. Geometry Preliminaries • Signed areas and Pythagoras diﬀerences are used to describe some basic geometry relations: collinearity, parallelism, perpendicularity, congruence of line segments, and congruence of full-angles. 1. Three points A, B, and C are collinear if and only if S ABC = 0. 2. PQ ∥AB if and only if S PAQB = S PAB −S QAB = S BPQ −S APQ = 0. 3. PQ⊥AB if and only if PPAQB = PPAB −PQAB = PBPQ −PAPQ = 0. 4. ∠[AB, PQ] = ∠[XY, UV] if and only if S APBQ PAPBQ = S XUYV PXUYV . • The following results are powerful tools for solving diﬃcult geometry problems. 1. (The Co-side Theorem) Let M be the intersection of the lines AB and PQ and Q , M. Then we have S PAB S QAB = PM QM. 2. (The Co-angle Theorem) Let X, Y and Z be three non-collinear points. Then ∠ABC = ∠XYZ or ∠ABC + ∠XYZ = 180◦if and only if ▽ABC ▽XYZ = AB·BC XY·YZ. 3. (The Pythagoras Diﬀerence Theorem) Let ∠ABC , 90◦. Then ∠ABC = ∠XYZ if and only if PABC PXYZ = AB·BC XY·YZ; ∠ABC + ∠XYZ = 180◦if and only if PABC PXYZ = −AB·BC XY·YZ. 4. (The Visual Angle Theorem) Emanating from P, there are three rays PA, PB, and PC such that ∠APC = α, ∠CPB = β, and ∠APB = γ = α + β < 180◦. Then A, B, and C are collinear if and only if sin(∠γ)/PC = sin(∠α)/PB +sin(∠β)/PA. • The concept of oriented angles is introduced to represent the signed areas and Pythago-ras diﬀerences. S ABC = 1 2 · AB · BC · sin() ∠ABC) PABC = 2 · AB · BC · cos() ∠ABC) S ABCD = 1 2 · AC · BD · sin() ∠(AC, BD)) PABCD = 2 · AC · BD · cos() ∠(AC, DB)) • The Herron-Qin formulas give connections between the signed area and the Pythago-ras diﬀerence. 16S 2 ABC = 4AB 2 · CB 2 −P2 ABC 16S 2 ABCD = 4AC 2 · BD 2 −P2 ABCD. Chapter 2 The Area Method In this chapter, we will present the area method in the narrow sense, that is, as a method of mechanical theorem proving for constructive geometry statements only involving two ge-ometry relations: collinearity and parallel. In other words, we are dealing with constructive geometry statements in aﬃne geometry. 2.1 Traditional Proofs Versus Machine Proofs We start this section with a comment on the traditional Euclidean proof method from . One of the main defects in the traditional Euclidean proof is its almost com-plete disregard of such notions as the two sides of a line and the interior of an angle. Without clariﬁcation of these ideas, absurd consequences result. The following example shows that this defect may occur even in very simple proofs. A B C D E Figure 2-1 Example 2.1 Let ABCD be a parallelogram (i.e, AB ∥CD, BC ∥AD), E be the inter-section of the diagonals AC and BD. Show that AE = CE. The traditional proof of this theorem is ﬁrst to prove △ACB △CAD (hence AB = CD), then to prove △AEB △CED (hence AE = CE). In proving the congruence of these triangles, we have repeatedly used the fact ∠CAB = ∠ACD. This fact is quite evident 51 52 Chapter 2. The Area Method because the two angles are the alternative angles with respect to parallels AB and CD. However, here we have implicitly assumed the “trivial fact” that points D and B are on opposite sides of line AC. The last fact is harder to prove than the original statement. (Please try it!) This extremely simple example reveals the diﬃculty in implementing a powerful and sound geometry theorem prover based on congruence and similarity of triangles. Of course, one may develop an interactive prover so that the user can input some trivial facts such as the one in the previous paragraph. These facts can be stored by the program in a data base. Then one would face a much more severe problem of the consistency of proofs. Example 2.2 Every triangle is isosceles. Let ABC be a triangle as shown in Figure 2-2. We want to prove CA = CB. A B C D O E F Figure 2-2 Proof. Let D be the intersection of the per-pendicular bisector of AB and the internal bi-sector of angle ACB. Let DE ⊥AC and DF ⊥CB. It is easy to see that △CDE △CDF and △ADE △BDF. Hence CE + EA = CF + FB, i.e., CA = CB. Try to solve this paradox. Another defect in the traditional Euclidean proofs is the lack of non-degenerate condi-tions. Each geometry theorem is valid only under some auxiliary conditions which are not stated explicitly in the theorem. For instance, in Example 2.1 we need to assume that A, B, and C are not collinear. We call such kinds of conditions the non-degenerate conditions of the theorem, which may become complicated for diﬃcult theorems. Without explicitly stated non-degenerate conditions, the traditional proofs of geometry theorems are gener-ally not strict. First, in each step of the proof, we use some lines or triangles which are implicitly assumed to be in normal positions, i.e., each line is uniquely determined and each triangle does not degenerate to a line. However, in a machine proof these conditions should be explicitly stated or justiﬁed. Second, during a proof, we need to use other known theorems; but in the statement of the cited theorems the nondegenerate conditions are usu-ally not given explicitly. Hence the correctness of the use of these known theorems is not fully justiﬁed. Partially due to these defects, it is very diﬃcult to incorporate the traditional Euclidean proof methods into a computer program so that skillful proofs can be automatically pro-duced by computers. Researchers have been studying automated generation of traditional proofs using computer programs since the work by H. Gelernter, J. R. Hanson, and D. W. Loveland in the early 60s. In spite of the enormous amount of research and great 2.1 Traditional Proofs Versus Machine Proofs 53 improvements [43, 83, 106, 141, 129, 144], the successes in this direction have been lim-ited in the sense that no program has been developed which can prove non–trivial geometry theorems eﬃciently. On the other hand, A. Tarski, introduced a decision procedure for what he called ele-mentary geometry, based on the algebraic method in the 1930s . Tarski’s quantiﬁer elimination method was later improved and redesigned by A. Seidenberg , G. Collins and others. In particular, Collins’ cylindrical decomposition algorithm is the ﬁrst Tarski type algorithm which has been implemented on a computer. Solutions of several nontrivial problems of elementary geometry and algebra have been obtained using the im-plementation [45, 46, 114]. Meanwhile, Wen–Ts¨ un Wu introduced a highly successful algebraic method of me-chanical geometry theorem proving . Inspired by Wu’s work, many researchers have developed eﬃcient computer programs for proving geometry theorems [36, 12, 55, 92, 120, 126, 133, ?, 156]. Nearly one thousand theorems from Euclidean geometry, non-Euclidean geometry, diﬀerential geometry, and mechanics have been proved by various provers based on Wu’s method and its variants [12, 69, 70, 135, 157]. Many hard theorems whose traditional proofs need an enormous amount of human intelligence, such as Feuer-bach’s theorem, Morley’s trisector theorem etc., can be proved by computer programs based on algebraic methods within seconds. In addition, Wu ﬁrst recognized the importance of the non-degenerate conditions in mechanical geometry theorem proving. However, algebraic methods can only tell whether a statement is true or not. If we want to know the proofs, we usually have to look at tedious computations of polynomials in the coordinates of the related points. So the goal of automatically producing readable proofs for geometry theorems has not been achieved. The goal of this book is to present a method which can produce short and readable proofs for geometry statements eﬃciently. The starting point of this book is the mechanization of a special case of the area method discussed in Chapter 1. Our machine proof method has the following advantages. • Auxiliary points and lines will be added automatically if they are needed. • Suﬃcient non-degenerate conditions can be generated automatically. • The proof produced according to the method is independent of the diagram. A key fact behind the success of our method (or Wu’s algebraic method) is that the validity of most elementary geometry theorems involving equalities only is independent of the rel-ative order positions of the points involved. Such geometry theorems belong to unordered geometry. In unordered geometry, the proofs of these theorems can be very simple. How-ever, the ordinary proofs of these theorems involve the order relation (among points and lines), hence are not only complicated, but also not strict, see for example, Example 2.1. 54 Chapter 2. The Area Method Here we list several other examples. In the statement of Ceva’s theorem (Example 1.7 on page 11) we usually assume that P is inside the triangle ABC. But this restriction on P is not necessary: the statement is true regardless of whether P is inside or outside triangle ABC. The proof produced by the area method is valid for all cases. Also see Examples 1.9 on page 11, 1.115 on page 47, and 1.116 on page 48. F O C E D G N M F O C E D G N M F O C E D G N M Figure 2-3 For the Butterﬂy theorem (Example 1.99 on page 42), as shown in Figure 2-3, we have three diﬀerent diagrams, which are often treated as diﬀerent theorems in geometry text-books. The proof for this theorem based on the area method is valid for all three cases. A B C O D E F Figure 2-4 Remark 2.3 1. For a strict proof of Example 2.1 using the area, see Example 1.24 on page 17. To prove it using congruence triangles, you may start from points A, B, and C to construct point D as follows: E is the midpoint of AC and D is the symmetry of B with respect to E. 2. The problem in the “proof” of Example 2.2 is that we use a wrong diagram. The correct diagram for Example 2.2 is Figure 2-4. 2.2 Signed Areas of Oriented Triangles We will formally deﬁne two geometry quantities: the ratio of the signed lengths of directed parallel line segments and the signed areas of oriented triangles. Properties of these two quantities will serve as the basis of our area method. Those who are mainly interested in machine proofs may skip the next subsection and read Subsection 2.2.2 directly. 2.2 Signed Areas of Oriented Triangles 55 2.2.1 The Axioms We use capital English letters with or without subscripts to denote points. The only basic geometry relation is a trinary relation collinear, i.e., three points A, B, and C are collinear. The exact meaning of collinear is given by Axioms A.1–A.6. Let R be the ﬁeld of the real numbers. Axiom A.1 For three collinear points P, A, and B such that A , B, AP AB is an element in R and satisﬁes AP AB = −PA AB = PA BA = −AP BA and PA AB = 0 iﬀ(abbr. if and only if) P = A. AP AB is called the ratio of the directed segments AP and AB. Let r = AP AB. We sometimes also write AP = rAB. Axiom A.2 Let A and B be two distinct points. For r ∈R, there exists a unique point P which is collinear with A and B and satisﬁes (1) AP AB = r and (2) AP AB + PB AB = 1. Three points A, B, and C determine an oriented triangle ABC. We use the order of the vertices of a triangle to represent its orientation. Thus triangles △ABC, △BCA, and △CAB have the opposite orientation, whereas △ACB, △CBA, and △BAC have the same orientation, i.e., a triangle has two orientations. The signed area of an oriented triangle ABC, denoted by S ABC, is an element in R which satisﬁes the following four basic properties. Axiom A.3 S ABC = S CAB = S BCA = −S BAC = −S CBA = −S ACB. If A, B, and C are three non-collinear points, we have S ABC , 0. Axiom A.4 There exist at least three points A, B, and C such that S ABC , 0. Axiom A.5 For any four points A, B, C, and D, we have S ABC = S ABD + S ADC + S DBC. Axioms A.4 and A.5 are called the dimension axioms. Axiom A.4 ensures that not all the points are collinear. Axiom A.5 ensures that all the points are in one plane. As a consequence of Axiom A.5, we can deﬁne the signed area of oriented quadrilater-als. The signed area of an oriented quadrilateral ABCD is deﬁned to be S ABCD = S ABC + S ACD. By Axioms A.3 and A.5, it is clear that 56 Chapter 2. The Area Method S ABCD = S ABD −S CBD; S ABCD = S BCDA = S CDAB = S DABC; S ABCD = −S ADCB = −S DCBA = −S CBAD = −S BADC. Axiom A.6 Let A, B, and C be three collinear points such that AB = λAC. Then for any point P, we have S PAB = λS PAC. Axiom A.6 is the most important property of the area. We will see in the next section that most of the interesting and nontrivial properties of area come from this axiom. It is convenient to extend the notion of collinearity to be a geometry relation among any set of points: one or two points are always collinear, and a set of points are collinear if any three points in it are collinear. We can thus introduce a new geometry object: the line. Deﬁnition 2.4 A line is a maximal set of collinear points. Let l be a line and P ∈l. Then we say that P is on line l (instead of in line l). Proposition 2.5 Three points A, B, and C are collinear iﬀS ABC = 0. Proof. If S ABC = 0, then by Axiom A.3 A, B, and C are collinear. Now let us assume that A, B, and C are collinear. If A = C, since CC = 2CC = 0, by Axiom A.6 we have S ACC = 2S ACC = 0. If A , C and λ = AB AC, by Axiom A.6, S ABC = λS ACC = 0. Corollary 2.6 Two distinct points A and B determine a unique line AB which is the set of all points P satisfying S ABP = 0. Proof. Let C, D, and E be three distinct points on line AB. We need to show that C, D, and E are collinear, i.e., S CDE = 0. By Proposition 2.5 S ADE = AD AB · S ABE = 0. Then A, D, and E are collinear. Similarly C, D, and A are collinear. By Proposition 2.5 S CDE = DE DA · S CDA = 0. In what follows, when speaking about a line AB, we always assume that A , B. A point P on line AB is determined uniquely by AP AB or PB AB. We thus call xP = AP AB , yP = PB AB the position ratio or position coordinates of the point P with respect to AB. It is clear that xP + yP = 1. 2.2 Signed Areas of Oriented Triangles 57 2.2.2 Basic Propositions The basic propositions presented in this section are the basis of our area method. We ﬁrst extend Axiom A.6 to the following convenient form. A B P D C Figure 2-5 Proposition 2.7 If points C and D are on line AB and P is any point not on line AB (Figure 2-5), then S PCD S PAB = CD AB . Proof. Without loss of generality, let us assume C , A. Then S PCD S PAB = S PCD S PCA · S PCA S PAB = CD CA · CA AB = CD AB . Proposition 2.8 (The Co-side Theorem) Let M be the intersection of two lines AB and PQ. Then PM QM = S PAB S QAB ; PM PQ = S PAB S PAQB ; QM PQ = S QAB S PAQB . Proof. See Proposition 1.5 on page 9. Proposition 2.9 Let R be a point on line PQ. Then for any two points A and B S RAB = PR PQ S QAB + RQ PQ S PAB. Proof. See Proposition 1.18 on page 15. Similar to Chapter 1, we use the notation AB ∥PQ to denote the fact that A, B, P, and Q satisfy one of the following conditions: (1) A = B or P = Q; (2) A, B, P and Q are on the same line; or (3) line AB and line PQ do not have a common point. Proposition 2.10 PQ ∥AB iﬀS PAB = S QAB, i.e., iﬀS PAQB = 0. Proof. If S PAB , S QAB, it is clear that A , B, P , Q, and A, B, P, and Q are not collinear. Let O be a point on line PQ such that PO PQ = S PAB S PAQB . Thus OQ PQ = −S QAB S PAQB . By Proposition 2.9, S OAB = PO PQS QAB + OQ PQS PAB = 0. By Proposition 2.5, point O is also on line AB, i.e., AB is not parallel to line PQ. Conversely, if PQ ∦AB then A , B, P , Q and lines AB and PQ 58 Chapter 2. The Area Method intersection at a unique point O. By Proposition 2.8, OP OQ = S PAB S QAB = 1. Thus P = Q which is a contradiction. A parallelogram is a quadrilateral ABCD such that AB ∥CD, BC ∥AD, and no three vertices of it are on the same line. Let ABCD be a parallelogram and P, Q be two points on CD. We deﬁne the ratio of two parallel line segments as follows PQ AB = PQ DC . In our machine proofs, auxiliary parallelograms are often added automatically and the following two propositions are used frequently. Proposition 2.11 Let ABCD be a parallelogram. Then for two points P and Q, we have S APQ + S CPQ = S BPQ + S DPQ or S PAQB = S PDQC. Proof. See Proposition 1.30 on page 20. P A B C D Figure 2-6 Proposition 2.12 Let ABCD be a parallelo-gram and P be any point. Then S PAB = S PDC −S ADC = S PDAC. Proof. By Proposition 2.11, S PAB = S PDB −S PCB = S DBCP = S DBC + S DCP = S PDC −S ADC. So far, we do not mention the existence of parallel lines. The following statement shows that Euclid’s parallel postulate is a consequence of our axioms. Example 2.13 (Euclid’s Parallel Axiom) Passing through a point not on a line l, there exists a unique line which is parallel to l. Proof. Let P be a point not on the line AB. By Axiom A.2, we can take points O and Q such that O is the midpoint of PA and Q is the symmetric point of B with O, i.e., QO = OB. By the co-side theorem S QAB = 2S OAB = S PAB. By Proposition 2.10, PQ ∥AB. To show the uniqueness, let T be another point such that TP ∥AB. By Proposition 2.12, S T PQ = S T AB −S PAB = 0, i.e., T is on line PQ. Example 2.14 If PR ∥AC and QS ∥BD then S PQRS S ABCD = PR AC · QS BD. 2.3 Hilbert Intersection Point Statements 59 Proof. Let X, Y be points such that PR = AX, QS = BY. By Proposition 2.11, S PQRS = S PBRY = S PBY −S RBY = BY BD (S PBD −S RBD) = QS BD S PBRD. Similarly, we have S PBRD = PR ACS ABCD. Example 2.15 If line PQ is parallel to line AB then AB PQ = S PAB S AQP Proof. Let R be a point such that AR = PQ. By Propositions 2.7 and 2.11, AB PQ = AB AR = S PAB S PAR = S PAB S PAQ . 2.3 The Hilbert Intersection Point Statements In Chapter VI of Hilbert’s classic book, “Foundations of Geometry”, he introduced a class of geometry statements which he called the “pure point of intersection theorems.” Ac-cording to Hilbert, every “pure point of intersection theorem” can be put in the following form: Choose an arbitrary set of a ﬁnite number of points and lines. Then draw in a prescribed manner any parallels to some of these lines. Choose any points on some of the lines and draw any lines through some of these points. Then, if connecting lines, points of intersection and parallels are constructed through the points existing already in the prescribed manner, a deﬁnite set of ﬁnitely many lines is eventually reached, about which the theorem asserts that they either pass through the same point or are parallel. Hilbert also gave a mechanical proving method for statements of this type. His result can be summarized as follows. Theorem 62 in . Every pure point of intersection theorem that holds in aﬃne geometry takes, through the construction of suitable auxiliary points and lines, the form of a combination of ﬁnite number of Pascal’s conﬁgurations. By Pascal’s conﬁguration, he meant the diagram of Example 1.27 on page 19. The above result called Hilbert’s Mechanization theorem by Wu , is the ﬁrst me-chanical theorem proving method for a class of geometry statements, i.e., class CH in this book. 60 Chapter 2. The Area Method Hilbert’s mechanization theorem works as follows. First, we prove a theorem using algebraic calculations (see and Chapter 3 of ). Since each arithmetic operation of numbers, e.g. a + b = b + a, a ∗b = b ∗a, can be represented by Pascal conﬁgurations, the algebraic proof can thus be converted into a series of Pascal conﬁgurations. But the geometric proofs produced in this way are expected to be very long and cumbersome, and as far as we know no single theorem has been proved in this way. The aim of this chapter is to provide an eﬃcient method of producing short and readable proofs for the Hilbert pure point of intersection statements. 2.3.1 Description of the Statements To describe the Hilbert intersection point statements precisely, we need the concepts of geometry quantities and constructions. In this chapter, by a geometric quantity we mean • the ratio of two oriented segments on one line or on two parallel lines, or • the signed area of an oriented triangle or an oriented quadrilateral. A construction is used to introduce a new point from some known points. We need the following constructions. C1 Take arbitrary points Y1, · · · , Ym on the plane. Yi are free points, i.e., Yi can move freely on the plane. C2 Take a point Y on line PQ. Point Y is a semi-free point, i.e. point Y can move freely on the line PQ. To make sure that point Y can be taken properly, we will introduce a nondegenerate (abbr. ndg) condition P , Q, i.e., the line PQ is well deﬁned. C3 Take a point Y on line PQ such that PY = λPQ where λ can be a rational number, a rational expression in geometry quantities, or a variable. Notice that λ is the position ratio of point Y with regard to PQ. If λ is a ﬁxed quantity then Y is a ﬁxed point; if λ is a variable then Y is a semifree point. The ndg conditions for this construction are P , Q and λ is meaningful, i.e., its denominator does not vanish. C4 Take the intersection Y of line PQ and line UV. Point Y is a ﬁxed point. The ndg condition is that P , Q, U , V, and the lines PQ and UV have one and only one common point, i.e., PQ ∦UV. C5 Take a point Y on the line passing through point R and parallel to line PQ. Here Y is a semi-free point. The ndg condition is P , Q. 2.3 Hilbert Intersection Point Statements 61 C6 Take a point Y on the line passing through R and parallel to line PQ such that RY = λPQ, where λ can be a rational number, a rational expression in geometry quantities, or a variable. If λ is a ﬁxed quantity then Y is a ﬁxed point; if λ is a variable then Y is a semifree point. The ndg conditions are the same as those of C3. C7 Take the intersection Y of line UV and the line passing through R and parallel to line PQ. Point Y is a ﬁxed point. The ndg condition is that PQ ∦UV. C8 Take the intersection Y of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. Point Y is a ﬁxed point. The ndg condition is that PQ ∦UV. Point Y in each of the above constructions is said to be introduced by that construction. We need to show that the above constructions are always possible. That is the introduced points do exist. C1 and C2 are trivial. The existence of the point Y in C3 comes from Axiom A.2. C5 and C6 come from Example 2.13 and C3. By Example 2.13, C7 and C8 can be reduced to C4. For C4, since PQ ∦UV, line PQ and UV have a unique common point. Deﬁnition 2.16 A Hilbert intersection point statement can be represented by a list S = (C1, C2, . . . , Ck, G) where 1. Each construction Ci introduces a new point from the points which are introduced by the previous C j, j = 1, · · ·i −1; and 2. G = (E1, E2) where E1 and E2 are polynomials in some geometric quantities about the points introduced by the constructions Ci and E1 = E2 is the conclusion of S . The ndg condition of S is the set of ndg conditions of Ci and the condition that the denomi-nators of the length ratios in E1 and E2 are not zero. The set of all the Hilbert intersection point statements is denoted by CH. As indicated by the deﬁnition, the ndg conditions of a statement in CH can be generated automatically. Take Ceva’s theorem (Example 1.7 on page 11) as an example. Example 2.17 (Ceva’s Theorem) We describe the statement in the following constructive way. Take four arbitrary points A, B, C, and P. Take the intersection D of BC and AP. 62 Chapter 2. The Area Method Take the intersection E of AC and BP. Take the intersection F of AB and CP. Show that AF FB · BD DC · CE EA = 1. According to the deﬁnition, the ndg conditions for Ceva’s Theorem are BC ∦AP; AC ∦BP; AB ∦CP; F , B; D , C; E , A, i.e., point P can not be on the three sides of △ABC and the three dotted lines in Figure 2-7. A B C P D E F Figure 2-7 You may wonder that the condition “A, B, and C not collinear” is not in the ndg con-ditions. Indeed, when A, B, and C are three diﬀerent points (this comes from the ndg condition) on the same line, Ceva’s theorem is still true (now F = C, D = A, and E = B) and the proofs based on the area method is still valid in this case. The ndg conditions pro-duced according to our method guarantee that we can produce a proof for the statement. Certainly, we can avoid this seemingly unpleasant fact by introducing a new construction, TRIANGLE, which introduces three noncollinear points. But theoretically, this is not nec-essary. Also the ndg conditions are not unique for a geometry statement: they depend on the constructive description of the statements. For instance, Ceva’s theorem can be described as follows. Take three arbitrary points A, B, C. Take a point E on line AC. Take a point F on line AB. Take the intersection P of BE and CF. Take the intersection D of BC and AP. Show that AF FB · BD DC · CE EA = 1. Now the ndg conditions of Ceva’s theorem are A , C, A , B, BE ∦CG, BC ∦AP, B , F, C , D, and A , P. The ndg conditions generated according to Deﬁnition 2.16 are suﬃcient, i.e., if a ge-ometric statement is true in the usual sense then it must be true strictly under these ndg conditions. For more details, see Algorithm 2.32 on page 70. 2.3 Hilbert Intersection Point Statements 63 2.3.2 The Predicate Form A Hilbert intersection point statement can be transformed into predicate form. We ﬁrst introduce some basic predicates. 1. (POINT P): P is a point in the plane. 2. (COLL P1, P2, P3): points P1, P2, and P3 are on the same line. It is equivalent to S P1P2P3 = 0. 3. (PARA P1, P2, P3, P4): P1P2 ∥P3P4. It is equivalent to S P1P3P2P4 = 0. Each construction is equivalent to the conjunction of several predicates. C2 Take a point Y on line PQ. The predicate form is (COLL Y P Q) and P , Q. C3 Take a point Y on line PQ such that PA = λPQ. The predicate form is (COLL Y P Q), λ = PY PQ, and P , Q. C4 Take the intersection Y of line PQ and line UV. The predicate form is (COLL Y P Q), (COLL Y U V), and ¬(PARA U V P Q). C5 Take a point on the line passing through point R and parallel to line PQ. The predicate form is (PARA Y R P Q) and P , Q. C6 Take a point Y on the line passing through R and parallel to line PQ such that RY = λPQ. The predicate form is (PARA Y R P Q), λ = RY PQ, and P , Q. C7 Take the intersection Y of line UV and the line passing through R and parallel to line PQ. The predicate form is (COLL Y U V), (PARA Y R P Q), and ¬(PARA P Q U V). C8 Take the intersection Y of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. The predicate form is (PARA Y R P Q), (PARA Y W U V), and ¬(PARA P Q U V). The predicate form of each construction C has two parts: the equation part E(C) and the ndg condition ¬D(C). Now a constructive statement S = (C1, · · · , Cr, (E, F)) can be transformed into the fol-lowing predicate form ∀Pi[(E(C1) ∧· · · ∧E(Cr) ∧¬D(C1) ∧· · · ∧¬D(Cr)) ⇒(E = F)] where Pi is the point introduced by Ci. 64 Chapter 2. The Area Method It is clear that the predicate form of a statement depends on how we describe the state-ment constructively. For the ﬁrst constructive description of Ceva’s theorem (Example 2.17 on page 61), its predicate form is ∀A, B, C, P, E, F, D(HYP ⇒CONC) where HYP = (COLL D B C) ∧(COLL D A O) ∧¬(PARA B C A O) ∧ (COLL E A C) ∧(COLL E B O) ∧¬(PARA A C B O) ∧ (COLL F A B) ∧(COLL F C D) ∧¬(PARA A B C O) ∧ B , F ∧D , C ∧A , E CONC = (AF FB · BD DC · CE EA = 1). Exercises 2.18 1. We deﬁne a new predicate (CONC P1 P2 P3 P4 P5 P6) which means that the lines P1P2, P3P4, and P5P6 are concurrent. Use an equation in areas to represent this predicate. 2. Construction C3 is to take a point on a line with position ratio λ. Show that if point Y is introduced by one of the eight constructions then Y can also be introduced by constructions C1 and C3. The reason we use more constructions is that we want to describe geometry statements using fewer constructions, and as a consequence to obtain short proofs for the statements. 2.4 The Area Method Before presenting the method, let us re-examine the proof of Ceva’s theorem (Example 1.7 on page 11). By describing Ceva’s theorem constructively (Example 2.17 on page 61), we can introduce an order among the points naturally: A, B, C, P, D, E, and F, i.e., the order according to which the points are introduced. The proof is actually to eliminate the points from the conclusions according to the reverse order: F, E, D, P, C, B, and A. We thus have the proof: AF FB = −S ACP S BCP Eliminate point F. CE EA = S BCP S ABP Eliminate point E. BD DC = −S ABP S ACP Eliminate point D. Then AF FB · BD DC · CE EA = S ACPS BCPS ABP S BCPS ACPS ABP = 1. Thus the key step of the method is to eliminate points from geometry quantities. We will show how this is done in the following three subsections. 2.4 The Area Method 65 2.4.1 Eliminating Points from Areas We ﬁrst show that construction C2 is a special case of construction C3. This is because taking an arbitrary point Y on line UV is equivalent to taking a point Y on UV such that UA = λUV for an indeterminate λ. Similarly, construction C5 is a special case of construc-tion C6: taking an arbitrary point on the line passing through point W and parallel to UV is equivalent to taking a point Y such that WA = λUV for an indeterminate λ. We will discuss C1 in Section 2.4.3. Thus we need only to consider ﬁve constructions C3, C4, C6, C7, and C8. Lemma 2.19 Point Y is introduced by construction C3, i.e., Y satisﬁes PY = λPQ. To eliminate point Y from S ABY, we have S ABY = λS ABQ + (1 −λ)S ABP. Proof. This is Proposition 2.9. Lemma 2.20 Point Y is introduced by construction C4, i.e., Y = PQ ∩UV. To eliminate point Y from S ABY, we have S ABY = 1 S PUQV (S PUVS ABQ + S QVUS ABP). Proof. By Proposition 2.9, we have S ABY = PY PQ S ABQ + YQ PQ S ABP. By the co-side theorem, we have PY PQ = S PUV S PUQV , YQ PQ = S QVU S PUQV . Substituting this into the previous equation, we obtain the conclusion. Since PQ ∦UV, we have S PUQV , 0. Lemma 2.21 Point Y is introduced by construction C6, i.e., Y satisﬁes RY = λPQ. To eliminate point Y from S ABY, we have S ABY = S ABR + λS APBQ A B P Q R Y S Figure 2-8 Proof. We take a point S such that RS = PQ. By Lemma 2.19, S ABY = λS ABS + (1 −λ)S ABR. By Proposition 2.11, we have S ABS = S ABR + S ABQ −S ABP = S ABR + S APBQ. Substituting this into the previous formula, we obtain the conclusion. 66 Chapter 2. The Area Method Lemma 2.22 Point Y is introduced by construction C7, i.e., Y is the intersection of line UV and the line passing through R and parallel to line PQ. To eliminate point Y from S ABY, we have S ABY = 1 S PUQV (S PUQRS ABV −S PVQRS ABU). Proof. Take a point S such that RS = PQ. By Lemma 2.20, we have S ABY = 1 S RUS V · (S US RS ABV + S VRSS ABU). (1) We also have S RUS V = S PUQV by Proposition 2.11. S US R = S UQP −S RQP = S PUQR by Proposition 2.12. S VS R = S VQP −S RPQ = S PRQV by Proposition 2.12. Substituting these into (1), we obtain the conclusion. Lemma 2.23 Let point Y be introduced by construction C8. To eliminate point Y from S ABY, we have S ABY = S PWQR S PUQV · S AUBV + S ABW. Proof. By Lemma 2.21, S ABY = S ABW + WY UV S AUBV. Now the lemma comes from Lemma 2.26 below. 2.4.2 Eliminating Points from Length Ratios Y A P Q S C D Figure 2-9 Lemma 2.24 Let point Y be introduced by construc-tion C3. To eliminate Y from AY CD, we have AY CD =            AP PQ +λ CD PQ if A ∈PQ; S APQ S CPDQ otherwise. Proof. If A ∈PQ then AY CD = AP+PY CD = AP PQ + PY PQ CD PQ = AP PQ +λ CD PQ . Otherwise, take a point S such that AS = CD. Then Y is the intersection of PQ and AS and AS ∥CD. By Propositions 2.8 and 2.11, AY CD = AY AS = S APQ S APS Q = S APQ S CPDQ . 2.4 The Area Method 67 Lemma 2.25 Let point Y be introduced by construction C4. To eliminate point Y from AY CD, we have AY CD =        S AUV S CUDV if A is not on UV S APQ S CPDQ otherwise Proof. The proof is the same as the second case of Lemma 2.24. Y A R S T C D P Q Figure 2-10 Lemma 2.26 Let point Y be introduced by construction C6. Then we have AY CD =            AR PQ +r CD PQ if A ∈RY. S APRQ S CPDQ if A < RY. Proof. The ﬁrst case is obvious. For the second case, take points T and S such that RT PQ = 1 and AS CD = 1. By the co-side theorem, AY CD = AY AS = S ART S ARS T = S APRQ S CPDQ . Lemma 2.27 Let Y be introduced by construction C7. To eliminate point Y from G = AY CD, we have AY CD =        S AUV S CUDV if A is not on UV S APRQ S CPDQ if A is on UV Proof. If A is not on UV then the proof is the same as the second case of Lemma 2.24. If A is on UV, then the proof is the same as the second case of Lemma 2.26. Since PQ ∦UV, we have S CPDQ , 0. Lemma 2.28 Let point Y be introduced by construction C8. To eliminate point Y from G = AY CD, we have AY CD =        S APRQ S CPDQ if AY is not parallel to PQ S AUWV S CUDV otherwise. Proof. The proof is the same as the proof of the second case of Lemma 2.26. 2.4.3 Free Points and Area Coordinates In Subsections 2.4.1 and 2.4.2, we present methods of eliminating ﬁxed or semi-free points from geometry quantities. For a geometry statement S = (C1, C2, . . . , Ck, (E, F)), we can 68 Chapter 2. The Area Method use these lemmas to eliminate all the nonfree points introduced by Ci. Now the new E and F are rational expressions in indeterminates, areas and Pythagoras diﬀerences of free points. These geometric quantities are generally not independent, e.g. for any four points A, B, C, D we have S ABC = S ABD + S ADC + S DBC. In order to reduce E and F to expressions of independent variables, we introduce the con-cept of area coordinates. Deﬁnition 2.29 Let A, O, U, and V be four points such that O, U, and V are not collinear. The area coordinates of A with respect to OUV are xA = S OUA S OUV , yA = S OAV S OUV , zA = S AUV S OUV . It is clear that xA +yA +zA = 1. Since xA, yA, and zA are not independent, we also call xA, yA the area coordinates of Q with respect to OUV. O U V T A Figure 2-11 Proposition 2.30 The points in the plane are in a one to one correspondence with the triples (x, y, z) satisfying x + y + z = 1. Proof. Let O, U, and V be three non-collinear points. Then for each point A, its area coordinates satisfy xA + yA + zA = 1. Conversely, for any x, y, and z such that x + y + z = 1 we will ﬁnd a point A whose area coordinates are x, y, and z. If z = 1, take a point A such that OA UV = x. Then by Lemma 2.21, xA = S OUA S OUV = x, yA = −x = y, and zA = 1. If z , 1, take a point T on UV such that UT UV = x 1−z; take a point A on OT such that AT OT = z. By the co-side theorem, zA = S AUV S OUV = AT OT = z. By the co-side theorem again, we have xA = S OUA S OUV = (1 −z)S OUT S OUV = xS OUV S OUV = x. Similarly, yA = y. The following lemma reduces any area to an expression of area coordinates with respect to three given reference points. Lemma 2.31 Let O, U, and V be three noncollinear points. Then for points A, B, and Y, we have S ABY = 1 S OUV S OUA S OVA 1 S OUB S OVB 1 S OUY S OVY 1 . 2.4 The Area Method 69 O U V Y W O U V Y Figure 2-12 Proof. Since by Axiom A.5 S ABY = S OAB + S OBY −S OAY, we need only to compute S OBY and S OAY. Let W be the intersection of UV and OY. Then by Lemma 2.20 we have S OBW = 1 S OUYV (S OBV · S OUY + S OBU · S OYV). By Proposition 2.8, we have S OBY S OBW = S OUYV S OUV . Thus S OBY = 1 S OUV · (S OBV · S OUY + S OBU · S OYV). (1) If OY ∥UV, (1) can be proved as follows. By Example 2.15, OY UV = S OUY S OUV . By Lemma 2.21 S OBY = OY UV · S OUBV = S OUY S OUV (S OBV + S OUB) = S OBV · S OUY + S OBU · S OYV S OUV . Now we have proved that (1) is true under the condition that O, U, V are not collinear. Similarly, we have S OAY = 1 S OUV · (S OAV · S OUY + S OAU · S OYV); S OAB = 1 S OUV · (S OAV · S OUB + S OAU · S OBV). Substituting (1) and the above formulas into S ABY = S OAB + S OBY −S OAY, we obtain the conclusion. Use the same notations as in Lemma 2.31, let xA = S OUA S OUV , yA = S OVA S OUV ; xB = S OUB S OUV , yB = S OVB S OUV ; xY = S OUY S OUV , yY = S OVY S OUV . Then the formula in Lemma 2.31 becomes S ABY = S OUV xA yA 1 xB yB 1 xY yY 1 which is quite similar to the formula of the area of a triangle in terms of the Cartesian coordinates of its three vertices. 70 Chapter 2. The Area Method Algorithm 2.32 (AFFINE) INPUT: S = (C1, C2, . . . , Ck, (E, F)) is a statement in CH. OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof for S . S1. For i = k, · · · , 1, do S2, S3, S4 and ﬁnally do S5. S2. Check whether the ndg conditions of Ci are satisﬁed. The ndg conditions of a statement have two forms: A , B and PQ ∦UV. For the ﬁrst case, we check whether AB XY = 0 where X, Y are two arbitrary points on AB. For the second case, we check whether S PUV = S QUV. If the ndg condition of a geometry statement is not satisﬁed, the statement is trivially true. The algorithm terminates. S3. Let G1, · · · , Gs be the geometric quantities occurring in E and F. For j = 1, · · ·, s do S4. S4. Let H j be the result obtained by eliminating the point introduced by construction Ci from G j using Lemmas 2.24–2.31 and replace G j by H j in E and F to obtain the new E and F. S5. Finally, E and F are expressions of free parameters. If E is identical to F, S is true under the ndg conditions. Otherwise S is false in the Euclidean plane geometry. Proof of the correctness. If E = F then it is clear that S is true. Notice that all the elimi-nation lemmas in this section have the property that after applying a lemma to a geometry quantity that has geometric meaning (i.e., its denominator is not zero), the expression ob-tained also has geometric meaning under the ndg conditions of this statement. Therefore all the geometric quantities occurring in the proof have geometric meaning. The geometric quantities in E and F are all free parameters, i.e., in the geometric con-ﬁguration of S they can take arbitrary values. Since E , F, by Proposition 2.33 below we can take some concrete values for these quantities such that when replacing these quantities by the corresponding values in E and F, we obtain two diﬀerent numbers. In other words, we obtain a counter example for S . Proposition 2.33 Let P be a nonzero polynomial of indeterminates x1, · · · , xn with real num-bers as coeﬃcients. Show that we can ﬁnd rational numbers e1, · · · , en such that P(e1, · · · , en) , 0. The proof is left as an exercise. 2.4 The Area Method 71 For the complexity of the algorithm, let m and n be the number of free and non-free points in a statement respectively. To eliminate each non-free point, we need to apply the lemmas in Subsections 2.4.1 and 2.4.2 to each geometry quantity involving this point once. Also note that each lemma will replace a geometric quantity by a rational expression with degree less than or equal to two. Then if the conclusion of the geometry statement is of degree d, the expression obtained after eliminating the n non-free points is at most degree 2nd. Also note that after eliminating the m free points using Lemma 2.31, each quantity will be replaced by an expression of degree two. Then the ﬁnal result is at most degree 2d2n = d2n+1. This simple exponential complexity of the algorithm seems discouraging. But we will see that on the contrary this method can produce short proofs for almost all statements in CH very eﬃciently. One reason is that during the proof, the common factors of E and F can be removed. This simple trick alone usually reduces the sizes of the polynomials occurring in a proof drastically. Also the algorithm still has much room for improvement in order to obtain short proofs, as shown in Section 2.5 below. Exercises 2.34 1. Let O, U, and V be three noncollinear points. Then for points A, B, and Y, we have S ABY = 1 S OUV S OUA S OVA S UVA S OUB S OVB S UVB S OUY S OVY S UVY . 2. Show that each polynomial P(x) ∈R[x] of degree d has at most d diﬀerent roots. Use this result to prove Proposition 2.33. 3. Let P(x) = xd + ad−1xd−1 + ... + a0 be a polynomial, and m = max(1, Pd i=1 \|ai\|). Then for any r > m we have P(r) , 0. Use this result to prove Proposition 2.33. 4. Prove the Examples in Sections 1.2 and 1.3 using Algorithm 2.32. 2.4.4 Working Examples Before going further, we want to explain a little bit about the meaning of the axioms, propo-sitions, lemmas, and algorithms in this book. Since the goal of this book is to provide a method of proving theorems, the algorithms are our ﬁnal goals. The input to an algorithm is a geometry statement. The output of the algorithm is a proof or a disproof of the state-ment. The algorithms use the lemmas to eliminate points from geometry quantities. In the proofs of the lemmas, only the basic propositions are used. Finally, the basic propositions are consequences of the axioms. We can thus divide the proofs produced by Algorithm 2.32 into three levels: 1. a proof is at the ﬁrst level or lemma level if in that proof we only use the lemmas; 72 Chapter 2. The Area Method 2. a proof is at the second level or proposition level if in that proof we not only give the result obtained by applying the lemmas but also the process of how the results are obtained by using the basic propositions; 3. a proof is at the third level or axiom level if in that proof we only use the axioms. Theoretically, proofs at all levels can be produced automatically. But only proofs at the ﬁrst or second level are relatively short. If we limit ourselves to the six axioms only, the proofs produced according to our algorithms are generally very long. Also, it is not reasonable to limit oneself to axioms only. The proofs of the geometry statements in Chapter 1 are all at the proposition level and most of the proofs given in this chapter are at the lemma level. Algorithm 2.32 has been implemented as a prover on a computer. At the present time, this prover can only produce proofs at the lemma level. In what follows, when speaking about a machine proof, we mean the proof (in LaTeX form) produced by this prover. For instance, the machine proof for Ceva’s theorem (Example 1.7) is as follows. The machine proof −CE AE · BD CD · AF BF F = −(−S ACP) −S BCP · CE AE · BD CD E = −S BCP·S ACP S BCP·(−S ABP) · BD CD simplify = S ACP S ABP · BD CD D = S ABP·S ACP S ABP·S ACP simplify = 1 The eliminants AF BF F =S ACP S BCP CE AE E = S BCP −S ABP BD CD D = S ABP S ACP In the proof, a P = b means that b is the result obtained by eliminating point P from a; a simplify = b means that b is obtained by canceling some common factors from the denomi-nator and numerator of a; “eliminants” are the results obtained by eliminating points from separate geometry quantities. The prover can also give the ndg conditions and the predicate form of the geometry statement. We use a sequence of consecutive equations to represent a proof. Some might argue that this proof looks diﬀerent from the usual form of proofs. It is actually very easy to rewrite a proof in consecutive equations as the usual form. For instance, the above machine proof of Ceva’s theorem is essentially the same as the proof of Ceva’s theorem on page 11. It is clear that the proofs produced according to Algorithm 2.32 depend on how we describe a geometry statement constructively. For the same statement, some descriptions will lead to long proofs while other descriptions will lead to short ones. Both the way of introducing points and the way of formulating the conclusions will aﬀect the output. We use some examples to show some heuristic rules in specifying the statement in constructive 2.4 The Area Method 73 form which may lead to short proofs. Example 2.35 (Menelaus’ Theorem) A transversal meets the sides AB, BC, and CA of a tri-angle ABC in F, D, and E. Show that AF FB · BD DC · CE EA = −1. First describe the statement in the constructive way. A B C D E F Figure 2-13 Take three arbitrary points A, B, C. Take a point D on line BC. Take a point E on line AC. Take the intersection F of line DE and line AB. Show that AF FB · BD DC · CE EA = −1. The ndg conditions are C , B, A , C, DE ∦BA, B , F, C , D, and A , E. The machine proof CE AE · BD CD · AF BF F = −S ADE −S BDE · CE AE · BD CD E = ( AE AC −1)·(−S ACD)· AE AC (−S ABD· AE AC +S ABD)· AE AC · BD CD simplify = S ACD S ABD · BD CD D = BD BC ·(S ABC· BD BC −S ABC) S ABC· BD BC ·( BD BC −1) simplify = 1 The eliminants AF BF F =S ADE S BDE S BDE E = −(( AE AC −1)·S ABD) S ADE E = −(S ACD· AE AC) CE AE E = AE AC −1 AE AC S ABD D =S ABC· BD BC S ACD D =( BD BC −1)·S ABC BD CD D = BD BC BD BC −1 The above proof produced according to our algorithm is not the simplest one. By de-scribing the example constructively as follows, we can obtain a much shorter proof. Take arbitrary points A, B, C, X, Y. D is the intersection of BC and XY. E is the intersection of AC and XY. F is the intersection of AB and XY. Show that AF FB · BD DC · CE EA = −1. 74 Chapter 2. The Area Method The machine proof CE AE · BD CD · AF BF F = S AXY S BXY · CE AE · BD CD E = S CXY ·S AXY S BXY·S AXY · BD CD simplify = S CXY S BXY · BD CD D = S CXY ·S BXY S BXY·S CXY simplify = 1 The eliminants AF BF F =S AXY S BXY CE AE E =S CXY S AXY BD CD D = S BXY S CXY Example 2.36 (Gauss-line Theorem) Let A0, A1, A2, and A3 be four points on a plane, X be the intersection of A1A2 and A0A3, and Y be the intersection of A0A1 and A2A3. Let M1, M2, and M3 be the midpoints of A1A3, A0A2, and XY respectively. Then M1, M2, and M3 are collinear. 3 M 2 M 1 M Y X 3 A 2 A 1 A 0 A Figure 2-14 The constructive description Take arbitrary points A0, A1, A2, and A3. X = A0A3 ∩A1A2. Y = A2A3 ∩A1A0. M1 is the midpoint of A1A3. M2 is the midpoint of A0A2. M3 is the midpoint of XY. Show that S M1M2M3 = 0 Here is the machine proof. S M1M2M3 n = 1 2 S YM1M2+ 1 2 S XM1 M2 n = ( 1 2 )·( 1 2 S A2YM1+ 1 2 S A2XM1 + 1 2 S A0YM1+ 1 2 S A0XM1 ) n = ( 1 4 )·(−1 2 S A2A3X+ 1 2 S A1A2Y−1 2 S A0A3Y−1 2 S A0A1X) n = (−1 8 )(S 2 A0A2A1A3S A2A3X+S 2 A0A2A1A3S A0A1X+S A0A2A1A3S A1A2A3S A0A1A2 −S A0A2A1A3S A0A2A3S A0A1A3) simplify = (−1 8 )·(S A0A2A1A3·S A2A3X+S A0A2A1A3·S A0A1X+S A1A2A3·S A0A1A2−S A0A2A3·S A0A1A3) n = (−1 8 )·(−S A0A2A1A3·S A0A1A3A2·S A1A2A3·S A0A2A3+S A0A2A1A3·S A0A1A3A2·S A0A1A3·S A0A1A2+ S 2 A0A1A3A2·S A1A2A3·S A0A1A2−S 2 A0A1A3A2·S A0A2A3·S A0A1A3) simplify = ( 1 8 )·(S A0A2A1A3·S A1A2A3·S A0A2A3−S A0A2A1A3·S A0A1A3·S A0A1A2− S A0A1A3A2·S A1A2A3·S A0A1A2+S A0A1A3A2·S A0A2A3·S A0A1A3) n = ( 1 8 )·(0) simplify = 0 2.4 The Area Method 75 Here a n = b means b is the numerator of a. To show a = 0, we need only to show its numerator is zero. If we describe the statement as follows, then we can obtain a shorter proof. Take arbitrary points A0, A1, A2, A3. X = A0A3 ∩A1A2. Y = A2A3 ∩A1A0. M1 is the midpoint of A1A3. M2 is the midpoint of A0A2. M3 is the midpoint of XY. Z = M2M1 ∩XY. Show that XM3 YM3 = XZ YZ. The machine proof ( XM3 YM3 )/( XZM3 YZM3 ) ZM3 = −S YM1M2 −S XM1 M2 · XM3 YM3 M3 = ( 1 2 )·S YM1M2 S XM1 M2 ·(−1 2) M2 = −( 1 2 S A2YM1+ 1 2 S A0YM1) 1 2 S A2XM1 + 1 2 S A0XM1 M1 = −( 1 2 S A1A2Y−1 2 S A0A3Y) −1 2 S A2A3X−1 2 S A0A1X Y = S A0A1A2A3 S A0A1A2A3 simplify = 1 The eliminants XZM3 YZM3 ZM3 = S XM1 M2 S YM1M2 XM3 YM3 M3 = −(1) S XM1 M2 M2 = 1 2(S A2XM1 +S A0XM1) S YM1M2 M2 = 1 2(S A2YM1+S A0YM1) S A0XM1 M1 = −1 2(S A0A1X) S A2XM1 M1 = −1 2(S A2A3X) S A0YM1 M1 = −1 2(S A0A3Y) S A2YM1 M1 = 1 2(S A1A2Y) S A2A3X+S A0A1X=S A0A1A2A3 So for a “nice” expression of the conclusion of a statement, we can obtain a short proof. The idea is to express the conclusion as a = b such that a and b are symmetric in some sense. For example, if we want to prove three points P, Q, and R are collinear, and if one of the three points, say R, is on (LINE E F) where E and F are points in the statement, then we usually introduce a new point N by (INTER N (LINE P Q) (LINE E F)) and prove the equivalent conclusion N = R or ER FR = EN FN. According to our experience with many examples, the proof for this new conclusion is shorter than the proof for S PQR = 0. The following example shows that this rule is also true for the length ratios. Example 2.37 A line parallel to the base of trapezoid ABCD meet its two sides and two diagonals at H, G, F, and E. Show that EF = GH. We may describe the statements as follows. A B C D E H F G Figure 2-15 Take arbitrary points A, B, C. 76 Chapter 2. The Area Method Take a point D such that DC ∥AB. Take a point E on line BC. Take the intersection H of line AD and the line passing through E and parallel AB. F = BD ∩EH. G = AC ∩EF. Prove that EF AB = −HG AB . The machine proof EF AB −HG AB G = S ABC −S ACH · EF AB F = S BDE·S ABC −S ACH·S ABD H = −S BDE·S ABC·(−S ABD) (−S ACD·S ABE)·S ABD simplify = −S BDE·S ABC S ACD·S ABE E = −(−S BCD· BE BC )·S ABC S ACD·S ABC· BE BC simplify = S BCD S ACD D = −S ABC· CD AB −S ABC· CD AB simplify = 1 The eliminants HG AB G =S ACH S ABC EF AB F =S BDE S ABD S ACH H =S ACD·S ABE S ABD S ABE E =S ABC· BE BC S BDE E = −(S BCD· BE BC) S ACD D = −(S ABC· CD AB) S BCD D = −(S ABC· CD AB) By changing the conclusion to EF GH = 1, the proof will become much longer. 2.5 More Elimination Techniques The eleven lemmas (2.19–2.31) provide a complete method of eliminating points from geometry quantities. But if only these lemmas are used, the proofs for some geometry theorems are still too long to be readable. In order to produce short and readable proofs for geometry theorems, we will provide more elimination techniques. 2.5.1 Reﬁned Elimination Techniques Lemmas 2.19–2.31 only give the elimination result in the general case. In some special cases, the elimination results could be much simpler, e.g., see Excises 2.38, 2.39, and 2.40 below. By only using these reﬁned elimination results, we can obtain much shorter proofs 2.5 More Elimination Techniques 77 for many geometry theorems. Exercise 2.38 Prove the following version of Lemma 2.20. Let Y = PQ ∩UV. To eliminate point Y from S ABY, we have S ABY =                                            S ABU if AB ∥UV; S ABP if AB ∥PQ; S UBV·S APQ S UPVQ if U, V, A are collinear; S AUV ·S BPQ S UPVQ if U, V, B are collinear; S APQ·S BUV S PUQV if P, Q, A are collinear; S PBQ·S AUV S PUQV if P, Q, B are collinear; 1 S UPVQ (S UPQ · S ABV −S VPQ · S ABU) if U or V is on AB; 1 S PUQV (S PUV · S ABQ + S QVU · S ABP) otherwise. Exercise 2.39 Prove the following version of Lemma 2.22. Let Y be the intersection of the line UV and the line passing through R and parallel to line PQ. To eliminate point Y from S ABY, we have S ABY =                                      S ABU if AB ∥UV; S ABR if AB ∥PQ; S UBV·S APRQ S UPVQ if U, V, A are collinear; S AUV ·S BPRQ S UPVQ if U, V, B are collinear; S AUV ·S BQRP S PUQV if AY ∥PQ; S BUVS APRQ S PUQV if BY ∥PQ; 1 S PUQV (S PUQR · S ABV + S PRQV · S ABU) otherwise. Exercise 2.40 Prove the following version of Lemma 2.23. Let Y be the intersection of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. We have S ABY =                                      S ABW if AB ∥UV; S ABR if AB ∥PQ; S AUWV ·S BQRP S PUQV if AY ∥PQ; S BUWV·S APRQ S PUQV if BY ∥PQ; S BVWU·S APRQ S UPVQ if AY ∥UV; S AUWV ·S BPRQ S UPCQ if BY ∥UV; S ABY = S PWQR S PUQV · S AUBV + S ABW otherwise. 78 Chapter 2. The Area Method To use these elimination methods, we have ﬁrst to decide whether three points are collinear, or whether two lines are parallel. We can use Algorithm 2.32 to do so. But this process is too time consuming. A faster way is to collect all the obvious collinear and parallel relations from the constructions and use them as criteria. For instance, in Ceva’s theorem (Example 2.17 on page 61) we can ﬁnd the following collinear point sets easily: {A, B, F}, {A, C, E}, {B, C, D}, {A, D, P}, {B, E, P}, and {C, F, P}. Once we obtain all the lines and parallels for a geometry statement, we can use them to simplify some geometry quantities. For instance, S P1P2P3P4 =                    0 if P1P3 ∥P2P4; S P1P3P4 if P1, P2, P3 are collinear; S P1P2P4 if P2, P3, P4 are collinear; S P1P2P3 if P1, P3, P4 are collinear; S P2P3P4 if P1, P2, P4 are collinear. Example 2.41 Let A, B, and P be three noncollinear points, and C be a point on line PA. The line passing through C and parallel to AB intersects PB at D. Q is the intersection of AD and BC. M is the intersection of AB and PQ. Show that M is the midpoint of AB. The example can be described in the following constructive way. A B D C Q P M Figure 2-16 Take three arbitrary points A, B, and P. Take a point C on line AP. D is the intersection of line BP and the line passing through C and parallel to AB. Q is the intersection of lines AD and BC. M is the intersection of lines AB and PQ. Show that AM BM = −1. The ndg conditions: A , P, P < AB, AD ∦BC, and AB ∦PQ. The machine proof −AM BM M = −S APQ S BPQ Q = −S APD·S ABC·(−S ABDC) (−S BPC·S ABD)·S ABDC simplify = −S APD·S ABC S BPC·S ABD D = −S BPC·S ABC −S BPC·S ABC simplify = 1 The eliminants AM BM M = S APQ S BPQ S BPQ Q =S BPC·S ABD S ABDC S APQ Q =S APD·S ABC S ABDC S ABD D =S ABC S APD D = −(S BPC) 2.5 More Elimination Techniques 79 Our prover ﬁrst collects the collinear point sets: {M, P, Q}; {Q, A, D}; {Q, B, C}; {M, A, B}; {D, B, P}; {C, A, P}; and the parallel lines: DC ∥MAB. To eliminate D from S ABD, we use the ﬁrst case of Exercise 2.39: S ABD = S ABC. To elim-inate D from S APD, since P, B, and D are collinear we can use the fourth case of Exercise 2.39: S APD = S ABPS PACB S BAPB = −S PACB Since P, A, and C are collinear, S PACB = S PAC + S PCB = S BPC. 2.5.2 The Two-line Conﬁguration Another commonly used technique is the two-line conﬁguration. This trick works only when there exist at least ﬁve free or semi-free points and these points are on two lines l1 and l2. If l1 ∥l2, let α be the oriented distance from line l1 to line l2. If l1 ∦l2 let β = sin() ∠(l1, l2)) and O be the intersection of l1 and l2. Let l1 and l2 be two lines satisfying the above conditions. Then we have the following elimination procedure. Case 1. G = S ABC. S ABC =                      0 if A, B, C are collinear 1 2αBC if l1 ∥l2, A ∈l1, and B, C ∈l2 −1 2αBC if l1 ∥l2, A ∈l2, and B, C ∈l1 1 2βOA · BC if l1 ∦l2, A ∈l1, and B, C ∈l2 −1 2βOA · BC if l1 ∦l2, A ∈l2, and B, C ∈l1. Case 2. G = AB CD. We have G = ( AB CD) = AB/CD, i.e., we break one geometry quantity into the ratio of two quantities. Case 3. G = AB AB =            OB −OA if l1 and l2 intersects at O O1B −O1A if l1 ∥l2, A ∈l1 and B ∈l1 O2B −O2A if l1 ∥l2, A ∈l2 and B ∈l2 where O1 and O2 are ﬁxed points on l1 and l2 respectively. For the constructive description of Menelaus’ theorem on page 73, we have the following machine proof using the two-line trick. 80 Chapter 2. The Area Method CE AE · BD CD · AF BF F = −S ADE −S BDE · CE AE · BD CD 2lines = CE·BD·(−CD·AE·β)·(2) (−CE·BD·β)·(2)·CD·AE simplify = 1 The eliminants AF BF F =S ADE S BDE S BDE= −1 2(CE·BD·β) S ADE= −1 2(CD·AE·β) Example 2.42 (Pappus’ Theorem) Let points A, B and C be on one line, and A1, B1 and C1 be on another line. Let P = AB1 ∩A1B, Q = AC1 ∩A1C, and S = BC1 ∩B1C. Show that P, Q, and S are collinear. A B 1 A 1 B C 1 C P Q S Figure 2-17 The input to the program. Take arbitrary points A, A1, B, B1. Take a point C on line AB. Take a point C1 on line A1B1. P = A1B ∩AB1. Q = AC1 ∩A1C. S = B1C ∩BC1. T = B1C ∩PQ. Prove that B1S CS = B1T CT . The machine proof ( B1S CS )/( B1T CT ) T = S CPQ S B1PQ · B1S CS S = (−S BB1C1)·S CPQ S B1PQ·(−S BCC1 ) Q = S BB1C1·S A1CP·S ACC1 ·S AA1C1C (−S B1C1P·S AA1C)·S BCC1 ·(−S AA1C1C) simplify = S BB1C1·S A1CP·S ACC1 S B1C1P·S AA1C·S BCC1 P = S BB1C1·S A1BC·S AA1B1·S ACC1 ·S AA1 B1B (−S A1BB1·S AB1C1)·S AA1C·S BCC1 ·(−S AA1 B1B) simplify = S BB1C1·S A1BC·S AA1B1·S ACC1 S A1BB1·S AB1C1·S AA1C·S BCC1 2lines = B1C1·OB·β·(−BC·OA1·β)·A1B1·OA·β·(−AC·OC1·β)·((2))4 (−A1B1·OB·β)·B1C1·OA·β·AC·OA1·β·(−BC·OC1·β)·((2))4 simplify = 1 The eliminants B1T CT T = S B1PQ S CPQ B1S CS S = S BB1C1 S BCC1 S B1PQ Q = −S B1C1P·S AA1C S AA1C1C S CPQ Q = S A1CP·S ACC1 −S AA1C1C S B1C1P P = −S A1BB1·S AB1C1 S AA1B1B S A1CP P = S A1BC·S AA1B1 −S AA1B1B S BCC1 = −1 2(BC·OC1·β) S AA1C=1 2(AC·OA1·β) S AB1C1=1 2(B1C1·OA·β) S A1BB1= −1 2(A1B1·OB·β) S ACC1 = −1 2(AC·OC1·β) S AA1 B1= 1 2(A1B1·OA·β) S A1BC= −1 2(BC·OA1·β) S BB1C1=1 2(B1C1·OB·β) 2.6 Aﬃne Geometry 81 2.6 Area Method and Aﬃne Geometry We shall ﬁrst discuss brieﬂy the relationship between geometry and algebra, beginning with passages from E. Artin’s book “Geometric Algebra”, : We are all familiar with analytic geometry where a point in a plane is de-scribed by a pair (x, y) of real numbers, a straight line by a linear, a conic by a quadratic equation. Analytic geometry enables us to reduce any elemen-tary geometric problem to a mere algebraic one. The intersection of a straight line and a circle suggests, however, enlarging the system by introducing a new plane whose points are pairs of complex numbers. An obvious generalization of this procedure is the following. Let k be a given ﬁeld; construct a plane whose “points” are the pairs (x, y) of elements of k and deﬁne lines by linear equations. ... A much more fascinating problem is, however, the converse. Given a plane geometry whose objects are the elements of two sets, the set of points and the set of lines; assume that certain axioms of geometric nature are true. Is it possible to ﬁnd a ﬁeld k such that the points of our geometry can be described by coordinates from k and the lines by linear equations? These passages suggest that there are two approaches to deﬁning geometry. The Algebraic Approach. Starting from a number system E (usually ﬁelds), we can deﬁne ge-ometry objects and relations between those objects in the Cartesian product En (or En/E∗ in projective geometry). In modern geometry, especially in algebraic geometry, this ap-proach indisputably prevails. If we take this approach, then there are only a few diﬀerences between algebra and geometry; geometry can be regarded as a part of algebra. However, the second approach suggested by Artin is more attractive from the point of view of traditional proofs of geometry theorems. The Geometric Approach. By this approach we mean the one that was used by Euclid and Hilbert. In the Euclid-Hilbert system, number systems are developed as parts of the ge-ometry. For each model of a theory of geometry, we can prove the existence of a number system (usually a ﬁeld) inherent to that geometry. This ﬁeld is called the ﬁeld associated with that geometry. That geometry then can be represented as the Cartesian product of its associated ﬁeld. Though beautiful and elegant, the Euclid-Hilbert approach is on the other hand a heavy burden to develop. The axiom systems which we have adopted in this chapter is a mixture of the above approaches. First we take the number systems for granted. On the other hand we use a geometric language instead of an algebraic one. This system is a modiﬁcation of an axiom system developed by J.Z. Zhang for the purpose of geometry education . It has the advantage of providing simple but also general methods of solving geometric problems, a virtue the algebraic and the geometric approaches do not possess. 82 Chapter 2. The Area Method 2.6.1 Aﬃne Plane Geometry Aﬃne geometry is the study of incidence and parallelism. There are two kinds of geometric objects: points and lines. The only basic relation in this geometry is that of incidence, i.e., a point A is on a line l, or equivalently, a line l passes through (contains) a point A. Two lines which do not have a point in common are called parallel lines. The following is a group of axioms of aﬃne plane geometry . Axiom H.1. Given two distinct points P and Q, there exists a unique line passing through both P and Q. Axiom H.2. Given a line l and a point P not on l, there exists one and only one line m such that P lies on m and such that m is parallel to l. Axiom H.3. There exist three distinct points A, B, C such that C does not lie on the line passing through A and B. Axiom H.4 (Desargues’ Axiom). Let l1, l2, l3 be distinct lines which are either parallel or meet in a point S . Let A, A1 be points on l1, B, B1 points on l2 and C, C1 points on l3 which are distinct from S if our lines meet. We assume line AB ∥A1B1 and BC ∥B1C1. Show that AC ∥A1C1. Axiom H.5 (Pascalian Axiom). Let l and l1 be two distinct lines, and A, B, C and A1, B1, C1 be distinct points on l and l1, respectively. If BC1 ∥B1C and AB1 ∥A1B, then AC1 ∥A1C. A geometry in which all the above ﬁve axioms hold is called an aﬃne geometry. The above is a geometric approach for deﬁning aﬃne geometry. Now let us start at the other end and give a deﬁnition of aﬃne geometry based on the algebraic approach. Let E be a ﬁeld. From E we can construct a structure Ωas follows. Let ˜ L = {(a, b, c) \| a, b, c ∈E, a , 0 or b , 0}. We deﬁne a relation ∼in ˜ L as: (a, b, c) ∼(a′, b′, c′) if and only if there is a k ∈E such that k , 0 and (a, b, c) = (ka′, kb′, kc′). It is easy to see that ∼is an equivalence relation. Let ˜ L/ ∼(the set of all equivalence classes of ˜ L) be denoted by L. Deﬁne \|Ω\| to be E2 ∪L. An element p in \|Ω\| is a point if and only if p ∈E2 (i.e., p = (x, y), x, y ∈E); an element l in \|Ω\| is a line if and only if l ∈L. A point p = (x, y) is on a line l = (a, b, c) if and only if ax+by+c = 0. Two lines l1 = (a, b, c) and l2 = (a′, b′, c′) are parallel if there exists a k ∈E and k , 0 such that a = ka′, b = kb′. It is easy to check the following theorem. Theorem 2.43 Axioms H.1-H.5 are valid in the structure Ω. Proof. It can be easily checked that the ﬁve axioms are valid in Ω. H4 and H5, particularly, can be proved automatically using Wu’s method (Example 121 and Example 346 in ). 2.6 Aﬃne Geometry 83 The converse of the above theorem is a much deeper result. Theorem 2.44 Every geometry G of the theory H.1-H.5 is isomorphic to a structure Ωwith some ﬁeld E. The key step of the proof is to introduce the segment arithmetic and hence to introduce the ﬁeld E inherent to G. The ﬁeld E, uniquely determined by geometry G up to isomorphism, is called the ﬁeld associated with geometry G. Desargues’ axiom makes it possible to introduce a division ring E and Pascalian axiom makes E a commutative ﬁeld. Each alge-braic rule of operation (e.g., associativity of addition) corresponds to a geometry theorem. The process of introducing number systems in this way is the core of the Euclid-Hilbert approach. For details, see , , and . 2.6.2 Area Method and Aﬃne Geometry Suppose that the number ﬁeld E in the six axioms A.1–A.6 is not the real number ﬁeld R but an arbitrary ﬁeld. We shall show that these six axioms deﬁne an aﬃne geometry. Theorem 2.45 Show that all the ﬁve Axioms H.1-H.5 are consequences of Axioms A.1-A.6. Proof. Axiom H.1 follows from Corollary 2.6. Axiom H.3 is a consequence of Axioms A.3 and A.4. For Axiom H.2, see Example 2.13 on page 58. H.4 and H.5 can be proved automatically by our prover. For their proofs, see the following examples. Example 2.46 (Desargues’ Axiom) S AA1, S BB1, and SCC1 are three distinct lines. If AB ∥ A1B1 and AC ∥A1C1 then BC ∥B1C1. S A B C A B C Figure 2-18 1 1 1 Take arbitrary points S , A, B, and C. Take a point A1 on line S A. Take the intersection B1 of line S B and the line passing through A1 and parallel to AB. Take the intersection C1 of line SC and the line passing through A1 and parallel to AC. Prove that S B1BC = S C1BC. 84 Chapter 2. The Area Method The machine proof S BCB1 S BCC1 C1 = S BCB1·S S AC S ACA1 ·S S BC B1 = S ABA1·S S BC·S S AC S ACA1 ·S S BC·S S AB simplify = S ABA1·S S AC S ACA1 ·S S AB A1 = (−S S AB· S A1 S A +S S AB)·S S AC (−S S AC· S A1 S A +S S AC)·S S AB simplify = 1 The eliminants S BCC1 C1 = S ACA1 ·S S BC S S AC S BCB1 B1 = S ABA1·S S BC S S AB S ACA1 A1 = −(( S A1 S A −1)·S S AC) S ABA1 A1 = −(( S A1 S A −1)·S S AB) The ndg conditions are S , A, S, A, B and S, A, C are not collinear, which are conse-quences of the hypotheses of the statement. Example 2.47 (Pascalian Axiom) Let A, B and C be three points on one line, and A1, B1, C1 be three points on another line. If AB1 ∥A1B and AC1 ∥A1C then BC1 ∥B1C. A B A C B C Figure 2-19 1 1 1 The constructive description. Take arbitrary points A, B, and A1. Take a point C on line AB. Take a point B1 such that B1A ∥BA1. Take the intersection C1 of line A1B1 and the line passing through A and parallel to CA1. Prove that S BCB1 = S C1CB1. The machine proof S BCB1 S CB1C1 C1 = S BCB1 ·S A1CB1 −S AA1 B1C·S A1CB1 simplify = S BCB1 −S AA1 B1C B1 = −S BA1C· AB1 BA1 −S BA1C· AB1 BA1 simplify = 1 The eliminants S CB1C1 C1 =−S AA1 B1C S AA1B1C B1 =S BA1C· AB1 BA1 S BCB1 B1 = −(S BA1C· AB1 BA1 ) The ndg conditions are A , B, B , A1, and A1B1 ∦CA1 which are all in the statements of H.5. 2.6 Aﬃne Geometry 85 Now we have the converse theorem. Theorem 2.48 In the aﬃne geometry associated with any ﬁeld E, we can deﬁne length ratios and areas such that Axioms A.1-A.6 are valid. Proof. Let Pi = (xi, yi), i = 1, · · · , 4, be four points on a line l such that P3 , P4. Then P1P2 P3P4 = ( x1−x2 x3−x4 if y3 = y4. y1−y2 y3−y4 otherwise. Let Pi = (xi, yi), i = 1, 2, 3, be three points. Then deﬁne S P1P2P3 = k x1 y1 1 x2 y2 1 x3 y3 1 where k is any nonzero element in E. Axioms A.1-A.6 can be veriﬁed by direct calculation. Now it is clear that Algorithm 2.32 is for the constructive statements not only in Eu-clidean geometry but also in aﬃne geometry associated with any ﬁeld, even ﬁnite ﬁelds. In other words, the area method works also for ﬁnite geometries. For examples related to various ﬁelds, ﬁnite or inﬁnite, see Subsection 2.7.2. The completeness of Algorithm 2.32 is based on Proposition 2.33. For an arbitrary ﬁeld E, we have. Proposition 2.49 Let E be an inﬁnite ﬁeld and P a nonzero polynomial of indeterminates x1, · · · , xn with coeﬃcients in E. Show that we can ﬁnd elements e1, · · · , en in E such that P(e1, · · · , en) , 0. Proof. We prove the result by induction on n. If n = 1, let P(x1) be of degree d. Then P(x1) has at most d diﬀerent roots. Since E is inﬁnite in any d + 1 distinct elements of E there exists one which is not a root of P(x1). Suppose that the result is true for n −1. We write P as follows P(x1, ..., xn) = as(x1, ..., xn−1)xs n + ... + ao(x1, ..., xn−1). If s = 0, we need do nothing. If s > 0, by the induction hypotheses there are elements e1, .., en−1 in E such that as(e1, ..., en−1) , 0. Let Q(xn) = P(e1, ..., en−1, xn) , 0. Now the result can be proved similarly to the case n = 1. If E is a ﬁnite ﬁeld, then the above result is false and we do not know whether there exist eﬃcient algorithms to check the existence of such elements. Obviously, a slow algorithm exists: we can check all possible elements in En since E is ﬁnite. Note that the area is not an invariant in the aﬃne geometry. But due to the following fact the ratios of areas are invariants. 86 Chapter 2. The Area Method Exercise 2.50 Let M be a 2 × 2 matrix, Pi ∈E2, i = 1, 2, 3. Let Qi = PiM, i = 1, 2, 3. Then S Q1Q2Q3 = \|M\|S P1P2P3 where \|M\| is the determinant of M. So we can use ratios of areas instead of areas as geometry quantities. Also it is worth men-tioning that in the proofs of all the examples in this and the preceding chapters, the areas always occur in the form of ratios. This is not a coincidence. Let C(r1, · · · , rd, a1, · · · , as) = 0 be the conclusion of a geometry theorem where the ri are length ratios and the ai are areas of triangles. Let M = λI be the multiplication of an indeterminate λ and the unit matrix I. After transforming each point P in the plane to PM, C = 0 is still valid. By Exercise 2.50, C = (r1, · · · , rd, λ2a1, · · · , λ2as) = 0. Therefore, if E is an inﬁnite ﬁeld, each homogeneous component of P in the variables a1, · · · , as must be zero, i.e., without loss of generality we can assume P is homogeneous in the area variables. That is C can be expressed as a polynomial of the ratio of lengths and the ratio of areas. 2.7 Applications Besides theorem proving, the area method can be used to solve other geometry problems such as deriving unknown formulas automatically. In this section, we will show the appli-cation of the area method in three geometry topics: the formula derivation, the existence of n3 conﬁgurations, and the transversal problems. 2.7.1 Formula Derivation Algorithm 2.32 can be used to derive unknown formulas. We use a simple example to illustrate how this works. Example 2.51 Let L, M, and N be the midpoints of the sides AB, BC, and CA of triangle ABC respectively. Find the area of triangle LMN. Solution. Since N is the midpoint of AC, by Proposition 2.9, S LMN=1 2(S CLM + S ALM). By the co-side theorem, S ALM= −1 2(S ACL), S CLM=1 2(S BCL). Then S LMN= 1 2 S BCL−1 2 S ACL 2 = S ABC 4 . Example 2.52 Let A1, B1, and C1 be points on the sides BC, CA, and AB of a triangle ABC such that BA1/A1C = r1, CB1/B1A = r2, and AC1/C1B = r3. Show that S A1B1C1 S ABC = r3r2r1+1 (r1+1)(r2+1)(r3+1). 2.7 Applications 87 A B C A B C Figure 2-20 1 1 1 Constructive description Take arbitrary points A, B, and C. Take a point A1 such that BA1 A1C = r1. Take a point B1 such that CB1 B1A = r2. Take a point C1 such that AC1 C1B = r3. Compute S A1B1C1 S ABC The machine derivation. S A1B1C1 S ABC C1 = S BA1B1·r3+S AA1 B1 S ABC·(r3+1) B1 = −S ACA1 ·r2−S ACA1 +S ABA1 ·r3·r2 2+S ABA1 ·r3·r2 S ABC·(r3+1)·(r2+1)2 simplify = −(S ACA1 −S ABA1 ·r3·r2) S ABC·(r3+1)·(r2+1) A1 = −(−S ABC·r3·r2·r2 1−S ABC·r3·r2·r1−S ABC·r1−S ABC) S ABC·(r3+1)·(r2+1)·(r1+1)2 simplify = r3·r2·r1+1 (r3+1)·(r2+1)·(r1+1) The eliminants S A1B1C1 C1 = S BA1B1·r3+S AA1 B1 r3+1 S AA1B1 B1 = −S ACA1 r2+1 S BA1B1 B1 = S ABA1·r2 r2+1 S ABA1 A1 = S ABC·r1 r1+1 S ACA1 A1 = −S ABC r1+1 Remark. As a consequence of Example 2.52, we “discover” Menelaus’ theorem: A1, B1, and C1 are collinear iﬀr1r2r3 = −1. Example 2.53 Let A1, B1, C1, D1 be points on the sides CD, DA, AB, BC of a parallelogram ABCD such that CA1/CD = DB1/DA = AC1/AB = BD1/BC = r. Let A2B2C2D2 be the quadrilateral formed by the lines AA1 BB1, CC1, DD1. Compute S ABA2 S ABCD and S A2B2C2D2 S ABCD . 2 D 2 C 2 B 2 A 1 D 1 C 1 B 1 A D C B A Figure 2-21 The constructive description Take arbitrary points A, B, and C. Take a point D such that AB DC = 1. Take a point A1 such that CA1 CD = r. Take a point B1 such that DB1 DA = r. A2 = AA1 ∩BB1. Compute S ABA2 S ABCD . 88 Chapter 2. The Area Method The machine derivation. S ABA2 S ABCD A2 = S ABB1·S ABA1 S ABCD·S ABA1B1 B1 = (−S ABD·r+S ABD)·S ABA1 S ABCD·(S ADA1 ·r−S ADA1+S ABA1 ) simplify = −(r−1)·S ABD·S ABA1 S ABCD·(S ADA1 ·r−S ADA1+S ABA1 ) A1 = −(r−1)·S ABD·(S ABD·r−S ABC·r+S ABC) S ABCD·(S ACD·r2−2S ACD·r+S ACD+S ABD·r−S ABC·r+S ABC) D = −(r−1)·(S ABC)2 (2S ABC)·(S ABC·r2−2S ABC·r+2S ABC) simplify = −(r−1) (2)·(r2−2r+2) The eliminants S ABA2 A2 = S ABB1·S ABA1 S ABA1B1 S ABA1B1 B1 =S ADA1 ·r−S ADA1+S ABA1 S ABB1 B1 = −((r−1)·S ABD) S ADA1 A1 =(r−1)·S ACD S ABA1 A1 =S ABD·r−S ABC·r+S ABC S ACD D =S ABC S ABCD D =2(S ABC) S ABD D =S ABC Thus S ABA2 S ABCD = 1−r 2(r2−2r+2). To compute S A2B2C2D2 S ABCD , we have S A2B2C2D2 = S ABCD −S ABA2 −S BCB2 −S CDC2 −S DAD2 = (1 −4 · 1 −r 2(r2 −2r + 2))S ABCD = r2 r2 −2r + 2S ABCD Example 2.54 Let E, F, H, and G be points on sides AB, CD, AD, and BC such that AE AB = DF DC = r1 and AH AD = BG BC = r2. Let EF and HG meet in I. Compute EI EF and HI HG. A B C D E F H G I Figure 2-22 Constructive description Take arbitrary points A, B, C, D. Take a point E such that AE AB = r1. Take a point F such that DF DC = r1. Take a point H such that AH AD = r2. Take a point G such that BG BC = r2. I = EF ∩HG. Compute HI GI . The machine proof for this example is a little long. The following proof is the modiﬁ-cation of the machine proof. HI GI = S HEF S GEF = r2·S DEF+(1−r2)·S AEF r2·S CEF+(1−r2)·S BEF 2.7 Applications 89 = r2·r1·S DEC+(1−r2)·r1·S ABF r2·(1−r1)·S DEC+(1−r2)·(1−r1)·S ABF simplify = r1 1−r1 Thus HI EG = HI IG HI IG +1 = r1. Similarly EI EF = r2. Example 2.55 The sides AB and DC of a quadrilateral are cut into 2n + 1 equal segments by points P1, · · · , P2n and Q1, · · · , Q2n respectively. Show that (1) S PnPn+1Qn+1Qn = 1 2n+1S ABCD. (2) If sides BC and AD are cut into 2m + 1 equal segments by points R1, · · · , R2m and S 1, · · · , S 2m respectively, then the area of the quadrilateral formed by the lines PnQn, Pn+1Qn+1, RmS m, and Rm+1S m+1 is 1 (2n+1)(2m+1)S ABCD. Figure 2-23 shows the case n = m = 2. Note that in the following machine proof for (1), we use some diﬀerent names for points Pn, Pn+1, Qn+1, Qn. We also use a trick: point Qn+1 is introduced two times (Q and V) so that diﬀerent elimination methods will be used to eliminate Qn+1 in diﬀerent cases. Constructive description Take arbitrary points A, B, C, D. Take a point X such that AX AB = n 2n+1. Take a point U such that DU DC = n 2n+1. Take a point Q such that DQ DC = n+1 2n+1. Take a point V such that UV DC = 1 2n+1. Take a point Y such that XY AB = 1 2n+1. Compute (S AXY +S UXV ) S ABCD . The eliminants S XQY Y =−S ABQ 2n+1 S XUV V =−S CDX 2n+1 S ABQ Q =S ABD·n+S ABC·n+S ABC 2n+1 S CDX X =S BCD·n+S ACD·n+S ACD 2n+1 S ABCD=S ACD+S ABC S BCD=S ACD−S ABD+S ABC A B C D P P P P 1 3 2 4 Q Q Q Q 1 3 2 4 S S S S 1 3 2 4 R R R R 1 3 2 4 Figure 2-23 The machine proof −(S XQY+S XUV ) S ABCD Y = −(2S XUV ·n+S XUV −S ABQ) S ABCD·(2n+1) V = −(−2S CDX·n−S CDX−2S ABQ·n−S ABQ) S ABCD·(2n+1)2 90 Chapter 2. The Area Method simplify = S CDX+S ABQ S ABCD·(2n+1) Q = 2S CDX·n+S CDX+S ABD·n+S ABC·n+S ABC S ABCD·(2n+1)2 X = 2S BCD·n2+S BCD·n+2S ACD·n2+3S ACD·n+S ACD+2S ABD·n2+S ABD·n+2S ABC·n2+3S ABC·n+S ABC S ABCD·(2n+1)3 simplify = S BCD·n+S ACD·n+S ACD+S ABD·n+S ABC·n+S ABC S ABCD·(2n+1)2 area−co = 2S ACD·n+S ACD+2S ABC·n+S ABC (S ACD+S ABC)·(2n+1)2 simplify = 1 2n+1 By Example 2.54, PnQn and Pn+1Qn+1 are cut into 2m + 1 equal segments by RiS i, i = 1, ..., 2m respectively. Now (2) comes from (1) directly. For more examples of formula derivation, see Examples 1.11, 1.12, 1.13, and the exam-ples in the next subsection. 2.7.2 Existence of n3 Conﬁgurations We deﬁne a plane conﬁguration as a system of p points and l lines arranged in a plane in such a way that every point of the system is incident with a ﬁxed number λ of lines of the system and every straight line of the system is incident with a ﬁxed number π of points of the system. We characterize such a conﬁguration by the symbol (pλ, lπ). For example, the triangle forms the conﬁguration of (32, 32). The four numbers p, l, λ, and π may not be chosen arbitrarily. For, by the conditions we have stipulated, λp straight lines of the system, in all, pass through the p points; however, every line is counted π times because it passes through π points; thus the number of lines l is equal to λp π . Therefore, for every conﬁguration (pλ, lπ), we have λp = πl. We only discuss those conﬁgurations in which the number of points is equal to the number of lines, i.e., for which p = l. Then it follows from the relation λp = πl, that λ = π. The symbol for such a conﬁguration is always of the form (pλ, pλ). We shall introduce the more concise notation (pλ) for such a conﬁguration. We shall further limit the number λ. λ = 1 yields only the trivial conﬁguration con-sisting of a point and a line passing through it. The case λ = 2 is realized by the closed polygons in the plane. On the other hand, the case λ = 3 includes the most important con-ﬁgurations in projective geometry, the Fano conﬁguration, Desargues’ conﬁguration, and Pappus’ conﬁguration. In this case the number of points, p, must be at least seven. For through any given point of the conﬁguration there pass three lines, on each of which there must be two further points of the conﬁguration. If in a geometry, there exist p points and lines consisting of a (p3) conﬁguration, we say that the conﬁguration (p3) can be realized in that geometry. As an application of the area method, we obtain the suﬃcient and necessary conditions 2.7 Applications 91 for the existence of the (73), (83), and (93) conﬁgurations. For more complicated conﬁgu-rations, see . Example 2.56 There exists only one (73) conﬁguration and this conﬁguration can only be realized in the geometry whose associated ﬁeld E is of characteristic 2. (Fano plane) Proof. Let the seven points be Pi, i = 1, · · · , 7. The only possible (73) conﬁguration consists of the lines: P1 P1 P2 P2 P3 P3 P1 P2 P4 P4 P5 P4 P5 P6 P3 P5 P6 P7 P7 P6 P7 P P P P P P P Figure 2-24 1 2 7 4 5 3 6 Consider the following geometry problem. Take arbitrary points P1, P2, P4. Take a point P3 on line P1P2. Take a point P5 on line P1P4. P6 = P2P4 ∩P3P5. P7 = P2P5 ∩P3P4. Compute S P1P6P7. Using Algorithm 2.32, we have S P1P6P7 = (2) · ( P1P5 P1P4 −1) · ( P1P3 P1P2 −1) · P1P3 P1P2 · S P1P2P4 · P1P5 P1P4 ( P1P5 P1P4 · P1P3 P1P2 −1) · ( P1P5 P1P4 −P1P3 P1P2) . The (73) conﬁguration exists iﬀS P1P6P7 = 0, i.e. (2) · (P1P5 P1P4 −1) · (P1P3 P1P2 −1) · P1P3 P1P2 · S P1P2P4 · P1P5 P1P4 = 0. If P1P5 P1P4 −1 = 0, we have P4 = P5. If P1P3 P1P2 −1 = 0, we have P2 = P3. If P1P3 P1P2 = 0, we have P1 = P3. If S P1P2P4 = 0, we have P1, P2, and P4 are collinear. If P1P5 P1P4 = 0, we have P1 = P5. All the above cases will lead to degenerate conﬁgurations. Then the (73) conﬁguration exists iﬀ2 = 0, i.e., the associate ﬁeld of the geometry is of characteristic 2. Conﬁguration (83) also has only one possible table: P1 P1 P1 P2 P2 P3 P3 P4 P2 P4 P6 P3 P7 P4 P5 P5 P5 P8 P7 P6 P8 P7 P8 P6 92 Chapter 2. The Area Method Theorem 2.57 The (83) conﬁguration only exists in the geometry such that √ −3 belongs to E, the ﬁeld associated with the geometry. Proof. Consider the following geometry problem. Take arbitrary points P1, P2, P4. Take a point P5 such that P1P5 P1P2 = r1. Take a point P6 such that P4P6 P4P5 = r2. Take a point P8 such that P1P8 P1P4 = r3. P7 = P2P8 ∩P1P6. P3 = P2P6 ∩P5P8. Compute S P3P7P4. Using Algorithm 2.32, S P3P7P4 is found to be (r2 3(r2r2 + r2r1 −2r2 + r1r1 −r1 + 1) −r3r1(r2 −2r1 + 1) + r2 1)S P1P2P3(r1 −1)r2 (r3r2r1 −r2r1 + r2 −1)(r3(r2r1 −r2 −r1 + 1) + r1) . Then S P3P7P4 = 0 iﬀ r2 3(r2r2 + r2r1 −2r2 + r1r1 −r1 + 1) −r3r1(r2 −2r1 + 1) + r2 1 = 0 has solutions for r3. The discriminant of the quadratic equation is −3(r2 −1)2r2 1, and there-fore the result. Contrary to the (73) and (83) conﬁgurations which do not exist in the Euclidean plane, the case n = 9 gives rise to three essentially diﬀerent conﬁgurations, all of which can be realized in the Euclidean plane. The ﬁrst (93) conﬁguration is that related to the Pappus theorem, a machine proof of which can be found in Example 2.42 on page 80. Example 2.58 Prove the existence of the (93) conﬁguration as shown in Figure 2-25. Proof. Consider the following geometry problem. P P P P P P P P P Figure 2-25 1 2 3 4 5 6 7 8 9 Take arbitrary points P1, P3, and P5. Take a point P7 such that P1P7 P1P3 = r1. Take a point P8 such that P1P8 P1P5 = r2. Take a point P9 such that P3P9 P3P5 = r3. Take a point P2 such that P5P2 P5P7 = r4. P4 = P1P9 ∩P2P8. P6 = P3P8 ∩P2P9. Compute S P4P6P7. 2.7 Applications 93 Using Algorithm 2.32, S P4P6P7 is found to be (r4(r3r2 + r3r1 −2r3 −r2r1 −r2 + 2) −2r3r2 + 2r3 + 2r2 −2)r1r2(1 −r1)r3r4S P1P3P5 (r4r2r1 −r4 + r3r2 −r3 −r2 + 1)(r4r3r1 −r4r3 + r4 −r3r2 + r3 + r2 −1) . Then S P4P6P7 = 0 iﬀ r4(r3r2 + r3r1 −2r3 −r2r1 −r2 + 2) −2r3r2 + 2r3 + 2r2 −2 = 0, or equivalently, iﬀ r4 = (2r3r2 −2r3 −2r2 + 2)/(r3r2 + r3r1 −2r3 −r2r1 −r2 + 2) Example 2.59 Show the existence of the (93) conﬁguration as shown in Figure 2-26. Proof. Consider the following geometry problem. P P P P P P P P P Figure 2-26 1 4 7 3 6 8 9 5 2 Take arbitrary points P1, P4, and P7. Take a point P3 such that P1P3 P1P7 = r1. Take a point P6 such that P1P6 P1P4 = r2. Take a point P8 such that P3P8 P3P6 = r3. Take a point P9 such that P4P9 P4P7 = r4. P5 = P1P8 ∩P3P9. P2 = P4P8 ∩P6P9. Compute S P2P5P7. Using the program, we have S P2P5P7 = −f1· f2·S P1P4P7 d1·d2 where f1 = r4(r2 −r1) −r2r1 + r1 f2 = r4(r2 3r2 2 −r2 3r2r1 + r2 3r2 1 −r2 3r1 + r3r2r1 −r3r2 −2r3r2 1 + 2r3r1 + r2 1 −r1) −r2 3r2 1 + r2 3r1 + 2r3r2 1 −2r3r1 −r2 1 + r1 d1 = r4(r3(r2 −r1) + r1 −1) −r3r1(r2 + 1) + r2r1 −r1 d2 = r4(r3(r2 −r1) + r1) −r3r1(r2 + 1) −r1. Using the program again, we may check that f1 = 0 (i.e., r4 = (−r2r1 +r1)/(r1 −r2)) implies P3, P6, and P9 are collinear, which is a degenerate case. If f2 = 0, that is, r4 = −r2 3r2 1 + r2 3r1 + 2r3r2 1 −2r3r1 −r2 1 + r1 r2 3r2 2 −r2 3r2r1 + r2 3r2 1 −r2 3r1 + r3r2r1 −r3r2 −2r3r2 1 + 2r3r1 + r2 1 −r1 then S P2P5P7 = 0, and we obtain a realization for the conﬁguration shown in Figure 2-26. Remark 2.60 From the above two examples and Example 2.42, the three 93 conﬁgurations can be realized rationally, i.e., they can be realized in the geometry associated with the ﬁeld of rational numbers. 94 Chapter 2. The Area Method 2.7.3 Transversals for Polygons First, Example 1.8 on page 11 can be further generalized to the following form. Notice that in these theorems involving m points, the subscripts are understood to be mod m. Theorem 2.61 (Ceva’s Theorem for an m-polygon) Let V1...Vm be an m-polygon , and O a point. Let Pi be the intersection of line OVi and the side Vi+kVi+k+1. Then C(m, k) = Qm i=1 Vi+kPi PiVi+k+1 = 1 iﬀm is an odd number and k = m−1 2 . Proof. By the co-side theorem, Vi+kPi PiVi+k+1 = S OViVi+k S OVi+k+1Vi , i = 1, ..., m. Multiply the above equations together. We have that C(m, k) = 1 iﬀthe elements in the numerator are the same as the elements in the denominator. Let us assume that the i-th element in the numerator is the same as the j-th element in the denominator, i.e., S OViVi+k = S OVj+k+1Vj. Then i = j + k + 1 mod (m); i + k = j mod (m). The above two equations have solutions for i and j iﬀ 2k + 1 = 0 mod (m). The only nontrivial solution of the above equation is 2k + 1 = m which proves the theorem. By polygrams, we mean the ﬁgures formed by the diagonals of polygons. The Menelaus and Ceva type theorems are about the transversals for the sides of polygons. We will discuss some results involving the transversals of polygrams which were discovered by B. Gr¨ unbaum and G. C. Shephard using numerical searching (). Using the area method, we can not only prove these results easily but also strengthen some of them. Example 2.62 Let ABCD be a quadrilateral and O a point. Let E, F, G, and H be the in-tersections of lines AO, BO, CO, and DO with the corresponding diagonals of the quadri-lateral. Show that AH HC CF FA BE ED DG GB = 1. C 2.7 Applications 95 Constructive description Take arbitrary points A, B, C, D, and O. E = BD ∩AO. F = AC ∩BO. G = BD ∩CO. H = AC ∩DO. Prove that CF FA · BE ED · AH HC · DG GB = 1. The machine proof DG BG · CF AF · BE DE · AH CH H = S ADO S CDO · DG BG · CF AF · BE DE G = (−S CDO)·S ADO S CDO·S BCO · CF AF · BE DE simplify = −S ADO S BCO · CF AF · BE DE F = −(−S BCO)·S ADO S BCO·S ABO · BE DE simplify = S ADO S ABO · BE DE E = (−S ABO)·S ADO S ABO·(−S ADO) simplify = 1 The eliminants AH CH H = S ADO S CDO DG BG G =−S CDO S BCO CF AF F =−S BCO S ABO BE DE E = S ABO S ADO Example 2.62 is a special case of the following result. Theorem 2.63 Let an arbitrary polygon V1...Vm and a point O be given, together with a positive integer k such that 1 ≤k ≤m 2 . Let Pi,k be the intersection of line OVi and line Vi−kVi+k. Then m Y i=1 Vi+kPi,k Pi,kVi−k = 1. Proof. By the co-side theorem Vi+kPi,k Pi,kVi−k = S OViVi+k S OVi−kVi , i = 1, .., m. Multiplying the m equations together and noticing that the elements in the numerator and in the denominator are the same, we prove the result. The following more intricate extension of Ceva’s theorem contains the above example as the special case j = k. Theorem 2.64 Let an arbitrary polygon V1...Vm and a point O be given, and integers j and k that satisfy 1 ≤j ≤n−2, 1 ≤k ≤n−2 and j+k ≤n−1. Let P j,k,i denote the intersection point of the line Vi+kVi−j and the line OVi. We put C(m, j, k) = m Y i=1 Vi+kP j,k,i P j,k,iVi−j . Then C(m, j, k) = (−1)mC(n, j, m −k) = 1 C(m,k, j). 96 Chapter 2. The Area Method Proof. By the co-side theorem C(m, j, k, i) = Vi+kPi,k, j Pi,k, jVi−j = S OViVi+k S OVi−jVi . (1) Replacing k by m −k in (1), we have C(m, j, m −k, i) = Vi+m−kPi,m−k, j Pi,m−k, jVi−j = S OViVi+m−k S OVi−jVi = −S OVi−kVi S OVi−jVi . (2) Interchanging k and j in (1), we have C(m, k, j, i) = S OViVi+j S OVi−kVi . (3) From (1) and (2), it is clear that C(m, j, k) = (−1)mC(n, j, m −k). From (1) and (3), we have C(m, j, k) = 1 C(m,k, j). Example 2.65 As shown in Figure 2-28, ABCDE −PQRS T is a pentagram. Then AT PD DR S B BP QE ES TC CQ RA = AP T D DS RB BQ PE ET SC CR QA = 1. To prove AT PD DR S B BP QE ES TC CQ RA = 1, we formulate the problem as follows. A B C D E P Q R S T Figure 2-28 Constructive description Take arbitrary points A, B, C, D, and E. P = AD ∩BE. Q = AC ∩BE. R = BD ∩AC. S = BD ∩CE. T = AD ∩CE. Show that AT AD DR DB BP BE ES EC CQ CA = PD AD S B DB QE BE TC EC RA CA 2.7 Applications 97 The machine proof −1 EQ BE · DP AD · CT CE · BS BD · AR AC · ES CE · DR BD· CQ AC · BP BE · AT AD T = −S ACE·(−S ACDE) EQ BE · DP AD ·(−S ACD)· BS BD · AR AC ·S ACDE · ES CE · DR BD · CQ AC · BP BE simplify = −S ACE EQ BE · DP AD ·S ACD· BS BD · AR AC · ES CE · DR BD · CQ AC · BP BE S = −S BDE·S ACE·S BCDE EQ BE · DP AD ·S ACD·S BCE· AR AC ·(−S BCDE) · DR BD · CQ AC · BP BE simplify = S BDE·S ACE EQ BE · DP AD ·S ACD·S BCE· AR AC · DR BD · CQ AC · BP BE R = S BDE·S ACD·S ACE·S ABCD EQ BE · DP AD ·S ACD·S BCE·S ABD·(−S ABCD) · CQ AC · BP BE simplify = S BDE·S ACE −EQ BE · DP AD ·S BCE·S ABD · CQ AC · BP BE Q = S BDE·(−S BCE)·S ACE·(−S ABCE) −S ACE· DP AD ·S BCE·S ABD·S ABCE · BP BE simplify = −S BDE DP AD ·S ABD · BP BE P = −S BDE·(−S ABD)·S ABDE (−S BDE)·S ABD·(−S ABDE) simplify = 1 The eliminants CT CE T = S ACD S ACDE AT AD T = S ACE S ACDE BS BD S = S BCE S BCDE ES CE S = S BDE −S BCDE AR AC R = S ABD S ABCD DR BD R = S ACD −S ABCD EQ BE Q = S ACE −S ABCE CQ AC Q = −S BCE S ABCE DP AD P =−S BDE S ABDE BP BE P = S ABD S ABDE The second result in Example 2.65 is equivalent to the following statement. 98 Chapter 2. The Area Method B E A C D 1 A 1 B 1 C 1 D 1 E Figure 2-29 Example 2.66 (Theorem of Pratt-Kasapi) Let ABCDE be a pentagon. A1B1 ∥AC, B1C1 ∥ BD, C1D1 ∥CE, D1E1 ∥AD, E1A1 ∥EB. Show that A1B · B1C · C1D · D1E · E1A = BB1 · CC1 · DD1 · EE1 · AA1. The constructive description. Take arbitrary points A, B, C, D, and E. Take the intersection A1 of the line passing through B and parallel to CA and the line passing through A and parallel to EB. Take the intersection B1 of the line passing through C and parallel to BD and the line passing through B and parallel to AC. Take the intersection C1 of the line passing through D and parallel to CE and the line passing through C and parallel to BD. Take the intersection D1 of the line passing through E and parallel to AD and the line passing through D and parallel to CE. Take the intersection E1 of the line passing through A and parallel to BE and the line passing through E and parallel to AD. Show that A1B BB1 B1C CC1 C1D DD1 D1E EE1 E1A AA1 = 1. The machine proof −ED1 EE1 · DC1 DD1 · CB1 CC1 · BA1 BB1 · AE1 AA1 E1 = −(−S BED1)·(−S ADE) (−S ADA1 )·(−S ABE) · DC1 DD1 · CB1 CC1 · BA1 BB1 D1 = −(−S ABDE)·S CDE·(−S ADC1 )·S ADE S ADA1 ·S ABE·(−S ADE)·S ACDE · CB1 CC1 · BA1 BB1 simplify = S ABDE·S CDE·S ADC1 S ADA1·S ABE·S ACDE · CB1 CC1 · BA1 BB1 C1 = S ABDE·S CDE·S ACDE·S BCD·(−S CEB1) S ADA1 ·S ABE·S ACDE·S CDE·(−S BCDE) · BA1 BB1 simplify = S ABDE·S BCD·S CEB1 S ADA1 ·S ABE·S BCDE · BA1 BB1 B1 = S ABDE·S BCD·(−S BCDE·S ABC)·(−S BDA1) S ADA1 ·S ABE·S BCDE·S BCD·(−S ABCD) simplify = −S ABDE·S ABC·S BDA1 S ADA1 ·S ABE·S ABCD A1 = −S ABDE·S ABC·S ABCD·S ABE·(−S ABCE) S ABDE·S ABC·S ABE·S ABCD·S ABCE simplify = 1 The eliminants AE1 AA1 E1 = S ADE S ADA1 ED1 EE1 E1 = S BED1 S ABE DC1 DD1 D1 = S ADC1 S ADE S BED1 D1 = −S ABDE·S CDE S ACDE CB1 CC1 C1 = −S CEB1 S CDE S ADC1 C1 = S ACDE·S BCD −S BCDE BA1 BB1 B1 = −S BDA1 S BCD S CEB1 B1 = S BCDE·S ABC S ABCD S ADA1 A1 = S ABDE·S ABC −S ABCE S BDA1 A1 = S ABCD·S ABE S ABCE 2.7 Applications 99 Examples 2.65 and 2.66 are special cases of the following general results. Note that the proofs of these general results are a natural extension of the machine proofs for the two examples. Let V1...Vm be a polygram and 1 ≤d ≤m 2 , 1 ≤j ≤m 2 integers. We denote by Pd, j,i the intersections of lines ViVi+d and lines Vi+jVi+j+d, i = 1, ..., m. Then Pd, j,i−j is the intersection of line Vi−jVi−j+d and line ViVi+d. Let T(m, d, j) = m Y i=1 ViPd, j,i Pd, j,i−jVi+d ; S (m, d, j) = m Y i=1 ViPd, j,i−j Pd, j,iVi+d . The result in Example 2.65 is T(5, 2, 1) = S (5, 2, 1) = 1. In general, we have Theorem 2.67 T(m, d, j) = 1 iﬀone of the following cases are true • d + 2 j = m; • 2d + j = m. Proof. By the co-side theorem, ViPd, j,i Pd, j,i−jVi+d = ViPd, j,i ViVi+d · ViVi+d Pd, j,i−jVi+d = S ViVi+jVi+j+d S ViVi+jVi+dVi+j+d S Vi−jViVi−j+dVi+d S Vi−jVi−j+dVi+d . Multiplying the m equations together, we see that T(d, j, m) = 1 iﬀthe areas of triangles and the areas of quadrilaterals in the numerator are the same as the ones in the denominator respectively. Let us assume that the i-th area in the numerator is the same as the x-th area in the denominator, i.e. S ViVi+jVi+j+d = ±S Vx−jVx−j+dVx+d. Then the point sets {Vi, Vi+j, Vi+j+d} and {Vx−j, Vx−j+d, Vx+d} should be the same. Considering the order, there are six possible matches. One of the matches is i + j = x + d; i + j + d = x −j; i = x −j + d where the = is understood to be mod(m). Then it is easy to see that the three equations are true for all i and x iﬀ 2d + j = 0 mod(m). Since 1 ≤d ≤m 2 , 1 ≤j ≤m 2 , the only possible solution is 2d + j = m. The other ﬁve cases can be treated similarly and d + 2 j = m is the only nontrivial solution. Theorem 2.68 S (m, d, j) = 1 iﬀone of the following cases are true • d + 2 j = m; • 2d = j; 100 Chapter 2. The Area Method • 2 j = d. Proof. By the co-side theorem, ViPd, j,i−j Pd, j,iVi+d = ViPd, j,i−j ViVi+d · ViVi+d Pd, j,iVi+d = S ViVi−jVi−j+d S Vi+jVi+j+dVi+d S Vi+jViVi+j+dVi+d S ViVi−jVi+dVi−j+d . Now, we can prove it in a similar way as Theorem 2.67. At the present time, our prover can not deal with geometry statements about m-polygons for an arbitrary number m. But our prover can prove these results for any concrete m, e.g. quadrilaterals, pentagons, etc. Summary of Chapter 2 • The following basic propositions are the deductive basis of the area method. 1. If points C and D are on line AB and P is any point not on line AB, then S PCD S PAB = CD AB . 2. (The co-side theorem) Let M be the intersection of two non parallel lines AB and PQ and M , Q. Then PM QM = S PAB S QAB ; PM PQ = S PAB S PAQB ; QM PQ = S QAB S PAQB . 3. Let R be a point on line PQ. Then S RAB = PR PQ S QAB + RQ PQ S PAB. 4. PQ ∥AB iﬀS PAQB = S PAB −S QAB = S BPQ −S APQ = 0. 5. Let ABCD be a parallelogram, P and Q be two points. Then S APQ + S CPQ = S BPQ + S DPQ or S PAQB = S PDQC. 6. Let ABCD be a parallelogram and P be any point . Then S PAB = S PDC −S ADC = S PDAC. • The Hilbert intersection point statements are geometry statements whose hypothe-ses can be described constructively and whose conclusions can be represented by polynomial equations in two geometry quantities: the ratio of collinear or parallel segments and the signed areas of triangles or quadrilaterals. 2.7 Applications 101 • The area method can eﬃciently produce short and readable proofs for the Hilbert in-tersection point statements. The proving process is to eliminate points from geometry quantities using Lemmas 2.19 - 2.31. • The area method works for constructive statements in the aﬃne geometry associated with any ﬁeld. • The area method is used to solve the following geometry problems: deriving un-known geometry formulas, ﬁnding the necessary and suﬃcient conditions for the existence of n3 conﬁgurations, and proving theorems about the transversals for arbi-trary polygons. 102 Chapter 2. The Area Method Chapter 3 Machine Proof in Plane Geometry In Chapter 2, we presented an automated theorem proving method for constructive state-ments involving collinearity and parallelism. In this chapter, we will discuss constructive statements involving perpendicular lines and circles. The key tool for dealing with perpen-dicularity is the Pythagoras diﬀerence, which is essentially the same as the inner product. Therefore, the method presented in this chapter is actually for constructive statements in metric geometry. 3.1 The Pythagoras Diﬀerence For three points A, B, and C, the Pythagoras diﬀerence PABC is deﬁned as PABC = AB 2 + CB 2 −AC 2. Note that in the above deﬁnition, we use a new geometry object: the square distance be-tween two points A and B, i.e., AB 2. For four points A, B, C, and D, we deﬁne PABCD = AB 2 + CD 2 −BC 2 −DA 2. For basic properties of the Pythagoras diﬀerence, see Section 1.7. 3.1.1 Pythagoras Diﬀerence and Perpendicular Besides collinear and parallel considered in Chapter 2, we now have a new basic geometry relation: line l is perpendicular to line l′, denoted by l⊥l′. Following are some basic properties of the perpendicularity. 103 104 Chapter 3. Machine Proof in Plane Geometry 1. If l⊥l′ then l′⊥l. 2. Let P be a point and l a line. Then there exists a unique line l′ which passes through point P and is perpendicular to line l. 3. If two distinct lines l′ and l′′ are both perpendicular to line l then l′ is parallel to line l′′. 4. (Pythagorean Theorem) AB⊥BC iﬀAC 2 = AB 2 + BC 2, i.e., iﬀPABC = 0. For the proof of the Pythagorean theorem, see Example 1.42 on page 24 and Proposi-tion 1.55 on page 28. But here, we take Pythagorean theorem as a basic property of the Pythagoras diﬀerence. Other propositions in this subsection can be derived from it. For four points A, B, C, and D, the notation AB⊥CD implies that one of the following conditions is true: A = B, or C = D, or the line AB is perpendicular to line CD. Proposition 3.1 AB⊥CD iﬀPACD =PBCD or PACBD = 0. Proof. See Proposition 1.62 on page 30. The above generalized Pythagorean proposition is one of the most useful tools in our mechanical theorem proving method. Proposition 3.2 Let D be the foot of the perpendicular drawn from point P to a line AB. Then we have AD DB = PPAB PPBA , AD AB = PPAB 2AB 2, DB AB = PPBA 2AB 2. Proof. See Proposition 1.65 on page 31. Proposition 3.3 Let AB and PQ be two nonperpendicular lines and Y be the intersection of line PQ and the line passing through A and perpendicular to AB. Then PY QY = PPAB PQAB , PY PQ = PPAB PPAQB , QY PQ = PQAB PPAQB . Proof. See Proposition 1.66 on page 31. Proposition 3.4 Let A, B, and C be three diﬀerent collinear points. Then for any point P if PPAC , 0 we have PPAB PPAC = AB AC. 3.1 The Pythagoras Diﬀerence 105 A B P C Q Figure 3-1 Proof. Let Q be the orthogonal projection from P to line AB. By Proposition 3.1, PPAB = PQAB = 2AQ · AB; PPAC = PQAC = 2AQ · AC. Now the result is clear. A B P Q R P Q R Figure 3-2 1 1 1 Proposition 3.5 Let R be a point on line PQ with position ratios r1 = PR PQ, r2 = RQ PQ with respect to PQ. Then for any points A and B, we have PRAB = r1PQAB + r2PPAB PARB = r1PAQB + r2PAPB −r1r2PPQP. Proof. We ﬁrst assume RA 2 = r1QA 2 + r2PA 2 −r1r2PQ 2 (1) RB 2 = r1QB 2 + r2PB 2 −r1r2PQ 2. (2) Then PRAB = RA 2 + AB 2 −RB 2 = r1(QA 2 + AB 2 −QB 2) + r2(PA 2 + AB 2 −PB 2) = r1PQAB + r2PPAB. The second one can be proved similarly. To prove (1), let us ﬁrst notice that by Proposition 3.4, PAPR PAPQ = PR PQ = r1. Then r1QA 2 + r2PA 2 −r1r2PQ 2 = r1QA 2 + (1 −r1)PA 2 −r1(1 −r1)PQ 2 = PA 2 + r1(QA 2 −PA 2 −PQ) + r2 1PQ 2 = PA 2 + PR 2 −r1PAPQ = PA 2 + PR 2 −PAPR = AR 2. 3.1.2 Pythagoras Diﬀerence and Parallel Proposition 3.6 For a parallelogram ABCD, we have AC 2 + BD 2 = 2AB 2 + 2BC 2, i.e., PABC = −PBAD. Proof. Let O be the intersection of the diagonals AC and BD. By Proposition 3.5, AC 2 = 4AO 2 = 4(1 2AB 2 + 1 2AD 2 −1 4BD 2) = 2AB 2 + 2AD 2 −BD 2. 106 Chapter 3. Machine Proof in Plane Geometry The following proposition shows how the Pythagoras diﬀerence changes under a paral-lel translation. Proposition 3.7 Let ABCD be a parallelogram. Then for any points P and Q, we have PAPQ + PCPQ = PBPQ + PDPQ or PAPBQ = PDPCQ PPAQ + PPCQ = PPBQ + PPDQ + 2PBAD Proof. Let O be the intersection of AC and BD. By the ﬁrst equation of Proposition 3.5, 2POPQ = PAPQ + PCPQ = PBPQ + PDPQ. By the second equation of Proposition 3.5, 2PPOQ = PPAQ + PPCQ −1 2PACA = PPBQ + PPDQ −1 2PBDB. We need only to show that 2PBAD = 1 2(PACA −PBDB) which is a consequence of Proposition 3.6. Proposition 3.8 Let ABCD be a parallelogram and P be any point. Then PPAB = PPDC −PADC = PPDAC PAPB = PAPA −PPDAC Proof. By Proposition 3.7, PPAB = PPAC −PPAD = PCADP = PPDAC = PPDC −PADC. For the second equation, PAPB = PAPA + PAPC −PAPD = PAPA + PCPDA = PAPA −PPDAC. Proposition 3.9 If PR ∥AC and QS ∥BD then PPQRS PABCD = PR AC · QS BD. Proof. Let X and Y be points such that PR = AX, QS = BY. By Propositions 3.7 and 3.4, PPQRS = PPBRY = PPBY −PRBY = QS BD (PPBD −PRBD) = QS BD PPBRD. Similarly we have PPBRD = PR ACPABCD. Proposition 3.10 Let AB ∥CD. Then AB CD = PADBC 2CD 2 . Proof. By Proposition 3.9, AB CD = AB CD · CD CD = PACBD PCCDD = PADBC 2CD 2 . Exercises 3.11 3.1 The Pythagoras Diﬀerence 107 1. Let ABCD be a parallelogram. Then 1. PABC = PADC = −PBAD = −PBCD. 2. PABD = PBDC; PCBD = PADB. 3. PADB −PADC = 2AD 2. 2. Let ABCD be a parallelogram and O be the intersection of its diagonals. For any point P, show that 1. PPAB + PPBC + PPCD + PPDA = 2(AB 2 + BC 2). 2. PAPD + PBPC −PCPD −PDPD = 2(AB 2 −BC 2). 3. PAPD + PBPC + PCPD + PDPD = 8PO 2. 3.1.3 Pythagoras Diﬀerence and Area In this subsection, we will prove the Herron-Qin formula which connects the area and the Pythagoras diﬀerences of a triangle. Deﬁnition 3.12 Let F be the foot of the perpendicular drawn from point R to line PQ. The signed distance from R to PQ, denoted by hR,PQ, is a real number which has the same sign as S RPQ and \|hR,PQ\| = \|RF\|. Proposition 3.13 For any two triangles ABC and RPQ, let hA = hA,BC, hR = hR,PQ. Show that S ABC \|BC\|hA = S RPQ \|PQ\|hR. B C A R P Q F M Figure 3-3 Proof. Without loss of generality, we assume that points B, C, P, and Q are on the same line. As in Figure 3-3, let RF be the altitude of tri-angle RPQ and M be a point on RF such that AM ∥BC. Then S ABC = S MBC. By Proposi-tions 2.7 and 2.8, S ABC S APQ = BC PQ ; S MPQ S RPQ = MF RF . Multiplying the two formulas together and noticing that hA and hR have the same sign as S ABC and S RPQ, we have S ABC \|BC\|hA = S RPQ \|PQ\|hR. Corollary 3.14 For a triangle ABC, we have hA,BC\|BC\| = hB,CA\|AC\| = hC,AB\|AB\|. 108 Chapter 3. Machine Proof in Plane Geometry Proof. Putting △RPQ to be △BCA and △CAB in Proposition 3.13 respectively, we obtain the result. By Proposition 3.13, we have S ABC = khA\|BC\| = khB\|AC\| = khC\|AB\| where k is a constant which is independent of the triangle ABC. Setting k = 1/2, we obtain the usual formula for the areas of triangles. Proposition 3.15 For a triangle ABC, S ABC = 1 2hA\|BC\| = 1 2hB\|AC\| = 1 2hC\|AB\|. Proposition 3.16 (The Herron-Qin Formula) For a triangle ABC, we have 16S 2 ABC = 4AB 2AC 2− P2 BAC. C A B F Figure 3-4 Proof. Let F be the foot of the perpendicular line drawn from point A to lien BC. By Proposition 3.4, PABC PABF = BC BF. Then PABC = BC BF PABF = BC BF PFBF = 2BC · BF. Then 16S 2 ABC = 4AF 2 · BC 2 = 4(AB 2 −BF 2)BC 2 = 4AB 2 · AC 2 −P2 BAC. Proposition 3.17 (The Herron-Qin Formula for Quadrilaterals) For any quadrilateral ABCD, we have 16S 2 ABCD = 4AC 2 · BD 2 −P2 ABCD. Proof. Take a point X such that CXDB is a parallelogram. Then CX = BD. By Propositions 2.11, 3.7, and the Herron-Qin formula for triangles, S 2 ABCD = S 2 AACX = S 2 XAC = 1 16(4AX 2 · AC 2 −P2 XAC) = 1 16(4BD 2 · AC 2 −P2 XAAC) = 1 16(4BD 2 · AC 2 −P2 BADC). You may compare the proofs for the two the Herron-Qin formulas based on trigonometric functions on page 38. The proof given here is independent of the concept of angles. Exercises 3.18 3.2 Constructive Geometry Statements 109 1. Prove the following forms of the Herron-Qin formula. • 16S 2 ABC = PACBPABC + PBCBPBAC. • 16S 2 ABC = PBACPACB + PACAPABC. • 16S 2 ABC = PCABPCBA + PABAPACB. 2. The absolute value of the area of a square is equal to the square of its side. Use this result to prove the Pythagorean theorem. (Hint. Use Figure 3-5.) Figure 3-5 3.2 Constructive Geometry Statements By constructive geometry statements, we mean statements which are assertions about con-ﬁgurations that can be drawn using only a ruler and a pair of compasses. More precisely, these conﬁgurations can be constructed by ﬁrst taking arbitrary points, lines and circles and then taking the intersections of two lines and of lines and circles in a prescribed manner. From the constructed points, we can form new lines and circles. By forming the intersec-tions of these new lines and circles, we can obtain new points, etc. Finally, we obtain a conﬁguration consisting of points, lines and circles. The class of constructive statements is denoted by C. It is clear that the Hilbert intersection point statements belong to class C. In this sec-tion, we introduce a new subset of C, i.e., the linear constructive geometry statement CL, which is larger than CH and has the advantage that on one hand it contains most of the com-monly used geometry theorems and on the other hand its readable proofs can be obtained eﬃciently by a mechanical method. 3.2.1 Linear Constructive Geometry Statements Now we have three basic geometric quantities: • the area of a triangle or a quadrilateral, • the Pythagoras diﬀerence of a triangle or a quadrilateral, and • the ratio of parallel line segments. 110 Chapter 3. Machine Proof in Plane Geometry By Proposition 3.10, the ratio of parallel line segments can be represented as expressions in Pythagoras diﬀerences. Points are the basic geometry objects. From points, we can introduce two other basic geometric objects: lines and circles. A straight line can be given in one of the following four forms. (LINE U V) is the line passing through two points U and V. (PLINE W U V) is the line passing through point W and parallel to (LINE U V). (TLINE W U V) is the line passing through point W and perpendicular to (LINE U V). (BLINE U V) is the perpendicular-bisector of UV. To make sure that the four kinds of lines are well deﬁned, we need to assume U , V which is called the nondegenerate condition (ndg) of the corresponding line. A circle with point O as its center and passing through point U is denoted by (CIR O U). A construction is one of the following ways of introducing new points. For each con-struction, we also give its ndg condition and the degree of freedom for the constructed point. C1 (POINT[S] Y1, · · · , Yl). Take arbitrary points Y1, · · · , Yl in the plane. Each Yi has two degrees of freedom. C2 (ON Y ln). Take a point Y on a line ln. The ndg condition of C2 is the ndg condition of the line ln. Point Y has one degree of freedom. C3 (ON Y (CIR O P)). Take a point Y on a circle (CIR O P). The ndg condition is O , P. Point Y has one degree of freedom. C4 (INTER Y ln1 ln2). Point Y is the intersection of line ln1 and line ln2. Point Y is a ﬁxed point. The ndg condition is ln1 ∦ln2. More precisely, we have 1. If ln1 is (LINE U V) or (PLINE W U V) and ln2 is (LINE P Q) or (PLINE R P Q), then the ndg condition is UV ∦PQ. 2. If ln1 is (LINE U V) or (PLINE W U V) and ln2 is (BLINE P Q) or (TLINE R P Q), then the ndg condition is ¬(UV⊥PQ). 3. If ln1 is (BLINE U V) or (TLINE W U V) and ln2 is (BLINE P Q) or (TLINE R P Q), then the ndg condition is UV ∦PQ. C5 (INTER Y ln (CIR O P)). Point Y is the intersection of line ln and circle (CIR O P) other than point P. Line ln could be (LINE P U), (PLINE P U V), and (TLINE P U V). The ndg conditions are O , P, Y , P, and line ln is not degenerate. Point Y is a ﬁxed point. 3.2 Constructive Geometry Statements 111 C6 (INTER Y (CIR O1 P) (CIR O2 P)). Point Y is the intersection of the circle (CIR O1 P) and the circle (CIR O2 P) other than point P. The ndg condition is O1, O2, and P are not collinear. Point Y is a ﬁxed point. C7 (PRATIO Y W U V r). Take a point Y on the line passing through W and parallel to line UV such that WY = rUV, where r can be a rational number, a rational expression in geometric quantities, or a variable. If r is a ﬁxed quantity, then Y is a ﬁxed point; if r is a variable then Y has one degree of freedom. The ndg condition is U , V. If r is a rational expression in geometry quantities then we will further assume that the denominator of r could not be zero. C8 (TRATIO Y U V r). Take a point Y on line (TLINE U U V) such that r = 4S UVY PUVU (= UY UV), where r can be a rational number, a rational expression in geometric quantities, or a variable. If r is a ﬁxed quantity then Y is a ﬁxed point; if r is a variable then Y has one degree of freedom. The ndg condition is the same as that of C7. The point Y in each of the above constructions is said to be introduced by that construction. Since there are four kinds of lines, constructions C2, C4, and C5 have 4, 10, and 3 possible forms respectively. Thus, in total, we have 22 diﬀerent forms of constructions. Now class CL, the class of the linear constructive geometry statements, can be deﬁned similarly as CH, i.e., a statement in class CL is a list S = (C1, C2, . . . , Ck, G) where Ci, i = 1, . . . , k, are constructions such that each Ci introduces a new point from the points introduced before; and G = (E1, E2) where E1 and E2 are polynomials in geometric quantities of the points introduced by the Ci and E1 = E2 is the conclusion of the statement. Let S = (C1, C2, . . . , Ck, (E1, E2)) be a statement in CL. The ndg condition of S is the set of ndg conditions of the Ci plus the condition that the denominators of the length ratios in E1 and E2 are not equal to zero. We call the statements in CL linear, because each of the constructions C1–C8 introduces a unique point. For the constructions involving circles, this fact may not be obvious. See the next subsection for more discussions. Example 3.19 The orthocenter theorem on page 32 can be described in the following con-structive way. ((POINTS A B C) (INTER E (LINE A C) (TLINE B A C)) 112 Chapter 3. Machine Proof in Plane Geometry (INTER F (LINE C B) (TLINE A C B)) (INTER H (LINE A F) (LINE B E)) (PACH = PBCH)) AB⊥CH. The ndg condition: A , C, C , B, AF ∦BE. 3.2.2 A Minimal Set of Constructions There are a total of 22 constructions and three kinds of geometry quantities. So to provide an elimination method for each construction and each geometry quantity, we need to con-sider 22 3 = 66 cases. Instead of considering all these cases, we introduce a minimal set of constructions which are equivalent to all the 22 constructions but much fewer in number. A minimal set of constructions consists of C1, C7, C8 and the following two construc-tions. C41 (INTER Y (LINE U V) (LINE P Q)). C42 (FOOT Y P U V), or equivalently (INTER Y (LINE U V) (TLINE P U V))). The ndg condition is U , V. We ﬁrst show how to represent the four kinds of lines by one kind: (LINE U V). For ln = (PLINE W U V), we ﬁrst introduce a new point N by (PRATIO N W U V 1). Then ln = (LINE W N). For ln = (TLINE W U V), we have two cases: if W, U, V are collinear, ln = (LINE N W) where N is introduced by (TRATIO N W U 1); otherwise ln = (LINE N W) where N is given by (FOOT N W U V). (BLINE U V) can be written as (LINE N M) where N and M are introduced as follows (MIDPOINT M U V) (i.e., (PRATIO M U U V 1/2)), (TRATIO N M U 1). Since now there is only one kind of line, to represent all the 22 constructions by the constructions in the minimal set we need only to consider the following cases. • (ON Y (LINE U V)) is equivalent to (PRATIO Y U U V r) where r is an indetermi-nate. • (INTER Y (LINE U V) (CIR O U)) is equivalent to two constructions: (FOOT N O U V), (PRATIO Y N N U -1). • C6 can be reduced to (FOOT N P O1 O2) and (PRATIO Y N N P -1). • For C3, i.e., to take an arbitrary point Y on a circle (CIR O P), we ﬁrst take an arbitrary point Q. Then Y is introduced by (INTER Y (LINE P Q) (CIR O P)). 3.2 Constructive Geometry Statements 113 Proposition 3.20 The existence of the point introduced by each of the 22 constructions fol-lows from Axiom A.2 on page 55. Proof. We can limit ourselves to the ﬁve minimal constructions. Constructions C1, C7, and C41 have been discussed on page 61. Let Y be introduced by (FOOT Y P U V). Then by Proposition 3.2 point Y is a point on UV with the position ratio PPUV PUVU . Hence Y does exist by Axiom A.2. Let Y be introduced by (TRATIO Y U V r). Then Y can also be introduced as follows (check this). (POINT M); (FOOT N M U V); (PRATIO B U M N 1); (PRATIO Y U U B rPUVU 4S UVB ). Thus Y exists. Exercises 3.21 1. Show that constructions C1, C7, and C8 can also serve as a minimal set of constructions. The reason we use a larger minimal set is that constructions C41 and C42 are used frequently and special treatment of them will lead to short proofs. 2. We introduce a new construction (LRATIO Y U V r) which means taking a point Y on UV such that UY = rUV. Show that C1, C8 and the above construction also form a minimal set of constructions. (See Example 2.13). 3. Show that constructions C1, C7, and C42 could form a minimal set of constructions. 3.2.3 The Predicate Form The constructive description of geometry statements can be transformed into the commonly used predicate form. In addition to the three predicates POINT, COLL, and PARA intro-duced on page 63, we introduce two new predicates. 1. Perpendicular (PERP P1, P2, P3, P4): [(P1 = P2)∨(P3 = P4)∨(P1P2 is perpendicular to P3P4)]. It is equivalent to PP1P3P2P4 = 0. 2. Congruence (CONG P1, P2, P3, P4): Segment P1P2 is congruent to P3P4. It is equiv-alent to PP1P2P1 = PP3P4P3. To transform constructions into predicate forms, we need only to consider the minimal set of constructions introduced in the preceding subsection. Also constructions C1, C41, and C7 have been discussed in Section 2.3.2. We thus need only to consider C42 and C8. C42 (FOOT Y P U V) is equivalent to (COLL Y U V), (PERP Y P U V), and U , V. C8 (TRATIO Y U V r) is equivalent to (PERP Y U U V), r = 4S UVY PUVU , and U , V. 114 Chapter 3. Machine Proof in Plane Geometry Now a constructive statement S = (C1, · · · , Ck, (E, F)) can be transformed into the follow-ing predicate form ∀Pi[(P(C1) ∧· · · ∧P(Ck)) ⇒(E = F)] where P(Ci) is the predicate form for Ci and Pi is the point introduced by Ci. We will now discuss what geometry properties can be the conclusion of a geometry statement in CL, i.e., what geometry properties can be represented by polynomial equations of geometry quantities. To illustrate that, let us give an algebraic interpretation for the area and Pythagoras diﬀerence. Let A, B, C, and D be four points in the Euclidean plane. Then S ABCD and PABCD are propositional to the exterior and inner product of the diagonals AC and BD of the quadrilateral ABC: (see Chapter 5 for details) S ABCD = 1 2[− − → AC, − − → BD], PABCD = 2⟨− − → AC, − − → BD⟩. So any geometry property that can be represented by an equation of the inner and exterior products can be the conclusion of a geometry statement. As examples, we show how to represent several often used geometry properties by the geometry quantities. (COLLINEAR A B C). Points A, B, and C are collinear iﬀS ABC = 0. For other variants, see the comments after Example 2.36 on page 74. (PARALLEL A B C D). AB is parallel to CD iﬀS ACD = S BCD. (PERPENDICULAR A B C D). AB is perpendicular to CD iﬀPACD = PBCD. (MIDPOINT O A B). O is the midpoint of AB iﬀAO OB = 1. (EQDISTANCE A B C D). AB has the same length as CD iﬀPABA = PCDC. (HARMONIC A B C D). A, B and C, D are harmonic points iﬀAC CB = DA DB. (EQ-PRODUCT A B C D P Q R S ). The product of AB and CD is equal to the product of PQ and RS , which is equivalent to AB PQ = ± RS CD if AB ∥PQ and RS ∥CD; PACBD = ±PPRQS if AB ∥CD and PQ ∥RS ; PABAPCDC = PPQPPRS R otherwise. (TANGENT O1 A O2 B). Circle (CIR O1 A) is tangent to circle (CIR O2 B) iﬀd2 + r2 1 + r2 2 −2dr1 −2dr2 −2r1r2 = 0 where d = O1O2 2, r1 = O1A 2, r2 = O2B 2. 3.3 Machine Proof for Class CL 3.3.1 The Algorithm In Chapter 2, we have seen that the process of proving geometry theorems using the area method actually eliminates points from geometry quantities. To prove geometry theorems 3.3 A Proof Method for CL 115 in class CL, we need to eliminate points introduced by constructions: C1, C7, C8, C41, C42 from three geometry quantities: the area, the Pythagoras diﬀerence, and the length ratio. Let G(Y) be one of the following geometry quantities: S ABY, S ABCY, PABY, or PABCY for distinct points A, B, C, and Y. For three collinear points Y, U, and V, by Propositions 2.9 and 3.5 we have (I) G(Y) = UY UV G(V) + YV UV G(U). We call G(Y) a linear geometry quantity for variable Y. Elimination procedures for all linear geometry quantities are similar for constructions C7, C41, and C42. Lemma 3.22 Let G(Y) be a linear geometry quantity and point Y be introduced by construc-tion (PRATIO Y W U V r). Then we have G(Y) = ( (UW UV + r)G(V) + (WV UV −r)G(U) if W is on line UV. G(W) + r(G(V) −G(U)) otherwise. Proof. If W, U, and V are collinear, we have UY UV = UW UV + r; YV UV = WV UV −r. Substituting these into (I), we obtain the ﬁrst formula. For the second one, take a point S such that WS = UV. By (I) G(Y) = WY WS G(S ) + YS WS G(W) = rG(S ) + (1 −r)G(W). By Propositions 2.11 and 3.7, G(S ) = G(W) + G(V) −G(U). Substituting this into the above equation, we obtain the result. Notice that in both cases, we need the ndg condition U , V. Lemma 3.23 Let G(Y) be a linear geometry quantity and Y be introduced by (INTER Y (LINE U V) (LINE P Q)). Then G(Y) = S UPQG(V) −S VPQG(U) S UPVQ . Proof. By the co-side theorem, UY UV = S UPQ S UPVQ , YV UV = −S VPQ S UPVQ . Substituting these into (I), we prove the result. Lemma 3.24 Let G(Y) be a linear geometry quantity and Y be introduced by (FOOT Y P U V). Then G(Y) = PPUVG(V) + PPVUG(U) 2UV 2 . 116 Chapter 3. Machine Proof in Plane Geometry Proof. By Proposition 3.2, UY UV = PPUV PUVU , YV UV = PPVU PUVU . Substituting these into (I), we prove the result, Let G(Y) = PAYB. By Proposition 3.5, for three collinear points Y, U, and V (II) G(Y) = UY UV G(V) + YV UV G(U) −UY UV · YV UV PUVU. We call PAYB a quadratic geometry quantity for variable Y. Since in the above three lemmas we have obtained the position ratios UY UV, YV UV for Y when it is introduced by constructions C7, C41, C42, we can substitute them into (II) to eliminate point Y from G(Y). Notice that in the case of construction C7, we need to use the second formula of Proposition 3.7. The result is as follows. Lemma 3.25 Let Y be introduced by (PRATIO Y W U V r). Then we have PAYB = PAWB + r(PAVB −PAUB + PWUV) −r(1 −r)PUVU. Construction C8 needs special treatment. Lemma 3.26 Let Y be introduced by (TRATIO Y P Q r). Then we have S ABY = S ABP − r 4PPAQB. A B P Q Y A Figure 3-6 1 Proof. Let A1 be the orthogonal projection from A to PQ. Then by Propositions 2.10 and 3.2 S PAY S PQY = S PA1Y S PQY = PA1 PQ = PA1PQ PQPQ = PAPQ PQPQ . Thus S PAY = PAPQ PQPQ S PQY = r 4PAPQ. Similarly, S PBY = PBPQ PQPQ S PQY = r 4PBPQ. Now S ABY = S ABP + S PBY −S PAY = S ABP −r 4PPAQB. Lemma 3.27 Let Y be introduced by (TRATIO Y P Q r). Then we have PABY = PABP − 4rS PAQB. A B P Q Y A B Figure 3-7 1 1 Proof. Let the orthogonal projections from A and B to PY be A1 and B1. Then PBPAY PYPY = PB1PA1Y PYPY = A1B1 PY = S PA1QB1 S PQY = S PAQB S PQY . 3.3 A Proof Method for CL 117 By the Herron-Qin formula, S 2 PQY = 1 4PQ 2 · PY 2. Then PYPY = 2PY 2 = 4rS PQY. Therefore PABY = PABP −PBPAY = PABP −4rS PAQB. Lemma 3.28 Let Y be introduced by (TRATIO Y P Q r). Then we have PAYB = PAPB + r2PPQP −4r(S APQ + S BPQ). Proof. By Lemma 3.27, PAPY = 4rS APQ, PBPY = 4rS BPQ. By the Herron-Qin formula, PYPY = 2PY 2 = 4rS PQY = r2PPQP. Then PAYB = PAPB −PAPY −PBPY + PYPY = PAPB + r2PPQP −4r(S APQ + S BPQ). By Proposition 3.10 the ratios of parallel line segments can be represented by Pythagoras diﬀerences. Thus, we have given a complete method of eliminating points from geometry quantities. But usually, we consider the length ratios separately in order to obtain short proofs. The methods of eliminating point Y introduced by C41 and C7 from length ratios have been given by Lemmas 2.25 and 2.26. For other constructions, we have Lemma 3.29 Let Y be introduced by (FOOT Y P U V). We assume D , U; otherwise interchange U and V. G = DY EF = ( PPEDF PEFE if D ∈UV. S DUV S EUFV if D < UV. Proof. If D ∈UV, let T be a point such that DT = EF. By Propositions 3.2 and 3.8 G = DY EF = DY DT = PPDT PDT D = PPEDF PEFE . The second equation is a direct consequence of the co-side theorem. Lemma 3.30 Let Y be introduced by (TRATIO Y P Q r). G = DY EF =        PDPQ PEPFQ if D < PY. S DPQ−r 4 PPQP S EPFQ if D ∈PY. Proof. The ﬁrst case is a direct consequence of Proposition 3.3. If D ∈PY, then DY EF = DP EF −YP EF. By the co-side theorem, DP EF = S DPQ S EPFQ ; YP EF = S YPQ S EPFQ = rPPQP 4S EPFQ . 118 Chapter 3. Machine Proof in Plane Geometry Now the second result follows immediately. For a geometry statement S = (C1, C2, . . . , Ck, (E, F)), after eliminating all the non-free points introduced by Ci from E and F using the above lemmas, we obtain two rational expressions E′ and F′ in indeterminates, areas and Pythagoras diﬀerences of free points. These geometric quantities are generally not independent, e.g., for any three points A, B, and C we have the Herron-Qin formula (Proposition 3.16): 16S 2 ABC = 4AB 2AC 2 −(AC 2 + AB 2 −BC 2)2. We thus need to reduce E′ and F′ to expressions in independent variables. To do that, we ﬁrst introduce three new points O, U, and V such that UO⊥OV. We will reduce E′ and F′ to expressions in the area coordinates of the free points with respect to OUV. Lemma 3.31 For three points A, B, and C, we have 1. S ABC = 1 S OUV S OUA S OVA 1 S OUB S OVB 1 S OUC S OVC 1 . O M U V A B Figure 3-8 2. PABC = AB 2 + CB 2 −AC 2. 3. AB 2 = OU 2(S OVA−S OVB)2 S 2 OUV + OV 2(S OUA−S OUB)2 S 2 OUV . 4. S 2 OUV = 1 4OU 2OV 2. Proof. Case 1 is Lemma 2.31. Case 2 is the deﬁnition of the Pythagoras diﬀerence. For case 3, we introduce a new point M by construction (INTER M (PLINE A O U) (PLINE B O V)). Then by the Pythagorean theorem, AB 2 = AM 2 +BM 2. By the second case of Lemma 2.26, AM OU = S AOBV S OOUV = S AOV−S BOV S OUV ; BM OV = S AOU−S BOU S OUV . We have proved case 3. Case 4 is a consequence of Proposition 3.13. Using Lemma 3.31, E and F can be written as expressions of OU, OV, and the area coordinates of the free points. Remark 3.32 In Lemma 3.31, we actually use the Cartesian coordinates of points to repre-sent areas and Pythagoras diﬀerences. For a point P, let xP = 2S OUP \|OU\| , yP = 2S OVP \|OV\| . Then the formulas in Lemma 3.31 become 1’. S ABY = 1 2 xA yA 1 xB yB 1 xY yY 1 . 3.3 A Proof Method for CL 119 3’. AB 2 = (xA −xB)2 + (yA −yB)2. Algorithm 3.33 (PLANE) INPUT: S = (C1, C2, . . . , Ck, (E, F)) is a statement in CL. OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof for S . S1. For i = k, · · · , 1, do S2, S3, S4 and ﬁnally do S5. S2. Check whether the ndg conditions of Ci are satisﬁed. The ndg condition of a construc-tion has three forms: A , B, PQ ∦UV, or PQ ̸⊥UV. For the ﬁrst case, we check whether PABA = 2AB 2 = 0. For the second case, we check whether S PUV = S QUV. For the third case, we check whether PPUV = PQUV. If a ndg condition of a geometry statement is not satisﬁed, the statement is trivially true. The algorithm terminates. S3. Let G1, · · · , Gs be the geometric quantities occurring in E and F. For j = 1, · · ·, s do S4. S4. Let H j be the result obtained by eliminating the point introduced by construction Ci from G j using the lemmas in this section and replace G j by H j in E and F to obtain the new E and F. S5. Now E and F are rational expressions in independent variables. Hence if E = F, S is true. Otherwise S is false. Proof of the correctness. Only the last step needs explanation. If E = F, the statement is obviously true. Otherwise, by Proposition 2.33 we can ﬁnd speciﬁc values for the free parameters in E and F such that when substituting them into E and F, we obtain two diﬀerent numbers, i.e., we have found a counterexample. For the complexity of the algorithm, let n be the number of the non-free points in a statement which is described using constructions C1–C8. By the analysis in Section 3.2, we will use at most 5n constructions in the minimal set to represent the hypotheses (we need ﬁve minimal constructions to represent construction (INTER A (BLINE U V) (BLINE P Q))). Then we will use at most 5n minimal constructions to describe the statement. Notice that each lemma will replace a geometric quantity by a rational expression with degree less than or equal to three. Then if the conclusion of the geometry statement is of degree d, the output of our algorithm is at most degree 35nd. In the last step, we need to represent the area and Pythagoras diﬀerence by area coordinates. In the worst case, a geometry quantity (Pythagoras diﬀerence) will be replaced by an expression of degree ﬁve. Thus the degree of the ﬁnal polynomial is at most 5d35n. 120 Chapter 3. Machine Proof in Plane Geometry Remark 3.34 In Section 2.6, we showed that the area method is for aﬃne geometry over any ﬁeld. Algorithm 3.33 is actually for metric geometry over any ﬁeld whose characteristic is not 2. We have to exclude the case of characteristic 2. Otherwise, for three collinear points A, B, and C we have PABC = 2AB · CB = 0, i.e., AB⊥BC. For more details, see Chapter 5. 3.3.2 Reﬁned Elimination Techniques We have presented a complete method for proving geometry statements in class CL by considering a minimal set of constructions. But if we use only those ﬁve constructions, we must introduce many auxiliary points in the description of geometry statements. More points usually mean longer proofs. In this section we will introduce more constructions and more elimination techniques which can be used to obtain shorter proofs. The elimination lemmas in Section 3.3.1 can be reﬁned in two ways: ﬁrst we may consider more constructions instead of the minimal set; second for each elimination lemma we may give the elimination results in some special cases of the conﬁguration. As an example of the second way of reﬁnement, prove the following result. Exercise 3.35 Let point Y be introduced by construction (FOOT Y P U V). Prove the fol-lowing results. S ABY =                S ABU if AB ∥UV; S ABP if AB⊥UV; S UBV PPUAV PUVU if U, V, and A are collinear; S AUV S PUBV S UVU if U, V, and B are collinear. PABY =            PABP if AB ∥UV; PABU if AB⊥UV; PABUPPBU PUBU if U, V, and B are collinear. PAYB =                      16S 2 PUV PUVU if A = B = P; P2 PUV PUVU if A = B = U; P2 PVU PUVU if A = B = V; −PPVU PPUV PUVU if A = U, B = V. To use the above elimination technique, similar to in Section 2.5.1, we need to ﬁnd the collinear point sets, parallel lines, and perpendicular lines which are obvious from the constructive description of the geometry statement. See the following example. Example 3.36 The following machine proof of the orthocenter theorem on page 32 uses the above exercise. 3.3 A Proof Method for CL 121 Constructive description ( (POINTS A B C) (FOOT E B A C) (FOOT F A B C) (INTER H (LINE A F) (LINE B E)) (PERPENDICULAR A B C H) ) The machine proof PACH PBCH H = PACB PACB simplify = 1 The eliminants PBCH H =PACB PACH H =PACB From the description of the statement, we can ﬁnd the sets of collinear points: {H, A, F}, {F, C, B}, {H, E, B}, {E, A, C}, and the sets of perpendicular lines: HAF⊥BCF, HEB⊥EAC. Then by Exercise 3.35, PBCH = PACB and PACH = PACB since AB⊥CH and CA⊥BH. Exercise 3.37 Let G(Y) be a linear geometry quantity of Y, and Q(Y) a quadratic geometry quantity of Y. If Y is on line UV then G(Y) = UY UVG(V) + YV UVG(U); Q(Y) = UY UV Q(V) + YV UV Q(U) −2 · UY UV · YV UV · UV 2. DY EF =                S DUV S EUFV if D < UV; DU UV + UY UV EF UV if D ∈UV. For several constructions, we need to compute the position ratio of Y with respect to UV and substitute them into the above formulas to eliminate Y. Precisely, we have 1. If Y is introduced by (INTER Y (LINE U V) (PLINE R P Q)) then UY UV = S UPRQ S UPVQ , YV UV = −S VPRQ S UPVQ . 2. If Y is introduced by (INTER Y (LINE U V) (TLINE P P Q)) then UY UV = PUPQ PUPVQ , YV UV = −PVPQ PUPVQ . 3. If Y is introduced by (INTER Y (LINE U V) (BLINE P Q)) then UY UV = PUPQ−PQ 2 PUPVQ , YV UV = −PVPQ−PQ 2 PUPVQ . 4. If Y is introduced by (INTER Y (LINE U V) (CIR O U)) then UY UV = 2 · POUV PUVU , YV UV = POVO−POUO PUVU . Exercise 3.38 If Y is introduced by (INTER Y (PLINE W U V) (PLINE R P Q)) then 122 Chapter 3. Machine Proof in Plane Geometry G(Y) = G(W) + r(G(V) −G(U)); Q(Y) = Q(W) + r(G(V) −G(U)) −2r(1 −r)UV 2 where r = S WPRQ S UPVQ . Exercise 3.39 Of the 22 constructions, there are still eight which are not discussed (Take points on a BLINE, a PLINE, or a TLINE; Take the intersections of two TLINEs, two BLINEs, a TLINE and a PLINE, a TLINE and a BLINE, and a PLINE and a BLINE; Take the intersection of a circle and a PLINE or a TLINE.) Try to eliminate a point introduced by one the eight constructions from a geometry quantity. 3.4 The Ratio Constructions By the ratio constructions, we mean the constructions PRATIO and TRATIO and other constructions which can be reduced to them. These constructions are the most subtle con-structions; appropriate use of the ratio constructions may lead to elegant proofs for geom-etry statements. 3.4.1 More Ratio Constructions We ﬁrst introduce the following constructions for convenience. C9 (MIDPOINT Y U V). Y is the midpoint of UV. It is equivalent to (PRATIO Y U U V 1/2). C10 (SYMMETRY Y U V). Y is the symmetry of point V with respect to point U. It is equivalent to (PRATIO Y U U V -1). C11 (LRATIO Y U V r). Y is a point on UV such that UY UV = r. It is equivalent to (PRATIO Y U U V r). C12 (MRATIO Y U V r). Y is a point on UV such that UY YV = r. It is equivalent to (PRATIO Y U U V r 1+r). Four collinear points A, B, C, and D are said to form a harmonic sequence if CA CB = −DA DB . Given two points A and B, we have the following ways of introducing points C and D such that A, B, C, and D form a harmonic sequence. 3.4 The Ratio Constructions 123 (ON C (LINE A B)), (LRATIO D A B CA CB CA CB +1); (MRATIO C A B r), (MRATIO D A B −r). For convenience, we can introduce a new construction C13 (HARMONIC D C A B) introduces a point D such that, for three collinear points A, B, and C, A, B, C, and D form a harmonic consequence. Another important geometry concept related to ratios is inversion. Suppose that we have given a circle with a center O and a radius r , 0. Let P and Q be any two points collinear with O such that OP · OQ = r2. Then P is said to be the inversion of Q with regard to the circle and O is called the inversion center. We introduce a new construction as follows C14 (INVERSION P Q O A) means that P is the inversion of Q with regard to circle (CIR O A). This construction is equivalent to (LRATIO P O A OA OQ) if Q ∈OA, (LRATIO P O Q POAO POQO) otherwise. The ratio r in the ratio constructions could be rational numbers, variables, or expressions in geometry quantities. Now we allow r to be any algebraic numbers by adding a new special construction. C15 (CONSTANT p(r)) where p(r) is an irreducible polynomial in the variable r. This construction introduces an algebraic number r which is a root of p(r) = 0. With the help of construction CONSTANT, we can deal with statements involving special angles such as 30◦, 45◦, and 60◦, etc. In the rest of this section, we will use several examples to show how to solve geometry problems using the ratio constructions. The construction TRATIO can be used to express geometry statements involving squares easily. Example 3.40 The following is the machine proof of Example 1.69 on page 32. Constructive description ((POINTS A B C) (TRATIO E A B 1) (TRATIO G A C −1) (PERPENDICULAR E C G B) ) The machine proof PBGE PBGC G = PACA+4S ACE−4S ABC PBAC+PACA−4S ABC E = PBAC+PACA−4S ABC PBAC+PACA−4S ABC simplify = 1 The eliminants PBGC G =PBAC+PACA−4S ABC PBGE G =PACA+4S ACE−4S ABC S ACE E =1 4(PBAC) 124 Chapter 3. Machine Proof in Plane Geometry Example 3.41 On the two sides AC and BC of triangle ABC, two squares ACDE and BCFG are drawn. M is the midpoint of AB. Show that CM is perpendicular to DF. (Figure 3-9) Constructive description ((POINTS A B C) (TRATIO D C A 1) (TRATIO F C B −1) (MIDPOINT M A B) (PERPENDICULAR D F C M) ) The machine proof PDCM PFCM M = 1 2 PBCD 1 2 PACF F = PBCD 4S ABC D = 4S ABC (4)·S ABC simplify = 1 The eliminants PFCM M = 1 2(PACF) PDCM M = 1 2(PBCD) PACF F =4(S ABC) PBCD D =4(S ABC) A B M C D E F V N Figure 3-10 M G F E D C B A Figure 3-9 Example 3.42 1 Let M be a point on line AB. Two squares AMCD and BMEF are drawn on the same side of AB. Let U and V be the center of the squares AMCD and BMEF. Line BC and circle VB meet in N. Show that A, E, and N are collinear. (Figure 3-10) Constructive description ((POINTS A B) (ON M (LINE A B)) (TRATIO C M A 1) (TRATIO E M B −1) (MIDPOINT V E B) (INTER N (LINE B C) (CIR V B)) (INTER T (LINE B C) (LINE A E)) ( BN CN = BT CT ) ) 1This is a problem from the 1959 International Mathematical Olympiad. 3.4 The Ratio Constructions 125 The machine proof ( BN CN)/( BT CT ) T = −S ACE −S ABE · BN CN N = PCBV·S ACE S ABE·(PCBV−1 2 PBCB) V = ( 1 2 PCBE)·S ACE S ABE·( 1 2 PCBE−1 2 PBCB) E = (PMBC+4S BMC)·( 1 4 PMABC−S AMC) (−1 4 PABM)·(PMBC−PBCB+4S BMC) C = −(PBMB−PAMB)·(PBMB+PAMA−PABM) PABM·(−PAMB−PAMA) M = (−PABA· AM AB +PABA)·(2PABA·( AM AB )2−PABA· AM AB ) (−PABA· AM AB +PABA)·(2PABA·( AM AB )2−PABA· AM AB ) simplify = 1 The eliminants BT CT T = S ABE S ACE BN CN N = PCBV ( 1 2 )·(2PCBV−PBCB) PCBV V =1 2(PCBE) S ABE E = −1 4(PABM) S ACE E = 1 4(PMABC−4S AMC) PCBE E =PMBC+4S BMC PBCB C =PBMB+PAMA S AMC C = −1 4(PAMA) PMABC C =PBMB−PABM S BMC C = −1 4(PAMB) PMBC C =PBMB PABM M = −(( AM AB −1)·PABA) PAMA M =PABA·( AM AB )2 PAMB M =( AM AB −1)·PABA· AM AB PBMB M =( AM AB −1)2·PABA For more examples involving squares, see Section 9.4. Example 3.43 Given four points A, B, C, and D which form a harmonic sequence and a point O outside the line AB, any transversal cuts the four lines OA, OB, OC, and OD in four harmonic points. 126 Chapter 3. Machine Proof in Plane Geometry Constructive description ( (POINTS O A B X Y) (MRATIO C A B r) (MRATIO D A B −r) (INTER P (LINE O A) (LINE X Y)) (INTER Q (LINE O B) (LINE X Y)) (INTER R (LINE O C) (LINE X Y)) (INTER S (LINE O D) (LINE X Y)) (HARMONIC P Q S R) ) The machine proof (−PS QS )/( PR QR) S = −S ODP PR QR·S ODQ R = −S ODP·S OCQ S OCP·S ODQ Q = −S ODP·(−S OXY·S OBC)·S OXBY S OCP·(−S OXY·S OBD)·S OXBY simplify = −S ODP·S OBC S OCP·S OBD P = −(−S OXY·S OAD)·S OBC·S OXAY (−S OXY·S OAC)·S OBD·S OXAY simplify = −S OAD·S OBC S OAC·S OBD D = −(−S OAB·r)·S OBC·(−r+1) S OAC·(−S OAB)·(−r+1) simplify = −r·S OBC S OAC C = −r·(−S OAB)·(r+1) S OAB·r·(r+1) simplify = 1 The eliminants PS QS S = S ODP S ODQ PR QR R = S OCP S OCQ S ODQ Q =−S OXY·S OBD S OXBY S OCQ Q =−S OXY·S OBC S OXBY S OCP P =−S OXY·S OAC S OXAY S ODP P =−S OXY·S OAD S OXAY S OBD D =S OAB r−1 S OAD D =S OAB·r r−1 S OAC C =S OAB·r r+1 S OBC C =−S OAB r+1 For more examples involving harmonic sequences, see Section 6.3. Example 3.44 The inverse of a circle passing through the center of inversion is a line. O A X P Q U R G Figure 3-12 Constructive description ((points O A X) (lratio P O A r1) (inversion Q P O A) (midpoint U P O) (inter R (l O X) (cir U O)) (inversion G R O A) (perpendicular G Q O A) ) The machine proof PAOG PAOQ = PAOR·POAO PAOQ·PORO R = (2PXOU·PAOX)·POAO·POXO PAOQ·(4P2 XOU)·POXO simplify = PAOX·POAO (2)·PAOQ·PXOU U = PAOX·POAO (2)·PAOQ·( 1 2 PXOP) = PAOX·POAO OA OP·POAO·PXOP simplify = PAOX OA OP·PXOP P = PAOX·r1 PAOX·r1 simplify = 1 The eliminants PAOG= PAOR·POAO PORO PORO R =(4)·(PXOU)2 POXO PAOR R = (2)·PXOU·PAOX POXO PXOU U = 1 2(PXOP) PAOQ= OA OP ·POAO PXOP P =PAOX·r1 OA OP P = 1 r1 3.4 The Ratio Constructions 127 For more examples involving inversions, see Section 10.3. The ratio constructions are used extensively in the following example. Also notice that the construction CONSTANT is used to describe equilateral triangles. Example 3.45 Three equilateral triangles A1BC, AB1C, ABC1 are erected on the three sides of triangle ABC. Show that CA1C1B1 is a parallelogram. 1 B 1 A 1 C C B A G E F Figure 3-13 Constructive description ( (points A B C) (constant r2−3) (midpoint E A C) (tratio B1 E A r) (midpoint F B C) (tratio A1 F C r) (midpoint G A B) (tratio C1 G B −r) (parallel A1 C1 C B1) ) The machine proof S CB1A1 S CB1C1 C1 = S CB1A1 1 4 PB1BCG·r+S CB1G G = (4)·S CB1A1 −1 2 PBCB1·r+ 1 2 PACB1·r+2S BCB1+2S ACB1 A1 = (−8)·(−1 4 PB1CF·r+S CB1F) PBCB1·r−PACB1·r−4S BCB1−4S ACB1 F = (2)·( 1 2 PBCB1·r−2S BCB1) PBCB1·r−PACB1·r−4S BCB1−4S ACB1 B1 = PCABE·r+PBCE·r−4S BCE+4S ABE·r2 PCABE·r+PCAE·r+PBCE·r−PACE·r−4S BCE+4S ABE·r2 E = 1 2 PBCB·r+ 1 2 PACB·r−1 2 PABC·r+2S ABC·r2−2S ABC 1 2 PBCB·r+ 1 2 PACB·r−1 2 PABC·r+2S ABC·r2−2S ABC simplify = 1 The eliminants S CB1C1 C1 = 1 4(PB1BCG·r+4S CB1G) S CB1G G =1 2(S BCB1+S ACB1) PB1BCG G = −1 2(PBCB1−PACB1) S CB1A1 A1 = −1 4(PB1CF·r−4S CB1F) S CB1F F =1 2(S BCB1) PB1CF F =1 2(PBCB1) S ACB1 B1 = −1 4(PCAE·r) PACB1 B1 = PACE S BCB1 B1 = −1 4(PCABE·r−4S BCE) PBCB1 B1 = PBCE+4S ABE·r PACE E = 1 2(PACA) PCAE E = 1 2(PACA) S ABE E =1 2(S ABC) S BCE E = 1 2(S ABC) PBCE E = 1 2(PACB) PCABE E =1 2(PBCB−PABC) Note that the condition r2 = 3 is not needed in the proof, i.e., the result is true if triangles B1AC, A1BC, and C1AB are similar isosceles triangles. The above proof is used to illustrate the use of ratio constructions. For a much shorter proof of this example, see Example 5.61 on page 253. 128 Chapter 3. Machine Proof in Plane Geometry 3.4.2 Mechanization of Full-Angles As an application of the construction TRATIO, we will present an automated theorem proving method for geometry theorems involving full-angles. The formal deﬁnition of full-angles is as follows. Deﬁnition 3.46 An ordered pair of lines AB and CD determines a full-angle, denoted by ∠[AB, CD], which satisﬁes 1. ∠[AB, CD] = ∠[PQ, UV] if and only if S ACBDPPUQV = S PUQVPACBD. Thus the tangent function for the full-angle, tan(∠[AB, CD]) = 4S ACBD PADBC is a well deﬁned geometry quantity. 2. For all parallel lines AB ∥PQ, ∠ = ∠[AB, PQ] is a constant. 3. For all perpendicular lines AB⊥PQ, ∠ = ∠[AB, PQ] is a constant. 4. There exists an operation “+” for full-angles which is associative and commutative. Furthermore, we have • ∠ + ∠ = ∠. • If PQ ∥UV then ∠[AB, PQ] + ∠[UV, CD] = ∠[AB, CD]. • The tangent function of the sum of two full-angles is deﬁned as follows tan(∠[AB, CD] + ∠[PQ, UV]) = tan(∠[AB, CD]) + tan(∠[PQ, UV]) 1 −tan(∠[AB, CD]) tan(∠[PQ, UV]). You can ﬁnd the geometric background for the above deﬁnition in Section 1.10. For three points A, B, and C, let ∠[ABC] = ∠[AB, BC]. Remark 3.47 According to the above deﬁnition, ∠[AB, CD] = ∠[PQ, UV] if and only if one of the following conditions holds. 1. (PARA A B C D) and (PARA P Q U V); 2. (PERP A B C D) and (PERP P Q U V); 3. A , B, C , D, P , Q, U , V and the full-angle ∠[AB, CD] is equal to the full-angle ∠[PQ, UV]. 3.4 The Ratio Constructions 129 Proposition 3.48 (The Co-angle Theorem) In triangles ABC and XYZ, if ∠[ABC] = ∠[XYZ], ∠[ABC] , ∠, and ∠[ABC] , ∠, then S ABC S XYZ = PABC PXYZ = λ where λ2 = AB 2 · BC 2 XY 2 · ZY 2 . Proof. By Deﬁnition 3.46 if ∠[ABC] = ∠[XYZ] then S ABC S XYZ = PABC PXYZ = λ. By the Herron-Qin formula 16S 2 ABC + P2 ABC = 4AB 2 · CB 2, 16S 2 XYZ + P2 XYZ = 4XY 2 · ZY 2. Set S ABC = λS XYZ, PABC = λPXYZ in the ﬁrst equation we have λ2 = 4AB 2 · CB 2 16S 2 AYX + P2 AYZ = 4AB 2 · CB 2 4XY 2 · ZY 2 . With the concept of full-angles, the constructive geometry statements can be extended as follows. First we have a new geometry quantity: the tangent function of full-angles. Since the tangent function can be represented by the area and Pythagoras diﬀerence, we do not need to introduce new elimination techniques to eliminate points from it. Its main contri-bution is that we may now prove assertions like ∠[AB, CD] = ∠[PQ, UV] and ∠[AB, CD] = ∠[PQ, UV] + ∠[XY, WZ]. The second extension is more interesting: we can introduce a new representation for lines. (ALINE P Q U W V) which is the line l passing through P such that ∠[PQ, l] = ∠[UW, WV]. With this new type of lines, we can introduce seven new constructions: 1. (ON Y (ALINE P Q L M N)). Take an arbitrary point on an ALINE; the ndg condi-tions are P , Q, L , M, and N , M. 2. (INTER Y ln (ALINE P Q L M N)). Take the intersection of ln and (ALINE P Q L M N). • If ln = (LINE U V) or ln = (PLINE W U V), then the ndg condition is ∠[PQ, UV] , ∠[LM, MN]. • If ln = (BLINE U V) or ln = (TLINE W U V), then the ndg condition is ∠[UV, PQ] + ∠[LM, MN] , ∠. • If ln = (ALINE U V X Y Z), then the ndg condition is ∠[UV, PQ] , ∠[NM, ML]+ ∠[XY, YZ]. 130 Chapter 3. Machine Proof in Plane Geometry 3. (INTER Y (CIR O P) (ALINE P Q L M N)). The ndg conditions are Y , P, P , O, P , Q, L , M, and N , M. To provide methods of eliminating points introduced by these new constructions from ge-ometry quantities, we need only to reduce an ALINE to a LINE. P R X Q W U V Figure 3-14 Proposition 3.49 If UW is not perpendicular to WV, line l =(ALINE P Q U W V) is the same as (LINE P R) where R is introduced by construction (TRATIO R Q P 4S UWV PUWV ). If UW⊥WV, line l is (TLINE P P Q). Proof. Let the line passing through point Q and perpendicular to PQ meet line l in R. Then R is introduced by construction (TRATIO R Q P r), where r = 4S RQP PQPQ = 4S QPR PQPR = tan(∠[RPQ]) = tan(∠[VWU]) = 4S UWV PUWV . Remark 3.50 From the above proposition we see that the construction (TRATIO Y P Q r) is actually to take a point Y such that YP⊥PQ and tan(∠[YPQ]) = r. Now, we introduce the following predicates. (EQANGLE A B C D E F). ∠[ABC] = ∠[DEF] iﬀS ABCPDEF = S DEFPABC. (COCIRCLE A B C D). Points A, B, C, and D are co-circle iﬀ∠[CAD] = ∠[CBD], or equivalently, S CADPCBD = PCADPCBD. Example 3.51 If N, M are points on the sides AC, AB of a triangle ABC and the lines BN, CM intersect at a point J which is on the altitude AD, show that AD is the bisector of the angle MDN. B C A D J M N Figure 3-15 Constructive description ( (POINTS A B C) (FOOT D A B C) (ON J (LINE A D)) (INTER M (LINE A B) (LINE C J)) (INTER N (LINE A C) (LINE B J)) (EQANGLE M D A A D N) ) 3.4 The Ratio Constructions 131 The ndg conditions: B , C, A , D, AB ̸∥CJ, AC ̸∥BJ. The machine proof (−S ADM)·PADN S ADN·PADM N = (−S ADM)·(−PADJ·S ABC)·S ABCJ (−S ACD·S ABJ)·PADM·(−S ABCJ) simplify = S ADM·PADJ·S ABC S ACD·S ABJ·PADM M = (−S ACJ·S ABD)·PADJ·S ABC·(−S ACBJ) S ACD·S ABJ·PADJ·S ABC·S ACBJ simplify = S ACJ·S ABD S ACD·S ABJ J = S ACD· AJ AD ·S ABD S ACD·S ABD· AJ AD simplify = 1 The eliminants S ADN N =−S ACD·S ABJ S ABCJ PADN N = PADJ·S ABC S ABCJ PADM M = PADJ·S ABC −S ACBJ S ADM M = −S ACJ·S ABD S ACBJ S ABJ J =S ABD· AJ AD S ACJ J =S ACD· AJ AD Example 3.52 (The Inscribed Angle Theorem) Let A, B, C, and D be four points on a circle with center O. Then ∠[ACB] = ∠[ADB] and ∠[AOB] = 2∠[ACB]. Proof. We ﬁrst use our program to compute tan(∠[ACB]). A B C O P Figure 3-16 Constructive description ( (POINTS A B) (ON O (BLINE A B)) (TRATIO P B A r) (INTER C (LINE A P) (CIR O A)) ((TANGENT A C B)) ) The machine proof (−4)·S ABC PACB C = (−4)·(2POAP·S ABP)·PAPA (−2POPO·POAP+2POAP·PAPB+2POAP·PAOA)·PAPA simplify = (4)·S ABP POPO−PAPB−PAOA P = (4)·(−1 4 PABA·r) PBOB−PAOA+8S ABO·r O = −PABA·r 8S ABO·r simplify = −PABA (8)·S ABO The eliminants PACB C =(−2)·(POPO−PAPB−PAOA)·POAP PAPA S ABC C =(2)·POAP·S ABP PAPA PAPB P =PABA·(r)2 POPO P =PBOB+PABA·r2+8S ABO·r S ABP P = −1 4(PABA·r) 132 Chapter 3. Machine Proof in Plane Geometry From the above computation, it is clear that tan(∠[ACB]) is independent of point P and C, i.e., tan(∠[ACB]) = tan(∠[ADB]). tan(2∠[ABC]) = 2 tan(∠[ABC]) 1 −tan(∠[ABC])2 = 8AB 2S AOB 16S 2 AOB −AB 4 = 8AB 2S AOB 4OA 2 · OB 2 −P2 AOB −AB 4 = 4S AOB 2AO 2 −AB 2 = 4S AOB PAOB = tan(∠[AOB]) i.e., 2∠[ABC] = ∠[AOB]. Example 3.53 (Morley’s Theorem) The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle. B C L A M N Figure 3-17 Constructive description ( (POINTS B C L) (ON X (ALINE B L L B C)) (ON Y (ALINE B X L B C)) (ON Z (ALINE C L L C B)) (ON W (ALINE C Z L C B)) (INTER A (LINE B Y) (LINE C W)) (MIDPOINT O L C) (CONSTANT r2 3) (TRATIO T O C r) (INTER Q (LINE B L) (LINE T C)) (ON H (ALINE A B T L Q)) (INTER M (LINE B X) (LINE A H)) (ON G (ALINE A C Q L T)) (INTER N (LINE C Z) (LINE A G)) ((TANGENT M L N) = r) ) Morley’s theorem is among the most diﬃcult problems proved by our method. The proof for it is still too long to be considered readable. Our method need further improvement to produce a readable proof for this theorem. Exercises 3.54 1. Note that the sine and cosine functions for full-angles are meaningless. But we can deﬁne their squares. For a full-angle α, let sin2(α) = tan2(α) 1+tan2(α), cos2(α) = 1 1+tan2(α). Then S 2 ABCD = 1 4AC 2 · BD 2 sin2(∠[AC, BD]), P2 ABCD = 4AC 2 · BD 2 cos2(∠[AC, BD]). (Use the Herron-Qin formula.) 3.5 Area Coordinates 133 2. Try to eliminate Y from S ABY if Y is introduced by (INTER Y (LINE U V) (ALINE P Q L M N)). Show that the ndg conditions ensure that all the geometric quantities occurring in the elimination have geometric meaning. 3. If Y is introduced by (INTER Y (TLINE U U O) (TLINE V V O)) then O U V Y Figure 3-18 S ABY =                      S ABU if AB⊥OU; S ABV if AB⊥OV; POUV POVU −16S OUV if A = U, B = V; POVUPOUO 16S OUV if A = O, B = U; POUV POVO 16S OVU if A = O, B = V; PABY =          PABU if AB ∥OU; PABV if AB ∥OV; POUV if A = U, B = V; PAYB =                        −PUOV POVUPOUV 16S 2 OUV if A = U, B = V; POUOP2 OVU 16S 2 OUV if A = O, B = U or A = B = U; POVOP2 OUV 16S 2 OUV if A = O, B = V or A = B = V; POUOPOVOPUVU 16S 2 OUV if A = B = O. In the general case, this construction can be reduced to the following construction (TRATIO Y U O POVU 4S OVU ). Notice that this construction is actually to introduce the antipodal point of O with respect to the circumcircle of triangle OUV. 3.5 Area Coordinates 3.5.1 Area Coordinate Systems In Lemma 3.31, we use an orthogonal coordinate system. In order to do this, we have to introduce three auxiliary points O, U, and V. In this section, we will develop a skew area coordinate system in which any three free points can be used as the reference points. As a consequence, we obtain a new version of Lemma 3.31 and consequently a new version of Algorithm 3.33. We will also prove some interesting features of the skew area coordinate system. 134 Chapter 3. Machine Proof in Plane Geometry Let O, U, and V be three non-collinear points. For any point A, let xA = S OUA S OUV , yA = S VOA S OUV , zA = S UVA S OUV be the area coordinates of A with respect to OUV. It is clear that xA + yA + zA = 1. Below are some results proved before. Proposition 3.55 The points in the plane are in a one to one correspondence with the triples (x, y, z) such that x + y + z = 1. Proposition 3.56 For any points A, B, and C, we have S ABC = −S OUV xA yA 1 xB yB 1 xC yC 1 = −S OUV xA yA zA xB yB zB xC yC zC . As a consequence of Proposition 3.56, we can give the line equation in the area coordinate system. Let P be a point on line AB. Then the area coordinates of P must satisfy xA yA 1 xB yB 1 xP yP 1 = 0 which is the equation for line AB. Notice that this is the same as the line equation in the Cartesian coordinate system. Another interesting fact is the position ratio formula. Proposition 3.57 Let R be a point on line PQ and r1 = PR PQ and r2 = RQ PQ. Then xR = r1xQ + r2xP; yR = r1yQ + r2yP; zR = r1zQ + r2zP. Proof. This is a consequence of Proposition 2.9. We will now prove the distance formula between two points. Proposition 3.58 For two points A and B, we have AB 2 = OV 2(xB −xA)2 + OU 2(yB −yA)2 + (xB −xA)(yB −yA)PUOV. O U V A B M N Figure 3-19 Proof. Let M and N be points such that MA ∥ OU, MB ∥OV, NA ∥OV, and NB ∥OU. Then by Proposition 3.6. (1) AB 2 = AM 2 + BM 2 −PAMB = OU 2( AM OU )2 + OV 2( BM OV )2 + PNAM. 3.5 Area Coordinates 135 By Lemma 2.28, AM OU = S BOAV S UOV = yB −yA, BM OV = S BOAU S OVU = xA −xB. By Example 3.9, PNAM = AN OV · AM OU · PUOV = (yB −yA)(xB −xA)PUOV. Substituting these into (1), we prove the result. Corollary 3.59 Show that 2AB 2 = POVU(xB −xA)2 + POUV(yB −yA)2 + PUOV(zB −zA)2. Proof. We need only to observe that zA −zB = (xB −xA) + (yB −yA). In Algorithm 3.33, let E be an expression in areas and Pythagoras diﬀerences of free points. Instead of using Lemma 3.31, we can use the following procedure to transform E into an expression in independent variables. If there are fewer than three points involved in E then we need do nothing; if there are more than two points, choose three free points O, U, and V from the points occurring in E, and apply Propositions 3.56 and 3.58 to E to transform the areas and Pythagoras diﬀerences into area coordinates with respect to OUV. Now the new E is an expression in area coordinates of free points, OU 2, OV 2, UV 2, and S OUV. The only algebraic relation among these quantities is the Herron-Qin formula (Proposition 3.16): 16S 2 OUV = 4OU 2OV 2 −(OV 2 + OU 2 −UV 2)2 = 4OU 2OV 2 −P2 UOV. Substituting S 2 OUV into E, we obtain an expression in independent variables. If we assume that OU ⊥OV and \|OU\| = \|OV\| = 1 then the area coordinate system becomes the Cartesian coordinate system. Exercises 3.60 1. The process of transforming the areas and Pythagoras diﬀerences of free points into ex-pressions in independent variables presented in this section becomes particularly simple if there are only three free points in a geometry statement. Let the three free points be O, U, and V. Show that any polynomial g in the areas and Pythagoras diﬀerences involving O, U, and V can be transformed into a polynomial of the following form g′ = f + S OUVh where f and h are polynomials of OU 2, OV 2, and UV 2. We also have that g = 0 iﬀh = f = 0. 2. If in a geometry statement there are only two free points U and V, and the third point O is introduced by the construction (ON O (BLINE U V)) or (ON O (TLINE U V)), design a simple method of transforming a polynomial in the areas and Pythagoras diﬀerences involving points O, U, and V into a polynomial of independent variables. (Chose OU 2 and UV 2 as independent variables.) 136 Chapter 3. Machine Proof in Plane Geometry 3.5.2 Area Coordinates and Special Points of Triangles We introduce a new construction. C16 (ARATIO A O U V rO rU rV). Take a point A such that rO = S AUV S OUV , rU = S OAV S OUV , rV = S OUA S OUV are the area coordinates of A with respect to OUV. The rO, rU, and rV could be rational numbers, rational expressions in geometric quantities, indeterminates, or al-gebraic numbers. The ndg condition is that O, U, and V are not collinear. The degree of freedom for A is dependent on the number of indeterminates in {rO, rU, rV}. Lemma 3.61 Let G(Y) be a linear geometry quantity and Y be introduced by (ARATIO Y O U V rO rU rV). Then G(Y) = rOG(O) + rUG(U) + rVG(V). O U V Y T Figure 3-20 Proof. Without loss of generality, let OY intersect UV at T. If OY is parallel to UV, we may consider the intersection of UY and OV or the intersection of VY and OU since one of them must exist. By Propo-sition 2.10, G(Y) = OY OT G(T) + YT OT G(O) = OY OT (UT UV G(V) + TV UV G(U)) + YT OT G(O). By the co-side theorem, YT OT = rO; OY OT = S OUYV S OUV ; UT UV = S OUY S OUYV ; TV UV = S OYV S OUYV . Substituting these into the above formula, we obtain the desired result. Lemma 3.62 Let G(Y) be a quadratic geometry quantity and Y be introduced by (ARATIO Y O U V rO rU rV). Then G(Y) = rOG(O) + rUG(U) + rVG(V) −2(rOrUOU 2 + rOrVOV 2 + rUrVUV 2). Proof. Continue from the proof of Lemma 3.61, By (II) on page 116 G(Y) = OY OT G(T) + YT OT G(O) −OY OT YT OT POTO G(T) = UT UVG(V) + TV UVG(U) −UT UV TV UV PUVU. Substituting G(T) into G(Y), we have G(Y) −r = −OY OT UT UV TV UV PUVU −OY OT YT OT POTO = −rV TV UV PUVU −rA OY OT POTO, 3.5 Area Coordinates 137 where r = rOG(O) + rUG(U) + rVG(V). By (II), POTO = UT UV POVO + TV UV POUO −UT UV TV UV PUVU. Then G(Y) −r = −rV TV UV PUVU −rO OY OT UT UV POVO −rO OY OT TV UV POUO + rO OY OT UT UV TV UV PUVU = −rOrVPOVO −rOrUPOUO −rUrV(−S YUV S OUYV + S OUV S OUYV )PUVU = −rOrVPOVO −rOrUPOUO −rUrVPUVU. If Y is introduced by construction ARATIO, then we need rarely to eliminate Y from G = DY EF. Because points D, E, and F are introduced before point Y, we generally do not know whether DY is parallel to EF or not. But in some special cases, we still need to eliminate point Y from G. This can be done as follows. One of O, U, and V, say O, satisﬁes the condition that D, Y, and O are not collinear. Then G = S ODY S OEDF . Now, we can use Lemma 3.61 to eliminate Y. By using the construction ARATIO, we can treat some special points of triangles easily. Proposition 3.63 The area coordinate of the centroid of a triangle ABC is (1 3 1 3 1 3). Proof. Let G be the centroid of △ABC and M be the midpoint of BC. Since M is the midpoint of BC, we have S ABM = S AMC and S GBM = S GMC. Then S GAB = S ABM −PGBM = S AMC −S GMC = S GCA. Similarly S GBC = S GAB = S GCA = 1 3S ABC. Proposition 3.64 The area coordinate of the orthocenter of triangle ABC is (PABCPACB 16S 2 ABC PBACPBCA 16S 2 ABC PCABPCBA 16S 2 ABC ). A B C E D H Figure 3-21 Proof. As in the ﬁgure, let H be the intersection of the altitudes CD and AE. Then rB rA = S AHC S HBC = AD DB = PCAB PABC rB rC = S AHC S ABH = CE EB = PBCA PABC . Then rA : rB : rC = PABCPBCA : PCABPBCA : PABCPCAB. 138 Chapter 3. Machine Proof in Plane Geometry By the Herron-Qin Theorem (Exercise 3.18), PABCPBCA+PCABPBCA+PABCPCAB = 2AB 2PBCA+ PABCPCAB = 16S 2 ABC. Now the result is clear. Exercises 3.65 1. The area coordinate of the circumscribed center of triangle ABC is (PBCBPBAC 32S 2 ABC PACAPABC 32S 2 ABC PABAPACB 32S 2 ABC ). (Use full-angles.) 2. Let I be the incenter or one of the excenters of △ABC. The area coordinate of C with respect to IBC is (−2PIABPIBA PAIBPABA PIABPIBI PAIBPABA PIBAPIAI PAIBPABA ). (Use full-angles.) 3. Construction C6 on page 111 is equivalent to (ARATIO Y P O1 O2 −1 2PPO2O1 PO1O2O1 2PPO1O2 PO1O2O1 ) Now we can use the following new constructions: C17 (CENTROID G A B C). G is the centroid of triangle ABC. C18 (ORTHOCENTER H A B C). H is the orthocenter of the triangle ABC. C19 (CIRCUMCENTER O A B C). O is the circumcenter of triangle ABC. C20 (INCENTER C I A B). I is the center of the inscribed circle of triangle ABC. This construction is to construct point C from points A, B, and I. Construction C20 needs some explanation. If three vertices of a triangle are given and we need to ﬁnd the coordinates of the incenter, we generally have an equation of degree four in the coordinates of the incenter. The reason is that we can not distinguish the incenter and the three excenters without using inequalities. What we do here is to reverse the problem: when the incenter (or an excenter) and two vertices of a triangle are given, the third vertex is uniquely determined and can be introduced using the constructions given before. Remark 3.66 In this section, we actually use the centroid theorem (Example 1.12 on page 13), the orthocenter theorem (Example 1.67 on page 32), and the incenter theorem (Ex-ample 6.144 on page 333) in the proof of more complicated theorems. The four theorems 3.5 Area Coordinates 139 themselves can be proved using the basic propositions. In general, we can use diﬀerent constructions to describe the same geometry statement, and the more basic the construc-tions used the less prerequisite is needed in the proof of the theorem, and generally, the longer the proof is. On the contrary, the more complicated the constructions used, the more prerequisite is needed, and generally, the shorter the proof is. For constructions C17, C18, C19, and C20, the eliminating results for some special cases are very simple. As an example, we have Exercise 3.67 For (CIRCUMCENTER O A B C), we have 1. POAO = PABAPACAPBCB 64S 2 ABC . 2. PABO = AB 2 3. PAOB = PABA(PACAPBCB−32S 2 ABC) 64S 2 ABC . 4. S PQO = S PQA 2 + S PQB 2 , if PQ⊥AB. Example 3.68 Let H be the orthocenter of triangle ABC. Then the circumcenters of the four triangles ABC, ABH, ACH, and BCH are such that each is the orthocenter of the triangle formed by the remaining three. 1 C 1 B 1 A O H C B A Figure 3-22 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (circumcenter A1 B C H) (circumcenter B1 A C H) (circumcenter C1 A B H) (parallel B1 C1 B C) ) The machine proof S BCB1 S BCC1 C1 = S BCB1·(2) S BCH+S ABC B1 = (2)·(S BCH+S ABC) (S BCH+S ABC)·(2) simplify = 1 The eliminants S BCC1 C1 = 1 2(S BCH+S ABC) S BCB1 B1 = 1 2(S BCH+S ABC) 140 Chapter 3. Machine Proof in Plane Geometry The above machine proof uses the fourth result of Exercise 3.67. A B C X G D E F Figure 3-23 B C I A D A I Figure 3-24 M 2 M 1 M O H C B A Figure 3-25 Example 3.69 A line is drawn through the centroid of a triangle. Show that the sum of the distances of the line from the two vertices of the triangle situated on the same side of the line is equal to the distance of the line from the third vertex. (Figure 3-23) Constructive description ( (points A B C X) (centroid G A B C) (foot D A G X) (foot E B G X) (foot F C G X) ( EB DA + FC DA = −1) ) The machine proof −( CF AD + BE AD) F = −BE AD ·S AXG−S CXG −(−S AXG) E = −(−S CXG·S AXG−S BXG·S AXG) S AXG·(−S AXG) simplify = −(S CXG+S BXG) S AXG G = −(3S ACX+3S ABX)·(3) (−S ACX−S ABX)·((3))2 simplify = 1 The eliminants CF AD F = S CXG S AXG BE AD E =S BXG S AXG S AXG G = −1 3(S ACX+S ABX) S BXG G = −1 3(S BCX−S ABX) S CXG G =1 3(S BCX+S ACX) Example 3.70 Two tritangent centers divide the bisector on which they are located, harmon-ically (Figure 3-24). Constructive description ( (POINTS B C I) (INCENTER A I C B) (INTER D (LINE A I) (LINE B C)) (INTER IA (LINE A I) (TLINE B B I)) (HARMONIC A D I IA) ) The machine proof (−IA ID)/( AIA DIA) IA = PIBD PIBA · −IA ID D = −(−S BICA)·PCBI·S BIA PIBA·S BICA·S BCI simplify = PCBI·S BIA PIBA·S BCI A = PCBI·(−PBIB·PBCI·S BCI)·PBIC·PBCB (−PCBI·PBIB·PBCI)·S BCI·PBIC·PBCB simplify = 1 The eliminants AIA DIA IA = PIBA PIBD PIBD D = PCBI·S BIA S BICA IA ID D = −S BICA S BCI PIBA A =−PCBI·PBIB·PBCI PBIC·PBCB S BIA A =−PBIB·PBCI·S BCI PBIC·PBCB 3.6 Co-Circle Points 141 Example 3.71 (Euler’s Theorem) The centroid of a triangle is on the segment determined by the circumcenter O and the orthocenter H of the same triangle, and divides OH in the ratio of 1:2 (Figure 3-25). Constructive description ( (POINTS A B C) (CIRCUMCENTER O A B C) (CENTROID M A B C) (LRATIO H M O −2) (PERPENDICULAR A H B C) ) The machine proof PABC PCBH H = PABC 3PCBM−2PCBO M = PABC·(3) −6PCBO+3PBCB+3PABC O = −PABC·(2) −2PABC simplify = 1 The eliminants PCBH H =3PCBM−2PCBO PCBM M = 1 3(PBCB+PABC) PCBO O =1 2(PBCB) 3.6 Trigonometric Functions and Co-Circle Points The aim of this section is to provide an eﬃcient method for dealing with co-circle points. We will prove the co-circle theorems given in Section 1.9 without using trigonometric func-tions; on the contrary, we will develop properties of trigonometric functions by using the area and Pythagoras diﬀerence only. The readers who are interested primaryly in machine proof may skip the following subsection and proceed directly to Subsection 3.6.2. 3.6.1 The Co-circle Theorems Let J, A, and B be three points on a circle with center O. In what follows, we will ﬁx J as a reference point. An oriented chord, f AB, is a directed line segment such that f JA (A , J) is always positive and f AB > 0 iﬀS JAB > 0. Proposition 3.72 Let δ be the diameter of the circumcircle of the triangle ABC. Then S ABC = f AB· f CB· f CA 2δ . C E O B J A Figure 3-26 Proof. Let CE be a diameter of circle O. By Propo-sition 3.48, S 2 ABC S 2 ACE = AB 2 · CB 2 AE 2 · δ2 . Since S 2 ACE = 1 4AC 2 · AE 2, we have S 2 ABC = AB 2 · BC 2 · AC 2 4δ2 , i.e., \|S ABC\| = \|f AB\| · \| f CB\| · \| f CA\| 2δ . 142 Chapter 3. Machine Proof in Plane Geometry We still need to check whether the signs of both sides of the conclusion equation are the same. At ﬁrst, it is easy to see that when we interchange two vertices of the triangle ABC, the signs of both sides of the equation will change. Therefore, we need only to check a particular position for A, B, and C, e.g., the case shown in Figure 3-26. In this case, we have S ABC > 0, f AB > 0, f CB > 0, and f CA > 0. Let f BC be an oriented chord on circle O and BB′ be a diameter. We deﬁne the cochord of f BC to be c BC whose absolute value is equal to \|CB′\| and has the same sign as PBJC. It is clear that f BC 2 + c BC 2 = δ2. Proposition 3.73 For points A, B, and C on circle O, we have PABC = 2f AB· f CB·c CA δ . Proof. By the Herron-Qin formula and Proposition 3.72, we have P2 ABC = 4AB 2 · CB 2 −16S 2 ABC = 4AB 2 · CB 2(δ2 −AC 2) δ2 . Then \|PABC\| = 2\|f AB\|·\| f CB\|·\|c CA\| δ . As in Proposition 1.100, we can check that the signs of the two sides of the equation are equal. O D C B A E Figure 3-27 Proposition 3.74 Let A, B, C, and D be four co-circle points, E the intersection of the circle, and the line passing through D and parallel to AC. Then S ABCD = f AC·g BD· f EB 2δ where δ is the diameter of the circle. Proof. If AC ∥BD, we have E = B and S ABCD = 0. The result is clearly true. Otherwise, since DE ∥AC, we have S ABCD = f AC g ED S EBD = f AC · g BD · f EB 2δ . Proposition 3.75 (Ptolemy’s Theorem) Let A, B, C, and D be four co-circle points. Then f AB · g CD + f BC · g AD = f AC · g BD. Proof. Let E be the intersection of the circle and the line passing through D and parallel to AC (Figure 3-27). Then by Propositions 3.72 and 3.74, f AC · g BD · f EB 2δ = S ABCD = S BCE + S EAB = f BC · g EC · f EB + f EA · f BA · f BE 2δ . 3.6 Co-Circle Points 143 Let B be the reference point. Notice that f EB = −f BE, f AE = −g CD, and g CE = −g AD, we prove the result. J A B O Q P S Figure 3-28 Proposition 3.76 Let AB = d be the diameter of the circle, and P, Q be two points on the circle. Then d · g PQ = g AQ · c AP −f AP · c AQ. Proof. Apply Ptolemy’s theorem to A, P, B, and Q: f AB · g PQ = f AP · g BQ + g AQ · f PB. Let J be the reference point. If S JAB < 0 (Figure 3-28), we have f AB = −d, c AQ = g BQ, and c AP = f BP. If S JAB > 0 or J = A, we have f AB = d, c AQ = g QB, and c AP = f PB. The result is always true. Proposition 3.77 Let AB = d be the diameter of the circle, and P, Q be two points on the circle. Then d · c PQ = f AP · g AQ + c AP · c AQ. Proof. Let S be the antipodal of Q (Figure 3-28). By Ptolemy’s theorem, we have f AB · f PS = f AP · f BS + f AS · f PB. Let J be the reference point. If f AB and g QS have the same sign, we have f AB · f PS = d c PQ, f BS = g AQ, f AS · f PB = c AQ · c AP. If f AB and g QS have diﬀerent signs, we have f AB · f PS = −d · c PQ, f BS = −g AQ, f AS · f PB = −c AQ · c AP. The result is true in both cases. Propositions 3.76 and 3.77 are called decomposition formulas for chords. They can be used to reduce any chord or cochord to a polynomial in chords and cochords with a ﬁxed point (A) as an end point. For points A and B on the circle, let us deﬁne • sin(A) = f JA δ ; cos(A) = c JA δ . • sin(AB) = f AB δ ; cos(AB) = c AB δ . 144 Chapter 3. Machine Proof in Plane Geometry Then we have derived the following properties of trigonometric functions. sin(A)2 + cos(A)2 = 1 sin(AB) = sin(B) cos(A) −sin(A) cos(B) cos(AB) = cos(B) cos(A) + sin(A) sin(B). The sin(AB) and cos(AB) are actually the sine and cosine of the inscribed angle in arc AB. But in the above deﬁnition of trigonometric functions, we did not mention the the concept of angles. 3.6.2 Eliminating Co-Circle Points We introduce a new construction. C21 (CIRCLE Y1 · · · Ys), (s ≥3). Points Y1 · · · Ys are on the same circle. There is no ndg condition for construction C21. The degree of freedom of this construction is s + 3. By Propositions 3.72–3.77, we have Lemma 3.78 Let A, B, and C be points on a circle with center O and diameter δ. Then S ABC = f AB· f CB· f CA 2δ , PABC = 2f AB· f CB·c CA δ . f AC = δ sin(AC), c AC = δ cos(AC). Using Lemma 3.78, an expression in the areas and Pythagoras diﬀerences of points on a circle can be reduced to an expression in the diameter δ of the circle and trigonometric functions of independent angles. Two such expressions have the same value iﬀwhen sub-stituting, for each angle α, (sin α)2 by 1 −(cos α)2 the resulting expression should be the same. We thus have a complete method for this construction. The reader may have noticed that this construction must always come ﬁrst in the description of a statement. Otherwise, in the next step, we do not know how to eliminate these trigonometric functions. The proofs of many interesting geometry theorems use this construction. Example 3.79 (Simson’s Theorem) Let D be a point on the circumscribed circle of triangle ABC. From D three perpendiculars are drawn to the three sides BC, AC, and AB of triangle ABC. Let E, F, and G be the three feet respectively. Show that E, F and G are collinear. A B C O D E F G Figure 3-29 Here is the input to our program. ((CIRCLE A B C D) 3.6 Co-Circle Points 145 (FOOT E D B C) (FOOT F D A C) (FOOT G D A B) (INTER G1 (LINE E F) (LINE A B)) ( AG BG = AG1 BG1 ) ) The ndg conditions: B , C, A , C, A , B, EF ∦AB, B , G, B , G1. Here is the machine proof. The last step of the proof uses Lemma 3.78. The machine proof ( AG BG)/( AG1 BG1 ) G1 = S BEF S AEF · AG BG G = PBAD·S BEF S AEF·(−PABD) F = −PBAD·PACD·S ABE·PACA (−PCAD·S ACE)·PABD·PACA simplify = PBAD·PACD·S ABE PCAD·S ACE·PABD The eliminants AG1 BG1 G1 = S AEF S BEF AG BG G = PBAD −PABD S AEF F =−PCAD·S ACE PACA S BEF F = PACD·S ABE PACA S ACE E = −PBCD·S ABC PBCB S ABE E = PCBD·S ABC PBCB PABD= −2(g BD·f AB·cos(AD)) E = PBAD·PACD·PCBD·S ABC·PBCB PCAD·(−PBCD·S ABC)·PABD·PBCB simplify = PBAD·PACD·PCBD −PCAD·PBCD·PABD co−cir = (2g AD·f AB·cos(BD))·(−2g CD· f AC·cos(AD))·(2g BD· f BC·cos(CD)) −(2g AD· f AC·cos(CD))·(−2g CD· f BC·cos(BD))·(−2g BD·f AB·cos(AD)) simplify = 1 The eliminants PBCD= −2(g CD· f BC·cos(BD)) PCAD=2(g AD· f AC·cos(CD)) PCBD=2(g BD· f BC·cos(CD)) PACD= −2(g CD· f AC·cos(AD)) PBAD=2(g AD·f AB·cos(BD)) Example 3.80 (Pascal’s Theorem on a Circle) Let A, B, C, D, E, and F be six points on a circle. Let P = AB ∩DF, Q = BC ∩EF, and S = CD ∩EA. Show that P, Q, and S are collinear. Here is the input to the program. S Q P E F D O C B A Figure 3-30 ( (CIRCLE A B C D F E) (INTER P (LINE D F) (LINE A B)) (INTER Q (LINE F E) (LINE B C)) (INTER S (LINE E A) (LINE C D)) (INTER S 1 (LINE P Q) (LINE C D)) ( CS DS = CS1 DS1 ) ) The ndg conditions: DF ∦AB, EF ∦BC, AE ∦CD, PQ ∦CD, D , S , D , S 1. 146 Chapter 3. Machine Proof in Plane Geometry The machine proof ( CS DS )/( CS1 DS1 ) S 1 = S DPQ S CPQ · CS DS S = S ACE·S DPQ S CPQ·S ADE Q = S ACE·(−S DEP·S BCF)·S BFCE (−S CFE·S BCP)·S ADE·(−S BFCE) simplify = −S ACE·S DEP·S BCF S CFE·S BCP·S ADE P = −S ACE·(−S DFE·S ABD)·S BCF·S ADBF S CFE·(−S BDF·S ABC)·S ADE·(−S ADBF) simplify = S ACE·S DFE·S ABD·S BCF S CFE·S BDF·S ABC·S ADE The eliminants CS1 DS1 S1 = S CPQ S DPQ CS DS S = S ACE S ADE S CPQ Q =−S CFE·S BCP S BFCE S DPQ Q = S DEP·S BCF S BFCE S BCP P =−S BDF·S ABC S ADBF S DEP P =S DFE·S ABD S ADBF S ADE= g DE· f AE·g AD (−2)·d S ABC= f BC· f AC·f AB (−2)·d S BDF= g DF· f BF·g BD (−2)·d S CFE= g FE·g CE·g CF (−2)·d S BCF= g CF· f BF· f BC (−2)·d The machine proof co−cir = (−g CE· f AE· f AC)·(−g FE·g DE·g DF)·(−g BD·g AD·f AB)·(−g CF· f BF· f BC)·((2d))4 (−g FE·g CE·g CF)·(−g DF· f BF·g BD)·(−f BC· f AC·f AB)·(−g DE· f AE·g AD)·((2d))4 simplify = 1 The eliminants S ABD= g BD·g AD·f AB (−2)·d S DFE= g FE·g DE·g DF (−2)·d S ACE= g CE· f AE· f AC (−2)·d Example 3.81 (The General Butterﬂy Theorem.) As in the ﬁgure, A, B, C, D, E, F are six points on a circle. M = AB ∩CD; N = AB ∩EF; G = AB ∩CF; H = AB ∩DE. Show that MG AG BH NH AN MB = 1. H N M E O C B A F G D Figure 3-31 Constructive description ( (CIRCLE A B C D E F) (INTER M (LINE D C) (LINE A B)) (INTER N (LINE E F) (LINE A B)) (INTER G (LINE A B) (LINE C F)) (INTER H (LINE D E) (LINE A B)) ( MG AG BH NH = BM AB BA AN ) ) 3.6 Co-Circle Points 147 The machine proof MG AG · BH NH −BM AB · AB AN H = S BDE −BM AB · AB AN ·S DEN · MG AG G = −S CFM·S BDE BM AB · AB AN ·S DEN·S ACF N = −S CFM·S BDE·(−S AEBF)·S AEF BM AB ·S AEBF·S DEF·S ABE·S ACF simplify = S CFM·S BDE·S AEF BM AB ·S DEF·S ABE·S ACF M = (−S CDF·S ABC)·S BDE·S AEF·(−S ACBD) (−S BCD)·S DEF·S ABE·S ACF·(−S ACBD) simplify = S CDF·S ABC·S BDE·S AEF S BCD·S DEF·S ABE·S ACF co−cir = (−g DF·g CF·g CD)·(−f BC· f AC·f AB)·(−g DE· f BE·g BD)·(−g EF· f AF· f AE)·((2d))4 (−g CD·g BD· f BC)·(−g EF·g DF·g DE)·(−f BE· f AE·f AB)·(−g CF· f AF· f AC)·((2d))4 simplify = 1 The eliminants BH NH H = S BDE S DEN MG AG G =S CFM S ACF S DEN N =S DEF·S ABE −S AEBF AB AN N =S AEBF S AEF BM AB M = S BCD S ACBD S CFM M = S CDF·S ABC S ACBD S ACF= g CF· f AF· f AC (−2)·d S ABE= f BE· f AE·f AB (−2)·d S DEF= g EF·g DF·g DE (−2)·d S BCD= g CD·g BD· f BC (−2)·d S AEF= g EF· f AF· f AE (−2)·d S BDE= g DE· f BE·g BD (−2)·d S ABC= f BC· f AC·f AB (−2)·d S CDF= g DF·g CF·g CD (−2)·d Example 3.82 (Cantor’s Theorem) The perpendiculars from the midpoints of the sides of a cyclic quadrilateral to the respectively opposite sides are concurrent. B C D O A E F G N Figure 3-32 Constructive description ((CIRCLE A B C D) (CIRCUMCENTER O A B C) (MIDPOINT G A D) (MIDPOINT F A B) (MIDPOINT E C D) (PRATIO N E O F 1) (PERPENDICULAR G N B C) ) The machine proof PCBG PCBN N = PCBG PCBE+PCBF−PCBO E = PCBG PCBF−PCBO+ 1 2 PCBD+ 1 2 PBCB F = (2)·PCBG −2PCBO+PCBD+PBCB+PABC G = (−2)·( 1 2 PCBD+ 1 2 PABC) 2PCBO−PCBD−PBCB−PABC O = −(PCBD+PABC)·(2) −2PCBD−2PABC simplify = 1 The eliminants PCBN N =PCBE+PCBF−PCBO PCBE E = 1 2(PCBD+PBCB) PCBF F =1 2(PABC); PCBO O =1 2(PBCB) PCBG G =1 2(PCBD+PABC) The ndg conditions: A, B, C are not collinear; A , D; A , B; C , D; O , F. 148 Chapter 3. Machine Proof in Plane Geometry 3.7 Machine Proof for Class C The class C, or the class of constructive geometry statements, are the geometry statements which are assertions about the conﬁgurations that can be drawn by rulers and compasses only. To describe the statements in class C, we need two new constructions. We ﬁrst introduce a new kind of circle: (CIR O r) is the the circle with center O and radius √r. As before, r could be an algebraic number, a rational expression in geometry quantities, or a variable. Thus (CIR O P) is the same as (CIR O OP 2). C22 (INTER Y (LINE U V) (CIR O r)). Point Y is one of the intersections of line (LINE U V) and circle (CIR O r). The ndg conditions are r , 0, U , V. Point Y is a ﬁxed point and has two possibilities. C23 (INTER Y (CIR O1 r1) (CIR O2 r2)). Point Y is one of the intersections of the circle (CIR O1 r1) and the circle (CIR O2 r2). The ndg conditions are O1 , O2, r1 , 0, and r2 , 0. Point Y is a ﬁxed point and has two possibilities. Each of constructions C22 and C23 actually introduces two points, and we generally can not distinguish the two points. This is the main diﬃculty of dealing with these two con-structions. 3.7.1 Eliminating Points from Geometry Quantities Proposition 3.83 Let Y be one of the intersection points of line UV and circle (CIR O r). Then we have (III) (UY UV )2 −POUV UV 2 UY UV + OU 2 −r UV 2 = 0. O X U V M Y Figure 3-33 Proof. Let X be another intersection point of the line UV and the circle (CIR O r), and M the midpoint of XY. By Proposition 3.2, UY UV + UX UV = 2UM UV = PMUV UV 2 = POUV UV 2 . (1) By Proposition 3.5, OU 2 = XU XY OY 2 + UY XY OX 2 −XU XY · UY XY XY 2 = OX 2 + UY · UX = r + UY · UX. (2) 3.7 Machine Proof for Class C 149 From (1) and (2), it is easy to obtain the result. Now we can give the elimination methods for construction C22 easily. Lemma 3.84 Let Y be introduced by (INTER Y (LINE U V) (CIR O r)). Then S ABY = UY UVS VAUB + S ABU, PABY = UY UV PVAUB + PABU, PAYB = UY UV PVAUB + PABU −UY UV(1 −UY UV)PUVU where UY UV satisﬁes (III). Proof. By Proposition 2.9, S ABY = UY UVS ABV +(1−UY UV)S ABU = UY UVS VAUB+S ABU. The second and third cases are consequences of Proposition 3.5. Lemma 3.85 Let Y be introduced by (INTER Y (LINE U V) (CIR O r)). Then DY EF =        DU EF + UY UV UV EF if D ∈UV. S DUV S EUFV otherwise. Proof. The ﬁrst case is trivial. The second case is a consequence of the co-side theorem. Proposition 3.86 Construction C23 is equivalent to the following two constructions (LRATIO O O1 O2 r) (TRATIO Y O O1 s) where r = O1O2 2+r1−r2 2O1O2 2 , s2 = r1 r2O1O2 2 −1. O O Y O Figure 3-34 1 2 Proof. Let O be the foot of the perpendicular dropped from point Y upon the line O1O2. By Proposition 3.2, O1O O1O2 = PYO1O2 PO1O2O1 = O1O2 2 + r1 −r2 2O1O2 2 . For s, we have s2 = OY 2 OO1 2 = r1 OO1 2 −1 = r1 r2O1O2 2 −1. Lemma 3.87 Let Y be introduced by (INTER Y (CIR O1 r1) (CIR O2 r2)). Then S ABY = S ABO1 + rS O2AO1B −rs 4 PO2AO1B where r and s are the same as in Proposition 3.86. 150 Chapter 3. Machine Proof in Plane Geometry Proof. Let O be the foot of the perpendicular dropped from point Y upon the line O1O2. By Lemma 3.26, S ABY = S ABO −s 4POAO1B = rS ABO2 + (1 −r)S ABO1 −s 4(rPO2AB + (1 −r)PO1AB −PO1AB) = S ABO1 + rS O2AO1B −rs 4 PO2AO1B. Lemma 3.88 Let Y be introduced by (INTER Y (CIR O1 r1) (CIR O2 r2)). Then PABY = PABO1 + rPO2AO1B −4rsS O2AO1B where r and s are the same as in Proposition 3.86. Proof. Let O be the foot of the perpendicular dropped from point Y upon the line O1O2. By Lemma 3.27, PABY = PABO −4sS OAO1B = rPABO2 + (1 −r)PABO1 −4s(rS O2AB + (1 −r)S O1AB −S O1AB) = PABO1 + rPO2AO1B −4rsS O2AO1B. The methods of eliminating Y from DY EF and PAYB can be found in Lemmas 2.26, 3.30, 3.25, and 3.28. 3.7.2 Pseudo Divisions and Triangular Forms In this section, we introduce some algebraic tools which will be used in the machine proof for statements in class C. These tools are actually parts of Wu’s method of automated ge-ometry theorem proving, more details of which can be found in [36, 12]. Let K be a computable ﬁeld of characteristic zero (e.g., Q) and A = K[x1, ..., xn] be the polynomial ring of the variables x1, · · · , xn. For P ∈K[x] −K, we can write P = cdxd p + ... + c1xp + c0 where ci ∈B[x1, ..., xp−1], p > 0, and cd , 0. We call p the class, cd the initial, xp the leading variable, and d the leading degree of P respectively, or class(P) = p, init(P) = cd, lv(P) = xp, ld(P) = d. If P ∈K, we have class(P) = 0. Let p = class(P) > 0. A polynomial Q is said to be reduced with respect to P if deg(Q, xp) < ld(P). Let f = anvn + · · · + a0 and h = bkvk + · · · + b0 be two polynomials in A[v], where v is a new indeterminate. Suppose k, the leading degree of h in v, is greater than 0. Then the pseudo division proceeds as follows: 3.7 Machine Proof for Class C 151 Deﬁnition 3.89 (Pseudo Division) First let r = f . Then repeat the following process until m = deg(r, v) < k: r = bkr −cmvm−kh, where cm is the leading coeﬃcient of r. It is easy to see that m strictly decreases after each iteration. Thus the process terminates. At the end, we have the pseudo remainder prem(f, h, v) = r = r0. Proposition 3.90 Continuing from the above deﬁnition, we have the following formula, bs k f = qh + r0, where s ≤n −k + 1 and deg(r0, v) < deg(h, v). (1) Proof. We ﬁx polynomial h and use induction on n = deg(f, v). If n < k, then we have r0 = f and f = 0 · h + r0. Suppose n ≥k and formula (1) is true for those polynomials f , for which deg(f, v) < n. After the ﬁrst iteration, we have r = bk f −anvn−kh, where an is the leading coeﬃcient of f . Since deg(r, v) < n, we have bt kr = q1h + r0 by the induction hypothesis. Substituting r = bk f −anvn−kh in the last formula, we have (1). Deﬁnition 3.91 A sequence of polynomials TS = A1, ..., Ap in K[X] is said to be a triangular form, if either p = 1 and A1 , 0 or 0 < class(Ai) < class(A j) for 1 ≤i < j. For a triangular form TS = A1, ..., Ap, we make a renaming of the variables. If Ai is of class mi, we rename xmi as xi, other variables are renamed as u1, ..., uq, where q = n −p. The variables u1, ..., uq are called the parameter set of TS . Then TS is like A1(u1, · · · , uq, x1) A2(u1, · · · , uq, x1, x2) (IV) · · · Ap(u1, · · · , uq, x1, · · · , xp). For another polynomial G, we can deﬁne the successive pseudo division: Rp = prem(G, Ap, xp), . . . , R1 = prem(R1, A1, x1). R = R1 is called the ﬁnal remainder and is denoted by prem(G, A1, . . . , Ar). It is easy to prove the following important proposition: Proposition 3.92 (The Remainder Formula) Let TS and R be the same as the above. There are some non-negative integers s1, . . . , sp and polynomials Q1, . . . , Qp such that 1. Is1 1 · · · I sp p G = Q1A1 + · · · + QpAp + R where the Ii are the initials of the Ai. 2. deg(R, xi) < deg(Ai, xi), for i = 1, · · · , p. Proof. We use induction on p. The case p = 1 is actually Proposition 3.90. Suppose that p > 1 and the proposition is true for p −1. Thus we have: Is1 1 · · · I sp−1 p−1 Rp−1 = Q1A1 + · · · + Qp−1Ap−1 + R, with deg(R, xi) < deg(Ai, xi), for i = 1, . . . , p −1. Combining this with Rp−1 = IspG −QAp, we have 1 and 2. 152 Chapter 3. Machine Proof in Plane Geometry Theorem 3.93 Let TS = A1, ..., Ap be a triangular set, G a polynomial. If prem(G, TS ) = 0 then ∀xi[(A1 = 0 ∧· · · ∧Ap = 0 ∧I1 , 0 ∧· · · ∧Ip , 0) ⇒G = 0]. Proof. Since prem(G, TS ) = 0, by the remainder formula Is1 1 · · · I sp p G = Q1A1 + · · · + QpAp. Now it is clear that Ai = 0 and Ii , 0 imply G = 0. Deﬁnition 3.94 A triangular set A1, ..., Ap of the form (IV) is called irreducible if for each i 1. the initial Ii of Ai does not vanish in the polynomial ring Ri = K(u)[x1, · · · , xi]/(A1, ..., Ai−1), and 2. Ai is irreducible in Ri. Thus the sequence F0 = K(u), F1 = A0[x1]/(A1), ..., Fp = Ap−1[xp]/(Ap) = A0[x]/(A1, ..., Ap) is a tower of ﬁeld extensions. Example 3.95 Let TS be the triangular set A1 = x2 1 −u1, A2 = x2 2 −2x1x2 + u1. A1 is irreducible over A0 = Q[u1]; but A2 is reducible over A1 = A0[x1]/(A1) because A2 = (x2 −x1)2 under x2 1 −u1 = 0. Thus TS is reducible. Proposition 3.96 Let TS = A1, ..., Ap of the form (IV) be irreducible, and G be a polynomial in K[u, x]. Then the following conditions are equivalent: (i) prem(G, TS ) = 0. (ii) Let E be an extension ﬁeld of K. If µ = (η1, · · · , ηq, ζ1, · · · , ζp) in Ed+r is a common zero of A1, ..., Ap with ηi transcendental over K, then µ is also a zero of G, i.e., G(µ) = 0. Proof. For any polynomial h, let ˜ h be the polynomial obtained from h by substituting ui, xi for ηi, ζi. We use induction on k to prove the following assertions (for 0 < k ≤p): (U) For any polynomial P = asxs k +· · ·+a0 (0 < k ≤p, 1 ≤s, ai ∈K[u, x1, ..., xk−1], as , 0) reduced with respect to A1, ..., Ap, if µ is a zero of P, then P = 0. If k = 1, then P = asxs k + · · · + a0, with all a j ∈K[u]. µ is a zero of P means ˜ P = ˜ asζ1 s + · · · + ˜ a0 = 0. Since P is reduced with respect to A1, s < deg(A1, x1). By the 3.7 Machine Proof for Class C 153 uniqueness of representation in algebraic extension, we have all ˜ a j = 0. Since ηi are transcendental over K, all a j = 0. Hence P = 0. Now we want to prove (U) is true for k assuming it is true for k −1. Since µ is a zero of P, ˜ P = ˜ asζk s + · · · + ˜ a0 = 0. Since s < deg(Ak, xk), by the uniqueness of representation in algebraic extension again, all ˜ a j = 0. Thus µ is also a zero of all a j. Since all a j are also reduced with respect to A1, ..., Ap, a j = 0 by the induction hypothesis. Hence P = 0. (ii) ⇒(i). Suppose µ is a zero of G. Let R = prem(G, A1, ..., Ap). We have the remainder formula Is1 1 · · · I sp p G = Q1A1 + · · · + QpAp + R. Hence µ is a zero of R. Since R is reduced with respect to A1, ..., Ap, R = 0. (i) ⇒(ii). Suppose prem(G, A1, ..., Ap) = 0. Then by the remainder formula, we have Is1 1 · · · I sp p G = Q1A1 + · · · + QpAp. where the Ik the are initials of the Ak. Since prem(Ik, TS ) , 0, µ is not a zero of Ik (by (ii) ⇒(i)). Hence µ is a zero of G. We call µ in (ii) a generic point of that irreducible triangular form in ﬁeld E. The theorem is no longer true if A1, ..., Ap is reducible. We can ﬁnd such an example by letting A1, A2 be the same as in example 3.95 and G = x2 −x1. Theorem 3.97 Let TS = A1, ..., Ap be an irreducible triangular form, G a polynomial. If ∀xi[(A1 = 0 ∧· · · ∧Ap = 0 ∧I1 , 0 ∧· · · ∧Ip , 0) ⇒G = 0] is true in any extension ﬁeld of K then prem(G, TS ) = 0. Proof. Let η1, · · · , ηq be some elements which are transcendental over K. By the deﬁnition of irreducible triangular set, we can ﬁnd a generic zero µ = (η1, · · · , ηq, ζ1, · · · , ζp) of TS such that Ii(µ) , 0, i = 1, · · · , p. Since Ai(µ) = 0, Ii(µ) , 0, i = 1, · · · , p, we have G(µ) = 0. By Proposition 3.96, prem(G, TS ) = 0. Exercise 3.98 Let K be the ﬁeld of the rational numbers, TS = A1, ..., Ap an irreducible triangular form, and G a polynomial. If ∀xi[(A1 = 0 ∧· · · ∧Ap = 0 ∧I1 , 0 ∧· · · ∧Ip , 0) ⇒G = 0] is true in the ﬁeld of complex numbers then prem(G, TS ) = 0. 154 Chapter 3. Machine Proof in Plane Geometry 3.7.3 Machine Proof for Class C We now return to the theory of machine proof. First, let us restate Algorithm 3.33 using the language of triangular forms and pseudo divisions. Let S = (C1, · · · , Cr, (E, F)) be a statement in CL. We denote by u1, · · · , uq, x1, · · · , xp the geometry quantities occurring in the proof of S such that the ui are the free parameters and the xi are those quantities from which a point will be eliminated. We arrange the subscripts such that xi = Ui(u1, · · · , uq, x1, · · · , xi−1) Ii(u1, · · · , uq, x1, · · · , xi−1) , i = 1, · · ·, p. Let Ai = Iixi −Ui. Then TS = A1, · · · , Ap is a triangular form with Ii as the initials of Ai. Furthermore TS is irreducible since Ii , 0 is true under the ndg conditions of S . Theorem 3.99 Use the same notations as above. The statement S is true iﬀprem(E − F, TS ) = 0. Proof. Notice that the proving process of S is as follows: ﬁrst replace xi by Ui Ii , i = p, · · · , 1, in E and F to obtain two polynomials E′ and F′ in the ui only. Since the ui are free parameters, the statement S is true iﬀE′ = F′. The above process is equivalent to taking the pseudo remainder of E −F with respect to TS . Therefore S is true iﬀprem(E −F, TS ) = 0. If S = (C1, · · · , Cr, (E, F)) is a statement in C, then for each i, Ai has two possible forms: (1) either Ai = Iixi + Ui, or (2) Ai = Iix2 i + Uixi + Vi where Ii, Ui, and Vi are polynomials in u1, · · · , uq, x1, · · · , xi−1. TS = A1, · · · , Ap is still a triangular form. Let R = prem(E −F, TS ). Then we have Theorem 3.100 1. If R = 0, S is true. 2. If R , 0 and TS is irreducible then S is not a theorem in the complex plane. Proof. For the ﬁrst case, by Theorem 3.93 we have ∀uixi[(A1 = 0 ∧· · · ∧Ap = 0 ∧I1 , 0 ∧· · · ∧Ip , 0) ⇒E −F = 0]. Under the ndg conditions of S , we have Ii , 0, i = 1, · · · , p. Then E = F is true. The second case is a consequence of Theorem 3.97. 3.7 Machine Proof for Class C 155 Remark 3.101 In practice, we do not have to take the pseudo remainder of E −F with respect to TS . The better way is to eliminate xi from E and F separately as usual, so that we can take the advantage of removing the common factors from E and F during the proof. To eliminate xi from E, if Ai = Iixi −Ui, we need only to replace xi in E by Ui Ii ; if Ai = Iix2 i + Uixi + Vi, we need to keep replacing x2 i in E by −Uixi+Vi Ii until the degree of xi in E is less that two. If R , 0 and TS is reducible, we need to factorize TS into irreducible triangular forms. We will not discuss the factorization method in this book; those who are interested in this topic may refer to [36, 12]. Let us assume that for some Ai we have Ai = Iix2 i + Uixi + Vi = (Ii,1xi −Ui,1)(Ii,2xi −Ui,2) which is true under the condition Ak = 0, k = 1, · · ·, i −1. Then TS is factored into two triangular forms TS 1 and TS 2 as follows TS 1 = A1, ..., Ai,1, ..., Ap; TS 2 = A1, ..., Ai,2, ..., Ap where Ai,1 = Ii,1xi −Ui,1, Ai,2 = Ii,2xi −Ui,2. Geometrically, this means that the two points introduced by a construction of type C22 or C23 can be distinguished and the two triangular forms TS 1 and TS 2 correspond to the two intersections. If R , 0 and TS is reducible, let TS be factored into several irreducible triangular forms TS 1, · · · , TS m. Then we have three possible cases: • For each TS i, prem(E −F, TS i) , 0, i.e., E = F is not valid for all triangular forms. In this case, we say that the statement S is generally false. • For some TS i, prem(E −F, TS i) , 0, while for other TS j, prem(E −F, TS j) = 0. In this case, the statement S is true only for some conﬁgurations. • prem(E −F, TS i) = 0 for all i. In this case, S is still true in the Euclidean geometry. This will happen only when we want to introduce the intersection points of two circles or a line and a circle which are tangent to each other. With the help of algebraic tools, we have a complete method of machine proof for geometry statements in class C. 156 Chapter 3. Machine Proof in Plane Geometry A B D C G E F A B D C G E F Figure 3-35 Example 3.102 Let ABCD be a square. CG is parallel to the diagonal BD. Point E is on CG such that BE = BD. F is the intersection of BE and DC. Show that DF = DE. Proof. As shown in Figure 3-35, let G be a point on line AD such that AD = DG. Then point E has two possible positions. The following proof shows that PDFD = PDED is true for both positions of E. By Lemmas 3.22 and 3.27, we have PBCG= −PDBC PDBC=PADB=PDGD=PBCB=PADA = PDCD=PABA PCGC=PBDB = 2PABA Let r = CE CG. By the co-side theorem, we have PDFD = PDCDS 2 BDE S 2 BDEC = PABAS 2 BDC (S BDC −rS DCG)2 = PABAS 2 BDC (S BDC + rS BDC)2 = PABA (1 + r)2. By (II) on page 116, PDED = (1 −r)PDCD + rPDGD −(1 −r)rPCGC = ((1 −r) + r −2(1 −r)r)PABA. Then PDFD = PDED is true iﬀ (1 + r)2((1 −r) + r −2(1 −r)r) = 1 and this is the case since by Proposition 3.83, r2 = 2rPBCG −PBCB + PBDB PCGC = −2r + 1 2 . Example 3.103 Let ABC be a triangle such that AC = BC. D is a point on AC; E is a point on BC such that AD = BE. F is the intersection of DE and AB. Show that DF = EF. 3.8 Geometry Information Bases 157 If we describe the statement as follows, it becomes reducible. A B C D E E F Figure 3-36 1 ((POINTS A B) (ON C (BLINE A B)) (ON D (LINE A C)) (INTER E (LINE B C) (CIR B AD 2)) (INTER F (LINE A B) (LINE D E)) (MIDPOINT F D E)) By Proposition 3.83, (BE BC )2 · PBCB −PADA = 0. (1) Since PADA=PACA · ( AD AC)2 = PBCB · ( AD AC)2, (1) becomes (BE BC )2 · PBCB −PBCB · (AD AC )2 = (BE BC −AD AC ) · (BE BC + AD AC )PBCB = 0. Then we have BE BC = AD AC or BE BC = −AD AC which correspond to points E1 and E in Figure 3-36. In the ﬁrst case, we have AB ∥DE1; the nondegenarate condition needed to construct point F is not satisﬁed. In the second case, the conclusion is true. Here is the proof of the example. DF FE F = −S ABD S ABE E = −S ABD BE BC S ABC E = S ABD AD AC S ABC D = S ABC AD AC AD AC S ABC simplify = 1 DF FE F = −S ABD S ABE S ABE E = BE BC · S ABC BE BC = −AD AC S ABD D = S ABC · AD AC 3.8 Geometry Information Bases and Machine Proofs Based on Full-Angles In many traditional proofs, the relationships among angles are always used directly if pos-sible. This may be one of the main reasons that traditional geometric proofs are very short, skillful and interesting. But the angle is a concept involving the relation of orders, and 158 Chapter 3. Machine Proof in Plane Geometry is thus very diﬃcult to ﬁt into our machine proof system. In Section 1.10, we introduce the concept of full-angles as the basis of machine proof of geometry statements involving angles. In Subsection 3.4.2, we present a machine proof method for geometry statements involving full-angles based on the property of the tangent function of full-angles. But this approach loses some of the unique character of the traditional proofs based on angles. This section will be devoted to another approach to mechanical generation of proofs based on full-angles. The basic idea for this new approach is that we will build a geometry information base (GIB) based on the constructive description of the statement. The GIB for a statement contains some basic geometry relations about the conﬁguration of the state-ment such as collinear points, parallel lines, perpendicular lines, and cyclic points, etc. In Subsections 2.5.1 (see the paragraph after Exercise 2.40) and 3.3.2, we have touched unon the idea of building some kind of geometry information bases. The purpose of building the GIB in those two cases is that the reﬁned elimination techniques need these geometry relations. The GIB actually has much potential in the automated production of traditional proofs for geometry statements. In the following subsections, we will show that elegant proofs for many geometry theorems can be obtained by merely checking a good GIB. This section reports our initial study of this promising approach. 3.8.1 Building the Geometry Information Base We will use Example 1.118 to illustrate our method. First, we will check every step in the proof of Example 1.118 to ﬁnd out how to eliminate points O, D and E from full-angles ∠[AD, AO] and ∠[AC, CE]. We ﬁrst use rules Q7 and Q9 (on page 46) to eliminate point E from ∠[AC, CE]. A B C O D E M Figure 3-37 (1) ∠[AC, CE] = ∠[AC, BC] + ∠[BC, CE]. Since BE = CE and that E is on line AB, by Q9 we have (2) ∠[BC, CE] = ∠[BE, BC] = ∠[BA, BC]. To eliminate the points O and D, we ﬁrst use Q7 to divide ∠[AD, AO] into two parts: (3) ∠[AD, AO] = ∠[AD, AC] + ∠[AC, AO]. Since AD ⊥BC, by Q7 we can eliminate D. (4) ∠[AD, AC] = ∠[AD, BC] + ∠[BC, AC] = ∠ + ∠[BC, AC]. The next step is to eliminate O from ∠[AC, AO]. 3.8 Geometry Information Bases 159 (5)∠[AC, AO] = ∠[CO, AC] (Q9 and AO = CO) = ∠[CO, MO] + ∠[MO, AC] (Q7) = ∠[AC, AB] + ∠[MO, AC] (Q12, MB = MC and AO = BO = CO) = ∠[AC, AB] + ∠[MO, BC] + ∠[BC, AC] (Q7) = ∠[BC, AB] + ∠. (Q7, MB = MC and BO = BC) Finally, replacing ∠[AD, AC], ∠[AC, AO] in (3) by (4),(5) and ∠[BC, CE] in (1) by (2), we have the conclusion: (6)∠[AD, AO] + ∠[AC, CE] = ∠[AC, BC] + ∠[BA, BC] + ∠ + ∠[BC, AC] + ∠[BC, AB] + ∠ = ∠[AC, AC] + ∠[BA, AB] + ∠ + ∠ (Q7) = ∠ (Q1,Q3 and Q4) Suppose that a program named QAP could prove the geometry theorem as above. Now we are going to check what kinds of information would be needed for designing the pro-gram. Obviously, the basic properties Q1–Q12 (on page 46) about full-angles are necessary in each step of proof. But we will soon see that it is not enough to prove this geometry theorem by using these rules only. Much more geometric information is needed. Let us check the above proof step by step. Step (1) seems very easy. But from the point of view of mechanization, it is actually diﬃcult to start this step. The question is why the full-angle ∠[AC, CE] should be divided into ∠[AC, BC] + ∠[BC, CE] but not anything else, for example, ∠[AC, AD] + ∠[AD, CE]. Why could our program QAP foresee that the point E will be eliminated from ∠[BC, CE]? To make QAP work like this, we can imagine that a geometric information base (GIB) will be generated automatically before QAP proves the theorem. The program will know that ∠[BC, CE] = ∠[BA, BC] by checking GIB. So it chooses step (1) so that point E will be eliminated in the next step. How do we generate the GIB? Of course, the hypotheses of the proposition should be put into the GIB ﬁrst. There are four conditions in Example 1.118: (G1) The circumcenter of triangle ABC is O. (OA = OB = OC) (G2) AD is an altitude of triangle ABC. (AD ⊥BC and D ∈BC) (G3) M is the midpoint of BC. (BM = MC and M ∈BC) 160 Chapter 3. Machine Proof in Plane Geometry (G4) E is the intersection of AB and MO. (E ∈AB and E ∈MO) But here (G1)–(G4) are only a small part of the GIB. We must put more information into the GIB. To see what is still needed in the GIB, we check step (2). In this step, the rules on full-angles are not suﬃcient. We need the hypotheses of the proposition (i.e., conditions (G1)–(G4)). Furthermore, we need some geometric facts which are derived by applying some geometry knowledge to the hypotheses of the statement, (such as deriving BE = CE from MB = MC, OB = OC and E is on the line OM). So we also need a geometry knowledge base (GKB) to build the GIB. As an example, in order to obtain BE = CE, our GKB should include the following propositions: (a) If PB = PC and QB = QC then PQ is the perpendicular bisector of BC. (ndg condition: B , C and P , Q). (b) If P is on the perpendicular bisector of BC, then BP = CP. (c) If PC = PB then ∠[PC, BC] = ∠[BC, PB]. (d) If P is on line AB, then ∠[PB, XY] = ∠[AB, XY]. Applying (a) to G1 and G3, QAP obtains a new information and puts it into the GIB: (G5) MO is the perpendicular bisector of BC. Applying (b) to G4 and G5, we have (G6) BE = CE. Applying (c) to G6, we have (G7) ∠[BC, CE] = ∠[BE, BC]. Finally, applying (d) to G4 and G7, we have (G8) ∠[BE, BC] = ∠[AB, BC]. G8 is the deductive basis of step (2). For step (3), QAP has to foresee that points D and O can be eliminated as in steps (4) and (5) and as a consequence, it decides to split the full-angle ∠[AD, AO] into ∠[AD, AC] and ∠[AC, AO]. The following additional geometry knowledge should be included in GKB: (e) If PQ is perpendicular to UV, then ∠[PQ, XY] = ∠ + ∠[UV, XY]. (f) (Another form of Q12) If O is the circumcenter of triangle ABC, then ∠[AC, AO] = ∠[BC, AB] + ∠. Proof of (f). Let D be the intersection of AO and the circle. By the inscribed angle theorem, ∠[AB, BC] = ∠[AD, CD]. Since AC⊥CD, we have ∠[AC, AO] = ∠ + ∠[DC, AD] = ∠ + ∠[AB, BC]. 3.8 Geometry Information Bases 161 Applying (e) to G2 and (f) to G1, QAP will put the following information into GIB: (G9) ∠[AD, AC] = ∠ + ∠[BC, AC]. (G10) ∠[AC, AO] = ∠[BC, AB] + ∠. When QAP ﬁnds the information G1-G10, step (6) will be done according to rules of full-angle (Q1-Q12) easily. Here we mentioned only the information G1–G10 which are useful for proving the state-ment. In fact, much more information about this statement will be put into the GIB, because QAP does not know what information will be used when it generates the GIB. It will keep applying every rule in the GKB to all the information in GIB to get new information and to put the new information into GIB until nothing new can be obtained. What geometry knowledge should be included in the GKB? Is it complete? At the present stage, the choices of the rules in the GKB are based on our experience of proving geometry theorems. Its completeness is still not considered. But if the method in this section fails to prove or disprove a statement, we can always use Algorithm 3.33 which is complete for constructive geometry statements. Thus, the GIB-GKB method is actually an expert system of proving geometry statements. Up to now, the following rules have been put into our GKB: K1 Two points A and B determine one line. (ndg condition: A , B) K2 Three points A, B and C determine one circle. (ndg condition: S ABC , 0) K3 ∠[PQ, XY] = ∠[UV, XY] if and only if ∠[PQ, UV] = ∠. (ndg condition: X , Y) K4 Four points A, B, C and D are cyclic if and only if ∠[AC, BC] = ∠[AD, BD]. (ndg condition: A, B, C and D are not collinear) K5 AB = AC if and only if ∠[AB, BC] = ∠[BC, AC]. (ndg condition: S ABC , 0) K6 ∠[AB, XY] + ∠[XY, UV] = ∠[AB, UV]. (ndg condition: X , Y) K7 AB ⊥BC if and only if AC is the diameter of the circumcircle of triangle ABC. (ndg condition: S ABC , 0) K8 AB is the perpendicular bisector of XY if and only if AX = AY and BX = BY. (ndg condition: A , B and X , Y) K9 If point O is the circumcenter of triangle ABC then ∠[OA, AB] = ∠ + ∠[AC, BC] (ndg condition: S ABC , 0) 162 Chapter 3. Machine Proof in Plane Geometry Suppose that for a geometry statement, the GIB has been generated by using GKB. The next step is how to generate a proof based on the GIB. The key idea is still to eliminate points introduced by constructions of the given statement. But here the eliminating rules are based mainly on the GIB rather than construction. Following are the rules for eliminating point X from the full-angle ∠[AB, PX]: QE1 If X is on line PQ, then ∠[AB, PX] = ∠[AB, PQ]. QE2 If PX is parallel to UV, then ∠[AB, PX] = ∠[AB, UV]. QE3 If PX is perpendicular to UV, then ∠[AB, PX] = ∠ + ∠[AB, UV]. QE4 If X is on line UV and U, P, Q and X are cyclic, then ∠[AB, PX] = ∠[AB, UV] + ∠[UQ, PQ]. (because ∠[AB, PX] = ∠[AB, UV]+∠[UV, PX], ∠[UV, PX] = ∠[UX, PX] = ∠[UQ, PQ].) QE5 If X is on line UV and PX = PU then ∠[AB, PX] = ∠[AB, UV]+∠[PU, UV]. (because ∠[PU, UV] = ∠[PU, UX] = ∠[UX, PX] = ∠[UV, PX].) QE6 If X is on line UV and PU is the perpendicular bisector of QX, then ∠[AB, PX] = ∠[AB, UV] + ∠[PQ, QU]. (because ∠[PQ, QU] = ∠[UX, PX] = ∠[UV, PX].) QE7 If X is the circumcenter of triangle PQU, then ∠[AB, PX] = ∠[AB, PQ]+∠[UQ, UP]+ ∠. (by K9, ∠[PQ, PX] = ∠ + ∠[UQ, UP]) QE8 If ∠[UV, PX] = ∠[f ] is known, then ∠[AB, PX] = ∠[AB, UV] + ∠[f ]. Here we assume that points A, B, P, Q, U and V are introduced before point X. Exercise 3.104 Prove propositions QE1-QE8 based on Q1-Q12. 3.8.2 Machine Proof Based on the Geometry Information Base We will use the following new version of Example 1.118 to illustrate how the program works. Example 3.105 The circumcenter of triangle ABC is O. AD is the altitude on side BC. Show that ∠[AO, DA] = ∠[BA, BC] −∠[BC, CA]. The constructive description is A B N C O D E M Figure 3-38 ((POINTS B C A) 3.8 Geometry Information Bases 163 (FOOT D A B C) (MIDPOINT M B C) (MIDPOINT N A B) (INTER E (LINE A B) (PLINE M A D)) (INTER O (LINE M E) (TLINE N N B))) The conclusion is ∠[AO, DA] + ∠[BC, CA] + ∠[BC, BA] = ∠ which is equivalent to ∠[AO, DA] = ∠[BA, BC] −∠[BC, CA]. The GIB for this example contains several groups of geometry relations which will be explained separately below. (I1) p-list: includes all the points in the statement, listed in the introducing order, i.e., p-list = (B C A D M N E O). (I2) free-points: includes all the free points in the statement, i.e., free-points= (B C A) (I3) all-lines: GIB can list all2 the lines in the statement. ((O N) (O E M) (E N A B) (D A) (M D B C)) which means that points O, E and M are collinear, etc. (I4) p-lines: contains all the parallel lines: ((O E M) (D A)) which means that OEM ∥DA. (I5) t-lines: contains all the perpendicular lines: ((O N) (E N A B)) (((O E M) (D A)) (M D B C)) which means that line ON is perpendicular to line ENAB; and lines OEM and DA are both perpendicular to line MDBC. (I6) circles: contains all the circles in the statement; ((E O) N E O) ((A O) N A O) ((B O) N M B O) ((D O) M D O) ((C O) M C O) ((D E) M D E) ((B E) M B E) ((C E) M C E) ((A M) D A M) ((B A) D B A (N)) ((C A) D C A) ((B C) B C (M)) (B C (E)) (C B A (O)) which means EO is a diameter of the circumcircle of triangle NEO; ...; BA is a diameter and N is the center of circumcircle of triangle DBA; 2As we mentioned before, these lines are those which can be obtained directly from the hypotheses of the geometry statement. 164 Chapter 3. Machine Proof in Plane Geometry BC is a diameter and M is the center of the circle through points B and C; E is the center of the circle through points B and C; and O is the circumcenter of triangle CBA. Using this information, QAP gives the following proof. ∠[BC, BA] + ∠[BC, CA] + ∠[AO, AD] = ∠[BC, BA] + ∠[BC, CA] + ∠ + ∠[BA, BC] + ∠[CA, AD] (By Q7, ∠[AO, AD] = ∠[AO, CA] + ∠[CA, AD]. Since O is the circumcenter of ABC, ∠[AO, CA] = ∠ + ∠[BA, BC].) = ∠ + ∠[BC, CA] + ∠ + ∠[CA, BC] + ∠ (∠[BC, BA] + ∠[BA, BC] = ∠[BC, BC] = 0. Since AD ⊥BC, ∠[CA, AD] = ∠[CA, BC] + ∠[BC, AD] = ∠[CA, BC] + ∠.) = ∠. (∠ + ∠ = ∠ and ∠[BC, CA] + ∠[CA, BC] = ∠.) In what follows, we will use more examples to illustrate the GIB-GKB method. The examples in Section 1.10 were all produced according to the above method by our program. Example 3.106 (Simson’s Theorem) The same as Example 3.79 on page 144. A B C O D E F G Figure 3-39 Constructive description ((CIRCLE A B C D) (FOOT E D B C) (FOOT F D A C) (FOOT G D A B) The conclusion: ∠[EF, FG] = ∠. Proof. The machine proof ∠[EF, GF] = ∠[EF, DF] + ∠[DF, GF] (Q7) = ∠[EC, DC] + ∠[DA, GA] (Q10 (D, C, E, F; A, D, G, F cyclic.)) = ∠[BC, DC] + ∠[DA, BA] (Q8 (E ∈BC; G ∈AB)). = ∠[BA, DA] + ∠[DA, BA] (Q10 (A, B, C, D cyclic.)) = ∠[BA, BA] = ∠. There is a traditional proof which proves the theorem by showing that ∠EFC = ∠GFA. This proof is not strict: the fact that points E and G are in diﬀerent sides of AC is used but not proved. Before presenting the next example, we introduce a new geometry object: (CIRC A B C) which stands for the circle passing through points A, B, and C. Example 3.107 (Miquel Point) Four lines form four triangles. Show that the circumcircles of the four triangles passes through a common point. 3.8 Geometry Information Bases 165 A B E Q D C P Figure 3-40 Constructive description ( (POINT A D E Q) (INTER B (LINE D E) (CIRC A Q E)) (INTER C (LINE A E) (CIRC D Q E)) (INTER P (LINE A B) (LINE C D)) ∠[QC,CP]+∠[AP,AQ]=∠ The machine proof ∠[QC, CP] + ∠[AP, AQ] = ∠[QC, DC] + ∠[AB, AQ] (because CP ∥DC and AP ∥AB.) =∠[EQ, ED] + ∠[EB, EQ] = ∠[EB, ED] = ∠. (because E, Q, D, C and A, B, E, Q are cyclic points respectively.) Example 3.108 In a circle, the lines joining the midpoints of two arcs AB and AC meet line AB and AC at D and E. Show that AD = AE. A B C O Q P M N D E Figure 3-41 Constructive description ( (POINTS A M N) (CIRCUMCENTER O A M N) (FOOT P A O N) (FOOT Q A O M) (INTER D (LINE N M) (LINE A Q)) (INTER E (LINE N M) (LINE A P)) ∠[AD,DE]+∠[AE,ED]=∠ The machine proof ∠[AE, DE] + ∠[AD, DE] = ∠[AP, MN] + ∠[AQ, MN] (because AE ∥AP, DE ∥MN, and AD ∥AQ.) = ∠[AP, MN] + ∠ + ∠[MO, MN] (because AQ⊥MO.) = ∠ + ∠[NO, MN] + ∠ + ∠ + ∠[AM, AN] (AP⊥NO. ∠[MO, MN] = ∠ + ∠[AM, AN], because O is the circumcenter of △AMN) = ∠ + ∠ + ∠[AN, AM] + ∠[AM, AN] (since O is the circumcenter of △AMN, ∠[NO, MN] = ∠ + ∠[AN, AM].) = 0 166 Chapter 3. Machine Proof in Plane Geometry Example 3.109 From the midpoint C of arc AB of a circle, two secants are drawn meeting line AB at F, G, and the circle at D and E. Show that F, D, E, and G are on the same circle. C O A M B D E F G Figure 3-42 Constructive description ( (CIRCLE A C D E) (CIRCUMCENTER O A C D) (FOOT M A O C) (INTER F (LINE A M) (LINE D C)) (INTER G (LINE A M) (LINE C E)) ∠[CE,FG]+∠[CD,DE]=∠ The Machine proof ∠[CE, FG] + ∠[CD, DE] = ∠[CE, AM] + ∠[AC, AE] (FG ∥AM; since A, C, D, and E are cyclic, ∠[CD, DE] = ∠[AC, AE].) = ∠ + ∠[CE, CO] + ∠[AC, AE] (since AM⊥CO, ∠[CE, AM] = ∠ + ∠[CE, CO].) = ∠ + ∠ + ∠[AE, AC] + ∠[AC, AE] (since A, C, D, and E are on the circle with center O, ∠[CE, CO] = ∠+∠[AE, AC].) = 0 Example 3.110 Let Q, S and Y be three collinear points and (O, P) be a circle. Circles S PQ and YPQ meet circle (O, P) again at points R and X, respectively. Show that XY and RS meet on the circle (O, P). P S O Q R Y X I Figure 3-43 Constructive description ((CIRCLE R P Q S ) (POINT X) (INTER Y (LINE Q S )(CIRC P Q X)) (INTER I (LINE X Y) (LINE R P)) ∠[XI,RI]+∠[RP,XP]=∠ The machine proof ∠[PR, PX] + ∠[IX, RI] = ∠[PR, PX] + ∠[XY, RS ] (because XI ∥XY and RI ∥RS .) = ∠[PR, PX] + ∠[XY, QS ] + ∠[QS, RS ] = ∠[PR, PX] + ∠[XP, QP] + ∠[QS, RS ] (because X, P, Y, and Q are cyclic, and QS ∥QY.) (because X, P, Y, and Q are cyclic, and IX ∥YI.) 3.8 Geometry Information Bases 167 = ∠[PR, PQ] + ∠[QS, RS ] = ∠. (Points R, Q, P, and S are cyclic.) Example 3.111 Let ABC be a triangle. Show that the six feet obtained by drawing perpen-diculars through the foot of each altitude upon the other two sides are co-circle. A B C F E D G H I K Figure 3-44 Constructive description ( (POINTS A B C) (FOOT F C A B) (FOOT D A B C) (FOOT E B A C) (FOOT G F B C) (FOOT I D A B) (FOOT H F A C) (FOOT K E A B) ∠[GH,GI]+∠[AK,HK]=∠ The machine proof ∠[GH, GI] + ∠[AK, HK] = ∠[GH, GI] + ∠[AB, FI] + ∠[FE, EH] (because AK ∥AB, and K, H, F, and E are cyclic.) = ∠[FE, EH] + ∠[GH, GI] (because AB ∥FI.) = ∠[FE, AC] + ∠[CE, GI] + ∠[FG, CF] (because EH ∥AC, and H, G, C, and F are cyclic.) = ∠[FE, AC] + ∠[CE, BF] + ∠[FD, DG] + ∠[FG, CF] (because I, G, F, and D are cyclic.) = ∠[FG, CF] + ∠[FE, AC] (because A, D, C, and F are cyclic.) = ∠[AD, CF] + ∠[FE, AC] (because FG ∥AD.) = ∠[AD, CF] + ∠[BF, BC] (because E, F, C, and B are cyclic.) = ∠ + ∠[BC, CF] + ∠[BF, BC] (because AD⊥BC.) = ∠ + ∠[BF, CF] = ∠ (because BF⊥CF.) Example 3.112 The nine-point circle cuts the sides of the triangle at angles \|B −C\|, \|C −A\|, and \|A −B\|. 168 Chapter 3. Machine Proof in Plane Geometry A B C M Q P S F N L Figure 3-45 Constructive description ((POINTS A B C) (FOOT F C A B) (MIDPOINT M B C) (MIDPOINT Q A C) (MIDPOINT P B A) (MIDPOINT L F P) (MIDPOINT S Q P) (INTER N (TLINE L L P) (TLINE S S P)) ∠[BC,AB]+∠[AC,AB]+∠[FN,LN]=∠ The machine proof ∠[FN, LN] + ∠[AC, AB] + ∠[BC, AB] = ∠ + ∠[FN, PF] + ∠[AC, AB] + ∠[BC, AB] (because LN ∥CF and CF⊥PF.) = ∠ + ∠ + ∠[FQ, PQ] + ∠[AC, AB] + ∠[BC, AB] (Points F, Q, P are on the circle with center N.) = ∠[FQ, CF] + ∠[CF, PQ] + ∠[AC, AB] + ∠[BC, AB] =∠ + ∠[FP, QP] + ∠[FQ, CF] + ∠[AC, AB] + ∠[BC, AB] (because LN ∥CF and CF⊥PF.) =∠ + ∠[MQ, BC] + ∠[FQ, CF] + ∠[AC, AB] + ∠[BC, AB] =∠ + ∠[AC, AB] + ∠[FQ, CF] (because FP ∥MQ,QP ∥BC, and MQ ∥AB.) = ∠ + ∠[AC, AB] + ∠ + ∠[CF, AC] + ∠[AF, CF] (because F, A, C are on the circle with center Q.) = ∠[AC, AB] + ∠[AF, AC] = ∠[AC, AB] + ∠[AB, AC] = ∠ Summary of Chapter 3 • We have the following formulas for the areas of triangles. 1. S ABC = 1 2\|BC\|hA = 1 2\|AC\|hB = 1 2\|AB\|hC. 2. S ABC = S OUV xA yA 1 xB yB 1 xC yC 1 where xA, yA, xB, yB, xC, and yC are the area coordinates of points A, B, and C with respect to OUV. 3. 16S 2 ABC = 4AB 2AC 2 −(AC 2 + AB 2 −BC 2)2 = 4AB 2AC 2 −P2 BAC. 3.8 Geometry Information Bases 169 • The following basic propositions and the ones on page 100 are the basis of the area method. 1. AB⊥CD iﬀPACD = PBCD or PACBD = 0. 2. Let R be a point on line PQ with position ratios r1 = PR PQ, r2 = RQ PQ with respect to PQ. Then PRAB = r1PQAB + r2PPAB PARB = r1PAQB + r2PAPB −r1r2PPQP. 3. Let D be the foot of the perpendicular drawn from point P upon a line AB. Then we have AD AB = PPAB 2AB 2, DB AB = PPBA 2AB 2. 4. Let AB and PQ be two nonperpendicular lines and Y be the intersection of line PQ and the line passing through A and perpendicular to AB. Then PY QY = PPAB PQAB , PY PQ = PPAB PPAQB , QY PQ = PQAB PPAQB . 5. Let ABCD be a parallelogram. Then for any points P and Q, we have PAPQ + PCPQ = PBPQ + PDPQ or PAPBQ = PDPCQ PPAQ + PPCQ = PPBQ + PPDQ + 2PBAD. 6. Let ABCD be a parallelogram and P be any point. Then PPAB = PPDC −PADC = PPDAC PAPB = PAPA −PPDAC. • A constructive conﬁguration is a ﬁgure which can be drawn using a ruler and a pair of compasses only. In other words, a constructive conﬁguration can be obtained by ﬁrst taking some arbitrary points, lines, and circles in the plane, and then taking the intersections of these lines and circles in a prescribed way. A constructive geometry statement is an assertion about a constructive conﬁguration, and this assertion can be represented by a polynomial equation of three geometry quantities: the ratios of par-allel line segments, the signed areas of triangles or quadrilaterals, and the Pythagoras diﬀerences. The set of all constructive statements is denoted by C. We also intro-duced a subclass of C, i.e., the class of the linear constructive geometry statements, which is denoted by CL. • A mechanical theorem proving method for class C was presented. The key idea of the method is to eliminate points from geometry quantities. The method can be used to produce short and readable proofs for geometry statements eﬃciently. 170 Chapter 3. Machine Proof in Plane Geometry • We report some initial results on how to use the geometry information base to gener-ate readable proofs for geometry statements using full-angles. Chapter 4 Machine Proof in Solid Geometry This chapter deals with the machine proof in solid geometry. Similar to plane geometry, we will consider those geometry statements that can be described constructively using lines, planes, circles, and spheres. In the ﬁrst three sections, we deal with geometry statements in-volving collinear and parallel of lines and planes. More precisely, we deal with constructive statements in aﬃne geometry of dimension three. Starting from Section 4, the Pythagoras diﬀerence is employed to deal with constructive statements involving perpendicular lines, circles, and spheres. 4.1 The Signed Volume As before, we denote by AB the signed length of the oriented segment from A to B; we denote by S ABC the signed area of the oriented triangle ABC. In solid geometry, we have a new basic fourfold relation among points, coplanar, which will be characterized by Axioms S.1–S.5 about signed volumes. We assume that Axioms A.1–A.6 (on page 55) are still valid, provided that all the points involved are coplanar. The signed areas of coplanar triangles can be compared, added, or subtracted. For instance, if A, O, U, and V are four coplanar points, by Axiom A.5 (4.1) S OUV = S OUA + S OAV + S AUV. A tetrahedron ABCD has two possible orientations. We use the order of its vertices to represent its orientation. If we interchange two neighbor vertices, the orientation of the tetrahedron will be changed. The signed volume VABCD of an oriented tetrahedron ABCD is a real1 number which satisﬁes the following properties. 1Here, we can use any number ﬁeld and the results in this chapter are still valid. 171 172 Chapter 4. Machine Proof in Solid Geometry Axiom S.1 When two neighbor vertices of an oriented tetrahedron are interchanged, the signed volume of the tetrahedron will change signs, e.g., VABCD = −VABDC. Axiom S.2 If A, B, C, and D are four non-coplanar points, we have VABCD , 0. Axiom S.3 There exist at least four points A, B, C, and D such that VABCD , 0. Axiom S.4 For ﬁve points A, B, C, D, and O (Figure 4-1), we have VABCD = VABCO + VABOD + VAOCD + VOBCD. Axioms S.3 and S.4 are called dimension axioms. They ensure that we are dealing with a proper three dimensional space: Axiom S.3 says that there are at least four non-coplanar points; Axiom S.4 says that all points must be in the same three dimensional space. A B C D T Figure 4-2 A B C D O Figure 4-1 Axiom S.5 If A, B, C, and D are four coplanar triangles and S ABC = λS ABD then for any point T we have VT ABC = λVT ABD. (Figure 4-2) We extend the coplanar to be a geometry relation among any set of points: a set con-taining fewer than four points is always coplanar, and a set of points is coplanar if any four points in it are coplanar. We thus can introduce a new geometry object, the plane, which is a maximal set of coplanar points. Proposition 4.1 Four points A, B, C, and D are coplanar iﬀVABCD = 0. Proof. If VABCD = 0, by Axiom S.2 A, B, C, and D are coplanar. Let us assume A, B, C, and D to be coplanar points. If A, B, C, and D are collinear, we have S ABC = 0. Let X be a point not on line AB. Then by Axiom S.5, VABCD = S ABC S ABX VABXD = 0. If A, B, and C are not collinear, we have VABCD = S BCD S ABC VAABC = 0. Corollary 4.2 For three non-collinear points A, B, and C, the set of all the points D satisfying VABCD = 0 is a plane and is denoted by plane ABC. Proof. Let P, Q, R, and S be four points in plane ABC. We need to show that VPQRS = 0. We ﬁrst show that two of the points, say P and Q, are coplanar with A, B, C. By Axiom 4.1 The Signed Volume 173 S.5, VABPQ = S ABP S ABC VABCQ = 0, i.e., A, B, P, and Q are coplanar. Similarly, we can show that three of P, Q, R, and S are coplanar with A, B, C, and ﬁnally P, Q, R, and S are coplanar. In what follows, when speaking about a plane ABC, we always assume that A, B, and C are not collinear. Similarly, when speaking about a line AB, we assume A , B. 4.1.1 Co-face Theorem In this and the next subsections, we will derive some basic properties about volumes which will serve as the basis of the volume method. First, Axiom S4 can be written in the follow-ing convenient way. Proposition 4.3 (The Co-vertex Theorem) Let ABC and DEF be two proper triangles in the same plane and T be a point not in the plane. Then we have VTABC VTDEF = S ABC S DEF . Proof. By Axiom S.5, VT ABC VT DEF = VT ABC VT ABF VT ABF VT AEF VT AEF VT DEF = S ABC S ABF S ABF S AEF S AEF S DEF = S ABC S DEF . Before proving the co-face theorem, we need to deﬁne the signed volume of a special polyhedron with ﬁve vertices. P A B C Q P A B C Q P A B C Q Figure 4-3 The polyhedron formed by ﬁve points in space is complicated. By PABCQ, we denote the one with faces PAB, PBC, PAC, QAB, QBC, and QAC. Figure 4-3 shows that several possible shapes of PABCQ. The volume of PABCQ is deﬁned to be (4.2) VPABCQ = VPABC −VQABC. By Axiom S4, we have (4.3) VPABCQ = VPABQ + VPCAQ + VPBCQ. 174 Chapter 4. Machine Proof in Solid Geometry Proposition 4.4 (The Co-face Theorem) A line PQ and a plane ABC meet at M. If Q , M, we have PM QM = VPABC VQABC ; PM PQ = VPABC VPABCQ ; QM PQ = VQABC VPABCQ . P A B C Q M P A B C Q M P A B C Q M P A B C Q M Figure 4-4 Proof. Figure 4-4 shows that several possible conﬁgurations of this proposition. Take points A′ and B′ such that MA′ = CA, MB′ = CB. Then S ABC = S A′B′M. By Propositions 4.3 and 2.8, we have, VPABC VQABC = VPA′B′M VQA′B′M = S PB′M S QB′M = PM QM. Other equations are consequences of the ﬁrst one. The above proof is a dimension reduction process. A quantity in the space ( VPA′B′M VQA′B′M ) is reduced to a quantity in the plane ( S PB′M S QB′M ) which is further reduced to a quantity in a line ( PM QM). A B C P Q R Figure 4-5 Proposition 4.5 Let R be a point on a line PQ and ABC be a triangle. Then we have VRABC = PR PQ VQABC + RQ PQ VPABC. Proof. By Proposition 4.4, VPRBC VPQBC = PR PQ , VPARC VPAQC = PR PQ , VPABR VPABQ = PR PQ . By Axiom S4, VRABC = VPABC −VPRBC −VPARC −VPABR = VPABC −PR PQ (VPQBC + VPAQC + VPABQ) 4.1 The Signed Volume 175 = VPABC −PR PQ VPABCQ = (1 −PR PQ )VPABC + PR PQ VQABC = PR PQ VQABC + RQ PQ VPABC. A B C P Q R S M Figure 4-6 Proposition 4.6 Let R be a point in the plane PQS . Then for three points A, B, and C we have VRABC = S PQR S PQS VS ABC + S RQS S PQS VPABC + S PRS S PQS VQABC. Proof. For any point X, let VX = VXABC. Without loss of generality, let M be the intersection of PR and QS . By Proposition 4.5, VR = PR PM VM + RM PM VP = PR PM (QM QS VS + MS QS VQ) + RM PM VP. (1) By the co-side theorem, RM PM = S RQS S PQS , QM QS = S PQR S PQRS , MS QS = S PRS S PQRS , PR PM = S PQRS S PQS . Substituting these into (1), we obtain the result. 4.1.2 Volumes and Parallels Two planes or a line and a plane, are said to be parallel if they have no point in common. Two lines are said to be parallel if they are in the same plane and do not have a common point. By the notation PQ ∥ABC, we mean that A, B, C, P, and Q satisfy one of the following conditions: (1) P = Q, (2) A, B, and C are collinear, or (3) A, B, C, P, and Q are on the same plane, or (4) line PQ and plane ABC are parallel. According to the above deﬁnition, if PQ ̸∥ABC then line PQ and plane ABC have a normal intersection. For six points A, B, C, P, Q, and R, ABC ∥PQR iﬀAB ∥PQR, BC ∥PQR, and AC ∥PQR. Proposition 4.7 PQ ∥ABC iﬀVPABC = VQABC or equivalently VPABCQ = 0. Proof. If VPABC , VQABC, then P , Q and A, B, and C are not collinear. Let O be a point on line PQ such that PO PQ = VPABC VPABCQ . Thus OQ PQ = −VQABC VPABCQ . By Proposition 4.5, VOABC = PO PQVQABC + OQ PQVPABC = 0. By Axiom S2, point O is also in plane ABC, i.e., line PQ is not 176 Chapter 4. Machine Proof in Solid Geometry parallel to ABC. Conversely, if PQ ∦ABC then P , Q and A, B, and C are not collinear. Let O be the intersection of PQ and ABC. By Proposition 4.4, OP OQ = VPABC VQABC = 1. Thus P = Q, which is a contradiction. Proposition 4.8 PQR ∥ABC iﬀVPABC = VQABC = VRABC. Proof. By Proposition 4.7, VPABC = VQABC = VRABC iﬀlines PQ and PR are parallel to plane ABC. We must show that for any point D in plane PQR, line PD is also parallel to ABC. By Proposition 4.6, VDABC = S PQD S PQR VRABC + S PDR S PQR VQABC + S DQR S PQR VPABC = VPABC(S PQD S PQR + S PDR S PQR + S DQR S PQR ) = VPABC i.e., PD ∥ABC. A ﬁgure P1P2...Pn is said to be a translation of Q1Q2...Qn if PiPi+1 = QiQi+1. Let triangle XYZ be a translation of triangle ABC. Then for any points P, Q, and R in plane XYZ, we deﬁne S PQR S ABC = S PQR S XYZ . For convenience, we use the symbol S PQR S ABC = λ or S PQR = λS ABC to denote the fact that plane PQR is the same as or parallel to plane ABC, and λ is the ratio of the signed areas S PQR and S ABC. The following propositions about translations of line segments and triangles are often used in the machine proof method, in order to add auxiliary translations of line segments and triangles. Proposition 4.9 Let PQTS be a parallelogram. Then for points A, B, and C, we have VPABC + VT ABC = VQABC + VS ABC or VPABCQ = VS ABCT. Proof. This is a consequence of Proposition 4.5, because both sides of the equation are equal to 2VOABC where O is the intersection of PT and S Q. Proposition 4.10 Let triangle ABC be a translation of triangle DEF. Then for any point P we have VPABC = VPDEFA. 4.1 The Signed Volume 177 Proof. By Proposition 4.9 and (4.3), VPABC = VPAEC −VPADC = VPAEF −VPAED −VPADC = VPAEF −VPAED −VPADF = VPDEFA. Corollary 4.11 1. For two parallel planes ABC and PQR and a point T not in ABC we have S ABC S PQR = VTABC VTPQRA. 2. For two diﬀerent parallel planes ABC and PQR we have S ABC S PQR = −VPABC VAPQR . Proof. Let XYZ be a translation of RPQ to plane ABC. By the co-vertex theorem and the preceding proposition, S ABC S PQR = S ABC S XYZ = VT ABC VT XYZ = VT ABC VT PQRA . Replacing T by P in the above equation, we prove the second result. Proposition 4.12 Let triangle ABC be a translation of triangle DEF. Then for two points P and Q we have VPABC + VQDEF = VQABC + VPDEF or VPABCQ = VPDEFQ. In other words, when ABC moves by a translation the volume of PABCQ remains the same. Proof. By Proposition 4.10, VPABC = VPDEF −VADEF; VQABC = VQDEF −VADEF from which we obtain the result immediately. From Propositions 4.9 and 4.12, we have the following interesting property for the vol-ume VPABCQ. Corollary 4.13 Let PQ = rS T and S ABC = sS EFG then VPABCQ = rsVS EFGT. 4.1.3 Volumes and Aﬃne Geometry of Dimension Three This subsection has two purposes. First, we will show how to prove geometry theorems using the basic propositions about volumes. Second, we will derive some basic properties of lines and planes in space using the volume method. Example 4.14 If points P and Q are in plane ABC then line PQ is also in plane ABC. Proof. For any point R on line PQ, by Proposition 4.5, VRABC = PR PQVQABC + RQ PQVPABC = 0. By Proposition 4.1, R is in ABC. Example 4.15 If two planes have a point in common then they have a line in common. 178 Chapter 4. Machine Proof in Solid Geometry Proof. Let planes ABC and RPQ have a point X in common. Without loss of generality, we assume that A, B, C, R, P, and Q are not common to both planes. By Proposition 4.8, two of AB, BC, and AC, say AB and BC could not be parallel to RPQ. Let AB and AC meet RPQ in two distinct points Y and Z respectively. We must show that X, Y, and Z are collinear, i.e., S XYZ = 0. By Propositions 4.3 and 4.1, S XYZ S ABC = VRXYZ VRABC = 0, i.e., S XYZ = 0. Remark 4.16 The above two examples and Corollary 4.2 are the incidence axioms in the usual axiom system for solid geometry . Therefore, the geometry deﬁned by Axioms A.1-A.6 and S.1-S.5 is an aﬃne geometry of dimension three. The results related to aﬃne plane geometry proved in Section 2.6 are also true for aﬃne geometry of dimension three. So the volume method presented in Section 4.3 below is for constructive statements in aﬃne geometry of dimension three associated with any ﬁeld. We will now prove some basic properties for the parallel. Example 4.17 Through any point P not in plane ABC there is one and only one plane par-allel to plane ABC. Proof. By the Euclidean parallel axiom (Example 2.13 on page 58), there exist points Q and R such that PABQ and PACR are parallelograms. By Propositions 4.7 and 4.8, PQR ∥ABC. To prove the uniqueness, let PTS be another plane parallel to ABC. By Proposition 4.10, VT PQR = VT ABC −VPABC = 0, i.e., T ∈PAR. Similarly we have S ∈PAR. Example 4.18 If any two lines are cut by a number of parallel planes, their intercepts are proportional. A C P R B Q Figure 4-7 Proof. Let the parallel planes α, β, γ cut two lines in the sets of points A, B, C and P, Q, R respectively. Let X, Y, and Z be three noncollinear points in plane β. By the co-face theorem, AB CB = VAXYZ VCXYZ = VPXYZ VRXYZ = PQ RQ . Example 4.19 If a straight line is parallel to a straight line in a plane, it is in or parallel to the plane. Conversely, if a line in one plane is parallel to another plane, it is parallel to their line of intersection. 4.1 The Signed Volume 179 C D A E B F Figure 4-8 Proof. Let AB be parallel to CD and E another point in plane CDE. By Propositions 4.3 and 2.10, VEACD VEBCD = S ACD S BCD = 1, i.e., AB ∥CDE. Conversely, let AB be parallel to plane CDE and CD be the intersection of the two planes. By Proposition 4.3, S ACD S BCD = VEACD VEBCD = 1, i.e., AB ∥CD. Example 4.20 If two lines are each parallel to a third, they are parallel to one another. A B C E D F Figure 4-9 Proof. Let ABCD and ABEF be two parallelograms. We ﬁrst prove that C, D, E, and F are coplanar. By Proposition 4.9, VCDEF = VBDEF −VADEF = VBDAF − VADBF = 0, i.e., C, D, E, and F are coplanar. By Proposition 4.3, S CEF S DEF = VACEF VADEF = 1, i.e., CD ∥EF. Example 4.21 If a plane cuts two parallel planes, the lines of intersection are parallel. A B P C R Q D S Figure 4-10 Proof. Let plane ABPQ cut planes ABC and RPQ at lines AB and PQ respectively. By Proposition 4.3, S APQ S BPQ = VAPQR VBPQR = 1, i.e., AB ∥PQ. Example 4.22 (Co-trihedral Theorem) If OW ∥DA, OU ∥DB, and OV ∥DC then VOWUV VDABC = OW DA · OU DB · OV OC. Proof. Let R, P, Q be points such that DR = OW, DP = OU, DQ = OV. By the co-face theorem, VDABC = DA DR VDRBC = DA OW VDRBC = DA OW · DB OU · DC OV VDRPQ. By Propositions 4.9 and 4.10, VDRPQ = VORPQ −VWRPQ = VOWUV −VRWUV −VWRPQ. 180 Chapter 4. Machine Proof in Solid Geometry Since RW ∥PU ∥QV, VRWUV = VRWPV = VRWPQ which proves the result. The above examples are about the basic properties of lines and planes. In what fol-lows, we will prove some relatively non-trivial theorems. The proofs of these theorems are actually modiﬁcations of the proofs produced by our program. X Y Z A B C D H G E F Figure 4-11 A B C A B C Figure 4-12 1 1 1 A B C D H F E G Figure 4-13 Example 4.23 (Menelaus’ Theorem for Skew Quadrilaterals) If the sides AB, BC, CD, and DA of any skew quadrilateral are cut by a plane XYZ in the points E, F, G, and H respectively, then AE EB · BF FC · CG GD · DH HA = 1. (Figure 4-11) Proof. By the co-face theorem DH AH = VDXYZ VAXYZ , CG DG = VCXYZ VDXYZ , BF CF = VBXYZ VCXYZ , AE BE = VAXYZ VBXYZ . Then it is clear that AE EB · BF FC · CG GD · DH HA = 1. For the non-degenerate conditions of this example, see Section 4.2 Example 4.24 Let A1B1C1 be the parallel projection of any triangle ABC in any plane. Show that the tetrahedra ABCA1 and A1B1C1A are equal in volume. (Figure 4-12) Proof. Since CC1 is parallel to plane AA1B1, by Proposition 4.7, VAA1B1C1 = VAA1B1C. Simi-larly, VAA1B1C = VAA1BC. Example 4.25 If a plane divides proportionally one pair of opposite sides of a skew quadri-lateral, it also divides proportionally the other two sides. (Figure 4-13) Proof. This statement is a direct consequence of Example 4.23. The following is a proof produced by our program.2 Let r1 = AE EB = DF FC and r2 = AH AD. We need to show that BG BC = r2. By the co-face theorem, BG BC = VBEFH VBEFH−VCEFH . By Proposition 4.5 2This may serve as an example to show the diﬀerence between people and the computer in proving theorems. People may use all results available to make the proof short. But the computer does only prescribed steps according to the input. 4.2 Constructive Geometry Statements 181 VCEFH = (r2 −1)VACEF = (r2 −1)(r1 −1)VACDE = (r2 −1)(r1 −1)r1VABCD; VBEFH = r2VBDEF = r2r1VBCDE = r2r1(r1 −1)VABCD. Then BG BC = r2r1(r1−1) r2r1(r1−1)−r1(r1−1)(r2−1) = r2. 4.2 Constructive Geometry Statements In this section, we will introduce a class of constructive geometry statements which is a generalization of the constructive statements in plane geometry. 4.2.1 Constructive Geometry Statements In this chapter, by a geometric quantity we mean one of the following quantities: 1. the ratio of the signed lengths of two oriented segments on one line or on two parallel lines; 2. the ratio of the signed areas of two oriented triangles in the same plane or in two parallel planes; or 3. the signed volume of an oriented tetrahedron. In Section 4.4, we will introduce more geometry quantities. We now introduce constructions in space. First, it is clear that most plane constructions can still be used if all related points are in the same plane. We will choose several of them as basic constructions in space. Deﬁnition 4.26 A construction is one of the following ways of introducing new points in space. S1 (POINTS Y1, · · · , Yl). Take arbitrary points Y1, · · · , Yl in the space. Each Yi has three degrees of freedom. S2 (PRATIO Y W U V r). Take a point Y on the line passing through W and parallel to line UV such that WY = rUV, where r can be a rational number, a rational expression in geometric quantities, or a variable. If r is a ﬁxed quantity, Y is a ﬁxed point; if r is a variable, Y has one degree of freedom. The non-degenerate (ndg) condition is U , V. If r is a rational expression of geometry quantities then we will further assume that the denominator of r could not equal to zero. 182 Chapter 4. Machine Proof in Solid Geometry S3 (ARATIO Y L M N r1 r2 r3), where r1 = S YMN S LMN , r2 = S LYN S LMN , and r3 = S LMY S LMN are the area coordinates of point Y with respect to LMN. The r1, r2 and r3 could be rational numbers, rational expressions in geometric quantities, or indeterminates satisfying r1 + r2 + r3 = 1. The degree of freedom of Y is equal to the number of indeterminates in {r1, r2, r3}. The ndg conditions are that L, M, and N are not collinear and the denominators of r1, r2, and r3 are not equal to zero. S4 (INTER Y (LINE U V) (LINE P Q)). Point Y is the intersection of line PQ and line UV which are in the same plane. The ndg condition is PQ ∦UV. Point Y is a ﬁxed point. S5 (INTER Y (LINE U V) (PLANE L M N)). Point Y is the intersection of a line UV and a plane LMN. The ndg condition is that UV ∦LMN. Point Y is a ﬁxed point. S6 (FOOT2LINE Y P U V) Point Y is the foot from point P to line UV. The ndg condition is U , V. Point Y is a ﬁxed point. Proposition 4.27 Let Y be introduced by one of the six constructions S1-S6. Show that the existence of Y follows from Axiom A.2. Proof. Constructions S1, S2, S4, and S6 have been discussed in Proposition 3.20 on page 113. The existence of the point introduced by S3 follows from Proposition 2.30 on page 68. Let Y be introduced by S5. By the co-face theorem UY UV = VULMN VULMNV . Since UV ∦LMN, we have VULMNV , 0. By Axiom A.2, Y does exist. Deﬁnition 4.28 A constructive statement is a list S = (C1, C2, . . . , Ck, G) where 1. Each Ci, introduces a new point from the points introduced by the previous construc-tions; and 2. G = (E1, E2) where E1 and E2 are polynomials in geometric quantities about the points introduced by the Ci, and E1 = E2 is the conclusion of S . The non-degenerate condition of S is the set of non-degenerate conditions of the construc-tions Ci plus the condition that the geometry quantities in E1 and E2 have geometry mean-ings, i.e., their denominators are not zero. If the constructions are limited to S1–S5, the corresponding statements are called Hilbert’s intersection point statements in space. The set of all Hilbert’s intersection point statements is denoted by SH. The constructive description of geometry statements can be transformed into the com-monly used predicate form. Following are several basic predicates. 4.2 Constructive Geometry Statements 183 1. Point (POINT P): P is a point in the space. 2. Collinear (COLL P1 P2 P3): points P1, P2, and P3 are on the same line. 3. Coplanar (COPL P1 P2 P3 P4): P1, P2, P3, and P4 are in the same plane. 4. Parallel between two lines. (PRLL P1 P2 P3 P4): (COPL P1 P2 P3 P4) and P1P2 ∥ P3P4. 5. Parallel between a line and a plane. (PRLP P1 P2 P3 P4 P5): P1P2 ∥P3P4P5. 6. Perpendicular (PERP P1 P2 P3 P4): [(P1 = P2) ∨(P3 = P4) ∨(P1P2 is perpendicular to P3P4)] We will now transform constructions into predicate forms. S2 (PRATIO Y W U V r) is equivalent to (PRLL Y W U V), r = WY UV , and U , V. S3 (ARATIO Y L M N r1 r2 r3) is equivalent to (COPL Y L M N), r1 = S YMN S LMN , r2 = S LYN S LMN , r3 = S LMY S LMN , and ¬(COLL L M N). S4 (INTER Y (LINE U V) (LINE P Q)) is equivalent to (COLL Y U V), (COLL Y P Q), and ¬(PRLL U V P Q). S5 (INTER Y (LINE U V) (PLANE L M N)) is equivalent to (COLL Y U V), (COPL Y L M N), and ¬(PRLP U V L M N). S6 (FOOT2LINE Y P U V) is equivalent to (COLL Y U V), (PERP Y P U V), and U , V. Now a constructive statement S = (C1, · · · , Cr, (E, F)) can be transformed into the follow-ing predicate form ∀P1 · · · ∀Pr((P(C1) ∧· · · ∧P(Cr)) ⇒E = F) where Pi is the point introduced by Ci and P(Ci) is the predicate form of Ci. Example 4.23 can be described in the following constructive way. ((POINTS A B C D X Y Z) (INTER E (LINE A B) (PLANE X Y Z)) (INTER F (LINE B C) (PLANE X Y Z)) (INTER G (LINE C D) (PLANE X Y Z)) (INTER H (LINE A D) (PLANE X Y Z)) ( AE BE BF CF CG DG DH AH = 1)) The ndg conditions: 184 Chapter 4. Machine Proof in Solid Geometry AB ∦XYZ, BC ∦XYZ, CD ∦XYZ, AD ∦XYZ, B , E, C , F, D , G, and A , H. The predicate form of this example is: ∀A, B, · · · , H(HYP ⇒CONC) where HYP = ((COLL E A B) ∧(COPL E X Y Z) ∧¬(PRLP A B X Y Z) ∧ (COLL F C B) ∧(COPL F X Y Z) ∧¬(PRLP C B X Y Z) ∧ (COLL G C D) ∧(COPL G X Y Z) ∧¬(PRLP C D X Y Z) ∧ (COLL H A D) ∧(COPL H X Y Z) ∧¬(PRLP D A X Y Z) ∧ B , E ∧C , F ∧D , G ∧A , H); CONC = (AE BE BF CF CG DG DH AH = 1). 4.2.2 Constructive Conﬁgurations A geometric ﬁgure which can be described by constructions S1–S6 is called a constructive conﬁguration. Constructions S1–S6, though simple, can be used to describe most of the commonly used conﬁgurations about lines, planes, circles, and spheres. To illustrate, we will introduce more geometry objects. • We will consider four kinds of lines: 1. (LINE P Q). 2. (PLINE R P Q). 3. (OLINE S P Q R): the line passing through point S and perpendicular to plane PQR. The ndg condition is ¬ (COLL P Q R). • We will consider six kinds of planes: 1. (PLANE L M N). 2. (PPLANE W L M N): the plane passing through a point W and parallel to plane LMN. The ndg condition is ¬(COPL L M N). 3. (TPLANE W U V): the plane passing trough a point W and perpendicular to line UV. The ndg condition is U , V. 4. (BPLANE U V): the perpendicular-bisector of line UV. The ndg condition is U , V. 4.2 Constructive Geometry Statements 185 5. (CPLANE A B P Q R): the plane passing through line AB and perpendicular to plane PQR. The ndg condition is ¬(AB⊥PQR). 6. (DPLANE A B P Q): the plane passing through line AB and parallel to line PQ. The ndg condition is AB ∦PQ. Now we can consider more constructions: • (ON Y ln). Take an arbitrary point Y on a line ln. Line ln could be one of the three kinds of lines. • (ON Y pl). Take an arbitrary point Y in a plane pl. Plane pl could be one of the 6 kinds of planes. • (INTER Y ln1 ln2). Take the intersection of two lines ln1 and ln2 in the same plane. • (INTER Y ln pl). Take the intersection of line ln and plane pl. • (INTER Y pl1 pl2 pl3). Take the intersection Y of three planes pl1, pl2, and pl3. Combining all the possible cases, there are totally (3 + 6 + 6 + 18 + 56 =) 89 constructions. Conveniently, all these constructions can be described by constructions S1–S6. To show that, we need only to reduce all kinds of lines to the form (LINE P Q) and reduce all kinds of planes to the form (PLANE R P Q). We ﬁrst introduce a construction frequently used. S7 (FOOT2PLANE Y P L M N) Point Y is the foot of the perpendicular from point P to plane LMN, i.e., Y is the intersection of line (OLINE P L M N) and plane (PLANE L M N). The nondegenerate condition is that L, M, and N are not collinear. Example 4.29 Construction S7 can be represented by constructions S1-S6. Point Y can be introduced by the following sequence of constructions. L M N P T F S Y Figure 4-14 (FOOT2LINE T P M N) (FOOT2LINE F L M N) (PRATIO S T F L 1) (FOOT2LINE Y P T S ) Example 4.30 Find two points P and Q in plane (TPLANE W U V) such that W, P, and Q are not collinear. Then plane (TPLANE W U V) is the same as plane (PLANE W P Q). 186 Chapter 4. Machine Proof in Solid Geometry U V P W Q T R Figure 4-15 Proof. If W < UV, let P be introduced by (FOOT2LINE P W U V). Take an arbitrary point R in the space. Then Q can be introduced by con-structions: (FOOT2PLANE T R W U V); (PRATIO Q P T R 1). It is clear that we need a nondegenerate condition R , T, or R is not in plane WUV. If W ∈UV, let R be an arbitrary point. We introduce point P as follows (FOOT2LINE T R U V); (PRATIO P W T R 1). Now point Q can be introduced as in the ﬁrst case. Exercises 4.31 1. For an OLINE ln, ﬁnd two distinct points U and V such that ln =(LINE U V). 2. Let pl be a plane of the form PPLANE, BPLANE, CPLANE, or DPLANE. Find three noncollinear points W, U, and V such that pl =(PLANE W U V). Example 4.32 The following results are clear. 1. Construction (ON Y (LINE U V)) is equivalent to (PRATIO Y U U V r) where r is an indeterminate. 2. Construction (ON Y (PLANE L M N)) is equivalent to (ARATIO Y L M N r1 r2 1 −r1 −r2) where r1 and r2 are indeterminates. 3. Construction (INTER Y (PLANE L M N) (PLANE W U V) (PLANE R P Q)) is equivalent to (INTER Y (LINE A B) (PLANE R P Q)) where A and B are introduced as follows (INTER A (LINE L M) (PLANE W U V)) (INTER B (LINE L N) (PLANE W U V)) The ndg conditions are LM ∦WUV, LN ∦WUV, AB ∦RPQ. From Exercise 4.31 and Example 4.32 it is clear that all 89 constructions can be described by constructions S1–S6. We may also consider circles and spheres. We deﬁne (CIR O P Q) to be the circle in the plane OPQ which has O as its center and passes through point P. We deﬁne (SPHERE O P) to be the sphere with center O and passing through point P. Then we can introduce the following new constructions. S8 (ON Y (CIR O U V)). Take an arbitrary point on the circle. S9 (ON Y (SPHERE O U)). Take an arbitrary point on the sphere. 4.3 Machine Proof for Class SH 187 S10 (INTER Y ln (CIR O W P)). Take the intersection of line ln and circle (CIR O W P) which is diﬀerent from W. Line ln could be (LINE W V), (PLINE W U V), and (OLINE W L M N). We assume that line ln and the circle are in the same plane. S11 (INTER Y ln (SPHERE O W)). Take the intersection of line ln and sphere (SPHERE O W) which is diﬀerent from W. Line ln could be (LINE W V), (PLINE W U V), and (OLINE W R P Q). S12 (INTER Y (CIR O1 W U) (CIR O2 W V)). Take the intersection of circle (CIR O1 W U) and circle (CIR O2 W V) which is diﬀerent from W. We assume that the two circles are in the same plane. S13 (INTER Y (CIR O1 U V) (SPHERE O2 U)). Take the intersection of circle (CIR O1 U V) and sphere (SPHERE O2 U) which is diﬀerent from U. Here, we introduce another 10 new constructions. In total, we introduced 100 new con-structions including 89 constructions about lines and planes, 10 constructions about circles and spheres, and construction S7. Example 4.33 All ten constructions involving circles and spheres can be represented by constructions S1-S6. Proof. For constructions S8, S10, and S12, see Section 3.2.2. By what we have discussed above, we may assume that a line is always of the form (LINE W U). For construction S11, let Y be introduced by (INTER Y (LINE U V) (SPHERE O U)). Then point Y can also be introduced as follows (FOOT2LINE N O U V); (PRATIO Y N U N 1). Let Y be introduced by construction S13. Then Y can be constructed as follows (FOOT2PLANE M O2 O1 U V); (FOOT2LINE N U M O1); (PRATIO Y N U N 1). For S9, we need to take an arbitrary point Y on (SPHERE O U). To do that, we ﬁrst take an arbitrary point P, and then Y can be introduced by construction S11: (INTER Y (LINE P U) (SPHERE O U)). In summary, we have Proposition 4.34 All 100 constructions introduced in this subsection can be reduced to con-structions S1-S6. 4.3 Machine Proof for Class SH In this section, we will present a mechanical proving method for Hilbert’s intersection statements in the aﬃne space of dimension three. It is clear that the volume method is a 188 Chapter 4. Machine Proof in Solid Geometry natural generalization of the area method presented in Chapter 2. As with the area method, we must eliminate points from geometry quantities. 4.3.1 Eliminating Points from Volumes The method of eliminating points from volumes is the basis of the volume method. Two other geometry quantities, the area ratio and the length ratio, will ultimately be reduced to volumes. In this subsection, we will discuss four constructions S2–S5. S1 will be discussed in Subsection 4.3.4. Lemma 4.35 Let Y be introduced by (PRATIO Y W U V r). Then we have VABCY = ( (UW UV + r)VABCV + (WV UV −r)VABCU if W is on line UV. VABCW + r(VABCV −VABCU) in all cases. Proof. If W, U, and V are collinear, by Proposition 4.5 we have VABCY = UY UV VABCV + YV UV VABCU = (UW UV + r)VABCV + (WV UV −r)VABCU. Otherwise, take a point S such that WS = UV. Then we have VABCY = WY WS VABCS + YS WS VABCW = rVABCS + (1 −r)VABCW. By Proposition 4.9, we have VABCS = VABCW + VABCV −VABCU. Substituting this into the above equation, we obtain the result. Notice that in both cases, we need the ndg condition U , V. Lemma 4.36 Let Y be introduced by (ARATIO Y L M N r1 r2 r3). Then we have VABCY = r1VABCL + r2VABCM + r3VABCN. Proof. This lemma is a direct consequence of Proposition 4.6. Lemma 4.37 Let Y be introduced by (INTER Y (LINE U V) (LINE I J)). Then we have VABCY = S UIJ S UIVJ VABCV −S VIJ S UIVJ VABCU. Proof. By Propositions 4.5 and the co-side theorem, VABCY = UY UVVABCV + YV UVVABCU = S UIJVABCV−S VIJVABCU S UIVJ . Since UV ∦IJ, we have S UIVJ , 0. 4.3 Machine Proof for Class SH 189 Lemma 4.38 Let Y be introduced by (INTER Y (LINE U V) (PLANE L M N)). Then we have VABCY = 1 VULMNV (VULMNVABCV −VVLMNVABCU) Proof. By Proposition 4.5 and the co-face theorem, VABCY = UY UVVABCV + YV UVVABCU = VULMNVABCV−VVLMNVABCU VULMNV . Since UV ∦LMN, we have VULMNV , 0. Example 4.39 Let Y be the intersection point of three planes WUV, LMN and RPQ. Then Y can be constructed as follows (INTER X (LINE L M) (PLANE R P Q)) (INTER Z (LINE L N) (PLANE R P Q)) (INTER Y (LINE X Z) (PLANE W U V)) By Proposition 4.38, we have VABCY = VXWUV VXWUVZ VABCZ −VZWUV VXWUVZ VABCX VABCZ = VLRPQ VLRPQN VABCN −VNRPQ VLRPQN VABCL VABCX = VLRPQ VLRPQM VABCM −VMRPQ VLRPQM VABCL VXWUV = VLRPQ VLRPQM VMWUV −VMRPQ VLRPQM VLWUV VZWUV = VLRPQ VLRPQN VNWUV −VNRPQ VLRPQN VLWUV From the above formulas, we can express VABCY as a rational expression of volumes formed by the known points. Example 4.40 (Steiner’s Theorem) If two opposite edges of a tetrahedron move on two ﬁxed skew lines in any way whatever but remain ﬁxed in length, the volume of the tetrahedron remains constant. X Z B A D C Y W Figure 4-16 Constructive description ((POINTS A B C D) (ON X (LINE A C)) (PRATIO Z X A C 1) (ON Y (LINE B D)) (PRATIO W Y B D 1) (VXYZW = VABCD)) The ndg conditions: A , C, B , D. Proof. By Lemma 4.35, we can eliminate W VXYZW=VDXZY −VBXZY. 190 Chapter 4. Machine Proof in Solid Geometry By Lemma 4.35 again, VBXZY=VBDXZ · BY BD; VDXZY=( BY BD −1) · VBDXZ. Then VXYZW= −VBDXZ = VXBZD. Similarly we can prove VXBZD = VABCD. Example 4.41 Show that a plane which bisects two opposite edges of a tetrahedron bisects its volume. A B C D P S Q R Figure 4-17 Constructive description ((POINTS A B C D) (MIDPOINT P A D) (MIDPOINT S B C) (LRATIO Q B D t) (INTER R (LINE A C) (PLANE P S Q)) (VPCSR−VPDCS −VPDS Q = 1 2 VABCD)) The ndg conditions: A , D; B , C; B , D; AC ∦PS Q. Proof. Using Lemma 4.38, we can eliminate point R VPCS R = RC AC VPCS A = VACPS · VCPS Q VCPS Q −VAPS Q We can eliminate the remaining points by using Lemma 4.35: VPCSR = −VCPS Q · VACPS −VCPS Q + VAPS Q = VCDPS · r · VACPS VCDPS · r + VABPS · r −VABPS = (−1 2VBCDP) · r · (1 2VABCP) −1 2VBCDP · r −1 2VABCP · r + 1 2VABCP = (−1 2VABCD) · r · (1 2VABCD) (2) · (−1 2VABCD) = 1 4(r·VABCD) Similarly, we can compute VPDCS = −1 4VABCD, VPDS Q = 1 4(r−1)·VABCD. Then VPCS R −VPDCS −VPDS Q = ( r 4 + 1 4 −r−1 4 )VABCD = 1 2VABCD. This proof seems to be a little complicated, but the idea behind it is quite simple: one need only to eliminate the points from volumes using Lemmas 4.35–4.38. 4.3.2 Eliminating Points from Area Ratios The following lemmas provide methods of eliminating Y from G = S ABY S CDE . The proofs of the lemmas are similar: if all the points are in one plane, methods of eliminating Y have been given in Chapter 2. Otherwise, there is a point T which is not in the plane ABY. By 4.3 Machine Proof for Class SH 191 Corollary 4.11, (4.4) G = S ABY S CDE = VT ABY VTCDEA . Now the lemmas in Section 4.3.1 can be used to eliminate Y from VT ABY. Lemma 4.42 Let Y be introduced by (PRATIO Y W U V r). Then we have ABY CDE =            VUABWV VUCDEV if W is not in plane ABY VUABW+rVUABV VUCDEA if W is in plane ABY but line UV is not. S ABW+r(S ABV−S ABU) S CDE if W, U, V, A, B, and Y are coplanar. A B Y P Q R W S U V C D E Figure 4-18 Proof. If W < ABY, let WS be a parallel translation of UV to line WY. By (4.4), ABY CDE = VWABY VWCDEY = 1 r VWABY VUCDEV . By Propositions 4.5 and 4.9, VWABY = WY WS VWS AB = rVUABWV. We prove the ﬁrst case. The second case can be proved in a similar manner by replacing W by U. The third case is from Lemma 2.21 on page 65. Remark 4.43 The ﬁrst case of Lemma 4.42 is rarely used in practice, since in this case Y is actually the intersection of (PLINE W U V) and (PPLANE A C D E), i.e., r is a ﬁxed quantity and is not easy to ﬁnd. Lemma 4.44 Let Y be introduced by (ARATIO Y L M N r1 r2 r3). Then we have ABY CDE = ( r2VLABM+r3VLABN VLCDEA if one of L, M, and N, say L, is not in ABY r1S ABL+r2S ABM+r3S ABN S CDE if L, M, and N are in plane ABY. Proof. If L is not in ABY, ABY CDE = VLABY VLCDEA . Now the result comes from Lemma 4.36. The second case is Lemma 2.31. Lemma 4.45 Let Y be introduced by (INTER Y (LINE U V) (LINE I J)). Then we have ABY CDE = ( S UIJVUABV S UIVJVUCDEA if one of U, V, I, and J, say U, is not in ABY. S IUVS ABJ−S JUV S ABI S CDES IUJV if U, V, I, J, A, B, and Y are coplanar. 192 Chapter 4. Machine Proof in Solid Geometry Proof. If U is not in ABY, ABY CDE = VUABY VUCDEA = UY UV VUABV VUCDEA = S UIJ S UIVJ VUABV VUCDEA . The second case is Lemma 2.20. Lemma 4.46 Let Y be introduced by (INTER Y (LINE U V) (PLANE L M N)). Then we have ABY CDE = ( VULMN VULMNV VUABV VUCDEA if one of U and V, say U, is not in ABY. VULMNS ABV−VVLMNS ABU S CDEVULMNV if U and V are in ABY. Proof. If U is not in ABY, ABY CDE = VUABY VUCDEA = UY UV VUABV VUCDEA = VULMN VULMNV VUABV VUCDEA . The second case is a consequence of Proposition 2.9 and the co-face theorem. Example 4.47 (Centroid Theorem for Tetrahedra) The four medians of a tetrahedron meet in a point. A B C D S Z Y G H Figure 4-19 Constructive description ((POINTS A B C D) (MIDPOINT S B C) (LRATIO Y D S 2/3) (Y is the centroid of △DBC.) (LRATIO Z A S 2/3) (Z is the centroid of △ABC.) (INTER G (LINE D Z) (LINE A Y)) (INTER H (LINE C G) (PLANE A B D)) ( S ABH S ABD = 1/3)) The ndg conditions: B , C, D , S , A , S , DZ ∦AY, CG ∦ABD, and S ABD , 0. Proof. Points will be eliminated in the order H, G, Z, Y, S, D, C, B, and A. By Lemma 4.46, S ABH S ABD = VCABD VCABDG VCABG VCABDA = VABCG VABDG + VABCD . By Lemma 4.37, VABDG = S DAY S DAZY VABDZ, VABCG = S ZAY S ZADY VABCD. By Lemma 4.37 again, VABDZ = 2/3VABDS = −1/3VABCD. Now S ABH S ABD = S ZAY S ZADY 1 −1/3 S DAY S ZADY . Now all points are in plane ADS . Then by Lemma 4.42, S ZAY = 1 3S S AY = 2 9S S AD; S DAY = 1 3S DAS = −1 3S S AD; S ZADY = S ZAY −S DAY = 5 9S S AD. Then S ABH S ABD = 1/3. Exercise 4.48 We can also use the following general method to eliminate points from G = S ABY S CDE . If A, B, Y, C, D, and E are coplanar, we need to ﬁnd a point T that is not in plane ABY. Then G = S ABY S CDE = VT ABY VTCDE . 4.3 Machine Proof for Class SH 193 Otherwise, by Corollary 4.11 G = VCABY VACDE . The advantage of this method is that we do not need to use the volume of the polyhedron involving ﬁve points in all cases. Prove Lemmas 4.42-4.46 using the above method. 4.3.3 Eliminating Points from Length Ratios The following lemmas present methods of eliminating point Y introduced by construction C from the length ratio G = DY EF. Y W D T U V S E F Figure 4-20 Lemma 4.49 Let G = DY EF, C =(PRATIO Y W U V r). Then G =                      DW UV +r EF UV if D ∈WY. VDWUV VEWUVF if D < WY, U < DWY. −VUEDWV VUEFWV if D < WY, E < DWY. S DUWV S EUFV if all points are coplanar. Proof. The ﬁrst and the last cases have been proved in Lemma 2.26. If U < DWY, take a point S such that DS = EF. By the co-face theorem G = DY DS = VDWUV VDWUVS = VDWUV VEWUVF . If E < DWY, take a point T such that WT = UV. By the co-side and co-face theorems G = DY DS = S DWT S DWST = VDWTE VDWTES . By Propositions 4.9 and 4.12 VDWT E = VDWVE −VDWUE = VUDEWV, VDWT ES = VEWT EF = −VFWT E = −VFWVE + VFWUE = VUEFWV. Then G = −VUEDWV VUEFWV . Lemma 4.50 Let G = DY EF, C =(ARATIO Y L M N r1 r2 r3). Then we have G =            VDLMN VELMNF if D < LMN. VDMNE−r1VLMNE VEMNEF . if D ∈LMN, E < LMN,and DY ∦NM. S DMN−r1S LMN S EMFN if all points are coplanar and DY ∦NM. Proof. If D is not in plane LMN, the result is a direct consequence of Propositions 4.4 and 4.12. For the second case, by Corollary 4.13, G = VDMNEY VDMNES = VDMNE −r1VLMNE VEMNEF . The third case can be proved similarly. 194 Chapter 4. Machine Proof in Solid Geometry Lemma 4.51 Let G = DY EF, C =(INTER Y (LINE U V) (LINE I J)). Then we have G =            VDUVI VEUVIF D < UVIJ and ¬(COLL U V I). VEDUV VEFVU D ∈UVIJ, EF < UVIJ, and D < UV. S DUV S EUFV D, E, F are in UVIJ, and D < UV. Proof. The ﬁrst case is a consequence of the co-face theorem. For the second case, by Corollary 4.13, G = DY EF = VDUVEY VEUVEF = −VDUVE VFUVE . If all points are coplanar, see Lemma 2.25. Lemma 4.52 Let G = DY EF, C =(INTER Y (LINE U V) (PLANE L M N)). Then we have G =            VDLMN VELMN−VFLMN If D is not in plane LMN. VDUVL VEUVL−VFUVL If D ∈LMN and one of L,M, N, say L < DUV. Proof. If D is not in plane LMN, the result is a direct consequence of the co-face theorem. For the second case, take a point S such that DS = EF. Then we have G = DY DS = VDUVL VDUVLS = VDUVL VEUVLF . Example 4.53 (Ceva’s Theorem for Skew Quadrilaterals) The planes passing through a point O and the sides AB, BC, CD, and DA of any skew quadrilateral meet the opposite sides of the quadrilateral at G, H, E, and F respectively. Show that AE EB · BF FC · CG GD · DH HA = 1. A B C D O H G F E Figure 4-21 Constructive description ((POINTS A B C D O) (INTER E (LINE A B) (PLANE O C D)) (INTER F (LINE B C) (PLANE O A D)) (INTER G (LINE C D) (PLANE O A B)) (INTER H (LINE D A) (PLANE O B C)) ( AE EB BF FC CG GD DH HA = 1)) The ndg conditions: AB ∦OCD; BC ∦OAD; CD ∦OAB; AD ∦OBC; B , E; C , F; D , G, A , H. Proof. By Lemma 4.52 or just by Proposition 4.4, we have AE BE = VACDO VBCDO ; BF CF = −VABDO −VACDO ; CG DG = VABCO VABDO ; DH AH = VBCDO VABCO . Therefore AE EB BF FC CG GD DH HA = 1. 4.3 Machine Proof for Class SH 195 Example 4.54 (Centroid of Tetrahedra) The four medians of a tetrahedron meet in a point which divides each median in the ratio 3:1, the longer segment being on the side of the vertex of the tetrahedron. A B C D S Z Y G Figure 4-22 Constructive description ((POINTS A B C D) (MIDPOINT S B C) (LRATIO Z A S 2/3) (LRATIO Y D S 2/3) (INTER G (LINE D Z) (LINE A Y)) ( AG GY = 3)) The ndg conditions: B , C, A , S , D , S , DZ ∦AY, G , Y. Proof. By Lemma 4.51, AG YG = S ADZ S DZY . By Lemma 4.42, S ADZ S DZY = S ADZ S DSZ −2/3 ; S ADZ S DSZ = AZ ZS S ADS S ADS = 2 1. Then AG YG = 3. Exercise 4.55 Let point Y be introduced by construction (INTER Y (LINE U V) (PLINE R P Q)). Show that • VABCY =        S UPRQ S UPVQ VABCV −S VRPQ S UPVQ VABCU if P, Q, R, U, V are coplanar. VPURQVABCV−VPVRQVABCU VURQPV otherwise. • If D is on UV DY EF =              S DPRQ S EPFQ If all points are coplanar. VDPQR VEPQRF If D < PAR. VPEDRQ VQEFRP If D ∈PQR and E, F < RUV. • S ABY S CDE =                  r′ S ABV S CDE + (1 −r′) S ABU S CDE AB ∈RUV and PQ ∈RUV. r′′ S ABV S CDE + (1 −r′′) S ABU S CDE AB ∈RUV and PQ < RUV. r′ VXABV VXCDEA + (1 −r′) VXABU VXCDEA AB < RUV and PQ ∈RUV. r′′ VXABV VXCDEA + (1 −r′′) VXABU VXCDEA AB < RUV and PQ < RUV. where r′ = VUPRQ VUPVQ , r′′ = VUPRQ VVPRQU and X is U, V or R depending on which is not in plane ABY. Exercise 4.56 Try to eliminate Y from the three geometry quantities if Y is introduced by the following constructions. 196 Chapter 4. Machine Proof in Solid Geometry (INTER Y (PLINE W U V) (PLINE R P Q)) (INTER Y (PLINE W U V) (PLANE L M N)) 4.3.4 Free Points and Volume Coordinates After applying the above lemmas to any rational expression E in geometric quantities, we can eliminate the non-free points introduced by all constructions from E. Now the new E is a rational expression in indeterminates and volumes of free points in space. For more than ﬁve free points in the space, the volumes of the tetrahedra formed by them are not independent, e.g. the following equation is always true: VABCD = VABCO + VABOD + VAOCD + VOBCD. To express E and F as expressions in free parameters, we introduce the volume coordinates. Deﬁnition 4.57 Let X be a point in the space. For four noncoplanar points O, W, U, and V, the volume coordinates of X with respect to OWUV are r1 = VOWUX VOWUV , r2 = VOWXV VOWUV , r3 = VOXUV VOWUV , r4 = VXWUV VOWUV . It is clear that r1 + r2 + r3 + r4 = 1. Since the sum of the volume coordinates of a point is one, we sometimes omit the last one to obtain an independent set of coordinates. Exercise 4.58 Show that the points in the space are in a one to one correspondence with the four-tuples (x, y, z, w) satisfying x + y + w + z = 1. Lemma 4.59 Let G = VABCY, and O, W, U, V be four noncoplanar points. Then we have G = VABCO + VOABCVVOWUY + VOABCUVOVWY + VOABCWVOUVY VOWUV Proof. We have VABCY = VABCO + VABOY + VAOCY + VOBCY. (1) Without loss of generality, we assume that YO meets plane WUV at X. (Otherwise, let YW meet plane OUV at X, and so on.) By Proposition 4.4, we have VOABY = OY OX VOABX = VOWUVYVOABX VOWUV (2) 4.3 Machine Proof for Class SH 197 By Proposition 4.6, VOABX = S WUX S WUV VOABV + S WXV S WUV VOABU + S XUV S WUV VOABW (3) By Lemma 4.46, we have S WUX S WUV = VOWUY VOWUVY ; S WXV S WUV = VOVWY VOWUVY ; S XUV S WUV = VOUVY VOWUVY . Substituting them into (3) and (2), we have VOABY = VOWUYVOABV + VOVWYVOABU + VOUVYVOABW VOWUV (4) Similarly, we have VOBCY = VOWUYVOBCV + VOVWYVOBCU + VOUVYVOBCW VOWUV VOCAY = VOWUYVOCAV + VOVWYVOCAU + VOUVYVOCAW VOWUV Substituting them into (1) and with (4.3) VOABV + VOBCV + VOCAV = VOABCV, VOABU + VOBCU + VOCAU = VOABCU, and VOABW + VOBCW + VOCAW = VOABCW, we can obtain the result. Corollary 4.60 Use the same notations as in Lemma 4.59. For any point P let xP = VOWUP VOWUV , yP = VOWPV VOWUV , zP = VOPUV VOWUV . Then the formula in Lemma 4.59 can be written as VABCY = VOWUV xA yA zA 1 xB yB zB 1 xC yC zC 1 xY yY zY 1 which is quite similar to the formula of the volumes in terms of the Cartesian coordinates of its vertices. Now we can describe the volume method as follows: for a geometry statement in SH: S = (C1, · · · , Cr, (E1, E2)), we can use the above lemmas to eliminate all the non-free points and express the volumes of free points as their volume coordinates with respect to four ﬁxed 198 Chapter 4. Machine Proof in Solid Geometry points. Finally, we obtain two rational expressions R1 and R2 in independent parameters respectively. S is a correct geometry statement if R1 is identical to R2. For the precise description of the algorithm, see Algorithm 4.86 on page 210. We end this section by a methodological comment. One of the most important ideas in the traditional method of solving problems in solid geometry is reducing a problem of higher dimension to a problem of lower dimension and then solving it using knowledge from plane geometry. This is the so-called “dimension reduction method.” But our volume method always does the converse, i.e., it reduces length ratios and area ratios to volumes, because in space the volume is easy to deal with. This can be seen from the three groups of lemmas in Sections 4.3.1, 4.3.2, and 4.3.3. This is also the reason why the method is called the volume method. 4.3.5 Working Examples Example 4.61 For a tetrahedron ABCD and a point O, let P = AO∩BCD, Q = BO∩ACD, R = CO ∩ABD, and S = DO ∩ABC. Show that OP AP + OQ BQ + OR CR + OS DS = 1. A B C D O P S Q R Figure 4-23 Constructive description ((POINTS A B C D O) (INTER P (LINE A O) (PLANE B C D)) (INTER Q (LINE B O) (PLANE A C D)) (INTER R (LINE C O) (PLANE A B D)) (INTER S (LINE D O) (PLANE A B C)) ( OP AP + OQ BQ + OR CR + OS DS = 1)) The ndg conditions: AO ∦BCD, BO ∦ACD, CO ∦ABD, DO ∦ABC, A , P, B , Q, C , R, D , S . Proof. By the co-face theorem OS DS = VABCO VABCD ; OR CR = VABOD VABCD ; OQ BQ = VAOCD VABCD ; OP AP = VOBCD VABCD . By Lemma 4.59, OP AP + OQ BQ + OR CR + OS DS = VOBCD + VAOCD + VABOD + VABCO VABCD = 1. Example 4.62 Let ABCD be a tetrahedron and O a point. Let line DO and plane ABC meet in S ; line AD and plane OBC meet in P; line BD and plane OAC meet in Q; and line CD and plane OAB meet in R. Show that DO OS = DP PA + DQ QB + DR RC . 4.3 Machine Proof for Class SH 199 Constructive description ((POINTS A B C D O) (INTER S (LINE D O) (PLANE A B C)) (INTER P (LINE D A) (PLANE O B C)) (INTER Q (LINE D B) (PLANE O A C)) (INTER R (LINE D C) (PLANE O A B)) ( DO OS = DP PA + DQ QB + DR RC )) The eliminants DR CR R = VABDO VABCO DQ BQ Q = VACDO −VABCO DP AP P =VBCDO VABCO DO OS S =VABCO−VABCD −VABCO VBCDO=VACDO−VABDO+VABCO−VABCD A B C D P O S R Q Figure 4-24 The machine proof DO OS −( DR CR + DQ BQ + DP AP ) R = VABCO −(VABDO+VABCO· DQ BQ +VABCO· DP AP ) · DO OS Q = VABCO·(−VABCO) −(VACDO·VABCO−VABDO·VABCO−V2 ABCO· DP AP ) · DO OS simplify = VABCO VACDO−VABDO−VABCO· DP AP · DO OS P = (VABCO)2 −VBCDO·VABCO+VACDO·VABCO−VABDO·VABCO · DO OS simplify = VABCO −(VBCDO−VACDO+VABDO) · DO OS S = (VABCO−VABCD)·VABCO −(VBCDO−VACDO+VABDO)·(−VABCO) simplify = VABCO−VABCD VBCDO−VACDO+VABDO volume−co = VABCO−VABCD VABCO−VABCD simplify = 1 In the above proof, a volume−co = b means that b is the result obtained by replacing each volume using the volume coordinates with respect to four ﬁxed points. Example 4.63 Let a line l meet four coaxial planes in A, B, C, and D respectively. The cross-ratio of l for the four planes is deﬁned to be (ABCD) = AC AD · BD BC. Show that for any line l the cross ratio is ﬁxed. 200 Chapter 4. Machine Proof in Solid Geometry Constructive description ((POINTS X Y A B C1 D1) (INTER C (LINE A B) (PLANE C1 X Y)) (INTER D (LINE A B) (PLANE D1 X Y)) (INTER A1 (LINE C1 D1) (PLANE A X Y)) (INTER B1 (LINE C1 D1) (PLANE B X Y)) ( AC AD BD BC = A1C1 A1D1 B1D1 B1C1 )) The eliminants D1B1 C1B1 B1 = VXYBD1 VXYBC1 C1A1 D1A1 A1 = VXYAC1 VXYAD1 AC AD D = VXYD1C+VXYAD1 VXYAD1 BD BC D = VXYBD1 VXYD1C+VXYBD1 VXYD1C C = VXYBD1·VXYAC1 −VXYBC1·VXYAD1 VXYBC1−VXYAC1 X Y A B C D A B C D Figure 4-25 1 1 1 1 The machine proof BD BC · AC AD D1B1 C1B1 · C1A1 D1A1 B1 = −VXYBC1 (−VXYBD1)· C1A1 D1A1 · BD BC · AC AD A1 = VXYBC1·(−VXYAD1) VXYBD1·(−VXYAC1 ) · BD BC · AC AD D = VXYBD1·(VXYD1C+VXYAD1 )·VXYBC1·VXYAD1 VXYBD1·VXYAC1 ·VXYAD1·(VXYD1C+VXYBD1) simplify = (VXYD1C+VXYAD1)·VXYBC1 VXYAC1 ·(VXYD1C+VXYBD1) C = (−VXYBD1·VXYAC1 +VXYAD1·VXYAC1 )·VXYBC1·(−VXYBC1+VXYAC1 ) VXYAC1 ·(−VXYBD1·VXYBC1+VXYBC1·VXYAD1)·(−VXYBC1+VXYAC1 ) simplify = 1 A B C D G P R Q Figure 4-26 Example 4.64 3 Let ABCD be a tetrahedron and G the centroid of triangle ABC. The lines passing through points A, B, and C and parallel to line DG meet their opposite face in P, Q, and R respectively. Show that VGPQR = 3VABCD. 3This is a problem from the 1964 International Mathematical Olympiad. 4.3 Machine Proof for Class SH 201 Constructive description ((POINTS A B C D) (CENTROID G A B C) (INTER P (PLINE A D G) (PLANE B C D)) (INTER Q (PLINE B D G) (PLANE A C D)) (INTER R (PLINE C D G) (PLANE A B D)) (3VABCD = VGPQR)) The eliminants VCDGP=−VACDG VBDGP=−VABDG VBCGP=VABCD VDGPQ=−VBDGP VCGPQ=VCDGP·VABCD−VBCGP·VACDG VACDG VGPQR=VDGPQ·VABCD−VCGPQ·VABDG VABDG The machine proof (3)·VABCD VGPQR R = (3)·VABCD·VABDG VDGPQ·VABCD−VCGPQ·VABDG Q = (3)·VABCD·VABDG·(VACDG)2 −VCDGP·VACDG·VABDG·VABCD−VBDGP·V2 ACDG·VABCD+VBCGP·V2 ACDG·VABDG simplify = (−3)·VABCD·VABDG·VACDG VCDGP·VABDG·VABCD+VBDGP·VACDG·VABCD−VBCGP·VACDG·VABDG P = (−3)·VABCD·VABDG·VACDG·(VBCDG)3 −3V3 BCDG·VACDG·VABDG·VABCD simplify = 1 In the above proof the fact that G is the centroid of triangle ABC is not used. We thus have the following extension of Example 4.64. Example 4.65 The result of Example 4.64 is still true if point G is any point in plane ABC. We further ask whether the result of Example 4.64 is true or not if point G is an arbitrary point. Constructive description ((POINTS A B C D G) (INTER P (PLINE A D G) (PLANE B C D)) (INTER Q (PLINE B D G) (PLANE A C D)) (INTER R (PLINE C D G) (PLANE A B D)) (3VABCD = VGPQR)) The eliminants VCDGP=−VACDG VBDGP=−VABDG VBCGP=−(VABCG−VABCD) VDGPQ=−VBDGP VCGPQ=VCDGP·VABCD−VBCGP·VACDG VACDG VGPQR=VDGPQ·VABCD−VCGPQ·VABDG VABDG The machine proof (3)·VABCD VGPQR R = (3)·VABCD·VABDG VDGPQ·VABCD−VCGPQ·VABDG Q = (3)·VABCD·VABDG·(VACDG)2 −VCDGP·VACDG·VABDG·VABCD−VBDGP·V2 ACDG·VABCD+VBCGP·V2 ACDG·VABDG simplify = (−3)·VABCD·VABDG·VACDG VCDGP·VABDG·VABCD+VBDGP·VACDG·VABCD−VBCGP·VACDG·VABDG 202 Chapter 4. Machine Proof in Solid Geometry P = (−3)·VABCD·VABDG·VACDG·(VBCDG)3 V3 BCDG·VACDG·VABDG·VABCG−3V3 BCDG·VACDG·VABDG·VABCD simplify = (−3)·VABCD VABCG−3VABCD We thus obtain the following extension of Example 4.65: VGPQR = 3VABC iﬀG is in plane ABC. 4.4 Pythagoras Diﬀerences in Space 4.4.1 Pythagoras Diﬀerence and Perpendicularity From now on, the concept of the square-distance between two points will be used. The deﬁnition for the Pythagoras diﬀerence is the same as in the plane geometry, i.e., for △ABC PABC = AB 2 + CB 2 −AC 2. For a skew quadrilateral ABCD, we deﬁne PABCD = PABD −PCBD = AB 2 + CD 2 −BC 2 −DA 2. Properties of the Pythagoras diﬀerence in space are quite similar to those in a plane. Propo-sitions 3.4, 3.2, and 3.3 are true in space, since all the points involved are in the same plane. Surprisingly, Propositions 3.1, 3.5, and 3.7 are also true in space even if the involving points are not coplanar. Proposition 4.66 Let R be a point on line PQ with position ratio r1 = PR PQ, r2 = RQ PQ with respect to PQ. Then for any points A and B in space, we have PRAB = r1PQAB + r2PPAB PARB = r1PAQB + r2PAPB −r1r2PPQP. Proof. Since points P, Q, R, A and P, Q, R, B are two groups of coplanar points, we may use Proposition 3.5 on them separately: RA 2 = r1QA 2 + r2PA 2 −r1r2PQ 2 RB 2 = r1QB 2 + r2PB 2 −r1r2PQ 2. Then PRAB = RA 2 + AB 2 −RB 2 = r1(QA 2 + AB 2 −QB 2) + r2(PA 2 + AB 2 −PB 2) = r1PQAB + r2PPAB. The second equation can be proved similarly. 4.4 Pythagoras Diﬀerences in Space 203 Proposition 4.67 Let R be a point in the plane PQS , and r1 = S PQR S PQS , r2 = S RQS S PQS , and r3 = S PRS S PQS . Then for points A and B we have PRAB = r1PS AB + r2PPAB + r3PQAB PARB = r1PAS B + r2PAPB + r3PAQB −2(r1r2PS 2 + r1r3QS 2 + r2r3PQ 2) Proof. The proof for proposition is similar to that for Propositions 3.61 and 3.62. Proposition 4.68 Let ABCD be a parallelogram. Then for any points P and Q, we have PAPQ + PCPQ = PBPQ + PDPQ or PAPBQ = PDPCQ PPAQ + PPCQ = PPBQ + PPDQ + 2PBAD Proof. The proof is the same as for Proposition 3.7. As in plane geometry, we use the notation AB⊥CD to denote the fact that four points A, B, C, and D satisfy one of the following conditions: A = B, or C = D, or line AB is perpendicular to line CD. Proposition 4.69 AB⊥CD iﬀPACD = PBCD or PACBD = 0. Proof. Let E be a point such that AE = CD. By Proposition 4.68, PACBD = PAABE = PBAE. By the Pythagorean theorem, AB⊥CD iﬀPACBD = PBAE = 0. As consequences, Proposition 3.10 and Example 3.9 are still true in space. Just as the volume makes the machine proof for geometry theorems involving collinear and parallel possible, the Pythagoras diﬀerence will make the machine proof for geome-try theorems involving perpendicular possible. Before presenting the method of machine proof, let us ﬁrst get acquainted with the Pythagoras diﬀerences by proving some basic properties of the perpendicular. Example 4.70 If a given line is perpendicular to a pair of non-parallel lines in a plane, it is perpendicular to every lines in the given plane, i.e., the line is perpendicular to the plane. Proof. As in Figure 4-27, line PQ is perpendicular to line OU and OV. Let W be a point in plane OUV. We need to show that PQ⊥OW. By Propositions 4.67 and 4.69 P Q O U W V Figure 4-27 PPQW = S OUW S OUV PPQV + S OWV S OUV PPQU + S WUV S OUV PPQO = (S OUW S OUV + S OWV S OUV + S WUV S OUV )PPQO = PPQO. By Proposition 4.69 again, PQ⊥OW. 204 Chapter 4. Machine Proof in Solid Geometry Example 4.71 (The Theorem of Three Perpendiculars) A line PQ is perpendicular to a line AB iﬀPQ is perpendicular to the orthogonal projection of AB to a plane containing PQ. P Q A O B Figure 4-28 Proof. Let O be the orthogonal projection of point A to plane PQB. Then AO⊥PQ. If PQ⊥BO, we have PPQA = PPQO = PPQB, i.e., PQ⊥AB. Conversely if PQ⊥AB, we have PPQB = PPQA = PPQO, i.e., PQ⊥BO. Example 4.72 (The Orthocenter Theorem for Tetrahedra) If two pairs of opposite edges of a tetrahedron are at right angles to one another, the third pair are at right angles; and the altitudes are concurrent, and pass through the orthocenters of the opposite faces. A B C D R E F P Q H Figure 4-29 Proof. Let AB⊥CD and AC⊥BD. Then PABC = PABD = PCBD = PDBC, i.e., AD⊥BC. Let the two altitudes AQ and DR of triangle ACD meet in F. Then PBAQ −PFAQ = PBAD −PFAC = PCAD −PDAC = 0, i.e., BF⊥AQ. Similarly, BF⊥DR. Therefore BF is the altitude from B to plane ACD. To prove that the four altitudes are concurrent, we ﬁrst show that BR⊥AC which follows from PACR = PACD = PACB. Let altitudes AP and BR of triangle ABC meet in E and DE and BF meet in H. We need to show that AH⊥BCD. By Proposition 4.69, PHDC = PBDC = PADC, i.e., AH⊥DC. Similarly AH⊥BC. Thus AH⊥BCD. Example 4.73 Show that the construction (FOOT2PLANE A P L M N) is equivalent to the construction (ARATIO A L M N r1 r2 r3) where L M N P T F S Y Figure 4-30 r1 = S YMN S LMN = −PPMN PLMN+2MN 2PPML 4LN 2·LM 2−P2 MLN r2 = S LYN S LMN = −PPNLPLNM+2NL 2PPNM 4LN 2·LM 2−P2 MLN r3 = S LMY S LMN = −PPLMPMLN+2ML 2PPLN 4LN 2·LM 2−P2 MLN Proof. We construct point A as in Example 4.29. Then S YMN S LMN = YT S T = PPTS PTS T = PPTS PLFL . 4.4 Pythagoras Diﬀerences in Space 205 By Propositions 4.66 and 4.69, PPTS = PPT L = PPMNPPNL + PPNMPPML −PPMNPPNM PMNM = PPMNPPNL + (2MN 2 −PPMN)PPML −PPMNPPNM PMNM = −PPMNPNML + 2MN 2PPML PMNM PLFL = 16S 2 LMN PMNM = 4LN 2 · LM 2 −P2 MLN PMNM . We have proved the ﬁrst case. Other cases are similar. 4.4.2 Pythagoras Diﬀerence and Volume With the concept of perpendicularity, we can obtain the exact measurement for the volume of a tetrahedron. Deﬁnition 4.74 Let F be the foot of the perpendicular dropped from the point R upon the plane LMN. The distance from R to LMN, denoted by hR,LMN, is a real number which has the same sign as VRLMN and \|hR,LMN\| = \|RF\|. Proposition 4.75 For any two tetrahedra ABCD and RLMN, let hA = hA,BCD, hR = hR,LMN. Show that VABCD \|S BCD\|hA = VRLMN \|S LMN\|hR. A B C D R L M N F S Figure 4-31 Proof. Without loss of generality, we assume that points B, C, D, L, M, and N are in the same plane. As in Figure 4-31, let RF be the altitude of the tetrahedron RLMN and S be a point on RF such that AS ∥BCD. Then VABCD = VS BCD. By Propositions 4.3 and 4.4, VABCD VALMN = S BCD S LMN ; VS LMN VRLMN = S F RF . Multiplying the two formulas together and noting that hA and hR have the same sign as VABCD and VRLMN, we prove the result. Corollary 4.76 For a tetrahedron ABCD, we have hA,BCD\|S BCD\| = hB,CDA\|S CDA\| = hC,DAB\|S DAB\| = hD,ABC\|S ABC\|. 206 Chapter 4. Machine Proof in Solid Geometry Proof. Replacing the tetrahedron RLMN in the Proposition 4.75 by BCDA, we obtain the ﬁrst equation, etc. By Proposition 4.75, we have VABCD = khA,BCD\|S BCD\| = khB,CDA\|S CDA\| = khC,DAB\|S DAB\| = khD,ABC\|S ABC\| where k is a constant which is independent of the tetrahedron ABCD. Setting k = 1/3, we obtain the usual formula for the volumes of tetrahedra. Proposition 4.77 We have VABCD = 1 3hA,BCD\|S BCD\| = 1 3hB,CDA\|S CDA\| = 1 3hC,DAB\|S DAB\| = 1 3hD,ABC\|S ABC\|. Proposition 4.78 (The Herron-Qin Formula for Tetrahedra) Prove the following formula 144V2 ABCD = 4AB 2 · AC 2 · AD 2 −AB 2P2 DAC −AC 2P2 BAD −AD 2P2 BAC + PBACPBADPCAD. D A C B H F G O Figure 4-32 Proof. Similar to Example 4.29, we construct the altitude BO as follows. (FOOT2LINE F D A C) (FOOT2LINE H B A C) (PRATIO G H F D 1) (FOOT2LINE O B H G) Then BO 2 = BH 2 −HO 2 = 4S 2 ABC AC 2 −HO 2. (1) By Propositions 3.1, 3.2, and 4.66 OH 2 =       OH HG       2 HG 2 = PBHG PHGH !2 HG 2 = P2 BHG 4HG 2 = P2 BHD 4DF 2 = (PBACPBCD + PBCAPBAD −PBACPBCA)2/(4AC 4 · 4DF 2) = (PBACPBCD + (2AC 2 −PBAC)PBAD −PBACPBCA)2/(64AC 2S 2 DAC) = (−PBACPCAD + 2AC 2PBAD)2/(64AC 2S 2 DAC) Substitute this into (1). By the Herron-Qin formula for triangles (on page 108), we have 144V2 ABCD = 144(1/3)2BO 2S 2 ACD 4.5 The Volume Method 207 = (16)2S 2 BACS 2 DAC −(−PBACPCAD + 2AC 2PBAD)2 4AC 2 = (4AB 2 · AC 2 −P2 BAC)(4AD 2 · AC 2 −P2 DAC) −(−PBACPCAD + 2AC 2PBAD)2 4AC 2 = 4AB 2 · AC 2 · AD 2 −AB 2P2 DAC −AC 2P2 BAD −AD 2P2 BAC + PBACPBADPCAD. Corollary 4.79 (The Cayley-Menger Formula) We have the following commonly used version of the Herron-Qin formula. 288VP1P2P3P4 = 0 r2 12 r2 13 r2 14 1 r2 21 0 r2 23 r2 24 1 r2 31 r2 32 0 r2 34 1 r2 41 r2 42 r2 43 0 1 1 1 1 1 0 where rij = \|PiP j\|. Proof. In the above determinant, subtracting the ﬁrst row from the second, the third, and the fourth rows and subtracting the ﬁrst column from the second, the third, and the fourth columns, the determinant becomes 0 r2 12 r2 13 r2 14 1 r2 21 −2r2 12 −P213 −P214 0 r2 31 −P312 −2r2 13 −P314 0 r2 41 −P412 −P413 −2r2 14 0 1 0 0 0 0 = 2r2 12 P213 P214 P312 2r2 13 P314 P412 P413 2r2 14 . Expanding the last determinant and comparing the formula in Proposition 4.78, we prove the result. 4.5 The Volume Method Since we have a new geometry quantity, the constructive statements can be enlarged in the following way: the conclusion of a geometry statement could be the equation of two polynomials of length ratios, area ratios, volumes and Pythagoras diﬀerences. 4.5.1 The Algorithm Now we have six constructions S1–S6 and four geometry quantities. We need to give a method to eliminate the point introduced by each of the constructions S1–S6 from each of 208 Chapter 4. Machine Proof in Solid Geometry the four quantities. This section deals with the cases which are not discussed in Section 4.3. Lemma 4.80 Let G = PABY. Then G =                                                  PABW + r(PABV −PABU) if Y is introduced by (PRATIO Y W U V r) r1PABL + r2PABM + r3PABN if Y is introduced by (ARATIO Y L M N r1 r2 r3) S UIJ S UIVJ PABV −S VIJ S UIVJ PABU if Y is introduced by (INTER Y (LINE U V) (LINE I J)) 1 VULMNV (VULMNPABV −VVLMNPABU) if Y is introduced by (INTER Y (LINE U V) (PLANE L M N)) PPUV PABV+PPVU PABU 2UV 2 if Y is introduced by (FOOT2LINE Y P U V) Proof. We only need to ﬁnd the position ratio of Y with respect to UV and substitute it into the ﬁrst equation of Proposition 4.66. For the second case, see Lemma 4.36. For other cases, see Subsection 4.3.1 for details. From the above lemma and Proposition 4.66, it is easy to eliminate Y from PAYB. We leave this as an exercise. Exercise 4.81 Try to eliminate Y from PAYB if Y is introduced by each of the constructions S2-S6. Lemma 4.82 If Y is introduced by (FOOT2LINE Y P U V) then VABCY = PPUV PUVU VABCV + PPVU PUVU VABCU. Proof. This is a consequence of Propositions 4.5 and 3.2. Lemma 4.83 Let Y be introduced by (FOOT2LINE Y P U V). Then DY EF =                  PPUDV PEUFV if D ∈UV. VDPUV VEPUVF D < PUV VDUVE VEUVF if D ∈PUV and E < PUV S DUV S EUFV if all points are coplanar In all cases, we assume P is not on line UV; otherwise P = Y and DY EF = DP EF. 4.5 The Volume Method 209 Proof. The ﬁrst and last cases are from Lemma 3.29. The second case is a consequence of the co-face theorem. For the third case, let T be a point such that DT = EF. Then DY EF = DY DT = S DUV S DUTV = VDUVE VDUVET = VDUVE VEUVEF = −VDUVE VFUVE . Lemma 4.84 Let Y be introduced by (FOOT2LINE Y P U V). Then S ABY S CDE =              PPUVVPABV+PPVUVPABU 2UV 2VPCDEA if P is not in ABY. VUABV VUCDEV if UV ∦ABY. PPUVS ABV+PPVUS ABU 2UV 2S CDE if P, U, and V are in ABY. Proof. If P is not in ABY, by Proposition 4.3, S ABY S CDE = VPABY VPCDEA . Now the result comes from Lemma 4.82. For the second case S ABY S CDE = VUABYV VUCDEV = VUABV VUCDEV . The last case is Lemma 3.24. By now, we have given methods of eliminating points introduced by constructions S2-S6. We still need to deal with free points. By Lemma 4.59, volumes of tetrahedra can be reduced to volume coordinates with respect to four non-coplanar points. The following lemma will reduce the Pythagoras diﬀerence of free points to volume coordinates. Lemma 4.85 Let O, W, U, and V be four points such that OW⊥OUV, OU⊥OWV, and OV⊥OWU. Then (1) AB 2 = OW 2(VAOUVB VOWUV )2 + OU 2(VAOWVB VOWUV )2 + OV 2(VAOWUB VOWUV )2. (2) V2 OWUV = 1 36OW 2OU 2OV 2. O U V W D A E F B Figure 4-33 Proof. (2) is from Propositions 4.77 and 3.15. For (1), let D, E, and F be points such that AD ∥OW, BE ∥OV, BF ∥OU, DE ∥OU, and DF ∥OV. Then AB 2 = AD 2 + BD 2 = AD 2 + BE 2 + BF 2 = OW 2( AD OW )2 + OU 2( BF OU )2 + OV 2( BE OV )2. By the co-face theorem, AD OW = −VAOUVD VWOUV = VAOUVB VOWUV , BE OV = VBOWUA VOWUV , and BF OU = VBOWVA VOWUV . Now we have the main algorithm. 210 Chapter 4. Machine Proof in Solid Geometry Algorithm 4.86 (Solid) INPUT: S = (C1, C2, . . . , Ck, (E, F)) is a constructive geometric statement. OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof for S . S1. For i = k, · · · , 1, do S2, S3, S4 and ﬁnally do S5. S2. Check whether the nondegenerate conditions of Ci are satisﬁed. The nondegenerate condition of a construction has three forms: A , B, PQ ∦UV, PQ ∦WUV. For the ﬁrst case, we check whether PABA = 2AB 2 = 0. For the second case, we check whether VPQUV = 0 and S PUV = S QUV. For the third case, we check whether VPWUV = VQWUV. If a nondegenerate condition of a geometry statement is not satisﬁed, the statement is trivially true. The algorithm terminates. S3. Let G1, · · · , Gs be the geometric quantities occurring in E and F. For j = 1, · · ·, s do S4 S4. Let H j be the result obtained by eliminating the point introduced by construction Ci from G j using the lemmas in this chapter and replace G j by H j in E and F to obtain the new E and F. S5. Now E and F are rational expressions of independent variables. Hence if E = F, S is true under the nondegenerate conditions. Otherwise S is false in Euclidean solid geometry. Proof. The algorithm is correct, because the volume coordinates of free points are indepen-dent parameters. For the complexity of the algorithm, let n be the number of the non-free points in a statement. If the conclusion of the geometry statement is of degree d, the output of our algorithm is at most of degree 5d3n. 4.5.2 Working Examples Example 4.87 If a line divides two opposite sides of a skew quadrilateral in direct propor-tion, and a second line divides the other two opposite sides in direct proportion, the two lines are coplanar. A B C D E F H G I Figure 4-34 Constructive description ((POINTS A B C D) 4.5 The Volume Method 211 (LRATIO E A B r1) (LRATIO F D C r1) (LRATIO H A D r2) (LRATIO G B C r2) (VEFHG = 0)) The machine proof VEFHG n = −VCEFH·r2+VBEFH·r2−VBEFH n = −(−VBDEF·r2 2+VBDEF·r2+VACEF·r2 2−VACEF·r2) simplify = (r2−1)·(VBDEF−VACEF)·r2 n = (r2−1)·(VBCDE·r1−VACDE·r1+VACDE)·r2 n = (r2−1)·(0)·r2 simplify = 0 The eliminants VEFHG G = −(VCEFH·r2−VBEFH·r2+VBEFH) VBEFH H =VBDEF·r2 VCEFH H =(r2−1)·VACEF VACEF F =(r1−1)·VACDE VBDEF F =VBCDE·r1 VACDE E =VABCD·r1 VBCDE E =(r1−1)·VABCD Example 4.88 Continue from Example 4.87. Let the two lines EF and GH intersect at I. Then EI EF = AH AD. Constructive description ((POINTS A B C D) (LRATIO E A B r1) (LRATIO F D C r1) (LRATIO H A D r2) (LRATIO G B C r2) (INTER I (LINE E F) (LINE H G)) ( EI EF = r2)) The machine proof EI EF r2 I = 1 r2·(−S FHG S EHG +1) G = −(−VBCEH) r2·(−VBCFH+VBCEH) H = −(−VBCDE·r2) r2·(VBCDE·r2+VABCF·r2−VABCF) simplify = VBCDE VBCDE·r2+VABCF·r2−VABCF F = VBCDE VBCDE·r2−VABCD·r2·r1+VABCD·r2+VABCD·r1−VABCD E = VABCD·r1−VABCD VABCD·r1−VABCD simplify = 1 The eliminants EI EF I = 1 −( S FHG S EHG −1) S FHG S EHG G = VBCFH VBCEH VBCFH H =(r2−1)·VABCF VBCEH H = −(VBCDE·r2) VABCF F = −((r1−1)·VABCD) VBCDE E =(r1−1)·VABCD Example 4.89 The sides AB and DC of a skew quadrilateral are cut into 2n + 1 equal seg-ments by points P1, · · · , P2n and Q1, · · · , Q2n respectively. Show that (1) VPnPn+1Qn+1Qn = 1 (2n+1)2 VABCD. 212 Chapter 4. Machine Proof in Solid Geometry (2) If sides BC and AD are cut into 2m + 1 equal segments by points R1, · · · , R2m and S 1, · · · , S 2m respectively, then the area of the quadrilateral formed by the lines PnQn, Pn+1Qn+1, RmS m, and Rm+1S m+1 is 1 (2n+1)2(2m+1)2 VABCD. Figure 4-35 shows the case n = m = 2. Note that in the following machine proof for (1), we use some diﬀerent names for points Pn, Pn+1, Qn+1, Qn. Constructive description ((POINTS A B C D) (LRATIO X A B n 2n+1) (LRATIO Y A B n+1 2n+1) (LRATIO U D C n 2n+1) (LRATIO V D C n+1 2n+1 ) (VXYVU = VABCD)) The eliminants VXYUV V =−(VDXYU·n+VCXYU·n+VCXYU) 2n+1 VCXYU U =(n+1)·VCDXY 2n+1 VDXYU U =−VCDXY ·n 2n+1 VCDXY Y =−(VBCDX·n+VBCDX+VACDX·n) 2n+1 VACDX X =VABCD·n 2n+1 VBCDX X =−(n+1)·VABCD 2n+1 A B C D P P P P 1 3 2 4 Q Q Q Q 1 3 2 4 S S S S 1 3 2 4 R R R R 1 3 2 4 Figure 4-35 The machine proof −VXYUV VABCD V = −(−VDXYU·n−VCXYU·n−VCXYU) VABCD·(2n+1) U = 4VCDXY ·n2+4VCDXY ·n+VCDXY VABCD·(2n+1)3 simplify = VCDXY VABCD·(2n+1) Y = −VBCDX·n−VBCDX−VACDX·n VABCD·(2n+1)2 X = −(−4VABCD·n2−4VABCD·n−VABCD) VABCD·(2n+1)4 simplify = 1 (2n+1)2 By Example 4.88, PnQn and Pn+1Qn+1 are cut into 2m + 1 equal segments by RiS i, i = 1, ..., 2m respectively. Now (2) comes from (1) directly. A line joining the mid-points of two opposite edges of a tetrahedron will be called a bimedian of the tetrahedron relative to the pair of edges considered. The common perpen-dicular to the two opposite edges of a tetrahedron is called the bialtitude of the tetrahedron relative to these edges. Example 4.90 The bialtitude relative to one pair of opposite edges of a tetrahedron is per-pendicular to the two bimedians relative to the two other pairs of opposite edges. 4.5 The Volume Method 213 A B C D S T X Y N Q Figure 4-36 Constructive description ((POINTS X Y A C) (FOOT2LINE S A X Y) (ON B (LINE S A)) (FOOT2LINE T C X Y) (ON D (LINE T C)) (MIDPOINT N B C) (MIDPOINT Q A D) (PERPENDICULAR N Q X Y)) The machine proof PYXN PYXQ Q = PYXN 1 2 PYXD+ 1 2 PYXA P = (2)·PYXN PYXD+PYXA N = (2)·( 1 2 PYXB+ 1 2 PYXC) PYXD+PYXA M = PYXB+PYXC PYXD+PYXA D = PYXB+PYXC −PYXT · TD TC +PYXT+PYXC· TD TC +PYXA = −(PYXB+PYXC) −PYXC−PYXA The eliminants PYXS =PYXA PYXB= −(PYXS · S B S A −PYXS −PYXA· S B S A) PYXT=PYXC PYXD= −(PYXT· TD TC −PYXT−PYXC· TD TC) PYXN=1 2(PYXB+PYXC) PYXQ=1 2(PYXD+PYXA) B = −PYXS · S B S A +PYXS +PYXC+PYXA· S B S A PYXC+PYXA = −(−PYXC−PYXA) PYXC+PYXA simplify = 1 Example 4.91 4 A plane parallel to AB and CD meets the edges AD, AC, BD, and BC in P, Q, S , and R respectively. The plane divides the tetrahedron into two parts. Let r be the ratio of the distances between AB, CD and the plane PQS . Find the ratio of the volumes of the two parts. First by the co-face theorem, r = VAPQS VDPQS = AP PD . Since the volume V1 of AB −PQRS is equal to VABS R + VAPRQ + VAPS R, we will compute VABSR VABCD , VAPRQ VABCD , and VAPSR VABCD separately. 4This is a problem from the 1965 International Mathematical Olympiad. 214 Chapter 4. Machine Proof in Solid Geometry A B C D P Q S R Figure 4-37 Constructive description ((POINTS A B C D) (LRATIO P A D r 1+r ) (INTER Q (LINE A C) (PLINE P C D)) (INTER S (LINE B D) (PLINE P A B)) (INTER R (LINE B C) (PLANE P Q S )) ( VABSR VABDC )) The eliminants S ACD SCDP =r+1 VBCDP=−VABCD r+1 S ABD S ABP =r+1 r VABCP=VABCD·r r+1 VBDPQ= ( S ACD SCDP −1)·VBCDP S ACD SCDP VBCPQ=−VABCP S ACD SCDP VBPQS =VBDPQ S ABD S ABP VABCS =VABCD S ABD S ABP VCPQS = ( S ABD S ABP −1)·VBCPQ −S ABD S ABP VABSR= VBPQS ·VABCS VCPQS −VBPQS The machine proof VABSR −VABCD R = −VBPQS ·VABCS −VABCD·(−VCPQS +VBPQS ) S = −(−VBDPQ)·(−VABCD)·((−S ABD S ABP ))2 VABCD·(−VBDPQ· S ABD S ABP −VBCPQ· S ABD S ABP 2+VBCPQ· S ABD S ABP )·((−S ABD S ABP ))2 simplify = VBDPQ (VBDPQ+VBCPQ· S ABD S ABP −VBCPQ)· S ABD S ABP Q = (VBCDP· S ACD SCDP −VBCDP)·( S ACD SCDP )2 (VBCDP· S ACD SCDP 2−VBCDP· S ACD SCDP −VABCP· S ACD SCDP · S ABD S ABP +VABCP· S ACD SCDP )· S ABD S ABP · S ACD SCDP simplify = ( S ACD SCDP −1)·VBCDP (VBCDP· S ACD SCDP −VBCDP−VABCP· S ABD S ABP +VABCP)· S ABD S ABP P = (r)3·(−VABCD)·(r+1)2 (−VABCD·r3−2VABCD·r2−VABCD·r)·(r+1)2 simplify = (r)2 (r+1)2 Similarly, we can compute VAPRQ VABCD = (r)2 (r + 1)3, VAPS R VABCD = (r)2 (r + 1)3. Thus V1 = ( (r)2 (r + 1)2 + (r)2 (r + 1)3 + (r)2 (r + 1)3)VABCD = (r)2(r + 3) (r + 1)3 VABCD. Let V2 = VABCD −V1 = 3r+1 (r+1)3VABCD. Finally we have V1 V2 = r2(r+3) 3r+1 . 4.6 Volume Coordinate System 215 Example 4.92 (Monge’s Theorem) The six planes through the midpoints of the edges of a tetrahedron and perpendicular to the edges respectively opposite have a point in common. This point is called the Monge point of the tetrahedron. A B C D L L A L Figure 4-38 1 1 2 Constructive description ((POINTS A B C D) (MIDPOINT L A B) (FOOT2LINE L1 L C D) (FOOT2LINE A1 A C D) (PRATIO L2 L1 A1 A 1) (MIDPOINT R A C) (FOOT2LINE R1 R B D) (FOOT2LINE A2 A B D) (PRATIO R2 R1 A2 A 1) (INTER P (LINE L L1) (PLANE R R1 R2)) (INTER Q (LINE L L2) (PLANE R R1 R2)) (MIDPOINT S B C) (FOOT2LINE S 1 S A D) (FOOT2LINE B1 B A D) (PRATIO S 2 S 1 B1 B 1) (INTER M (LINE P Q) (PLANE S S 1 S 2)) (MIDPOINT N C D) (PERPENDICULAR N M A B)) The proof for this theorem is too long to print here. We need more elimination tech-niques to produce short and readable proofs for problems like this one. 4.6 Volume Coordinate System In Lemma 4.85, we use an orthogonal coordinate system, which is essentially the same as the usual Cartesian coordinate system. In order to do this, we have to introduce four auxiliary points O, W, U, and V. In this section, we will develop some properties for a skew volume coordinate system in which any four free points can be selected as the reference points. As a consequence, we obtain a new proof of Lemma 4.85 and hence a new version of Algorithm 4.86. Let O, W, U, and V be four non-coplanar points. Then for any point A, we will denote its volume coordinates with respect to OWUV as xA = VOWUA VOWUV , yA = VOWAV VOWUV , zA = VOAUV VOWUV , wA = VAWUV VOWUV . It is clear that xA + yA + zA + wA = 1. Following are some known results. Proposition 4.93 The points in the space are in a one to one correspondence with the four-tuples (x, y, z, w) such that x + y + z + w = 1. 216 Chapter 4. Machine Proof in Solid Geometry Proposition 4.94 For any points A, B, C, and D, we have VABCD = VOWUV xA yA zA 1 xB yB zB 1 xC yC zC 1 xD yD zD 1 . As a consequence of Proposition 4.94, we can give the equation for planes in the volume coordinate system. Let P be a point in plane ABC. Then the volume coordinates of P must satisfy xA yA zA 1 xB yB zB 1 xC yC zC 1 xP yP zP 1 = 0 which is the equation for the plane ABC. The position ratio formula is still true Proposition 4.95 Let R be a point on line PQ and r1 = PR PQ and r2 = RQ PQ be the position ration of R with respect to PQ. Then xR = r1xQ + r2xP; yR = r1yQ + r2yP; zR = r1zQ + r2zP; wR = r1wQ + r2wP. Proof. This is a consequence of Proposition 4.5. We now develop the formula for the square distance between two points. Proposition 4.96 Let OXZ1Y −ZY1O1X1 be a parallelepiped. We have OO1 2 = OX 2 + OY 2 + OZ 2 + PXOY + PXOZ + PYOZ. O X Y Z Z X Y O C Figure 4-39 1 1 1 1 Proof. By Proposition 3.6, OX1 2 = 2OZ 2 + 2OY 2 −ZY 2. OY1 2 = 2OX 2 + 2OZ 2 −XZ 2. OZ1 2 = 2OX 2 + 2OY 2 −XY 2. (1) XX1 2 + YY1 2 = 2OZ 2 + 2XY 2. OO1 2 + ZZ1 2 = 2OZ 2 + 2OZ1 2 = 4OX 2 + 4OY 2 + 2OZ 2 −2XY 2. 4.6 Volume Coordinate System 217 Let C be the center of the parallelepiped. By Proposition 4.66, OO1 2 = 4OC 2 = 4(1 2OY 2 + 1 2OY1 2 −1 4YY1 2). Then OO1 2 + YY1 2 = 4OX 2 + 2OY 2 + 4OZ 2 −2XY 2. (2) Similarly, we have OO1 2 + XX1 2 = 2OX 2 + 4OY 2 + 4OZ 2 −2ZY 2. (3) Solving the linear system (1), (2), (3), we have OO1 2 = 3OX 2 + 3OY 2 + 3OZ 2 −ZY 2 −ZX 2 −XY 2 = OX 2 + OY 2 + OZ 2 + PXOY + PXOZ + PYOZ. Proposition 4.97 Let O, W, U, and V be four free points. Then AB 2 = OU 2(xB −xA)2 + OV 2(yB −yA)2 + OW 2(zB −zA)2 + (yB −yA)(xB −xA)PUOV + (zB −zA)(yB −yA)PWOV + (zB −zA)(xB −xA)PWOU. O U V W A R M N P Q B L Figure 4-40 Proof. We form a parallelepiped AMLN − RPBQ such that AR ∥OW, AM ∥OU, and AN ∥OV. By Proposition 4.96, AB 2 = AM 2 + AN 2 + AR 2 +PRAN + PRAM + PNAM. By Lemma 4.49, AR 2 = OW 2( AR OW )2 = OW 2(VBOUVA VWOUV )2 = OW 2(zB −zA)2. AN 2 and AM 2 can be computed similarly. By Proposition 3.9, PNAM = AM OU AN OV PUOV = (yB −yA)(xB −xA)PUOV. We can compute PRAN and PRAM similarly. Once again with Algorithm 4.86, let E be an expression in volumes and Pythagoras dif-ferences of free points. Instead of using Lemma 4.85, we can use the following procedure to transform E into an expression in independent variables: if there are fewer than four points occurring in E then we need do nothing. Otherwise choose four free points O, W, U, and V from the points occurring in E and apply Propositions 4.94 and 4.97 to E to transform the 218 Chapter 4. Machine Proof in Solid Geometry volumes and Pythagoras diﬀerences into volume coordinates with respect to OWUV. Now the new E is an expression in volume coordinates of free points, OW 2, OU 2, OV 2, UV 2, and VOWUV. The only algebraic relation among these quantities is the Herron-Qin formula (Proposition 4.78). Substituting V2 OWUV into E, we obtain an expression in independent variables. Example 4.98 The above process of transforming an expression in volumes and Pythagoras diﬀerences of free points into an expression in free parameters becomes very simple when there exist exactly four free points O, W, U, and V in a geometry statement. In this case, we ﬁrst transform the square of the volume VOWUV into Pythagoras diﬀerences using the Herron-Qin formula (Proposition 4.78), and then express the Pythagoras diﬀerences in terms of the six square-distances, OW 2, OU 2, OV 2, UV 2, UW 2, and WV 2, which form a set of free parameters for this geometry statement. Exercises 4.99 1. If OV⊥OU, OV⊥OW, OW⊥OU, and OU 2 = OV 2 = OW 2 = 1, the volume coordinate system developed in this section becomes the standard Cartesian coordinate system. Prove the following formulas in the Cartesian coordinate system. 1. AB 2 = (xB −xA)2 + (yB −yA)2 + (zB −zA)2. 2. PABC = 2((xB −xA)(xB −xC) + (yB −yA)(yB −yC) + (zB −zA)(zB −zC)). 3. VABCD = 1 6 xA yA zA 1 xB yB zB 1 xC yC zC 1 xD yD zD 1 . 2. Prove the Herron-Qin formula for tetrahedra (see Propositions 4.78 and 4.79) using the three formulas in the preceding exercise. (First notice that VABCD = −1 6 xB −xA yB −yA zB −zA xC −xA yC −yA zC −zA xD −xA yD −yA zD −zA . Let M be the matrix in the above formula. Then V2 ABCD = 1 36\|M × M∗\| where M∗is the transpose of M.) Summary of Chapter 4 • Signed volumes and Pythagoras diﬀerences are used to describe some basic geometry relations in solid geometry: collinear, coplanar, parallel, perpendicular, and congru-ence of line segments. 4.6 Volume Coordinate System 219 1. Four points A, B, C, and D are coplanar iﬀVABCD = 0. 2. PQ ∥ABC iﬀVPABC = VQABC or equivalently iﬀVPABCQ = 0. 3. PQR ∥ABC iﬀVPABC = VQABC = VRABC. 4. PQ⊥AB iﬀPPAQB = PPAB −PQAB = 0. • We have the following formulas for the volumes of tetrahedra. 1. VABCD = 1 3hA,BCD\|S BCD\| where hA,BCD is the signed altitude from point A to plane BCD. 2. VABCD = VOWUV xA yA zA 1 xB yB zB 1 xC yC zC 1 xD yD zD 1 where xA, yA, zA are the volume coordinates of point A with respect to points O, W, U, and V. 3. (The Herron-Qin Formula) 144V2 P1P2P3P4 = 4r2 12r2 13r2 14 −r2 12P314 −r2 13P214 −r2 14P312 + P314P214P312 where rij = \|PiP j\|, Pijk = PPiPjPk. 4. (The Cayley-Menger Formula) 288V2 P1P2P3P4 = 0 r2 12 r2 13 r2 14 1 r2 21 0 r2 23 r2 24 1 r2 31 r2 32 0 r2 34 1 1 1 1 1 0 . • The following basic propositions are the basis of the volume method. 1. (The Co-vertex Theorem) Let ABC and DEF be two proper triangles in the same plane and T be a point not in the plane. Then we have VTABC VTDEF = S ABC S DEF . 2. (The Co-face Theorem) A line PQ and a plane ABC meet at M. If Q , M, we have PM QM = VPABC VQABC ; PM PQ = VPABC VPABCQ ; QM PQ = VQABC VPABCQ . 3. Let R be a point on line PQ. Then for any points A, B, and C, we have VRABC = PR PQ VQABC + RQ PQ VPABC. 220 Chapter 4. Machine Proof in Solid Geometry 4. Let R be a point in the plane PQS . Then for any points A, B, and C we have VRABC = S PQR S PQS VS ABC + S RQS S PQS VPABC + S PRS S PQS VQABC. 5. Let PQTS be a parallelogram. Then for points A, B, and C, we have VPABC + VT ABC = VQABC + VS ABC, or VPABCQ = VS ABCT. 6. Let triangle ABC be a parallel translation of triangle DEF. Then for points P and Q we have VPABC = VPDEFA and VPABCQ = VPDEFQ. Also notice that all the propositions on page 168 about Pythagoras diﬀerences are also valid if the points involved are in the space. • We present a mechanical proving method which can produce short and readable proofs for many constructive geometry statements in the space. Chapter 5 Vectors and Machine Proofs In Section 2.6, we mentioned that there are two approaches to deﬁning geometries: the geometric approach and the algebraic approach. In this chapter, we will show how to prove geometry theorems automatically in a geometry that is deﬁned using the algebraic approach. The modern source for aﬃne and metric geometries based on the algebraic approach is linear algebra or the theory of vector spaces (see [5, 16, 33]). In this approach, the metric is introduced by the inner product of vectors, while the areas and volumes are represented by the exterior product of vectors. It is interesting to note that two of the most important concepts in this linear algebra approach to geometry, the inner and exterior products of vectors, are essentially the same as the two basic geometry quantities used by us: the Pythagoras diﬀerence and the area (or volume). This strongly suggests that the method based on areas, volumes, and Pythagoras diﬀerences can be stated in the language of vectors. Moreover, the vector approach based on inner and exterior products has the advantage that it is easy to develop; it uses more geometry quantities such as the vector itself; and it covers more geometries such as the Minkowskian geometry. But on the other hand, the vector approach needs more algebraic prerequisites. Also, the proofs produced by the vector approach generally do not have such clear geometric meaning as do those produced by the volume-Pythagoras diﬀerence approach. 5.1 Metric Vector Spaces of Dimension Three Let E be a ﬁeld with characteristic diﬀerent from two. A vector space over E is a set V with two structures: • V × V −→V, denoted by (x, y) −→x + y and • E × V −→V, denoted by (α, x) −→αx which satisfy the following properties: 221 222 Chapter 4. Machine Proof in Solid Geometry V1 x + y = y + x (commutative law) V2 (x + y) + z = x + (y + z) (associate law) V3 There exists a zero-element o such that x+ o = x for every x ∈E. o is called the origin of V. V4 To every element x there exists an inverse element: −x such that x + (−x) = o. V5 (αβ)x = α(βx) V6 (α + β)x = αx + βx; α(x + y) = αx + αy (distributive laws) V7 1 · x = x. The elements of V are called vectors and are denoted by x, y, z, · · ·; the elements of E are called scalars and are denoted by α, β, a, b, . . .. Scalars are always written on the left of the vectors. The vector space V is called n-dimensional if there exist n elements e1, · · · , en in V such that • for any x ∈V there exist scalars α1, · · · , αn such that x = α1e1 + · · · + αnen and • if α1e1 + · · · + αnen = 0 then αi = 0, i = 1, · · · , n. The n elements e1, · · · , en satisfying the above property form a basis for the vector space V. If x = α1e1 + · · · + αnen, we say that x is a linear combination of the vectors e1, . . . , en and (α1, . . . , αn) are the coordinates of x with respect to the basis e1, · · · , en. Given an ordered basis for V, we can associate with each vector x the unique n-tuple of coordinates (α1, . . . , αn) such that x = α1e1 + · · · + αnen. This establishes a one-to-one correspondence between the vectors in V and the elements in En which is the Cartesian product of E with itself n times. It is easy to show that this correspondence preserves the structure of the vector spaces, i.e., it is an isomorphism between vector spaces. So we may safely assume that V is actually En. A nonempty subset W of a vector space V is called a subspace of V if the following conditions hold: 1. If x ∈W and y ∈W, then x + y ∈W. 2. If x ∈W, then αx ∈W for α ∈E. Let f1, · · · , fm be vectors in V. Then the set of all the vectors like m X i=1 αi fi, αi ∈E 5.1. Metric Vector Spaces of Dimension Three 223 is a subspace of V and is called the subspace generated by vectors f1, · · · , fm. This chapter uses some basic knowledge from linear algebra, e.g., the ﬁrst ﬁve chapters of . 5.1.1 Inner Products and Metric Vector Space In what follows, we assume that n = 3 and an ordered basis for V has been given. For x = (x1, x2, x3), y = (y1, y2, y3), and α ∈E, we thus have αx = (αx1, αx2, αx3) and x + y = (x1 + y1, x2 + y2, x3 + y3). Deﬁnition 5.1 An inner product on V is a map V × V −→E, denoted by (x, y) −→⟨x, y⟩ which satisﬁes the following properties I1 ⟨x, y⟩= ⟨y, x⟩. I2 ⟨αx + βy, z⟩= α⟨x, z⟩+ β⟨y, z⟩where α and β are scalars. Proposition 5.2 Let (e1, e2, e3) be a basis of V, x = x1e1 + x2e2 + x3e3, and y = y1y1 + y2y2 + y3y3. Show that ⟨x, y⟩= (x1, x2, x3)M           y1 y2 y3           where M = (⟨ei, ej⟩) is a symmetric matrix. Proof. ⟨x, y⟩= 3 X i, j=1 xiyj⟨ei, ej⟩= (x1, x2, x3)M           y1 y2 y3           where M = (⟨ei, ej⟩) is a symmetric matrix. Deﬁnition 5.3 A vector space with an inner product is called a metric vector space. Deﬁnition 5.4 Two vectors x and y are perpendicular if ⟨x, y⟩= 0. A metric vector space is called nonsingular if its origin is the only vector which is orthogonal to all vectors. 224 Chapter 4. Machine Proof in Solid Geometry Proposition 5.5 Metric vector space V is nonsingular if and only if \|M\| = det(M) , 0. Proof. If x = (x1, x2, x3) and y = (y1, y2, y3), ⟨x, y⟩= (x1, x2, x3)M           y1 y2 y3          . Then x is perpendicular to all vectors in V iﬀ (x1, x2, x3)M           y1 y2 y3          = 0 for all possible yi, i.e., iﬀ (x1, x2, x3)M = (0, 0, 0). The above linear system has nonzero solution iﬀ\|M\| = 0. A vector x is called isotropic if x is perpendicular to itself, or equivalently if ⟨x, x⟩= 0. The origin o is always isotropic. Even if V is nonsingular, there may be many nonzero isotropic vectors. This is seen by observing that a vector is isotropic iﬀits coordinates satisfy the equation n X i, j=1 mi, jxixj = 0 where M = (mi, j). The solutions for the above equation (if exist) consist of a cone which is referred as the light cone. The word light cone is originated from physics. More physics background can be found in . For x = (x1, x2, x3), let the square of x be x2 = ⟨x, x⟩. Suppose that V is a metric vector space, but that we only know the value of x2 for each vector x ∈V. Can we compute the inner product ⟨x, y⟩for all x and y? The answer is aﬃrmative. Proposition 5.6 In a metric vector space, the square function x2 determines the inner prod-uct completely. Proof. For x, y ∈V, by I1 and I2, (x + y)2 = x2 + 2⟨x, y⟩+ y2. Since E is not of characteristic 2, we have ⟨x, y⟩= 1 2(x2 + y2 −(x + y)2). 5.1. Metric Vector Spaces of Dimension Three 225 Corollary 5.7 (Pythagorean Theorem) x⊥y iﬀx2 + y2 −(x + y)2 = 0. Deﬁnition 5.8 A coordinate system e1, e2, e3 of V is called a rectangular coordinate system if ei⊥ej for j , i. For a rectangular basis e1, e2, e3, the matrix deﬁning the inner product is diagonal, i.e., M =           ⟨e1, e1⟩ ⟨e2, e2⟩ ⟨e3, e3⟩          . For x = (x1, x2, x3) and y = (y1, y2, y3), we have ⟨x, y⟩= ⟨e1, e1⟩x1y1 + ⟨e2, e2⟩x2y2 + ⟨e3, e3⟩x3y3. Proposition 5.9 A metric vector space V always has a rectangular coordinate system. Proof. We prove the proposition using induction on the dimension of V. If V is of dimen-sion one, any basis of it is a rectangular basis. Suppose that the result is true for all vector spaces with dimension less than n. Let V be a vector space of dimension n. If all the vectors in V are isotropic, by the Pythagorean theorem any basis is a rectangular basis. Otherwise, let e1, · · · , en be a basis of V such that e1 is a non-isotropic vector. Let e′ i = ei −⟨ei, e1⟩ e12 e1, i = 2, · · · , n. Then it is clear that e1, e′ 2, · · · , e′ n are also a basis of V and e1⊥e′ i, i = 2, · · ·, n. By the induction hypothesis, the vector space generated by e′ i, i = 2, · · · , n, has a rectangular basis fi, i = 2, · · · , n. Then it is easy to check that e1, f2, · · · , fn form a rectangular basis for V. Exercises 5.10 1. If E is the ﬁeld of real numbers and the inner product of x = (x1, x2, x3) and y = (y1, y2, y3) is deﬁned to be ⟨x, y⟩= x1y1 + x2y2 + x3y3, the resulting space is the Euclidean space of dimension three. Show that the Euclidean space as deﬁned above is a nonsingular metric vector space satisfying I3 ⟨x, x⟩≥0 and ⟨x, x⟩= 0 iﬀx = (0, 0, 0). 2. The n-dimensional (n ≤3) Minkowskian space is a metric vector space whose inner product for x = (x1, . . . , xn) and y = (y1, . . . , yn) is ⟨x, y⟩= x1y1 + . . . + xn−1yn−1 −xnyn. Show that the Minkowskian space is a nonsingular metric vector space in which there exist nonzero isotropic vectors. 226 Chapter 4. Machine Proof in Solid Geometry 3. Show that if E is the ﬁeld of real numbers and n = 3, every nonsingular metric vector space has a coordinate system such that its matrix is one of the following form. M1 =           1 0 0 0 1 0 0 0 1          , M2 =           1 0 0 0 1 0 0 0 −1           M3 =           1 0 0 0 −1 0 0 0 −1          , M4 =           −1 0 0 0 −1 0 0 0 −1          . The matrices M1 and M2 determine, respectively, the Euclidean and the Minkowskian spaces. We call the geometries determined by M4 and M3 the negative Euclidean space and the negative Minkowskian space respectively. 4. Let (e1, ..., en) and (f1, ..., fn) be two diﬀerent bases for V. Then there is a nonsingular matrix P such that (f1, ..., fn) = (e1, ..., en)P. Let M and M′ be the matrices of the inner products corresponding to the bases (e1, ..., en) and (f1, ..., fn). Show that M′ = P∗MP where P∗is the transpose of P. The two matrices in the above proposition are called congruent. Thus two matrices are congruent iﬀthey represent the same metric of V relative to dif-ferent coordinates systems. Therefore, the study of the metric vector space is equivalent to the study of the symmetric matrices under the equivalent relation of congruence. 5. In the algebraic language, Proposition 5.9 is equivalent to the following fact. Let G be an n × n symmetric matrix. Then there exists an n × n nonsingular matrix P such that P∗GP is a diagonal matrix. Prove the above fact directly. 5.1.2 Exterior Products in Metric Vector Space In what follows, we always assume that V is a nonsingular metric vector space with a rectangular basis (e1, e2, e3). Thus the matrix that deﬁnes the inner product is M =           ⟨e1, e1⟩ ⟨e2, e2⟩ ⟨e3, e3⟩          . Deﬁnition 5.11 An exterior product on V is a map V × V −→V, denoted by (x, y) −→[x, y] which satisﬁes the following properties 5.1. Metric Vector Spaces of Dimension Three 227 E1 [x, y] = −[y, x]. E2 [αx + βy, z] = α[x, z] + β[y, z] where α and β are scalars. E3 x⊥[x, y]. Note that property E3 is not in the deﬁnition for the exterior product in the general case. We add it to make the relation between the inner and exterior products simple. From E1 and the fact that E is not of characteristic two, we have [x, x] = 0. Proposition 5.12 Let (e1, e2, e3) be a rectangular basis of a nonsingular metric vector space V. Then [e1, e2] = α ⟨e3, e3⟩e3, [e2, e3] = α ⟨e1, e1⟩e1, [e3, e1] = α ⟨e2, e2⟩e2 where α = ⟨e1, [e2, e3]⟩= ⟨e2, [e3, e1]⟩= ⟨e3, [e1, e2]⟩. Proof. Since [e1, e2]⊥e1 and [e1, e2]⊥e2, it is clear that [e1, e2] = s1e3. Thus s1 = ⟨e3,[e1,e2]⟩ ⟨e3,e3⟩ . Similarly [e2, e3] = s2e1, [e3, e1] = s3e2 where s2 = ⟨e1,[e2,e3]⟩ ⟨e1,e1⟩ , s3 = ⟨e2,[e3,e1]⟩ ⟨e2,e2⟩ . Adding the above two equations, we have [e2 −e1, e3] = s2e1 + s3e2. Taking the exterior products of e2 −e1 and the vectors on both sides of the above equation, we have s3⟨e2, e2⟩= s2⟨e1, e1⟩, i.e., ⟨e2, [e3, e1]⟩= ⟨e1, [e2, e3]⟩. Similarly, we can prove that ⟨e3, [e1, e2]⟩= ⟨e2, [e3, e1]⟩. Remark 5.13 The constant α = ⟨e1, [e2, e3]⟩is a basic quantity related to the exterior prod-uct. We always assume that α , 0. Proposition 5.14 Let (e1, e2, e3) be a rectangular basis of a nonsingular metric vector space V, x = x1e1 + x2e2 + x3e3, and y = y1e1 + y2e2 + y3e3. Then [x, y] = α( 1 m1 x2 x3 y2 y3 , 1 m2 x3 x1 y3 y1 , 1 m3 x1 x2 y1 y2 ) where α = ⟨e1, [e2, e3]⟩, m1 = ⟨e1, e1⟩, m2 = ⟨e2, e2⟩, and m3 = ⟨e3, e3⟩. Proof. By E1 and E2, [x, y] = 3 X i, j=1 xiyj[ei, ej] = x2 x3 y2 y3 [e2, e3] + x3 x1 y3 y1 [e3, e1] + x1 x2 y1 y2 [e1, e2]. Now the result follows immediately from Proposition 5.12. 228 Chapter 4. Machine Proof in Solid Geometry Proposition 5.15 If the metric vector space is not singular, then x = αy iﬀ[x, y] = 0. Proof. By Proposition 5.9, we can chose a rectangular basis for V. If x = αy then [x, y] = α[y, y] = 0. Conversely, let us assume [x, y] = 0. By Proposition 5.14,we have x1 y1 = x2 y2 = x3 y3 = λ for a scalar λ. Thus x = λy. Deﬁnition 5.16 The triple scalar product for three vectors x, y, and z in V is deﬁned as follows. (x, y, z) = ⟨[x, y], z⟩. Proposition 5.17 Let x = (x1, x2, x3),y = (y1, y2, y3), and z = (z1, z2, z3). We have (x, y, z) = ⟨e1, [e2, e3]⟩ x1 x2 x3 y1 y2 y3 z1 z2 z3 . Proof. Since V has a rectangular basis, this is a direct consequence of Proposition 5.14. We thus have T1 (x, y, z) = (y, z, x) = (z, x, y) = −(x, z, y) = −(z, y, x) = −(y, x, z). T2 In a nonsingular metric space, (x, y, z) = 0 iﬀvectors x, y, and z are coplanar, i.e., iﬀ there exist scalars α1, α2, α3 not all zero such that α1x + α2y + α3z = 0. Proposition 5.18 • (The Lagrange Identity) ⟨[x, y], [u, v]⟩= α(⟨x, u⟩⟨y, v⟩−⟨x, v⟩⟨y, u⟩). • [[x, y], z] = α(⟨x, z⟩y −⟨y, z⟩x) where α = ⟨e1,[e2,e3]⟩2 ⟨e1,e1⟩⟨e2,e2⟩⟨e3,e3⟩. Proof. The two formulas can be obtained by direct computation. We leave them as exer-cises. Exercise 5.19 Show that = α(⟨r4, [r1, r2]⟩r3−⟨r3, [r1, r2]⟩r4) where α is the same as in Proposition 5.18. 5.2. Metric Geometries of Dimension Three 229 5.2 The Solid Metric Geometry Let E be a ﬁeld with characteristic diﬀerent from two. The vector space E3 is also called the aﬃne space associated with ﬁeld E. Deﬁnition 5.20 A non-singular metric vector space E3 is called a solid metric geometry. As usual, elements in E3 are called points. Let A and B be two points. Then the line passing through A and B is the set {αA + βB\| α + β = 1}. The plane passing through three points A, B, and C is the set {αA + βB + γC\| α + β + γ = 1}. Two points A and B in E3 determine a new vector, − − → AB = B −A A being the origin and B being the endpoint. Thus two vectors − − → AB and − − → PQ are equal if and only if A + Q = P + B. Let O be the origin of V. For any point A, let − → A = − − → OA. Thus we also have − − → AB = − → B −− → A. It is easy to show that line AB can also be written as follows {A + β− − → AB\| β ∈E}. Line AB is called isotropic if − − → AB 2 = 0. Similarly, the plane ABC can be written as {A + β− − → AB + γ− − → AC\| β, γ ∈E}. Line AB is called parallel to line PQ if there is a scalar λ such that − − → AB = λ− − → PQ. If − − → AB = λ− − → PQ, we say that the ratio of the parallel line segments AB and PQ is λ, i.e., AB PQ = λ. O U V W A R M N P Q B L Figure 5-2 A B C P Q Figure 5-1 230 Chapter 4. Machine Proof in Solid Geometry To see the geometric meaning of the addition of two vectors − − → AB and − − → PQ. Let C be a point such that − − → BC = − − → PQ. Then (Figure 5-1) − − → AB + − − → PQ = − − → AB + − − → BC = − − → AC. We will now give a geometric interpretation of the coordinates of vectors with respect to a basis. Let O, W, U, and V be four points not in the same plane. For any vector − − → AB, we form a parallepiped AMLN −RPBQ (Figure 5-2) such that AR ∥OW, AM ∥OU, and AN ∥OV. Then − − → AB = − − → AR + − − → AM + − − → AN = AR OW − − → OW + AM OU − − → OU + AN OV − − → OV, i.e., − − → OW, − − → OU, and − − → OV form a basis for E3 and the coordinates of − − → AB with respect to this basis are ( AR OW, AM OU , AN OV ). Exercise 5.21 Show that point Y is on line AB iﬀthere is a scalar α such that − − → AY = α− − → AB; Point Y is on plane LMN iﬀthere are two scalars α and β such that − → LY = α− − → LM + β− − → LN. 5.2.1 Inner Products and Exterior Products The inner product of vectors − − → AB and − − → CD satisﬁes 1. ⟨− − → AB, − − → CD⟩= 0 if and only if AB⊥CD. 2. ⟨− − → AB, − − → CD⟩= ⟨− − → CD, − − → AB⟩, 3. ⟨α− → A + β− → B, − − → CD⟩= α⟨− → A, − − → CD⟩+ β⟨− → B, − − → CD⟩where α and β are scalars. The square distance between two points A and B, or the square length of the vector − − → AB, is deﬁned to be AB2 = − − → AB 2 = ⟨− − → AB, − − → AB⟩. By Proposition 5.6, 2⟨− − → AB, − − → BC⟩= AC2 −AB2 −BC2 = −PABC. Then it is easy to check that PABCD = −2⟨− − → AC, − − → BD⟩. Proposition 5.22 (Pythagorean Theorem) For any points A, B, C, and D 5.2. Metric Geometries of Dimension Three 231 • AB⊥BC iﬀAB2 + BC2 −AC2 = 0. • AB⊥CD iﬀPACBD = AC2 −CB2 + BD2 −AD2 = 0. If four points A, B, C, and D are collinear or AB ∥CD, then the product of the oriented segments is AB · CD = ⟨− − → AB, − − → CD⟩ and the ratio of the oriented segments is AB CD = ⟨− − → AB, − − → CD⟩ ⟨− − → CD, − − → CD⟩ . The exterior product [− − → AB, − − → CD] of − − → AB and − − → CD satisﬁes the following properties 1. [− − → AB, − − → CD] = 0 iﬀAB ∥CD. 2. [− − → AB, − − → CD] = −[− − → CD, − − → AB]. 3. [α− → A + β− → B, − − → CD] = α[− → A, − − → CD] + β[− → B, − − → CD] where α and β are scalars. By Lagrange’s identity, [− − → AB, − − → AC] 2 = α(AB2 · AC2 −⟨− − → AB, − − → AC⟩2) = α 4(4AB2 · AC2 −P2 BAC) where α is the constant deﬁned in Proposition 5.18. Comparing with the Herron-Qin for-mula on page 108, we see that the length of [− − → AB, − − → AC] is proportional to the area of triangle ABC. Remark 5.23 We can determine the exact relation between the area and the exterior product as follows. In Euclidean geometry, α = 1. By the Herron-Qin formula, [− − → AB, − − → AC] 2 = 4S 2 ABC. Thus [− − → AB, − − → AC] is a vector − − → AD such that − − → AD is perpendicular to the plane ABC, and pointed in such direction as to make (− − → AB, − − → AC, − − → AD) a right handed triple and \|AD\| = 2\|S ABC\|. We thus deﬁne the signed area of triangle ABC to be a quantity with the same sign of [− − → AB, − − → AC] and S 2 ABC = 1 4[− − → AB, − − → AC] 2. Then Heron-Qin’s formula in any metric geometry is S 2 ABC = α 16(4AB2 · AC2 −P2 BAC). 232 Chapter 4. Machine Proof in Solid Geometry Two planes ABC and PQR are parallel if [− − → AB, − − → AC] ∥[− − → PQ, − − → PR]. Let λ be the scalar ratio of these two parallel vectors, i.e., [− − → AB, − − → AC] = λ[− − → PQ, − − → PR]. Then λ = S ABC S PQR . We thus have λ = S ABC S PQR = ⟨[− − → AB, − − → AC], [− − → PQ, − − → PR]⟩ ⟨[− − → PQ, − − → PR], [− − → PQ, − − → PR]⟩ . The volume of the tetrahedron ABCD is deﬁned to be one sixth of the triple scalar product: VABCD = 1 6⟨− − → AD, [− − → AB, − − → AC]⟩. As a consequence VABCD = 1 6(⟨− → D, [− → A, − → B]⟩+ ⟨− → D, [− → B, − → C]⟩+ ⟨− → D, [− → C, − → A]⟩−⟨− → A, [− → B, − → C]⟩). Starting from several points in the space, we can form vectors and inner and exterior products of these vectors. Since exterior products of vectors are still vectors, we can further form the inner and exterior products of these new vectors. Expressions thus obtained are called recursive expressions in inner and exterior products of vectors. It is clear that a recursive expression in inner and exterior products of vectors can be a scalar or a vector. A vector like − − → AB for points A and B is called a simple vector. Proposition 5.24 Any recursive expression in inner and exterior products of vectors can be represented as a polynomial in inner products of simple vectors, exterior products of simple vectors, and triple scalar products of simple vectors. Proof. By Proposition 5.18, for any vectors r1, r2, r3 and r4 1. [[r1, r2], r3] = α(⟨r1, r3⟩r2 −⟨r2, r3⟩r1). 2. (The Lagrange Identity) ⟨[r1, r2], [r3, r4]⟩= α(⟨r1, r3⟩⟨r2, r4⟩−⟨r1, r4⟩⟨r2, r3⟩). By repeated use of the above two identities, any recursive expression in inner and exterior products of vectors can be represented as a polynomial of inner products of simple vectors, exterior products of simple vectors, and triple scalar products of simple vectors. Exercises 5.25 1. Show that with the above deﬁnition for the ratio of lengths, signed areas, signed vol-umes, and the Pythagoras diﬀerences, Axioms A.1-A.6, S.1-S.5, and the properties of the Pythagoras diﬀerence are true. 5.2. Metric Geometries of Dimension Three 233 2. Prove the following formula of the distance from a point A to a line PQ d(A, PQ)2 = AP2 −⟨− − → PA, − − → PQ⟩2 PQ2 . 3. Prove the following formula of the distance from a point A to a plane LMN. d2 A,LMN = ⟨− → LA, [− − → LM, − − → LN]⟩2 [− − → LM, − − → LN] 2 . 4. Prove the following formula of the distance between two skew lines UV and PQ d2 UV,PQ = 9(⟨− − → UV, [− − → UP, − − → UQ]⟩)2 [− − → PQ, − − → UV] 2 . 5.2.2 Constructive Geometry Statements The constructive statement deﬁned in Section 4.2 can be generalized by considering more constructions and more geometry quantities. Deﬁnition 5.26 By geometric quantities we mean vectors, the inner or exterior products of vectors, or the quantities which can be represented by the inner and exterior products of vectors. With the geometry concepts introduced in the preceding subsection, constructions S1– S7 on page 181 are still meaningful in our metric geometry, except for constructions S6 and S7 whose ndg conditions need modiﬁcation. We will also introduce a new construction S8. S6 (FOOT2LINE Y P U V) Point Y is the foot from point P to line UV. Point Y is a ﬁxed point. The ndg condition is − − → UV 2 , 0. Notice that in the general metric geometry − − → UV 2 , 0 is not equivalent to U , V. S7 (FOOT2PLANE Y P L M N) Point Y is the foot of the perpendicular from point P to plane LMN. The nondegenerate condition is [− − → LM, − − → LN]2 , 0. S8 (SRATIO A L M N r) Take a point A such that − → LA = r[− − → LM, − − → LN], where r can be a rational number, a rational expression in geometric quantities, or variables. If r is a ﬁxed quantity, A is a ﬁxed point; otherwise, A has one degree of freedom. The ndg condition is [− − → LM, − − → LN]2 , 0. 234 Chapter 4. Machine Proof in Solid Geometry Two basic geometric relations, parallel and perpendicular, can be easily described by the exterior and inner products. For instance, to represent AB ∥CD we need two equations VABCD = 0 and S ACD = S BCD. But using exterior product, we need only one equation [− − → AB, − − → CD] = 0. Proposition 5.27 We have 1. AB ⊥CD ⇐⇒⟨− − → AB, − − → CD⟩= 0. 2. AB ∥CD ⇐⇒[− − → AB, − − → CD] = 0. 3. A, B, and C are collinear ⇐⇒[− − → AB, − − → AC] = 0. 4. AB ⊥PQR ⇐⇒[− − → AB, [− − → PQ, − − → PR]] = 0. 5. AB ∥PQR ⇐⇒⟨− − → AB, [− − → PQ, − − → PR]⟩= 0. 6. ABC ⊥PQR ⇐⇒⟨[− − → AB, − − → AC], [− − → PQ, − − → PR]⟩= 0. 7. ABC ∥PQR ⇐⇒ = 0. 8. A, B, C, and D are coplanar ⇐⇒⟨− − → DA, [− − → AB, − − → AC]⟩= 0. Proof. The ﬁrst two cases are from the deﬁnition. The other cases are consequences of the ﬁrst two cases. Example 5.28 (The Centroid Theorem for Tetrahedra) Let G be the centroid of tetrahedra ABCD. Show that − → G = − → A+− → B+− → C +− → D 4 . This example can be described constructively as follows. 5.3. Machine Proof by Vector Calculation 235 A B C D S Z Y G Figure 5-3 Constructive description ((POINTS A B C D) (MIDPOINT S B C) (LRATIO Z A S 2/3) (LRATIO Y D S 2/3) (INTER G (LINE D Z) (LINE A Y)) (− → G = − → A +− → B +− → C +− → D 4 ) ) The ndg conditions: B , C, A , S , D , S , and DZ ∦AY. 5.3 Machine Proof by Vector Calculation As before, we will give methods of eliminating points from the geometry quantities. 5.3.1 Eliminating Points From Vectors We ﬁrst consider how to eliminate points from vectors. Proposition 5.29 Let R be a point on line PQ (P , Q). Then − → R = PR PQ − → Q + RQ PQ − → P. Proof. We have PR PQ − → Q + RQ PQ − → P = PR PQ(− → Q −− → P) + − → P = PR PQ − − → PQ + − → P = − − → PR + − → P = − → R. Lemma 5.30 Let Y be introduced by (PRATIO Y W U V r). Then − → Y = − → W + r(− → V −− → U) Proof. Since − − → WY = − → W −− → Y = r− − → UV, we have − → Y = − → W −r− − → UV. Lemma 5.31 Let Y be introduced by (ARATIO Y L M N r1 r2 r3). Then − → Y = r1− → L + r2− → M + r3− → N. Proof. Since Y is in the plane LMN, we have − → LY = c1− − → LM + c2− − → LN 236 Chapter 4. Machine Proof in Solid Geometry where c1 and c2 are some scalars. Then [− − → LN, − → LY] = c1[− − → LN, − − → LM] + c2[− − → LN, − − → LN] = c1[− − → LN, − − → LM]. Thus c1 = S LNY S LNM = r2. Similarly c2 = r3. Thus − → Y = − → L + r2(− → M −− → L) + r3(− → N −− → L) = r1− → L + r2− → M + r3− → N. Lemma 5.32 Let Y be introduced by (INTER Y (LINE U V) (LINE P Q)). Then − → Y = S UPQ S UPVQ − → V −S VPQ S UPVQ − → U. Proof. By Proposition 5.29, − → Y = UY UV − → V −VY UV − → U. Let r = UY UV . Then − − → UY = r− − → UV and [− − → UY, − − → PQ] = r[− − → UV, − − → PQ]. Since [− − → UY, − − → PQ] = [− − → UQ, − − → PQ] + [− − → QY, − − → PQ] = [− − → UQ, − − → PQ], we have r = S UPQ S UPVQ . Similarly VY UV = S VPQ S UPVQ . Lemma 5.33 Let Y be introduced by (INTER Y (LINE U V) (PLANE L M N)). Then − → Y = VULMN VULMNV − → V −VVLMN VULMNV − → U. Proof. By Proposition 5.29, − → Y = UY UV − → V + YV UV − → U. (1) Let r = UY UV . Then − − → UY = r− − → UV and [− − → UY, [− − → LM, − − → LN]] = r[− − → UV, [− − → LM, − − → LN]]. Since [− − → UY, [− − → LM, − − → LN]] = [− − → UL, [− − → LM, − − → LN]] + [− → LY, [− − → LM, − − → LN]] = VULMN, we have r = VULMN VULMNV . Similarly VY UV = VVLMN VULMNV . Lemma 5.34 Let Y be introduced by (FOOT2LINE Y P U V). Then − → Y = PPUV PUVU − → V + PPVU PUVU − → U. 5.3. Machine Proof by Vector Calculation 237 Proof. By Proposition 5.29, − → Y = UY UV − → V −VY UV − → U. Let r = UY UV, or − − → UY = r− − → UV. We have r⟨− − → UV, − − → UV⟩= ⟨− − → UY, − − → UV⟩= ⟨− − → UP, − − → UV⟩+ ⟨− − → PY, − − → UV⟩= ⟨− − → UP, − − → UV⟩. Then r = PPUV PUVU . Similarly YV UV = PPVU PUVU . Lemma 5.35 Let Y be introduced by (SRATIO Y L M N r). Then − → Y = − → L + r[− − → LM, − − → LN]. Proof. This is the deﬁnition of the construction SRATIO. Lemma 5.36 Let Y be introduced by (FOOT2PLANE Y P L M N). Then − → Y = − → P + 6VPLMN [− − → LM, − − → LN]2[− − → LM, − − → LN]. Proof. Let − − → PY = r[− − → LM, − − → LN]. Then ⟨− → LP, − − → PY⟩= r⟨− → LP, [− − → LM, − − → LN]⟩= 6rVLMNP. ⟨− → LP, − − → PY⟩= −PY2 = − 36V2 PLMN [− − → LM,− − → LN]2. Thus r = 6VPLMN [− − → LM,− − → LN]2. Example 5.37 Let Y be introduced by (INTER Y (PLINE W U V) (PLINE R P Q)). Show that − → Y = − → W + S WPRQ S UPVQ (− → V −− → U). Proof. Take points X and S such that WX UV = 1 and RS PQ = 1. By Proposition 5.29, − → Y = r− → X + (1 −r)− → W = r(− → X −− → W) + − → W = r(− → V −− → U) + − → W where r = WY WX = S WPRQ S UPVQ . Example 5.38 Continue from Example 5.28. We can actually derive the result of this exam-ple without knowing it previously. Constructive description ((POINTS A B C D) (MIDPOINT S B C) (LRATIO Z A S 2/3) (LRATIO Y D S 2/3) (INTER G (LINE D Z) (LINE A Y)) (− → G ) ) The machine proof. − → G G = −− → Z ·S ADY+− → D ·S AZY −S ADYZ Y = 2 3− → Z ·S ADS + 1 3− → D ·S ADZ 2 3 S DSZ+S ADZ Z = 4 3− → S ·S ADS + 2 3− → D ·S ADS + 2 3− → A ·S ADS 8 3 S ADS simplify = 1 4(2− → S +− → D +− → A ) S = − → D +− → C +− → B +− → A 4 The eliminants − → G G =− → Z ·S ADY−− → D ·S AZY S ADYZ S ADYZ Y =1 3(2S DSZ+3S ADZ) S AZY Y = −1 3(S ADZ) S ADY Y =2 3(S ADS) S DSZ Z = 1 3(S ADS) S ADZ Z =2 3(S ADS) − → Z Z =1 3(2− → S +− → A ) − → S S =1 2(− → C +− → B ) 238 Chapter 4. Machine Proof in Solid Geometry 5.3.2 Eliminating Points from Inner and Exterior Products Let Y be introduced by one of the constructions S1–S8. By Lemmas 5.30–5.36, (I) − → Y = αr1 + βr2 for vectors r1 and r2 and scalars α and β. To eliminate Y from the inner product, let us note that ⟨− − → AB, − − → CY⟩= ⟨− → B, − → Y ⟩+ ⟨− → A, − → C⟩−⟨− → B, − → C⟩−⟨− → A, − → Y ⟩. Then we need only to consider how to eliminate Y from ⟨− → A, − → Y ⟩and ⟨− → Y , − → Y ⟩. If A , Y, ⟨− → A, − → Y ⟩= α⟨− → A, r1⟩+ β⟨− → A, r2⟩. For ⟨− → Y , − → Y ⟩, we have ⟨− → Y , − → Y ⟩ = < αr1 + βr2, αr1 + βr2 >= α2⟨r1, r1⟩+ β2⟨r2, r2⟩+ 2αβ⟨r1, r2⟩. To eliminate point Y from the exterior product [− − → AB, − − → CY], let us note that [− − → AB, − − → CY] = [− → A, − → C] + [− → B, − → Y ] −[− → A, − → Y ] −[− → B, − → C]. If A = Y we have [− → A, − → Y ] = 0; otherwise we have [− → A, − → Y ] = α[− → A, r1] + β[− → A, r2]. Since other geometry quantities can always be represented as a rational expression in inner and exterior products, we can eliminate points introduced by constructions S1–S8 from them. Here are some examples. Example 5.39 Let Y be introduced by (SRATIO Y L M N r). By Lemma 5.35, we have VYBCD = ⟨− − → BY, [− − → BC, − − → BD]⟩= VLBCD −r⟨[− − → LM, − − → LN], [− − → BC, − − → BD]⟩ ⟨− − → YB, − − → CD⟩ = ⟨− → B, − − → CD⟩−⟨− → Y , − − → CD⟩ = ⟨− → LB, − − → CD⟩−r⟨− − → CD, [− − → LM, − − → LN]⟩. ⟨− − → YB, − − → YC⟩ = ⟨− → B, − → C⟩+ ⟨− → Y , − → Y ⟩−⟨− → Y , − → B⟩−⟨− → Y , − → C⟩ = ⟨− → LB, − − → LC⟩+ r⟨− → BL + − − → CL, [− − → LM, − − → LN]⟩+ r2[− − → LM, − − → LN] 2 [− − → YB, − − → CD] = [− → B, − − → CD] −[− → Y , − − → CD] = [− → LB, − − → CD] + r[− − → CD, [− − → LM, − − → LN]]. 5.3. Machine Proof by Vector Calculation 239 Example 5.40 Let Y be introduced by (FOOT2PLANE Y P L M N). By Lemma 5.36, VABCD = VPBCD + r⟨[− − → PM, − − → PN], [− − → BC, − − → BD]⟩. ⟨− − → AB, − − → CD⟩ = ⟨− − → PB, − − → CD⟩−r⟨− − → CD, [− − → PM, − − → PN]⟩. ⟨− − → AB, − − → AC⟩ = ⟨− − → PB, − − → PC⟩+ r⟨− − → BP + − − → CP, [− − → PM, − − → PN]⟩+ r2[− − → PM, − − → PN] 2 where r = 6VPLMN [− − → LM,− − → LN]2 . To eliminate points introduced by constructions S4, S5, and S6, we do not need to break the inner and exterior products into the sum of several components. In these three constructions, Y is always on a line UV. Remember that a geometry quantity G(Y) is called a linear quantity of point Y if G(Y) = UY UV G(V) + YV UV G(U). A geometry quantity G(Y) is called a quadratic geometry quantity of point Y if G(Y) = UY UV G(V) + YV UV G(U) −UY UV YV UV UV 2. Example 5.41 Show that − − → YB, [− − → YB, − − → CD], and ⟨− − → YB, − − → CD⟩are all linear in Y; and ⟨− − → YB, − − → YC⟩is quadratic in Y. Proof. By Proposition 5.29, − → Y = UY UV − → V + YV UV − → U. − − → YB = − → B −− → Y = − → B −UY UV − → V −YV UV − → U = UY UV (− → B −− → V) + YV UV (− → B −− → U) = UY UV − − → VB + YV UV − − → UB Now it is clear that [− − → YB, − − → CD] and ⟨− − → YB, − − → CD⟩are also linear in Y. For G(Y) = ⟨− − → YB, − − → YC⟩, let r1 = UY UV and r2 = YV UV. Then G(Y) = ⟨r1− − → VB + r2− − → UB, r1− − → VC + r2− − → UC⟩ = r2 1G(V) + r2 2G(U) + r1r2(⟨− − → VB, − − → UC⟩+ ⟨− − → UB, − − → VC⟩) = r1(r1 + r2)G(V) + r2(r1 + r2)G(U) −r1r2⟨− − → UV, − − → UV⟩ = r1G(V) + r2G(U) −r1r2⟨− − → UV, − − → UV⟩. Therefore to eliminate point Y from the geometry quantities, we need only to ﬁnd the position ratios of Y with UV, which has been done in Lemmas 5.32, 5.33, and 5.34. 240 Chapter 4. Machine Proof in Solid Geometry 5.3.3 The Algorithm Algorithm 5.42 (VECTOR) INPUT: S = (C1, C2, . . . , Ck, (E, F)) is a constructive geometry statement. OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof for S . S1. For i = k, · · · , 1, do S2, S3, S4 and ﬁnally do S5. S2. Check whether the nondegenerate conditions of Ci are satisﬁed. The nondegenerate conditions of a statement have ﬁve forms: A , B, AB2 , 0, PQ ∦UV, PQ ∦WUV, and [− − → LM, − − → LN]2 , 0. For the ﬁrst case, we check whether − → A = − → B. For the second case, we check whether ⟨− − → AB, − − → AB⟩= 0. For the third case, we check whether [− − → PQ, − − → UV] = 0. For the fourth case, we check whether VPWUV = VQWUV. For the ﬁfth case, we check whether [− − → LM, − − → LN]2 = 0. If one of the nondegenerate conditions of a geometry statement is not satisﬁed, the statement is trivially true. The algorithm terminates. S3. Let G1, · · · , Gs be the geometric quantities occurring in E and F. For j = 1, · · ·, s do S4 S4. Let H j be the result obtained by eliminating the point introduced by construction Ci from G j using the lemmas in this section, and replace G j by H j in E and F to obtain the new E and F. S5. Now there are only free points left. E and F are rational expressions in indeterminates, inner and exterior products of free points. Replacing the inner and exterior products by their coordinate expressions, we obtain E′ and F′. Then if E′ = F′, S is true under the nondegenerate conditions. Otherwise S is false. In the above algorithm, we represent the ratio of lengths and the ratio of areas as ex-pressions of inner and exterior products, and then eliminate points from the inner and ex-terior products. In order to obtain short proofs, we may also use the lemmas in Section 4.3 to eliminate points directly from the ratios of lengths or areas. We actually use a hy-brid method: inner products, exterior products, length ratios, area ratios, volumes, and the Pythagoras diﬀerences are all used in the proof. Remark 5.43 Since the inner product, the exterior product, and the triple scalar product are proportional with the Pythagoras diﬀerence, the area, and the volume, the volume (area) and the Pythagoras diﬀerence method developed in Chapters 3 and 4 are valid for construc-tive geometry statements in metric geometries associated with any ﬁeld with characteristic diﬀerent from 2. Thus our method works not only for Euclidean geometry but also for non-Euclidean geometries such as the Minkowskian geometry. 5.3. Machine Proof by Vector Calculation 241 The vector approach has the advantages that it is easy to develop, and more geometry quantities such as the vector itself can be used. But the proofs produced by the vector approach generally do not have the clear geometric meaning of those produced by the volume-Pythagoras diﬀerence approach. Notice that the elimination results (Lemmas 5.30-5.34) for points introduced by S1–S6 are exactly the same for all metric geometries. From this, one might wonder is there any connection among these metric geometries? But at the last step of the algorithm, we need to replace the inner and exterior products of free points by their coordinate expressions. This step depends on the speciﬁc geometries. Thus in order to obtain some meta theorems for diﬀerent geometries, we need to limit the class of geometry statements. Deﬁnition 5.44 A constructive geometry statement is called pure constructive if it can be described by constructions S1-S7 and its conclusion can be one of the geometry relations in Proposition 5.27. We further assume that the ratio r in the ratio constructions PRATIO and ARATIO can only be scalars or variables. In the predicate form for a pure constructive geometry statement, there are only aﬃne in-variants like the ratio of parallel line segments and geometry predicates like COLL, PRLL, PERP, etc. Proposition 5.45 Let G1 and G2 be two geometries over the same base ﬁeld E. Then a pure constructive geometry statement is true in G1 iﬀit is true in geometry G2. Proof. By Proposition 5.9, we can assume that the matrices for the inner products of G1 and G2 are M1 =           a1 a2 a3          , M2 =           b1 b2 b3          . Let x, y, z and x′, y′, z′ be vectors in G1 and G2 with the same coordinates respectively. Any geometry predicate can be represented by the following three quantities. ⟨x, y⟩= a1x1y1 + a2x2y2 + a3x3y3 in G1. [x, y] = α1(a2a3(x2y2 −x3y2), a1a3(x3y1 −x1x3), a1a2(x1y2 −x2x1)) in G1. where α1 is the constant in Proposition 5.18 for G1. (x, y, z) = α1 x1 x2 x3 y1 y2 y3 z1 z2 z3 . in G1. ⟨x′, y′⟩= b1x1y1 + b2x2y2 + b3x3y3 in G2. [x′, y′] = α2(b2b3(x2y2 −x3y2), b1b3(x3y1 −x1x3), b1b2(x1y2 −x2x1)) in G2. where α2 is the constant in Proposition 5.18 for G2. 242 Chapter 4. Machine Proof in Solid Geometry (x′, y′, z′) = α2 x1 x2 x3 y1 y2 y3 z1 z2 z3 . in G2. Let F be the algebraic closure of ﬁeld E. Consider the following automorphism of the polynomial ring F [x1, x2, x3, ..., z1, z2, z3] TR : (x1, x2, x3, ..., z3) −→( ra1 b1 x1, ra2 b2 x2, ra3 b3 x3, ..., ra3 b3 z3). We have TR(⟨x′, y′⟩) = ⟨x, y⟩, TR([x′, y′]) = 0 ⇐⇒[x, y] = 0, TR(x′, y′, z′) = 0 ⇐⇒(x, y, z) = 0. Also note that an aﬃne invariant is not changed under TR. Since each pure constructive statement can be described by aﬃne invariants and the above geometry quantities, it is clear that a pure constructive geometric statement is true in G1 iﬀit is true in G2. 5.4 Machine Proof in Metric Plane Geometries Since a metric plane can be seen as a subset of a metric space, the method presented in the preceding section is also valid for plane metric geometries. For an independent study of this topic, see . But for metric plane geometry, the method can be greatly simpliﬁed. Without loss of generality, we assume that the metric plane consists of all the points x = (x1, x2, x3) such that x3 = 0. Let M =           a1 0 0 0 a2 0 0 0 a3           be the matrix deﬁning the inner product. Then for vectors x = (x1, x2, 0), y = (y1, y2, 0), and z = (z1, z2, 0) in the plane, we have ⟨x, y⟩= a1x1y1 + a2x2y2. [x, y] = (0, 0, αa1a2(x1y2 −x2y1)). α is a constant. (x, y, z) = 0. Since all the exterior products of vectors in the same plane are parallel, we can just deﬁne [x, y] = αa1a2(x1y2 −x2y1). It is easy to check that some of the basic properties of the exterior products such as E1 and E2 are still true. For points A, B, and C, the signed area of triangle ABC is deﬁned to be 1 2[− − → AB, − − → AC]. 5.4. Machine Proof in Metric Plane Geometries 243 5.4.1 Vector Approach for Euclidean Plane Geometry For Euclidean geometry, the matrix M deﬁning the inner product is the unit matrix. Thus if the conclusion of a geometry statement is a polynomial of inner and exterior products, the proofs produced according to the vector approach and the area-Pythagoras diﬀerence approach are actually the same. But in the vector approach, we can use a new geometry quantity: the vector itself. Example 5.46 (The Centroid Theorem) Show that the three medians of a triangle are concur-rent and the medians are divided by their common point in the ratio 2:1. A B E F G C Figure 5-4 Constructive description ((POINTS A B C) (MIDPOINT F A C) (MIDPOINT E B C) (INTER G (LINE A E) (LINE B F)) (− → AG = 2 3− → AE) ) The eliminants − → AGG =− → AE·S ABF S ABEF S ABEF E = 1 2(S BCF+2S ABF) S BCF F =1 2(S ABC) S ABF F = 1 2(S ABC) The machine proof. − → AG ( 2 3 )·− → AE G = − → AE·S ABF ( 2 3 )·− → AE·S ABEF simplify = (3)·S ABF (2)·S ABEF E = (3)·S ABF (2)·( 1 2 S BCF+S ABF) F = (3)·( 1 2 S ABC) 3 2 S ABC simplify = 1 Example 5.47 Let G be the centroid of a triangle ABC. Show that − → G = 1 3(− → A + − → B + − → C ). Proof. Using our method, we can actually compute this result without knowing it previ-ously. The centroid G of the triangle ABC can be introduced as follows. (POINTS A B C) (MIDPOINT F A C) (LRATIO G B F 2/3) By Lemma 5.30, − → G = 2/3− → F + 1/3− → B = 1 3(− → C + − → A + − → B ). You might think that the above introduction of G is too tricky. The usual way of intro-ducing G is as follows. 244 Chapter 4. Machine Proof in Solid Geometry Constructive description ((POINTS A B C) (MIDPOINT F A C) (MIDPOINT E B C) (INTER G (LINE A E) (LINE B F)) (− → G ) ) The machine proof. − → G G = −− → F ·S ABE+− → B ·S AFE −S ABEF E = 1 2− → F ·S ABC+ 1 2− → B ·S ABF 1 2 S BCF+S ABF F = 1 2− → C ·S ABC+ 1 2− → B ·S ABC+ 1 2− → A ·S ABC 3 2 S ABC simplify = 1 3(− → C +− → B +− → A ) The eliminants − → G G =− → F ·S ABE−− → B ·S AFE S ABEF S ABEF E =1 2(S BCF+2S ABF) S AFE E = −1 2(S ABF) S ABE E =1 2(S ABC) S BCF F =1 2(S ABC) S ABF F =1 2(S ABC) − → F F =1 2(− → C +− → A ) Example 5.48 The triangle having for its vertices the midpoints of the sides of a given tri-angle has the same centroid as the given triangle. A B C D E F G K Figure 5-5 Constructive description ((POINTS A B C X Y) (MIDPOINT D B C) (MIDPOINT E A C) (MIDPOINT F A B) (CENTROID G A B C) (CENTROID K D E F) (− → G = − → K ) ) The machine proof. − → G − → K K = − → G ·(3) − → F +− → E +− → D G = (3)·(− → C +− → B +− → A ) (− → F +− → E +− → D )·(3) F = − → C +− → B +− → A − → E +− → D + 1 2− → B + 1 2− → A E = (2)·(− → C +− → B +− → A ) 2− → D +− → C +− → B +2− → A D = (2)·(− → C +− → B +− → A ) 2− → C +2− → B +2− → A simplify = 1 The eliminants − → K K =1 3(− → F +− → E +− → D ) − → G G =1 3(− → C +− → B +− → A ) − → F F =1 2(− → B +− → A ) − → E E =1 2(− → C +− → A ) − → D D =1 2(− → C +− → B ) Example 5.49 The triangle formed by the three lines passing through the three vertices and parallel to the opposite sides of a triangle is called the anticomplementary triangle of the given triangle. Show that a triangle and its anticomplementary triangle have the same centroid. 5.4. Machine Proof in Metric Plane Geometries 245 Constructive description ((POINTS A B C) (PRATIO P A C B 1) (PRATIO Q A B C 1) (PRATIO R B A C 1) (CENTROID G A B C) (CENTROID K P Q R) (− → G = − → K ) ) The machine proof. − → G − → K K = − → G ·(3) − → R +− → Q +− → P G = (3)·(− → C +− → B +− → A ) (− → R +− → Q +− → P )·(3) R = − → C +− → B +− → A − → Q +− → P +− → C +− → B −− → A Q = − → C +− → B +− → A − → P +2− → C P = − → C +− → B +− → A − → C +− → B +− → A simplify = 1 The eliminants − → K K =1 3(− → R +− → Q +− → P ) − → G G =1 3(− → C +− → B +− → A ) − → R R =− → C +− → B −− → A − → Q Q =− → C −− → B +− → A − → P P = −(− → C −− → B −− → A ) With the help of the vector method, we can prove the following theorem about n-polygons. The centroid of n points P1, · · · , Pn is deﬁned to be 1 n(− → P1 + · · · + − → Pn). Example 5.50 (Cantor’s First Theorem) For n points on the circle O, perpendiculars from the centroids of any n −1 points taken from the n points to the tangent lines of circle O at the remaining n-th points are concurrent. Proof. Let the n points be P1, · · · , Pn and − → Pi = − − → OPi. Let G1 be the centroid of P2, · · · , Pn, G2 the centroid of P1, P3, · · · , Pn, and Y the intersection of (PLINE G1 O P1) and (PLINE G2 O P2). By Example 5.37, − → Y = − → G1 + r(− → P1 −− → O) = − → G1 + r− → P1. where r = S G1OP2−S G2OP2 S OOP1P2 = S P1OP2 (n−1)S P1OP2 = 1 n−1. Therefore − → Y = 1 n −1(− → P1 + · · · + − → Pn) which is a ﬁxed point. Example 5.51 (Cantor’s Second Theorem) For n points on the circle O, perpendiculars from the centroids of any n −2 points taken from the n points to the lines joining the remaining two points are occurrent. Proof. Let the n points be P1, · · · , Pn and − → Pi = − − → OPi. Let G1 be the centroid of P3, · · · , Pn, G2 the centroid of P1, P2, P5 · · · , Pn, M the midpoint of P1P2, N the midpoint of P3P4, and Y the intersection of (PLINE G1 O M) and (PLINE G2 N P2). By Example 5.37, − → Y = − → G1 + r(− → M −− → O) = − → G1 + r 2(− − − − − − → P1 + P2). 246 Chapter 4. Machine Proof in Solid Geometry where r = S G1ON−S G2ON S OOMN = 2S MON (n−2)S MON = 2 n−2. Therefore − → Y = 1 n −2(− → P1 + · · · + − → Pn) which is a ﬁxed point. 5.4.2 Machine Proof in Minkowskian Plane Geometry In Minkowskian plane geometry, M = 1 0 0 −1 ! . Then the inner product of x = (x1, x2) and y = (y1, y2) is ⟨x, y⟩= x1y1 −x2y2. Thus x⊥y iﬀ x1y1 −x2y2 = 0. We deﬁne the exterior product of x and y to be [x, y] = −x1y2 + x2y1. For points A, B, C, and D in the Minkowskian plane [− − → AC, − − → BD] is also interpreted as twice the area for the quadrilateral ABCD, i.e., S ABCD = 1 2[− − → AC, − − → BD]. We thus have the Herron-Qin formula in the Minkowskian geometry 16S 2 ABCD = P2 ABCD −4AC2 · BD2. In the Minkowskian plane, there exist isotropic lines (vectors). Vector x = (x1, x2) is isotropic iﬀ x2 1 −x2 2 = (x1 −x2)(x1 + x2) = 0, i.e., the isotropic lines are those which are parallel to one of the following lines x1 −x2 = 0 or x1 + x2 = 0. As a consequence of Proposition 5.45, we have Proposition 5.52 A pure constructive geometry statement is true in Euclidean geometry if and only if it is true in Minkowskian geometry. 5.4. Machine Proof in Metric Plane Geometries 247 As a consequence, most of the geometry theorems proved in this book are also valid in Minkowskian geometry. But there are do exist geometry statements in Euclidean geome-try which are not true or do not have geometric meaning in Minkowskian geometry. For instance, in Minkowskian geometry there exist no equilateral triangles. If a geometry statement is in the aﬃne geometry, i.e., the only geometry relations in the statements are incidence and parallel, then it is obvious that this statement is true in Eu-clidean geometry iﬀit is true in Minkowskian geometry. The reason is that both geometries are developed from the same aﬃne geometry by adding diﬀerent metric structures. In other words, the aﬃne part of the two geometries are the same. But for a geometry statement involving geometry relations like perpendicular or measurement, it is not obvious that its validities for both geometries are the same. Example 5.53 (Orthocenter Theorem in Minkowskian Geometry) The same as Example 3.19 on page 111. The following proof is essentially the same as the proof of Example 3.19. But it is for a diﬀerent geometry theorem. Compare Figure 1-42 (on page 32) and Figure 5-7. A B C D E H Figure 5-7 Constructive description ((POINTS A B C) (FOOT E B A C) (FOOT D C A B) (INTER H (LINE C D) (LINE B E)) (PERPENDICULAR B C A H) ) The machine proof. ⟨− − → AB,− − → AH⟩ ⟨− − → AC,− − → AH⟩ H = ⟨− − → AB,− − → AC⟩ ⟨− − → AB,− − → AC⟩ simplify = 1 The eliminants ⟨− − → AC,− − → AH⟩H =⟨− − → AB,− − → AC⟩ ⟨− − → AB,− − → AH⟩H =⟨− − → AB,− − → AC⟩ Before giving examples involving circles, let us note that the method presented in Sub-section 3.6.2 to eliminate co-circle points is still valid in Minkowskian geometry except that we need to use the hyperbolic trigonometric functions instead of the trigonometric functions. A “circle” in Minkowskian geometry is actually a hyperbola: x2 1 −x2 2 = r2. The diameter of the above circle is δ = 2r. 248 Chapter 4. Machine Proof in Solid Geometry Lemma 5.54 Let A, B, and C be points on a circle with diameter δ in the Minkowskian plane. Then S ABC = AB · CB · CA 2δ , PABC = 2AB · CB · cosh(CA) δ , AC = δ sinh(AC). The proof for the above lemma can be carried out as in Section 3.6. Example 5.55 (Simson’s Theorem in Minkowskian Geometry) The geometry statement of Sim-son’s theorem is exactly the same as in Example 3.79 on page 144. A B C O D E F G Figure 5-8 Constructive description ((CIRCLE A B C D) (FOOT E D B C) (FOOT F D A C) (FOOT G D A B) ( AG GB BE EC CF FA = −1) ) The machine proof. CF AF · BE CE · AG BG G = ⟨− − → AB,− − → AD⟩ −⟨− − → BA,− − → BD⟩ · CF AF · BE CE F = −⟨− − → CA,− − → CD⟩·⟨− − → AB,− − → AD⟩ ⟨− − → BA,− − → BD⟩·(−⟨− − → AC,− − → AD⟩) · BE CE E = ⟨− − → CA,− − → CD⟩·⟨− − → BC,− − → BD⟩·⟨− − → AB,− − → AD⟩ ⟨− − → BA,− − → BD⟩·⟨− − → AC,− − → AD⟩·(−⟨− − → CB,− − → CD⟩) co−cir = −(−2CD·AC·cosh(AD))·(2BD·BC·cosh(CD))·(2AD·AB·cosh(BD)) (−2BD·AB·cosh(AD))·(2AD·AC·cosh(CD))·(−2CD·BC·cosh(BD)) simplify = 1 The eliminants AG BG G = ⟨− − → AB,− − → AD⟩ −⟨− − → BA,− − → BD⟩ CF AF F = ⟨− − → CA,− − → CD⟩ −⟨− − → AC,− − → AD⟩ BE CE E = ⟨− − → BC,− − → BD⟩ −⟨− − → CB,− − → CD⟩ ⟨− − → CB,− − → CD⟩= −2CD·BC·cosh(BD) ⟨− − → AC,− − → AD⟩=2AD·AC·cosh(CD) ⟨− − → BA,− − → BD⟩= −2BD·AB·cosh(AD) ⟨− − → AB,− − → AD⟩=2AD·AB·cosh(BD) ⟨− − → BC,− − → BD⟩=2BD·BC·cosh(CD) ⟨− − → CA,− − → CD⟩= −2CD·AC·cosh(AD) By Menelaus’ theorem, E, F, and G are collinear. Example 5.56 (Cantor’s Theorem) The same as Example 3.82 on page 147. Constructive description ((CIRCLE A B C D) (CIRCUMCENTER O A B C) (MIDPOINT G A D) (MIDPOINT F A B) (MIDPOINT E C D) (PRATIO N E O F 1) (PERPENDICULAR G N B C) ) The eliminants ⟨− − → BC,− − → BN⟩N =⟨− − → BC,− − → BE⟩+⟨− − → BC,− − → BF⟩−⟨− − → BC,− − → BO⟩ ⟨− − → BC,− − → BE⟩E =1 2(⟨− − → BC,− − → BD⟩+⟨− − → CB,− − → CB⟩) ⟨− − → BC,− − → BF⟩F = 1 2(⟨− − → BA,− − → BC⟩) ⟨− − → BC,− − → BG⟩G =1 2(⟨− − → BC,− − → BD⟩+⟨− − → BA,− − → BC⟩) ⟨− − → BC,− − → BO⟩O =1 2(⟨− − → CB,− − → CB⟩) 5.5. Machine Proof Using Complex Numbers 249 A B C D O G E F N Figure 5-9 The machine proof. ⟨− − → BC,− − → BG⟩ ⟨− − → BC,− − → BN⟩ N = ⟨− − → BC,− − → BG⟩ ⟨− − → BC,− − → BE⟩+⟨− − → BC,− − → BF⟩−⟨− − → BC,− − → BO⟩ E = ⟨− − → BC,− − → BG⟩ ⟨− − → BC,− − → BF⟩−⟨− − → BC,− − → BO⟩+ 1 2 ⟨− − → BC,− − → BD⟩+ 1 2⟨− − → CB,− − → CB⟩ F = (2)·⟨− − → BC,− − → BG⟩ −2⟨− − → BC,− − → BO⟩+⟨− − → BC,− − → BD⟩+⟨− − → CB,− − → CB⟩+⟨− − → BA,− − → BC⟩ G = (−2)·( 1 2 ⟨− − → BC,− − → BD⟩+ 1 2 ⟨− − → BA,− − → BC⟩) 2⟨− − → BC,− − → BO⟩−⟨− − → BC,− − → BD⟩−⟨− − → CB,− − → CB⟩−⟨− − → BA,− − → BC⟩ O = −(⟨− − → BC,− − → BD⟩+⟨− − → BA,− − → BC⟩)·(2) −2⟨− − → BC,− − → BD⟩−2⟨− − → BA,− − → BC⟩ simplify = 1 In the last step, let T be the midpoint of BC. Then we have BC⊥TO, and hence ⟨− − → BC, − − → BO⟩= ⟨− − → BC, − − → BT⟩= 1 2(⟨− − → CB, − − → CB⟩). 5.5 Machine Proof Using Complex Numbers In addition to vectors, complex numbers may be used to generate readable proofs for ge-ometry theorems. Complex numbers are often used to solve diﬃcult geometry theorems and to interpret various diﬀerent geometries. See [11, 38]. They have also been used to mechanical geometry theorem proving in based on the Gr¨ obner basis computation. In this section, we will show that it is possible to obtain readable proofs for many geometry theorems using complex numbers. Complex numbers may be looked as vectors. But we can also multiply two complex numbers. We will start our discussion with this special property of complex numbers. Let x = x1 + x2i and y = y1 + y2i be two complex numbers, where i = √ −1. The conjugate of y = y1 + y2i is ˜ y = y1 −y2i. Then x · ˜ y = x1y1 + x2y2 −(x1y2 −x2y1)i. If the complex number x = x1 + x2i is seen as the same as the vector x = (x1, x2), then it is clear that x · ˜ y = ⟨x, y⟩−[x, y]i. 250 Chapter 4. Machine Proof in Solid Geometry Thus ⟨x, y⟩= 1 2(x · ˜ y + ˜ x · y); [x, y] = 1 2i(˜ x · y −x · ˜ y). That is, the inner product and the exterior product can be expressed by the multiplications of complex numbers. In geometric language, for each point P, let − → P be the corresponding complex number, and ˜ P the conjugate of − → P. Then the above two equations become PABC = (− → B −− → A)( ˜ C −˜ A) + ( ˜ B −˜ A)(− → C −− → A) S ABC = 1 4i(( ˜ B −˜ A)(− → C −− → A) −(− → B −− → A)( ˜ C −˜ A)). Therefore, the vector approach for Euclidean geometry can be translated into the lan-guage of complex numbers. But the proofs thus produced are generally longer than those produced by the vector approach. The reason is that in the vector approach the area and Pythagoras diﬀerence are treated like one-term variables, while in the complex number ap-proach they are expressions of several terms. This complex number approach is essentially the same as the Wu’s method in the case of constructive geometry statements [56, 63]. But in many cases, the complex number approach does give short proofs. Example 5.57 As shown in Figure 5-10, on two sides AC and BC of triangle ABC, two similar triangles PAC and QCB are drawn. RPCQ is a parallelogram. Show that triangle RAB is similar with triangle PAC. For two points A and B, let − − → AB = − → B −− → A. Two triangles PAB and QCB are similar and have the same orientation iﬀ (5.1) − − → PA − − → AC = − − → QC − − → CB or − − → PA · − − → CB = − − → AC · − − → QC. We thus can use a new construction (SIM-TRIANGLE Q C B P A C) which introduces a point Q such that (5.1) is true. Now Example 5.57 can be described constructively as follows. P Q R C B A Figure 5-10 Constructive description ((POINTS A B C P) (SIM-TRIANGLE Q B C P C A) (PRATIO R Q C P 1) (− → PA·− → AB = − → RA·− → AC) ) 5.5. Machine Proof Using Complex Numbers 251 The machine proof. − → AP·− → AB − → AR·− → AC R = − → AP·− → AB (− → AQ+− → AP−− → AC)·− → AC Q = − → AP·− → AB·(−− → AC) (− → CP·− → BC−− → B ·− → AC−− → AP·− → AC+− → AC 2 +− → AC·− → A )·− → AC simplify = −− → AP·− → AB − → CP·− → BC−− → B ·− → AC−− → AP·− → AC+− → AC 2 +− → AC·− → A = −(− → P −− → A )·(− → B −− → A ) −− → P ·− → B +− → P ·− → A +− → B ·− → A −− → A 2 simplify = 1 The eliminants − → ARR =− → AQ+− → AP−− → AC − → AQ Q =− → CP·− → BC−− → B ·− → AC+− → AC·− → A −− → AC − → AC=− → C −− → A − → BC=− → C −− → B − → CP=− → P −− → C − → AB=− → B −− → A − → AP=− → P −− → A Let ω = e 2iπ 3 . Then the three points corresponding to the three complex numbers 1, ω, and ω2 form an equilateral triangle with positive orientation (Figure 5-11). w 2 w A B 1 C Figure 5-11 Thus ABC is an equilateral triangle with positive orientation (Figure 5-11) iﬀ − → AB − → AC = ω −1 ω2 −1. From the above equation, we have (ω3 −ω2)− → B + (ω −1)− → C + (ω2 −ω)− → A = 0. Dividing ω −1 from both sides of the above equation, we have that ABC is an equilateral triangle with positive orientation iﬀ − → C + ω2− → B + ω− → A = 0. Similarly ABC is an equilateral triangle with negative orientation iﬀ − → C + ω− → B + ω2− → A = 0. We thus introduce two new constructions. 252 Chapter 4. Machine Proof in Solid Geometry (PE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with positive orientation, i.e., − → C + ω2− → B + ω− → A = 0. (NE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with negative orientation, i.e, − → C + ω− → B + ω2− → A = 0. Example 5.58 (Echols’ First Theorem1) If ABC and PQR are equilateral triangles with the same orientation, then the triangle formed by the midpoints of AP, BQ, and CR is also an equilateral triangle. A B P Q C R L M N Figure 5-12 Constructive description ((POINTS A B P Q) (PE-TRIANGLE C B A) (PE-TRIANGLE R Q P) (MIDPOINT L A P) (MIDPOINT M B Q) (MIDPOINT N C R) (PE-TRIANGLE N M L) ) The machine proof. − → N +− → M ·w+− → L ·w2 n = − → M ·w+− → L ·w2+ 1 2− → R + 1 2− → C n = ( 1 2 )·(2− → L ·w2+− → R +− → C +− → Q ·w+− → B ·w) n = ( 1 2 )·(− → R +− → C +− → Q ·w+− → P ·w2+− → B ·w+− → A ·w2) n = ( 1 2 )·(− → C +− → B ·w+− → A ·w2) n = ( 1 2 )·(0) simplify = 0 The eliminants − → N N =1 2(− → R +− → C ) − → M M = 1 2(− → Q +− → B ) − → L L =1 2(− → P +− → A ) − → R R = −((− → Q +− → P ·w)·w) − → C C = −((− → B +− → A ·w)·w) Example 5.59 (Echols’ Second Theorem) If ABC, PQR, and XYZ are equilateral triangles with the same orientation, then the triangle formed by the centroids of triangle APX, BQY, and CRZ is an equilateral triangle. 1American Mathematical Monthly, 39, 1932, p.46 5.5. Machine Proof Using Complex Numbers 253 Constructive description ((POINTS A B P Q X Y) (PE-TRIANGLE C B A) (PE-TRIANGLE R Q P) (PE-TRIANGLE Z Y X) (CENTROID L A P X) (CENTROID M B Q Y) (CENTROID N C R Z) (PE-TRIANGLE N M L) ) The eliminants − → N N = 1 3(− → Z +− → R +− → C ) − → M M = 1 3(− → Y +− → Q +− → B ) − → L L = 1 3(− → X +− → P +− → A ) − → Z Z = −((− → Y +− → X ·w)·w) − → R R = −((− → Q +− → P ·w)·w) − → C C = −((− → B +− → A ·w)·w) The machine proof. − → N +− → M ·w+− → L ·w2 n = 3− → M ·w+3− → L ·w2+− → Z +− → R +− → C n = 9− → L ·w2+3− → Z +3− → R +3− → C +3− → Y ·w+3− → Q ·w+3− → B ·w n = (3)·(3− → Z +3− → R +3− → C +3− → Y ·w+3− → X ·w2+3− → Q ·w+3− → P ·w2+3− → B ·w+3− → A ·w2) n = (9)·(− → R +− → C +− → Q ·w+− → P ·w2+− → B ·w+− → A ·w2) n = (9)·(− → C +− → B ·w+− → A ·w2) simplify = 0 Example 5.60 (Echols’ Theorem in General Form) Let Pi,1...Pi,n, i = 1, ..., m be m regular n-polygons with the same orientation. Then the centroids of the m-polygons P1, j...Pm, j, j = 1, ..., n, form a regular n-polygon. Proof. From the proofs of Examples 5.58 and 5.59, we need only to show that A1...An is a regular n-polygon iﬀpoints A1, ..., An satisfy some linear relations Rk(A1, ..., An) = 0, k = 1, ...s. We leave the details to the reader. Example 5.61 The same as Example 3.45 on page 127. The following proof is much shorter. (Figure 5-13) 254 Chapter 4. Machine Proof in Solid Geometry Constructive description ((POINTS A B C) (CONSTANT w2+w+1) (PE-TRIANGLE B1 A C) (PE-TRIANGLE A1 C B) (NE-TRIANGLE C1 B A) (− − → A1C1−− → CB1 = 0) ) The machine proof. − − → A1C1−− → CB1 n = −− → A1−− → CB1−− → B ·w2−− → A ·w n = −(− → CB1−− → C ·w+− → A ·w) n = −(−− → C ·w2−− → C ·w−− → C ) simplify = (w2+w+1)·− → C n = 0 simplify = 0 The eliminants − − → A1C1 C1 = −(− → A1 +− → B ·w2+− → A ·w) − → A1 A1 = −((− → C +− → B ·w)·w) − → CB1 B1 = −(− → C ·w2+− → C +− → A ·w) 1 B 1 A 1 C C B A Figure 5-13 A B C D M X N Y Figure 5-14 Example 5.62 The converse of Example 5.61. (Figure 5-14) Constructive description ((POINTS A B C) (CONSTANT w2+w+1) (PRATIO D A B C 1) (PE-TRIANGLE X B C) (PE-TRIANGLE Y C D) (PE-TRIANGLE A Y X) ) The machine proof. − → Y ·w+− → X ·w2+− → A n = − → X ·w2−− → D ·w3−− → C ·w2+− → A n = −− → D ·w3−− → C ·w4−− → C ·w2−− → B ·w3+− → A n = −(− → C ·w4+− → C ·w3+− → C ·w2+− → A ·w3−− → A ) simplify = −(w2+w+1)·(− → C ·w2+− → A ·w−− → A ) n = 0 simplify = 0 The eliminants − → Y Y = −((− → D ·w+− → C )·w) − → X X = −((− → C ·w+− → B )·w) − → D D =− → C −− → B +− → A Using complex numbers, we can also deal with some theorems involving squares easily. For two points A and B, C is a point introduced by (TRATIO C A B 1), or equivalently C satisﬁes CA = AB, CA ⊥AB, and S CAB > 0, iﬀ − − → AC = i · − − → AB. 5.5. Machine Proof Using Complex Numbers 255 Geometrically, the above equation means that − − → AC is obtained by rotating − − → AB by 90◦coun-terclockwise, or CAB is an isosceles right triangle with positive orientation. Similarly, C is a point introduced by (TRATIO C A B -1) iﬀ − − → AC = −i · − − → AB. We thus can introduce two new constructions (PE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with positive orientation, i.e., − → C = − → A + i · − − → AB. (NE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with negative orientation, i.e, − → C = − → A −i · − − → AB. Example 5.63 On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that FC⊥BH and FC = BH. Constructive description: ((POINTS A B C) (CONSTANT i2−1) (PE-SQUARE F A B) (NE-SQUARE H A C) (− → FC−i·− → BH = 0) ) A C B G H E F M Figure 5-15 The machine proof −(− → CF+− → BH·i) n = −(− → CF−− → AC·i2−− → AB·i) n = −(−− → AC·i2−− → AC) simplify = (i2+1)·− → AC n = 0 The eliminants − → BHH = −(− → AC·i+− → AB) − → CF F = −(− → AC−− → AB·i) Example 5.64 On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. M is the midpoint of BC. Show that FH = 2AM and FH⊥AM. (Figure 5-15) Constructive description ((POINTS A B C) (PE-SQUARE F A B) (NE-SQUARE H A C) (MIDPOINT M B C) (− → HF−2i·− → AM = 0) ) The machine proof −(− → FH+2− → AM·i) n = −(− → FH+− → AC·i+− → AB·i) n = −(−− → AF+− → AB·i) n = 0 The eliminants − → AMM = 1 2(− → AC+− → AB) − → FHH = −(− → AF+− → AC·i) − → AFF =− → AB·i 256 Chapter 4. Machine Proof in Solid Geometry A B C F G D E K H Q P Figure 5-16 Example 5.65 2 Starting with any triangle ABC, construct the exterior (or interior) squares BCDE, ACFG, and BAHK; then construct parallelograms FCDQ and EBKP. Show that PAQ is an isosceles right triangle. Constructive description ((POINTS A B C) (CONSTANT i2 −1) (NE-SQUARE F C A) (PE-SQUARE D C B) (PRATIO E D C B 1) (PE-SQUARE K B A) (PRATIO Q F C D 1) (PRATIO P K B E 1) (− → AQ−i·− → AP = 0) ) The machine proof −(− → AP·i−− → AQ) n = −(−− → AQ+− → AK·i+− → AE·i−− → AB·i) n = −− → AK·i−− → AE·i+− → AD+− → AF−− → AC+− → AB·i n = −(− → AE·i−− → AD−− → AF+− → AC−− → AB·i2) n = −(− → AD·i−− → AD−− → AF−− → AC·i+− → AC−− → AB·i2+− → AB·i) n = −(−− → BC·i2+− → BC·i−− → AF−− → AB·i2+− → AB·i) n = − → BC·i2−− → BC·i+− → AC·i+− → AC+− → AB·i2−− → AB·i n = −− → BC·i−− → BC+− → AC·i+− → AC−− → AB·i−− → AB simplify = −(i+1)·(− → BC−− → AC+− → AB) n = −(i+1)·(0) simplify = 0 The eliminants − → APP =− → AK+− → AE−− → AB − → AQ Q =− → AD+− → AF−− → AC − → AKK = −((i−1)·− → AB) − → AEE =− → AD−− → AC+− → AB − → ADD = −(− → BC·i−− → AC) − → AFF =(i+1)·− → AC − → AB=− → B −− → A − → AC=− → C −− → A − → BC=− → C −− → B Summary of Chapter 5 • The metric vector space is a vector space with inner and exterior products. The metric geometry of dimension three associated with ﬁeld E is the nonsingular metric vector space E3. • Some basic geometry quantities can be described by the inner and exterior products: 1. PABC = 2⟨− − → AB, − − → CB⟩; PABCD = 2⟨− − → AC, − − → DB⟩. 2. If four points A, B, C, and D are collinear or AB ∥CD, − − → AB = AB CD − − → CD, AB CD = ⟨− − → AB, − − → CD⟩ ⟨− − → CD, − − → CD⟩ . 2This example is from Amer. Math. Mon. 75(1968), p.899. 5.5. Machine Proof Using Complex Numbers 257 3. VABCD = 1 6⟨− → AD, [− − → AB, − − → AC]⟩. 4. If six points A, B, C, P, Q, and R are coplanar or ABC ∥PQR, [− − → AB, − − → AC] = S ABC S PQR [− − → PQ, − − → PR], S ABC S PQR = ⟨[− − → AB, − − → AC], [− − → PQ, − − → PR]⟩ ⟨[− − → PQ, − − → PR], [− − → PQ, − − → PR]⟩ . • We have the following criteria for parallel and perpendicularity. 1. AB ⊥CD ⇐⇒⟨− − → AB, − − → CD⟩= 0. 2. AB ∥CD ⇐⇒[− − → AB, − − → CD] = 0. 3. AB ⊥PQR ⇐⇒[− → AB, [− − → PQ, − − → PR]] = 0. 4. AB ∥PQR ⇐⇒⟨− → AB, [− − → PQ, − − → PR]⟩= 0. 5. ABC ⊥PQR ⇐⇒⟨[− − → AB, − − → AC], [− − → PQ, − − → PR]⟩= 0. 6. ABC ∥PQR ⇐⇒ = 0. • A mechanical proving method is presented for constructive geometry statements in metric geometry of dimensions two and three. This method works similarly to the area-volume-Pythagoras diﬀerence method developed in Chapters 3 and 4. The basis of the method is to eliminate points from vectors and inner and exterior products of vectors. • We have proved the following meta theorem. A pure constructive geometry statement is true in one metric geometry iﬀit is true in all the metric geometries associated with the same ﬁeld. 258 Chapter 4. Machine Proof in Solid Geometry Chapter 6 A Collection of 400 Mechanically Proved Theorems This chapter is a collection of 400 geometry theorems in plane geometry proved auto-matically by a computer program1 based on the method developed in the ﬁrst part of this book, including 205 machine proofs (in LaTeX form) produced entirely automatically by the program. In addition, there are another 78 problems solved by our computer program in Chapters 2–5. We include a collection like this for two reasons. First, it would show the power of the method/program: most theorems involving equalities only in geometry textbooks are in the collection. Second, many proofs produced by the program are very beautiful. Following the spirit of E. W. Dijkstra (p. 174 ), we believe that such beautiful proofs deserve special attention. Among the work consulted to collect these examples, special mention must be made of the following [1, 3, 4, 13, 23, 39]. The ﬁgures and inputs for many of the examples in this collection are from directly. The examples are classiﬁed according to the types of constructions needed to describe them. 6.1 Notation Convention As an example of how to read the examples in this collection, let us consider Ceva’s the-orem. As shown, a typical entry in this collection includes a description of the theorem in English, a diagram, a constructive description of the theorem as the input to the program, and a machine proof produced by the program. 1The program is available via ftp at emcity.cs.twsu.edu: pub/geometry. 259 260 Chapter 6. Topics from Geometry A B C O D E F Figure 6-0 Example 6.0 (Ceva’s Theorem) (0.01,1,3) The lines joining the vertices of a triangle to a given point determine on the sides of the tri-angle six segments such that the product of three nonconsecutive segments is equal to the product of the remaining three segments. Constructive description ((points A B C O) (inter D (l B C) (l O A)) (inter E (l A C) (l O B)) (inter F (l A B) (l O C)) ( AF FB BD DC CE EA = 1) ) The machine proof −CE AE · BD CD · AF BF F = −(−S ACO) −S BCO · CE AE · BD CD E = −S BCO·S ACO S BCO·(−S ABO) · BD CD simplify = S ACO S ABO · BD CD D = S ABO·S ACO S ABO·S ACO simplify = 1 The eliminants AF BF F =S ACO S BCO CE AE E = S BCO −S ABO BD CD D = S ABO S ACO Notation Convention. 1. We use a triple (time,maxt,lems) to measure how diﬃcult a machine proof is: time is the time needed to complete the machine proof in a NexT Turbo workstation (25 MIPS); maxt is the number of terms of the maximal polynomial occurring in the ma-chine proof; and lems is the number of elimination lemmas used to eliminate points from geometry quantities, i.e., the number of the eliminants. For Ceva’s theorem, time = 0.01 second, maxterm = 1, and lems = 3. 2. We use some abbreviations in the constructive description of the geometry statements in order to save printing space. For instance, LINE is represented by l; PLINE is represented by p; BLINE is represented by b; and ALINE is represented by a. 3. The ndg conditions and the predicate forms of the geometry statements are not given, since they can be generated directly from the constructive description. See page 110. 4. For explanation of the machine proof and the eliminants, see page 72. In Part I of the book, we have introduced 28 constructions for plane geometry, which are listed here for your convenience. 1. (ARATIO A O U V rO rU rV). Take a point A such that rU, rV, and rO are the area coordinates of A with respect to OUV. For the exact deﬁnition of this construction, see page 136. 2. (CENTROID G A B C). G is the centroid of triangle ABC. See page 138. 6.1. Notation Convention 261 3. (CIRCLE Y1 · · · Ys), (s ≥3). Points Y1 · · · Ys are on the same circle. See page 144. 4. (CIRCUMCENTER O A B C). O is the circumcenter of triangle ABC. See page 138. 5. (CONSTANT p(r)) where p(r) is an irreducible polynomial in r. This construction introduces an algebraic number r which is a root of p(r) = 0. See page 123. 6. (HARMONIC D C A B) means that D is a point such that A, B, C, and D form a harmonic consequence. See page 123. 7. (INCENTER C I A B) I is the center of the inscribed circle of triangle ABC. This construction is to construct point C from points A, B, and I. See page 138. 8. (INTER Y ln1 ln2). Point Y is the intersection of lines ln1 and ln2. See page 110. 9. (INTER Y ln (CIR O P)). Point Y is the intersection of line ln and circle (CIR O P) other than point P. Line ln could be (LINE P U), (PLINE P U V), and (TLINE P U V). See page 110. 10. (INTER Y (LINE U V) (CIR O r)). Point Y is one of the intersections of line (LINE U V) and circle (CIR O r). See page 148. 11. (INTER Y (CIR O1 P) (CIR O2 P)). Point Y is the intersection of the circle (CIR O1 P) and the circle (CIR O2 P) other than point P. See page 110. 12. (INTER Y (CIR O1 r1) (CIR O2 r2)). Point Y is one of the intersections of the circle (CIR O1 r1) and the circle (CIR O2 r2). See page 148. 13. (INVERSION P Q O A) means that P is the inversion of Q with regard to circle (CIR O A). See page 123. 14. (LRATIO Y U V r). Y is a point on UV such that UY UV = r. See page 122. 15. (ON Y ln). Take a point Y on a line ln. Line ln could be one of the forms below. (LINE A B) is the line passing through two points A and B. (PLINE C A B) is the line passing through C and parallel to (LINE A B). (TLINE C A B) is the line passing through C and perpendicular to (LINE A B). (BLINE A B) is the perpendicular-bisector of AB. (ALINE P Q U W V) is the line l passing through P such that ∠[PQ, l] = ∠[UW, WV]. See pages 110 and 129. 16. (ON Y (CIR O P)). Take a point Y on a circle (CIR O P). See page 110. 17. (MIDPOINT Y U V). Y is the midpoint of UV. See page 122. 262 Chapter 6. Topics from Geometry 18. (MRATIO Y U V r). Y is a point on UV such that UY YV = r. See page 122. 19. (NE-SQUARE C B A) introduces a point C such that CAB is an isosceles right trian-gle with negative orientation, i.e, − → C = − → A −i · − − → AB. See page 255. 20. (NE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with negative orientation. See page 251. 21. (ORTHOCENTER H A B C). H is the orthocenter of the triangle ABC. See page 138. 22. (PE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with positive orientation, i.e., − → C = − → A + i · − − → AB. See page 255. 23. (PE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with positive orientation. See page 251. 24. (POINT[S] Y1, · · · , Yl). Take arbitrary points Y1, · · · , Yl in the plane. See page 110. 25. (PRATIO Y W U V r). Take a point Y on the line passing through W and parallel to line UV such that WY = rUV. See page 110. 26. (SIM-TRIANGLE Q C B P A C) introduces a point Q such that triangle QCB and triangle PAC are similar and have the same orientation. See page 250. 27. (SYMMETRY Y U V). Y is the symmetry of point V with respect to point U. See page 122. 28. (TRATIO Y U V r). Take a point Y on line (TLINE U U V) such that r = 4S UVY PUVU (= UY UV). See page 110. The following are the predicates accepted by the program as conclusions. 1. (COCIRCLE A B C D). Points A, B, C, and D are co-circle iﬀ∠[CAD] = ∠[CBD], or equivalently, S CADPCBD = PCADPCBD. See page 130. 2. (COLLINEAR A B C). Points A, B, and C are collinear iﬀS ABC = 0. Also see the comments after Example 2.36 on page 74. 3. (EQANGLE A B C D E F). ∠[ABC] = ∠[DEF] iﬀS ABCPDEF = S DEFPABC. See page 114. 4. (EQDISTANCE A B C D). AB has the same length as CD iﬀPABA = PCDC. See page 114. 5. (EQ-PRODUCT A B C D P Q R S ). The product of AB and CD is equal to the product of PQ and RS . See page 114. 6.2. Geometry of Incidence 263 6. (HARMONIC A B C D). A, B and C, D are harmonic points iﬀAC CB = DA DB. See page 123. 7. (INVERSION P Q O A). P is the inversion of Q with regard to circle (CIR O A). See page 123. 8. (MIDPOINT O A B). O is the midpoint of AB iﬀAO OB = 1. See page 114. 9. (NE-TRIANGLE C B A) ABC is an equilateral triangle with negative orientation, i.e, − → C + ω− → B + ω2− → A = 0. See page 251. 10. (ON-RADICAL P O1 A O2 B). P is on the axis of circles O1A and O2B iﬀPPO1P − PAO1A = PPO2P −PBO2B. 11. (PARALLEL A B C D). AB is parallel to CD iﬀS ACD = S BCD. See page 114. 12. (PE-TRIANGLE C B A). ABC is an equilateral triangle with positive orientation, i.e., − → C + ω2− → B + ω− → A = 0. See page 251. 13. (PERPENDICULAR A B C D). AB is perpendicular to CD iﬀPACD = PBCD. See page 114. 14. (PERP-BISECT O P Q). O is on the perpendicular bisector of PQ iﬀPOPO = POQO. 15. (TANGENT O1 A O2 B). Circle (CIR O1 A) is tangent to circle (CIR O2 B). See page 114. 6.2 Geometry of Incidence In this section, we include those geometric statements that can be formulated and proved without the measurement or comparison of distances or of angles, i.e., geometric facts involving incidence only. Such problems include the transversal, properties of cross-ratios, projective conﬁgurations, etc. These problems belong essentially to aﬃne geometry. To prove statements involving incidence only, we need only to use Algorithm 2.32. Actually, the co-side theorem would be enough for the proofs of most of the examples in this section. 6.2.1 Menelaus’ Theorem For the machine proof of Menelaus’ theorem, see Example 2.35 on page 73. 264 Chapter 6. Topics from Geometry Example 6.1 (Converse of Menelaus’ Theorem) (0.033, 2, 6) If three points are taken, one on each side of a triangle, so that these points divide the sides into six segments such that the products of the segments in each of the two sets of nonconsecutive segments are equal in magnitude and opposite in sign, the three points are collinear. A B C D E F Figure 6-1 Constructive description ((points A B C) (mratio D B C r1) (mratio E C A r2) (inter F (l D E) (l A B)) ( BF FA = −r1·r2) ) The machine proof BF AF r2·r1 F = S BDE r2·r1·S ADE E = S ABD·r2·(r2+1) r2·r1·(−S ACD)·(r2+1) simplify = −S ABD r1·S ACD D = −S ABC·r1·(r1+1) r1·(−S ABC)·(r1+1) simplify = 1 The eliminants BF AF F =S BDE S ADE S ADE E =−S ACD r2+1 S BDE E =S ABD·r2 r2+1 S ACD D = −S ABC r1+1 S ABD D =S ABC·r1 r1+1 Example 6.2 (Menelaus’ Theorem for a Quadrilateral) (0.033, 1, 4) A line XY meets the sides AB, BC, CD, and DA of a quadrilateral at A1, B1, C1, and D1 respectively. Show that AA1 BA1 · BB1 CB1 · CC1 DC1 · DD1 AD1 = 1. 1 D 1 C 1 B 1 A D C B A X Y Figure 6-2 Constructive description ( (points A B C D X Y) (inter A1 (l A B) (l X Y)) (inter B1 (l B C) (l X Y)) (inter C1 (l C D) (l X Y)) (inter D1 (l A D) (l X Y)) ( AA1 BA1 · BB1 CB1 · CC1 DC1 · DD1 AD1 = 1) ) The machine proof DD1 AD1 · CC1 DC1 · BB1 CB1 · AA1 BA1 D1 = S DXY S AXY · CC1 DC1 · BB1 CB1 · AA1 BA1 C1 = S DXY·S CXY S AXY ·S DXY · BB1 CB1 · AA1 BA1 simplify = S CXY S AXY · BB1 CB1 · AA1 BA1 B1 = S CXY ·S BXY S AXY·S CXY · AA1 BA1 simplify = S BXY S AXY · AA1 BA1 A1 = S BXY·S AXY S AXY·S BXY simplify = 1 The eliminants DD1 AD1 D1 = S DXY S AXY CC1 DC1 C1 = S CXY S DXY BB1 CB1 B1 = S BXY S CXY AA1 BA1 A1 = S AXY S BXY Example 6.3 (0.033, 2, 7) The converse of Example 6.2. The ﬁgure of this example is the same as Example 6.2. 6.2. Geometry of Incidence 265 Constructive description ((points A B C D) (mratio B1 B C r1) (mratio C1 C D r2) (inter D1 (l D A) (l B1 C1)) (inter A1 (l A B) (l B1 C1)) ( BA1 A1A = r1·r2· DD1 D1A ) ) The machine proof BA1 AA1 DD1 AD1 ·r2·r1 A1 = S BB1C1 DD1 AD1 ·r2·r1·S AB1C1 D1 = S BB1C1·S AB1C1 S DB1C1 ·r2·r1·S AB1C1 simplify = S BB1C1 S DB1C1·r2·r1 C1 = (−S BDB1·r2)·(r2+1) S CDB1 ·r2·r1·(r2+1) simplify = −S BDB1 S CDB1·r1 B1 = −(−S BCD·r1)·(r1+1) S BCD·r1·(r1+1) simplify = 1 The eliminants BA1 AA1 A1 = S BB1C1 S AB1C1 DD1 AD1 D1 = S DB1C1 S AB1C1 S DB1C1 C1 = S CDB1 r2+1 S BB1C1 C1 = −S BDB1·r2 r2+1 S CDB1 B1 = S BCD r1+1 S BDB1 B1 = −S BCD·r1 r1+1 Deﬁnition. Two points on a side of a triangle are said to be isotomic to each other, if they are equidistant from the midpoint of this side. Example 6.4 2 (0.083, 2, 9) The isotomic points of three collinear points are collinear. A B C D E F D E F Figure 6-4 1 1 1 Constructive description ( (points A B C) (lratio D B C r1) (lratio E A C r2) (inter F (l A B) (l D E)) (pratio D1 B C D −1) (pratio E1 C A E −1) (pratio F1 A B F −1) ( AF1 F1B BD1 D1C CE1 E1A = −1) ) The machine proof CE1 AE1 · BD1 CD1 · AF1 BF1 F1 = −1 −AB BF −1 · CE1 AE1 · BD1 CD1 E1 = −1 ( AB BF +1)·( AC AE −1) · BD1 CD1 D1 = −(−1) ( AB BF +1)·( AC AE −1)·(−BC CD −1) F = −S BDE (S ADBE+S BDE)·( AC AE −1)·( BC CD +1) E = −(−S ABD·r2+S ABD)·r2 (−S ABD·r2+S ABC·r2)·(−r2+1)·( BC CD +1) simplify = S ABD (S ABD−S ABC)·( BC CD +1) D = S ABC·r1·(r1−1) (S ABC·r1−S ABC)·r1 simplify = 1 The eliminants AF1 BF1 F1 = 1 AB BF +1 CE1 AE1 E1 = −1 AC AE −1 BD1 CD1 D1 = 1 BC CD +1 AB BF F =S ADBE S BDE AC AE E = 1 r2 S ADBE E = −(S ABD−S ABC·r2) S BDE E = −((r2−1)·S ABD) BC CD D = 1 r1−1 S ABD D =S ABC·r1 2In this chapter, the ﬁgure indices are always the same with the indices of the examples that they belong to. 266 Chapter 6. Topics from Geometry 6.2.2 Ceva’s Theorem For the machine proof of Ceva’s theorem, see page 72. For the generalizations of Ceva’s theorem to arbitrary polygons, see Subsection 2.7.3. Example 6.5 (The Converse of Ceva’s Theorem) (0.066, 2, 8) If three points taken on the sides of a triangle determine on these sides six segments such that the products of the segments in the two nonconsecutive sets are equal, both in magnitude and in sign, the lines joining these points to the respectively opposite vertices are concurrent. A B D E O C F Figure 6-5 Constructive description ( (points A B C) (mratio D B C r1) (mratio E C A r2) (mratio F B A r1·r2) (inter O (l B E) (l A D)) (inter Z (l B A) (l O C)) ( BZ AZ = BF AF ) ) The machine proof ( BZ AZ)/( BF AF) Z = −S BCO BF AF ·(−S ACO) O = (−S BCE·S ABD)·S ABDE BF AF ·S ACD·S ABE·(−S ABDE) simplify = S BCE·S ABD BF AF ·S ACD·S ABE F = S BCE·S ABD·(−1) r2·r1·S ACD·S ABE E = −S ABC·r2·S ABD·(r2+1) r2·r1·S ACD·S ABC·(r2+1) simplify = −S ABD r1·S ACD D = −S ABC·r1·(r1+1) r1·(−S ABC)·(r1+1) simplify = 1 The eliminants BZ AZ Z =S BCO S ACO S ACO O =S ACD·S ABE S ABDE S BCO O =S BCE·S ABD S ABDE BF AF F = −(r2·r1) S ABE E =S ABC r2+1 S BCE E =S ABC·r2 r2+1 S ACD D = −S ABC r1+1 S ABD D =S ABC·r1 r1+1 Deﬁnition. If a line is drawn through a vertex of a triangle, the segment included between the vertex and the opposite side is called a cevian. The triangle DEF in Figure 6-5 is called the cevian triangle of the point O for the triangle ABC. Example 6.6 (0.050, 3, 12) If LMN is the cevian triangle of the point S for the triangle ABC, we have S L/AL + S M/BM + S N/CN = 1. 6.2. Geometry of Incidence 267 B C A S L N M Figure 6-6 Constructive description ( (points A B C S ) (inter L (l B C) (l A S )) (inter N (l B A) (l C S )) (inter M (l A C) (l B S )) ( S L AL + S M BM + S N CN = 1) ) The machine proof S M BM + S N CN + S L AL M = −S N CN ·S ABC−S L AL ·S ABC+S ACS −S ABC N = −S L AL ·S 2 ABC+S ACS ·S ABC−S ABS ·S ABC S ABC·(−S ABC) simplify = S L AL ·S ABC−S ACS +S ABS S ABC L = S BCS ·S ABC−S ACS ·S ABC+S ABS ·S ABC (S ABC)2 simplify = S BCS −S ACS +S ABS S ABC area−co = S ABC S ABC simplify = 1 The eliminants S M BM M = S ACS −S ABC S N CN N = S ABS S ABC S L AL L =S BCS S ABC S BCS =S ACS −S ABS +S ABC Example 6.7 If LMN is the cevian triangle of the point S for the triangle ABC (Figure 6-6), we have S AMLS BNMS CLN S ANLS BLMS CNM = 1. Constructive description: ((points A B C S ) (inter L (l B C) (l S A)) (inter M (l A C) (l S B)) (inter N (l A B) (l S C)) (S AMLS BNMS CLN = S ANLS BLMS CNM) ) The machine proof S CLN·S BMN·S ALM S CMN·S BLM·S ALN N = (−S CS L·S ABC)·(−S BCS ·S ABM)·S ALM·S ACBS ·(−S ACBS ) (−S CS M·S ABC)·S BLM·(−S ACS ·S ABL)·S ACBS ·(−S ACBS ) simplify = S CS L·S BCS ·S ABM·S ALM S CS M·S BLM·S ACS ·S ABL M = S CS L·S BCS ·S ABS ·S ABC·(−S ACL·S ABS )·(−S ABCS )·S ABCS S BCS ·S ACS ·S BS L·S ABC·S ACS ·S ABL·(S ABCS )2 simplify = S CS L·(S ABS )2·S ACL (S ACS )2·S BS L·S ABL L = S BCS ·S ACS ·(S ABS )2·(−S ACS ·S ABC)·((−S ABSC))2 (S ACS )2·S BCS ·S ABS ·(−S ABS ·S ABC)·((−S ABSC))2 simplify = 1 The eliminants S ALN N = −S ACS ·S ABL S ACBS S CMN N =S CS M·S ABC S ACBS S BMN N = −S BCS ·S ABM S ACBS S CLN N =S CS L·S ABC S ACBS S BLM M = S BS L·S ABC −S ABCS S CS M M = S BCS ·S ACS S ABCS S ALM M = −S ACL·S ABS S ABCS S ABM M = S ABS ·S ABC S ABCS S ABL L =S ABS ·S ABC S ABSC S BS L L =S BCS ·S ABS −S ABSC S ACL L = S ACS ·S ABC S ABSC S CS L L = S BCS ·S ACS −S ABSC Example 6.8 (0.033, 3, 4) The lines AP, BQ, CR through the vertices of a triangle ABC parallel, respectively, to the lines OA1, OB1, OC1 joining any point O to the points A1, B1, C1 marked in any manner whatever, on the sides of BC, CA, AB meet these sides in the points P, Q, R. Show that OA1/AP + OB1/BQ + OC1/CR = 1. 268 Chapter 6. Topics from Geometry A B C O 1 A 1 B 1 C P Q R Figure 6-8 Constructive description ( (points A B C O) (on A1 (l B C)) (on B1 (l C A)) (on C1 (l B A)) (inter P (l B C) (p A A1 O)) (inter Q (l A C) (p B B1 O)) (inter R (l A B) (p C C1 O)) ( OA1 AP + OB1 BQ + OC1 CR = 1) ) The machine proof OC1 CR + OB1 BQ + OA1 AP R = OB1 BQ ·S ABC+ OA1 AP ·S ABC+S ABO S ABC Q = −OA1 AP ·S 2 ABC+S ACO·S ABC−S ABO·S ABC S ABC·(−S ABC) simplify = OA1 AP ·S ABC−S ACO+S ABO S ABC P = S BCO·S ABC−S ACO·S ABC+S ABO·S ABC (S ABC)2 simplify = S BCO−S ACO+S ABO S ABC area−co = S ABC S ABC simplify = 1 The eliminants OC1 CR R =S ABO S ABC OB1 BQ Q = S ACO −S ABC , OA1 AP P = S BCO S ABC S BCO=S ACO−S ABO+S ABC Example 6.9 (0.083, 2, 9) If LMN is the cevian triangle of the point S for the triangle ABC, we have AS/S L = AM/MC + AN/NB. M N L S A C B Figure 6-9 Constructive description ( (points A B C S ) (inter L (l B C) (l A S )) (inter N (l B A) (l C S )) (inter M (l A C) (l B S )) ( AM MC + AN NB = AS S L ) ) The machine proof −( AM CM + AN BN ) AS S L M = −(−AN BN ·S BCS +S ABS ) AS S L ·(−S BCS ) N = −(S BCS ·S ACS −S BCS ·S ABS ) AS S L ·(S BCS )2 simplify = −(S ACS −S ABS ) AS S L ·S BCS L = −(S ACS −S ABS )·S BCS S ABSC·S BCS simplify = −(S ACS −S ABS ) S ABSC area−co = −(S ACS −S ABS ) −S ACS +S ABS simplify = 1 The eliminants AM CM M = S ABS −S BCS AN BN N =S ACS S BCS AS S L L =S ABSC S BCS S ABSC= −(S ACS −S ABS) Example 6.10 (0.083, 1, 5) The Ceva’s theorem for a pentagon. 6.2. Geometry of Incidence 269 A B C D E O A B C D E Figure 6-10 1 1 1 1 1 Constructive description ( (points A B C D E O) (inter A1 (l O A) (l C D)) (inter B1 (l O B) (l D E)) (inter C1 (l O C) (l E A)) (inter D1 (l O D) (l A B)) (inter E1 (l O E) (l B C)) ( CA1 A1D DB1 B1E EC1 C1A AD1 D1B BE1 E1C = 1) ) The machine proof −EC1 AC1 · DB1 EB1 · CA1 DA1 · BE1 CE1 · AD1 BD1 E1 = −(−S BEO) −S CEO · EC1 AC1 · DB1 EB1 · CA1 DA1 · AD1 BD1 D1 = −S BEO·(−S ADO) S CEO·(−S BDO) · EC1 AC1 · DB1 EB1 · CA1 DA1 C1 = −S CEO·S BEO·S ADO S CEO·S BDO·(−S ACO) · DB1 EB1 · CA1 DA1 simplify = S BEO·S ADO S BDO·S ACO · DB1 EB1 · CA1 DA1 B1 = S BDO·S BEO·S ADO S BDO·S ACO·S BEO · CA1 DA1 simplify = S ADO S ACO · CA1 DA1 A1 = S ACO·S ADO S ACO·S ADO simplify = 1 The eliminants BE1 CE1 E1 = S BEO S CEO AD1 BD1 D1 = S ADO S BDO EC1 AC1 C1 = S CEO −S ACO DB1 EB1 B1 = S BDO S BEO CA1 DA1 A1 = S ACO S ADO Example 6.11 (0.067, 3, 16) If the three lines joining three points marked on the sides of a tri-angle to the respectively opposite vertices are concurrent, the same is true of the isotomics of the given points. A B C O D E F 1 D 1 E 1 F H Figure 6-11 Constructive description ( (points A B C O) (inter D (l A O) (l C B)) (inter E (l B O) (l A C)) (inter F (l C O) (l A B)) (pratio D1 B C D −1) (pratio E1 C A E −1) (pratio F1 A B F −1) ( AF1 F1B BD1 D1C CE1 E1A = 1) ) The machine proof −CE1 AE1 · BD1 CD1 · AF1 BF1 F1 = −(−1) −AB BF −1 · CE1 AE1 · BD1 CD1 E1 = −(−1) ( AB BF +1)·( AC AE −1) · BD1 CD1 D1 = −1 ( AB BF +1)·( AC AE −1)·(−BC CD −1) F = S BCO (S ACBO+S BCO)·( AC AE −1)·( BC CD +1) The eliminants AF1 BF1 F1 = 1 AB BF +1 CE1 AE1 E1 = −1 AC AE −1 BD1 CD1 D1 = 1 BC CD +1 AB BF F =S ACBO S BCO AC AE E =S ABCO S ABO BC CD D = S ABOC S ACO 270 Chapter 6. Topics from Geometry The machine proof E = S BCO·S ABO (S ACBO+S BCO)·(S ABCO−S ABO)·( BC CD +1) D = S BCO·S ABO·(−S ACO) (S ACBO+S BCO)·(S ABCO−S ABO)·(−S ABOC−S ACO) area−co = (S ACO−S ABO+S ABC)·S ABO·S ACO S ACO·(S ACO−S ABO+S ABC)·S ABO simplify = 1 The eliminants S ABOC= −(S ACO−S ABO) S ABCO=S ACO+S ABC S ACBO=S ABO−S ABC S BCO=S ACO−S ABO+S ABC 6.2.3 The Cross-ratio and Harmonic Points Deﬁnition Let A, B, C, and D be four collinear points. The cross ratio of them, denoted by (ABCD), is deﬁned to be (ABCD) = (CA CB )/(DA DB ). Example 6.12 (0.066, 2, 10) The cross ratio of four points on a line is unchanged under a projection. A B O C D 1 A 1 B 1 C 1 D Figure 6-12 Constructive description ((points O A B C1 D1) (inter C (l A B) (l O C1)) (inter D (l A B) (l O D1)) (inter A1 (l C1 D1) (l O A)) (inter B1 (l C1 D1) (l O B)) ( AC BC BD AD = A1C1 A1D1 B1D1 B1C1 ) ) The machine proof BD AD · AC BC D1B1 C1B1 · C1A1 D1A1 B1 = S OBC1 S OBD1· C1A1 D1A1 · BD AD · AC BC A1 = S OBC1 ·S OAD1 S OBD1·S OAC1 · BD AD · AC BC D = (−S OBD1)·S OBC1·S OAD1 S OBD1·S OAC1 ·(−S OAD1) · AC BC simplify = S OBC1 S OAC1 · AC BC C = (−S OAC1 )·S OBC1 S OAC1 ·(−S OBC1) simplify = 1 The eliminants D1B1 C1 B1 B1 = S OBD1 S OBC1 C1A1 D1A1 A1 = S OAC1 S OAD1 BD AD D = S OBD1 S OAD1 AC BC C = S OAC1 S OBC1 6.2. Geometry of Incidence 271 Deﬁnition. For any ﬁve diﬀerent points O, A, B, C, and D, the cross-ratio of any four collinear points on lines OA, OB, OC, and OD is a constant and is denote by O(ABCD). Example 6.13 (0.050, 2, 10) Two lines OABC and OA1B1C1 intersect at point O. If (OABC) = (OA1B1C1) then AA1, BB1, and CC1 are concurrent. O A A B C B C X Figure 6-13 1 1 1 Constructive description ( (points O A A1) (mratio B A O r1) (mratio C O A r2) (mratio B1 O A1 r3) (mratio C1 O A1 r1·r2·r3) (inter X (l A A1) (l B B1)) (inter Y (l A A1) (l C C1)) ( AX XA1 = AY YA1 ) ) The machine proof ( AX A1X)/( AY A1Y) Y = S A1CC1 S ACC1 · AX A1X X = S ABB1·S A1CC1 S ACC1 ·S A1BB1 C1 = S ABB1·S OA1C·(r3·r2·r1+1) (−S AA1C·r3·r2·r1)·S A1BB1·(r3·r2·r1+1) simplify = S ABB1·S OA1C −S AA1C·r3·r2·r1·S A1BB1 B1 = (−S AA1 B·r3)·S OA1C·(r3+1) −S AA1C·r3·r2·r1·S OA1B·(r3+1) simplify = S AA1 B·S OA1C S AA1C·r2·r1·S OA1B C = S AA1B·(−S OAA1 ·r2)·(r2+1) S OAA1 ·r2·r1·S OA1B·(r2+1) simplify = −S AA1B r1·S OA1B B = −S OAA1 ·r1·(r1+1) r1·(−S OAA1)·(r1+1) simplify = 1 The eliminants AY A1Y Y = S ACC1 S A1CC1 AX A1X X = S ABB1 S A1BB1 S ACC1 C1 = −S AA1C·r3·r2·r1 r3·r2·r1+1 S A1CC1 C1 = S OA1C r3·r2·r1+1 S A1BB1 B1 = S OA1B r3+1 S ABB1 B1 = −S AA1 B·r3 r3+1 S AA1C C = S OAA1 r2+1 S OA1C C = −S OAA1 ·r2 r2+1 S OA1B B = −S OAA1 r1+1 S AA1B B = S OAA1 ·r1 r1+1 Deﬁnition. If (ABCD) = −1, we call A, B, C, D a harmonic sequence, or C, D divide the segment AB harmonically. Four lines passing through a point O is said to form a harmonic pencil if another line meets them in a harmonic sequence A, B, C, D. Each of the line is called a ray of the the pencil and O is called the center of the pencil. OA and OB, OC and OD are said to be a pair of conjugate rays, or OC and OD are harmonically separated by OA and OB. Example 6.14 (0.016, 2, 4) Let A, B, C, D be four harmonic points and O be the midpoint of AB. Then OC · OD = OA 2. 272 Chapter 6. Topics from Geometry Constructive description ( (points A B) (mratio C A B r) (mratio D A B −r) (midpoint O A B) ( OC OA = OA OD ) ) The machine proof ( CO AO)/( AO DO) O = (−AC AB + 1 2 )·(−AD AB + 1 2 ) (( 1 2 ))2 D = (2 AC AB −1)·(−r−1) −r+1 C = (r−1)·(r+1) (r−1)·(r+1) simplify = 1 The eliminants AO DO O = 1 −(2 AD AB −1) CO AO O = −(2 AC AB −1) AD AB D = r r−1 AC AB C = r r+1 Example 6.15 (0.050, 2, 3) The converse theorem. If O is the midpoint of AB, and if C, D are points of the line AB such that OC · OD = OA 2, then A, B, C, D are harmonic points. Constructive description ( (points A B) (midpoint O A B) (lratio C O A r) (lratio D O A 1 r ) (harmonic A B C D) ) The machine proof (−AC BC)/( AD BD) D = −BO AO ·r+1 −r+1 · −AC BC C = −(r−1)·( BO AO ·r−1) (r−1)·(−BO AO +r) simplify = BO AO ·r−1 BO AO −r O = (−1 2 r−1 2)·( 1 2 ) (−1 2 r−1 2)·( 1 2 ) simplify = 1 The eliminants AD BD D = r−1 BO AO ·r−1 AC BC C = r−1 −( BO AO −r) BO AO O = −(1) Example 6.16 (0.266, 8, 6) The sum of the squares of two harmonic segments is equal to four times the square of the distance between the midpoints of these segments. Constructive description ( (points A B) (mratio C A B r) (mratio D A B −r) (midpoint O A B) (midpoint M C D) (( AB MO )2+( CD MO 2) = 4) ) The eliminants AB OM M = AB CD (−1 2 )·(2 CO CD −1) CD OM M = 1 (−1 2 )·(2 CO CD −1) CO CD O = (−1 2)·(2 AC AB −1) CD AB CD AB D = AC AB ·r−AC AB −r −(r−1) AB CD D = −(r−1) AC AB ·r−AC AB −r AC AB C = r r+1 6.2. Geometry of Incidence 273 The machine proof 1 4( CD OM 2 + AB OM 2) M = CO CD 2· AB CD 2+ CO CD 2−CO CD · AB CD 2−CO CD + 1 4 AB CD 2+ 1 4 (4)·(−CO CD + 1 2 )4 simplify = AB CD 2+1 (2 CO CD −1)2 O = AB CD 2 +1 (−CD AB −2 AC AB +1)2 · ( CD AB )2 D = ( AC AB 2·r2−2 AC AB 2·r+ AC AB 2−2 AC AB ·r2+2 AC AB ·r+2r2−2r+1)·( AC AB ·r−AC AB −r)2·(−r+1)2 (−AC AB ·r+ AC AB −1)2·(−r+1)2·( AC AB ·r−AC AB −r)2 simplify = AC AB 2 ·r2−2 AC AB 2 ·r+ AC AB 2 −2 AC AB ·r2+2 AC AB ·r+2r2−2r+1 ( AC AB ·r−AC AB +1)2 C = (r4+2r2+1)·(r+1)2 (r2+1)2·(r+1)2 simplify = 1 Example 6.17 (0.083, 1, 10) Let A, B, R be three points on a plane, Q and P be points on line AR and BR respectively. S is the intersection of QB and AP. C is the intersection of RS and AB. F is the intersection of QP and AB. Show that (ABCF) = −1. A B R P Q S C F Figure 6-17 Constructive description ((points A B R) (on P (l B R)) (on Q (l A R)) (inter S (l A P) (l B Q)) (inter C (l R S ) (l A B)) (inter F (l P Q) (l A B)) ( AC BC = −AF BF ) ) The machine proof AC BC −AF BF F = S BPQ −S APQ · AC BC C = S ARS ·S BPQ −S APQ·S BRS S = −S ARP·S ABQ·S BPQ·(−S ABPQ) S APQ·(−S BRQ·S ABP)·S ABPQ simplify = −S ARP·S ABQ·S BPQ S APQ·S BRQ·S ABP 2lines = −RP·AR·β·BR·AQ·β·(−RQ·BP·β)·((2))3 (−RP·AQ·β)·(−RQ·BR·β)·BP·AR·β·((2))3 simplify = 1 The eliminants AF BF F =S APQ S BPQ AC BC C =S ARS S BRS S BRS S =S BRQ·S ABP S ABPQ S ARS S =S ARP·S ABQ S ABPQ S ABP=1 2(BP·AR·β) S BRQ= −1 2(RQ·BR·β) S APQ= −1 2(RP·AQ·β) S BPQ= −1 2(RQ·BP·β) S ABQ=1 2(BR·AQ·β) S ARP=1 2(RP·AR·β) Example 6.18 (0.033, 7, 6) Given (ABCD) = −1 and a point O outside the line AB, if a parallel through B to OA meets OC, OD in P, Q, we then have PB = BQ. 274 Chapter 6. Topics from Geometry Q P D C O B A Figure 6-18 Constructive description ( (points A B O) (mratio C A B r) (mratio D A B −r) (inter P (l O C) (p B O A)) (inter Q (l O D) (p B O A)) (midpoint B P Q) ) The machine proof −BP BQ Q = −S BOPD S BOD P = −(−S OCD·S ABO+S BOD·S AOC) S BOD·S AOC D = (S BOC·S ABO·r2−S BOC·S ABO·r)·(−r+1) S ABO·S AOC·(−r+1)2 simplify = −S BOC·r S AOC C = −S ABO·r·(r+1) (−S ABO·r)·(r+1) simplify = 1 The eliminants BP BQ Q =S BOPD S BOD S BOPD P =−(S OCD·S ABO−S BOD·S AOC) S AOC S BOD D = S ABO −(r−1) S OCD D =S BOC·r−S AOC r−1 S AOC C =−S ABO·r r+1 S BOC C =S ABO r+1 Example 6.19 (0.033, 2, 8) The converse theorem of Example 6.18. Constructive description ( (points A B O) (on C (l A B)) (inter P (l O C) (p B O A)) (lratio Q B P −1) (inter D (l O Q) (l A B)) (harmonic A B C D) ) The machine proof (−AC BC)/( AD BD) D = S BOQ S AOQ · −AC BC Q = −(−S BOP) −S AOP−2S ABO · AC BC P = −(−S BOC)·S ABO S ABO·S AOC · AC BC simplify = S BOC S AOC · AC BC C = AC AB ·(−S ABO· AC AB +S ABO) (−S ABO· AC AB )·( AC AB −1) simplify = 1 The eliminants AD BD D =S AOQ S BOQ S AOQ Q = −(S AOP+2S ABO) S BOQ Q = −(S BOP) S AOP P = −(S ABO) S BOP P =−S BOC·S ABO S AOC S AOC C = −(S ABO· AC AB) S BOC C = −(( AC AB −1)·S ABO) AC BC C = AC AB AC AB −1 6.2.4 Pappus’ Theorem and Desargues’ Theorem For the background about n3 conﬁgurations, see Subsection 2.7.2. 6.2. Geometry of Incidence 275 Example 6.20 (Pappus’ Theorem) (0.166, 1, 14) Let ABC and A1B1C1 be two lines, and P = AB1 ∩A1B, Q = AC1 ∩A1C, S = BC1 ∩B1C. Then P, Q and S are collinear. For a machine proof of this example, see Example 2.42. The following proof is shorter. A B 1 A 1 B C 1 C P Q S Figure 6-20 Constructive description ((points A A1 B B1) (on C (l A B)) (on C1 (l A1 B1)) (inter P (l A1 B) (l A B1)) (inter Q (l A C1) (l A1 C)) (inter S (l B1 C) (l B C1)) (inter T (l B1 C) (l P Q)) ( B1S CS = B1T CT ) ) The machine proof ( B1S CS )/( B1T CT ) T = S QCP −S B1QP · B1S CS S = (−S BB1C1)·S QCP −S B1QP·(−S BCC1 ) P = −S BB1C1·(−S BQC·S AA1B1)·S AA1B1B (−S A1BB1·S AB1Q)·S BCC1 ·(−S AA1B1B) simplify = S BB1C1 ·S BQC·S AA1B1 S A1BB1·S AB1Q·S BCC1 C1 = S A1BB1·S AB1Q·S BQC·S AA1B1·S AA1QB1 S A1BB1·S AB1Q·(−S BQC·S AA1 B1)·(−S AA1QB1) simplify = 1 The eliminants B1T CT T = −S B1QP S QCP B1S CS S = S BB1C1 S BCC1 S B1QP P = −S A1BB1·S AB1Q S AA1 B1B S QCP P = S BQC·S AA1B1 S AA1 B1B S BCC1 C1 = −S BQC·S AA1B1 S AA1QB1 S BB1C1 C1 = S A1BB1·S AB1Q −S AA1QB1 Line PQ is called the Pappus’ line and is denoted by . Example 6.21 (0.116, 1, 6) The dual of Pappus’ theorem. I 1 C 1 B 1 A 1 O C B A O Figure 6-21 Constructive description ( (points A B C O O1) (inter A1 (l O1 B) (l O C)) (inter B1 (l O A) (l O1 C)) (inter C1 (l O B) (l O1 A)) (inter I (l B B1) (l A A1)) (inter ZI (l C1 C) (l A A1)) ( AI A1I = AZI A1ZI ) ) The machine proof ( AI A1I)/( AZI A1ZI ) ZI = S CA1C1 −S ACC1 · AI A1I I = S ABB1·S CA1C1 −S ACC1 ·(−S BA1B1) C1 = S ABB1·S BCA1 ·S AOO1·S ABO1O S ACO1 ·S ABO·S BA1B1·(−S ABO1O) simplify = S ABB1·S BCA1 ·S AOO1 −S ACO1 ·S ABO·S BA1B1 B1 = S ACO1 ·S ABO·S BCA1 ·S AOO1·(−S ACOO1 ) −S ACO1 ·S ABO·S BCA1·S AOO1·S ACOO1 simplify = 1 The eliminants AZI A1ZI ZI = −S ACC1 S CA1C1 AI A1I I = S ABB1 −S BA1B1 S ACC1 C1 = S ACO1 ·S ABO S ABO1O S CA1C1 C1 = S BCA1 ·S AOO1 −S ABO1O S BA1B1 B1 = S BCA1 ·S AOO1 −S ACOO1 S ABB1 B1 = S ACO1 ·S ABO S ACOO1 Example 6.22 (Pappus Point Theorem) (0.550, 2, 26) The three Pappus lines , and are concurrent. So are the Pappus lines , and . 276 Chapter 6. Topics from Geometry K J I H G F E C 1 C B A 1 B 1 A Figure 6-22 Constructive description ( (points A A1 B C1) (on B1 (l A1 C1)) (on C (l A B)) (inter E (l A1 B) (l A B1)) (inter F (l A1 C) (l A C1)) (inter G (l B C1) (l A A1)) (inter H (l B B1) (l C A1)) (inter I (l B B1) (l A C1)) (inter J (l B A1) (l C C1)) (inter K (l G H) (l E F)) (collinear I J K) ) Example 6.23 (Leisening’s Theorem) (1.233, 4, 32) Continuing from Example 6.20, let O = AB∩A1B1, and L1, L2, and L3 be the intersections of lines OP and CC1, lines OQ and BB1, and lines OS and AA1 respectively. Show that L1, L2, L3 are collinear. A B A B C C P Q S O 1 L L L Figure 6-23 1 1 1 2 3 Constructive description ( (points A B A1 B1) (on C (l A B)), (on C1 (l A1 B1)) (inter P (l A B1) (l B A1)) (inter Q (l A1 C) (l A C1)) (inter S (l B C1) (l B1 C)) (inter O (l A B) (l A1 B1)) (inter L1 (l O P) (l C C1)) (inter L2 (l O Q) (l B B1)) (inter Z1 (l L1 L2) (l A A1)) (inter Z2 (l L1 L2) (l O S )) ( L1Z1 L2Z1 = L1Z2 L2Z2 ) ) Example 6.24 (Desargues’ Theorem) (0.250, 2, 18) Given two triangle ABC, A1B1C1, if the three lines AA1, BB1, CC1 meet in a point, S , the three points P = BC ∩B1C1, Q = CA ∩C1A1, R = AB ∩A1B1 lie on a line. S A B C 1 A 1 B 1 C P Q R Figure 6-24 Constructive description ( (points A B C S ) (on A1 (l S A)) (on B1 (l S B)) (on C1 (l S C)) (inter P (l B C) (l B1 C1)) (inter Q (l A C) (l A1 C1)) (inter R (l A B) (l A1 B1)) (inter Z2 (l A1 B1) (l P Q)) (inter Z1 (l A B) (l P Q)) ( PZ1 QZ1 · QZ2 PZ2 = 1) ) 6.2. Geometry of Incidence 277 The machine proof PZ1 QZ1 · QZ2 PZ2 Z1 = S ABP S ABQ · QZ2 PZ2 Z2 = S ABP·S A1B1Q S ABQ·S A1B1P Q = S ABP·S A1B1C1·S ACA1 ·S AA1CC1 S AA1C1 ·S ABC·S A1B1P·(−S AA1CC1 ) simplify = −S ABP·S A1B1C1·S ACA1 S AA1C1·S ABC·S A1B1P P = −S BB1C1·S ABC·S A1B1C1·S ACA1 ·(−S BB1CC1 ) S AA1C1·S ABC·S A1B1C1 ·S BCB1·S BB1CC1 simplify = S BB1C1·S ACA1 S AA1C1·S BCB1 C1 = (−S BCB1· SC1 SC )·S ACA1 (−S ACA1 · SC1 SC )·S BCB1 simplify = 1 The eliminants PZ1 QZ1 Z1 = S ABP S ABQ QZ2 PZ2 Z2 = S A1B1Q S A1B1P S ABQ Q = S AA1C1·S ABC S AA1CC1 S A1B1Q Q = S A1B1C1·S ACA1 −S AA1CC1 S A1B1P P = S A1B1C1 ·S BCB1 −S BB1CC1 S ABP P = S BB1C1·S ABC S BB1CC1 S AA1C1 C1 = −(S ACA1 · SC1 SC ) S BB1C1 C1 = −(S BCB1· SC1 SC ) Example 6.25 (0.216, 2, 12) The converse of Desargues’ theorem. Constructive description ( (points A B C A1) (on P (l B C)) (on Q (l A C)) (inter R (l A B) (l P Q)) (on B1 (l R A1)) (inter C1 (l A1 Q) (l B1 P)) (inter S (l A A1) (l B B1)) (inter Z (l A A1) (l C C1)) ( AS S A1 = AZ ZA1 ) ) The machine proof ( AS A1S )/( AZ A1Z) Z = −S CA1C1 S ACC1 · AS A1S S = −S ABB1·S CA1C1 S ACC1 ·(−S BA1B1) C1 = S ABB1·S A1PB1·S CA1Q·S A1PQB1 S PQB1·S ACA1 ·S BA1B1·S A1PQB1 simplify = S ABB1·S A1PB1·S CA1Q S PQB1·S ACA1 ·S BA1B1 B1 = S ABA1· RB1 RA1 ·(−S A1PR· RB1 RA1 +S A1PR)·S CA1Q S A1PQ· RB1 RA1 ·S ACA1 ·(−S BA1R· RB1 RA1 +S BA1R) simplify = S ABA1·S A1PR·S CA1Q S A1PQ·S ACA1 ·S BA1R R = S ABA1 ·S A1PQ·S ABP·S CA1Q·S APBQ S A1PQ·S ACA1 ·(−S BPQ·S ABA1 )·(−S APBQ) simplify = S ABP·S CA1Q S ACA1 ·S BPQ Q = S ABP·(−S ACA1 · AQ AC +S ACA1 ) S ACA1 ·(−S ABP· AQ AC +S ABP) simplify = 1 The eliminants AZ A1Z Z = S ACC1 −S CA1C1 AS A1S S = S ABB1 −S BA1B1 S ACC1 C1 = S PQB1·S ACA1 S A1PQB1 S CA1C1 C1 = S A1PB1·S CA1Q S A1PQB1 S BA1B1 B1 = −(( RB1 RA1 −1)·S BA1R) S PQB1 B1 =S A1PQ· RB1 RA1 S A1PB1 B1 = −(( RB1 RA1 −1)·S A1PR) S ABB1 B1 =S ABA1 · RB1 RA1 S BA1R R = −S BPQ·S ABA1 S APBQ S A1PR R = S A1PQ·S ABP −S APBQ S BPQ Q = −(( AQ AC −1)·S ABP) S CA1Q Q = −(( AQ AC −1)·S ACA1) 278 Chapter 6. Topics from Geometry 6.2.5 Miscellaneous Example 6.26 (0.133, 1, 8) In a hexagon AC1BA1CB1, BB1, C1A, A1C are concurrent and CC1, A1B, B1A are concurrent. Prove that AA1, B1C, C1B are also concurrent. A 1 C B 1 A C O H 1 B I Figure 6-26 Constructive description ( (points A A1 B C C1) (inter O (l A C1) (l A1 C)) (inter H (l A1 B) (l C C1)) (inter B1 (l A H) (l B O)) (inter I (l B1 C) (l A A1)) (inter ZI (l C1 B) (l A A1)) ( AI A1I = AZI A1ZI ) ) The machine proof ( AI A1I)/( AZI A1ZI ) ZI = −S A1BC1 −S ABC1 · AI A1I I = (−S ACB1 )·S A1BC1 S ABC1 ·(−S A1CB1) B1 = S ACH·S ABO·S A1BC1·(−S ABHO) S ABC1 ·(−S A1BC·S AOH)·S ABHO simplify = S ACH·S ABO·S A1BC1 S ABC1 ·S A1BC·S AOH H = S A1BC·S ACC1 ·S ABO·S A1BC1·(−S A1CBC1 ) S ABC1 ·S A1BC·S A1BC1·S ACO·(−S A1CBC1 ) simplify = S ACC1 ·S ABO S ABC1 ·S ACO O = S ACC1 ·S ABC1 ·S AA1C·(−S AA1C1C) S ABC1 ·(−S ACC1 ·S AA1C)·S AA1C1C simplify = 1 The eliminants AZI A1ZI ZI = S ABC1 S A1BC1 AI A1I I = S ACB1 S A1CB1 S A1CB1 B1 = S A1BC·S AOH S ABHO S ACB1 B1 = S ACH·S ABO S ABHO S AOH H = S A1BC1·S ACO −S A1CBC1 S ACH H = S A1BC·S ACC1 −S A1CBC1 S ACO O = S ACC1 ·S AA1C S AA1C1C S ABO O = S ABC1 ·S AA1C S AA1C1C Example 6.27 (Nehring’s Theorem (1942)) (0.533, 2, 14) Let AA1, BB1, CC1 be three con-current cevian lines for triangle ABC. Let X1 be a point on BC, X2 = X1B1 ∩BA, X3 = X2A1 ∩AC, X4 = X3C1 ∩CB, X5 = X4B1 ∩BA, X6 = X5A1 ∩AC, X7 = X6C1 ∩CB. Show X7 = X1. 6 X 5 X 4 X 3 X 2 X 1 X 1 C 1 B 1 A O A C B Figure 6-27 Constructive description ((points B C A O) (inter A1 (l A O) (l B C)) (inter B1 (l B O) (l C A)) (inter C1 (l C O) (l B A)) (on X1 (l B C)) (inter X2 (l B1 X1) (l B A)) (inter X3 (l A1 X2) (l C A)) (inter X4 (l C1 X3) (l B C)) (inter X5 (l B1 X4) (l B A)) (inter X6 (l A1 X5) (l C A)) (inter Z2 (l C A) (l C1 X1)) (inter Z1 (l A1 X5) (l C1 X1)) 6.2. Geometry of Incidence 279 ( C1Z1 X1Z1 · X1Z2 C1Z2 = 1) ) The machine proof C1Z1 X1Z1 · X1Z2 C1Z2 Z1 = −S A1C1X5 −S A1X1X5 · X1Z2 C1Z2 Z2 = S A1C1X5·S CAX1 S A1X1X5·S CAC1 X5 = (−S B1C1X4·S BAA1)·S CAX1 ·S BB1AX4 S AA1X1·S BB1X4·S CAC1 ·S BB1AX4 simplify = −S B1C1X4·S BAA1·S CAX1 S AA1X1·S BB1X4·S CAC1 X4 = −S B1C1X3·S BCC1 ·S BAA1·S CAX1 ·S BC1CX3 S AA1X1·(−S BC1X3·S BCB1)·S CAC1 ·(−S BC1CX3 ) simplify = −S B1C1X3·S BCC1 ·S BAA1·S CAX1 S AA1X1·S BC1X3·S BCB1·S CAC1 X3 = −S A1B1X2·S CAC1 ·S BCC1 ·S BAA1·S CAX1 ·S CA1AX2 S AA1X1·S AA1X2·S BCC1 ·S BCB1·S CAC1 ·S CA1AX2 simplify = −S A1B1X2·S BAA1·S CAX1 S AA1X1·S AA1X2·S BCB1 X2 = −S A1B1X1·S BAB1·S BAA1·S CAX1 ·S BB1AX1 S AA1X1·(−S AB1X1·S BAA1)·S BCB1·(−S BB1AX1 ) simplify = −S A1B1X1·S BAB1·S CAX1 S AA1X1·S AB1X1·S BCB1 X1 = −( BA1 BC ·S BCB1−S BCB1 · BX1 BC )·S BAB1·(−S BCA· BX1 BC +S BCA) (−BA1 BC ·S BCA+S BCA· BX1 BC )·(−S BAB1· BX1 BC +S BAB1)·S BCB1 simplify = 1 The eliminants C1Z1 X1Z1 Z1 = S A1C1X5 S A1X1X5 X1Z2 C1Z2 Z2 = S CAX1 S CAC1 S A1X1X5 X5 = S AA1X1·S BB1X4 S BB1AX4 S A1C1X5 X5 = −S B1C1X4·S BAA1 S BB1AX4 S BB1X4 X4 = −S BC1X3·S BCB1 S BC1CX3 S B1C1X4 X4 = S B1C1X3·S BCC1 −S BC1CX3 S BC1X3 X3 = S AA1X2·S BCC1 S CA1AX2 S B1C1X3 X3 = S A1B1X2·S CAC1 S CA1AX2 S AA1X2 X2 = −S AB1X1·S BAA1 S BB1AX1 S A1B1X2 X2 = S A1B1X1·S BAB1 −S BB1AX1 S AB1X1 X1 = −(( BX1 BC −1)·S BAB1) S AA1X1 X1 = −(( BA1 BC −BX1 BC )·S BCA) S CAX1 X1 = −(( BX1 BC −1)·S BCA) S A1B1X1 X1 =( BA1 BC −BX1 BC )·S BCB1 Example 6.28 (0.733, 2, 12) Let three triangles ABC, A1B1C1, A2B2C2 be given such that lines AB, A1B1, A2B2 intersect in a point P, lines AC, A1C1, A2C2 intersect in a point Q, lines BC, B1C1, B2C2 intersect in a point R, and P, Q, R are collinear. In view of Desargues’ theorem, the lines in each of the triads AA1, BB1, CC1; AA2, BB2, CC2; A1A2, B1B2, C1C2; intersect in a point. Prove that these three points are collinear. A B C 1 A 1 B 2 B P Q R 1 C 2 A 2 C I J K Figure 6-28 Constructive description ( (points A A1 B B1 B2 C) (inter P (l A1 B1) (l A B)) (on Q (l A C)) (inter R (l P Q) (l B C)) (inter C1 (l B1 R) (l A1 Q)) (on A2 (l B2 P)) (inter C2 (l B2 R) (l A2 Q)) (inter I (l B B1) (l A A1)) 280 Chapter 6. Topics from Geometry (inter J (l B B2) (l A A2)) (inter K (l B1 B2) (l A1 A2)) (inter ZK (l J I) (l B1 B2)) ( B1K B2K = B1ZK B2ZK ) ) The machine proof ( B1K B2K)/( B1ZK B2ZK ) ZK = −S B2IJ −S B1IJ · B1K B2K K = (−S A1B1A2)·S B2IJ S B1IJ·(−S A1B2A2) J = S A1B1A2·S BB2I·S AB2A2·(−S ABA2B2) S B1B2I·S ABA2·S A1B2A2·(−S ABA2B2) simplify = S A1B1A2·S BB2I·S AB2A2 S B1B2I·S ABA2·S A1B2A2 I = S A1B1A2·(−S BB1B2·S AA1B)·S AB2A2·(−S ABA1B1) (−S BB1B2·S AA1B1)·S ABA2·S A1B2A2·(−S ABA1B1) simplify = S A1B1A2·S AA1 B·S AB2A2 S AA1 B1·S ABA2·S A1B2A2 A2 = (−S A1B1B2· B2A2 B2P +S A1B1B2)·S AA1B·S AB2P· B2A2 B2P S AA1B1·(−S ABB2· B2A2 B2P +S ABB2)·S A1B2P· B2A2 B2P simplify = S A1B1B2·S AA1 B·S AB2P S AA1 B1·S ABB2·S A1B2P P = S A1B1B2·S AA1 B·(−S ABB2·S AA1B1)·(−S AA1 BB1) S AA1 B1·S ABB2·S A1B1B2·S AA1B·S AA1BB1 simplify = 1 The eliminants B1ZK B2ZK ZK = S B1IJ S B2IJ B1K B2K K = S A1B1A2 S A1B2A2 S B1IJ J = S B1B2I·S ABA2 −S ABA2B2 S B2IJ J = S BB2I·S AB2A2 −S ABA2B2 S B1B2I I = S BB1B2·S AA1 B1 S ABA1B1 S BB2I I = S BB1B2·S AA1 B S ABA1B1 S A1B2A2 A2 =S A1B2P· B2A2 B2P S ABA2 A2 = −(( B2A2 B2P −1)·S ABB2) S AB2A2 A2 =S AB2P· B2A2 B2P S A1B1A2 A2 = −(( B2A2 B2P −1)·S A1B1B2) S A1B2P P = S A1B1B2·S AA1B −S AA1 BB1 S AB2P P = −S ABB2·S AA1B1 S AA1 BB1 Example 6.29 (0.250, 1, 18) If (Q) is the cevian triangle of a point M for the triangle (P), show that the triangle formed by the parallels through the vertices of (P) to the correspond-ing sides of (Q) is perspective to (P). A B C M 1 A 1 B 1 C 2 C 2 A 2 B P Figure 6-29 Constructive description ( (points A B C M) (inter A1 (l B C) (l A M)) (inter B1 (l A C) (l B M)) (inter C1 (l A B) (l C M)) (inter C2 (p B A1 C1) (p A B1 C1)) (inter A2 (l C2 B) (p C A1 B1)) (inter B2 (l C2 A) (l A2 C)) (inter P (l B B2) (l A A2)) (inter ZP (l C2 C) (l A A2)) ( AP A2P = AZP A2ZP ) ) 6.2. Geometry of Incidence 281 The machine proof ( AP A2P)/( AZP A2ZP) ZP = −S CC2 A2 −S ACC2 · AP A2P P = S ABB2·S CC2A2 S ACC2 ·(−S BA2B2) B2 = −S ACA2 ·S ABC2 ·S CC2A2·(−S ACC2 A2) S ACC2 ·S BCA2·S AC2A2·S ACC2 A2 simplify = S ACA2 ·S ABC2 ·S CC2A2 S ACC2 ·S BCA2·S AC2A2 A2 = S BCC2 ·S ACA1 ·S ABC2·S CA1C2B1·S BCC2 ·(S BA1C2B1)2 S ACC2 ·S BCC2 ·S BCB1·(−S CA1C2B1·S ABC2)·S BA1C2B1·(−S BA1C2 B1) simplify = S ACA1 ·S BCC2 S ACC2 ·S BCB1 C2 = S ACA1 ·S BCC1 ·S ABB1·(−S A1B1C1) (−S ACC1 ·S ABA1)·S BCB1·S A1B1C1 simplify = S ACA1 ·S BCC1 ·S ABB1 S ACC1 ·S ABA1·S BCB1 C1 = S ACA1 ·(−S BCM·S ABC)·S ABB1·S ACBM (−S ACM·S ABC)·S ABA1·S BCB1·S ACBM simplify = S ACA1 ·S BCM·S ABB1 S ACM·S ABA1·S BCB1 B1 = S ACA1 ·S BCM·S ABM·S ABC·S ABCM S ACM·S ABA1·S BCM·S ABC·S ABCM simplify = S ACA1 ·S ABM S ACM·S ABA1 A1 = (−S ACM·S ABC)·S ABM·(−S ABMC) S ACM·(−S ABM·S ABC)·(−S ABMC) simplify = 1 The eliminants AZP A2ZP ZP = S ACC2 S CC2A2 AP A2P P = S ABB2 −S BA2B2 S BA2B2 B2 = S BCA2 ·S AC2A2 −S ACC2 A2 S ABB2 B2 = S ACA2 ·S ABC2 S ACC2 A2 S AC2A2 A2 = −S CA1C2B1·S ABC2 S BA1C2 B1 S BCA2 A2 = S BCC2 ·S BCB1 S BA1C2B1 S CC2 A2 A2 = S CA1C2B1·S BCC2 S BA1C2 B1 S ACA2 A2 = S BCC2 ·S ACA1 −S BA1C2B1 S ACC2 C2 = S ACC1 ·S ABA1 S A1B1C1 S BCC2 C2 = S BCC1 ·S ABB1 S A1B1C1 S ACC1 C1 = −S ACM·S ABC S ACBM S BCC1 C1 = −S BCM·S ABC S ACBM S BCB1 B1 = S BCM·S ABC S ABCM S ABB1 B1 = S ABM·S ABC S ABCM S ABA1 A1 = S ABM·S ABC S ABMC S ACA1 A1 = S ACM·S ABC S ABMC Example 6.30 (0.066, 2, 7) If a hexagon ABCDEF has two opposite sides BC and EF par-allel to the diagonal AD and two opposite sides CD and FA parallel to the diagonal BE, while the remaining sides DE and AB also are parallel, then the third diagonal CF is parallel to AB. H G 2 N 1 N F A D E C B Figure 6-30 Constructive description ( (points B C E) (on D (p C B E)) (inter A (p B D E) (p D B C)) (inter F (p A C D) (p E B C)) (midpoint N1 A C) (midpoint N2 C E) (inter G (l A N2) (l E N1)) (inter H (l D F) (l G B)) 282 Chapter 6. Topics from Geometry (parallel C F A B) ) The machine proof S BCA −S BAF F = S BCA·(−S BCD) −S BECA·S BCAD A = (S BCD)2·(−S BECD) (S BCD−S BCE)·(−S BECD·S BCD+S BED·S BCD) simplify = S BCD·S BECD (S BCD−S BCE)·(S BECD−S BED) D = S BCE· CD BE ·(S BCE· CD BE −S BCE) (S BCE· CD BE −S BCE)·S BCE· CD BE simplify = 1 The eliminants S BAF F =S BECA·S BCAD −S BCD S BCAD A =(S BECD−S BED)·S BCD S BECD S BECA A =S BCD−S BCE S BCA A =S BCD S BED D = −(S BCE) S BECD D =( CD BE −1)·S BCE S BCD D =S BCE· CD BE Example 6.31 (0.683, 4, 37) Prove that the lines joining the midpoints of three concurrent cevians to the midpoints of the corresponding sides of the given triangle are concurrent. I 1 C 1 B 1 A 1 F 1 E 1 D F E D O C B A Figure 6-31 Constructive description ( (points A B C O) (inter D (l B C) (l A O)) (inter E (l A C) (l B O)) (inter F (l A B) (l C O)) (midpoint D1 A D) (midpoint E1 B E) (midpoint F1 C F) (midpoint A1 B C) (midpoint B1 C A) (midpoint C1 A B) (inter I (l B1 E1) (l A1 D1)) (collinear I C1 F1) ) Example 6.32 (1.583, 8, 28) Let O and U be two points in the plane of triangle ABC. Let AO, BO, CO intersect the opposite sides BC, CA, AB in P, Q, R. Let PU, QU, RU intersect QR, RP, PQ respectively in X, Y, Z. Show that AX, BY, CZ are concurrent. A B C O U Z P Q R X Y I Figure 6-32 Constructive description ( (points A B C O U) (inter P (l B C) (l A O)) (inter Q (l C A) (l B O)) (inter R (l A B) (l C O)) (inter X (l Q R) (l P U)) (inter Y (l R P) (l Q U)) (inter Z (l P Q) (l R U)) (inter I (l B Y) (l A X)) (collinear C Z I) ) 6.2. Geometry of Incidence 283 Example 6.33 (0.667, 4, 27) Let O be a point in the plane of a triangle ABC, and let A1, B1, C1 be the points of intersection of the lines AO, BO, CO with the sides of the triangle opposite A, B, C. Prove that if the points A2, B2, C2 on sides B1C1, C1A1, A1B1 of △A1B1C1 are collinear, then the points of intersection of the liens AA2, BB2, CC2 with the opposite sides of △ABC are collinear. 3 C 3 B 3 A 2 C 2 B 2 A 1 C 1 B 1 A O C B A Figure 6-33 Constructive description ( (points A B C O) (inter A1 (l B C) (l A O)) (inter B1 (l A C) (l B O)) (inter C1 (l A B) (l C O)) (on A2 (l B1 C1)) (on B2 (l A1 C1)) (inter C2 (l A2 B2) (l A1 B1)) (inter A3 (l B C) (l A A2)) (inter B3 (l A C) (l B B2)) (inter C3 (l A B) (l C C2)) (collinear A3 B3 C3) ) Example 6.34 (0.533, 5, 20) The sides BC, CA, AB of a triangle ABC are met by two transversal PQR, P1Q1R1 in the pairs of points P, P1; Q, Q1; R, R1. Show that the points X = BC ∩QR1, Y = CA ∩RP1, Z = AB ∩PQ1 are collinear. Z Y X 1 R 1 Q 1 P R Q P A C B Figure 6-34 Constructive description ( (points A B C) (on P (l B C)) (on Q (l C A)) (inter R (l P Q) (l A B)) (on P1 (l B C)) (on Q1 (l C A)) (inter R1 (l P1 Q1) (l A B)) (inter X (l Q R1) (l B C)) (inter Y (l R P1) (l C A)) (inter Z (l P Q1) (l A B)) (collinear X Y Z) ) Example 6.35 (0.516, 4, 28) Through the vertices of a triangle ABC lines are drawn inter-secting in O and meeting the opposite sides in D, E, F. Prove that the lines joining A, B, C to the midpoints of EF, FD, DE are concurrent. I 1 C 1 B 1 A F E D O C B A Figure 6-35 Constructive description ( (points A B C O) (inter D (l B C) (l A O)) (inter E (l A C) (l B O)) (inter F (l A B) (l C O)) (midpoint A1 E F) (midpoint B1 F D) (midpoint C1 D E) (inter I (l B B1) (l A A1)) (collinear I C C1) ) 284 Chapter 6. Topics from Geometry Example 6.36 (0.500, 4, 25) A transversal cuts the sides BC, CA, AB of triangle ABC in D, E, F. P, Q, R are the midpoints of EF, FD, DE, and AP, BQ, CR intersect BC, CA, AB in X, Y, Z. Show that X, Y, Z are collinear. Z Y X R Q P F E D C B A Figure 6-36 Constructive description ((points A B C) (on D (l B C)) (on E (l A C)) (inter F (l E D) (l A B)) (midpoint P E F) (midpoint Q F D) (midpoint R D E) (inter X (l A P) (l B C)) (inter Y (l B Q) (l A C)) (inter Z (l C R) (l A B)) (collinear X Y Z) ) Example 6.37 (0.200, 2, 19) The lines AL, BL, CL joining the vertices of a triangle ABC to a point L meet the respectively opposite sides in A1, B1, C1. The parallels through A1 to BB1 CC1 meet AC, AB in P, Q, and the parallels through A1 to AC, AB meet BB1, CC1 in R, S . Show that the four points P, Q, R, S are collinear. B C A L 1 A 1 B 1 C P Q R Figure 6-37 Constructive description ( (points A B C L) (inter A1 (l B C) (l A L)) (inter B1 (l A C) (l B L)) (inter C1 (l A B) (l C L)) (inter P (l A C) (p A1 B B1)) (inter Q (l A B) (p A1 C C1)) (inter R (l B B1) (p A1 A C)) (inter ZR (l Q P) (l B1 B)) ( B1R BR = B1ZR BZR ) ) Example 6.38 (0.533, 3, 18) Starting from ﬁve points A, B, C, D and E with A, B, C collinear, new lines and points of intersection are formed as in Figure 6-38. Show that AB, GJ and HI are collinear. O H K G L J I F C E D B A Figure 6-38 Constructive description for (1) ( (points A B D E) (on C (l A B)) (inter F (l E D) (l A B)) (inter I (l D B) (l A E)) (inter J (l C D) (l A E)) (inter L (l C D) (l B E)) (inter G (l B E) (l A D)) (inter K (l C E) (l D B)) (inter H (l E C) (l A D)) (inter O (l G J) (l A B)) (inter ZO (l H I) (l F C)) ( FO CO = FZO CZO ) ) 6.3. Triangles 285 6.3 Triangles 6.3.1 Medians and Centroids A B C M N Figure 6-39 Example 6.39 (0.001, 1, 2) The line joining the midpoints of two sides of a triangle is parallel to the third side and is equal to one-half its length. We have to prove two results. Constructive description ( (points A B C) (midpoint M A B) (midpoint N A C) (parallel M N B C) ) The machine proof S BCM S BCN N = S BCM 1 2 S ABC M = (2)·( 1 2 S ABC) S ABC simplify = 1 The eliminants S BCN N =1 2(S ABC) S BCM M = 1 2(S ABC) Constructive description ( (points A B C) (midpoint M A B) (midpoint N A C) ( MN BC = 1/2) ) The machine proof 2( MN BC ) N = S ACM ( 1 2 )·(−S ABC) M = (−2)·(−1 2 S ABC) S ABC simplify = 1 The eliminants MN BC N = S ACM −S ABC S ACM M = −1 2(S ABC) Deﬁnition. The line joining a vertex of a triangle and the midpoint of the opposite side is called a median of the triangle. Example 6.40 (Theorem of Centroid) (0.016, 1, 4) The three medians of a triangle meet in a point, and each median is trisected by this point. A B C E F D O Figure 6-40 Constructive description ( (points A B C) (midpoint F A C) (midpoint E B C) (inter O (l A E) (l B F)) ( AO OE = 2) ) The machine proof 1 2(−AO EO) O = −S ABF (2)·S BFE E = −S ABF (2)·(−1 2 S BCF) F = 1 2 S ABC 1 2 S ABC simplify = 1 The eliminants AO EO O = S ABF S BFE S BFE E = −1 2(S BCF) S BCF F = 1 2(S ABC) S ABF F =1 2(S ABC) 286 Chapter 6. Topics from Geometry For other forms of the centroid theorem, see Examples 5.46 and 5.47 on page 243. A B C M N K P L Figure 6-41 Example 6.41 (0.083, 2, 9) With the medians of a triangle a new triangle is constructed. The medians of the second triangle are equal to the three-fourth of the respective sides of the given triangle. Constructive description ( (points A B C) (midpoint M A B) (midpoint N A C) (midpoint K B C) (inter P (p A C M) (p K B N)) (midpoint L A K) ( LP BC = 3/4) ) The machine proof 1 3((−4)· PL BC) L = (−4)·S AKP (3)·(−S ABC) P = (4)·S ACKM·S ABKN (3)·S ABC·(−S BCNM) K = (−4)·(−1 2 S BCM+S ACM)·( 1 2 S BCN+S ABN) (3)·S ABC·S BCNM N = (S BCM−2S ACM)·( 3 2 S ABC) (3)·S ABC·(S BCM−1 2 S ACM) simplify = S BCM−2S ACM 2S BCM−S ACM M = 3 2 S ABC 3 2 S ABC simplify = 1 The eliminants PL BC L = S AKP −S ABC S AKP P = S ACKM·S ABKN −S BCNM S ABKN K =1 2(S BCN+2S ABN) S ACKM K = −1 2(S BCM−2S ACM) S BCNM N =1 2(2S BCM−S ACM) S ABN N =1 2(S ABC) S BCN N =1 2(S ABC) S ACM M = −1 2(S ABC) S BCM M = 1 2(S ABC) Example 6.42 (0.050, 2, 8) The area of the triangle having for sides the medians of a given triangle is equal to three-fourth of the given triangle (Figure 6-41). Constructive description ( (points A B C) (midpoint M A B) (midpoint N A C) (midpoint K B C) (inter P (p A C M) (p K B N)) (midpoint L A K) (4S AKP = 3S ABC) ) The machine proof (4)·S AKP (3)·S ABC P = (4)·S ACKM·S ABKN (3)·S ABC·(−S BCNM) K = (−4)·(−1 2 S BCM+S ACM)·( 1 2 S BCN+S ABN) (3)·S ABC·S BCNM N = (S BCM−2S ACM)·( 3 2 S ABC) (3)·S ABC·(S BCM−1 2 S ACM) simplify = S BCM−2S ACM 2S BCM−S ACM M = 3 2 S ABC 3 2 S ABC simplify = 1 The eliminants S AKP P = S ACKM·S ABKN −S BCNM S ABKN K =1 2(S BCN+2S ABN) S ACKM K = −1 2(S BCM−2S ACM) S BCNM N =1 2(2S BCM−S ACM) S ABN N =1 2(S ABC) S BCN N =1 2(S ABC) S ACM M = −1 2(S ABC) S BCM M = 1 2(S ABC) 6.3. Triangles 287 Example 6.43 (0.016, 1, 4) Show that the line joining the midpoint of a median to a vertex of the triangle trisects the side opposite the vertex considered. A B C F N K Figure 6-43 Constructive description ( (points A B C) (midpoint F A B) (midpoint N C F) (inter K (l B C) (l A N)) ( BK KC = 2) ) The machine proof 1 2(−BK CK) K = −(−S ABN) (2)·(−S ACN) N = −( 1 2S ABC) (2)·( 1 2 S ACF) F = −S ABC (2)·(−1 2 S ABC) simplify = 1 The eliminants BK CK K = S ABN S ACN S ACN N =1 2(S ACF) S ABN N =1 2(S ABC) S ACF F = −1 2(S ABC) Example 6.44 (0.033, 2, 6) Show that a parallel to a side of a triangle through the centroid divides the area of the triangle into two parts, in the ratio 4:5. A B C G P Q Figure 6-44 Constructive description ( (points A B C) (centroid G A B C) (inter P (l A B) (p G B C)) (inter Q (l A C) (p G B C)) (4S PQCB = 5S AQP) ) The machine proof (4)·S BCQP (5)·S APQ Q = (4)·(S BCP·S ABC−S BCG·S ACP)·S ABC (5)·(−S ABGC·S ACP)·S ABC simplify = (−4)·(S BCP·S ABC−S BCG·S ACP) (5)·S ABGC·S ACP P = (−4)·(S ABGC·S BCG·S ABC+S BCG·S 2 ABC)·S ABC (5)·S ABGC·(−S ABGC·S ABC)·S ABC simplify = (4)·(S ABGC+S ABC)·S BCG (5)·(S ABGC)2 G = (4)·(5S ABC)·S ABC·((3))2 (5)·((2S ABC))2·((3))2 simplify = 1 The eliminants S APQ Q = −S ABGC·S ACP S ABC S BCQP Q =S BCP·S ABC−S BCG·S ACP S ABC S ACP P =−S ABGC S BCP P =S BCG S BCG G =1 3(S ABC) S ABGC G =2 3(S ABC) Example 6.45 (0.001, 2, 5) If L is the harmonic conjugate of the centroid G of a triangle ABC with respect to the ends A, D of the median AD, show that LD = AD. 288 Chapter 6. Topics from Geometry A B C D G L Figure 6-45 Constructive description ( (points A B C) (midpoint D B C) (centroid G A B C) (harmonic L G D A) (midpoint D L A) ) The machine proof DL AD L = 1 −DG AG −1 · DG AG G = −S ACD·S ABC (S ACD+S ABC)·S ABC simplify = −S ACD S ACD+S ABC D = −(−1 2 S ABC) 1 2 S ABC simplify = 1 The eliminants DL AD L = DG AG −( DG AG +1) DG AG G =S ACD S ABC S ABD D = 1 2(S ABC) S ACD D = −1 2(S ABC) Example 6.46 (0.033, 1, 6) Show that the distances of a point on a median of triangle from the sides including the median are inversely proportional to these sides. J K N F C B A Figure 6-46 Constructive description ( (points A B C) (midpoint F A B) (on N (l C F)) (foot K N A C) (foot J N B C) (eq-product N K A C N J B C) ) The machine proof PNKN·PACA PNJN·PBCB J = PNKN·PACA·PBCB (16S 2 BCN)·PBCB simplify = PNKN·PACA (16)·(S BCN)2 K = (16S 2 ACN)·PACA (16)·(S BCN)2·PACA simplify = (S ACN)2 (S BCN)2 N = (S ACF· CN CF )2 (S BCF· CN CF )2 simplify = (S ACF)2 (S BCF)2 F = ((−1 2 S ABC))2 (( 1 2 S ABC))2 simplify = 1 The eliminants PNJN J =(16)·(S BCN)2 PBCB PNKN K =(16)·(S ACN)2 PACA S BCN N =S BCF· CN CF S ACN N =S ACF· CN CF S BCF F = 1 2(S ABC) S ACF F = −1 2(S ABC) Example 6.47 (0.066, 5, 7) Show that, if a line through the centroid G of the triangle ABC meets AB in M and AC in N, we have, both in magnitude and in sign, AN · MB+AM·NC = AM · AN. 6.3. Triangles 289 B C A D G M N Figure 6-47 Constructive description ((points A B C) (centroid G A B C) (on M (l A B)) (inter N (l G M) (l A C)) ( MB AM + NC AN = 1) ) The eliminants CN AN N =S CGM S AGM S CGM M =S BCG· AM AB −S ACG· AM AB +S ACG S AGM M = −(S ABG· AM AB ) BM AM M = AM AB −1 AM AB S ABG G =1 3(S ABC) S ACG G = −1 3(S ABC) S BCG G =1 3(S ABC) The machine proof −( CN AN + BM AM ) N = −( BM AM ·S AGM+S CGM) S AGM M = −(S BCG· AM AB 2−S ACG· AM AB 2+S ACG· AM AB −S ABG· AM AB 2+S ABG· AM AB ) (−S ABG· AM AB )· AM AB simplify = S BCG· AM AB −S ACG· AM AB +S ACG−S ABG· AM AB +S ABG S ABG· AM AB G = (9S ABC· AM AB )·(3) S ABC· AM AB ·((3))3 simplify = 1 Example 6.48 (0.033, 1, 8) Two equal segments AE, AF are taken on the sides AB, AC of the triangle ABC. Show that median issued from A divides EF in the ratio of the sides AC, AB. D 1 A F E C B A Figure 6-48 Constructive description ((points F E) (on A (b F E)) (on C (l A F)) (on B (l A E)) (midpoint A1 B C) (inter D (l E F) (l A A1)) ( ED FD ED FD PABA = PACA) ) The machine proof PABA PACA · ( ED FD )2 D = (S EAA1 )2·PABA PACA·(S FAA1)2 A1 = (( 1 2 S EAC))2·PABA PACA·(( 1 2 S FAB))2 B = (S EAC)2·PEAE· AB AE 2 PACA·((−S FEA· AB AE ))2 simplify = (S EAC)2·PEAE PACA·(S FEA)2 C = (S FEA· AC AF )2·PEAE PFAF· AC AF 2·(S FEA)2 simplify = PEAE PFAF A = PFAF PFAF simplify = 1 The eliminants ED FD D = S EAA1 S FAA1 S FAA1 A1 = 1 2(S FAB) S EAA1 A1 = 1 2(S EAC) S FAB B = −(S FEA· AB AE) PABA B =PEAE·( AB AE )2 PACA C =PFAF·( AC AF )2 S EAC C =S FEA· AC AF PEAE A =PFAF 290 Chapter 6. Topics from Geometry Example 6.49 (0.250, 4, 8) Show that the (algebraic) sum of the distances of the vertices of a triangle from any line in the plane is equal to the sum of the distances of the midpoints of the sides of the triangle from this line. A B C X Y D E F 1 A 1 B 1 C 1 D 1 E 1 F Figure 6-49 Constructive description ( (points A B C X Y) (midpoint D B C) (midpoint E A C) (midpoint F A B) (foot A1 A X Y) (foot B1 B X Y) (foot C1 C X Y) (foot D1 D X Y) (foot E1 E X Y) (foot F1 F X Y) (1+ BB1 AA1 + CC1 AA1 = DD1 AA1 + EE1 AA1 + FF1 AA1 ) ) The eliminants FF1 AA1 F1 = S XYF S AXY EE1 AA1 E1 = S XYE S AXY DD1 AA1 D1 = S XYD S AXY CC1 AA1 C1 = S CXY S AXY BB1 AA1 B1 = S BXY S AXY S XYF F =1 2(S BXY+S AXY) S XYE E = 1 2(S CXY +S AXY) S XYD D =1 2(S CXY +S BXY) The machine proof CC1 AA1 + BB1 AA1 +1 FF1 AA1 + EE1 AA1 + DD1 AA1 F1 = ( CC1 AA1 + BB1 AA1 +1)·S AXY EE1 AA1 ·S AXY+ DD1 AA1 ·S AXY+S XYF E1 = ( CC1 AA1 + BB1 AA1 +1)·(S AXY)2 DD1 AA1 ·S 2 AXY+S XYF·S AXY+S XYE·S AXY simplify = ( CC1 AA1 + BB1 AA1 +1)·S AXY DD1 AA1 ·S AXY+S XYF+S XYE D1 = ( CC1 AA1 + BB1 AA1 +1)·(S AXY)2 S XYF·S AXY+S XYE·S AXY+S XYD·S AXY simplify = ( CC1 AA1 + BB1 AA1 +1)·S AXY S XYF+S XYE+S XYD C1 = ( BB1 AA1 ·S AXY+S CXY +S AXY)·S AXY (S XYF+S XYE+S XYD)·S AXY simplify = BB1 AA1 ·S AXY+S CXY +S AXY S XYF+S XYE+S XYD B1 = S CXY ·S AXY+S BXY·S AXY+S 2 AXY (S XYF+S XYE+S XYD)·S AXY simplify = S CXY +S BXY+S AXY S XYF+S XYE+S XYD F = S CXY +S BXY+S AXY S XYE+S XYD+ 1 2 S BXY+ 1 2 S AXY E = (2)·(S CXY +S BXY+S AXY) 2S XYD+S CXY +S BXY+2S AXY D = (2)·(S CXY +S BXY+S AXY) 2S CXY +2S BXY+2S AXY simplify = 1 Example 6.50 (0.016, 3, 1) Compute the square of the lengths of the medians. Constructive description ( (points A B C) (midpoint M B C) (MA 2) ) The machine proof 1 2(PAMA) M = −1 4 PBCB+ 1 2 PACA+ 1 2 PABA 2 The eliminants PAMA M = −1 4(PBCB−2PACA−2PABA) Example 6.51 (0.066, 2, 14) If K, K1 are two isotomic points on the side BC of the triangle ABC, and the line AK meets the line NM of in K2, where N and M are the midpoints of AB and AC. Show that line K1K2 passes through the centroid G of ABC. 6.3. Triangles 291 A B C N M G K 2 K 1 K Figure 6-51 Constructive description ( (points A B C) (midpoint N A B) (midpoint M A C) (inter G (l C N) (l B M)) (lratio K B C r) (lratio K1 C B r) (inter K2 (l M N) (l A K)) (inter Z (l B C) (l K2 G)) ( BZ ZC = BK1 K1C ) ) The machine proof ( BZ CZ)/( BK1 CK1 ) Z = −S BGK2 BK1 CK1 ·(−S CGK2 ) K2 = (−S BNG·S AMK)·(−S ANKM) BK1 CK1 ·S CMG·S ANK·(−S ANKM) simplify = −S BNG·S AMK BK1 CK1 ·S CMG·S ANK K1 = −S BNG·S AMK·r (r−1)·S CMG·S ANK K = −S BNG·(S ABM·r−S ABM)·r (r−1)·S CMG·(−S ACN·r) simplify = S BNG·S ABM S CMG·S ACN G = (−S BNM·S BCN)·S ABM·(−S BCMN) S CNM·S BCM·S ACN·(−S BCMN) simplify = −S BNM·S BCN·S ABM S CNM·S BCM·S ACN M = −(−1 2 S BCN)·S BCN·( 1 2 S ABC) ( 1 2 S ACN)·( 1 2S ABC)·S ACN simplify = (S BCN)2 (S ACN)2 N = (( 1 2 S ABC))2 ((−1 2 S ABC))2 simplify = 1 The eliminants BZ CZ Z = S BGK2 S CGK2 S CGK2 K2 = S CMG·S ANK −S ANKM S BGK2 K2 = S BNG·S AMK S ANKM BK1 CK1 K1 = r−1 r S ANK K = −(S ACN·r) S AMK K =(r−1)·S ABM S CMG G =S CNM·S BCM −S BCMN S BNG G =S BNM·S BCN S BCMN S BCM M = 1 2(S ABC) S CNM M = 1 2(S ACN) S ABM M = 1 2(S ABC) S BNM M = −1 2(S BCN) S ACN N = −1 2(S ABC) S BCN N =1 2(S ABC) Example 6.52 (0.083, 5, 3) The sum of the squares of the medians of a triangle is equal to three-fourth the sum of the squares of the sides. A B C E F D Figure 6-52 Constructive description ( (points A B C) (midpoint E A C) (midpoint F A B) (midpoint D B C) (4AD 2+4FC 2+4BE 2 = 3AC 2+3AB 2+3BC 2) ) 292 Chapter 6. Topics from Geometry The machine proof (4)·(PCFC+PBEB+PADA) (3)·(PBCB+PACA+PABA) D = (4)·(PCFC+PBEB−1 4 PBCB+ 1 2 PACA+ 1 2 PABA) (3)·(PBCB+PACA+PABA) F = 4PBEB+PBCB+4PACA+PABA (3)·(PBCB+PACA+PABA) E = 3PBCB+3PACA+3PABA (3)·(PBCB+PACA+PABA) simplify = 1 The eliminants PADA D = −1 4(PBCB−2PACA−2PABA) PCFC F =1 4(2PBCB+2PACA−PABA) PBEB E = 1 4(2PBCB−PACA+2PABA) Example 6.53 (0.050, 4, 6) If two points are equidistant from the centroid of a triangle, show that the sums of the squares of their distances from the vertices of a triangle are equal. F A B C G N M Figure 6-53 Constructive description ((points A B N M) (on G (b N M)) (midpoint F A B) (lratio C F G 3) (AN 2+NB 2+NC 2 = AM 2+MB 2+MC 2) ) The machine proof. PNCN+PBNB+PANA PMCM+PBMB+PAMA C = 6PGFG−2PNFN+3PNGN+PBNB+PANA 6PGFG−2PMFM+3PMGM+PBMB+PAMA F = 3PNGN+3PBGB+3PAGA−PABA 3PMGM+3PBGB+3PAGA−PABA G = 3PNGN+3PBGB+3PAGA−PABA 3PNGN+3PBGB+3PAGA−PABA simplify = 1 The eliminants PMCM C =6PGFG−2PMFM+3PMGM PNCN C =6PGFG−2PNFN+3PNGN PMFM F =1 4(2PBMB+2PAMA−PABA) PNFN F =1 4(2PBNB+2PANA−PABA) PGFG F =1 4(2PBGB+2PAGA−PABA) PMGM G =PNGN Deﬁnition. The triangle having for its vertices the midpoints of the sides of a given triangle is called the medial triangle of the given triangle. The triangle formed by the parallels to the sides of a given triangle through the opposite vertices is called the anticomplementary triangle of the given triangle. Example 6.54 (0.050, 2, 6) The median AA1 of the triangle ABC meets the side B1C1 of the medial triangle A1B1C1 in P, and CP meets AB in Q. Show that AB = 3AQ. 6.3. Triangles 293 B C A 1 A 1 C P Q Figure 6-54 Constructive description ( (points A B C) (midpoint A1 B C) (midpoint C1 A B) (inter P (l A A1) (p C1 B C)) (inter Q (l A B) (l P C)) ( QA AB = −1/3) ) The machine proof (3)· AQ AB Q = (3)·(−S ACP) −S ACBP P = (3)·(−S ACC1 ·S ACA1 )·S ABC (−S ACC1 ·S ABA1−S 2 ABC)·S ABC simplify = (3)·S ACC1 ·S ACA1 S ACC1 ·S ABA1+S 2 ABC C1 = (3)·(−1 2 S ABC)·S ACA1 −1 2 S ABA1·S ABC+S 2 ABC simplify = (3)·S ACA1 S ABA1−2S ABC A1 = (3)·(−1 2 S ABC) −3 2 S ABC simplify = 1 The eliminants AQ AB Q = S ACP S ACBP S ACBP P = −(S ACC1 ·S ABA1+S 2 ABC) S ABC S ACP P = −S ACC1 ·S ACA1 S ABC S ACC1 C1 = −1 2(S ABC) S ABA1 A1 = 1 2(S ABC) S ACA1 A1 = −1 2(S ABC) Example 6.55 (0.066, 4, 5) The sum of the squares of the distances of the centroid of a trian-gle from the vertices is equal to one-third the sum of the squares of the sides. A B C E F D G Figure 6-55 Constructive description ((points A B C) (midpoint E A C) (lratio G B E 2/3) (3AG 2+3GC 2+3BG 2 = AC 2+AB 2+BC 2) ) The machine proof (3)·(PCGC+PBGB+PAGA) PBCB+PACA+PABA G = (3)·( 2 3 PCEC+ 1 3 PBCB+ 2 3 PAEA+ 1 3 PABA) PBCB+PACA+PABA E = PBCB+PACA+PABA PBCB+PACA+PABA simplify = 1 The eliminants PAGA G = −1 9(2PBEB−6PAEA−3PABA) PBGB G =4 9(PBEB) PCGC G =1 9(6PCEC−2PBEB+3PBCB) PAEA E =1 4(PACA), PCEC E = 1 4(PACA) Example 6.56 (0.050, 7, 8) If M is any point in the plane of the triangle ABC, and G is the centroid of ABC, we have MA 2 + MB 2 + MC 2 = GA 2 + GB 2 + GC 2 + 3MG 2. A B C M E F D G Figure 6-56 Constructive description ((points A B C M) (midpoint E A C) (lratio G B E 2/3) 294 Chapter 6. Topics from Geometry (3MG 2+AG 2+GC 2+BG 2 = AM 2+MB 2+MC 2) ) The machine proof 3PMGM+PCGC+PBGB+PAGA PCMC+PBMB+PAMA G = 2PMEM+ 2 3 PCEC−2 3 PBEB+PBMB+ 1 3 PBCB+ 2 3 PAEA+ 1 3 PABA PCMC+PBMB+PAMA E = 3PCMC+3PBMB+3PAMA (3)·(PCMC+PBMB+PAMA) simplify = 1 The eliminants PAGA G = −1 9(2PBEB−6PAEA−3PABA) PBGB G =4 9(PBEB) PCGC G =1 9(6PCEC−2PBEB+3PBCB) PMGM G =1 9(6PMEM−2PBEB+3PBMB) PAEA E =1 4(PACA) PBEB E =1 4(2PBCB−PACA+2PABA) PCEC E =1 4(PACA) PMEM E = 1 4(2PCMC+2PAMA−PACA) Example 6.57 (0.067, 7, 10) If the pairs of points D, D1; E, E1; F, F1 are isotomic on the sides BC, CA, AB of the triangle ABC, the areas of two triangles DEF, D1E1F1 are the same. A B C D E F 1 M 2 M 3 M 1 D 1 E 1 F Figure 6-57 Constructive description ((points A B C) (lratio D B C r1) (lratio E C A r2) (lratio F A B r3) (lratio D1 C B r1) (lratio E1 A C r2) (lratio F1 B A r3) (S D1E1F1 = S DEF) ) The eliminants S D1E1F1 F1 = −(S BD1E1·r3−S BD1E1−S AD1E1·r3) S AD1E1 E1 = −(S ACD1 ·r2) S BD1E1 E1 = −((r2−1)·S ABD1) S ABD1 D1 = −((r1−1)·S ABC) S ACD1 D1 = −(S ABC·r1) S DEF F =S BDE·r3−S ADE·r3+S ADE S ADE E =(r2−1)·S ACD S BDE E =S ABD·r2 S ABD D =S ABC·r1 S ACD D =(r1−1)·S ABC The machine proof S D1E1F1 S DEF F1 = −S BD1E1·r3+S BD1E1+S AD1E1·r3 S DEF E1 = −(S ACD1 ·r3·r2−S ABD1 ·r3·r2+S ABD1·r3+S ABD1 ·r2−S ABD1) S DEF D1 = −(−S ABC·r3·r2−S ABC·r3·r1+S ABC·r3−S ABC·r2·r1+S ABC·r2+S ABC·r1−S ABC) S DEF simplify = (r3·r2+r3·r1−r3+r2·r1−r2−r1+1)·S ABC S DEF F = (r3·r2+r3·r1−r3+r2·r1−r2−r1+1)·S ABC S BDE·r3−S ADE·r3+S ADE E = (r3·r2+r3·r1−r3+r2·r1−r2−r1+1)·S ABC −S ACD·r3·r2+S ACD·r3+S ACD·r2−S ACD+S ABD·r3·r2 D = −(r3·r2+r3·r1−r3+r2·r1−r2−r1+1)·S ABC −S ABC·r3·r2−S ABC·r3·r1+S ABC·r3−S ABC·r2·r1+S ABC·r2+S ABC·r1−S ABC simplify = 1 6.3. Triangles 295 Example 6.58 (0.200, 3, 11) Show that the parallels through the vertices A, B, C of the trian-gle ABC to the medians of this triangle issued from the vertices B, C, A, respectively, form a triangle whose area is three times the area of the given triangle. Constructive description ((points A B C) (midpoint A1 C B) (midpoint B1 A C) (midpoint C1 B A) (inter C2 (p B C C1) (p A B B1)) (inter B2 (p A B B1) (p C A A1)) (inter A2 (p C A A1) (p B C C1)) (S A2B2C2 = 3S ABC) ) The eliminants S C2B2A2 A2 = S BCB2C1·S ACA1C2 −S ACA1C1 S BCB2C1 B2 = −(S ACA1C1 ·S ABC−S ABA1 B1·S BCC1 ) S ABA1 B1 S ACA1C2 C2 = S BCB1C1 ·S ACA1 −S ABA1B1·S ABC S BCB1C1 S BCB1C1 C1 = 1 2(2S BCB1+S ABB1) S BCC1 C1 = 1 2(S ABC) S ACA1C1 C1 = 1 2(2S ACA1 −S ABA1) S ABB1 B1 = 1 2(S ABC) S BCB1 B1 = 1 2(S ABC) S ABA1B1 B1 = −1 2(S ACA1 −2S ABA1) S ABA1 A1 = 1 2(S ABC) S ACA1 A1 = −1 2(S ABC) 2 A 2 B 2 C 1 C 1 B 1 A C B A Figure 6-58 The machine proof −S C2 B2A2 (3)·S ABC A2 = −S BCB2C1 ·S ACA1C2 (3)·S ABC·(−S ACA1C1) B2 = (−S ACA1C1 ·S ABC+S ABA1 B1·S BCC1 )·S ACA1C2 (3)·S ABC·S ACA1C1·S ABA1B1 C2 = −(S ACA1C1 ·S ABC−S ABA1 B1·S BCC1 )·(S BCB1C1·S ACA1 −S ABA1B1·S ABC) (3)·S ABC·S ACA1C1·S ABA1B1·S BCB1C1 C1 = −(−1 2 S ABA1B1 ·S ABC +S ACA1 ·S ABC −1 2 S ABA1 ·S ABC)·(−S ABA1B1 ·S ABC +S BCB1 ·S ACA1 + 1 2 S ACA1 ·S ABB1 ) (3)·S ABC·(S ACA1 −1 2 S ABA1 )·S ABA1B1 ·(S BCB1 + 1 2 S ABB1 ) simplify = −(S ABA1 B1−2S ACA1 +S ABA1 )·(2S ABA1B1·S ABC−2S BCB1 ·S ACA1 −S ACA1 ·S ABB1) (3)·(2S ACA1 −S ABA1)·S ABA1B1·(2S BCB1+S ABB1) B1 = −(−5 2 S ACA1 +2S ABA1 )·(−5 2 S ACA1 ·S ABC+2S ABA1 ·S ABC) (3)·(2S ACA1 −S ABA1)·(−1 2 S ACA1 +S ABA1 )·( 3 2 S ABC) simplify = (5S ACA1 −4S ABA1 )2 (9)·(2S ACA1 −S ABA1 )·(S ACA1 −2S ABA1 ) A1 = ((−9 2 S ABC))2 (9)·((−3 2 S ABC))2 simplify = 1 296 Chapter 6. Topics from Geometry Deﬁnition. If the corresponding sides of two similar polygons are parallel, the two polygons are said to be homothetic. The lines joining the corresponding vertices of two homothetic polygons are concurrent at the homothetic center. Example 6.59 (0.500, 5, 29) Let L, L1 and M, M1 be two pairs of isotomic points on the two sides AC, AB of the triangle ABC, and L2, M2 the traces of the lines BL, CM on the sides A1C1, A1B1 of the medial triangle A1B1C1 of ABC. Show that the triangles AL1M1, A1L2M2 are homothetic. A B C 1 B 1 A 1 C M L 1 L 1 M 2 L 2 M O Figure 6-59 Constructive description ( (points A B C) (midpoint B1 A C) (inter A1 (l B C) (p B1 A B)) (midpoint C1 A B) (on M (l A B)) (on L (l A C)) (lratio L1 B1 L −1) (lratio M1 C1 M −1) (inter L2 (l A1 C1) (l B L)) (inter M2 (l A1 B1) (l C M)) (inter O (l L1 L2) (l A A1)) (collinear O M1 M2) ) Example 6.60 (0.283, 6, 16) A parallel to the median AA1 of the triangle ABC meets BC, CA, AB in the points H, N, D. Prove that the symmetries of H with respect to the midpoints of NC, BD are symmetrical with respect to the vertex A. 2 H 1 H L K D N 1 A H A C B Figure 6-60 Constructive description ( (points A B C) (on H (l B C)) (midpoint A1 B C) (inter N (l C A) (p H A A1)) (inter D (l H N) (l A B)) (midpoint K N C) (midpoint L B D) (lratio H1 K H −1) (lratio H2 L H −1) (midpoint A H1 H2) ) Example 6.61 (0.366, 4, 16) The parallels to the sides of a triangle ABC through the same point, M, meet the respective medians in the points P, Q, R. Prove that we have, both in magnitude and in sign, (GP/GA) + (GQ/GB) + (GR/GC) = 0. R Q P G F E D M C B A Figure 6-61 Constructive description ( (points A B C M) (midpoint D B C) (midpoint E C A) (midpoint F A B) (inter P (l A D) (p M B C)) 6.3. Triangles 297 (inter Q (l B E) (p M A C)) (inter R (l C F) (p M A B)) (inter G (l A D) (l C F)) (collinear G B E) ( GQ GB + GR GC = GP AG ) ) 6.3.2 Altitudes and Orthocenters For machine proofs of the orthocenter theorem, see Example 3.36 on page 120 and Example 5.53 on page 247. Example 6.62 (0.033, 1, 6) The dual of the orthocenter theorem. F E D O C B A Figure 6-62 Constructive description ( (points A B C O) (inter D (l B C) (t O O A)) (inter E (l C A) (t O O B)) (inter F (l A B) (t O O C)) (inter ZF (l E D) (l A B)) ( AF BF = AZF BZF ) ) The machine proof ( AF BF)/( AZF BZF ) ZF = −S BDE −S ADE · AF BF F = PAOC·S BDE S ADE·PBOC E = PAOC·PBOC·S ABD·(−PCBAO) PAOB·S ACD·PBOC·(−PCBAO) simplify = PAOC·S ABD PAOB·S ACD D = PAOC·PAOB·S ABC·PCABO PAOB·PAOC·S ABC·PCABO simplify = 1 The eliminants AZF BZF ZF = S ADE S BDE AF BF F = PAOC PBOC S ADE E = PAOB·S ACD −PCBAO S BDE E = PBOC·S ABD −PCBAO S ACD D = PAOC·S ABC PCABO S ABD D = PAOB·S ABC PCABO Example 6.63 (0.050, 1, 8) In a given triangle the three products of the segments into which the orthocenter divides the altitudes are equal. A B C D E H Figure 6-63 Constructive description ( (points A B C) (foot D C A B) (foot E B A C) (inter H (l C D) (l B E)) (eq-product C H H D B H H E) ) 298 Chapter 6. Topics from Geometry The machine proof −PCHD −PBHE H = (−PCDC·S BDE·S BCE)·S 2 BCED (−PBEB·S CDE·S BCD)·S 2 BCED simplify = PCDC·S BDE·S BCE PBEB·S CDE·S BCD E = PCDC·(−PBAC·S BCD)·PACB·S ABC·(PACA)2 (16S 2 ABC)·PACB·S ACD·S BCD·(PACA)2 simplify = −PCDC·PBAC (16)·S ABC·S ACD D = −(16S 2 ABC)·PBAC·PABA (16)·S ABC·(−PBAC·S ABC)·PABA simplify = 1 The eliminants PBHE H = PBEB·S CDE·S BCD (S BCED)2 PCHD H = PCDC·S BDE·S BCE (S BCED)2 S CDE E = PACB·S ACD PACA PBEB E = (16)·(S ABC)2 PACA S BCE E = PACB·S ABC PACA S BDE E =−PBAC·S BCD PACA S ACD D =−PBAC·S ABC PABA PCDC D =(16)·(S ABC)2 PABA Example 6.64 (0.066, 2, 8) The product of the segments into which a side of a triangle is divided by the foot of the altitude is equal to this altitude multiplied by the distance of the side from the orthocenter. A B C F E H Figure 6-64 Constructive description ( (points A B C) (foot F C A B) (foot E B A C) (inter H (l C F) (l B E)) (eq-product A F F B C F H F) ) The eliminants PCFH H = PCFE·S BCF S BCEF PCFE E = PCFC·PBAC PACA S BCEF E =−(PACB·S ACF−PACA·S BCF) PACA PCFC F =(16)·(S ABC)2 PABA S BCF F = PABC·S ABC PABA S ACF F = −PBAC·S ABC PABA PAFB F =−PBAC·PABC PABA 16S 2 ABC=PBAC·PACB+PACA·PABC The machine proof −PAFB PCFH H = (−PAFB)·S BCEF PCFE·S BCF E = −PAFB·(−PACB·S ACF+PACA·S BCF)·PACA PCFC·PBAC·S BCF·PACA simplify = PAFB·(PACB·S ACF−PACA·S BCF) PCFC·PBAC·S BCF F = (−PBAC·PABC)·(−PBAC·PACB·PABA·S ABC−PACA·PABC·PABA·S ABC)·(PABA)2 (16S 2 ABC)·PBAC·PABC·S ABC·(PABA)3 6.3. Triangles 299 simplify = PBAC·PACB+PACA·PABC (16)·(S ABC)2 herron = (PBAC·PACB+PACA·PABC)·(16) 16PBAC·PACB+16PACA·PABC (This step uses Herron-Qin’s formula.) simplify = 1 Example 6.65 (0.050, 3, 4) If p, q, r are the distances of a point inside a triangle ABC from the sides of the triangle, show that (p/ha) + (q/hb) + (r/hc) = 1. A B C O D E F 1 D 1 E 1 F Figure 6-65 Constructive description ( (points A B C O) (foot D A B C) (foot E B A C) (foot F C A B) (foot D1 O B C) (foot E1 O A C) (foot F1 O A B) ( OD1 AD + OE1 BE + OF1 CF = 1) ) The machine proof OF1 CF + OE1 BE + OD1 AD F1 = OE1 BE ·S ABC+ OD1 AD ·S ABC+S ABO S ABC E1 = −OD1 AD ·S 2 ABC+S ACO·S ABC−S ABO·S ABC S ABC·(−S ABC) simplify = OD1 AD ·S ABC−S ACO+S ABO S ABC D1 = S BCO·S ABC−S ACO·S ABC+S ABO·S ABC (S ABC)2 simplify = S BCO−S ACO+S ABO S ABC area−co = S ABC S ABC simplify = 1 The eliminants OF1 CF F1 = S ABO S ABC OE1 BE E1 = S ACO −S ABC OD1 AD D1 = S BCO S ABC S BCO=S ACO−S ABO+S ABC Example 6.66 (0.050, 2, 3) Show that the sum of distances of a point on the base of an isosceles triangle to its two sides is equal to the altitudes on the sides. The eliminants: DK AH K = S BCD −S BAC , DF BG F =S ACD S BAC , S ACD−S BCD=S BAC. 300 Chapter 6. Topics from Geometry G F K H D A C B Figure 6-66 Constructive description ((points B A) (on C (b A B)) (on D (l A B)) (foot G B A C) (foot F D A C) (foot H A B C) (foot K D B C) ( DK AH + DF BG = 1) ) The machine proof DK AH + DF BG K = −DF BG ·S BAC+S BCD −S BAC F = S ACD·S BAC−S BCD·S BAC (S BAC)2 simplify = S ACD−S BCD S BAC G = S BAC S BAC simplify = 1 Deﬁnition. The triangle having for its vertices the feet of the altitudes of a given triangle is called the orthic triangle of the given triangle. A B C F E Figure 6-67 Example 6.67 (0.016, 1, 2) The three triangles cut oﬀfrom a given triangle by the sides of its orthic triangles are similar to the given trian-gle. Constructive description ( (points A B C) (foot F C A B) (foot E B A C) (eq-product A F A B A E A C) ) The machine proof PBAF PCAE E = PBAF PBAC F = PBAC PBAC simplify = 1 The eliminants PCAE E =PBAC PBAF F =PBAC Example 6.68 (0.083, 1, 12) The sides of the orthic triangle meet the sides of the given trian-gle in three collinear points. 6.3. Triangles 301 Constructive description ( (points A B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter A1 (l B C) (l E F)) (inter B1 (l A C) (l D F)) (inter C1 (l A B) (l D E)) (inter ZC1 (l B1 A1) (l D E)) ( DC1 EC1 = DZC1 EZC1 ) ) The machine proof ( DC1 EC1 )/( DZC1 EZC1 ) ZC1 = −S EA1 B1 −S DA1B1 · DC1 EC1 C1 = S ABD·S EA1B1 S DA1B1·S ABE B1 = S ABD·S DEF·S ACA1 ·(−S ADCF) (−S DFA1·S ACD)·S ABE·S ADCF simplify = S ABD·S DEF·S ACA1 S DFA1·S ACD·S ABE A1 = S ABD·S DEF·S CEF·S ABC·S BECF (−S DEF·S BCF)·S ACD·S ABE·S BECF simplify = S ABD·S CEF·S ABC −S BCF·S ACD·S ABE F = S ABD·PBAC·S BCE·S ABC·PABA −PABC·S ABC·S ACD·S ABE·PABA simplify = S ABD·PBAC·S BCE −PABC·S ACD·S ABE E = S ABD·PBAC·PACB·S ABC·PACA −PABC·S ACD·PBAC·S ABC·PACA simplify = S ABD·PACB −PABC·S ACD D = PABC·S ABC·PACB·PBCB −PABC·(−PACB·S ABC)·PBCB simplify = 1 The eliminants DZC1 EZC1 ZC1 = S DA1B1 S EA1B1 DC1 EC1 C1 = S ABD S ABE S DA1 B1 B1 = S DFA1·S ACD S ADCF S EA1B1 B1 = S DEF·S ACA1 S ADCF S DFA1 A1 = −S DEF·S BCF S BECF S ACA1 A1 = S CEF·S ABC S BECF S BCF F = PABC·S ABC PABA S CEF F = PBAC·S BCE PABA S ABE E = PBAC·S ABC PACA S BCE E = PACB·S ABC PACA S ACD D =−PACB·S ABC PBCB S ABD D = PABC·S ABC PBCB Example 6.69 (0.067, 4, 11) The altitudes of a triangle bisect the internal angles of its orthic triangle. A B C F E D Figure 6-69 Constructive description ((points A B C) (foot F C A B) (foot E B A C) (foot D A B C) (eqangle E D C B D F) ) 302 Chapter 6. Topics from Geometry The machine proof. S CED·PBDF (−S BFD)·PCDE D = PACB·S BCE·(PBCF·PABC−PACB·PABC)·(PBCB)2 PABC·S BCF·(PCBE·PACB−PACB·PABC)·(PBCB)2 simplify = S BCE·(PBCF−PACB) S BCF·(PCBE−PABC) E = PACB·S ABC·(PBCF−PACB)·PACA S BCF·(PBCB·PBAC+PACB·PABC−PACA·PABC)·PACA simplify = PACB·S ABC·(PBCF−PACB) S BCF·(PBCB·PBAC+PACB·PABC−PACA·PABC) F = PACB·S ABC·(PBCB·PBAC+PACB·PABC−PACB·PABA)·PABA PABC·S ABC·(PBCB·PBAC+PACB·PABC−PACA·PABC)·PABA simplify = PACB·(PBCB·PBAC+PACB·PABC−PACB·PABA) PABC·(PBCB·PBAC+PACB·PABC−PACA·PABC) py = (PBCB+PACA−PABA)·(−2P2 BCB+4PBCB·PACA−2P2 ACA+2P2 ABA)·((2))4 (PBCB−PACA+PABA)·(−2P2 BCB+4PBCB·PABA+2P2 ACA−2P2 ABA)·((2))4 simplify = 1 The eliminants PCDE D =(PCBE−PABC)·PACB PBCB S BFD D =−PABC·S BCF PBCB PBDF D =(PBCF−PACB)·PABC PBCB S CED D = PACB·S BCE PBCB PCBE E = PBCB·PBAC+PACB·PABC PACA S BCE E = PACB·S ABC PACA S BCF F = PABC·S ABC PABA PBCF F = PBCB·PBAC+PACB·PABC PABA PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) PACB=1 2(PBCB+PACA−PABA) Deﬁnition. Let H be the orthocenter of triangle ABC. Then the four points A, B, C, H are such that each is the orthocenter of the triangle formed by the remaining three. Four such points will be revered to as an orthocentric group of points, or an orthocentric quadrilat-eral. Example 6.70 (0.067 4 10) Let H be the orthocenter of triangle ABC. Then the circumcenters of the four triangles ABC, ABH, ACH, and HBC form a triangle congruent to ABC; the sides of the two triangles are parallel. O c O b O a O H A C B Figure 6-70 Constructive description ((points A B C) (orthocenter H A B C) (circumcenter Oa B C H) (circumcenter Ob A C H) (circumcenter Oc A B H) (circumcenter O A B C) ( ObOc CB = 1) ) The eliminants ObOc BC Oc = PBAOb−1 2 PABA PABC PBAOb Ob = PBAH·PAHC·PACA+PBAC·PAHA·PACH (32)·(S ACH )2 S ACH H = PBAC·PACB (−16)·S ABC PACH H =PACB PAHA H = PBCB·(PBAC)2 (16)·(S ABC)2 PAHC H = PBAC·PACB·PABC (−16)·(S ABC)2 PBAH H =PBAC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) 6.3. Triangles 303 The machine proof −ObOc BC Oc = −(PBAOb−1 2 PABA) PABC Ob = −(PBAH·PAHC·PACA+PBAC·PAHA·PACH−16PABA·S 2 ACH) PABC·(32S 2 ACH) H = (- (−4096PBCB·P3 BAC·PACB·S 6 ABC+4096P2 BAC·P2 ACB·PABA·S 6 ABC+4096P2 BAC·PACB·PACA·PABC·S 6 ABC) · ( (16S 2 ABC) )2) / ( (32) · PABC · ( (−PBAC·PACB·S ABC) )2· ( (16S 2 ABC) )3· (−16S 2 ABC) ) simplify = −(PBCB·PBAC−PACB·PABA−PACA·PABC) (2)·PABC·PACB py = −(−4P2 BCB+4P2 ACA−8PACA·PABA+4P2 ABA)·((2))2 ((2))4·(PBCB−PACA+PABA)·(PBCB+PACA−PABA) simplify = 1 Example 6.71 (0.067, 1, 6) Continuing from Example 6.70, show that the point H is the circumcenter of the triangle OaObOc. Constructive description ((points A B C) (orthocenter H A B C) (circumcenter Oa B C H) (circumcenter Ob A C H) (circumcenter Oc A B H) (circumcenter O A B C) (eqdistance H Oa H Ob) ) The machine proof PHOaH PHObH Ob = PHOaH·(64S 2 ACH) PCHC·PAHA·PACA Oa = (64)·PCHC·PBHB·PBCB·(S ACH)2 PCHC·PAHA·PACA·(64S 2 BCH) simplify = PBHB·PBCB·(S ACH)2 PAHA·PACA·(S BCH)2 H = PACA ·P2 ABC ·PBCB·((−PBAC·PACB·S ABC ))2·((16S2 ABC))3 PBCB·P2 BAC ·PACA ·(PACB·PABC ·S ABC )2·((16S2 ABC))3 simplify = 1 The eliminants PHObH Ob = PCHC·PAHA·PACA (64)·(S ACH)2 PHOaH Oa = PCHC·PBHB·PBCB (64)·(S BCH)2 S BCH H = PACB·PABC (16)·S ABC PAHA H = PBCB·(PBAC)2 (16)·(S ABC)2 S ACH H = PBAC·PACB (−16)·S ABC PBHB H = PACA·(PABC)2 (16)·(S ABC)2 Example 6.72 (0.350, 10, 17) Show that the three perpendiculars to the sides of a triangle at the points isotomic to the feet of the respective altitudes are concurrent. I 1 F 1 E 1 D F E D C B A Figure 6-72 Constructive description ( (points A B C) (foot D A B C) (foot E B A C) (foot F C A B) (pratio D1 B C D −1) (pratio E1 A C E −1) (pratio F1 A B F −1) (inter I (t E1 A C) (t D1 B C)) (perpendicular I F1 A B) ) Example 6.73 (0.100, 4, 15) Show that the symmetries of the foot of the altitude to the base of a triangle with respect to the other two sides lie on the side of the orthic triangle relative to the base. 304 Chapter 6. Topics from Geometry J P F E D A C B Figure 6-73 Constructive description ( (points A B C) (foot D A B C) (foot E B A C) (foot F C A B) (foot P D B A) (inter J (l E F) (l D P)) (midpoint P D J) ) Example 6.74 (0.183, 3, 11) Show that the product of the segments into which a side of a triangle is divided by the corresponding vertex of the orthic triangle is equal to the product of the sides of the orthic triangle passing through the vertex considered. D E F A C B Figure 6-74 Constructive description ( (points A B C) (foot F C A B) (foot E B A C) (foot D A B C) (eq-product B D D C E D F D) ) Example 6.75 (0.100, 3, 14) If P, Q are the feet of the perpendiculars from the vertices B, C of the triangle ABC upon the sides DF, DE, respectively, of the orthic triangle DEF, show that EQ = FP. P Q D E F A C B Figure 6-75 Constructive description ( (points A B C) (foot F C A B) (foot E B A C) (foot D A B C) (foot Q C D E) (foot P B D F) (eqdistance E Q F P) ) Example 6.76 (0.167, 6, 22) The four projections of the foot of the altitude on a side of a triangle upon the other two sides and the other two altitudes are collinear. F A B C H P T Q Figure 6-76 Constructive description ( (points A B C) (orthocenter H A B C) (foot F C A B) (foot P F A C) (foot T F B C) (foot Q F A H) (collinear P Q T) ) 6.3. Triangles 305 Example 6.77 (0.083, 4, 9) DP, DQ are the perpendiculars from the foot D of the altitude AD of the triangle ABC upon the sides AC, AB. Prove that the points B, C, P, Q are cyclic. P Q D A C B Figure 6-77 Constructive description ( (points A B C) (foot D A B C) (foot Q D A B) (foot P D A C) (cocircle B C P Q) ) The machine proof (−S BCP)·PBQC (−S BCQ)·PBPC P = (−PACD·S ABC)·PBQC·PACA (−S BCQ)·(−PCAD·PACD+PBAC·PACD)·PACA simplify = −S ABC·PBQC S BCQ·(PCAD−PBAC) Q = −S ABC·(−PBAD·PABD+PBAC·PABD)·PABA PABD·S ABC·(PCAD−PBAC)·PABA simplify = PBAD−PBAC PCAD−PBAC D = (−PBCB·PBAC+PBAC·PABC+PACB·PABA)·PBCB (−PBCB·PBAC+PBAC·PACB+PACA·PABC)·PBCB simplify = PBCB·PBAC−PBAC·PABC−PACB·PABA PBCB·PBAC−PBAC·PACB−PACA·PABC py = (−2P2 BCB+2P2 ACA−4PACA·PABA+2P2 ABA)·((2))3 (−2P2 BCB+2P2 ACA−4PACA·PABA+2P2 ABA)·((2))3 simplify = 1 The eliminants PBPC P =−(PCAD−PBAC)·PACD PACA S BCP P = PACD·S ABC PACA S BCQ Q = PABD·S ABC PABA PBQC Q =−(PBAD−PBAC)·PABD PABA PCAD D = PBAC·PACB+PACA·PABC PBCB PBAD D = PBAC·PABC+PACB·PABA PBCB PACB=1 2(PBCB+PACA−PABA) PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.78 (0.450, 6, 29) The perpendiculars DP, DQ dropped from the foot D of the altitude AD of the triangle ABC upon the sides AB, AC meet the perpendiculars BP, CQ erected to BC at B, C in the points P, Q respectively. Prove that the line PQ passes through the orthocenter H of ABC. Q P H D A C B Figure 6-78 Constructive description ( (points A B C) (foot D A B C) (inter H (l A D) (t B A C)) (inter P (t D A B) (t B B C)) (inter Q (t D A C) (t C B C)) (collinear H Q P) ) Example 6.79 (2.067, 68, 61) The algebraic sum of the distances of the points of an ortho-centric group from any line passing through the nine-point center of the group is equal to zero. 306 Chapter 6. Topics from Geometry 1 H 1 C 1 B 1 A X N 3 M 2 M 1 M H A C B Figure 6-79 Constructive description ( (points A B C) (orthocenter H A B C) (midpoint M1 B C) (midpoint M2 C A) (midpoint M3 A B) (midpoint N1 M1 M2) (midpoint N2 M1 M3) (inter N (t N1 N1 M1) (t N2 N2 M1)) (on X (l B C)) (foot H1 H X N) (inter A1 (l X N) (p A H H1)) (inter B1 (l X N) (p B H H1)) (inter C1 (l X N) (p C H H1)) ( AA1 HH1 + BB1 HH1 = C1C HH1 −1) ) Example 6.80 (0.350, 6, 10) If through the midpoints of the sides of a triangle having its vertices on the altitudes of a given triangle, perpendiculars are dropped to the respective sides of the given triangle, show that the three perpendiculars are concurrent. I 1 P 1 Q 1 R Q P R F E D A C B Figure 6-80 Constructive description ((points A B C) (orthocenter H A B C) (on R (l A H)) (on P (l B H)) (on Q (l C H)) (midpoint R1 P Q) (midpoint Q1 R P) (midpoint P1 Q R) (inter I (p R1 A H) (p Q1 C H)) (perpendicular I P1 A C) ) Example 6.81 (2.833, 7, 43) Show that the perpendiculars dropped from the orthocenter of a triangle upon the lines joining the vertices to a given points meet the respectively opposite sides of the triangle in three collinear points. 1 C 1 B 1 A H O A C B Figure 6-81 Constructive description ( (points A B C O) (orthocenter H A B C) (inter A1 (l B C) (t H O A)) (inter B1 (l A C) (t H O B)) (inter C1 (l A B) (t H O C)) (inter C2 (l A B) (l A1 B1)) ( AC1 BC1 = AC2 BC2 ) ) Deﬁnition. Two lines passing through the vertex of a given angle and marking equal angles with the bisector of the given angle are said to be isogoal or isogonal conjugates. Example 6.82 (0.133, 5, 22) Show that the line joining a given point to the vertex of a given angle has for its isogonal line the mediator of the segment determined by the symmetries of the given point with respect to the sides of the angle. 6.3. Triangles 307 A B C M P Q 1 P 1 Q N Figure 6-82 Constructive description ( (points A B C M) (foot P M A C) (foot Q M A B) (lratio P1 P M −1) (lratio Q1 Q M −1) (midpoint N P1 Q1) (perpendicular A N P1 Q1) ) Example 6.83 (1.550, 5, 25) If circles are constructed on two cevians as diameters, their radical axis passes through the orthocenter H of the triangle. N M E D H P C B A Figure 6-83 Constructive description ( (points A B C P) (orthocenter H A B C) (inter D (l B C) (l A P)) (inter E (l A C) (l B P)) (midpoint M A D) (midpoint N B E) (on-radical H M A N B) ) Example 6.84 (0.183, 3, 12) In triangle ABC, let F the midpoint of the side BC, D and E the feet of the altitudes on AB and AC, respectively. FG is perpendicular to DE at G. Show that G is the midpoint of DE. G F E D A C B Figure 6-84 Constructive description ((points A B C) (foot D C A B) (foot E B A C) (midpoint F B C) (midpoint G D E) (perpendicular G F D E) ) The eliminants PEDG G =1 2(PDED) PEDF F =1 2(PCDE+PBDE) PBDE E = PADB·PACB PACA PCDE E = PCDC·PBAC PACA PDED E = PCDC·PBAC−PBAC·PACB+PADA·PACB PACA PADB D = −PBAC·PABC PABA PADA D =(PBAC)2 PABA PCDC D =(16)·(S ABC)2 PABA 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) The machine proof PEDG PEDF G = 1 2 PDED PEDF F = PDED (2)·( 1 2 PCDE+ 1 2 PBDE) E = (PCDC·PBAC−PBAC·PACB+PADA·PACB)·(PACA)2 (PCDC·PBAC·PACA+PADB·PACB·PACA)·PACA 308 Chapter 6. Topics from Geometry simplify = PCDC·PBAC−PBAC·PACB+PADA·PACB PCDC·PBAC+PADB·PACB D = (P2 BAC·PACB·PABA−PBAC·PACB·P2 ABA+16PBAC·PABA·S 2 ABC)·(PABA)2 (−PBAC·PACB·PABC·PABA+16PBAC·PABA·S 2 ABC)·(PABA)2 simplify = PBAC·PACB−PACB·PABA+16S 2 ABC −(PACB·PABC−16S 2 ABC) herron = (32PBAC·PACB−16PACB·PABA+16PACA·PABC)·(16) (16PBAC·PACB−16PACB·PABC+16PACA·PABC)·(16) py = ((2))4·(−2P2 BCB+2PBCB·PACA+2PBCB·PABA) (−4P2 BCB+4PBCB·PACA+4PBCB·PABA)·((2))3 simplify = 1 Example 6.85 (0.083, 3, 10) Let B1 and C1 be the midpoints of AC and AB, D the foot from A to BC. Show that the triangle DB1C1 is congruent to the Euler triangle of triangle ABC. B C A D H P Q 1 C Figure 6-85 Constructive description ( (points A B C) (foot D A B C) (foot FH B A C) (inter H (l A D) (l B FH)) (midpoint P A H) (midpoint Q B H) (midpoint C1 B A) (eqdistance C1 D P Q) ) The machine proof PDC1D PPQP C1 = 1 2 PBDB+ 1 2 PADA−1 4 PABA PPQP Q = 2PBDB+2PADA−PABA (4)·( 1 2 PHPH+ 1 2 PBPB−1 4 PBHB) P = 2PBDB+2PADA−PABA PABA D = −P2 BCB·PABA+2PBCB·P2 ABC+32PBCB·S 2 ABC PABA·(PBCB)2 simplify = −(PBCB·PABA−2P2 ABC−32S 2 ABC) PABA·PBCB herron = −16PBCB·PABA+32PBAC·PACB+32PACA·PABC+32P2 ABC PBCB·PABA·(16) py = −(−16PBCB·PABA) PBCB·PABA·((2))4 simplify = 1 The eliminants PDC1D C1 = 1 4(2PBDB+2PADA−PABA) PPQP Q =1 4(2PHPH+2PBPB−PBHB) PBPB P =1 4(2PBHB−PAHA+2PABA) PHPH P =1 4(PAHA) PADA D = (16)·(S ABC)2 PBCB , PBDB D =(PABC)2 PBCB 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) 6.3.3 The Circumcircle Deﬁnition The circle passing through the three vertices of a triangle is called the circum-circle of the triangle. The center of the circumcircle is called the circumcenter of the given triangle. 6.3. Triangles 309 Example 6.86 (0.033, 6, 8) The angle between the circumdiameter and the altitude issued from the same vertex of a triangle is bisected by the bisector of the angle of the triangle at the vertex considered. A B C O D Figure 6-86 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (eqangle B A D O A C) ) The machine proof (−S ABD)·PCAO S ACO·PBAD D = (−PABC·S ABC)·PCAO·PBCB S ACO·(PBAC·PABC+PACB·PABA)·PBCB simplify = −PABC·S ABC·PCAO S ACO·(PBAC·PABC+PACB·PABA) O = −PABC·S ABC·PACA·(−32S ABC) PACA·PABC·(PBAC·PABC+PACB·PABA)·(2) simplify = (16)·(S ABC)2 PBAC·PABC+PACB·PABA herron = 16PBAC·PACB+16PACA·PABC (PBAC·PABC+PACB·PABA)·(16) py = (−2P2 BCB+4PBCB·PACA+4PBCB·PABA−2P2 ACA+4PACA·PABA−2P2 ABA)·((2))3 (−2P2 BCB+4PBCB·PACA+4PBCB·PABA−2P2 ACA+4PACA·PABA−2P2 ABA)·((2))3 simplify = 1 The eliminants PBAD D = PBAC·PABC+PACB·PABA PBCB S ABD D = PABC·S ABC PBCB S ACO O = PACA·PABC (−32)·S ABC PCAO O =1 2(PACA) 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.87 (0.001, 1, 2) The product of two sides of a triangle is equal to the altitude to the third side multiplied by the circumdiameter. Constructive description ( (points A B C) (circumcenter O A B C) (foot F C A B) (AC 2BC 2 = 4OA 2·CF 2) ) The eliminants PCFC F =(16)·(S ABC)2 PABA PAOA O = PBCB·PACA·PABA (64)·(S ABC)2 A B C O F Figure 6-87 The machine proof ( 1 4 )·PBCB·PACA PCFC·PAOA 310 Chapter 6. Topics from Geometry F = ( 1 4 )·PBCB·PACA·PABA (16S 2 ABC)·PAOA O = PBCB·PACA·PABA·(64S 2 ABC) (64)·(S ABC)2·PBCB·PACA·PABA simplify = 1 Example 6.88 (0.050, 5, 14) Prove that the circumcenter of a triangle is the orthocenter of its medial triangle. A B C O 1 A 1 B 1 C Figure 6-88 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint A1 B C) (midpoint B1 A C) (midpoint C1 A B) (perpendicular O A1 B1 C1) ) The machine proof POB1C1 PA1B1C1 C1 = 1 2 PBB1O+ 1 2 PAB1O 1 2 PBB1A1+ 1 2 PAB1A1 B1 = 1 2 PBCO+ 1 2 PBAO+ 1 2 PACO−1 2 PACA 1 2 PBCA1+ 1 2 PBAA1+ 1 2 PACA1 −1 2 PACA A1 = PBCO+PBAO+PACO−PACA 1 2 PBCB+ 1 2 PBAC+ 1 2 PACB−PACA+ 1 2 PABA O = (2)·(4PBCB−4PACA+4PABA) (PBCB+PBAC+PACB−2PACA+PABA)·((2))3 py = (PBCB−PACA+PABA)·((2))2 4PBCB−4PACA+4PABA simplify = 1 The eliminants PA1B1C1 C1 = 1 2(PBB1A1+PAB1A1) POB1C1 C1 = 1 2(PBB1O+PAB1O) PAB1A1 B1 = 1 4(2PACA1 −PACA) PBB1A1 B1 = 1 4(2PBCA1+2PBAA1−PACA) PAB1O B1 = 1 4(2PACO−PACA) PBB1O B1 = 1 4(2PBCO+2PBAO−PACA) PACA1 A1 = 1 2(PACB) PBAA1 A1 = 1 2(PBAC+PABA) PBCA1 A1 = 1 2(PBCB), PACO O =1 2(PACA) PBAO O =1 2(PABA), PBCO O =1 2(PBCB) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.89 (0.150, 4, 11) The area of a triangle is equal to the product of its three sides divided by the double circumdiameter of the triangle. Constructive description ((points A B C) (midpoint N A C) (midpoint L A B) (midpoint M B C) (inter O (t L L B) (t M M B)) (AB 2AC 2CB 2 = 16S 2 ABC·OA 2) ) Example 6.90 (0.016, 1, 4) The radii of the circumcircle passing through the vertices of a triangle are perpendicular to the corresponding sides of the orthic triangle. Constructive description: ( (points A B C) (circumcenter O A B C) (foot F C A B) (foot E B A C) (perpendicular E F A O) ) 6.3. Triangles 311 O E F C B A Figure 6-90 The machine proof POAE POAF E = PCAO·PBAC POAF·PACA F = PCAO·PBAC·PABA PBAO·PBAC·PACA simplify = PCAO·PABA PBAO·PACA O = PACA·PABA·(2) PABA·PACA·(2) simplify = 1 The eliminants POAE E = PCAO·PBAC PACA POAF F = PBAO·PBAC PABA PBAO O =1 2(PABA) PCAO O =1 2(PACA) Example 6.91 (0.016, 1, 3) Let ABC be a triangle with AC = AB. D is a point on BC. Line AD meets the circumcircle of ABC at E. Show that AB2 = AD · AE. The eliminants: PDAE E =(2)·POAD. POAD D =PBAO. PBAO O =1 2(PBAB). D E O C B M A Figure 6-91 Constructive description ( (points B M) (on A (t M M B)) (inter O (l A M) (b A B)) (lratio C M B −1) (on D (l B M)) (inter E (l A D) (cir O A)) (eq-product A B A B A D A E) ) The machine proof PBAB PDAE E = PBAB·PADA 2POAD·PADA simplify = PBAB (2)·POAD D = PBAB (2)·PBAO O = PBAB·(2) (2)·PBAB simplify = 1 Example 6.92 (0.001, 1, 3) Let C be the midpoint of the arc AB of circle (O). D is a point on the circle. E = AB ∩CD. Show that CA2 = CE · CD. The eliminants: PECD D =(2)·POCE. POCE E =PACO. PACO O =1 2(PACA). A B O M C D E Figure 6-92 Constructive description ( (points A M) (on C (t M M A)) (inter O (l C M) (b A C)) (on E (l A M)) (inter D (l C E) (cir O C)) (eq-product C A C A C E C D) ) The machine proof PACA PECD D = PACA·PCEC 2POCE·PCEC simplify = PACA (2)·POCE E = PACA (2)·PACO O = PACA·(2) (2)·PACA simplify = 1 Example 6.93 (0.050, 3, 7) The distance of a side of a triangle from the circumcenter is equal to half the distance of the opposite vertex from the orthocenter. 312 Chapter 6. Topics from Geometry 1 A O H D A C B Figure 6-93 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (foot D A B C) (inter A1 (l B C) (p O A D)) ( AH OA1 = 2) ) The machine proof 1 2( AH OA1 ) A1 = S ABHC (2)·S BCO H = −PACB·PABC·S ABC+16S 3 ABC (2)·S BCO·(16S 2 ABC) simplify = −(PACB·PABC−16S 2 ABC) (32)·S BCO·S ABC O = −(PACB·PABC−16S 2 ABC)·(32S ABC) (32)·PBCB·PBAC·S ABC simplify = −(PACB·PABC−16S 2 ABC) PBCB·PBAC herron = 16PBAC·PACB−16PACB·PABC+16PACA·PABC PBCB·PBAC·(16) py = (−4P2 BCB+4PBCB·PACA+4PBCB·PABA)·(2) PBCB·(−PBCB+PACA+PABA)·((2))3 simplify = 1 The eliminants AH OA1 A1 = S ABHC S BCO S ABHC H = PACB·PABC−16S 2 ABC (−16)·S ABC S BCO O = PBCB·PBAC (32)·S ABC 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.94 (0.067, 3, 9) The ratio of a side of a triangle to the corresponding side of the orthic triangle is equal to the ratio of the circumradius to the distance of the side considered from the circumcenter. A B C E D O Q Figure 6-94 Constructive description ( (points A B C) (circumcenter O A B C) (foot Q O A B) (foot E B A C) (foot D A B C) (eq-product A B O Q O A E D) ) The eliminants PEDE D = PCEC·PABC+PBEB·PACB−PACB·PABC PBCB PBEB E = (16)·(S ABC)2 PACA PCEC E =(PACB)2 PACA POQO Q =(16)·(S ABO)2 PABA PAOA O = PBCB·PACA·PABA (64)·(S ABC)2 S ABO O = PACB·PABA (32)·S ABC 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) 6.3. Triangles 313 The machine proof PABA·POQO PAOA·PEDE D = PABA·POQO·PBCB PAOA·(PCEC·PABC+PBEB·PACB−PACB·PABC) E = PABA·POQO·PBCB·(PACA)2 PAOA·(P2 ACB·PACA·PABC−PACB·P2 ACA·PABC+16PACB·PACA·S 2 ABC) simplify = PABA·POQO·PBCB·PACA PAOA·(PACB·PABC−PACA·PABC+16S 2 ABC)·PACB Q = PABA·(16S 2 ABO)·PBCB·PACA PAOA·(PACB·PABC−PACA·PABC+16S 2 ABC)·PACB·PABA simplify = (16)·(S ABO)2·PBCB·PACA PAOA·(PACB·PABC−PACA·PABC+16S 2 ABC)·PACB O = (16)·(PACB·PABA)2·PBCB·PACA·(64S 2 ABC) PBCB·PACA·PABA·(PACB·PABC−PACA·PABC+16S 2 ABC)·PACB·((32S ABC))2 simplify = PACB·PABA PACB·PABC−PACA·PABC+16S 2 ABC herron = PACB·PABA·(16) 16PBAC·PACB+16PACB·PABC simplify = PABA PBAC+PABC py = PABA·((2))2 4PABA simplify = 1 Example 6.95 (0.083, 2, 6) AH 2 + BC 2 = 4OA 2. A B C O H Figure 6-95 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (AH 2+BC 2 = 4AO 2) ) The machine proof PBCB+PAHA (4)·PAOA O = (PBCB+PAHA)·(64S 2 ABC) (4)·PBCB·PACA·PABA H = (16)·(PBCB·P2 BAC+16PBCB·S 2 ABC)·(S ABC)2 PBCB·PACA·PABA·(16S 2 ABC) simplify = P2 BAC+16S 2 ABC PACA·PABA herron = 16P2 BAC+16PBAC·PACB+16PACA·PABC PACA·PABA·(16) py = 16PACA·PABA PACA·PABA·((2))4 simplify = 1 The eliminants PAOA O = PBCB·PACA·PABA (64)·(S ABC)2 PAHA H = PBCB·(PBAC)2 (16)·(S ABC)2 16S 2 ABC=PBAC·PACB+PACA·PABC PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) 314 Chapter 6. Topics from Geometry Example 6.96 (0.116, 4, 7) The circumdiameters AP, BQ, CR of a triangle ABC meet the sides BC, CA, AB in the points K, L, M. Show that (KP/AK) + (LQ/BL) + (MR/CM) = 1. A B C O P Q R K L M Figure 6-96 Constructive description ( (points A B C) (circumcenter O A B C) (lratio P O A −1) (lratio Q O B −1) (lratio R O C −1) (inter K (l B C) (l A O)) (inter L (l C A) (l B O)) (inter M (l A B) (l C O)) ( KP AK + LQ BL + MR CM = 1) ) The machine proof −( RM CM + QL BL + PK AK ) M = −( QL BL ·S ABC+ PK AK ·S ABC+S ABR) S ABC L = −( PK AK ·S 2 ABC−S ACQ·S ABC+S ABR·S ABC) (S ABC)2 simplify = −( PK AK ·S ABC−S ACQ+S ABR) S ABC K = −(S BCP·S ABC−S ACQ·S ABC+S ABR·S ABC) (S ABC)2 simplify = −(S BCP−S ACQ+S ABR) S ABC R = −(S BCP−S ACQ+2S ABO−S ABC) S ABC Q = −(S BCP−2S ACO+2S ABO−2S ABC) S ABC P = −(2S BCO−2S ACO+2S ABO−3S ABC) S ABC O = (−2)·(−1 2S ABC) S ABC simplify = 1 The eliminants RM CM M = S ABR S ABC QL BL L =−S ACQ S ABC PK AK K =S BCP S ABC S ABR R =2S ABO−S ABC S ACQ Q =2S ACO+S ABC S BCP P =2S BCO−S ABC S BCO−S ACO+S ABO−3 2 S ABC=−1 2 S ABC Example 6.97 (0.066, 2, 8) The mediators of the sides AC, AB of the triangle ABC meet the sides AB, AC in P and Q. Prove that the points B, C, P, Q lie on a circle. Constructive description ( (points A B C) (inter P (l A C) (b A B)) (inter Q (l A B) (b A C)) (cocircle B C P Q) ) The machine proof (−S BPQ)·PPCQ (−S CPQ)·PPBQ Q = (PBAC·S ABP−1 2 PACA·S ABP)·PACP·((−PBAC))2 ( 1 2 PACA·S BCP)·(−PBAC·PABP+ 1 2 PACA·PABP)·(2)·(−PBAC) simplify = S ABP·PACP·PBAC PACA·S BCP·PABP P = (−1 2 PABA·S ABC)·(−PBAC·PACA+ 1 2 PACA·PABA)·PBAC·(2)·(−PBAC) PACA·(−PBAC·S ABC+ 1 2 PABA·S ABC)·PABA·((−PBAC))2 simplify = 1 6.3. Triangles 315 B C A P Q O Figure 6-97 The eliminants PPBQ Q =(2PBAC−PACA)·PABP (2)·PBAC S CPQ Q = PACA·S BCP (2)·PBAC PPCQ Q =1 2(PACP) S BPQ Q =(2PBAC−PACA)·S ABP (2)·PBAC PABP P =1 2(PABA), S BCP P =(2PBAC−PABA)·S ABC (2)·PBAC PACP P =(2PBAC−PABA)·PACA (2)·PBAC , S ABP P = PABA·S ABC (2)·PBAC Example 6.98 (0.083, 4, 13) The two tangents to the circumcircle of ABC at A and C meet at E. The mediator of BC meet AB at D. Show that DE ∥BC. A B C O H D E Figure 6-98 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint H B C) (inter D (l A B) (l O H)) (inter E (t C C O) (t A A O)) (parallel D E B C) ) The machine proof S BCD S BCE E = S BCD·(−4S ACO) 1 4 PCAO·PBCO D = (−16)·(−S BOH·S ABC)·S ACO PCAO·PBCO·S AOBH H = (16)·(−1 2 S BCO)·S ABC·S ACO PCAO·PBCO·(−S ABO+ 1 2 S ABC) O = (16)·PBCB·PBAC·S ABC·PACA·PABC·(32S ABC)·((2))2 PACA·PBCB·(2PACB·PABA−32S 2 ABC)·(−32S ABC)·(32S ABC) simplify = −PBAC·PABC PACB·PABA−16S 2 ABC herron = (−PBAC·PABC)·(16) −16PBAC·PACB+16PACB·PABA−16PACA·PABC py = (−PBCB+PACA+PABA)·(PBCB−PACA+PABA)·((2))3 (−2P2 BCB+4PBCB·PACA−2P2 ACA+2P2 ABA)·((2))2 simplify = 1 The eliminants S BCE E = PCAO·PBCO (−16)·S ACO S BCD D =−S BOH·S ABC S AOBH S AOBH H = −1 2(2S ABO−S ABC) S BOH H = −1 2(S BCO) S ABO O = PACB·PABA (32)·S ABC PBCO O =1 2(PBCB) PCAO O =1 2(PACA) S ACO O = PACA·PABC (−32)·S ABC S BCO O = PBCB·PBAC (32)·S ABC 16S 2 ABC=PBAC·PACB+PACA·PABC PACB=1 2(PBCB+PACA−PABA) PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.99 (0.016, 1, 12) The lines tangent to the circumcircle of a triangle at the vertices meet opposite sides in three collinear points. (The Lemoine axis of the given triangle.) 316 Chapter 6. Topics from Geometry A B C O 1 A 1 B 1 C Figure 6-99 Constructive description ( (points A B C) (circumcenter O A B C) (inter A1 (l B C) (t A A O)) (inter B1 (l A C) (t B B O)) (inter C1 (l A B) (t C C O)) (inter ZC1 (l B1 A1) (l A B)) ( AC1 BC1 = AZC1 BZC1 ) ) The machine proof ( AC1 BC1 )/( AZC1 BZC1 ) ZC1 = −S BA1B1 −S AA1 B1 · AC1 BC1 C1 = PACO·S BA1B1 S AA1B1·PBCO B1 = PACO·(−PCBO·S ABA1 )·(−PCBAO) (−PABO·S ACA1 )·PBCO·(−PCBAO) simplify = PACO·PCBO·S ABA1 PABO·S ACA1 ·PBCO A1 = PACO·PCBO·PBAO·S ABC·(−PCABO) PABO·PCAO·S ABC·PBCO·(−PCABO) simplify = PACO·PCBO·PBAO PABO·PCAO·PBCO O = PACA·PBCB·PABA·((2))3 PABA·PACA·PBCB·((2))3 simplify = 1 The eliminants AZC1 BZC1 ZC1 = S AA1 B1 S BA1B1 AC1 BC1 C1 = PACO PBCO S AA1 B1 B1 = PABO·S ACA1 PCBAO S BA1B1 B1 = PCBO·S ABA1 PCBAO S ACA1 A1 = PCAO·S ABC −PCABO S ABA1 A1 = PBAO·S ABC −PCABO PBCO O =1 2(PBCB) PCAO O =1 2(PACA) PABO O =1 2(PABA) PBAO O =1 2(PABA) PCBO O =1 2(PBCB) PACO O =1 2(PACA) Deﬁnition. The symmetric of a median of a triangle with respect to the internal bisector issued from the same vertex is called a symmedian. Y X a T 1 A O A C B Figure 6-100 Example 6.100 (0.050, 1, 6) The distances from a point on a symmedian of a triangle to the two including sides are proportional to these sides. Constructive description ( (points A B C) (circumcenter O A B C) (midpoint A1 B C) (inter Ta (l B C) (t A A O)) (foot X Ta A C) (foot Y Ta A B) (eq-product Ta Y A C Ta X A B) ) The machine proof PTaYTa·PACA PTaXTa·PABA Y = (16S 2 ABTa)·PACA PTaXTa·(PABA)2 X = (16)·(S ABTa)2·(PACA)2 (16S 2 ACTa )·(PABA)2 Ta = (PBAO·S ABC)2·(PACA)2·((−PCABO))2 (PCAO·S ABC)2·(PABA)2·((−PCABO))2 simplify = (PBAO)2·(PACA)2 (PCAO)2·(PABA)2 O = (PABA)2·(PACA)2·((2))2 (PACA)2·(PABA)2·((2))2 simplify = 1 The eliminants PTaYTa Y =(16)·(S ABTa)2 PABA PTaXTa X = (16)·(S ACTa )2 PACA S ACTa Ta = PCAO·S ABC −PCABO S ABTa Ta = PBAO·S ABC −PCABO PCAO O =1 2(PACA) PBAO O =1 2(PABA) 6.3. Triangles 317 Example 6.101 (0.033, 1, 8) Show that the distances of the vertices of a triangle from the Lemoine axis are proportional to the squares of the respective altitudes. J K D E 1 C 1 B O C B A Figure 6-101 Constructive description ( (points A B C) (circumcenter O A B C) (foot E B A C) (foot D A B C) (inter B1 (l A C) (t B B O)) (inter C1 (l A B) (t C C O)) (foot K A B1 C1) (foot J B B1 C1) (AK 2BE 2BE 2 = BJ 2AD 2AD 2) ) The machine proof (PBEB)2·PAKA PBJB·(PADA)2 J = (PBEB)2·PAKA·PB1C1B1 (16S 2 BB1C1)·(PADA)2 K = (PBEB)2·(16S 2 AB1C1 )·PB1C1 B1 (16)·(S BB1C1)2·(PADA)2·PB1C1B1 simplify = (PBEB)2·(S AB1C1)2 (S BB1C1)2·(PADA)2 C1 = (PBEB)2·((−PACO·S ABB1))2·((−PBCAO))2 ((−PBCO·S ABB1))2·(PADA)2·((−PBCAO))2 simplify = (PBEB)2·(PACO)2 (PBCO)2·(PADA)2 D = (PBEB)2·(PACO)2·(PBCB)2 (PBCO)2·((16S 2 ABC))2 E = ((16S 2 ABC))2·(PACO)2·(PBCB)2 (256)·(PBCO)2·(S ABC)4·(PACA)2 simplify = (PACO)2·(PBCB)2 (PBCO)2·(PACA)2 O = (PACA)2·(PBCB)2·((2))2 (PBCB)2·(PACA)2·((2))2 simplify = 1 The eliminants PBJB J = (16)·(S BB1C1 )2 PB1C1B1 PAKA K = (16)·(S AB1C1)2 PB1C1B1 S BB1C1 C1 = PBCO·S ABB1 PBCAO S AB1C1 C1 = PACO·S ABB1 PBCAO PADA D = (16)·(S ABC)2 PBCB PBEB E = (16)·(S ABC)2 PACA PBCO O =1 2(PBCB) PACO O =1 2(PACA) Deﬁnition The tangents to the circumcircle at the vertices of a given triangle form a triangle called the tangential triangle of the given triangle. Example 6.102 (0.067, 3, 10) Show that the vertices of the tangential triangle of ABC are the isogonal conjugates of the vertices of the anticomplementary triangle of ABC. 318 Chapter 6. Topics from Geometry 2 A 1 A O C B A Figure 6-102 Constructive description ( (points A B C) (circumcenter O A B C) (pratio A1 B A C 1) (inter A2 (t B B O) (t C C O)) (eqangle B A A2 A1 A C) ) The eliminants PBAA2 A2 = PBCO·S ABO+PABA·S BCO S BCO S ABA2 A2 = PBCO·PABO (16)·S BCO S ACA1 A1 = −(S ABC) PCAA1 A1 = PBAC+PACA S BCO O = PBCB·PBAC (32)·S ABC S ABO O = PACB·PABA (32)·S ABC PABO O =1 2(PABA) PBCO O =1 2(PBCB) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) The machine proof (−S ABA2 )·PCAA1 S ACA1 ·PBAA2 A2 = (−1 4 PBCO·PABO)·PCAA1 ·(4S BCO) S ACA1 ·(4PBCO·S ABO+4PABA·S BCO)·(4S BCO) simplify = −PBCO·PABO·PCAA1 (16)·S ACA1 ·(PBCO·S ABO+PABA·S BCO) A1 = −PBCO·PABO·(PBAC+PACA) (16)·(−S ABC)·(PBCO·S ABO+PABA·S BCO) O = PBCB·PABA·(PBAC+PACA)·((32S ABC))2·(2) (16)·S ABC·(64PBCB·PBAC·PABA·S ABC+32PBCB·PACB·PABA·S ABC)·((2))2 simplify = PBAC+PACA 2PBAC+PACB py = (−PBCB+3PACA+PABA)·((2))2 (−2PBCB+6PACA+2PABA)·(2) simplify = 1 Example 6.103 (1.300 3 38) If two lines are antiparallel with respect to an angle, the per-pendiculars dropped upon them from the vertex are isogonal in the angle considered. B C A R O S M N Figure 6-103 Constructive description ( (circle B C R S ) (circumcenter O B C R) (inter A (l B R) (l C S )) (foot N A B C) (foot M A R S ) (eqangle B A M N A C) ) 6.3. Triangles 319 Example 6.104 (3.083, 9, 60) Show that the four perpendiculars to the sides of an angle at four cyclic points form a parallelogram whose opposite vertices lie on isogonal conjugate lines with respect to the given angle. A B C O D E F I Figure 6-104 Constructive description ( (circle A B C D) (circumcenter O A B C) (inter I (l A B) (l C D)) (midpoint X C D) (midpoint Y A B) (inter E (p D O X) (p A O Y)) (inter F (p C O X) (p B O Y)) (eqangle A I E F I C) ) Example 6.105 (0.001, 1, 3) The perpendicular at the orthocenter H to the altitude HC of the triangle ABC meets the circumcircle of HBC in P. Show that ABPH is a parallelogram. B C A H O P Figure 6-105 Constructive description ((points A B C) (orthocenter H A B C) (circumcenter O B C H) (pratio PP H A B 1) (inter P (l H PP) (cir O H)) ( AH BP = 1) ) The machine proof: AH BP P = S AHPP S BHPP PP = −S ABH −S ABH simplify = 1. The eliminants: AH BP P = S AHPP S BHPP . S BHPP PP = −(S ABH). S AHPP PP = −(S ABH). Example 6.106 (1.533, 10, 38) The tangential and the orthic triangles of a given triangle are homothetic. (also see Example 6.90) I 1 B 1 A 1 C F E D O C B A Figure 6-106 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter C1 (t B O B) (t A O A)) (inter A1 (l C1 B) (t C O C)) (inter B1 (l C A1) (l A C1)) (inter I (l B1 E) (l A1 D)) (inter J (l A1 D) (l C1 F)) ( A1I DI = A1J DJ ) ) 320 Chapter 6. Topics from Geometry Example 6.107 (0.083, 4, 11) Let P be the midpoint AH. Show that the segment OP is bi-sected by the median AA1. B C A O 1 A H P I Figure 6-107 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint A1 B C) (orthocenter H A B C) (midpoint P A H) (inter I (l O P) (l A A1)) (midpoint I O P) ) The machine proof −OI PI I = S AOA1 S AA1P P = S AOA1 1 2 S AA1H H = (2)·S AOA1 ·(16S 2 ABC) −PBAC·PACB·S ABA1−PBAC·PABC·S ACA1 simplify = (−32)·S AOA1 ·(S ABC)2 (PACB·S ABA1+PABC·S ACA1 )·PBAC A1 = (−32)·(−1 2 S ACO−1 2 S ABO)·(S ABC)2 ( 1 2 PACB·S ABC−1 2 PABC·S ABC)·PBAC simplify = (32)·(S ACO+S ABO)·S ABC (PACB−PABC)·PBAC O = (32)·(−32PACB·PABA·S ABC+32PACA·PABC·S ABC)·S ABC (PACB−PABC)·PBAC·(32S ABC)·(−32S ABC) simplify = PACB·PABA−PACA·PABC (PACB−PABC)·PBAC py = (−2PBCB·PACA+2PBCB·PABA+2P2 ACA−2P2 ABA)·((2))3 (4PACA−4PABA)·(−PBCB+PACA+PABA)·((2))2 simplify = 1 The eliminants OI PI I = −S AOA1 S AA1P S AA1P P =1 2(S AA1H) S AA1H H = (PACB·S ABA1+PABC·S ACA1 )·PBAC (−16)·(S ABC)2 S ACA1 A1 = −1 2(S ABC) S ABA1 A1 = 1 2(S ABC) S AOA1 A1 = −1 2(S ACO+S ABO) S ABO O = PACB·PABA (32)·S ABC S ACO O = PACA·PABC (−32)·S ABC PBAC= −1 2(PBCB−PACA−PABA) PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) Example 6.108 (4.866, 12, 15) Prove that HA1 passes through the diametric opposite of A on the circumcircle. I H 1 A O A C B Figure 6-108 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint A1 B C) (orthocenter H A B C) (inter I (l A O) (l H A1)) (perp-biesct O B I) ) 6.3. Triangles 321 Example 6.109 (0.250, 2, 31) Show that the product of the distances of a point of the circum-circle of a triangle from the sides of the triangle is equal to the product of the distances of the same point from the sides of the tangential triangle of the given triangle. A O B C D E F G 1 E 1 F 1 G Figure 6-109 Constructive description ( (circle A B C D) (circumcenter O A B C) (foot E D B C) (foot F D A C) (foot G D A B) (inter E1 (p D O A) (t A O A)) (inter F1 (p D O B) (t B O B)) (inter G1 (p D O C) (t C O C)) (DE 2DF 2DG 2 = DE1 2DF1 2DG1 2) ) Example 6.110 (0.200, 12, 16) If O is the circumcenter the triangle ABC, and G is the cen-troid of ABC, we have 3OA 2 = GA 2 + GB 2 + GC 2 + 3OG 2. A B C O E F D G Figure 6-110 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint E A C) (midpoint F A B) (midpoint D B C) (inter G (l A D) (l C F)) (3OG 2+AG 2+GC 2+BG 2 = 3OA 2) ) Example 6.111 (1.033, 28, 24) The internal and external bisectors of an angle of a triangle pass through the ends of the circumdiameters which is perpendicular to the side opposite the vertex considered. L O A C B E P Figure 6-111 Constructive description ( (points B E P) (on L (t E E B)) (lratio C E B −1) (inter O (l L E) (b B L)) (inter A (l B P) (cir O B)) (eqangle B A L L A C) ) 322 Chapter 6. Topics from Geometry Example 6.112 (0.866, 12, 15) The segment of the altitude extended between the orthocenter and the second point of intersection with the circumcircle is bisected by the corresponding side of the triangle. A B C O D E H K Figure 6-112 Constructive description ( (points A B C) (circumcenter O A B C) (foot D C A B) (foot E B A C) (inter H (l C D) (l B E)) (inter K (l C D) (cir O C)) (midpoint D H K) ) Example 6.113 (1.650, 5, 18) The circumcircle of the triangle formed by two vertices and the orthocenter of a given triangle is equal to the circumcircle of the given triangle. A B C O D E H 1 O Figure 6-113 Constructive description ( (points A B C) (circumcenter O A B C) (foot D C A B) (foot E B A C) (inter H (l C D) (l B E)) (circumcenter O1 A B H) (eqdistance O A O1 H) ) Example 6.114 (0.933, 17, 18) A vertex of a triangle is the midpoint of the arc determined on its circumcircle by the two altitudes, produced, issued from the other two vertices. 1 A 1 C O H A C B Figure 6-114 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (inter C1 (l C H) (cir O C)) (inter A1 (l A H) (cir O A)) (perp-biesct B C1 A1) ) Example 6.115 (0.833, 17, 27) If O is the circumcenter and H the orthocenter of a triangle ABC, and AH, BH, CH meet the circumcircle in D1, E1, F1, prove that parallels through D1, E1, F1 to OA, OB, OC, respectively, meet in a point. 6.3. Triangles 323 I 1 B 1 A 1 C O H A C B Figure 6-115 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (inter C1 (l C H) (cir O C)) (inter A1 (l A H) (cir O A)) (inter B1 (l B H) (cir O B)) (inter I (p A1 O A) (p B1 O B)) (parallel I C1 O C) ) Example 6.116 (0.133, 7, 15) Show that the foot of the altitude to the base of a triangle and the projections of the ends of the base upon the circumdiameter passing through the opposite vertex of the triangle determine a circle having for center the midpoint the base. E D M O C B A Figure 6-116 Constructive description ((points A B C) (circumcenter O A B C) (midpoint M A B) (foot D C A B) (foot E A O C) (perp-biesct M E D) ) Example 6.117 (0.333, 12, 16) Show that the symmetric of the orthocenter of a triangle with respect to a vertex, and the symmetric of that vertex with respect to the midpoint of the opposite side, are collinear with the circumcenter of the triangle. B C A H O 1 A S L Figure 6-117 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (midpoint A1 B C) (lratio S A H −1) (lratio L A1 A −1) (collinear O L S ) ) Example 6.118 (0.817, 26, 25) If D1 is the second point of intersection of the altitude ADD1 of the triangle ABC with the circumcircle, center O, and P is the trace on BC of the per-pendicular from D1 to AC, show that the lines AP, AO make equal angles with the bisector of the angle DAC. 324 Chapter 6. Topics from Geometry P 1 D O D A C B Figure 6-118 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (inter D1 (l A D) (cir O A)) (inter P (l B C) (t D1 A C)) (eqangle O A D C A P) ) Example 6.119 (3.900, 14, 40) Show that the triangle formed by the foot of the altitude to the base of a triangle and the midpoints of the altitudes to the lateral sides is similar to the given triangle; its circumcircle passes through the orthocenter of the given triangle and through the midpoint of its base. Q P H F E D 1 A A C B Figure 6-119 Constructive description ( (points A B C) (midpoint A1 B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter H (l B E) (l A D)) (midpoint P B E) (midpoint Q C F) (cocircle H P Q D) ) Example 6.120 (0.717, 44, 38) The sides of the anticomplementary triangle of the triangle ABC meet the circumcircle of ABC in the points P, Q, R. Show that the area of the triangle PQR is equal to four times the area of the orthic triangle of ABC. R Q P O F E D A C B Figure 6-120 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter P (p A B C) (cir O A)) (inter Q (p B A C) (cir O B)) (inter R (p C A B) (cir O C)) (4S DEF = S PQR) ) Example 6.121 (0.450, 5, 17) Through the orthocenter of the triangle ABC parallels are drawn to the sides AB, AC, meeting BC in D, E. The perpendiculars to BC at D, E meet AB, AC in two points D1, E1 which are collinear with the diametric opposites of B, C on the circumcircle of ABC. 6.3. Triangles 325 Q P 1 D D O H A C B Figure 6-121 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (inter D (l B C) (p H A B)) (inter D1 (l A B) (p D A H)) (inter P (l B O) (cir O B)) (inter Q (l C O) (cir O C)) (collinear P Q D1) ) Example 6.122 (0.150, 6, 19) If the altitudes AD, BE, CF of the triangle ABC meet the circumcircle of ABC again in P, Q, R, show that we have (AP/AD)+(BQ/BE)+(CR/CF) = 4. A B C O D E F H P Q R Figure 6-122 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter H (l A D) (l C F)) (inter P (l A D) (cir O A)) (inter Q (l B E) (cir O B)) (inter R (l C F) (cir O C)) ( AP AD + BQ BE + CR CF = 4) ) Example 6.123 (0.433, 9, 21) Through the point of intersection of the tangents DB, DC to the circumcircle (O) of the triangle ABC a parallel is drawn to the line touching (O) at A. If this parallel meets AB, AC in E, F, show that D bisects EF. B C A O D F E Figure 6-123 Constructive description ( (points A B C) (circumcenter O A B C) (inter D (t C O C) (t B O B)) (inter F (l A B) (t D O A)) (inter E (l A C) (l F D)) (midpoint D E F) ) Example 6.124 (0.466, 13, 28) In a triangle ABC, let p and q be the radii of two circles through A, touching side BC at B and C, respectively. Then pq = R2. 326 Chapter 6. Topics from Geometry P Q O A C B M Figure 6-124 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint M B C) (inter Q (p B O M) (b A B)) (inter P (p C O M) (b A C)) (eq-product Q B P C O B O B) ) Example 6.125 (0.400, 7, 18) The parallel to the side AC through the vertex B of the triangle ABC meets the tangent to the circumcircle (O) of ABC at C in B1, and the parallel through C to AB meets the tangent to (O) at B in C1. Prove that BC2 = BC1 · B1C. B C A O 1 B 1 C Figure 6-125 Constructive description ( (points A B C) (circumcenter O A B C) (inter B1 (p B A C) (t C O C)) (inter C1 (p C A B) (t B O B)) (eq-product B C B C B C1 B1 C) ) Example 6.126 (5.733, 22, 23) Show that the foot of the perpendicular from the orthocenter of a triangle upon the line joining a vertex to the point of intersection of the opposite side with the corresponding side of the orthic triangle lies on the circumcircle of the triangle. O S X H E D A C B Figure 6-126 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot E B A C) (inter H (l A D) (l B E)) (inter X (l E D) (l A B)) (foot S H C X) (perp-biesct O S A) ) 6.3.4 The Euler Line Deﬁnition. The circumcenter O, orthocenter H, and the centroid G of a given triangle are collinear and this line is called the Euler line of the triangle. For the machine proof of Euler’s theorem, see Example 3.71 on page 141. Example 6.127 (0.033, 4, 2) Let O and H be the circumcenter and the orthocenter of a trian-gle ABC. Show that OH2 = 9R2 −a2 −b2 −c2. 6.3. Triangles 327 B C A O H Figure 6-127 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (OH 2 = 9OB 2−AB 2−AC 2−BC 2) ) Example 6.128 (0.167, 6, 9) With the usual notations for the triangle ABC, we have 4AO 2 = 4AB 2 + 4AC 2 + 4BC 2 + GH 2. G 2 M 1 M O H C B A Figure 6-128 Constructive description ( (points A B C) (circumcenter O A B C) (centroid G A B C) (orthocenter H A B C) (4AO 2 = 4AB 2+4AC 2+4BC 2+GH 2) ) Example 6.129 (0.816, 9, 9) With the usual notations for the triangle ABC, we have 9AO 2 = AB 2 + AC 2 + BC 2 + OH 2. O H C B A Figure 6-129 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (9AO 2 = AB 2+AC 2+BC 2+OH 2) ) Example 6.130 (0.250, 6, 8) With the usual notations for the triangle ABC, we have 12AO 2 = AB 2 + AC 2 + BC 2 + AH 2 + BH 2 + CH 2. O H C B A Figure 6-130 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (12AO 2 = AB 2+AC 2+BC 2+AH 2+BH 2+CH 2) ) Example 6.131 (5.050, 8, 40) The homothetic center of the orthic and the tangential triangles of a given triangle lies on the Euler line of the given triangle. 328 Chapter 6. Topics from Geometry I 1 A 1 C 1 B F E D O H C B A Figure 6-131 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (inter D (l B C) (l A H)) (inter E (l A C) (l B H)) (inter F (l A B) (l C H)) (inter B1 (t C O C) (t A O A)) (inter C1 (l A B1) (t B B O)) (inter A1 (l C1 B) (l B1 C)) (inter I (l F C1) (l B1 E)) (collinear O H I) ) Example 6.132 (0.833, 12, 27) The Euler lines of the four triangles of an orthocentric group are concurrent. A B C H O 1 A 1 B I Figure 6-132 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (circumcenter A1 B C H) (circumcenter B1 A C H) (inter I (l B B1) (l A A1)) (collinear O H I) ) Example 6.133 (0.950, 4, 40) Show that the perpendiculars from the vertices of a triangle to the lines joining the midpoints of the respectively opposite sides to the orthocenter of the triangle meet these sides in three points of a straight line perpendicular to the Euler line of the triangle. 2 C 1 C 2 B 1 B 2 A 1 A H O A C B Figure 6-133 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (midpoint A1 B C) (inter A2 (l B C) (t A A1 H)) (midpoint B1 C A) (inter B2 (l A C) (t B H B1)) (midpoint C1 A B) (inter C2 (l A B) (t C C1 H)) (inter C3 (l A B) (l A2 B2)) ( AC2 BC2 = AC3 BC3 ) ) 6.3.5 The Nine-Point Circle Deﬁnition. The midpoints of the segments joining the orthocenter of a triangle to its vertices 6.3. Triangles 329 are called the Euler points of the triangle. The three Euler points determine the Euler triangle. A B C M D E F N I J Figure 6-134 Example 6.134 (The Nine-Point Circle Theorem) (0.050, 2, 5) In a triangle the midpoints of the sides, the feet of altitudes, and the Euler points lie on the same circle. This circle is called the nine point circle of the given triangle. Constructive description ( (points A B C) (foot M A B C) (midpoint D B C) (midpoint E A C) (midpoint F A B) (midpoint I E F) (midpoint J M D) (parallel I J A M) ) The machine proof S AMI S AMJ J = S AMI 1 2 S AMD I = (2)·( 1 2 S AMF+ 1 2 S AME) S AMD F = S AME−1 2 S ABM S AMD E = −S ACM−S ABM (2)·S AMD D = −(S ACM+S ABM) (2)·(−1 2 S ACM−1 2S ABM) simplify = 1 The eliminants S AMJ J = 1 2(S AMD) S AMI I =1 2(S AMF+S AME) S AMF F = −1 2(S ABM) S AME E = −1 2(S ACM) S AMD D = −1 2(S ACM+S ABM) Since the orthocenter and the three feet of the triangle form an orthocentric group, we conclude from the above machine proof that the circle also passes the three Euler points. Example 6.135 (Feuerbach’s Theorem) (55.500 102 46) The nine-point circle of a triangle touches each of the four tritangent circles of the triangle. I A B C 1 M 2 M 3 M N D Figure 6-135 Constructive description ( (points I A B) (incenter C I A B) (midpoint M1 A B) (midpoint M2 A C) (midpoint M3 B C) (midpoint P M1 M3) (midpoint Q M1 M2) (inter N (t P P M1) (t Q Q M1)) (foot D I A B) (cc-tangent I D N M1) ) Example 6.136 (0.683, 3, 19) The radius of the nine-point circle is equal to half the circum-radius of the triangle. 330 Chapter 6. Topics from Geometry N F E D O C B A Figure 6-136 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot E B A C) (foot F C A B) (circumcenter N D E F) (4NF 2 = OA 2) ) Example 6.137 (0.067, 5, 18) The nine-point center lies on the Euler line, midway between the circumcenter and the orthocenter. N O C H B A 1 M 2 M 3 M Figure 6-137 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (midpoint M1 B C) (midpoint M2 A B) (midpoint M3 A C) (circumcenter N M1 M2 M3) (midpoint N H O) ) The machine proof −HN ON N = 1 2 PM1M2M1 −PHM1M2 −1 2 PM1M2M1+POM1M2 M2 = −(−PBM1H+ 1 2 PBM1B−PAM1H+ 1 2 PAM1A−1 4 PABA) −PBM1O+ 1 2 PBM1B−PAM1O+ 1 2 PAM1A−1 4 PABA M1 = −(2PBCH−2PBCB+2PACH−PACA+2PABH) 2PBCO−2PBCB+2PACO−PACA+2PABO H = −(−2PBCB+4PACB−PACA+2PABC) 2PBCO−2PBCB+2PACO−PACA+2PABO O = (2PBCB−4PACB+PACA−2PABC)·((2))3 −8PBCB+8PABA py = −(−4PBCB+4PABA) (PBCB−PABA)·((2))2 simplify = 1 The eliminants HN ON N = PM1M2M1−2PHM1M2 PM1M2M1−2POM1M2 POM1M2 M2 = 1 2(PBM1O+PAM1O) PHM1M2 M2 = 1 2(PBM1H+PAM1H) PM1M2M1 M2 = 1 4(2PBM1B+2PAM1A−PABA) PAM1O M1 = −1 4(PBCB−2PACO−2PABO) PBM1O M1 = 1 4(2PBCO−PBCB) PAM1A M1 = −1 4(PBCB−2PACA−2PABA) PAM1H M1 = −1 4(PBCB−2PACH−2PABH) PBM1B M1 = 1 4(PBCB) PBM1H M1 = 1 4(2PBCH−PBCB) PABH H =PABC PACH H =PACB PBCH H =PACB PABO O =1 2(PABA) PACO O =1 2(PACA) PBCO O =1 2(PBCB) PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) 6.3. Triangles 331 Example 6.138 (4.383, 51, 42) Show that the foot of the altitude of a triangle on a side, the midpoint of the segment of the circumdiameter between this side and the opposite vertex, and the nine-point center are collinear. N 1 C 1 B 1 A M E O D A C B Figure 6-138 Constructive description ( (points A B C) (foot D A B C) (circumcenter O A B C) (inter E (l B C) (l A O)) (midpoint M A E) (midpoint A1 C B) (midpoint B1 A C) (midpoint C1 B A) (circumcenter N A1 B1 C1) (collinear M D N) ) Example 6.139 (1.667, 44, 38) The center of the nine-point circle is the midpoint of a Euler point and the midpoint of the opposite side. A B C 1 M 2 M 3 M H 1 H N Figure 6-139 Constructive description ( (points A B C) (orthocenter H A B C) (midpoint M1 B C) (midpoint M2 A C) (midpoint M3 A B) (circumcenter N M1 M2 M3) (inter H1 (l C H) (l M3 N)) (midpoint H1 C H) ) Example 6.140 (1.183, 24, 49) The two pairs of points O and N, G and H separate each other harmonically. A B H C 2 M 3 M G O D E F N Figure 6-140 Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter O A B C) (midpoint M2 A C) (midpoint M3 A B) (midpoint M1 B C) (circumcenter N M1 M2 M3) (inter G (l B M2) (l C M3)) (harmonic O N G H) ) 332 Chapter 6. Topics from Geometry Example 6.141 (0.850, 23, 38) If P is the symmetric of the vertex A with respect to the oppo-site side BC, show that HP is equal to four times the distance of the nine-point center from BC. B C A 1 A 1 B 1 C H D N P K Figure 6-141 Constructive description ( (points A B C) (orthocenter H A B C) (foot D A B C) (lratio P D A −1) (midpoint A1 B C) (midpoint B1 A C) (midpoint C1 B A) (circumcenter N A1 B1 C1) (foot K N B C) (PH 2 = 16NK 2) ) Example 6.142 (2.450, 56, 42) Show that the square of the tangent from a vertex of a triangle to the nine-point circle is equal to the altitude issued from that vertex multiplied by the distance of the opposite side from the circumcenter. B C A O 1 A 1 B 1 C D N T Figure 6-142 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint A1 B C) (midpoint B1 A C) (midpoint C1 B A) (midpoint M1 A1 B1) (midpoint M2 A1 C1) (inter N (t M1 M1 A1) (t M2 M2 A1)) (foot D A B C) (PANA−PA1NA1 = PA1AD−POAD) ) Example 6.143 (0.950, 50, 37) Show that the symmetric of the circumcenter of a triangle with respect to a side coincides with the symmetric of the vertex opposite the side considered with respect to the nine-point center of the triangle. A B C O 1 C 1 B 1 A N 1 O Figure 6-143 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint C1 A B) (midpoint B1 A C) (midpoint A1 B C) (circumcenter N A1 B1 C1) (inter O1 (l O C1) (l C N)) (midpoint N O1 C) ) 6.3. Triangles 333 6.3.6 Incircles and the Excircles Example 6.144 (Theorem of Incenter) (0.750, 14, 29) The three internal bisectors of the angels of a triangle meet in a point, the incenter I of the triangle. A B I A A B B C Figure 6-144 1 2 1 2 Constructive description ( (points A B I) (foot A1 B I A) (lratio A2 A1 B −1) (foot B1 A I B) (lratio B2 B1 A −1) (inter C (l A A2) (l B B2)) (eqangle A C I I C B) ) Similarly the internal bisector at vertex A (B, C) and the two external bisectors at vertices B (C, A) and C (A, B) meet in a point Ia (Ib, Ic). Deﬁnition. Each of I, Ia, Ib, Ic is the center of a circle tangent to the three sides of the triangle. The four circles with I, Ia, Ib, Ic as centers are called the inscribed circle and the exscribed circles or the four tritangent circles of the given triangle. Example 6.145 (0.067, 3, 10) Two tritangent centers of a triangle are the ends of a diameter of a circle passing through the two vertices of the triangle which are not collinear with the centers considered. B C I A a I O Figure 145 Constructive description ( (points I B C) (incenter A I C B) (inter Ia (l A I) (t B B I)) (midpoint O I Ia) (eqdistance O B O I) ) The eliminants PIOI O =1 4(PIIaI) PBOB O =1 4(2PBIaB−PIIaI+2PIBI) PIIaI Ia = PIAI·(PIBI)2 (PBIA)2 PBIaB Ia = −(P2 BIA−PBIA·PIAI−PIAI·PIBA)·PIBI (PBIA)2 PIBA A =−PICB·PIBC·PIBI PBCB·PBIC PBIA A =(16)·PIBI·(S IBC)2 PBCB·PBIC 16S 2 IBC=PBIC·PICB+PICI·PIBC PIBC= 1 2(PBCB−PICI+PIBI) PICB= 1 2(PBCB+PICI−PIBI) PBIC= −1 2(PBCB−PICI−PIBI) The machine proof PBOB PIOI O = 1 2 PBIaB−1 4 PIIaI+ 1 2 PIBI 1 4 PIIaI 334 Chapter 6. Topics from Geometry Ia = (2P3 BIA·PIAI·PIBI+2P2 BIA·PIAI·PIBA·PIBI−P2 BIA·PIAI·P2 IBI)·P2 BIA PIAI·P2 IBI·(P2 BIA)2 simplify = 2PBIA+2PIBA−PIBI PIBI A = −P2 BCB·P2 BIC·PIBI−2PBCB·PBIC·PICB·PIBC·PIBI+32PBCB·PBIC·PIBI·S 2 IBC PIBI·(PBCB·PBIC)2 simplify = −(PBCB·PBIC+2PICB·PIBC−32S 2 IBC) PBCB·PBIC herron = −16PBCB·PBIC+32PBIC·PICB−32PICB·PIBC+32PICI·PIBC PBCB·PBIC·(16) py = −(4P2 BCB−4PBCB·PICI−4PBCB·PIBI)·(2) PBCB·(−PBCB+PICI+PIBI)·((2))3 simplify = 1 Example 6.146 (0.300, 8, 9) The four tritangent centers of a triangle lie on six circles which pass through the pairs of vertices of the triangle and have for their centers the midpoints of the arcs subtended by the respective sides of the triangle on its circumcircle. B C I A a I K Figure 6-146 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B B I) (l A I)) (inter K (l A I) (b C B)) (midpoint K I Ia) ) Example 6.147 (0.067, 4, 11) Let incircle (with center I) of △ABC touch the side BC at X and let A1 be the midpoint of this side. Then the line A1I (extended) bisectors AX. Constructive description ( (points B C I) (incenter A I C B) (midpoint A1 B C) (foot X I B C) (inter O (l A X) (l A1 I)) (midpoint O A X) ) The eliminants AO XO O = S IAA1 −S IA1X S IA1X X = PBA1I·S BIA1 PBA1B S BIA1 A1 = −1 2(S BCI) PBA1I A1 = 1 4(2PBCI−PBCB) PBA1B A1 = 1 4(PBCB) S IAA1 A1 = 1 2(S CIA+S BIA) S BIA A =−PBIB·PBCI·S BCI PBIC·PBCB S CIA A = PCIC·PCBI·S BCI PBIC·PBCB PBIC=1 2(PCIC+PBIB−PBCB) PBCI=1 2(PCIC−PBIB+PBCB) PCBI= −1 2(PCIC−PBIB−PBCB) 6.3. Triangles 335 B C I A 1 A X O Figure 6-147 The machine proof −AO XO O = −S IAA1 −S IA1X X = S IAA1·PBA1B PBA1I·S BIA1 A1 = ( 1 2 S CIA+ 1 2 S BIA)·( 1 4 PBCB) ( 1 2 PBCI−1 4 PBCB)·(−1 2 S BCI) A = −(PCIC·PCBI·PBIC·PBCB·S BCI−PBIC·PBIB·PBCI·PBCB·S BCI)·PBCB (2PBCI−PBCB)·S BCI·(PBIC·PBCB)2 simplify = −(PCIC·PCBI−PBIB·PBCI) (2PBCI−PBCB)·PBIC py = −(−2P2 CIC+2PCIC·PBCB+2P2 BIB−2PBIB·PBCB)·((2))2 (2PCIC−2PBIB)·(PCIC+PBIB−PBCB)·((2))2 simplify = 1 Example 6.148 (0.033, 1, 6) The product of the distances of two tritangent centers of a tri-angle from the vertex of the triangle collinear with them is equal to the product of the two sides of the triangle passing through the vertex considered. B C I A a I Figure 6-148 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B B I) (l A I)) (eq-product A I A Ia A B A C) ) Example 6.149 (0.550, 8, 17) Show that an external bisector of an angle of a triangle is parallel to the line joining the points where the circumcircle is met by the external (internal) bisectors of the other two angles of the triangle. B C I A M 1 L Figure 6-149 Constructive description ( (points B C I) (incenter A I C B) (inter M (l I C) (b B A)) (inter L1 (t B B I) (b C A)) (parallel M L1 A I) ) Example 6.150 (3.933, 4, 20) The product of the four tritangent radii of a triangle is equal to the square of its area. 336 Chapter 6. Topics from Geometry B C I A a I c I b I X a X b X c X Figure 6-150 Constructive description ( (points B C I) (incenter A I B C) (inter Ia (l A I) (t B B I)) (inter Ic (l C I) (l B Ia)) (inter Ib (l B I) (l A Ic)) (foot X I B C) (foot Xa Ia B C) (foot Xb Ib B C) (foot Xc Ic B C) (IX 2IaXa 2IbXb 2IcXc 2 = S ABCS ABCS ABCS ABC) ) I B A C O L M K N Figure 6-151 Example 6.151 3 (0.717, 17, 28) In triangle ABC, the bisector of angle A meets BC at L and the circumcir-cle of triangle ABC at N. The feet of the perpendic-ulars from L to AB and AC are K and M. Show that S ABC = S AKNM. Constructive description: ( (points I B A) (incenter C I B A) (circumcenter O A B C) (inter L (l B C) (l A I)) (foot M L A C) (foot K L A B) (inter N (l A I) (cir O A)) (S ABC = S AKNM) ) Example 6.152 (0.333, 6, 10) The points of contact of a side of a triangle with the incircle and the excircle relative to this side are two isotomic points. a X X a I A I C B Figure 6-152 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B B I) (l A I)) (foot X I B C) (foot Xa Ia B C) (eqdistance B X C Xa) ) Example 6.153 (0.683, 6, 11) In Figure 6-153, ZZa = a. a Z Z a I A I C B Figure 6-153 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B B I) (l A I)) (foot Z I A B) (foot Za Ia A B) (eqdistance Z Za B C) ) 3This is a problem from the 1987 International Mathematical Olympiad. 6.3. Triangles 337 Example 6.154 (0.583, 3, 20) In Figure 6-154, YbYc = a. B C I A c I b I c Y b Y Figure 6-154 Constructive description ( (points B C I) (incenter A I C B) (inter Ic (t A A I) (l I C)) (inter Ib (l A Ic) (l I B)) (foot Yc Ic A C) (foot Yb Ib A C) (eqdistance Yc Yb B C) ) Example 6.155 (0.266, 2, 14) The ratio of the area of a triangle to the area of the triangle determined by the points of contact of the sides with the incircle is equal to the ratio of the circumdiameter of the given triangle to its inradius. O Z Y X A I C B Figure 6-155 Constructive description ( (points B C I) (incenter A I C B) (circumcenter O A B C) (foot X I B C) (foot Y I A C) (foot Z I A B) (4S 2 XYZ·OB 2 = S ABCS ABCIX 2) ) Example 6.156 (0.033, 1, 6) Show that a parallel through a tritangent center to a side of a triangle is equal to the sum, or diﬀerence, of the two segments on the other two sides of the triangle between the two parallel lines considered. N M A I C B Figure 6-156 Constructive description ( (points B C I) (incenter A I C B) (inter M (l A C) (p I A B)) (inter N (l I M) (l B C)) (eqdistance M A M I) ) The eliminants PIMI M = PBAB·(S CIA)2 (S BCA)2 PAMA M = PCAC·(S BIA)2 (S BCA)2 S CIA A = PCIC·PCBI·S BCI PBIC·PBCB PBAB A =(PBIB)2·(PBCI)2 (PBIC)2·PBCB S BIA A =−PBIB·PBCI·S BCI PBIC·PBCB PCAC A =(PCIC)2·(PCBI)2 (PBIC)2·PBCB The machine proof PAMA PIMI M = PCAC·S 2 BIA·S 2 BCA PBAB·S 2 CIA·S 2 BCA simplify = PCAC·(S BIA)2 PBAB·(S CIA)2 A = P2 CIC·P2 CBI·((−PBIB·PBCI·S BCI))2·(PBIC·PBCB)2·P2 BIC·PBCB P2 BIB·P2 BCI·(PCIC·PCBI·S BCI)2·(PBIC·PBCB)2·P2 BIC·PBCB simplify = 1 338 Chapter 6. Topics from Geometry Example 6.157 (0.100, 1, 9) Prove the formula: AZ · BX · CY = r ▽ABC. Z Y X A I C B Figure 6-157 Constructive description ( (points B C I) (incenter A I B C) (foot X I B C) (foot Y I A C) (foot Z I A B) (AZ 2BX 2CY 2 = XI 2S ABCS ABC) ) Example 6.158 (0.666, 10, 18) The projection of the vertex B of the triangle ABC upon the internal bisector of the angle A lies on the line joining the points of contact of the incircle with the sides BC and AC. B C I A X Y L Figure 6-158 Constructive description ( (points B C I) (incenter A I B C) (foot X I B C) (foot Y I A C) (foot L B A I) (inter J (l A I) (l X Y)) ( AL IL = AJ IJ ) ) Example 6.159 (8.250, 21, 34) The midpoint of a side of a triangle, the foot of the altitude on this side, and the projections of the ends of this side upon the internal bisector of the opposite angle are four cyclic points. B C I A X Y 1 A D Figure 6-159 Constructive description ( (points B C I) (incenter A I B C) (foot X B A I) (foot Y C A I) (midpoint A1 B C) (foot D A B C) (cocircle X Y A1 D) ) Example 6.160 (0.466, 8, 17) Show that the midpoint of an altitude of a triangle, the point of contact of the corresponding side with the excircle relative to that side, and the incenter of the triangle are collinear. 6.3. Triangles 339 M a X a I D A I C B Figure 6-160 Constructive description ( (points B C I) (incenter A I B C) (foot D A B C) (inter Ia (t B B I) (l I A)) (foot Xa Ia B C) (inter M (l A D) (l I Xa)) (midpoint M A D) ) Example 6.161 (1.633, 10, 18) The internal bisectors of the angles B, C of the triangle ABC meet the line AXa joining A to the point of contact of BC with the excircle relative to this side in the points L, M. Prove that AL/AM = AB/AC. B C I A a I a X L M Figure 6-161 Constructive description ( (points B C I) (incenter A I B C) (inter Ia (t B B I) (l I A)) (foot Xa Ia B C) (inter L (l B I) (l A Xa)) (inter M (l C I) (l A Xa)) (eq-product A M A B A L A C) ) Example 6.162 (0.050, 1, 6) Show that the product of the distances of the incenter of a trian-gle from the three vertices of the triangle is equal to 4Rr2. B C I A O X Figure 6-162 Constructive description ( (points B C I) (incenter A I B C) (circumcenter O A B C) (foot X I B C) (16OB 2·IX 4 = IA 2IB 2IC 2) ) The eliminants PIXI X =(16)·(S BCI)2 PBCB PBOB O = PCAC·PBAB·PBCB (64)·(S BCA)2 S BCA A =(−2)·PCBI·PBCI·S BCI PBIC·PBCB PIAI A = (16)·PCIC·PBIB·(S BCI)2 (PBIC)2·PBCB PBAB A =(PBIB)2·(PBCI)2 (PBIC)2·PBCB PCAC A =(PCIC)2·(PCBI)2 (PBIC)2·PBCB The machine proof (16)·(PIXI)2·PBOB PIAI·PCIC·PBIB X = (16)·((16S 2 BCI))2·PBOB PIAI·PCIC·PBIB·(PBCB)2 O = (4096)·(S BCI)4·PCAC·PBAB·PBCB PIAI·PCIC·PBIB·(PBCB)2·(64S 2 BCA) 340 Chapter 6. Topics from Geometry simplify = (64)·(S BCI)4·PCAC·PBAB PIAI·PCIC·PBIB·PBCB·(S BCA)2 A = (64)·(S BCI)4·P2 CIC·P2 CBI·P2 BIB·P2 BCI·(PBIC·PBCB)2·P2 BIC·PBCB (16PCIC·PBIB·S 2 BCI)·PCIC·PBIB·PBCB·((−2PCBI·PBCI·S BCI))2·(P2 BIC·PBCB)2 simplify = 1 Example 6.163 (0.433, 6, 10) If the lines AX, BY, CZ joining the vertices of a triangle ABC to the points of contact X, Y, Z of the sides BC, CA, AB with the incircle meet that circle again in the points X1, Y1, Z1, show that: AX · XX1 · BC = 4rS , where r, S are the inradius and the area of ABC. B C I A X Y Z 1 X 1 Y 1 Z Figure 6-163 Constructive description ( (points B C I) (incenter A I B C) (foot X I B C) (foot Y I A C) (foot Z I A B) (inter X1 (l A X) (cir I X)) (inter Y1 (l B Y) (cir I Y)) (inter Z1 (l C Z) (cir I Z)) (16IX 2·S 2 ABC = XA 2XX1 2BC 2) ) Example 6.164 (2.783, 16, 13) If h, m, t are the altitude, the median, and the internal bisector issued from the same vertex of a triangle whose circumradius is R, show that 4R2h2(t2 − h2) = t4(m2 −h2). T O 1 A D A I C B Figure 6-164 Constructive description ( (points B C I) (incenter A I B C) (circumcenter O A B C) (foot D A B C) (midpoint A1 B C) (inter T (l I A) (l B C)) (4T A 2·OB 2·AD 2−4AD 2·OB 2·AD 2 = AA1 2·T A 4−AD 2·T A 4) ) Example 6.165 (0.050, 1, 10) The external bisectors of the angles of a triangle meet the opposite sides in three collinear points. 1 C 1 B 1 A A I C B Figure 6-165 Constructive description ( (points B C I) (incenter A I B C) 6.3. Triangles 341 (inter A1 (l B C) (t A A I)) (inter B1 (l A C) (t B B I)) (inter C1 (l A B) (t C C I)) (inter C2 (l A1 B1) (l A B)) ( AC1 BC1 = AC2 BC2 ) ) The machine proof ( AC1 BC1 )/( AC2 BC2 ) C2 = S BA1B1 S AA1B1 · AC1 BC1 C1 = PICA·S BA1B1 S AA1B1·PBCI B1 = PICA·PCBI·S BAA1·PICBA PIBA·S CAA1 ·PBCI·PICBA simplify = PICA·PCBI·S BAA1 PIBA·S CAA1 ·PBCI A1 = PICA·PCBI·(−PBAI·S BCA)·PCIBA PIBA·(−PCAI·S BCA)·PBCI·PCIBA simplify = PICA·PCBI·PBAI PIBA·PCAI·PBCI A = (−PCIC·PCBI·PBCI)·PCBI·(16PBIB·PBCI·S 2 BCI)·P2 BIC·PBCB·PBIC·PBCB (−PCBI·PBIB·PBCI)·(16PCIC·PCBI·S 2 BCI)·PBCI·P2 BIC·PBCB·PBIC·PBCB simplify = 1 The eliminants AC2 BC2 C2 = S AA1 B1 S BA1B1 AC1 BC1 C1 = PICA PBCI S AA1 B1 B1 = PIBA·S CAA1 PICBA S BA1B1 B1 = PCBI·S BAA1 PICBA S CAA1 A1 = −PCAI·S BCA PCIBA S BAA1 A1 = −PBAI·S BCA PCIBA PCAI A =(16)·PCIC·PCBI·(S BCI)2 (PBIC)2·PBCB PIBA A =−PCBI·PBIB·PBCI PBIC·PBCB PBAI A =(16)·PBIB·PBCI·(S BCI)2 (PBIC)2·PBCB PICA A =−PCIC·PCBI·PBCI PBIC·PBCB Example 6.166 (1.850, 14, 28) Prove that the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides has the same area with the triangle formed by the points of contact of the sides of the triangle with the inscribed circle. c Z b Y c I b I a X Z Y X a I A I C B Figure 6-166 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B B I) (l A I)) (foot X I B C) (foot Y I A C) (foot Z I A B) (foot Xa Ia B C) (inter Ib (t C C I) (l B I)) (inter Ic (t B B I) (l C I)) (foot Yb Ib A C) (foot Zc Ic B A) (S XYZ = S XaYbZc) ) Example 6.167 (Gergonne Point) (0.050, 1, 7) The lines joining the vertices of a triangle to the points of contact of the opposite sides with the inscribed circle are concurrent (The Gergonne point). 342 Chapter 6. Topics from Geometry J Z Y X A I C B Figure 6-167 Constructive description ( (points B C I) (incenter A I C B) (foot X I B C) (foot Y I A C) (foot Z I A B) ( BX XC CY YA AZ ZB = 1) ) The eliminants AZ BZ Z = PBAI −PIBA CY AY Y = PICA −PCAI BX CX X = PCBI −PBCI PCAI A =(16)·PCIC·PCBI·(S BCI)2 (PBIC)2·PBCB PIBA A =−PCBI·PBIB·PBCI PBIC·PBCB PICA A =−PCIC·PCBI·PBCI PBIC·PBCB PBAI A =(16)·PBIB·PBCI·(S BCI)2 (PBIC)2·PBCB The machine proof −AZ BZ · CY AY · BX CX Z = −PBAI −PIBA · CY AY · BX CX Y = PBAI·PICA PIBA·(−PCAI) · BX CX X = −PBAI·PICA·PCBI PIBA·PCAI·(−PBCI) A = (16PBIB·PBCI·S 2 BCI)·(−PCIC·PCBI·PBCI)·PCBI·P2 BIC·PBCB·PBIC·PBCB (−PCBI·PBIB·PBCI)·(16PCIC·PCBI·S 2 BCI)·PBCI·PBIC·PBCB·P2 BIC·PBCB simplify = 1 Example 6.168 (Nagel Point) (0.650, 3, 31) The lines joining the vertices of a triangle to the points of contact of the opposite sides with the excircles relative to those sides are concur-rent (The Nagel point). J c Z b Y a X c I b I a I A I C B Figure 6-168 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B I B) (l A I)) (inter Ib (l B I) (l Ia C)) (inter Ic (l Ia B) (l C I)) (foot Xa Ia B C) (foot Yb Ib A C) (foot Zc Ic A B) (inter J (l A Xa) (l B Yb)) (inter K (l C Zc) (l B Yb)) ( BJ YbJ = BK YbK ) ) Example 6.169 (3.583, 12, 30) Show that the line joining the incenter of the triangle ABC to the midpoint of the segment joining A to the Nagel point of ABC is bisected by the median issued from A. B C I A a I b I a X b Y N 1 A S P Figure 6-169 Constructive description ( (points B C I) (incenter A I B C) (inter Ia (l A I) (t B B I)) (inter Ib (l C Ia) (l B I)) (foot Xa Ia B C) (foot Yb Ib A C) (inter N (l B Yb) (l A Xa)) 6.3. Triangles 343 (midpoint A1 B C) (midpoint S N A) (inter P (l I S ) (l A1 A)) (midpoint P I S ) ) Example 6.170 (0.816, 12, 13) The sides AB, AC intercept the segments DE, FG on the parallels to the side BC through the tritangent centers I and Ia. Show that: 2/BC = 1/DE + 1/FG. B C I A a I D E F G Figure 6-170 Constructive description ( (points B C I) (incenter A I B C) (inter Ia (t B B I) (l A I)) (inter D (p I B C) (l A B)) (inter E (l D I) (l A C)) (inter F (p Ia B C) (l A B)) (inter G (l F Ia) (l A C)) ( BC DE + BC FG = 2) ) Example 6.171 (1.817, 27, 20) Show that the perpendiculars to the internal bisectors of a tri-angle at the incenter meet the respective sides in three points lying on a line perpendicular to the line joining the incenter to the circumcenter of the triangle. B C I A O X Y Z Figure 6-171 Constructive description ( (points B C I) (incenter A I B C) (circumcenter O A B C) (inter X (t I I A) (l B C)) (inter Y (t I I B) (l A C)) (inter Z (t I I C) (l A B)) (perpendicular O I X Z) ) Example 6.172 (0.416, 4, 19) The side BC of the triangle ABC touches the incircle (I) in X and the excircle (Ia) relative to BC in Xa. Show that the line AXa passes through the diametric opposite X1 of X on (I). 1 X a X X a I A I C B Figure 6-172 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B I B) (l A I)) (foot X I B C) (foot Xa Ia B C) (inter X1 (l I X) (l A Xa)) (midpoint I X X1) ) 344 Chapter 6. Topics from Geometry Example 6.173 (0.816, 6, 15) With the notations of Example 6.172, show that if the line A1I meets the altitude AD of ABC in P, then AP is equal to the inradius of ABC. B C I A X D 1 A P Figure 6-173 Constructive description ( (points B C I) (incenter A I C B) (foot X I B C) (foot D A B C) (midpoint A1 B C) (inter P (l I A1) (l A D)) (eqdistance A P I X) ) Example 6.174 (0.267, 10, 16) With the notations of Example 6.172, if the parallels to AXa through B, C meet the bisectors CI, BI in L, M show that the line LM is parallel to BC. M L a X X a I A I C B Figure 6-174 Constructive description ( (points B C I) (incenter A I C B) (inter Ia (t B I B) (l A I)) (foot X I B C) (foot Xa Ia B C) (inter L (p B A Xa) (l C I)) (inter M (p C A Xa) (l B I)) (parallel L M B C) ) Example 6.175 (0.150, 2, 10) Show that the trilinear polar (see Example 6.203) of the incen-ter of a triangle passes through the feet of the external bisectors, and this line is perpen-dicular to the line joining the incenter to the circumcenter, and this lines is perpendicular to the line joining the incenter to circumcenter. O Z X D F E A I C B Figure 6-175 Constructive description ( (points B C I) (incenter A I B C) (circumcenter O A B C) (inter E (l B I) (l A C)) (inter F (l C I) (l B A)) (inter D (l A I) (l B C)) (inter X (l E F) (l B C)) (inter Z (l D E) (l A B)) (perpendicular X A A I) ) 6.3. Triangles 345 The machine proof PIAX n = PCAI·S BEF−PBAI·S CEF simplify = PCAI·S BEF−PBAI·S CEF n = S BCAI·PCAI·S BAE·S BCI−S BCAI·PBAI·S CIE·S BCA simplify = PCAI·S BAE·S BCI−PBAI·S CIE·S BCA n = −S BCIA·PCAI·S BIA·S BCA·S BCI−S BCIA·PBAI·S CIA·S BCA·S BCI simplify = −(PCAI·S BIA+PBAI·S CIA)·S BCA·S BCI n = −(0)·(−2PCBI·PBCI·S BCI)·S BCI simplify = 0 The eliminants PIAX X = PCAI·S BEF−PBAI·S CEF S BECF S CEF F =S CIE·S BCA −S BCAI S BEF F =−S BAE·S BCI S BCAI S CIE E =S CIA·S BCI S BCIA S BAE E =−S BIA·S BCA S BCIA S BCA A =(−2)·PCBI·PBCI·S BCI PBIC·PBCB S CIA A = PCIC·PCBI·S BCI PBIC·PBCB PBAI A =(16)·PBIB·PBCI·(S BCI)2 (PBIC)2·PBCB S BIA A =−PBIB·PBCI·S BCI PBIC·PBCB PCAI A = (16)·PCIC·PCBI·(S BCI)2 (PBIC)2·PBCB The second result has also been proved by the program. Example 6.176 (0.633, 3, 27) Show that the mediators of the internal bisectors of the angles of a triangle meet the respective sides of the triangle in three collinear points. S Q P W V U A I C B Figure 6-176 Constructive description ( (points B C I) (incenter A I B C) (inter U (l A I) (l B C)) (inter V (l B I) (l A C)) (inter W (l C I) (l A B)) (inter P (b A U) (l B C)) (inter Q (b B V) (l A C)) (inter S (b C W) (l A B)) (inter S 1 (l A B) (l P Q)) ( AS BS = AS1 BS1 ) ) Example 6.177 (0.450, 6, 25) Show that the lines joining the vertices of a triangle to the projections of the incenter upon the mediators of the respectively opposite sides meet in a point - the isotomic conjugate of the Gergonne point of the triangle. B C I A X 1 A 1 B 1 C D E F J Y Figure 6-177 Constructive description ( (points B C I) (incenter A I B C) (foot X I B C) (midpoint A1 B C) (midpoint B1 C A) (midpoint C1 A B) (inter D (t A1 A1 C) (p I B C)) (inter E (t B1 B2 C) (p I A C)) (inter F (t C1 C1 B) (p I A B)) (inter J (l A D) (l B E)) (inter Y (l A D) (l B C)) (midpoint A1 X Y) ) 346 Chapter 6. Topics from Geometry Example 6.178 (0.516, 3, 19) Show that the line AI meets the sides XY, XZ in two points P, Q inverse with respect to the incircle (I) = XYZ, and the perpendiculars to AI at P, Q pass through the vertices B, C of the given triangle ABC. M Q P Z Y X A I C B Figure 6-178 Constructive description ( (points B C I) (incenter A I B C) (foot X I B C) (foot Y I A C) (foot Z I A B) (inter P (l I A) (l X Y)) (inter Q (l I A) (l X Z)) (inversion I X P Q) ) Deﬁnition The symmetric of a median of a triangle with respect to the internal bisector issued from the same vertex is a symmedian of the triangle Example 6.179 (1.250, 6,18) The three symmedians of a triangle are concurrent (The Lemoine Point or the symmedian point). I 3 M 2 M 1 M C B A Figure 6-179 Constructive description ( (points A B C) (midpoint M1 B C) (midpoint M2 A C) (midpoint M3 A B) (on B1 (a B C M2 B A)) (on C1 (a C A M3 C B)) (inter I (l C C1) (l B B1)) (eqangle B A I M1 A C) ) Example 6.180 (0.566, 5, 19) The three symmetrics of the three lines joining a point and the three vertices of a triangle with respect to the internal bisectors issued from the same vertices are concurrent. (The isogonal conjugate point). A B C M N Figure 6-180 Constructive description ( (points A B C M) (on A1 (a A C M A B)) (on B1 (a B A M B C)) (inter N (l A A1) (l B B1)) (eqangle A C M N C B) ) 6.3. Triangles 347 6.3.7 Intercept Triangles Deﬁnition. Let L, M, N be three points on the sides BC, CA, AB of triangle ABC. Then triangle LMN and the triangle determined by lines AL, BM, and CN are called the intercept triangles of triangle ABC for points L, M, N. Example 6.181 (0.033, 2, 5) Let A1, B1, C1 be points on the sides BC, CA, AB of a triangle ABC such that BA1/BC = CB1/CA = AC1/AB = 1/3. Show that the area of the triangle determined by lines AA1, BB1 and CC1 is one seventh of the area of triangle ABC. We only need to show S ABA2 S ABC = 2/7. Constructive description: ( (points A B C) (lratio A1 B C 1/3) (lratio B1 C A 1/3) (inter A2 (l A A1) (l B B1)) (2S ABC = 7S ABA2 ) ). 2 C 2 B 2 A 1 C 1 B 1 A C B A Figure 6-181 The machine proof (2)·S ABC (7)·S ABA2 A2 = (2)·S ABC·S ABA1B1 (7)·S ABB1·S ABA1 B1 = (2)·S ABC·(−2 3 S ACA1 +S ABA1) (7)·( 2 3 S ABC)·S ABA1 simplify = −(2S ACA1 −3S ABA1) (7)·S ABA1 A1 = −(−7 3 S ABC) (7)·( 1 3 S ABC) simplify = 1 The eliminants S ABA2 A2 = S ABB1·S ABA1 S ABA1 B1 S ABB1 B1 = 2 3(S ABC) S ABA1 B1 B1 = −1 3(2S ACA1 −3S ABA1) S ABA1 A1 = 1 3(S ABC) S ACA1 A1 = −2 3(S ABC) Example 6.182 (0.017, 3, 5) Let A1, B1, C1 be points on the sides BC, CA, AB of a triangle ABC such that BA1/BC = CB1/CA = AC1/AB = r. The intercept triangle determined by lines AA1, BB1 and CC1 is A2B2C2 (Figure 6-181). Show that S A2B2C2 S ABC = (2r−1)2 r2−r+1. Constructive description: ( (points A B C) (lratio A1 B C r) (lratio B1 C A r) (inter A2 (l A A1) (l B B1)) ((r2−r+1)·S A2AB = (−r2+r)·S ABC) ). 348 Chapter 6. Topics from Geometry The machine proof (r2−r+1)·S ABA2 −(r−1)·r·S ABC A2 = (r2−r+1)·S ABB1·S ABA1 −(r−1)·r·S ABC·S ABA1B1 B1 = −(r2−r+1)·(−S ABC·r+S ABC)·S ABA1 (r−1)·r·S ABC·(S ACA1 ·r−S ACA1 +S ABA1) simplify = (r2−r+1)·S ABA1 r·(S ACA1 ·r−S ACA1 +S ABA1 ) A1 = (r2−r+1)·S ABC·r r·(S ABC·r2−S ABC·r+S ABC) simplify = 1 The eliminants S ABA2 A2 = S ABB1·S ABA1 S ABA1 B1 S ABA1B1 B1 =S ACA1 ·r−S ACA1 +S ABA1 S ABB1 B1 = −((r−1)·S ABC) S ACA1 A1 =(r−1)·S ABC S ABA1 A1 =S ABC·r Example 6.183 (0.466, 5, 16) Use the same notations as 6.182. If BA1/A1C = r1, CB1/B1A = r2, AC1/C1B = r3 then S A2B2C2 S ABC = (r3r2r1−1)2 (r2r1+r1+1)(r3r1+r3+1)(r3r2+r2+1). Constructive description: (Formula derivation) ( (points A B C) (mratio A1 B C r1) (mratio B1 C A r2) (mratio C1 A B r3) (inter A2 (l A A1) (l B B1)) (inter B2 (l B B1) (l C C1)) (inter C2 (l C C1) (l A A1)) ( S A2B2C2 S ABC ) ) Example 6.184 (0.033, 2, 4) Let A1, B1, C1 be points on the sides BC, CA, AB of a triangle ABC such that BA1/A1C = CB1/B1A = AC1/C1B = k. Furthermore, let A2, B2, C2 be points on the sides B1C1, C1A1, A1C1 of a triangle A1B1C1 such that C1A2/A2B1 = A1B2/B2C1 = B1C2/C2A1 = k. Show that triangles ABC and A2B2C2 are homothetic. D O 2 C 2 B 2 A 1 B 1 A 1 C C B A Figure 6-184 Constructive description ((points A B C) (lratio C1 A B YT) (lratio A1 B C YT ) (lratio B1 C A YT ) (lratio A2 C1 B1 YT ) (lratio B2 A1 C1 YT ) (lratio C2 B1 A1 YT ) (inter O (l A A2) (l B B2)) (inter D (l C C2) (l A B)) (parallel A2 B2 A B) ) The machine proof S ABA2 S ABB2 B2 = S ABA2 −S ABA1·YT +S ABA1 simplify = −S ABA2 (YT −1)·S ABA1 A2 = −S ABB1·YT (YT −1)·S ABA1 B1 = −(−S ABC·YT +S ABC)·YT (YT −1)·S ABA1 simplify = S ABC·YT S ABA1 A1 = S ABC·YT S ABC·YT simplify = 1 The eliminants S ABB2 B2 = −((YT −1)·S ABA1) S ABA2 A2 =S ABB1·YT S ABB1 B1 = −((YT −1)·S ABC) S ABA1 A1 =S ABC·YT Example 6.185 (0.083, 4, 11) Let M, N, and P be three points on the sides AB, BC and AC of a triangle ABC such that AM/MB = BN/NC = CP/PA. Show that the point of intersection of the medians of △MNP coincides with the point of intersection of the medians of △ABC. 6.3. Triangles 349 L G F E P N M C B A Figure 6-185 Constructive description ((points A B C) (lratio M A B YT) (lratio N B C YT ) (lratio P C A YT ) (centroid G A B C) (inter L (l N P) (l M G)) ( PL NL = −1) ) The machine proof −( PL NL) L = −S MPG −(−S MNG) G = −(S CMP+S BMP+S AMP)·(3) (S CMN +S BMN+S AMN)·(3) P = −(S BCM·YT−S BCM+2S ACM·YT−S ACM) S CMN+S BMN+S AMN N = −(S BCM·YT−S BCM+2S ACM·YT−S ACM) −2S BCM·YT+S BCM−S ACM·YT M = −3S ABC·Y2 T +3S ABC·YT−S ABC −3S ABC·Y2 T +3S ABC·YT−S ABC simplify = 1 The eliminants PL NL L = S MPG S MNG S MNG G =1 3(S CMN+S BMN+S AMN) S MPG G =1 3(S CMP+S BMP+S AMP) S AMP P =(YT −1)·S ACM S BMP P =(YT −1)·S BCM S CMP P =S ACM·YT S AMN N = −(S ACM·YT) S BMN N = −(S BCM·YT) S CMN N = −((YT −1)·S BCM) S ACM M = −(S ABC·YT) S BCM M = −((YT −1)·S ABC) Example 6.186 (0.266, 3, 13) Let M, N, and P be the same as in Example 6.185. Show that the point of intersection of the medians of the triangle formed by lines AN, BP and CM coincides with the point of intersection of the medians of △ABC. A B C M N P E F G X Y Z L Figure 6-186 Constructive description ((points A B C) (lratio M A B YT) (lratio N B C YT ) (lratio P C A YT ) (centroid G A B C) (inter X (l B P) (l A N)) (inter Y (l B P) (l C M)) (inter Z (l A N) (l C M)) (inter L (l Y Z) (l X G)) ( YL ZL = −1) ) 350 Chapter 6. Topics from Geometry The machine proof −( YL ZL) L = −S GXY −(−S GXZ) Z = −S GXY ·S ACNM S CMX·S ANG Y = −S CMX·S BPG·S ACNM S CMX·S ANG·S BCPM simplify = −S BPG·S ACNM S ANG·S BCPM G = −(−S BCP+S ABP)·S ACNM·(3) (−S ACN−S ABN)·S BCPM·(3) P = −(2S ABC·YT −S ABC)·S ACNM (S ACN+S ABN)·(S BCM−S ACM·YT) simplify = −(2YT −1)·S ABC·S ACNM (S ACN+S ABN)·(S BCM−S ACM·YT) N = −(2YT −1)·S ABC·(S BCM·YT −S BCM+S ACM) (2S ABC·YT−S ABC)·(S BCM−S ACM·YT ) simplify = −(S BCM·YT−S BCM+S ACM) S BCM−S ACM·YT M = −(−S ABC·Y2 T+S ABC·YT−S ABC) S ABC·Y2 T −S ABC·YT+S ABC simplify = 1 The eliminants YL ZL L = S GXY S GXZ S GXZ Z =S CMX·S ANG S ACNM S GXY Y =S CMX·S BPG S BCPM S ANG G = −1 3(S ACN+S ABN) S BPG G = −1 3(S BCP−S ABP) S BCPM P =S BCM−S ACM·YT S ABP P = −((YT −1)·S ABC) S BCP P =S ABC·YT S ABN N =S ABC·YT S ACN N =(YT −1)·S ABC S ACNM N =S BCM·YT−S BCM+S ACM S ACM M = −(S ABC·YT) S BCM M = −((YT −1)·S ABC) Example 6.187 (0.533, 2, 32) Through each of the vertices of a triangle ABC we draw two lines dividing the opposite side into three equal parts. These six lines determine a hexagon. Prove that the diagonals joining opposite vertices of this hexagon meet in a point. A B C 1 A 2 A 1 B 2 B 1 C 2 C X Y Z U V W O Figure 6-187 Constructive description ( (points A B C) (lratio A1 B C 1/3) (lratio A2 B C 2/3) (lratio B1 C A 1/3) (lratio B2 C A 2/3) (lratio C1 A B 1/3) (lratio C2 A B 2/3) (inter X (l A A1) (l B B2)) (inter Y (l C C1) (l A A2)) (inter Z (l B B1) (l C C2)) (inter U (l A A2) (l B B1)) (inter V (l C C2) (l A A1)) (inter W (l B B2) (l C C1)) (inter O (l Y V) (l U X)) (inter ZO (l W Z) (l U X)) ( UO XO = UZO XZO ) ) Example 6.188 (0.216, 4, 11) Let M, N, P be points on the sides AB, BC and AC of a triangle ABC. Show that if M1, N1 and P1 are points on sides AC, BA, and BC of a triangle ABC such that MM1 ∥BC, NN1 ∥CA and PP1 ∥AB, then triangles MNP and M1N1P1 have equal areas. 6.3. Triangles 351 Constructive description ( (points A B C) (on M (l A B)) (on N (l B C)) (on P (l A C)) (inter M1 (l A C) (p M B C)) (inter N1 (l A B) (p N A C)) (inter P1 (l B C) (p P A B)) (S MNP = S M1N1P1) ) The eliminants S M1N1P1 P1 = S CM1 N1·S ABP+S BM1N1·S BCP S ABC S BM1N1 N1 = S ABM1 ·S ABN S ABC S CM1 N1 N1 = −(S CNM1) S ABM1 M1 = −S ACM S CNM1 M1 = −(S CMN) S BCP P = −(( AP AC −1)·S ABC) S ABP P =S ABC· AP AC S MNP P =S CMN· AP AC −S AMN· AP AC +S AMN S ABN N =S ABC· BN BC S AMN N = −(S ACM· BN BC) S CMN N = −(( BN BC −1)·S BCM) The machine proof S MNP S M1N1P1 P1 = S MNP·(−S ABC) −S CM1 N1·S ABP−S BM1N1·S BCP N1 = S MNP·(S ABC)2 −S CNM1 ·S ABP·S ABC+S BCP·S ABM1·S ABN M1 = S MNP·(S ABC)3 −(−S CMN ·S ABP·S 2 ABC+S BCP·S ACM·S ABN·S ABC) simplify = S MNP·(S ABC)2 S CMN·S ABP·S ABC−S BCP·S ACM·S ABN P = (S CMN · AP AC −S AMN· AP AC +S AMN)·(S ABC)2 S CMN·S 2 ABC· AP AC +S ACM·S ABN·S ABC· AP AC −S ACM·S ABN·S ABC simplify = (S CMN · AP AC −S AMN· AP AC +S AMN)·S ABC S CMN·S ABC· AP AC +S ACM·S ABN· AP AC −S ACM·S ABN N = (−S BCM· BN BC · AP AC +S BCM· AP AC +S ACM· BN BC · AP AC −S ACM· BN BC )·S ABC −S BCM·S ABC· BN BC · AP AC +S BCM·S ABC· AP AC +S ACM·S ABC· BN BC · AP AC −S ACM·S ABC· BN BC simplify = 1 Example 6.189 (0.050, 2, 3) Three parallel lines drawn through the vertices of a triangle ABC meet the respectively opposite sides in the points X, Y, Z. Show that area XYZ / area ABC = 2 / 1. Z Y X C B A Figure 6-189 Constructive description ( (points A B C) (on X (l B C)) (inter Y (l A C) (p B A X)) (inter Z (l A B) (p C A X)) (2S BAC = S XYZ) ) The machine proof: (−2)·S ABC S XYZ Z = (−2)·S ABC·S ABX S BXY·S ACX+S AXY ·S ABC Y = (−2)·S ABC·S ABX·S ACX −2S ACX·S ABX·S ABC simplify = 1. The eliminants: S XYZ Z = S BXY·S ACX+S AXY ·S ABC S ABX . S AXY Y = −(S ABX). S BXY Y =−S ABX·S ABC S ACX . 352 Chapter 6. Topics from Geometry Example 6.190 (0.066, 1, 6) Two doubly perspective triangles are in fact triply perspective. O A B C 1 O 1 A 1 B 1 C 2 O Figure 6-190 Constructive description ( (points A B C O O1) (inter A1 (l C O1) (l A O)) (inter B1 (l A O1) (l B O)) (inter C1 (l B O1) (l C O)) (inter O2 (l B A1) (l A C1)) (inter ZO2 (l B1 C) (l B A1)) ( BO2 A1O2 = BZO2 A1ZO2 ) ) The machine proof ( BO2 A1O2)/( BZO2 A1ZO2 ) ZO2 = S CA1 B1 −S BCB1 · BO2 A1O2 O2 = (−S ABC1 )·S CA1 B1 −S BCB1·(−S AA1C1) C1 = −S BCO·S ABO1·S CA1 B1·(−S BCO1O) S BCB1 ·(−S BOO1·S ACA1 )·S BCO1O simplify = −S BCO·S ABO1·S CA1 B1 S BCB1 ·S BOO1·S ACA1 B1 = −S BCO·S ABO1·(−S BOO1·S ACA1 )·(−S ABO1O) (−S BCO·S ABO1)·S BOO1·S ACA1 ·S ABO1O simplify = 1 The eliminants BZO2 A1ZO2 ZO2 = −S BCB1 S CA1 B1 BO2 A1O2 O2 = S ABC1 S AA1C1 S AA1C1 C1 = S BOO1·S ACA1 S BCO1O S ABC1 C1 = S BCO·S ABO1 S BCO1O S BCB1 B1 = S BCO·S ABO1 S ABO1O S CA1 B1 B1 = −S BOO1·S ACA1 S ABO1O 6.3.8 Equilateral Triangles Example 6.191 (The Napoleon triangle) (0.433, 8, 22) If equilateral triangles are erected exter-nally (or internally) on the sides of any triangle, their centers form an equilateral triangle. The machine proof PO2O1O2 PO2O3O2 O3 = PO2O1O2 PO2GO2+ 1 9 PBGB·r2−8 3 S BO2G·r G = (9)·PO2O1O2 9 2 PBO2B+ 9 2 PAO2A+ 1 4 PABA·r2−9 4 PABA−12S ABO2 ·r O1 = (36)·(PO2FO2+ 1 9 PCFC·r2−8 3 S CO2F·r) 18PBO2B+18PAO2A+PABA·r2−9PABA−48S ABO2·r F = (4)·( 9 2 PCO2C+ 9 2 PBO2B+ 1 4 PBCB·r2−9 4 PBCB−12S BCO2 ·r) 18PBO2B+18PAO2A+PABA·r2−9PABA−48S ABO2·r O2 = 4PCABE·r2+18PCEC+18PBEB+PBCB·r2−9PBCB+4PAEA·r2−48S BCE·r−48S ABE·r 18PBEB+4PBAE·r2+4PAEA·r2+18PAEA+PABA·r2−9PABA−96S ABE·r E = 3PBCB·r2+PACA·r2−2PABC·r2+9PABA−48S ABC·r 9PBCB+2PBAC·r2+PACA·r2+PABA·r2−48S ABC·r cons = 9PBCB+3PACA−6PABC+9PABA−48S ABC·r 9PBCB+6PBAC+3PACA+3PABA−48S ABC·r py = (4PBCB+4PACA+4PABA−32S ABC·r)·(2) (4PBCB+4PACA+4PABA−32S ABC·r)·(2) simplify = 1 6.3. Triangles 353 Constructive description ( (points A B C) (constant r2−3), (midpoint E A C) (tratio O2 E A 1 3 r) (midpoint F B C) (tratio O1 F C 1 3 r) (midpoint G A B) (tratio O3 G B 1 3 r) (eqdistance O1 O2 O2 O3) ) The eliminants PO2O3O2 O3 = 1 9(9PO2GO2+PBGB·r2−24S BO2G·r) S BO2G G =1 2(S ABO2), PBGB G = 1 4(PABA) PO2GO2 G =1 4(2PBO2B+2PAO2A−PABA) PO2O1O2 O1 = 1 9(9PO2FO2+PCFC·r2−24S CO2F·r) S CO2F F =1 2(S BCO2), PCFC F = 1 4(PBCB) PO2FO2 F =1 4(2PCO2C+2PBO2B−PBCB) S ABO2 O2 = −1 12(PBAE·r−12S ABE) PAO2A O2 = 1 9((r2+9)·PAEA) S BCO2 O2 = −1 12(PCABE·r−12S BCE) PBO2B O2 = 1 9(9PBEB+PAEA·r2−24S ABE·r) PCO2C O2 = 1 9(9PCEC+PAEA·r2) PBAE E = 1 2(PBAC), S ABE E =1 2(S ABC), S BCE E =1 2(S ABC), PAEA E = 1 4(PACA) PBEB E = 1 4(2PBCB−PACA+2PABA) PCEC E =1 4(PACA), PCABE E = 1 2(PBCB−PABC) PBAC= −1 2(PBCB−PACA−PABA) PABC=1 2(PBCB−PACA+PABA) Example 6.192 (0.883, 15, 31) Continuing from the above example, the lines AO1, BO2, CO3 are concurrent. A B C 1 C 1 A 1 B 1 O 2 O 3 O O Figure 6-192 Constructive description ( (constant r2−3) (points A B C) (midpoint E A C) (tratio O2 E A 1 3 r) (midpoint F B C) (tratio O1 F C 1 3 r) (midpoint G A B) (tratio O3 G B 1 3 r) (inter N (l O2 B) (l O1 A)) (inter M (l O2 B) (l O3 C)) ( BN O2N = BM O2M ) ) Example 6.193 (0.900, 18, 31) In Example 6.192, if the three equilateral triangles become similar isosceles triangles, the conclusion is still true. Constructive description ( (points A B C) (midpoint E A C) (tratio O2 E A r) (midpoint F B C) (tratio O1 F C r) (midpoint G A B) (tratio O3 G B r) (inter N (l O2 B) (l O1 A)) (inter M (l O2 B) (l O3 C)) ( BN O2N = BM O2M ) ) Example 6.194 (0.367 26 54) Let the circumcenters of the equilateral triangles erected exter-nally (or internally) on the sides of any triangle be O1, O2 and O3 (X1, X2, and X3). Show that S O1O2O3 + S X1X2X3 = S ABC. 354 Chapter 6. Topics from Geometry Constructive description ( (points A B C) (constant r2−3) (midpoint E A C) (tratio O2 E A 1 3 r) (lratio X2 E O2 −1) (midpoint F B C) (tratio O1 F C 1 3 r) (lratio X1 F O1 −1) (midpoint G A B) (tratio O3 G B 1 3 r) (lratio X3 G O3 −1) (S O1O2O3+S X1X2X3 = S ABC) ) Example 6.195 (0.050, 5, 8) Let equilaterals BCD, ABF, and ACE are erected externally on the sides of triangle ABC. Show that AD = CF = BE. A B C F D E Figure 6-195 Constructive description ( (points A B C) (constant r2−3) (midpoint M A C) (tratio E M A r) (midpoint N B C) (tratio D N C r) (eqdistance B E A D) ) The machine proof PBEB PADA D = PBEB PCNC·r2+PANA+8S ACN·r N = PBEB 1 4 PBCB·r2−1 4 PBCB+ 1 2 PACA+ 1 2 PABA−4S ABC·r E = (4)·(PBMB+PAMA·r2−8S ABM·r) PBCB·r2−PBCB+2PACA+2PABA−16S ABC·r M = (4)·( 1 2 PBCB+ 1 4 PACA·r2−1 4 PACA+ 1 2 PABA−4S ABC·r) PBCB·r2−PBCB+2PACA+2PABA−16S ABC·r cons = 2PBCB+2PACA+2PABA−16S ABC·r 2PBCB+2PACA+2PABA−16S ABC·r simplify = 1 The eliminants PADA D =PCNC·r2+PANA+8S ACN·r S ACN N = −1 2(S ABC) PANA N = −1 4(PBCB−2PACA−2PABA) PCNC N = 1 4(PBCB) PBEB E =PBMB+PAMA·r2−8S ABM·r S ABM M = 1 2(S ABC) PAMA M = 1 4(PACA) PBMB M = 1 4(2PBCB−PACA+2PABA) 6.3.9 Pedal Triangles Deﬁnition. From a point P three perpendicular lines are drawn to the three sides of a triangle. The triangle whose vertices are the three feet of the three perpendicular lines is called the pedal triangle of point P with respect to the given triangle. Example 6.196 (0.783, 6, 23) The orthogonal projections from point D to BC, AC, and AB are E, F, and G respectively. Let O be the circumcenter of triangle ABC. Show that DO 2 = AO 2(1 −4S EFG S ABC ). A B C D O E F G Figure 6-196 Constructive description 6.3. Triangles 355 ( (points A B C D) (circumcenter O A B C) (foot E D B C) (foot F D A C) (foot G D A B) (−DO 2·S ABC = −OA 2·S ABC+4OA 2·S EFG) ) Example 6.197 (0.083, 2, 5) Let K be the area of the pedal triangle of the orthocenter of triangle ABC with respect to triangle ABC. Show that K S ABC = PABCPACBPBAC 4AB 2BC 2AC 2 . A B C E F G Figure 6-197 Constructive description ((points A B C) (foot E A B C) (foot F B A C) (foot G C A B) (PABC·PBAC·PACB·S ABC = 4AB 2·BC 2·AC 2·S EFG) ) The machine proof PBAC·PACB·PABC·S ABC ( 1 2 )·PBCB·PACA·PABA·S EFG G = PBAC·PACB·PABC·S ABC·PABA ( 1 2 )·PBCB·PACA·PABA·(PBAC·S BEF+PABC·S AEF) simplify = PBAC·PACB·PABC·S ABC ( 1 2 )·PBCB·PACA·(PBAC·S BEF+PABC·S AEF) F = PBAC·PACB·PABC·S ABC·(PACA)2 ( 1 2 )·PBCB·PACA·(PBAC·PACB·PACA·S ABE−PBAC·PACA·PABC·S ACE) simplify = PACB·PABC·S ABC ( 1 2 )·PBCB·(PACB·S ABE−PABC·S ACE) E = PACB·PABC·S ABC·(PBCB)2 ( 1 2 )·PBCB·(2PBCB·PACB·PABC·S ABC) simplify = 1 The eliminants S EFG G = PBAC·S BEF+PABC·S AEF PABA S AEF F =−PBAC·S ACE PACA S BEF F = PACB·S ABE PACA S ACE E = −PACB·S ABC PBCB S ABE E = PABC·S ABC PBCB Example 6.198 (0.233, 14, 16) Let K be the area of the pedal triangle of the centroid of triangle ABC with respect to triangle ABC. Show that K S ABC = AB 2+BC 2+AC 2 36R2 . A B C O G E F G Figure 6-198 Constructive description ((points A B C) (circumcenter O A B C) (centroid G A B C) (foot E G B C) (foot F G A C) (foot G G A B) (AB 2·S ABC+BC 2·S ABC+AC 2·S ABC = 36OA 2·S EFG) ) 356 Chapter 6. Topics from Geometry Example 6.199 (0.083, 2, 11) Let K be the area of the pedal triangle of the circumcenter of triangle ABC with respect to triangle ABC. Show that S ABC = 4K. A B C O E F G Figure 6-199 Constructive description ((points A B C) (circumcenter O A B C) (foot E O B C) (foot F O A C) (foot G O A B) (S ABC = 4S EFG) ) The machine proof S ABC (4)·S EFG G = S ABC·PABA (4)·(PBAO·S BEF+PABO·S AEF) F = S ABC·PABA·(PACA)2 (4)·(−PCAO·PACA·PABO·S ACE+PBAO·PACO·PACA·S ABE) simplify = S ABC·PABA·PACA (−4)·(PCAO·PABO·S ACE−PBAO·PACO·S ABE) E = S ABC·PABA·PACA·(PBCB)2 (−4)·(−PCBO·PBCB·PBAO·PACO·S ABC−PCAO·PBCO·PBCB·PABO·S ABC) simplify = PABA·PACA·PBCB (4)·(PCBO·PBAO·PACO+PCAO·PBCO·PABO) O = PABA·PACA·PBCB·((2))6 (4)·(16PBCB·PACA·PABA) simplify = 1 The eliminants S EFG G = PBAO·S BEF+PABO·S AEF PABA S AEF F =−PCAO·S ACE PACA S BEF F = PACO·S ABE PACA S ABE E = PCBO·S ABC PBCB S ACE E = −PBCO·S ABC PBCB PABO O =1 2(PABA) PBCO O =1 2(PBCB) PCAO O =1 2(PACA) PACO O =1 2(PACA) PBAO O =1 2(PABA) PCBO O =1 2(PBCB) Example 6.200 (0.300, 2, 14) Let K be the area of the pedal triangle of the incenter of triangle ABC with respect to triangle ABC. Show that K S ABC = r 2R. A B I C O E F G Figure 6-200 Constructive description ((points A B I) (incenter C I A B) (circumcenter O A B C) (foot E I B C) (foot F I A C) (foot G I A B) (IE 2·S 2 ABC = 4OA 2·S 2 EFG) ) 6.3.10 Miscellaneous Example 6.201 (0.016, 1, 4) AD, AA1 are the altitude and the median of the triangle ABC; the parallels through A1 to AB, AC meet AD in P, Q; show that (ADPQ) = -1. 6.3. Triangles 357 A B C D 1 A P Q Figure 6-201 Constructive description ( (points A B C) (foot D A B C) (midpoint A1 B C) (inter P (l A D) (p A1 A B)) (inter Q (l A D) (p A1 A C)) (harmonic A D P Q) ) The machine proof (−AP DP)/( AQ DQ) Q = −S ADA1 −S ACA1 · −AP DP P = −(−S ABA1 )·S ADA1 S ACA1 ·(−S ADA1 ) simplify = −S ABA1 S ACA1 A1 = −( 1 2 S ABC) −1 2 S ABC simplify = 1 The eliminants AQ DQ Q = S ACA1 S ADA1 AP DP P = S ABA1 S ADA1 S ACA1 A1 = −1 2(S ABC) S ABA1 A1 = 1 2(S ABC) Example 6.202 (0.067, 3, 13) If A1, B1, C1 are the midpoint of the sides of the triangle ABC, prove that A1A is the harmonic conjugate of A1C with respect to A1B1, A1C1. D G 1 C 1 B 1 A A C B Figure 6-202 Constructive description ((points A B C) (midpoint A1 C B) (midpoint B1 A C) (midpoint C1 B A) (centroid G A B C) (inter D (l C1 C) (l A1 B1)) (harmonic C1 D G C) ) The machine proof ( C1G GD )/( CC1 CD ) D = (−S A1C1B1G)·S CA1 B1 S CA1C1 B1·S A1B1G G = −(−3S A1 B1C1 +S CA1 B1+S BA1B1+S AA1 B1)·S CA1 B1·(3) S CA1C1B1·(S CA1 B1+S BA1B1+S AA1 B1)·(3) C1 = (−S CA1 B1+ 1 2 S BA1B1+ 1 2 S AA1B1)·S CA1 B1 (S CA1 B1−1 2 S BA1B1−1 2 S AA1B1)·(S CA1 B1+S BA1B1+S AA1 B1) simplify = −S CA1 B1 S CA1 B1+S BA1B1+S AA1 B1 B1 = −( 1 2 S ACA1 ) 1 2 S ABA1 A1 = −(−1 2 S ABC) 1 2 S ABC simplify = 1 The eliminants CC1 CD D = S CA1C1B1 S CA1 B1 C1G GD D = −S A1C1 B1G S A1B1G S A1B1G G =1 3(S CA1 B1+S BA1B1+S AA1 B1) S A1C1B1G G =−1 3(3S A1 B1C1−S CA1 B1−S BA1B1−S AA1 B1) S CA1C1B1 C1 = 1 2(2S CA1 B1−S BA1B1−S AA1 B1) S A1B1C1 C1 = 1 2(S BA1B1+S AA1 B1) S AA1 B1 B1 = −1 2(S ACA1) S BA1B1 B1 = 1 2(S ABA1) S CA1 B1 B1 = 1 2(S ACA1) S ABA1 A1 = 1 2(S ABC) S ACA1 A1 = −1 2(S ABC) 358 Chapter 6. Topics from Geometry Example 6.203 (0.250, 1, 12) Let P be a point in the plane of the triangle ABC. Let A1 = BC∩AP, B1 = AC∩BP, C1 = AB∩CP, A2 = BC∩B1C1, B2 = AC∩A1C1, C2 = AB∩A1B1. Show that A2, B2, C2 are collinear, and is called the trilinear polar of P for triangle ABC. A B C P 1 A 1 B 1 C 2 A 2 B 2 C Figure 6-203 Constructive description ( (points A B C P) (inter A1 (l B C) (l A P)) (inter B1 (l A C) (l B P)) (inter C1 (l A B) (l C P)) (inter A2 (l B C) (l B1 C1)) (inter B2 (l A C) (l A1 C1)) (inter C2 (l A B) (l A1 B1)) (inter ZC2 (l B2 A2) (l A1 B1)) ( A1C2 B1C2 = A1ZC2 B1ZC2 ) ) The machine proof ( A1C2 B1C2 )/( A1ZC2 B1ZC2 ) ZC2 = −S B1A2B2 −S A1A2B2 · A1C2 B1C2 C2 = S ABA1·S B1A2B2 S A1A2B2·S ABB1 B2 = S ABA1·S A1B1C1·S ACA2 ·(−S AA1CC1 ) (−S A1C1A2·S ACA1 )·S ABB1·S AA1CC1 simplify = S ABA1·S A1B1C1·S ACA2 S A1C1A2·S ACA1 ·S ABB1 A2 = S ABA1 ·S A1B1C1·S CB1C1·S ABC·S BB1CC1 (−S A1B1C1·S BCC1 )·S ACA1 ·S ABB1·S BB1CC1 simplify = S ABA1·S CB1C1 ·S ABC −S BCC1 ·S ACA1 ·S ABB1 C1 = S ABA1 ·(−S CPB1 ·S ABC)·S ABC·S ACBP −(−S BCP·S ABC)·S ACA1 ·S ABB1·(−S ACBP) simplify = S ABA1·S CPB1 ·S ABC S BCP·S ACA1 ·S ABB1 B1 = S ABA1·S BCP·S ACP·S ABC·S ABCP S BCP·S ACA1 ·S ABP·S ABC·S ABCP simplify = S ABA1·S ACP S ACA1 ·S ABP A1 = (−S ABP·S ABC)·S ACP·(−S ABPC) (−S ACP·S ABC)·S ABP·(−S ABPC) simplify = 1 The eliminants A1ZC2 B1ZC2 ZC2 = S A1A2B2 S B1A2B2 A1C2 B1C2 C2 = S ABA1 S ABB1 S A1A2B2 B2 = S A1C1A2·S ACA1 S AA1CC1 S B1A2B2 B2 = S A1B1C1 ·S ACA2 S AA1CC1 S A1C1A2 A2 = −S A1B1C1 ·S BCC1 S BB1CC1 S ACA2 A2 = S CB1C1·S ABC S BB1CC1 S BCC1 C1 = −S BCP·S ABC S ACBP S CB1C1 C1 = S CPB1 ·S ABC S ACBP S ABB1 B1 = S ABP·S ABC S ABCP S CPB1 B1 = S BCP·S ACP S ABCP S ACA1 A1 = S ACP·S ABC S ABPC S ABA1 A1 = S ABP·S ABC S ABPC Example 6.204 (0.083, 1, 6) Let P be a point in the plane of the triangle ABC. Let A1 = BC ∩AP, B1 = AC ∩BP, C1 = AB ∩CP, A2 = BC ∩B1C1. Show that A1, A2, B, C form a harmonic sequence. 6.3. Triangles 359 A B C P 1 A 1 B 1 C 2 A Figure 6-204 Constructive description ( (points A B C P) (inter A1 (l B C) (l A P)) (inter B1 (l A C) (l B P)) (inter C1 (l A B) (l C P)) (inter A2 (l B C) (l B1 C1)) (harmonic B C A1 A2) ) The machine proof (−BA1 CA1 )/( BA2 CA2 ) A2 = S CB1C1 S BB1C1 · −BA1 CA1 C1 = −(−S CPB1 )·S ABC·S ACBP (−S BCP·S ABB1)·(−S ACBP) · BA1 CA1 simplify = S CPB1 ·S ABC S BCP·S ABB1 · BA1 CA1 B1 = S BCP·S ACP·S ABC·S ABCP S BCP·S ABP·S ABC·S ABCP · BA1 CA1 simplify = S ACP S ABP · BA1 CA1 A1 = (−S ABP)·S ACP S ABP·(−S ACP) simplify = 1 The eliminants BA2 CA2 A2 = S BB1C1 S CB1C1 S BB1C1 C1 = −S BCP·S ABB1 S ACBP S CB1C1 C1 = S CPB1 ·S ABC S ACBP S ABB1 B1 = S ABP·S ABC S ABCP S CPB1 B1 = S BCP·S ACP S ABCP BA1 CA1 A1 = S ABP S ACP Example 6.205 (0.033, 1, 6) With the usual notations for the triangle ABC, if EF meets BC in M, show that (BCDM) = −1. M K H F E D C B A Figure 6-205 Constructive description ( (points A B C) (foot D A B C) (foot E B A C) (foot F C A B) (inter H (l A D) (l C F)) (inter K (l D F) (l B E)) (inter M (l B C) (l E F)) (harmonic B C D M) ) The machine proof (−BD CD)/( BM CM) M = S CEF S BEF · −BD CD F = −PBAC·S BCE·PABA PABC·S ABE·PABA · BD CD simplify = −PBAC·S BCE PABC·S ABE · BD CD E = −PBAC·PACB·S ABC·PACA PABC·PBAC·S ABC·PACA · BD CD simplify = −PACB PABC · BD CD D = −PABC·PACB PABC·(−PACB) simplify = 1 The eliminants BM CM M = S BEF S CEF , S BEF F = PABC·S ABE PABA S CEF F = PBAC·S BCE PABA S ABE E = PBAC·S ABC PACA S BCE E = PACB·S ABC PACA , BD CD D = PABC −PACB Example 6.206 (0.467, 13, 34) The circle having for diameter the median AA1 of the triangle ABC meets the circumcircle in L: show that A(LDBC) = −1, where AD is the altitude. 360 Chapter 6. Topics from Geometry B C A 1 A D O M L K Figure 6-206 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (midpoint A1 B C) (midpoint M A1 A) (inter L (cir M A) (cir O A)) (inter K (l B C) (l A L)) (harmonic K D B C) ) Example 6.207 (0.883, 10, 38) If L, M, N are the traces of the lines AP, BP, CP on the sides BC, CA, AB of the triangle ABC, and L1, M1, N1 the traces, on the same sides, of the trilinear polar of P for ABC, show that the midpoints of the segments LL1, MM1, NN1 are collinear. B C A P L M N 1 L 1 M 1 N 2 L 2 M 2 N Figure 6-207 Constructive description ( (points A B C P) (inter L (l A P) (l B C)) (inter M (l B P) (l A C)) (inter N (l C P) (l A B)) (inter L1 (l B C) (l M N)) (inter M1 (l A C) (l L N)) (inter N1 (l A B) (l L M)) (midpoint L2 L L1) (midpoint M2 M M1) (midpoint N2 N N1) (collinear L2 M2 N2) ) Example 6.208 (0.550, 2, 25) Let L, M, N be the feet of the cevians AP, BP, CP of the triangle ABC and let P1 be a point on the trilinear polar of P for ABC. If the lines AP1 BP1, CP1 meet MN, NL, LM in X, Y, Z, show that the triangle XYZ is circumscribed about the triangle ABC. Z Y 1 P 1 N 1 L N M L P A C B Figure 6-208 Constructive description ( (points A B C P) (inter L (l A P) (l B C)) (inter M (l B P) (l A C)) (inter N (l C P) (l A B)) (inter L1 (l B C) (l M N)) (inter N1 (l A B) (l L M)) (on P1 (l N1 L1)) (inter Y (l N L) (l B P1)) (inter Z (l L M) (l C P1)) (inter ZZ (l Y A) (l N1 L)) ( N1Z LZ = N1ZZ LZZ ) ) Example 6.209 (0.466, 4, 26) If A1 is the point of intersection of the side BC of the triangle ABC with the trilinear polar p of a point P on the circumcircle of ABC, show that the circle APA1 passes through the midpoint of BC. 6.3. Triangles 361 B C A O P M N 1 A G Figure 6-209 Constructive description ( (circle A B C P) (circumcenter O A B C) (inter M (l B P) (l A C)) (inter N (l C P) (l A B)) (inter A1 (l B C) (l M N)) (midpoint G B C) (cocircle A P A1 G) ) Example 6.210 (0.050, 1, 8) Let ABC be a triangle with ∠B = 2∠C, D the foot of the altitude on CB and M the midpoint of B and C. Show that AB = 2DM. M D A B C Figure 6-210 Constructive description ( (points B C) (midpoint M B C) (tratio X M B H) (foot Y C B X) (lratio Z Y C −1) (inter A (l B Z) (l C X)) (inter D (l B C) (p A M X)) ( BA BZ = 2 MD CB ) ) The eliminants MD BC D =−S MXA S BCX S MXA A =S CMX·S BXZ −S BCZX BA BZ A = S BCX S BCZX S BXZ Z =S BCX S CMX X = −1 4(PBMC·H) S BCX X = −1 4(PCBM·H) PBMC M = −1 4(PBCB) PCBM M = 1 2(PBCB) The machine proof: BA BZ (−2)· MD BC D = S BCX (−2)·(−S MXA) · BA BZ A = (S BCX)2·(−S BCZX) (2)·S CMX·S BXZ·S BCZX simplify = −(S BCX)2 (2)·S CMX·S BXZ Z = −(S BCX)2 (2)·S CMX·S BCX simplify = −S BCX (2)·S CMX X = −(−1 4 PCBM·H) (2)·(−1 4 PBMC·H) simplify = −PCBM (2)·PBMC M = −( 1 2 PBCB) (2)·(−1 4 PBCB) simplify = 1. Example 6.211 (1.117, 33, 38) The three adjoint circles of the direct group have a point in common. B C A 1 A 1 B 1 C N Figure 6-211 Constructive description ( (points A B C) (inter A1 (t C C A) (b B C)) (inter B1 (t A A B) (b A C)) (inter C1 (t B B C) (b A B)) (inter N (cir C1 B) (cir A1 B)) (perp-biesct B1 C N) ) 362 Chapter 6. Topics from Geometry Example 6.212 (2.383, 68, 47) The three adjoint circles of the indirect group have a point in common. B C A 1 A 1 B 1 C M Figure 6-212 Constructive description ( (points A B C) (inter A1 (t B B A) (b C B)) (inter B1 (t C C B) (b A C)) (inter C1 (t A A C) (b B A)) (inter M (cir C1 B) (cir A1 B)) (perp-biesct B1 C M) ) Deﬁnition. The points N, M in Examples 6.211 and 6.212 are called the Brocard points of the triangle. Example 6.213 (1.000, 12, 50) The Brocard points are a pair of isogonal points of the trian-gle. N M 2 B 2 A 1 C 1 A A C B Figure 6-213 Constructive description ( (points A B C) (inter A1 (t C C A) (b B C)) (inter C1 (t B B C) (b A B)) (inter A2 (t B B A) (b C B)) (inter B2 (t C C B) (b A C)) (inter M (cir C1 B) (cir A1 B)) (inter N (cir B2 C) (cir A2 C)) (eqangle A B M N B C) ) Example 6.214 (1.600, 27, 52) The two Brocard points of a triangle are equidistant from the circumcenter of the triangle. O N M 2 B 2 A 1 C 1 A A C B Figure 6-214 Constructive description ( (points A B C) (circumcenter O A B C) (inter A1 (t C C A) (b B C)) (inter C1 (t B B C) (b A B)) (inter A2 (t B B A) (b C B)) (inter B2 (t C C B) (b A C)) (inter M (cir C1 B) (cir A1 B)) (inter N (cir B2 C) (cir A2 C)) (perp-biesct O N M) ) Example 6.215 (0.683, 12, 34) For the Brocard point in Example 6.211 we have ∠NAB = ∠NBC = ∠NCA. Similar for the Brocard point in Example 6.212. 6.3. Triangles 363 N 1 C 1 A A C B Figure 6-215 Constructive description ( (points A B C) (inter A1 (t C C A) (b B C)) (inter C1 (t B B C) (b A B)) (inter N (cir C1 B) (cir A1 B)) (eqangle N A B N B C) ) Example 6.216 (0.016, 1, 6) Let D and E be two points on two sides AC and BC of triangle ABC such that AD = BE, F = DE ∩AB. Show that FD · AC = EF · BC. A B C D E F Figure 6-216 Constructive description ( (points A B C) (on D (l A C)) (inter E (l C B) (cir B AD 2)) (inter F (l E D) (l A B)) (eq-product F D A C E F B C) ) The machine proof PDFD·PACA PEFE·PBCB F = PDED·S 2 ABD·PACA·S 2 ADBE PDED·S 2 ABE·PBCB·S 2 ADBE simplify = (S ABD)2·PACA (S ABE)2·PBCB E = (S ABD)2·PACA·PBCB PADA·S 2 ABC·PBCB simplify = (S ABD)2·PACA PADA·(S ABC)2 D = (S ABC· AD AC )2·PACA PACA· AD AC 2·(S ABC)2 simplify = 1 The eliminants PEFE F = PDED·(S ABE)2 (S ADBE)2 PDFD F = PDED·(S ABD)2 (S ADBE)2 S ABE E = −( BE BC ·S ABC) BE BC 2 = PADA PBCB PADA D =PACA·( AD AC )2 S ABD D =S ABC· AD AC 364 Chapter 6. Topics from Geometry 6.4 Quadrilaterals 6.4.1 General Quadrilaterals Example 6.217 (0.050, 2, 4) The ﬁgure formed when the midpoints of the sides of a quadri-lateral are joined in order is a parallelogram. A B C D E F G H Figure 6-217 Constructive description ((points A B C D) (midpoint E A B) (midpoint F B C) (midpoint G C D) (midpoint H D A) ( HE GF = 1) ) The machine proof EH FG H = −S ADE S AFDG G = −S ADE −S ADF−1 2 S ACD F = (2)·S ADE −S ABD E = (−2)·(−1 2 S ABD) S ABD simplify = 1 The eliminants EH FG H = −S ADE S AFDG S AFDG G = −1 2(2S ADF+S ACD) S ADF F = −1 2(S ACD+S ABD) S ADE E = −1 2(S ABD) Example 6.218 (0.083, 5, 13) The area of the parallelogram whose vertices are the midpoints of the sides of a quadrilateral is equal to half the area of the given quadrilateral (Figure 6-217). Constructive descrip-tion ( (points A B C D) (midpoint E A B) (midpoint F B C) (midpoint G C D) (midpoint H D A) (2S EFGH = S ABCD) ) The machine proof (2)·S EFGH S ABCD H = (2)·(S EFG+ 1 2 S DEG+ 1 2 S AEG) S ABCD G = S DEF+S CEF+ 1 2 S CDE−1 2 S ADE−1 2 S ACE S ABCD F = 2S CDE+S BDE+S BCE−S ADE−S ACE (2)·S ABCD E = S BCD+S ACD+S ABD+S ABC (2)·S ABCD area−co = 2S ACD+2S ABC (2)·(S ACD+S ABC) simplify = 1 The eliminants S EFGH H = 1 2 (2S EFG+S DEG+S AEG) S AEG G =−1 2 (S ADE+S ACE) S DEG G = 1 2(S CDE) S EFG G = 1 2 (S DEF+S CEF) S CEF F = 1 2 (S BCE) S DEF F = 1 2 (S CDE+S BDE) S ACE E =−1 2 (S ABC) S ADE E =−1 2(S ABD) S BCE E = 1 2 (S ABC) S BDE E = 1 2 (S ABD) S CDE E = 1 2 (S BCD+S ACD) S ABCD=S ACD+S ABC S BCD=S ACD−S ABD+S ABC Example 6.219 (0.067, 3, 13) The lines joining the midpoints of the two pairs of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals are concurrent and are bisected by their common point. 6.4. Quadrilaterals 365 B C D A E Q G S M N O Figure 6-219 Constructive description ( (points A B C D) (midpoint E A B) (midpoint G C D) (midpoint N A C) (midpoint M B D) (inter O (l E G) (l N M)) (midpoint O N M) ) The machine proof −NO MO O = −S EGN S EGM M = −S EGN 1 2 S DEG+ 1 2 S BEG N = (−2)·( 1 2 S CEG+ 1 2 S AEG) S DEG+S BEG G = −(−1 2 S CDE−1 2 S ADE−1 2 S ACE) 1 2 S CDE−1 2 S BDE−1 2S BCE E = 1 2 S BCD+ 1 2 S ACD−1 2 S ABD−1 2 S ABC 1 2 S BCD+ 1 2 S ACD−1 2 S ABD−1 2 S ABC simplify = 1 The eliminants NO MO O = S EGN S EGM S EGM M = 1 2 (S DEG+S BEG) S EGN N = 1 2 (S CEG+S AEG) S BEG G =−1 2 (S BDE+S BCE) S DEG G = 1 2 (S CDE) S AEG G =−1 2 (S ADE+S ACE) S CEG G =−1 2 (S CDE) S BCE E = 1 2 (S ABC) S BDE E = 1 2 (S ABD) S ACE E =−1 2 (S ABC) S ADE E =−1 2(S ABD) S CDE E = 1 2 (S BCD+S ACD) Deﬁnition. The intersection of the lines joining the midpoints of two pairs of opposite sides of a quadrilateral is called the centroid of the quadrilateral. Example 6.220 (0.050, 2, 6) The four lines obtained by joining each vertex of a quadrilateral to the centroid of the triangle determined by the remaining three vertices are concurrent at the centroid of the given quadrilateral. A B C D A N M J Figure 6-220 1 Constructive description ( (points A B C D) (centroid A1 B C D) (midpoint N A C) (midpoint M B D) (inter J (l A A1) (l N M)) (midpoint J N M) ) The machine proof −NJ MJ J = −S AA1N S AA1M M = −S AA1N −1 2 S ADA1 −1 2 S ABA1 N = (2)·(−1 2 S ACA1 ) S ADA1 +S ABA1 A1 = −(S ACD−S ABC)·(3)2 (−3S ACD+3S ABC)·(3) simplify = 1 The eliminants NJ MJ J = S AA1N S AA1M S AA1M M = −1 2(S ADA1 +S ABA1) S AA1N N = −1 2(S ACA1) S ABA1 A1 = 1 3(S ABD+S ABC) S ADA1 A1 = −1 3(S ACD+S ABD) S ACA1 A1 = 1 3(S ACD−S ABC) Example 6.221 (0.083, 4, 3) The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals increased by four times the square of the segment joining the midpoints of the diagonals. 366 Chapter 6. Topics from Geometry Constructive description ( (points A B C D) (midpoint E A C) (midpoint F B D) (AB 2+CB 2+CD 2+DA 2 = AC 2+BD 2+4EF 2) ) The machine proof PCDC+PBCB+PADA+PABA 4PEFE+PBDB+PACA F = PCDC+PBCB+PADA+PABA 2PDED+2PBEB+PACA E = PCDC+PBCB+PADA+PABA PCDC+PBCB+PADA+PABA simplify = 1 The eliminants PEFE F = 1 4 (2PDED+2PBEB−PBDB) PBEB E = 1 4 (2PBCB−PACA+2PABA) PDED E = 1 4 (2PCDC+2PADA−PACA) Example 6.222 (0.033, 7, 6) The sum of the squares of the diagonals of a quadrilateral is equal to twice the sum of the squares of the two lines joining the midpoints of the two pairs of opposite sides of the quadrilateral. B C D A P Q S R Figure 6-222 Constructive description ( (points A B C D) (midpoint P A B) (midpoint Q B C) (midpoint S D A) (midpoint R C D) (AC 2+BD 2 = 2QS 2+2PR 2) ) The machine proof ( 1 2 )·(PBDB+PACA) PQS Q+PPRP R = ( 1 2)·(PBDB+PACA) PQS Q+ 1 2 PDPD+ 1 2 PCPC−1 4 PCDC S = (2)·(PBDB+PACA) 2PDQD+2PDPD+2PCPC−PCDC+2PAQA−PADA Q = (2)·(PBDB+PACA) 2PDPD+2PCPC+PBDB−PBCB−PADA+PACA+PABA P = (2)·(PBDB+PACA) 2PBDB+2PACA simplify = 1 The eliminants PPRP R = 1 4 (2PDPD+2PCPC−PCDC) PQS Q S = 1 4 (2PDQD+2PAQA−PADA) PAQA Q =−1 4 (PBCB−2PACA−2PABA) PDQD Q = 1 4 (2PCDC+2PBDB−PBCB) PCPC P = 1 4 (2PBCB+2PACA−PABA) PDPD P = 1 4(2PBDB+2PADA−PABA) Example 6.223 (0.050, 4, 10) If a quadrilateral ABCD has its opposite sides AD and BC (extended) meeting at W, while X and Y are the midpoints of the diagonals AC and BD, then 4S WXY = S ABCD. B C D A X Y W Figure 6-223 Constructive description ( (points B C D A) (midpoint X C A) (midpoint Y B D) (inter W (l B C) (l D A)) (S BCDA = 4S XYW) ) 6.4. Quadrilaterals 367 The machine proof S BCDA (4)·S XYW W = S BCDA·S BDCA (4)·(S CXY ·S BDA−S CDA·S BXY) Y = S BCDA·S BDCA (4)·(−1 2 S CDX·S BDA+ 1 2S CDA·S BDX+ 1 2 S BDA·S BCX) X = −S BCDA·S BDCA (2)·( 1 2 S CDA·S BCD−1 2 S BDA·S BCA) area−co = −(S BDA+S BCD)·(S BCA−S BCD) −S BDA·S BCA+S BDA·S BCD−S BCA·S BCD+S 2 BCD simplify = 1 The eliminants S XYW W = S CXY ·S BDA−S CDA·S BXY S BDCA S BXY Y = −1 2(S BDX) S CXY Y = −1 2(S CDX−S BCX) S BCX X =1 2(S BCA) S BDX X =1 2(S BDA−S BCD) S CDX X =1 2(S CDA) S CDA=S BDA−S BCA+S BCD S BDCA=S BCA−S BCD S BCDA=S BDA+S BCD Example 6.224 (0.216, 6, 12) Let P be a point on the line joining the midpoints of the diago-nals of the quadrilateral ABCD. Show that S PAB + S PCD = S PDA + S PBC. A B C D N M P Figure 6-224 Constructive description ( (points A B C D) (midpoint N A C) (midpoint M B D) (lratio P N M r) (S PAB+S PCD = S PDA+S PBC) ) The machine proof S CDP+S ABP S BCP−S ADP P = S CDM·r−S CDN·r+S CDN+S ABM·r−S ABN·r+S ABN S BCM·r−S BCN·r+S BCN−S ADM·r+S ADN·r−S ADN M = −S CDN·r+S CDN+ 1 2 S BCD·r−S ABN·r+S ABN+ 1 2S ABD·r −S BCN·r+S BCN+ 1 2 S BCD·r+S ADN·r−S ADN+ 1 2S ABD·r N = −S BCD·r+S ACD·r−S ACD−S ABD·r+S ABC·r−S ABC −S BCD·r+S ACD·r−S ACD−S ABD·r+S ABC·r−S ABC simplify = 1 The eliminants S ADP P =S ADM·r−S ADN·r+S ADN S BCP P =S BCM·r−S BCN·r+S BCN S ABP P =S ABM·r−S ABN·r+S ABN S CDP P =S CDM·r−S CDN·r+S CDN S ADM M = −1 2(S ABD), S BCM M = 1 2(S BCD) S ABM M = 1 2(S ABD), S CDM M = 1 2(S BCD) S ADN N = −1 2(S ACD), S BCN N =1 2(S ABC) S ABN N =1 2(S ABC), S CDN N =1 2(S ACD) Example 6.225 (0.083, 2, 10) Let ABCD be a quadrilateral, F and E be the midpoints of AD and BC respectively. M = BA ∩EF, N = CD ∩EF. Show that AM BM = DN CN 368 Chapter 6. Topics from Geometry B C A D E F M N Figure 6-225 Constructive description ( (points A B C D) (midpoint E B C) (midpoint F A D) (inter M (l A B) (l E F)) (inter N (l C D) (l E F)) ( AM BM = DN CN ) ) The machine proof ( AM BM)/( DN CN ) N = S CEF S DEF · AM BM M = S AEF·S CEF S DEF·S BEF F = (−1 2 S ADE)·(−1 2 S CDE+ 1 2 S ACE) ( 1 2 S ADE)·(−1 2 S BDE+ 1 2 S ABE) simplify = −(S CDE−S ACE) S BDE−S ABE E = −( 1 2 S BCD+ 1 2 S ABC) −1 2 S BCD−1 2 S ABC simplify = 1 The eliminants DN CN N = S DEF S CEF AM BM M = S AEF S BEF S BEF F = −1 2(S BDE−S ABE) S DEF F =1 2(S ADE) S CEF F = −1 2(S CDE−S ACE) S AEF F = −1 2(S ADE) S ABE E =1 2(S ABC) S BDE E = −1 2(S BCD) S ACE E = −1 2(S ABC) S CDE E =1 2(S BCD) Example 6.226 (1.333, 20, 37) Let ABCD be a quadrilateral with AB = CD, F and E be the midpoints of AD and BC respectively. M = BA ∩EF, N = CD ∩EF. Show that ∠BME = ∠ENC (Figure 6-225). Constructive description ( (points A B C X) (inter D (l C X) (cir C AB 2)) (midpoint F A D) (midpoint E B C) (inter M (l A B) (l E F)) (inter N (l C D) (l E F)) (eqangle B M E E N C) ) Example 6.227 (0.300, 3, 18) Let ABCD be a quadrilateral such that AB = CD. F and E are the midpoints of AD and BC respectively. M = BA ∩EF, N = CD ∩EF. Show that AM = DN. 6.4. Quadrilaterals 369 B C A D E F M N Figure 6-227 Constructive description ( (points A B C X) (inter D (l C X) (cir C AB 2)) (midpoint F A D) (midpoint E B C) (inter M (l A B) (l E F)) (inter N (l C D) (l E F)) (eqdistance A M D N) ) The eliminants PDND N = PCDC·S 2 DFE S 2 CFDE , PAMA M = PABA·S 2 AFE S 2 AFBE S AFBE E = −1 2(2S ABF−S ABC) S DFE E =1 2(S CDF+S BDF) S CFDE E = −1 2(2S CDF−S BCD) S AFE E = −1 2(S ACF+S ABF) S BDF F =1 2(S ABD), S CDF F =1 2(S ACD) S ABF F = 1 2(S ABD), S ACF F =1 2(S ACD) S ABD D = −( XD CX ·S ABC−CD CX ·S ABX) PCDC D = CD CX 2·PCXC S ACD D = −( CD CX ·S ACX), S BCD D = −( CD CX ·S BCX) CD CX 2 = PABA PCXC , CD CX 2 = PABA PCXC S ABC=S ABC, S BCX=S ACX−S ABX+S ABC The machine proof PAMA PDND N = PAMA·S 2 CFDE PCDC·S 2 DFE M = PABA·S 2 AFE·S 2 CFDE PCDC·S 2 DFE·S 2 AFBE E = PABA·(−1 2 S ACF−1 2 S ABF)2·(−S CDF+ 1 2 S BCD)2 PCDC·( 1 2 S CDF+ 1 2 S BDF)2·(−S ABF+ 1 2 S ABC)2 F = PABA·( 1 2 S ACD+ 1 2S ABD)2·(−S BCD+S ACD)2 PCDC·( 1 2 S ACD+ 1 2 S ABD)2·(S ABD−S ABC)2 simplify = PABA·(S BCD−S ACD)2 PCDC·(S ABD−S ABC)2 D = (PABA·S 2 BCX−2PABA·S BCX·S ACX+PABA·S 2 ACX)·PABA·P2 CXC (PCXC·P2 ABA·S 2 ABX−2PCXC·P2 ABA·S ABX·S ABC+PCXC·P2 ABA·S 2 ABC)·PCXC simplify = (S BCX−S ACX)2 (S ABX−S ABC)2 area−co = (−S ABX+S ABC)2 (S ABX−S ABC)2 simplify = 1 Example 6.228 (0.416, 12, 36) Let ABCD be a quadrilateral such that AB = CD. P and Q are the midpoints of AD and BC; N and M are the midpoints of AC and BD. Show that PQ⊥NM. A B C D P Q N M Figure 6-228 Constructive description ( (points A B C X) (inter D (l C X) (cir C AB 2)) (midpoint P A D) (midpoint Q B C) (midpoint N A C) (midpoint M B D) (perpendicular N M P Q) ) 370 Chapter 6. Topics from Geometry Example 6.229 (0.866, 10, 20) The sides BA, CD of the quadrilateral ABCD meet in O, and the sides DA, CB meet O1. Along OA, OC, O1A, O1C are measured oﬀ, respectively, OE, OF, O1E1, O1F1 equal to AB, DC, AD, BC. Prove that EF is parallel to E1F1. A B C D O 1 O E F 1 E 1 F Figure 6-229 Constructive description ( (points A B C D) (inter O (l A B) (l C D)) (inter O1 (l A D) (l B C)) (lratio E O B −AB BO) (lratio F O C −DC CO ) (lratio E1 O1 A −AD AO1 ) (lratio F1 O1 B −BC BO1 ) (S EFE1 = S EFF1) ) Example 6.230 (0.617, 13, 70) ABCD is a quadrilateral, P, Q, R, S the midpoints of its sides taken in order, U, V, the midpoints of the diagonals, O any point; OP, OQ, OR, OS , OU, OV are divided in the same ratio in P1, Q1, R1, S 1, U1, V1. Prove that P1R1, Q1S 1, U1V1, are concurrent. B C D A O P Q R S V U 1 P 1 Q 1 R 1 S 1 V 1 U J Figure 6-230 Constructive description ( (points A B C D O) (midpoint P A B) (midpoint Q B C) (midpoint R C D) (midpoint S D A) (midpoint U A C) (midpoint V D B) (lratio P1 O P YT) (lratio Q1 O Q YT) (lratio R1 O R YT ) (lratio S 1 O S YT) (lratio U1 O U YT ) (lratio V1 O V YT ) (inter J (l P1 R1) (l Q1 S 1)) (inter Z1 (l P1 R1) (l U1 V1)) (inter Z2 (l U1 V1) (l Q1 S 1)) ( U1Z1 V1Z1 = U1Z2 V1Z2 ) ) Example 6.231 (Theorem of Pratt-Wu) (5.133, 72, 123) Given a quadrilateral ABDC, let HE, EF, FG, GH be the tangents of circles CAB, ABD, BDC, DCA at A, B, D and C, respec-tively. Then HA · EB · FD · GC = AE · BF · DG · CH. A 1 D D B C 1 A 1 C E 1 B H F G Figure 6-231 Constructive description ( (points A B C D) (midpoint L A B) (midpoint M B D) (midpoint N C D) (midpoint P C A) (inter X (t L L A) (t P P A)) (inter Y (t L L B) (t M M B)) (inter Z (t M M D) (t N N D)) 6.4. Quadrilaterals 371 (inter W (t N N C) (t P P C)) (tratio O A X I) (inter H (t C C W) (l O A)) (tratio U D Z J) (inter F (t B B Y) (l U D)) (inter E (l B F) (l A O)) (inter G (l D U) (l C H)) ( HA AE · BE FB = DG FD · HC CG ) ) Example 6.232 (0.00, 1, 3) Let P, Q be the midpoints of the diagonals of a trapezoid ABCD. Then PQ is parallel to the two parallel sides of ABCD. C B A D P Q Figure 6-232 Constructive description ((points A B C) (on D (p C A B)) (midpoint Q A C) (midpoint P B D) (parallel P Q A B) ) The machine proof S ABP S ABQ P = 1 2 S ABD S ABQ Q = S ABD (2)·( 1 2 S ABC) D = S ABC S ABC simplify = 1 The eliminants S ABP P =1 2(S ABD) S ABQ Q =1 2(S ABC) S ABD D =S ABC Example 6.233 (0.050, 2, 3) Let P, Q be the midpoints of the diagonals of a trapezoid ABCD. Then PQ is half of the diﬀerence of the two parallel sides of ABCD. A B C D P Q Figure 6-233 Constructive description ( (points A B C) (on D (p A B C)) (midpoint P B D) (midpoint Q A C) ( BC PQ −AD PQ = 2) ) The machine proof 1 2( BC PQ −AD PQ) Q = S ACP·S ACD−S ACP·S ABC (2)·S 2 ACP simplify = S ACD−S ABC (2)·S ACP P = S ACD−S ABC (2)·( 1 2 S ACD−1 2S ABC) simplify = 1 The eliminants AD PQ Q =−S ACD S ACP BC PQ Q =−S ABC S ACP S ACP P =1 2(S ACD−S ABC) Example 6.234 (0.050, 3, 7) Let ABCD be a trapezoid and O be the intersection of its diag-onals AC and BD. The line passing through O and parallel to AB meet AD and BC at F and E. Show that O is the midpoint of EF. Constructive description 372 Chapter 6. Topics from Geometry ((points A B C) (on D (p C A B)) (inter O (l A C) (l B D)) (inter E (l B C) (p O A B)) (inter F (l A D) (p O A B)) (midpoint O E F) ) The machine proof −OE OF F = S AODE S ADO E = S BCO·S ABD+S ADO·S ABC+S ACD·S ABO S ADO·(−S ABC) O = −S 2 ABCD·S BCD·S ABD·S ABC·S ABCD (−S ACD·S ABD)·S ABC·(S ABCD)3 simplify = S BCD S ACD D = −S ABC· CD AB −S ABC· CD AB simplify = 1 The eliminants OE OF F = −S AODE S ADO S AODE E =S BCO·S ABD+S ADO·S ABC+S ACD·S ABO −S ABC S ABO O =S ABD·S ABC S ABCD S ADO O =−S ACD·S ABD S ABCD S BCO O =S BCD·S ABC S ABCD S ACD D = −(S ABC· CD AB) S BCD D = −(S ABC· CD AB) Other properties of the general quadrilateral can be found in Examples 2.62, 6.2, 6.3, and 6.17. 6.4.2 Complete Quadrilaterals Deﬁnition. By a complete quadrilateral, we mean the ﬁgure consisting of four points (any three of them are not collinear) and the six lines joining any two of them. As in Figure 6-235, A, B, C, D are the vertices of the complete quadrilateral. AB and CD, AC and BD, AD and BC are called opposite sides of the complete quadrilateral respectively. P Q B A C D R S Figure 6-235 Example 6.235 (0.083, 1, 10) The line joining the intersections of two pairs of opposite sides of a complete quadrilateral is divided harmonically by the remaining pair of opposite sides of the complete quadrilateral. In Figure 6-235, this means (PQRS ) = −1. For a machine proof of this example, see Example 6.17 on page 273. Example 6.235 can be used to deﬁne the concept of harmonic sequence: four collinear points P, Q, R, S are said to be a harmonic sequence if there exists a complete quadrilateral 6.4. Quadrilaterals 373 ABCD such that P = BC ∩AD, Q = AB ∩CD, R = BD ∩PQ, and S = AC ∩PQ. By the following example, the above deﬁnition is independent of the choices of ABCD. Example 6.236 (0.466, 1, 14) Let ABCD and EFGH be two complete quadrilaterals such that P = AB ∩CD = EF ∩HG, Q = AD ∩BC = EH ∩FG, and S = AC ∩PQ = EG ∩PQ are collinear. Then BD, FH, PQ are concurrent. A B C D E P Q S G F H R Figure 6-236 Constructive description ( (points A B C D E) (inter P (l A B) (l C D)) (inter Q (l B C) (l A D)) (inter S (l A C) (l P Q)) (on G (l E S )) (inter F (l P E) (l Q G)) (inter H (l E Q) (l P G)) (inter R (l B D) (l P Q)) (inter O (l F H) (l P Q)) ( PR QR = PO QO ) ) The machine proof ( PR QR)/( PO QO) O = S QFH S PFH · PR QR R = S BDP·S QFH S PFH·S BDQ H = S BDP·S PQG·S EQF·(−S EPQG) S PGF·S EPQ·S BDQ·S EPQG simplify = −S BDP·S PQG·S EQF S PGF·S EPQ·S BDQ F = −S BDP·S PQG·(−S EQG·S EPQ)·S EQPG (−S PQG·S EPG)·S EPQ·S BDQ·S EQPG simplify = −S BDP·S EQG S EPG·S BDQ G = −S BDP·S EQS · EG ES S EPS · EG ES ·S BDQ simplify = −S BDP·S EQS S EPS ·S BDQ S = −S BDP·S EPQ·S ACQ·(−S APCQ) S EPQ·S ACP·S BDQ·(−S APCQ) simplify = −S BDP·S ACQ S ACP·S BDQ Q = −S BDP·(−S ACD·S ABC)·(−S ABDC) S ACP·S BCD·S ABD·(−S ABDC) simplify = S BDP·S ACD·S ABC S ACP·S BCD·S ABD P = (−S BCD·S ABD)·S ACD·S ABC·S ACBD (−S ACD·S ABC)·S BCD·S ABD·S ACBD simplify = 1 The eliminants PO QO O = S PFH S QFH PR QR R = S BDP S BDQ S PFH H =S PGF·S EPQ −S EPQG S QFH H =S PQG·S EQF S EPQG S PGF F = −S PQG·S EPG S EQPG S EQF F =−S EQG·S EPQ S EQPG S EPG G =S EPS · EG ES S EQG G =S EQS · EG ES S EPS S =S EPQ·S ACP −S APCQ S EQS S =S EPQ·S ACQ −S APCQ S BDQ Q = S BCD·S ABD −S ABDC S ACQ Q = S ACD·S ABC S ABDC S ACP P =−S ACD·S ABC S ACBD S BDP P =−S BCD·S ABD S ACBD Example 6.237 (0.216, 2, 24) If the intersections of ﬁve corresponding sides of two complete quadrilaterals are on the same line l. Then the remaining sides also meet in l. A B C D P U V I J K S Q R L Figure 6-237 Constructive description ( (points A B C D P) (lratio U A D r1) (lratio V B C r2) 374 Chapter 6. Topics from Geometry (inter I (l D B) (l U V)) (inter J (l C D) (l U V)) (inter K (l A B) (l U V)) (lratio S U P r3) (inter Q (l I S ) (l K P)) (inter R (l V Q) (l S J)) (inter L (l P R) (l U V)) (inter N (l A C) (l U V)) ( UL VL = UN VN ) ) The machine proof ( UL VL)/( UN VN ) N = S ACV S ACU · UL VL L = (−S PUR)·S ACV S ACU ·(−S PVR) R = S VS Q·S PUJ·S ACV·S VJQS S ACU ·S VJS ·S PVQ·(−S VJQS ) simplify = S VS Q·S PUJ·S ACV −S ACU ·S VJS ·S PVQ Q = S VIS ·S PKS ·S PUJ·S ACV·S PIKS −S ACU ·S VJS ·S PIS ·S PVK·(−S PIKS ) simplify = S VIS ·S PKS ·S PUJ·S ACV S ACU ·S VJS ·S PIS ·S PVK S = S PVI·r3·(S PUK·r3−S PUK)·S PUJ·S ACV S ACU ·S PVJ·r3·(S PUI·r3−S PUI)·S PVK simplify = S PVI·S PUK·S PUJ·S ACV S ACU ·S PVJ·S PUI·S PVK K = S PVI·S PUV·S ABU·S PUJ·S ACV ·(−S AUBV) S ACU ·S PVJ·S PUI·S PUV·S ABV·(−S AUBV) simplify = S PVI·S ABU·S PUJ·S ACV S ACU ·S PVJ·S PUI·S ABV J = S PVI·S ABU·S PUV·S CDU·S ACV·(−S CUDV ) S ACU ·S PUV·S CDV·S PUI·S ABV·(−S CUDV ) simplify = S PVI·S ABU·S CDU·S ACV S ACU ·S CDV·S PUI·S ABV I = S PUV ·S BDV·S ABU·S CDU·S ACV·(−S BUDV) S ACU ·S CDV·S PUV ·S BDU·S ABV·(−S BUDV) simplify = S BDV·S ABU·S CDU·S ACV S ACU ·S CDV·S BDU·S ABV V = (−S BCD·r2)·S ABU·S CDU·(S ABC·r2−S ABC) S ACU ·(−S BCD·r2+S BCD)·S BDU·S ABC·r2 simplify = S ABU·S CDU S ACU ·S BDU U = S ABD·r1·(−S ACD·r1+S ACD) S ACD·r1·(−S ABD·r1+S ABD) simplify = 1 The eliminants UN VN N =S ACU S ACV UL VL L =S PUR S PVR S PVR R =S VJS ·S PVQ S VJQS S PUR R =S VS Q·S PUJ −S VJQS S PVQ Q = S PIS ·S PVK S PIKS S VS Q Q =S VIS ·S PKS −S PIKS S PIS S =(r3−1)·S PUI S VJS S =S PVJ·r3 S PKS S =(r3−1)·S PUK S VIS S =S PVI·r3 S PVK K = S PUV·S ABV −S AUBV S PUK K = S PUV·S ABU −S AUBV S PVJ J =S PUV ·S CDV −S CUDV S PUJ J =S PUV ·S CDU −S CUDV S PUI I =S PUV·S BDU −S BUDV S PVI I =S PUV·S BDV −S BUDV S ABV V =S ABC·r2 S CDV V = −((r2−1)·S BCD) S ACV V =(r2−1)·S ABC S BDV V = −(S BCD·r2) S BDU U = −((r1−1)·S ABD) S ACU U =S ACD·r1 S CDU U = −((r1−1)·S ACD) S ABU U =S ABD·r1 Deﬁnition. The line joining the midpoints of a pair of opposite sides of a complete quadri-6.4. Quadrilaterals 375 lateral bisects the segment between the intersections of the other two pairs of opposite sides of the quadrilateral. For a machine proof of this theorem see Example 2.36 on page 74. Let us call this line the Gauss line for the given complete quadrilateral. Example 6.238 (Gauss Point Theorem) (0.933, 26, 56) Given ﬁve points, we have ﬁve Gauss’ lines. These ﬁve Gauss’ lines are concurrent. Let us call this point of concurrency the Gauss point for the given ﬁve points. I 6 M 5 M 4 M 3 M 2 M 1 M Z Y X 2 A 4 A 3 A 0 A 1 A Figure 6-238 Constructive description ( (points A0 A1 A2 A3 A4) (inter X (l A3 A4) (l A1 A0)) (inter Y (l A2 A3) (l A0 A4)) (inter Z (l A4 A3) (l A1 A2)) (midpoint M1 A1 A3) (midpoint M2 X A2) (midpoint M3 A0 A2) (midpoint M4 Y A1) (midpoint M5 A1 A4) (midpoint M6 A0 Z) (inter I (l M3 M4) (l M1 M2)) (inter ZI (l M6 M5) (l M1 M2)) ( M1I M2I = M1ZI M2ZI ) ) Example 6.239 (0.033, 3, 6) The centroids of the four triangles determined by the vertices of a quadrilateral taken three at a time form a quadrilateral homothetic to the given quadri-lateral in the ratio 1/3. 376 Chapter 6. Topics from Geometry Constructive description ( (points A B C D) (midpoint Q B C) (midpoint P B A) (midpoint R C D) (inter A1 (l D Q) (l B R)) (inter D1 (l A Q) (l C P)) (inter J (l A A1) (l D D1)) ( AJ A1J = −3) ) The machine proof −1 3( AJ A1J) J = S ADD1 (−3)·(−S DA1D1) D1 = S ADQ·S ACP·S ACQP (3)·S CQP·S ADA1·S ACQP simplify = S ADQ·S ACP (3)·S CQP·S ADA1 A1 = S ADQ·S ACP·(−S BDRQ) (3)·S CQP·(−S BDR·S ADQ) simplify = S ACP·S BDRQ (3)·S CQP·S BDR R = S ACP·(−1 2 S CDQ+S BDQ) (3)·S CQP·(−1 2 S BCD) P = (−1 2 S ABC)·(S CDQ−2S BDQ) (3)·( 1 2 S ACQ)·S BCD Q = −S ABC·( 3 2 S BCD) (3)·(−1 2 S ABC)·S BCD simplify = 1 The eliminants AJ A1J J = S ADD1 −S DA1D1 S DA1D1 D1 = S CQP·S ADA1 S ACQP S ADD1 D1 = S ADQ·S ACP S ACQP S ADA1 A1 = S BDR·S ADQ S BDRQ S BDR R = −1 2(S BCD) S BDRQ R = −1 2(S CDQ−2S BDQ) S CQP P = 1 2(S ACQ) S ACP P = −1 2(S ABC) S ACQ Q = −1 2(S ABC) S BDQ Q = −1 2(S BCD) S CDQ Q =1 2(S BCD) Example 6.240 (5.866, 51, 69) The orthocenters of the four triangles formed by four lines taken three at a time are collinear. A B C D E F 1 G 1 H 2 G 2 H 3 G 3 H Figure 6-240 Constructive description ( (points A B C) (on D (l A B)) (on E (l A C)) (inter F (l E D) (l C B)) (foot G1 C A B) (foot FH1 A B C) (inter H1 (l C G1) (l A FH1 )) (foot G2 E A B) (foot FH2 A E D) (inter H2 (l E G2) (l A FH2)) (foot G3 F A B) (foot FH3 B E D) (inter H3 (l F G3) (l B FH3)) (inter H4 (l F G3) (l H1 H2)) ( FH3 G3H3 = FH4 G3H4 ) ) 6.4.3 Parallelograms The parallelogram has a special position in the area method. The reason is that when elim-inating points from length ratios, we need to add auxiliary parallelograms. Thus some properties (2.11, 2.12, 3.6, 3.7, 3.11) of parallelograms are often used in the proofs pro-duced by the area method. 6.4. Quadrilaterals 377 Example 6.241 (0.016, 1, 3) Let O be the intersection of the two diagonals AC and BD of a parallelogram ABCD. Show that O is the midpoint of AC. Constructive description ( (points A B C) (pratio D A B C 1) (inter O (l B D) (l A C)) (midpoint O A C) ) O D C B A Figure 6-241 The machine proof −AO CO O = −S ABD −S BCD D = S ABC S ABC simplify = 1 The eliminants AO CO O = S ABD −S BCD S BCD D =S ABC S ABD D =S ABC Example 6.242 (0.066, 3, 10) Let l be a line passing through the vertex of M of a paral-lelogram MNPQ and intersecting the lines NP, PQ, NQ in points R, S , T. Show that MT MR = S T S M (or 1/MR + 1/MS = 1/MT). M N P Q R S T Figure 6-242 Constructive description ( (points M N P) (pratio Q M N P 1) (lratio R N P r) (inter S (l M R) (l P Q)) (inter T (l M R) (l N Q)) ( MT MR = ST S M ) ) The machine proof MT MR −ST MS T = S MNQ·S MNS Q −S NQS ·S MNRQ S = S MNQ·(−S MPRQ·S MNQ+S NPQ·S MQR)·(−S MPRQ) −(−S NPQ·S MQR)·S MNRQ·(−S MPRQ) simplify = −S MNQ·(S MPRQ·S MNQ−S NPQ·S MQR) S NPQ·S MQR·S MNRQ R = −S MNQ·(S NPQ·S MPQ·r+S MPQ·S MNQ) S NPQ·(−S MPQ·r+S MNQ·r−S MNQ)·(S NPQ·r+S MNQ) simplify = S MNQ·S MPQ S NPQ·(S MPQ·r−S MNQ·r+S MNQ) Q = S 2 MNP S 2 MNP simplify = 1 The eliminants ST MS T = S NQS S MNS Q MT MR T = S MNQ S MNRQ S NQS S =S NPQ·S MQR S MPRQ S MNS Q S =S MPRQ·S MNQ−S NPQ·S MQR S MPRQ S MNRQ R =S NPQ·r+S MNQ S MQR R = −(S MPQ·r−S MNQ·r+S MNQ) S MPRQ R =S NPQ·r−S NPQ+S MPQ S NPQ Q =S MNP S MPQ Q =S MNP S MNQ Q =S MNP 378 Chapter 6. Topics from Geometry Example 6.243 (0.066, 2, 9) In the parallelogram ABCD, AE is drawn parallel to BD; show that A(ECBD) = −1. A B C D E F G Figure 6-243 Constructive description ( (points A B C) (pratio D A B C 1) (pratio E A B D 1) (inter F (l B E) (l A D)) (inter G (l E B) (l A C)) (harmonic E G F B) ) The machine proof ( EF FG)/( BE BG) G = (−S AECF)·(−S ABC) (−S ABCE)·S ACF F = −(−S ABDE·S ACE+S ACD·S ABE)·S ABC·S ABDE S ABCE·S ACD·S ABE·S ABDE simplify = (S ABDE·S ACE−S ACD·S ABE)·S ABC S ABCE·S ACD·S ABE E = (S ACD·S ABD+2S ABD·S ABC)·S ABC (S ACD+2S ABC)·S ACD·S ABD simplify = S ABC S ACD D = S ABC S ABC simplify = 1 The eliminants BE BG G = S ABCE S ABC EF FG G = −S AECF S ACF S ACF F =S ACD·S ABE S ABDE S AECF F =−(S ABDE·S ACE−S ACD·S ABE) S ABDE S ABCE E =S ACD+2S ABC S ABE E =S ABD S ACE E =S ACD+S ABC S ABDE E =(2S ABD) S ACD D =S ABC Example 6.244 (0.116, 2, 12) The diagonals of a parallelogram and those of its inscribed parallelogram are concurrent. 6.4. Quadrilaterals 379 Constructive description ( (points A B C E) (pratio D C B A 1) (inter F (l A E) (p D B E)) (inter G (l B E) (p C A E)) (inter H (l D F) (l C G)) (inter O (l A C) (l B D)) (inter Z2 (l B D) (l E H)) (inter Z1 (l A C) (l E H)) ( EZ1 HZ1 · HZ2 EZ2 = 1) ) The machine proof EZ1 HZ1 · HZ2 EZ2 Z1 = S ACE S ACH · HZ2 EZ2 Z2 = S ACE·S BDH S ACH·(−S BED) H = −S ACE·(−S CDG·S BDF)·S CDGF S CDF·S ACG·S BED·(−S CDGF ) simplify = −S ACE·S CDG·S BDF S CDF·S ACG·S BED G = −S ACE·S ACED·S BCE·S BDF·S ABE S CDF·(−S BCE·S ACE)·S BED·S ABE simplify = S ACED·S BDF S CDF·S BED F = S ACED·(−S BED·S AED)·(−S ABE) (−S BCED·S AED)·S BED·(−S ABE) simplify = S ACED S BCED D = S ABE S ABE simplify = 1 The eliminants EZ1 HZ1 Z1 = S ACE S ACH HZ2 EZ2 Z2 = S BDH −S BED S ACH H =S CDF·S ACG S CDGF S BDH H =S CDG·S BDF S CDGF S ACG G =−S BCE·S ACE S ABE S CDG G =S ACED·S BCE S ABE S CDF F = S BCED·S AED S ABE S BDF F =S BED·S AED S ABE S BCED D =S ABE S ACED D =S ABE Example 6.245 (0.066, 2, 8) Let ABCD be a parallelogram. Then the feet from A, B, C, D to the diagonals of the parallelogram form a parallelogram. A B C D F E H G Figure 6-245 Constructive descrip-tion ( (points A B C) (pratio D A B C 1) (foot F A B D) (foot E B A C) (foot H C B D) (foot G D A C) ( EH FG = 1) ) The machine proof EH FG G = −S ACH S ACF H = −(PCBD·S ACD−PBDC·S ABC) S ACF·PBDB F = −(PCBD·S ACD−PBDC·S ABC)·PBDB (−PADB·S ABC+PABD·S ACD)·PBDB simplify = PCBD·S ACD−PBDC·S ABC PADB·S ABC−PABD·S ACD D = PBCB·S ABC−PABA·S ABC PBCB·S ABC−PABA·S ABC simplify = 1 The eliminants EH FG G = −S ACH S ACF S ACH H = PCBD·S ACD−PBDC·S ABC PBDB S ACF F =−(PADB·S ABC−PABD·S ACD) PBDB PABD D =PABC+PABA PADB D =PBCB+PABC PBDC D =PABC+PABA S ACD D =S ABC PCBD D =PBCB+PABC 380 Chapter 6. Topics from Geometry Example 6.246 (0.083, 10, 8) If squares are erected externally (or internally) on the sides of any parallelogram, their centers form a square. L N T M A B C D O S Q P R Figure 6-246 Constructive description ((points A B C) (pratio D A B C 1) (constant i2 −1) (midpoint L A D) (pe-square S L D) (midpoint N B C) (pe-square Q N B) (midpoint T A B) (pe-square P T A) (− − → PS −i·− − → PQ = 0) ) The machine proof − → QP·i−− → S P n = −− → T ·i2+2− → T ·i−− → T −− → Q ·i+− → S +− → A ·i2−− → A ·i n = −(− → Q ·i−− → S + 1 2− → B ·i2−− → B ·i+ 1 2− → B −1 2− → A ·i2+ 1 2− → A ) n = (−1 2 )·(−2− → N ·i2+2− → N ·i−2− → S +3− → B ·i2−2− → B ·i+− → B −− → A ·i2+− → A ) n = ( 1 2 )·(2− → S +− → C ·i2−− → C ·i−2− → B ·i2+− → B ·i−− → B +− → A ·i2−− → A ) n = ( 1 2 )·(−2− → L ·i+2− → L +2− → D ·i+− → C ·i2−− → C ·i−2− → B ·i2+− → B ·i−− → B +− → A ·i2−− → A ) n = −− → L ·i+− → L +− → D ·i−1 2− → C ·i−1 2− → C + 1 2− → B ·i+ 1 2− → B −− → A n = (−1 2 )·(−− → D ·i−− → D +− → C ·i+− → C −− → B ·i−− → B +− → A ·i+− → A ) simplify = ( 1 2 )·(i+1)·(− → D −− → C +− → B −− → A ) simplify = 0 The eliminants − → S PP = −(− → T ·i−− → T +− → S −− → A ·i) − → QPP = −(− → T ·i−− → T +− → Q −− → A ·i) − → T T =1 2(− → B +− → A ) − → Q Q = −(− → N ·i−− → N −− → B ·i) − → N N =1 2(− → C +− → B ) − → S S = −(− → L ·i−− → L −− → D ·i) − → L L =1 2(− → D +− → A ) − → D D =− → C −− → B +− → A Example 6.247 (0.250, 7, 8) If squares are erected externally on two opposite sides and in-ternally on the other two sides of any parallelogram, their centers form a parallelogram. A B C D O S Q P R Figure 6-247 Constructive description: ((points A B C) (pratio D A B C 1) (midpoint L A D) (constant i2 −1) (pe-square S L D) (midpoint N B C) (pe-square Q N B) (midpoint T A B) (ne-square P T A) (midpoint M C D) (ne-square R M C) (− − → PS = − − → QR) ) 6.4. Quadrilaterals 381 The machine proof −− → S P − → QR R = −− → S P − → M ·i+− → M −− → Q −− → C ·i M = −− → S P −− → Q + 1 2− → D ·i+ 1 2− → D −1 2− → C ·i+ 1 2− → C P = (2)·(− → T ·i+− → T −− → S −− → A ·i) 2− → Q −− → D ·i−− → D +− → C ·i−− → C T = (2)·(−− → S + 1 2− → B ·i+ 1 2− → B −1 2− → A ·i+ 1 2− → A ) 2− → Q −− → D ·i−− → D +− → C ·i−− → C Q = −(2− → S −− → B ·i−− → B +− → A ·i−− → A ) −2− → N ·i+2− → N −− → D ·i−− → D +− → C ·i−− → C +2− → B ·i N = 2− → S −− → B ·i−− → B +− → A ·i−− → A − → D ·i+− → D −− → B ·i−− → B simplify = 2− → S −− → B ·i−− → B +− → A ·i−− → A (i+1)·(− → D −− → B ) S = −2− → L ·i+2− → L +2− → D ·i−− → B ·i−− → B +− → A ·i−− → A (i+1)·(− → D −− → B ) L = −(−− → D ·i−− → D +− → B ·i+− → B ) (i+1)·(− → D −− → B ) simplify = 1 The eliminants − → QRR =− → M ·i+− → M −− → Q −− → C ·i − → M M = 1 2(− → D +− → C ) − → S PP =− → T ·i+− → T −− → S −− → A ·i − → T T =1 2(− → B +− → A ) − → Q Q = −(− → N ·i−− → N −− → B ·i) − → N N =1 2(− → C +− → B ) − → S S = −(− → L ·i−− → L −− → D ·i) − → L L =1 2(− → D +− → A ) Example 6.248 (0.450, 5, 26) The diagonals of the square in Example 6.246 pass through the center of the parallelogram (Figure 6-246). Constructive description ((points A B C) (pratio D A B C 1) (midpoint L A D) (constant i2 −1) (pe-square S L D) (midpoint N B C) (pe-square Q N B) (midpoint O Q S ) (− → AO−− → OC = 0) ) The machine proof. − → CO+− → AO n = 1 2− → CQ+ 1 2− → CS + 1 2− → AQ+ 1 2− → AS n = ( 1 2 )·(−2− → N ·i+2− → N +− → CS −− → C +2− → B ·i+− → AS−− → A ) n = (−1 2 )·(−− → CS +− → C ·i−− → B ·i−− → B −− → AS+− → A ) n = ( 1 2 )·(−2− → L ·i+2− → L +2− → D ·i−− → C ·i−− → C +− → B ·i+− → B −2− → A ) n = (−1 2 )·(−− → D ·i−− → D +− → C ·i+− → C −− → B ·i−− → B +− → A ·i+− → A ) simplify = ( 1 2 )·(i+1)·(− → D −− → C +− → B −− → A ) simplify = 0 The eliminants − → AOO =1 2(− → AQ+− → AS) − → COO =1 2(− → CQ+− → CS) − → AQ Q = −(− → N ·i−− → N −− → B ·i+− → A ) − → CQ Q = −(− → N ·i−− → N +− → C −− → B ·i) − → N N =1 2(− → C +− → B ) − → AS S = −(− → L ·i−− → L −− → D ·i+− → A ) − → CS S = −(− → L ·i−− → L −− → D ·i+− → C ) − → L L =1 2(− → D +− → A ) − → D D =− → C −− → B +− → A Example 6.249 (0.250, 4, 15) If similar rectangles are erected externally on two opposite and internally on the other two sides of any parallelogram, their centers form a parallelogram. 382 Chapter 6. Topics from Geometry A B C D 1 A 1 D S N Q 2 A 1 B P M R Figure 6-249 Constructive description ((points A B C) (pratio D A B C 1) (tratio A1 A D r1) (pratio D1 A1 A D 1) (inter S (l A D1) (l D A1)) (midpoint N B C) (pratio Q N A1 A 1/2) (tratio A2 A B r2) (pratio B1 A2 A B 1) (inter P (l A B1) (l B A2)) (midpoint M C D) (pratio R M A2 A 1/2) ( PS QR = 1) ) The machine proof. −S P QR R = −S AS A2 P S AQA2M M = −S AS A2P S AQA2 −1 2 S ADA2 −1 2 S ACA2 P = (−2)·(−S ABB1A2·S AS A2 −S AA2 B1·S ABA2) (2S AQA2 −S ADA2 −S ACA2 )·(−S ABB1A2) B1 = (−2)·(2S AS A2 ·S ABA2−S 2 ABA2) (2S AQA2 −S ADA2 −S ACA2 )·(2S ABA2) simplify = −(2S AS A2 −S ABA2 ) 2S AQA2 −S ADA2 −S ACA2 A2 = −( 1 2 PBAS ·r2−1 4 PABA·r2) 1 2 PBAQ·r2−1 4 PBAD·r2−1 4 PBAC·r2 simplify = −(2PBAS −PABA) 2PBAQ−PBAD−PBAC Q = −(2PBAS −PABA) 2PBAN−PBAA1−PBAD−PBAC N = −(2PBAS −PABA) −PBAA1−PBAD+PABA S = −S ADD1A1·PABA+2PBAD1·S ADA1 (PBAA1+PBAD−PABA)·S ADD1A1 D1 = −(−2PBAA1 ·S ADA1−2PBAD·S ADA1 +2PABA·S ADA1) (PBAA1+PBAD−PABA)·(2S ADA1 ) simplify = 1 The eliminants S P QR R = S AS A2 P S AQA2 M S AQA2 M M = 1 2(2S AQA2 −S ADA2 −S ACA2) S AS A2P P = S ABB1A2·S AS A2 +S AA2B1·S ABA2 S ABB1A2 S AA2 B1 B1 = −(S ABA2) S ABB1A2 B1 =2(S ABA2) S ACA2 A2 = 1 4(PBAC·r2) S ADA2 A2 = 1 4(PBAD·r2) S AQA2 A2 = 1 4(PBAQ·r2) S ABA2 A2 = 1 4(PABA·r2) S AS A2 A2 = 1 4(PBAS ·r2) PBAQ Q =1 2(2PBAN−PBAA1) PBAN N =1 2(PBAC+PABA) PBAS S = PBAD1·S ADA1 S ADD1A1 PBAD1 D1 = PBAA1+PBAD S ADD1A1 D1 =2(S ADA1) Example 6.250 (0.083, 3, 8) Let A1, B1, C1, D1 be points on the sides CD, DA, AB, BC of a parallelogram ABCD such that CA1/CD = DB1/DA = AC1/AB = BD1/BC = 1/3. Show that the area of the quadrilateral formed by the lines AA1 BB1, CC1, DD1 is one thirteenth 6.4. Quadrilaterals 383 of the area of parallelogram ABCD. Note that we only need to show 3S ABCD = 13S ABA2. 2 D 2 C 2 B 2 A 1 D 1 C 1 B 1 A D C B A Figure 6-250 Constructive description ( (points A B C) (pratio D A B C 1) (lratio A1 C D 1/3) (lratio B1 D A 1/3) (inter A2 (l A A1) (l B B1)) (3S ABCD = 13S ABA2) ) The machine proof (3)·S ABCD (13)·S ABA2 A2 = (3)·S ABCD·S ABA1B1 (13)·S ABB1·S ABA1 B1 = (3)·S ABCD·(−2 3 S ADA1 +S ABA1) (13)·( 2 3 S ABD)·S ABA1 A1 = (−3)·S ABCD·(−4 3 S ACD−S ABD−2S ABC) (26)·S ABD·( 1 3 S ABD+ 2 3 S ABC) D = (3)·(2S ABC)·(13S ABC) (26)·S ABC·(3S ABC) simplify = 1 The eliminants S ABA2 A2 = S ABB1·S ABA1 S ABA1B1 S ABB1 B1 = 2 3(S ABD) S ABA1B1 B1 = (2S ADA1 −3S ABA1 ) −3 S ABA1 A1 = 1 3(S ABD+2S ABC) S ADA1 A1 = −2 3(S ACD) S ABD D =S ABC S ACD D =S ABC S ABCD D =2(S ABC) Example 6.251 (0.466, 7, 8) Use the same notations as Example 6.250. If CA1/CD = AC1/AB = r1, DB1/DA = BD1/BC = r2 then S A2B2C2D2 S ABCD = r1·r2 r2·r1−r2−r1+2. We only need to show that S A2AB S ABCD = 1−r2 2·(r2·r1−r2−r1+2). Constructive description (Formula derivation) ((points A B C) (pratio D A B C 1) (lratio A1 C D r1) (lratio B1 D A r2) (lratio C1 A B r1) (lratio D1 B C r2) (inter A2 (l B B1) (l A A1)) (inter B2 (l C C1) (l B B1)) (inter C2 (l D D1) (l C C1)) (inter D2 (l A A1) (l D D1)) (S A2AB = S ABCD) ) Example 6.252 (0.066, 2, 11) Let P and Q be two points on side BC and AD of a parallelo-gram such that PQ ∥AB; M = AP ∩BQ, N = DP ∩QC. Show that MN = 1 2AD. N M Q P D C B A Figure 6-252 Constructive description ( (points A B C) (pratio D C B A 1) (on P (l B C)) (inter Q (l A D) (p P A B)) (inter M (l A P) (l B Q)) (inter N (l D P) (l C Q)) (2S AMN = −S ADN) ) 384 Chapter 6. Topics from Geometry The machine proof (−2)·S AMN S ADN N = (−2)·(−S CPQ·S ADM)·(−S CDQP) (−S CDQ·S ADP)·(−S CDQP) simplify = (−2)·S CPQ·S ADM S CDQ·S ADP M = (−2)·S CPQ·S ADP·S ABQ S CDQ·S ADP·S ABPQ simplify = (−2)·S CPQ·S ABQ S CDQ·S ABPQ Q = (−2)·S ACP·S ABP·(−S ABD) S CDP·(S ADP·S ABP−S ABP·S ABD) simplify = (2)·S ACP·S ABD S CDP·(S ADP−S ABD) P = (2)·(S ABC· BP BC −S ABC)·S ABD (−S BCD· BP BC +S BCD)·(−S ACD· BP BC +S ABD· BP BC −2S ABD) simplify = (2)·S ABC·S ABD S BCD·(S ACD· BP BC −S ABD· BP BC +2S ABD) D = (2)·S 2 ABC S ABC·(2S ABC) simplify = 1 The eliminants S ADN N =S CDQ·S ADP S CDQP S AMN N = S CPQ·S ADM S CDQP S ADM M = S ADP·S ABQ S ABPQ S ABPQ Q =(S ADP−S ABD)·S ABP −S ABD S CDQ Q =S CDP S ABQ Q =S ABP S CPQ Q =S ACP S ADP P = −(S ACD· BP BC −S ABD· BP BC +S ABD) S CDP P = −(( BP BC −1)·S BCD) S ACP P =( BP BC −1)·S ABC S ACD D =S ABC S BCD D =S ABC S ABD D =S ABC Example 6.253 (0.033, 2, 4) Let ABCD be a rectangular, and EFGH a parallelogram in-scribed in ABCD such that the sides of EFGH are parallel to the diagonals of ABCD. Show that the perimeter of EFGH is ﬁxed. A B C D E F H G Figure 6-253 Constructive description ( (points A B) (tratio C B A r) (pratio D A B C 1) (on E (l A B)) (inter F (l B C) (p E A C)) (inter H (l A D) (p E B D)) ( EF AC + EH BD = 1) ) The machine proof EH BD + EF AC H = −EF AC ·S ABD+S ADE −S ABD F = S BCE·S ABD−S ADE·S ABC S ABD·S ABC E = S ABD·S ABC S ABD·S ABC simplify = 1 The eliminants EH BD H = S ADE −S ABD EF AC F =S BCE S ABC S ADE E = −(S ABD· AE AB) S BCE E = −(( AE AB −1)·S ABC) Example 6.254 (0.116, 3, 15) Three parallel lines are cut by three parallel transversals in the points A, B, C; A1, B1, C1; A2, B2, C2. Show that B2C, C1A2, AB1 are concurrent. Constructive description ( (points A B A1) (on C (l A B)) (pratio B1 B A A1 1) (inter C1 (l A1 B1) (p C A A1)) (on A2 (l A A1)) (inter B2 (l B B1) (p A2 A B)) (inter C2 (l C C1) (l A2 B2)) (inter I (l C B2) (l C1 A2)) (inter Z2 (l C1 A2) (l A B1)) 6.4. Quadrilaterals 385 (inter Z1 (l C B2) (l A B1)) ( AZ1 B1Z1 · B1Z2 AZ2 = 1) ) A B 1 A C 1 B 1 C 2 A 2 B 2 C I Figure 6-254 The machine proof AZ1 B1Z1 · B1Z2 AZ2 Z1 = S ACB2 −S CB1B2 · B1Z2 AZ2 Z2 = −S ACB2 ·S B1C1A2 S CB1B2·S AC1A2 B2 = −S ACA2 ·S B1C1A2·(−S ABB1) S AB1BA2·S BCB1·S AC1A2 A2 = (−S AA1C· AA2 AA1 )·(−S AB1C1· AA2 AA1 +S AB1C1 )·S ABB1 (−S ABB1+S ABA1 · AA2 AA1 )·S BCB1·(−S AA1C1· AA2 AA1 ) simplify = S AA1C·( AA2 AA1 −1)·S AB1C1·S ABB1 (S ABB1−S ABA1 · AA2 AA1 )·S BCB1·S AA1C1 C1 = S AA1C·( AA2 AA1 −1)·S ACA1 B1·S AA1B1·S ABB1 (S ABB1−S ABA1 · AA2 AA1 )·S BCB1·S AA1C·(−S AA1B1) simplify = −( AA2 AA1 −1)·S ACA1 B1·S ABB1 (S ABB1−S ABA1 · AA2 AA1 )·S BCB1 B1 = −( AA2 AA1 −1)·(−S AA1C−S ABA1 )·S ABA1 (−S ABA1· AA2 AA1 +S ABA1 )·(−S BA1C) simplify = S AA1C+S ABA1 S BA1C C = −S ABA1 · AC AB +S ABA1 −S ABA1 · AC AB +S ABA1 simplify = 1 The eliminants AZ1 B1Z1 Z1 = S ACB2 −S CB1 B2 B1Z2 AZ2 Z2 = S B1C1A2 S AC1A2 S CB1B2 B2 = S AB1BA2·S BCB1 −S ABB1 S ACB2 B2 =S ACA2 S AC1A2 A2 = −(S AA1C1 · AA2 AA1 ) S AB1BA2 A2 = −(S ABB1−S ABA1 · AA2 AA1 ) S B1C1A2 A2 = −(( AA2 AA1 −1)·S AB1C1) S ACA2 A2 = −(S AA1C· AA2 AA1 ) S AA1C1 C1 =S AA1C S AB1C1 C1 = −(S ACA1 B1) S BCB1 B1 = −(S BA1C) S ABB1 B1 =S ABA1 S ACA1 B1 B1 = −(S AA1C+S ABA1) S BA1C C = −(( AC AB −1)·S ABA1) S AA1C C = −(S ABA1· AC AB) Example 6.255 (0.233, 5, 11) A line passing through the intersection O of the diagonals of parallelogram ABCD meets the four sides at E, F, G, H. Show that EF = GH. H E F G O D C B A Figure 6-255 Constructive description ( (points A B C) (pratio D C B A 1) (inter O (l A C) (l B D)) (on G (l A B)) (inter F (l C D) (l O G)) (inter E (l A D) (l O G)) (inter H (l B C) (l O G)) ( FE GH = −1) ) 386 Chapter 6. Topics from Geometry Example 6.256 (0.050, 5, 5) Let ABCD be a parallelogram and P, Q, R, S be points on AB, BC, CD, DA such that AP = CR and BQ = DS . Show that PQRS is also a paral-lelogram Constructive description: ( (points A B C) (pratio D A B C 1) (lratio S D A r2) (lratio P A B r1) (lratio R C D r1) (lratio Q B C r2) ( QR PS = 1) ) A B C D S P R Q Figure 6-256 The machine proof RQ S P Q = S BCR −S BSCP R = −S BCD·r1 S BSCP P = −S BCD·r1 −S BCS −S ABC·r1+S ABC S = S BCD·r1 −S BCD·r2+S BCD+S ABC·r2+S ABC·r1−S ABC D = −S ABC·r1 −S ABC·r1 simplify = 1 The eliminants RQ S P Q = S BCR −S BSCP S BCR R =S BCD·r1 S BSCP P = −(S BCS +S ABC·r1−S ABC) S BCS S = −(S BCD·r2−S BCD−S ABC·r2) S BCD D =S ABC Example 6.257 (0.400, 20, 16) Let ABCD be a parallelogram and P, Q, R, S are points in AB, BC, CD, DA such that r1 = AP AB = CR CD and r2 = BQ BC = DS DA. Show that S PQRS S ABCD = 2r2r1 −r2 − r1 + 1. (Figure 6-256) Constructive description ( (points A B C) (pratio D A B C 1) (lratio S D A r2) (lratio P A B r1) (lratio R C D r1) (lratio Q B C r2) (S PQRS = (2r2·r1−r2−r1+1)·S ABCD) ) Example 6.258 (0.266, 6, 24) Let C and F be any points on the respective sides AE and BD of a parallelogram AEBD. Let M and N denote the points of intersection of CD and FA and of EF and BC. Let the line MN meet DA at P and EB at Q. Then AP = QB. A E B D C F M N P Q Figure 6-258 Constructive description ( (points A E B) (pratio D B E A 1) (on C (l A E)) (on F (l B D)) (inter M (l D C) (l A F)) (inter N (l E F) (l B C)) (inter P (l M N) (l A D)) (inter Q (l M N) (l E B)) ( AP DP = BQ EQ ) ) 6.4. Quadrilaterals 387 Example 6.259 (0.400, 10, 22) Let ABC be any triangle and ABDE, ACFG any parallelo-grams described on AB and AC. Let DE and FG meet in H and draw BL and CM equal and parallel to HA. Then area(BCML) = area(ABDE) + area(ACFG). M L S R H G E F D A C B Figure 6-259 Constructive description ( (points A B C D F) (pratio E A B D 1) (pratio G A C F 1) (inter H (l D E) (l F G)) (inter R (l B C) (l H A)) (pratio L B H A 1) (S ACF+S BAD = S CBL) ) Example 6.260 (0.467, 46, 18) The circle through the vertices A, B, C of a parallelogram ABCD meets DA, DC in the points A1, C1. Prove that A1D/A1C1 = A1C/A1B. A B C D O 1 A 1 C Figure 6-260 Constructive description ( (points A B C) (circumcenter O A B C) (pratio D A B C 1) (inter A1 (l A D) (cir O A)) (inter C1 (l C D) (cir O C)) (eq-product A1 D A1 B A1 C1 A1 C) ) Example 6.261 (0.383, 8, 16) Given the parallelogram MDOM1, the vertex O is joined to the midpoint C of MM1. If the internal and external bisectors of the angle COD meet MD in A and B, show that MD2 = MA · MB. B A I C D E M 1 M O Figure 6-261 Constructive description ( (points I O C) (incenter D I O C) (midpoint E O D) (pratio M D E C 1) (pratio M1 O E C 1) (inter A (l O I) (l D M)) (inter B (l D M) (t O O I)) ( MD MA = MB MD ) ) Example 6.262 (0.633, 46, 34) ABCD is a parallelogram with center O. E is the midpoint of OA. F is the midpoint of OB. G is the midpoint of OC. H is the midpoint of OD. P = DE ∩CF; Q = AF ∩DG; R = AH ∩BG; S = BE ∩CH. Prove that PQRS is a parallelogram. 388 Chapter 6. Topics from Geometry A B C O D E F G H P Q R S Figure 262 Constructive description ( (points A B C O) (pratio D A B C 1) (midpoint E O A) (midpoint F O B) (midpoint G O C) (midpoint H O D) (inter P (l D E) (l C F)) (inter Q (l A F) (l D G)) (inter R (l A H) (l B G)) (inter S (l B E) (l C H)) ( PQ SR = 1) ) Example 6.263 (0.200, 5, 11) ABCD is a parallelogram. O = AC ∩BD. Two perpendicular lines passing through O cut the sides AB, BC, CD, DA in points E, H, F, G. Prove that EHCG is a rhombus. A B C D O E F G H Figure 6-263 Constructive description ( (points A B C) (pratio D A B C 1) (inter O (l A C) (l B D)) (on E (l A B)) (inter F (l C D) (l E O)) (inter G (l A D) (t O O E)) (inter H (l B C) (l O G)) ( EH GF = 1) ) 6.4.4 Squares Example 6.264 (0.016, 1, 2) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that S ABC = S AHF. Constructive description ( (points A B C) (tratio F A B 1) (tratio H A C −1) (S ABC = S AHF) ) A C B G H E F M N Figure 6-264 The machine proof S ABC −S AFH H = S ABC −(−1 4 PCAF) F = (4)·S ABC 4S ABC simplify = 1 The eliminants S AFH H = −1 4(PCAF) PCAF F =4(S ABC) 6.4. Quadrilaterals 389 Example 6.265 (0.050, 3, 5) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. M is the midpoint of BC. Show that FH = 2AM (Figure 6-264). Constructive description ( (points A B C) (midpoint M B C) (tratio F A B 1) (tratio H A C −1) (PHFH = 4PMAM) ) The machine proof PFHF (4)·PAMA H = PAFA+PACA+8S ACF (4)·PAMA F = 2PBAC+PACA+PABA (4)·PAMA M = 2PBAC+PACA+PABA (4)·(−1 4 PBCB+ 1 2 PACA+ 1 2 PABA) py = −(−2PBCB+4PACA+4PABA) (PBCB−2PACA−2PABA)·(2) simplify = 1 The eliminants PFHF H =PAFA+PACA+8S ACF S ACF F =1 4(PBAC) PAFA F =PABA PAMA M = −1 4(PBCB−2PACA−2PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.266 (0.016, 3, 3) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Passing through A a perpendicular to BC is drawn which cuts FH in N. Show that N is the midpoint of FH. (Figure 6-264) Constructive description ( (points A B C) (tratio F A B 1) (tratio H A C −1) (midpoint N F H) (perpendicular N A B C) ) The machine proof PCBN PABC N = 1 2 PCBH+ 1 2 PCBF PABC H = PCBF+PABC−4S ABC (2)·PABC F = 2PABC (2)·PABC simplify = 1 The eliminants PCBN N =1 2(PCBH+PCBF) PCBH H =PABC−4S ABC PCBF F =PABC+4S ABC Example 6.267 (0.033, 3, 2) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that FC = BH. (Figure 6-264) Constructive description ( (points A B C) (tratio F A B 1) (tratio H A C −1) (eqdistance F C B H) ) The machine proof PCFC PBHB H = PCFC PACA+PABA−8S ABC F = PACA+PABA−8S ABC PACA+PABA−8S ABC simplify = 1 The eliminants PBHB H =PACA+PABA−8S ABC PCFC F =PACA+PABA−8S ABC 390 Chapter 6. Topics from Geometry Example 6.268 (0.066, 4, 6) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that FC⊥BH. (Figure 6-264) Constructive description ((points A B C) (tratio F A B 1) (tratio H A C −1) (perpendicular F C B H) ) The machine proof PFBH PCBH H = −4S ABCF+PABF PABC−4S ABC F = −(PBAC−PABA+4S ABC) PABC−4S ABC py = −(−PBCB+PACA−PABA+8S ABC)·(2) (PBCB−PACA+PABA−8S ABC)·(2) simplify = 1 The eliminants PCBH H =PABC−4S ABC PFBH H = −(4S ABCF−PABF) PABF F =PABA S ABCF F = 1 4(PBAC+4S ABC) PABC=(PBCB−PACA+PABA) 2 PBAC=(PBCB−PACA−PABA) −2 Example 6.269 (0.033, 2, 8) On the two sides AB and AC of triangle ABC, two squares ABDM and ACEN are drawn externally. F and G are the feet of the perpendiculars drawn from points D and E to BC. Show that S ABC = S DFB + S CGE. A B C D M F E N G Figure 6-269 Constructive description ( (points A B C) (tratio D B A 1) (foot F D B C) (tratio E C A −1) (foot G E B C) (S ABC = S BDF+S CGE) ) The machine proof S ABC −(S CEG−S BDF) G = S ABC·PBCB −(PBCE·S BCE−PBCB·S BDF) E = S ABC·PBCB −(−PBCB·S BDF−PACB·S ABC) F = S ABC·(PBCB)2 −PCBD·PBCB·S BCD+PBCB·PACB·S ABC simplify = S ABC·PBCB −(PCBD·S BCD−PACB·S ABC) D = S ABC·PBCB −(−PACB·S ABC−PABC·S ABC) simplify = PBCB PACB+PABC py = PBCB·((2))2 4PBCB simplify = 1 The eliminants S CEG G = PBCE·S BCE PBCB S BCE E =1 4(PACB) PBCE E = −4(S ABC) S BDF F =−PCBD·S BCD PBCB S BCD D = 1 4(PABC) PCBD D = −4(S ABC) PABC=(PBCB−PACA+PABA) 2 PACB=(PBCB+PACA−PABA) 2 Example 6.270 (0.050, 6, 8) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. P is the midpoint of EG. Show that BP = CP. A B C E F G H P Figure 6-270 Constructive description ( (points A B C) (tratio E B A 1) 6.4. Quadrilaterals 391 (tratio G C A −1) (midpoint P E G) (eqdistance B P C P) ) The machine proof PBPB PCPC P = −1 4 PEGE+ 1 2 PBGB+ 1 2 PBEB −1 4 PEGE+ 1 2 PCGC+ 1 2 PCEC G = PCEC−2PBEB−2PBCB−PACA−8S ACE−16S ABC −PCEC−PACA−8S ACE E = −(−PBCB+2PBAC−PACA−PABA) PBCB−2PBAC+PACA+PABA simplify = 1 The eliminants PCPC P = −1 4(PEGE−2PCGC−2PCEC) PBPB P = −1 4(PEGE−2PBGB−2PBEB) PCGC G =PACA PBGB G =PBCB+PACA+8S ABC PEGE G =PCEC+PACA−8S ACE S ACE E = −1 4(PBAC+4S ABC) PBEB E =PABA PCEC E =PBCB+PABA+8S ABC Example 6.271 (0.033, 3, 6) Let ABCD be a square and G a point on CD. A square GCEF is erected externally. Show that DE⊥BG and DE = BG. D C B A G E F Figure 6-271 Constructive description ( (points D C) (tratio B C D 1) (lratio G C D r) (lratio E C B −r) (perpendicular D E B G) ) The machine proof PDBG PGBE E = PDBG PCBG·r+PCBG simplify = PDBG (r+1)·PCBG G = −PDBC·r+PDBC+PDBD·r (r+1)·PCBC B = −(−PDCD·r−PDCD) (r+1)·PDCD simplify = 1 The eliminants PGBE E =(r+1)·PCBG PCBG G =PCBC PDBG G = −(PDBC·r−PDBC−PDBD·r) PCBC B =PDCD PDBD B =2(PDCD) PDBC B =PDCD Example 6.272 (0.183, 6, 17) On the two sides AB and AC of triangle ABC, two squares ABDE and ACFG are drawn externally. Let P and Q be the centers of squares ABDE and ACFG, M and L the midpoints of BC and EG. Show that PMQL is a square. A B C M E D P G F Q L Figure 6-272 Constructive description ( (points A B C) (midpoint M B C) (tratio E A B 1) (midpoint P B E) (tratio G A C −1) (midpoint Q C G) (midpoint L E G) 392 Chapter 6. Topics from Geometry ( PM LQ = 1) ) The machine proof MP QL L = S MEPG S EGQ Q = S MEPG 1 2 S CEG G = (2)·(−1 4 PCMAP+S MEP+S AMP) 1 4 PACE+S ACE P = (−2)·( 1 2 PCAE−PCAM+ 1 2 PBAC−2S BME−2S AME+2S ABM) PACE+4S ACE E = −(−2PCAM−PBAM+PBAC+PABM+4S ABC) PBAC+PACA−4S ABC M = 1 2 PBAC+PACA−1 2 PABC+ 1 2 PABA−4S ABC PBAC+PACA−4S ABC py = (−4PBCB+12PACA+4PABA−32S ABC)·(2) ((2))3·(−PBCB+3PACA+PABA−8S ABC) simplify = 1 The eliminants MP QL L =S MEPG S EGQ S EGQ Q =1 2(S CEG) S CEG G =1 4(PACE+4S ACE) S MEPG G = −1 4(PCMAP−4S MEP−4S AMP) S AMP P =1 2(S AME−S ABM) S MEP P =1 2(S BME) PCMAP P =1 2(PCAE−2PCAM+PBAC) S ACE E = 1 4(PBAC) PACE E =PACA−4S ABC S AME E =1 4(PBAM) S BME E = −1 4(PABM−4S ABM) PCAE E =4(S ABC) PABM M = 1 2(PABC) PBAM M = 1 2(PBAC+PABA) PCAM M = 1 2(PBAC+PACA) PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) Example 6.273 (0.016, 1, 3) Let ABCD be a square and l be a line passing through B. From A and C perpendiculars are drawn to l and the feet are A1 and C1 respectively. Show that AA1 = BC1. B C X A D A C Figure 6-273 1 1 Constructive description ( (points B C X) (tratio A B C 1) (foot A1 A B X) (foot C1 C B X) (eqdistance A A1 B C1) ) The machine proof PAA1A PBC1B C1 = PAA1A·PBXB P2 CBX A1 = (16S 2 BXA)·PBXB (PCBX)2·PBXB simplify = (16)·(S BXA)2 (PCBX)2 A = (16)·(( 1 4 PCBX))2 (PCBX)2 simplify = 1 The eliminants PBC1B C1 = (PCBX)2 PBXB PAA1A A1 = (16)·(S BXA)2 PBXB S BXA A = 1 4(PCBX) Example 6.274 4 (0.067, 5, 13) Starting with any triangle ABC, construct the exterior (or interior) squares BCDE, ACFG, and BAHK; then construct parallelograms FCDQ and EBKP. Show that PAQ is an isosceles. 6.4. Quadrilaterals 393 A B C F G D E K H Q P Figure 6-274 Constructive description ( (points A B C) (tratio F C A 1) (tratio D C B −1) (pratio E D C B 1) (tratio K B A −1) (pratio Q F C D 1) (pratio P K B E 1) (eqdistance A Q A P) ) The machine proof PAQA PAPA P = PAQA 2PEBK+PAKA+PAEA−PABA Q = 2PFCD+PADA+PAFA−PACA 2PEBK+PAKA+PAEA−PABA K = 2PFCD+PADA+PAFA−PACA PAEA+PABA+8S ABE E = 2PFCD+PADA+PAFA−PACA PADA−PACA+2PABA+8S ABD−8S ABC D = PBCB+PAFA+8S BCF−8S ABC PBCB−2PABC+2PABA−8S ABC F = PBCB−2PACB+2PACA−8S ABC PBCB−2PABC+2PABA−8S ABC py = (2PACA+2PABA−16S ABC)·(2) (2PACA+2PABA−16S ABC)·(2) simplify = 1 The eliminants PAPA P =2PEBK+PAKA+PAEA−PABA PAQA Q =2PFCD+PADA+PAFA−PACA PAKA K =2(PABA) PEBK K =4(S ABE) S ABE E =S ABD−S ABC PAEA E =PADA−PACA+PABA S ABD D = −1 4(PABC−4S ABC) PADA D =PBCB+PACA−8S ABC PFCD D =4(S BCF) S BCF F = −1 4(PACB) PAFA F =2(PACA) PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) Example 6.275 (0.516, 9, 32) On the four sides of a quadrilateral ABCD, four squares are drawn externally. Show that the two segments joining the two centers of the squares on two opposite sides are perpendicular and have the same length. A B C D L P N M Figure 6-275 Constructive description ( (points A B C D) (tratio E A B 1) (midpoint L B E) (tratio G A D −1) (midpoint P D G) (tratio F C B −1) (midpoint M B F) (tratio H C D 1) (midpoint N D H) (eqdistance M P L N) ) 4This example is from Amer. Math. Mon. 75(1968), p.899. 394 Chapter 6. Topics from Geometry Example 6.276 (0.067 2 14) On the hypotenuse AB of right triangle ABC a square ABFE is erected. Let P be the intersection of the diagonals AF and BE of ABFE. Show that ∠ACP = ∠PCB. C B A E F P Figure 6-276 Constructive description ((points B C) (tratio A C B r) (tratio F B A −1) (tratio E A B 1) (inter P (l B E) (l A F)) (eqangle A C P P C B) ) The machine proof (−S CAP)·PBCP (−S BCP)·PACP P = S CAF·S BAE·(−PBCF·S BAE)·(S BAEF)2 (−S BAF·S BCE)·PACE·S BAF·((−S BAEF))2 simplify = S CAF·(S BAE)2·PBCF (S BAF)2·S BCE·PACE E = S CAF·((−1 4 PBAB))2·PBCF (S BAF)2·(−1 4 PCBA+S BCA)·(PCAC−4S BCA) F = −(−1 4 PBAC+S BCA)·(PBAB)2·(PBCB−4S BCA) (4)·((−1 4 PBAB))2·(PCBA−4S BCA)·(PCAC−4S BCA) simplify = (PBAC−4S BCA)·(PBCB−4S BCA) (PCBA−4S BCA)·(PCAC−4S BCA) A = (PBCB·r2+PBCB·r)·(PBCB·r+PBCB) (PBCB·r+PBCB)·(PBCB·r2+PBCB·r) simplify = 1 The eliminants PACP P = PACE·S BAF S BAEF S BCP P =S BAF·S BCE S BAEF PBCP P = PBCF·S BAE S BAEF S CAP P =S CAF·S BAE S BAEF PACE E =PCAC−4S BCA S BCE E = −1 4(PCBA−4S BCA) S BAE E = −1 4(PBAB) S BAF F = −1 4(PBAB) PBCF F =PBCB−4S BCA S CAF F = −1 4(PBAC−4S BCA) PCAC A =PBCB·(r)2 PCBA A =PBCB S BCA A = −1 4(PBCB·r) PBAC A =PBCB·(r)2 Example 6.277 (0.617, 65, 21) ABCD is a square. G is a point on AD such that AG = AD. E, F are points on AB, BC such that EF ∥AC. H = EG ∩FD. Show that AH = BC. B C A D G F E H Figure 6-277 Constructive description ( (points B C) (tratio A B C 1) (pratio D A B C 1) (lratio G A D −1) (lratio F B C r) (inter E (l A B) (p F A C)) (inter H (l G E) (l F D)) (eqdistance A H B C) ) 6.4. Quadrilaterals 395 6.4.5 Cyclic Quadrilaterals Deﬁnition A quadrilateral whose vertices lie on the same circle is said to be cyclic. To prove theorems about cyclic quadrilaterals, we use the techniques developed in Sec-tion 3.6. Example 6.278 (Ptolemy’s Theorem) (0.050, 2, 12) Let A, B, C, D be four cyclic points. Then f AB · g CD + g AD · f BC = f AC · g BD. Constructive description ( (circle A B C D) (f AB·g CD+g AD· f BC = f AC·g BD) ) The machine proof g CD·f AB+ f BC·g AD g BD· f AC chord = sin(CD)·sin(AB)·d2+sin(BC)·sin(AD)·d2 sin(BD)·d·sin(AC)·d simplify = sin(CD)·sin(AB)+sin(BC)·sin(AD) sin(BD)·sin(AC) co−cir = S D·S C·CB−S C·S B·CD (S D·CB−S B·CD)·S C simplify = 1 The eliminants f AC=sin(AC)·d g BD=sin(BD)·d g AD=sin(AD)·d f BC=sin(BC)·d f AB=sin(AB)·d g CD=sin(CD)·d sin(AC)=S C sin(BD)=S D·CB−S B·CD sin(AD)=S D sin(BC)=S C·CB−S B·CC sin(AB)=S B sin(CD)=S D·CC−S C·CD Example 6.279 (Brahmagupta’s Formula) (0.267, 5, 16) Let ABCD be a cyclic quadrilateral. Then S 2 = (p −f AB)(p −f BC)(p −g CD)(p −g AD) = 1 16(4AC 2BD 2 −P2 ABCD) where p = 1 2(f AB + f BC + g CD + g AD). Constructive description ( (circle A B C D) (16S 2 ABCD = 4 f AC 2·g BD 2−P2 CBAD) ) Example 6.280 (0.033, 1, 10) Show that in a cyclic quadrilateral the distances of the point of intersection of the diagonals from two opposite sides are proportional to these sides. B C A O D I Q S Figure 6-280 Constructive description ( (circle A B C D) (inter I (l B D) (l A C)) (foot Q I B C) (foot S I A D) (eq-product I S B C I Q A D) ) 396 Chapter 6. Topics from Geometry The machine proof PIS I·PBCB PIQI·PADA S = (16S 2 ADI)·PBCB PIQI·P2 ADA Q = (16)·S 2 ADI·P2 BCB (16S 2 BCI)·P2 ADA I = (−S ACD·S ABD)2·P2 BCB·S 2 ABCD S BCD·S 2 ABC·P2 ADA·S 2 ABCD simplify = S ACD·S ABD·S ACD·S ABD·P2 BCB S BCD·S ABC·S BCD·S ABC·P2 ADA co−cir = (−g CD·g AD· f AC)·(−g BD·g AD·f AB)·(−g CD·g AD· f AC)·(−g BD·g AD·f AB)·(2 f BC 2)2·(2d)4 (−g CD·g BD· f BC)·(−f BC· f AC·f AB)·(−g CD·g BD· f BC)·(−f BC· f AC·f AB)·(2g AD 2)2·(2d)4 simplify = 1 The eliminants PIS I S = (16)·S 2 ADI PADA PIQI Q = (16)·S 2 BCI PBCB S BCI I =S BCD·S ABC S ABCD S ADI I =−S ACD·S ABD S ABCD PADA=2(g AD 2) S ABC= f BC· f AC·f AB (−2)·d S BCD= g CD·g BD· f BC (−2)·d PBCB=2( f BC 2) S ABD= g BD·g AD·f AB (−2)·d S ACD= g CD·g AD· f AC (−2)·d Example 6.281 (0.016, 1, 6) In the cyclic quadrilateral ABCD the perpendicular to AB at A meets CD in A1, and the perpendicular to CD at C meets AB in C1. Show that the line A1C1 is parallel to the diagonal BD. 1 C 1 A D O C B A Figure 6-281 Constructive description ( (circle A B C D) (inter A1 (l C D) (t A A B)) (inter C1 (l A B) (t C C D)) (parallel A1 C1 B D) ) The machine proof S BDA1 S BDC1 C1 = S BDA1·(−PBCAD) −PBCD·S ABD A1 = PBAD·S BCD·PBCAD PBCD·S ABD·(−PBCAD) simplify = −PBAD·S BCD PBCD·S ABD co−cir = −(2g AD·f AB·cos(BD))·(−g CD·g BD· f BC)·(2d) (−2g CD· f BC·cos(BD))·(−g BD·g AD·f AB)·(2d) simplify = 1 The eliminants S BDC1 C1 = PBCD·S ABD PBCAD S BDA1 A1 = PBAD·S BCD −PBCAD S ABD= g BD·g AD·f AB (−2)·d PBCD= −2(g CD· f BC·cos(BD)) S BCD= g CD·g BD· f BC (−2)·d PBAD=2(g AD·f AB·cos(BD)) Deﬁnition. The symmetric of the circumcenter of a cyclic quadrilateral with respect to the centroid is called the anticenter of the cyclic quadrilateral. Example 6.282 (0.083, 3, 9) The perpendicular from the midpoint of each diagonal upon the other diagonal also passes through the anticenter of a cyclic quadrilateral. 6.4. Quadrilaterals 397 M I R Q P D O A E C B Figure 6-282 Constructive description ((circle A B C D) (circumcenter O A B C) (midpoint T A D) (midpoint P A B) (midpoint Q B C) (midpoint R C D) (inter M (p P O R) (p Q O T)) (inter I (l P R) (l O M)) (midpoint I P R) ) The machine proof −PI RI I = S OPM −S ORM M = −(−S OPTQ·S OPR) (−S OPR)·(−S OTR) simplify = S OPTQ S OTR R = S OPTQ 1 2 S DOT + 1 2 S COT Q = (2)·(−S OTP+ 1 2 S COT + 1 2 S BOT ) S DOT +S COT P = −(−S COT +S AOT ) S DOT +S COT T = −1 2 S CDO+ 1 2 S ADO+ 1 2 S ACO −1 2 S CDO+ 1 2 S ADO+ 1 2 S ACO simplify = 1 The eliminants PI RI I =S OPM S ORM S ORM M = −(S OPR) S OPM M = S OPTQ·S OPR S OTR S OTR R =1 2(S DOT +S COT) S OPTQ Q =(2S OTP−S COT −S BOT ) −2 S OTP P = 1 2(S BOT+S AOT) S DOT T =1 2(S ADO) S AOT T = −1 2(S ADO) S COT T = −1 2(S CDO−S ACO) Example 6.283 (0.066, 4, 5) Show that the anticenter of a cyclic quadrilateral is the ortho-center of the triangle having for vertices the midpoints of the diagonals and the point of intersection of those two lines. B C A O D I Q S J M Figure 6-283 Constructive description ( (circle A B C D) (circumcenter O A B C) (inter I (l A D) (l B C)) (midpoint Q B C) (midpoint S A D) (midpoint J S Q) (lratio M J O −1) (perpendicular M S B C) ) The machine proof PCBM PCBS M = 2PCBJ−PCBO PCBS J = (2)·( 1 2 PCBS + 1 2 PCBQ−1 2 PCBO) PCBS S = PCBQ−PCBO+ 1 2 PCBD+ 1 2 PABC 1 2 PCBD+ 1 2 PABC Q = (2)·(−PCBO+ 1 2 PCBD+ 1 2 PBCB+ 1 2 PABC) PCBD+PABC O = (−2)·(−PCBD−PABC) (PCBD+PABC)·(2) simplify = 1 The eliminants PCBM M =2(PCBJ−1 2 PCBO) PCBJ J =1 2(PCBS +PCBQ) PCBS S =1 2(PCBD+PABC) PCBQ Q = 1 2(PBCB), PCBO O = 1 2(PBCB) Example 6.284 (0.050, 2, 5) Show that the anticenter of a cyclic quadrilateral is collinear with the two symmetrics of the circumcenter of the quadrilateral with respect to a pair of opposite sides. Constructive description 398 Chapter 6. Topics from Geometry ( (circle A B C D) (circumcenter O A B C) (midpoint Q B C) (midpoint S A D) (midpoint J S Q) (lratio Q1 Q O −1) (lratio S 1 S O −1) (inter M (l O J) (l Q1 S 1)) (midpoint J O M) ) B C A O D Q S J M 1 Q 1 S Figure 6-284 The machine proof OJ JM M = S OQ1JS1 S JQ1S1 S 1 = −S OJQ1−2S OS J 2S S JQ1 −S OJQ1 Q1 = −(2S OS J−2S OQJ) −2S OS J+2S OQJ simplify = 1 The eliminants OJ JM M = S OQ1JS1 S JQ1S1 S JQ1S1 S1 =2S S JQ1 −S OJQ1 S OQ1JS1 S1 = −(S OJQ1 +2S OS J) S S JQ1 Q1 = −(S OS J) S OJQ1 Q1 = −2(S OQJ) Example 6.285 (0.050, 1, 12) Show that the product of the distances of two opposite sides of a cyclic quadrilateral from a point on the circumcircles is equal to the product of the distances of the other two sides from the same point. B C A O D E P Q R S Figure 6-285 Constructive description ( (circle A B C D E) (foot P E A B) (foot Q E B C) (foot R E C D) (foot S E A D) (eq-product E S E Q E P E R) ) The machine proof PES E·PEQE PEPE·PERE S = (16S 2 ADE)·PEQE PEPE·PERE·PADA R = (16)·S 2 ADE·PEQE·PCDC PEPE·(16S 2 CDE)·PADA Q = S 2 ADE·(16S 2 BCE)·PCDC PEPE·S 2 CDE·PADA·PBCB P = (16)·S 2 ADE·S 2 BCE·PCDC·PABA (16S 2 ABE)·S 2 CDE·PADA·PBCB co−cir = (−g DE· f AE·g AD)2·(−g CE· f BE· f BC)2·(2g CD 2)·(2f AB 2)·(2d)4 (−f BE· f AE·f AB)2·(−g DE·g CE·g CD)2·(2g AD 2)·(2 f BC 2)·(2d)4 simplify = 1 The eliminants PES E S = (16)·S 2 ADE PADA , PERE R = (16)·S 2 CDE PCDC PEQE Q = (16)·S 2 BCE PBCB , PEPE P = (16)·S 2 ABE PABA PBCB=2( f BC 2), PADA=2(g AD 2) S CDE= g DE·g CE·g CD (−2)·d S ABE= f BE· f AE·f AB (−2)·d PABA=2(f AB 2) PCDC=2(g CD 2) S BCE= g CE· f BE· f BC (−2)·d S ADE= g DE· f AE·g AD (−2)·d 6.4. Quadrilaterals 399 Example 6.286 (0.050, 1, 12) Show that the projections of a point of the circumcircle of a cyclic quadrilateral upon the sides divide the sides into eight segments such that the product of four nonconsecutive segments is equal to the product of the remaining four. B C A O D P 1 A 1 B 1 C 1 D Figure 6-286 Constructive description ( (circle A B C D P) (circumcenter O A B C) (foot A1 P A B) (foot B1 P B C) (foot C1 P C D) (foot D1 P D A) ( AA1 A1B BB1 B1C CC1 C1D DD1 D1A = 1) ) The machine proof DD1 AD1 · CC1 DC1 · BB1 CB1 · AA1 BA1 D1 = PADP −PDAP · CC1 DC1 · BB1 CB1 · AA1 BA1 C1 = −PADP·PDCP PDAP·(−PCDP) · BB1 CB1 · AA1 BA1 B1 = PADP·PDCP·PCBP PDAP·PCDP·(−PBCP) · AA1 BA1 A1 = −PADP·PDCP·PCBP·PBAP PDAP·PCDP·PBCP·(−PABP) co−cir = (−2g DP·g AD·cos(AP))·(2 f CP·g CD·cos(DP)) (2f AP·g AD·cos(DP))·(−2g DP·g CD·cos(CP)) · (2f BP· f BC·cos(CP))·(2f AP·f AB·cos(BP)) (−2 f CP· f BC·cos(BP))·(−2f BP·f AB·cos(AP)) simplify = 1 The eliminants DD1 AD1 D1 = PADP −PDAP, CC1 DC1 C1 = PDCP −PCDP BB1 CB1 B1 = PCBP −PBCP AA1 BA1 A1 = PBAP −PABP PABP= −2(f BP·f AB·cos(AP)) PBCP= −2( f CP· f BC·cos(BP)) PCDP= −2(g DP·g CD·cos(CP)) PDAP=2(f AP·g AD·cos(DP)) PBAP=2(f AP·f AB·cos(BP)) PCBP=2(f BP· f BC·cos(CP)) PDCP=2( f CP·g CD·cos(DP)) PADP= −2(g DP·g AD·cos(AP)) Example 6.287 (0.866, 4, 29) The four lines obtained by joining each vertex of a cyclic quadrilateral to the orthocenter of the triangle formed by the remaining three vertices bisect each other. V U D O C B A Figure 6-288 M 1 A 1 D D O A C B Figure 6-287 Constructive description ( (circle A B C D) (orthocenter D1 A B C) (orthocenter A1 B C D) (inter M (l D D1) (l A A1)) (midpoint M A A1) ) Example 6.288 (1.750, 11, 35) A line AD through the vertex A meets the circumcircle of the triangle ABC in D. If U, V are the orthocenters of the triangle ABD, ACD, respectively, prove that UV is equal and parallel to BC. 400 Chapter 6. Topics from Geometry Constructive description ( (circle A B C D) (orthocenter U A B D) (orthocenter V A C D) (eqdistance V U B C) ) Example 6.289 (1.633, 12, 35) The sum of the squares of the distances of the anticenter of a cyclic quadrilateral from the four vertices is equal to the square of the circumdiameter of the quadrilateral. B C A O D 1 A M Figure 6-289 Constructive description ( (circle A B C D) (circumcenter O A B C) (orthocenter A1 B C D) (midpoint M A A1) (MA 2+MB 2+MC 2+MD 2 = 4OB 2) ) Example 6.290 (0.750, 6, 41) The line joining the centroid of a triangle to a point P on the circumcircle bisects the line joining the diametric opposite of P to the orthocenter. A B C 3 M 2 M G H O P Q I Figure 6-290 Constructive description ( (circle A B C P) (circumcenter O A B C) (orthocenter H A B C) (centroid G A B C) (lratio Q O P −1) (inter I (l H Q) (l P G)) (midpoint I Q H) ) Example 6.291 (0.616, 9, 48) Show that the perpendicular from the point of intersection of two opposite sides, produced, of a cyclic quadrilateral upon the line joining the midpoints of the two sides considered passes through the anticenter of the quadrilateral. B C A O D I Q S J M Figure 6-291 Constructive description ( (circle A B C D) (circumcenter O A B C) (inter I (l A D) (l B C)) (midpoint Q B C) (midpoint S A D) (midpoint J S Q) (lratio M J O −1) (perpendicular I M S Q) ) Example 6.292 (0.450, 3, 31) If Ha, Hb, Hc, Hd are the orthocenters of the four triangles determined by the vertices of the cyclic quadrilateral ABCD, show that the vertices of ABCD are the orthocenters of the four triangles determined by the points Ha, Hb, Hc, Hd. 6.4. Quadrilaterals 401 c H a H d H D O A C B Figure 6-292 Constructive description ( (circle A B C D) (orthocenter HD A B C) (orthocenter HA B C D) (orthocenter HC A B D) (perpendicular HC HA B HD) ) 6.4.6 Orthodiagonal Quadrilaterals Deﬁnition A quadrilateral is said to be orthodiagonal if its diagonals are perpendicular to each other. Example 6.293 (Theorem of Brahmagupta) (0.133, 4, 15) In a quadrilateral which is both or-thodiagonal and cyclic the perpendicular from the point of intersection of the diagonals to a side bisects the side opposite. F F F E D O B C A Figure 6-293 Constructive description ( (points A B C) (circumcenter O A B C) (foot E B A C) (inter D (l B E) (cir O B)) (foot FF E C D) (inter F (l A B) (l E FF)) (midpoint F A B) ) The eliminants AF BF F = S AEFF S BEFF S BEFF FF = −PCDE·S BCE PCDC S AEFF FF = PECD·S AED PCDC PCDE D =−(2POBE−PCBE)·(POEO−PBOB) PBEB S AED D = −(POEO−PBOB)·S ABE PBEB PECD D =PBCE S BCE E = PACB·S ABC PACA PCBE E = PBCB·PBAC+PACB·PABC PACA POBE E = PCBO·PBAC+PACB·PABO PACA S ABE E = PBAC·S ABC PACA PBCE E = P2 ACB PACA PABO O =1 2(PABA) PCBO O =1 2(PBCB) PABC=1 2(PBCB−PACA+PABA) PBAC= −1 2(PBCB−PACA−PABA) The machine proof −AF BF F = −S AEFF S BEFF FF = −PECD·S AED·PCDC (−PCDE·S BCE)·PCDC simplify = PECD·S AED PCDE·S BCE D = PBCE·(−POEO·S ABE+PBOB·S ABE)·PBEB (−2POEO·POBE+POEO·PCBE+2POBE·PBOB−PCBE·PBOB)·S BCE·PBEB 402 Chapter 6. Topics from Geometry simplify = PBCE·S ABE (2POBE−PCBE)·S BCE E = P2 ACB·PBAC·S ABC·P3 ACA (2PCBO·PBAC·PACA−PBCB·PBAC·PACA+2PACB·PACA·PABO−PACB·PACA·PABC)·PACB·S ABC·P2 ACA simplify = PACB·PBAC 2PCBO·PBAC−PBCB·PBAC+2PACB·PABO−PACB·PABC O = PACB·PBAC·(2)2 −4PACB·PABC+4PACB·PABA simplify = −PBAC PABC−PABA py = −(−PBCB+PACA+PABA)·(2) (PBCB−PACA−PABA)·(2) simplify = 1 Example 6.294 (0.333, 4, 34) Let E be the intersection of the two diagonals AC and BD of cyclic quadrilateral ABCD. Let I be the center of circumcircle of ABE. Show the IE ⊥DC. I E D C B O A Figure 6-294 Constructive description ( (circle A B C D) (inter E (l B D) (l A C)) (circumcenter I A B E) (perpendicular I E C D) ) Example 6.295 (0.600, 7, 16) In an orthodiagonal quadrilateral the two lines joining the midpoints of the pairs of opposite sides are equal. S R Q P D B C A Figure 6-295 Constructive description ( (points A B C) (foot FD B A C) (on D (l B FD)) (midpoint P A B) (midpoint Q B C) (midpoint R C D) (midpoint S D A) (eqdistance S Q P R) ) Example 6.296 (0.850, 7, 30) In an orthodiagonal quadrilateral the midpoints of the sides lie on a circle having for center the centroid of the quadrilateral. O S R Q P D B C A Figure 6-296 Constructive description ( (points A B C) (foot FD B A C) (on D (l B FD)) (midpoint P A B) (midpoint Q B C) 6.4. Quadrilaterals 403 (midpoint R C D) (midpoint S D A) (inter O (l Q S ) (l P R)) (perp-biesct O S R) ) Example 6.297 (3.300, 36, 29) If an orthodiagonal quadrilateral is cyclic, the anticenter coincides with the point of intersection of its diagonals. M A C B O D P R J Figure 6-297 Constructive description ( (points A B C) (circumcenter O A B C) (foot M B A C) (inter D (l B M) (cir O B)) (midpoint J O M) (midpoint R C D) (inter P (l A B) (l J R)) (midpoint P A B) ) Example 6.298 (0.967, 14, 15) In a cyclic orthodiagonal quadrilateral the distance of a side from the circumcenter of the quadrilateral is equal to half the opposite side. A C B O D P Figure 6-298 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint P A B) (foot FD B A C) (inter D (l B FD) (cir O B)) (PCDC = 4POPO) ) Example 6.299 (0.967, 15, 14) If a quadrilateral is both cyclic and orthodiagonal, the sum of the squares of two opposite sides is equal to the square of the circumdiameter of the quadrilateral. D O B C A Figure 6-299 Constructive description ( (points A B C) (circumcenter O A B C) (foot FD B A C) (inter D (l B FD) (cir O B)) (AB 2+DC 2 = 4OA 2) ) 404 Chapter 6. Topics from Geometry Example 6.300 (2.117, 26, 20) Show that the line joining the midpoints of the diagonals of a cyclic orthodiagonal quadrilateral is equal to the distance of the point of intersection of the diagonals from the circumcenter of the quadrilateral. A C B O D E U V Figure 6-300 Constructive description ( (points A B C) (circumcenter O A B C) (foot E B A C) (inter D (l B E) (cir O B)) (midpoint U A C) (midpoint V B D) (eqdistance U V O E) ) Example 6.301 (1.050, 16, 16) If the diagonals of a cyclic quadrilateral ABCD are orthogo-nal, and E is the diametric opposite of D on its circumcircle, show that AE = CB. A C B O D E Figure 6-301 Constructive description ( (points A B C) (circumcenter O A B C) (foot FD B A C) (inter D (l B FD) (cir O B)) (inter E (l D O) (cir O D)) (eqdistance A E C B) ) Example 6.302 (1.250, 4, 84) A1B1C1D1 is a quadrilateral with an inscribed circle O. Then O is on the line joining the midpoints of A1C1 and B1D1. O A B C D A B C D M N Figure 6-302 1 1 1 1 Constructive description ( (circle A B C D) (circumcenter O A B C) (on BT (t B B O)) (on AT (t A A O)) (on CT (t C C O)) (on DT (t D D O)) (inter A1 (l B BT) (l A AT)) (inter B1 (l C CT ) (l B BT)) (inter C1 (l D DT) (l C CT )) (inter D1 (l A AT) (l D DT )) (midpoint M A1 C1) (inter N (l B1 D1) (l O M)) (midpoint N B1 D1) ) 6.4. Quadrilaterals 405 6.4.7 The Butterﬂy Theorems For the machine proof of the general butterﬂy theorem for a circle, See Example 3.81 on page 146 Example 6.303 (0.516, 4, 31) The Butterﬂy theorem for a circle. G F A C B D O E Figure 6-303 Constructive description ( (circle A B C D) (circumcenter O A B C) (inter E (l A B) (l C D)) (inter G (l B C) (t E E O)) (inter F (l A D) (t E E O)) ( GF EF = 2) ) Example 6.304 (Butterﬂy Theorem for Quadrilaterals) (0.166, 3, 19) Let ABCD be a quadrilat-eral such that AB = BC and AD = CD. M is the intersection of AC and BD. Pass-ing through M two lines are drawn which meet the sides of ABCD at P, Q, S, R. Let G = PR ∩AC, H = S Q ∩AC. Show that GM = MH. A C B M D P Q S R G H Figure 6-304 Constructive description ( (points A C) (on B (b A C)) (midpoint M A C) (on D (l B M)) (on P (l A B)) (inter Q (l C D) (l P M)) (on S (l B C)) (inter R (l A D) (l S M)) (inter G (l A C) (l P R)) (inter H (l A C) (l S Q)) ( MG GA = MH HC ) ) The eliminants MH CH H =S MQS S CQS , MG AG G = S MPR S APR S APR R =−S ADP·S AMS S AMDS , S MPR R =S MPS ·S AMD S AMDS S AMS S =( BS BC −1)·S ABM, S CQS S =( BS BC −1)·S CBQ S MQS S = −(S BMQ· BS BC −S BMQ−S CMQ· BS BC) S MPS S = −(S BMP· BS BC −S BMP−S CMP· BS BC) S CMQ Q =S CMP·S CMD S CMDP , S BMQ Q =S BMP·S CMD S CMDP S CBQ Q =S CMP·S CBD S CMDP , S ADP P = −(S ABD· AP AB) S CMP P = −(S CBM· AP AB), S ABD D =S ABM· BD BM S CMD D =( BD BM −1)·S CBM, S CBD D =S CBM· BD BM S AMD D =( BD BM −1)·S ABM, S ABM M = −1 2(S ACB) S CBM M = 1 2(S ACB) The machine proof ( MG AG )/( MH CH ) H = −S CQS −S MQS · MG AG G = S MPR·S CQS S MQS ·S APR R = (−S MPS ·S AMD)·S CQS ·S AMDS S MQS ·(−S ADP·S AMS )·(−S AMDS ) simplify = −S MPS ·S AMD·S CQS S MQS ·S ADP·S AMS 406 Chapter 6. Topics from Geometry S = −(−S BMP· BS BC +S BMP+S CMP· BS BC )·S AMD·(S CBQ· BS BC −S CBQ) (−S BMQ· BS BC +S BMQ+S CMQ· BS BC )·S ADP·(S ABM· BS BC −S ABM) simplify = −(S BMP· BS BC −S BMP−S CMP· BS BC )·S AMD·S CBQ (S BMQ· BS BC −S BMQ−S CMQ· BS BC )·S ADP·S ABM Q = −(S BMP· BS BC −S BMP−S CMP· BS BC )·S AMD·S CMP·S CBD·S CMDP·(−S CMDP) (−S CMDP·S BMP·S CMD· BS BC +S CMDP·S BMP·S CMD+S CMDP·S CMP·S CMD· BS BC )·S ADP·S ABM·S CMDP simplify = −S AMD·S CMP·S CBD S CMD·S ADP·S ABM P = −S AMD·(−S CBM· AP AB)·S CBD S CMD·(−S ABD· AP AB)·S ABM simplify = −S AMD·S CBM·S CBD S CMD·S ABD·S ABM D = −(S ABM· BD BM −S ABM)·S CBM·S CBM· BD BM (S CBM· BD BM −S CBM)·S ABM· BD BM ·S ABM simplify = −S CBM S ABM M = −( 1 2S ACB) −1 2S ACB simplify = 1 Example 6.305 (Butterﬂy Theorem for Quadrilaterals) (0.066, 2, 14) Let ABCD be a quadrilat-eral such that the intersection M of its diagonals is the midpoint of AC. Passing through M two lines are drawn which meet the sides of ABCD at P, Q, S, R. Let G = PR ∩AC, H = S Q ∩AC. Show that GM = MH. 6.4. Quadrilaterals 407 Constructive description ( (points A C B) (midpoint M A C) (lratio D M B r1) (lratio P A B r2) (lratio R A D r3) (inter Q (l P M) (l C D)) (inter S (l R M) (l B C)) (inter G (l A C) (l P R)) (inter H (l A C) (l S Q)) ( MG GA = MH HC ) ) The machine proof ( MG AG )/( MH CH ) H = −S CQS −S MQS · MG AG G = S MPR·S CQS S MQS ·S APR S = S MPR·(−S CMR·S CBQ)·(−S CMBR) (−S MRQ·S CBM)·S APR·S CMBR simplify = −S MPR·S CMR·S CBQ S MRQ·S CBM·S APR Q = −S MPR·S CMR·S CMP·S CBD·(−S CMDP) S MPR·S CMD·S CBM·S APR·S CMDP simplify = S CMR·S CMP·S CBD S CMD·S CBM·S APR R = S CMD·r3·S CMP·S CBD S CMD·S CBM·(−S ADP·r3) simplify = −S CMP·S CBD S CBM·S ADP P = −(−S CBM·r2)·S CBD S CBM·(−S ABD·r2) simplify = −S CBD S ABD D = −(−S CBM·r1+S CBM) −S ABM·r1+S ABM simplify = −S CBM S ABM M = −( 1 2 S ACB) −1 2 S ACB simplify = 1 The eliminants MH CH H =S MQS S CQS MG AG G =S MPR S APR S MQS S =S MRQ·S CBM S CMBR S CQS S =−S CMR·S CBQ S CMBR S MRQ Q =S MPR·S CMD −S CMDP S CBQ Q = S CMP·S CBD S CMDP S APR R = −(S ADP·r3) S CMR R =S CMD·r3 S ADP P = −(S ABD·r2) S CMP P = −(S CBM·r2) S ABD D = −((r1−1)·S ABM) S CBD D = −((r1−1)·S CBM) S ABM M = −1 2(S ACB) S CBM M = 1 2(S ACB) Example 6.306 (0.100, 2, 15) Let ABCD be a quadrilateral and M the intersection of its diagonals. Passing through M two lines are drawn which meet the sides of ABCD at P, Q, S, R. Let G = PR ∩AC, H = S Q ∩AC. Show that MG AG = MH CH CM MA . 408 Chapter 6. Topics from Geometry Constructive description ( (points A B C) (lratio M A C r0) (lratio D M B r1) (lratio P A B r2) (lratio R A D r3) (inter Q (l P M) (l C D)) (inter S (l R M) (l B C)) (inter G (l A C) (l P R)) (inter H (l A C) (l S Q)) ( MG AG = MH CH CM MA ) ) The machine proof MG AG −MH CH · CM AM H = −S CQS −(−S MQS )· CM AM · MG AG G = −S MPR·S CQS S MQS · CM AM ·S APR S = −S MPR·(−S CMR·S BCQ)·(−S BMCR) (−S MRQ·S BCM)· CM AM ·S APR·S BMCR simplify = S MPR·S CMR·S BCQ S MRQ·S BCM· CM AM ·S APR Q = S MPR·S CMR·S CMP·S BCD·(−S CMDP) S MPR·S CMD·S BCM· CM AM ·S APR·S CMDP simplify = −S CMR·S CMP·S BCD S CMD·S BCM· CM AM ·S APR R = −S CMD·r3·S CMP·S BCD S CMD·S BCM· CM AM ·(−S ADP·r3) simplify = S CMP·S BCD S BCM· CM AM ·S ADP P = S BCM·r2·S BCD S BCM· CM AM ·(−S ABD·r2) simplify = −S BCD CM AM ·S ABD D = −(−S BCM·r1+S BCM) CM AM ·(−S ABM·r1+S ABM) simplify = −S BCM CM AM ·S ABM M = −(−S ABC·r0+S ABC)·r0 (r0−1)·S ABC·r0 simplify = 1 The eliminants MH CH H =S MQS S CQS MG AG G =S MPR S APR S MQS S =S MRQ·S BCM S BMCR S CQS S =−S CMR·S BCQ S BMCR S MRQ Q =S MPR·S CMD −S CMDP S BCQ Q = S CMP·S BCD S CMDP S APR R = −(S ADP·r3) S CMR R =S CMD·r3 S ADP P = −(S ABD·r2) S CMP P =S BCM·r2 S ABD D = −((r1−1)·S ABM) S BCD D = −((r1−1)·S BCM) S ABM M =S ABC·r0 CM AM M = r0−1 r0 S BCM M = −((r0−1)·S ABC) 6.5 Circles 409 6.5 Circles 6.5.1 Chords, Secants, and Tangents Example 6.307 (The Secant Theorem) (0.016, 1, 8) The product of the distances of a given point, from any two points which are collinear with the given point and lie on the circle, is a constant. This constant is called the power of the point with respect to the circle. A O B C D E Figure 6-307 Constructive description ( (circle A B C D) (inter E (l B D) (l A C)) (eq-product A E C E B E D E) ). The machine proof PAEC PBED E = (−PACA·S BCD·S ABD)·S 2 ABCD (−PBDB·S ACD·S ABC)·S 2 ABCD simplify = PACA·S BCD·S ABD PBDB·S ACD·S ABC = (2 f AC 2)·(−g CD·g BD· f BC)·(−g BD·g AD·f AB) (2g BD 2)·(−g CD·g AD· f AC)·(−f BC· f AC·f AB) simplify = 1 The eliminants PBED E =−PBDB·S ACD·S ABC S 2 ABCD PAEC E = −PACA·S BCD·S ABD S 2 ABCD S ABC= f BC· f AC·f AB (−2)·d S ACD= g CD·g AD· f AC (−2)·d PBDB=2(g BD 2) S ABD= g BD·g AD·f AB (−2)·d S BCD= g CD·g BD· f BC (−2)·d PACA=2( f AC 2) Example 6.308 (0.050, 4, 3) The power of a point with respect to a circle is equal, both in magnitude and in sign, to the square of the distance of the point from the center of the circle diminished by the square of the radius of the circle. Constructive description ( (points A B) (on O (b A B)) (on E (l A B)) ( 1 2 PAEB = OE 2−OA 2) ) The eliminants POEO E =PBOB· AE AB −PAOA· AE AB +PAOA+PABA· AE AB 2−PABA· AE AB PAEB E =( AE AB −1)·PABA· AE AB PBOB O =PAOA A B O E Figure 6-308 The machine proof PAEB POEO−PAOA E = PABA· AE AB 2 −PABA· AE AB PBOB· AE AB −PAOA· AE AB +PABA· AE AB 2−PABA· AE AB simplify = ( AE AB −1)·PABA PBOB−PAOA+PABA· AE AB −PABA O = PABA· AE AB −PABA PABA· AE AB −PABA simplify = 1 410 Chapter 6. Topics from Geometry Example 6.309 (The Tangent Theorem) (0.017 3 4) The product of the distances of a given point, from any two points which are collinear with the given point and lie on the circle, is equal to the square of the tangent line from the given point to the circle. A T O P B Figure 6-309 Constructive description ((points A T) (on O (b A T)) (tratio P T O r) (inter B (l A P) (cir O A)) (eq-product P T P T P A P B) ) The machine proof PTPT PAPB B = PTPT·PAPA POPO·PAPA−PAPA·PAOA simplify = PTPT POPO−PAOA P = PTOT·r2 PTOT·r2+PTOT−PAOA O = PAOA·r2 PAOA·r2 simplify = 1 The eliminants PAPB B =POPO−PAOA POPO P =(r2+1)·PTOT PTPT P =PTOT·(r)2 PTOT O =PAOA Example 6.310 (0.066, 3, 5) From the ends D and C of a diameter of circle (O) perpendic-ulars are drawn to chord AB. Let E and F be the feet of the perpendiculars. Show that OE = OF. A B M O C D E F Figure 6-310 Constructive description ( (points A B D) (circumcenter O A B D) (lratio C O D −1) (foot E C A B) (foot F D A B) (midpoint M A B) (midpoint M E F) ) The machine proof −EM FM M = AE AB −1 2 −AF AB + 1 2 F = −( AE AB −1 2 )·PABA PBAD−1 2 PABA E = −(PBAC−1 2 PABA)·PABA (PBAD−1 2 PABA)·PABA simplify = −(PBAC−1 2 PABA) PBAD−1 2 PABA C = −(2PBAO−PBAD−1 2 PABA) PBAD−1 2 PABA O = (−2)·(−PBAD+ 1 2 PABA) (PBAD−1 2 PABA)·(2) simplify = 1 The eliminants EM FM M = AE AB −1 2 AF AB −1 2 AF AB F = PBAD PABA AE AB E = PBAC PABA PBAC C =2(PBAO−1 2 PBAD) PBAO O =1 2(PABA) Example 6.311 Let A, B, C, D, E and P be six cyclic points. From P perpendicular lines are drawn to AB, BC, CD, DE, and EA respectively. Let the foot be L, M, N,, K, and S . Show that AL LB · BM MC · CN ND · DK KE · ES S A = −1. 6.5 Circles 411 A O B C D P E L M N K S Figure 6-311 Constructive description ((circle A B C D E P) (foot L P A B) (foot M P B C) (foot N P C D) (foot K P D E) (foot S P E A) ( AL LB BM MC CN ND DK KE ES S A = −1) ) The machine proof ES AS · DK EK · CN DN · BM CM · AL BL S = PAEP −PEAP · DK EK · CN DN · BM CM · AL BL K = −PAEP·PEDP PEAP·(−PDEP) · CN DN · BM CM · AL BL N = PAEP·PEDP·PDCP PEAP·PDEP·(−PCDP) · BM CM · AL BL M = −PAEP·PEDP·PDCP·PCBP PEAP·PDEP·PCDP·(−PBCP) · AL BL L = PAEP·PEDP·PDCP·PCBP·PBAP PEAP·PDEP·PCDP·PBCP·(−PABP) co−cir = −(−2 f EP· f AE·cos(AP))·(2g DP·g DE·cos(EP)) (2f AP· f AE·cos(EP))·(−2 f EP·g DE·cos(DP)) (2 f CP·g CD·cos(DP))·(2f BP· f BC·cos(CP))·(2f AP·f AB·cos(BP)) (−2g DP·g CD·cos(CP))·(−2 f CP· f BC·cos(BP))·(−2f BP·f AB·cos(AP)) simplify = 1 The eliminants ES AS S = PAEP −PEAP, DK EK K = PEDP −PDEP CN DN N = PDCP −PCDP , BM CM M = PCBP −PBCP AL BL L = PBAP −PABP PABP= −2(f BP·f AB·cos(AP)) PBCP= −2( f CP· f BC·cos(BP)) PCDP= −2(g DP·g CD·cos(CP)) PDEP= −2( f EP·g DE·cos(DP)) PEAP=2(f AP· f AE·cos(EP)) PBAP=2(f AP·f AB·cos(BP)) PCBP=2(f BP· f BC·cos(CP)) PDCP=2( f CP·g CD·cos(DP)) PEDP=2(g DP·g DE·cos(EP)) PAEP= −2( f EP· f AE·cos(AP)) Example 6.312 (0.416, 4, 22) If ABCD is a rectangle inscribed in a circle, center O, and if PX, PX1, PY PY1 are the perpendiculars from any point P upon the sides AB, CD, AD, BC, prove that PX · PX1 + PY · PY1 is equal to the power of P with respect to the circle O. A B P D C O X 1 X Y 1 Y Figure 6-312 Constructive description ( (points A B P) (on D (t A A B)) (inter C (p B A D) (p D A B)) (inter O (l B D) (l A C)) (foot X P A B) (inter X1 (l C D) (l P X)) (foot Y P A D) (inter Y1 (l B C) (l Y P)) (OP 2−OA 2 = 1 2 PXPX1+ 1 2 PYPY1) ) Example 6.313 (0.150, 3, 15) Let D be a point on the side CB of a right triangle ABC such that the circle (O) with diameter CD touches the hypotenuse AB at E. Let F = AC ∩DE. Show that AF = AE. 412 Chapter 6. Topics from Geometry F B A E O D C Figure 6-313 Constructive description ( (points C E) (on D (t E E C)) (midpoint O D C) (inter A (t E E O) (t C C O)) (inter B (l D C) (l A E)) (inter F (l D E) (l A C)) (eqdistance A F A E) ) The eliminants PAFA F = PCAC·(S EDA)2 (S CEAD)2 S CEAD A = PDEO·PECO−16S CEO·S CED (−16)·S CEO PEAE A = PEOE·(PECO)2 (16)·(S CEO)2 S EDA A = PDEO·PECO (16)·S CEO PCAC A = PCOC·(PCEO)2 (16)·(S CEO)2 S CEO O =1 2(S CED) PECO O =1 2(PECD) PEOE O = 1 4(2PEDE−PCDC+2PCEC) PDEO O =1 2(PEDE) PCEO O =1 2(PCEC) PCOC O =1 4(PCDC) S CED D = −1 4(PCEC· DE EC) PECD D =PCEC PEDE D =PCEC·( DE EC )2 PCDC D =( DE EC 2 +1)·PCEC The machine proof PAFA PEAE F = PCAC·S 2 EDA PEAE·S 2 CEAD A = PCOC·P2 CEO·((−1 4 PDEO·PECO))2·((−4S CEO))2·(16S 2 CEO) PEOE·P2 ECO·( 1 4 PDEO·PECO−4S CEO·S CED)2·((−4S CEO))2·(16S 2 CEO) simplify = PCOC·(PCEO)2·(PDEO)2 PEOE·(PDEO·PECO−16S CEO·S CED)2 O = ( 1 4 PCDC)·(( 1 2 PCEC))2·(( 1 2 PEDE))2 ( 1 2 PEDE−1 4 PCDC+ 1 2 PCEC)·( 1 4 PEDE·PECD−8S 2 CED)2 D = (PCEC· DE EC 2+PCEC)·(PCEC)2·(PCEC· DE EC 2)2 (PCEC· DE EC 2 +PCEC)·((−P2 CEC· DE EC 2 ))2 simplify = 1 Example 6.314 (3.250, 21, 23) Show that the sum of the powers, with respect to the circum-circle of a triangle, of the symmetries of the orthocenter with respect to the vertices of the triangle is equal to the sum of the squares of the sides of the triangle. A B C O D F E H 1 A 1 B 1 C Figure 6-314 Constructive description ( (points A B C) (circumcenter O A B C) (foot D A B C) (foot F C A B) (foot E B A C) 6.5 Circles 413 (inter H (l B E) (l A D)) (lratio A1 A H −1) (lratio B1 B H −1) (lratio C1 C H −1) (A1O 2−OB 2+B1O 2−OB 2+C1O 2−OB 2 = AB 2+BC 2+CA 2) ) Example 6.315 (0.700, 17, 17) Two unequal circles are tangent internally at A. The tangent to the smaller circle at a point B meets the larger circle in C, D. Show that AB bisects the angle CAD. A 1 O 2 O B C D Figure 6-315 Constructive description ( (points A B C) (inter O1 (t B B C) (b A B)) (inter O2 (l A O1) (b A C)) (inter D (l B C) (cir O2 C)) (eqangle C A B B A D) ) Example 6.316 (1.850, 65, 19) If G is the centroid of a triangle ABC, show that the powers of the vertices A, B, C for the circles GBC, GCA, GAB, respectively, are equal. 2 O 1 O G E D A B C Figure 6-316 Constructive description ( (points A B C) (centroid G A B C) (circumcenter O1 B C G) (circumcenter O2 A C G) (AO1 2−O1C 2 = BO2 2−O2C 2) ) Example 6.317 (0.083, 4, 10) Let C be a point on a chord AB of circle O. Let D and E be the intersections of perpendicular of OC through C with the two tangents of the circle at A and B, respectively. Show that CE = CD. O A B C D E Figure 6-317 Constructive description ((points A B) (on O (b A B)) (lratio C A B r) (tratio X B O s) (inter E (l B X) (t C C O)) (inter D (l C E) (t A A O)) (midpoint C D E) ) 414 Chapter 6. Topics from Geometry The machine proof −CD CE D = −POAC −POCAE E = POAC·PCBOX −PCBOX·POAC−POCX·PBAO+POAX·PBCO X = POAC·(4S BOC·s) −(4POAC·S BOC·s+4PBCO·S ABO·s−4PBAO·S BOC·s) simplify = −POAC·S BOC POAC·S BOC+PBCO·S ABO−PBAO·S BOC C = −PBAO·r·(−S ABO·r+S ABO) −PBAO·S ABO·r2+PBAO·S ABO·r+PABA·S ABO·r2−PABA·S ABO·r simplify = −PBAO PBAO−PABA py = −(−PBOB+PAOA+PABA)·(2) (−PBOB+PAOA−PABA)·(2) O = PABA PABA simplify = 1 The eliminants CD CE D = POAC −POCAE POCAE E = −(PCBOX ·POAC +POCX ·PBAO−POAX ·PBCO) PCBOX POAX X =PBAO−4S ABO·s POCX X =PBCO−4S BOC·s PCBOX X =4(S BOC·s) PBCO C = −((r−1)·(PBAO−PABA·r)) S BOC C = −((r−1)·S ABO) POAC C =PBAO·r, PBOB O =PAOA PBAO= −1 2(PBOB−PAOA−PABA) Example 6.318 (0.083, 5, 14) Let G be a point on the circle (O) with diameter BC, A be the midpoint of the arc BG. AD ⊥BC. E = AD ∩BG and F = AC ∩BG. Show that AE = BE(= EF). F E D A M G C B O Figure 6-318 Constructive description ( (points G B) (midpoint M B G) (tratio A M B r) (inter O (l A M) (b A B)) (lratio C O B −1) (foot D A B C) (inter E (l A D) (l B G)) (inter F (l A C) (l B G)) (eqdistance A E B E) ) Example 6.319 (0.050, 2, 6) Let D be the intersection of one of the bisectors of ∠A of triangle ABC with side BC, E be the intersection of AD with the circumcircle of ABC. Show that AB · AC = AD · AE. 6.5 Circles 415 B C A D O E Figure 6-319 Constructive description ( (circle A B C) (circumcenter O A B C) (lratio D B C f AB f AB+ f AC ) (inter E (l A D) (cir O A)) (PDAAE = 2f AB· f AC) ) The machine proof PDAE (2)· f AC·f AB E = 2POAD·PADA (2)· f AC·f AB·PADA simplify = POAD f AC·f AB D = f AC·PBAO+f AB·PCAO f AC·f AB·( f AC+f AB) O = 2 f AC·PABA+2f AB·PACA f AC·f AB·( f AC+f AB)·(2)2 co−cir = 2 f AC 2·f AB+2 f AC·f AB 2 (2)· f AC·f AB·( f AC+f AB) simplify = 1 The eliminants PDAE E =(2)·POAD POAD D = f AC·PBAO+f AB·PCAO f AC+f AB PCAO O =1 2(PACA) PBAO O =1 2(PABA) PACA=2( f AC 2) PABA=2(f AB 2) Example 6.320 (0.067, 5, 12) Let ABC be a triangle. Through A a line is drawn tangent to the circle with diameter BC at D. Let E be a point on AB such that AD = AE. The perpendicular to AB at E meets AC at F. Show that AE/AB = AC/AF. B C A O D E F Figure 6-320 Constructive description ( (points E D) (on A (b E D)) (on B (l A E)) (tratio OO D A 1) (inter O (l OO D) (b B D)) (lratio C O B −1) (inter F (l A C) (t E E A)) (eq-product A B A C A E A F) ) The eliminants AF AC F = PEAE PEAC PEAC C =2PEAO−PEAB PEAO O = 2PDBOO·PEAD−PDBD·PEAOO−PDBD·PEAD (−2)·PBDOO PEAOO OO = PEAD+4S EDA PDBOO OO = PDBD−4S DAB PBDOO OO = 4(S DAB) S DAB B =S EDA· AB AE PEAB B =PEAE· AB AE PDBD B = −(PDAD· AB AE −PDAD−PEAE· AB AE 2+PEAE· AB AE −PEDE· AB AE) AB EA B = −( AB AE) PEAD=1 2(PDAD+PEAE−PEDE) PDAD A =PEAE The machine proof (−AB EA)/( AF AC) F = PEAC PEAE · −AB EA C = −(2PEAO−PEAB) PEAE · AB EA O = −(PBDOO·PEAB+2PDBOO·PEAD−PDBD·PEAOO−PDBD·PEAD) PEAE·(−PBDOO) · AB EA 416 Chapter 6. Topics from Geometry OO = −4PDBD·S EDA+4PEAB·S DAB−8PEAD·S DAB PEAE·(4S DAB) · AB EA B = −AB AE ·(−PDAD·S EDA· AB AE +PDAD·S EDA+2PEAD·S EDA· AB AE −PEAE·S EDA· AB AE +PEDE·S EDA· AB AE ) PEAE·S EDA· AB AE ·(−1) simplify = −(PDAD· AB AE −PDAD−2PEAD· AB AE +PEAE· AB AE −PEDE· AB AE ) PEAE py = −(−2PDAD) PEAE·(2) A = PEAE PEAE simplify = 1 Example 6.321 (0.600, 2, 40) The cross ratio of four points on a circles with respect to any points on the circle is constant. A B C O D E G F 1 E 1 G 1 F Figure 6-321 Constructive description (circle A E E1 B C D) (inter G (l A B) (l C E)) (inter F (l D E) (l A B)) (inter G1 (l E1 C) (l A B)) (inter F1 (l E1 D) (l A B)) ( AF BF · BG AG = AF1 BF1 · BG1 AG1 ) ) The machine proof BG AG · AF BF BG1 AG1 · AF1 BF1 F1 = −S E1BD BG1 AG1 ·S AE1D · BG AG · AF BF G1 = −S E1BD·S AE1C (−S E1BC)·S AE1D · BG AG · AF BF F = (−S AED)·S E1BD·S AE1C S E1BC·S AE1D·S EBD · BG AG G = −S EBC·S AED·S E1BD·S AE1C S E1BC·S AE1D·S EBD·(−S AEC) co−cir = (−f BC·g EC· f EB)·(−g ED·g AD· f AE)·(−g BD· g E1D· g E1B)·(−g E1C· f AC· g AE1)·((2d))4 (−f BC· g E1C· g E1B)·(−g E1D·g AD· g AE1)·(−g BD·g ED· f EB)·(−g EC· f AC· f AE)·((2d))4 simplify = 1 The eliminants AF1 BF1 F1 = S AE1D −S E1BD BG1 AG1 G1 = −S E1BC S AE1C AF BF F =−S AED S EBD BG AG G = S EBC −S AEC S AEC= g EC· f AC· f AE (−2)·d S EBD= g BD·g ED· f EB (−2)·d S AE1D= g E1D·g AD· g AE1 (−2)·d S E1BC= f BC· g E1C· g E1B (−2)·d S AE1C= g E1C· f AC· g AE1 (−2)·d S E1BD= g BD· g E1D· g E1B (−2)·d S AED= g ED·g AD· f AE (−2)·d S EBC= f BC·g EC· f EB (−2)·d Example 6.322 (0.083, 1, 17) The tangent to a circle at the point C meets the diameter AB, produced, in T; Prove that the other tangent from T to the circle is divided harmonically by CA, CB, CT and its point of contact. 6.5 Circles 417 1 B 1 A T D E C B O A Figure 6-322 Constructive description ((points A C) (on B (t C C A)) (midpoint O A B) (foot E C A O) (lratio D E C −1) (inter T (l A O) (t C C O)) (inter A1 (l T D) (l C A)) (inter B1 (l T D) (l C B)) (harmonic T D A1 B1) ) The machine proof (−TA1 DA1 )/( T B1 DB1) B1 = S CBD S CBT · −TA1 DA1 A1 = −(−S ACT )·S CBD S CBT ·(−S ACD) T = −PACO·S ACO·S CBD·(−PAOC) (−PBCO·S ACO)·S ACD·(−PAOC) simplify = PACO·S CBD PBCO·S ACD D = PACO·(2S CBE) PBCO·(2S ACE) E = PACO·PABC·S ACB·PAOA PBCO·PCAO·S ACO·PABA O = ( 1 2 PACA)·PABC·S ACB·( 1 4 PABA) ( 1 2 PCBC)·( 1 2 PCAB)·( 1 2 S ACB)·PABA simplify = PACA·PABC PCBC·PCAB py = PACA·(PCBC+PABA−PACA)·(2) PCBC·(−PCBC+PABA+PACA)·(2) B = −2PCBC·PACA −2PCBC·PACA simplify = 1 The eliminants T B1 DB1 B1 = S CBT S CBD TA1 DA1 A1 = S ACT S ACD S CBT T = PBCO·S ACO PAOC S ACT T = PACO·S ACO −PAOC S ACD D =2(S ACE) S CBD D =2(S CBE) S ACE E = PCAO·S ACO PAOA S CBE E = PABC·S ACB PABA S ACO O =1 2(S ACB) PCAO O =1 2(PCAB) PBCO O =1 2(PCBC) PAOA O =1 4(PABA) PACO O =1 2(PACA) PCAB=(PCBC−PABA−PACA) −2 PABC=1 2(PCBC+PABA−PACA) PABA B =PCBC+PACA Example 6.323 (0.650, 10, 19) Let R be a point on the circle with diameter AB. At a point P of AB a perpendicular is drawn meeting BR at N, AR at M, and meeting the circle at Q. Show that PQ2 = PM · PN. O A B P Q R M N Figure 6-323 Constructive description ( (points A Q R) (circumcenter O A Q R) (lratio B O A −1) (foot P Q A O) (inter M (l P Q) (l A R)) (inter N (l P Q) (l R B)) (eq-product P M P N P Q P Q) ) Example 6.324 (0.066, 3, 12) Through point F on the circle with diameter AB a tangent to the circle is drawn meeting the two lines, perpendicular to AB at A and B, at D and E. Show the OA2 = DF · EF. 418 Chapter 6. Topics from Geometry O A B F D E Figure 6-324 Constructive description ( (points A F) (on B (t F F A)) (midpoint O A B) (on M (t F F O)) (inter D (l F M) (t A A B)) (inter E (l F M) (t B B A)) (eq-product D F F E O A O A) ) The machine proof −PDFE PAOA E = −PMFD·PABF PAOA·PBFAM D = −PFMF·PFAB·PABF PAOA·PBFAM·(−PBFAM) M = PFOF· MF FO 2·PFAB·PABF PAOA·((−4S AFB· MF FO ))2 simplify = PFOF·PFAB·PABF (16)·PAOA·(S AFB)2 O = ( 1 2 PFBF −1 4 PABA+ 1 2 PAFA)·PFAB·PABF (16)·( 1 4 PABA)·(S AFB)2 coor = (2u2 2+2u2 1)·(2u2 1)·(2u2 2)·((2))2 (16)·(2u2 2+2u2 1)·((−u2·u1))2 simplify = 1 The eliminants PDFE E = PMFD·PABF PBFAM PMFD D = PFMF·PFAB −PBFAM PBFAM M = −4(S AFB· MF FO) PFMF M =PFOF·( MF FO )2 PAOA O =1 4(PABA) PFOF O = 1 4 (2PFBF−PABA+2PAFA) S AFB= −1 2(u2·u1) PABF=2((u2)2) PFAB=2((u1)2) PAFA=2((u1)2) PABA=2(u2 2+u2 1) PFBF=2((u2)2) In the above machine proof, F = (0, 0), A = (u1, 0), B = (0, u2). Example 6.325 (0.066, 2, 12) Let AD be the altitude on the hypotenuse BC of right triangle ABC. A circle passing through C and D meets AC at E. BE meets the circle at another point F. Show that AF ⊥BE. A B C D M O E F Figure 6-325 Constructive description ( (points A B) (tratio C A B H) (foot D A B C) (midpoint M C D) (tratio O M C I) (inter E (l C A) (cir O C)) (inter F (l E B) (cir O E)) (perpendicular A F B E) ) The machine proof PABE PEBF F = PABE·PBEB −POEO·PBEB+PBEB·PBOB simplify = PABE −(POEO−PBOB) E = PABC −(PCOC−PBOB) O = −PABC PCMC−PBMB M = −PABC −BC CD 2·PCDC−BC CD ·PCDC simplify = PABC ( BC CD +1)· BC CD ·PCDC D = PABC·PBCB·(PACB)2 (−PBCB+PACB)·(−PBCB)·P2 ACB simplify = PABC PBCB−PACB C = PABA PABA simplify = 1 The eliminants PEBF F =−(POEO−PBOB) POEO E =PCOC PABE E =PABC PBOB O =PCMC·I2+PBMB PCOC O =(I2+1)·PCMC PBMB M = 1 4((2 BC CD +1)2·PCDC) PCMC M = 1 4(PCDC) PCDC D =(PACB)2 PBCB BC CD D = −PBCB PACB PACB C =PABA·(r)2 PBCB C =(r2+1)·PABA PABC C =PABA 6.5 Circles 419 Example 6.326 (0.067, 5, 9) Let M be the midpoint of the arc AB of circle (O), D be the midpoint of AB. The perpendicular through M is drawn to the tangent of the circle at A meeting that tangent at E. Show ME = MD. E M O B A D Figure 6-326 Constructive description ( (points A D) (on M (t D D A)) (lratio B D A −1) (inter O (l M D) (b A M)) (inter E (p M O A) (t A O A)) (perp-biesct M D E) ) Example 6.327 (0.200, 4, 11) The circle with the altitude AD of triangle ABC as a diameter meets AB and AC at E and F, respectively. Show that B, C, E and F are on the same circle. Constructive description ( (points A B C) (foot D A B C) (midpoint O A D) (inter E (l A B) (cir O A)) (inter F (l A C) (cir O A)) (cocircle E F C B) ) The eliminants S BCF F =(PCOC−PAOA)·S ABC PACA PBFC F =−(2PCAO−PBAC)·(PCOC−PAOA) PACA PBEC E = −(2PBAO−PBAC)·(PBOB−PAOA) PABA S BCE E = (PBOB−PAOA)·S ABC PABA PBAO O =1 2(PBAD) PCAO O =1 2(PCAD) PBAD D = PBAC·PABC+PACB·PABA PBCB PCAD D = PBAC·PACB+PACA·PABC PBCB PABC=1 2(PBCB−PACA+PABA) PACB=1 2(PBCB+PACA−PABA) PBAC= −1 2(PBCB−PACA−PABA) B C A D O E F Figure 6-327 The machine proof S BCE·PBFC S BCF·PBEC F = S BCE·(−2PCOC·PCAO+PCOC·PBAC+2PCAO·PAOA−PBAC·PAOA)·PACA (PCOC·S ABC−PAOA·S ABC)·PBEC·PACA simplify = −S BCE·(2PCAO−PBAC) S ABC·PBEC E = −(PBOB·S ABC−PAOA·S ABC)·(2PCAO−PBAC)·PABA S ABC·(−2PBOB·PBAO+PBOB·PBAC+2PBAO·PAOA−PBAC·PAOA)·PABA simplify = 2PCAO−PBAC 2PBAO−PBAC 420 Chapter 6. Topics from Geometry O = PCAD−PBAC PBAD−PBAC D = (−PBCB·PBAC+PBAC·PACB+PACA·PABC)·PBCB (−PBCB·PBAC+PBAC·PABC+PACB·PABA)·PBCB simplify = PBCB·PBAC−PBAC·PACB−PACA·PABC PBCB·PBAC−PBAC·PABC−PACB·PABA py = (−2P2 BCB+2P2 ACA−4PACA·PABA+2P2 ABA)·((2))3 (−2P2 BCB+2P2 ACA−4PACA·PABA+2P2 ABA)·((2))3 simplify = 1 Example 6.328 (0.033, 1, 11) Let A, B, C, D be four points on circle (O). E = CD ∩AB. CB meets the line passing through E and parallel to AD at F. GF is tangent to circle (O) at G. Show that FG = FE. By Example 6.309, we only need to prove the following statement. G F E D O C B A Figure 6-328 Constructive description ( (circle A B D C) (inter E (l A B) (l C D)) (inter F (l B C) (p E A D)) (eq-product F B F C E F E F) ) The machine proof PBFC PEFE F = PBCB·S BDE·S ACE·S 2 ABDC PADA·S 2 BCE·S 2 ABDC simplify = PBCB·S BDE·S ACE PADA·(S BCE)2 E = PBCB·(−S BDC·S ABD)·(−S ADC·S ABC)·(S ADBC)2 PADA·((−S BDC·S ABC))2·(S ADBC)2 simplify = PBCB·S ABD·S ADC PADA·S BDC·S ABC co−cir = (2 f BC 2)·(−g BD·g AD·f AB)·(−g DC· f AC·g AD)·((2d))2 (2g AD 2)·(−g DC· f BC·g BD)·(−f BC· f AC·f AB)·((2d))2 simplify = 1 The eliminants PEFE F = PADA·(S BCE)2 (S ABDC)2 PBFC F = PBCB·S BDE·S ACE (S ABDC)2 S BCE E = −S BDC·S ABC S ADBC S ACE E = −S ADC·S ABC S ADBC S BDE E =−S BDC·S ABD S ADBC S ABC= f BC· f AC·f AB (−2)·d S BDC= g DC· f BC·g BD (−2)·d PADA=2((g AD)2) S ADC= g DC· f AC·g AD (−2)·d S ABD= g BD·g AD·f AB (−2)·d PBCB=2(( f BC)2) Example 6.329 (1.783 25 21) The bisector of triangle ABC at vertex C bisects the arc AB of the circumcircle of triangle ABC. A B C O M N Figure 6-329 Constructive description ( (points A B C) (circumcenter O A B C) 6.5 Circles 421 (midpoint M A B) (inter N (l O M) (cir O A)) (eqangle A C N N C B) ) Example 6.330 (0.066, 2, 6) Let PT and PB be two tangents to a circle, AB the diameter through B, and TH the perpendicular from T to AB. Then AP bisects TH. I H A T O P B Figure 6-330 Constructive description ( (points B P) (on O (t B B P)) (inter T (cir O B) (cir P B)) (lratio A O B −1) (inter H (l B O) (p T B P)) (inter I (l T H) (l A P)) (midpoint I T H) ) The machine proof −TI HI I = −S PTA −S PAH H = S PTA·(−S BPO) −S BTPA·S BPO simplify = S PTA S BTPA A = −2S POT −S BPT −S BPT +2S BPO T = (2PPOP·S BPO−2PBPO·S BPO)·PPOP (−2PPOP·S BPO+2PBPO·S BPO)·PPOP·(−1) simplify = 1 The eliminants TI HI I = S PTA −S PAH S PAH H =S BTPA S BTPA A = −(S BPT −2S BPO) S PTA A = −(2S POT +S BPT) S BPT T =(2)·PBPO·S BPO PPOP S POT T = −(S BPO) Example 6.331 5 (0.083 3 15) Let M be a point on line AB. Two squares AMCD and BMEF are drawn on the same side of AB. Let U and V be the center of the squares AMCD and BMEF. Line BC and circle VB meet in N. Show that when point M moving on line AB, the line MN passes through a ﬁxed point. 5This is a problem from the 1959 International Mathematical Olympia. 422 Chapter 6. Topics from Geometry Constructive description ((points A B) (on M (l A B)) (tratio C M A 1) (midpoint U A C) (tratio E M B −1) (midpoint V E B) (midpoint O A B) (tratio R O A −1) (inter N (l B C) (cir V B)) (inter T (l B C) (l M R)) ( BN CN = BT CT ) ) The machine proof ( BN CN)/( BT CT ) T = −S MCR S BMR · BN CN N = −PCBV·S MCR S BMR·(PCBV−1 2 PBCB) R = −PCBV·S MCO ( 1 4 PMABO)·(PCBV−1 2 PBCB) O = (−4)·PCBV·( 1 2 S BMC+ 1 2 S AMC) (−1 2 PABM)·(PCBV−1 2 PBCB) V = (4)·( 1 2 PCBE)·(S BMC+S AMC) PABM·( 1 2 PCBE−1 2 PBCB) E = (4)·(PMBC+4S BMC)·(S BMC+S AMC) PABM·(PMBC−PBCB+4S BMC) C = (4)·(PBMB−PAMB)·(−1 4 PAMB−1 4 PAMA) PABM·(−PAMB−PAMA) simplify = PBMB−PAMB PABM M = −PABA· AM AB +PABA −PABA· AM AB +PABA simplify = 1 The eliminants BT CT T = S BMR −S MCR BN CN N = PCBV ( 1 2 )·(2PCBV−PBCB) S BMR R =1 4(PMABO) S MCR R =S MCO PMABO O = −1 2(PABM) S MCO O = 1 2(S BMC+S AMC) PCBV V =1 2(PCBE) PCBE E =PMBC+4S BMC PBCB C =PBMB+PAMA S AMC C = −1 4(PAMA) S BMC C = −1 4(PAMB) PMBC C =PBMB PABM M = −(( AM AB −1)·PABA) PAMB M =( AM AB −1)·PABA· AM AB PBMB M =( AM AB −1)2·PABA Example 6.332 (0.616, 22, 31) Let M be the midpoint of the hypotenuse of the right triangle ABC. A circle passing through A and M meet AB at E. F is the point on the circle such that EF ∥BC. Show that BC = 2EF. F E O M C B A Figure 6-332 Constructive description ( (points A B) (on C (t A A B)) (midpoint M B C) (midpoint X A M) (on O (t X X A)) (inter E (l A B) (cir O A)) (inter F (p E B C) (cir O E)) ( EF BC = 1/2) ) Example 6.333 (0.767, 14, 21) Let PA tangent to circle (O) at point A. M is the midpoint of PA. C is a point on the circle. PC and MC meet the circle at points E and B, respectively. PB meets the circle at D. Show that ED is parallel to AP. 6.5 Circles 423 E D C B M O P A Figure 6-333 Constructive description ( (points A B C) (circumcenter O A B C) (inter M (l B C) (t A A O)) (lratio P M A −1) (inter D (l B P) (cir O B)) (inter E (l C P) (cir O C)) (parallel E D A M) ) Example 6.334 (0.366, 6, 23) If P is any point on a semicircle, diameter AB, and BC, CD are two equal arcs, then if E = CA ∩PB, F = AD ∩PC, prove that AD is perpendicular to EF. O A B C D P E F Figure 6-334 Constructive description ( (points A P C) (circumcenter O A P C) (lratio B O A −1) (inter D (cir C B) (cir O B)) (inter E (l P B) (l C A)) (inter F (l P C) (l A D)) (perpendicular E F A D) ) Example 6.335 (1.900, 49, 18) From the point S the two tangents S A, S B and the secant S PQ are drawn to the same circle. Prove that AP/AQ = BP/BQ. Q P B M S A D O Figure 6-335 Constructive description ( (points A B P) (circumcenter O A B P) (midpoint M A B) (inter S (l O M) (t A A O)) (inter Q (l S P) (cir O P)) (eq-product A Q B P A P B Q) ) Example 6.336 (0.466, 6, 19) Show that the lines joining a point of a circle to the ends of a chord divide harmonically the diameter perpendicular to the chord. A D P O B F E Q S Figure 6-336 Constructive description ( (points A D P) (circumcenter O A D P) (lratio B O A −1) (foot F D O A) (lratio E F D −1) (inter Q (l A O) (l P E)) (inter S (l A O) (l P D)) (harmonic A B Q S ) ) 424 Chapter 6. Topics from Geometry Example 6.337 (0.133, 3, 16) From a point A two lines are drawn tangent to circle (O) at B and C. From a point P on the circle perpendiculars are drawn to BC, AB, and AC. Let D, F, E be the feet. Show that PD2 = PE · PF. B A O C P D E F Figure 6-337 Constructive description ( (points B C P) (circumcenter O B C P) (inter A (t B B O) (t C C O)) (foot D P B C) (foot E P A C) (foot F P A B) (eq-product P D P D P E P F) ) Example 6.338 (1.113, 14, 24) Through P a tangent PE and a secant PAB of circle (O) are drawn. The bisector of angle APE meets AE and BE at C and D. Show that EC = ED. E P O A B C D Figure 6-338 Constructive description ( (points P E C) (on Y (a P C E P C)) (inter A (l E C) (l P Y)) (midpoint M A E) (tratio Z E P H) (inter O (t M M E) (l E Z)) (inter B (l P A) (cir O A)) (inter D (l P C) (l B E)) (eqdistance E D E C) ) Example 6.339 (0.216, 3, 18) Let N be the traces of the internal bisectors of the triangle ABC on the circumscribed circle (O). Show that the Simson line of N is the external bisector of the medial triangle of ABC. K N F E D O C B A Figure 6-339 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint D B C) (midpoint E A C) (midpoint F B A) (inter N (l O F) (cir O A)) (foot K N A C) (eqangle E F K K F D) ) 6.5 Circles 425 6.5.2 Intersectional Circles Example 6.340 (0.216, 3, 18) Through the two common points A, B of two circles (O) and (O1) two lines are drawn meeting the circles at C and D, E and F, respectively. Show that CE ∥DF. For a proof of this theorem based on full-angles, see Example 1.115 on page 47. F E D C O B N A M 1 O X Figure 6-340 Constructive description ( (circle B A C E) (midpoint X A B) (tratio O X B H) (foot M O A C) (foot N O B E) (lratio D M A −1) (lratio F N B −1) (S CED = S CEF) ) The machine proof S CED S CEF F = S CED 2S CEN−S BCE D = 2S CEM−S ACE 2S CEN−S BCE N = (2S CEM−S ACE)·PBEB 2PBEO·S BCE−PBEB·S BCE simplify = (2S CEM−S ACE)·PBEB (2PBEO−PBEB)·S BCE M = (2PACO·S ACE−PACA·S ACE)·PBEB (2PBEO−PBEB)·S BCE·PACA simplify = (2PACO−PACA)·S ACE·PBEB (2PBEO−PBEB)·S BCE·PACA O = (2PACX−PACA−8S BCX·r)·S ACE·PBEB (2PBEX−PBEB−8S BEX·r)·S BCE·PACA X = (PBCA+4S BAC·r)·S ACE·PBEB (PBEA+4S BAE·r)·S BCE·PACA co−cir = (−4 f AC· f BC·f BA·r+4 f AC· f BC·cos(BA)·d)·(−g CE· f AE· f AC)·(2 f BE 2)·((2d))2 (−4 f AE· f BE·f BA·r+4 f AE· f BE·cos(BA)·d)·(−g CE· f BE· f BC)·(2 f AC 2)·((2d))2 simplify = 1 The eliminants S CEF F =2S CEN−S BCE S CED D =2S CEM−S ACE S CEN N = PBEO·S BCE PBEB S CEM M = PACO·S ACE PACA PBEO O =PBEX−4S BEX·r PACO O =PACX−4S BCX·r S BEX X = −1 2(S BAE) PBEX X =1 2(PBEA+PBEB) S BCX X = −1 2(S BAC) PACX X =1 2(PACA+PBCA) PACA=2(( f AC)2) S BCE= g CE· f BE· f BC (−2)·d S BAE= f AE· f BE·f BA (−2)·d PBEA=2( f AE· f BE·cos(BA)) PBEB=2(( f BE)2) S ACE= g CE· f AE· f AC (−2)·d S BAC= f AC· f BC·f BA (−2)·d PBCA=2( f AC· f BC·cos(BA)) Example 6.341 (1.583, 39, 25) Let A and B the two common points of two circles (O) and (O1). Through A a line is drawn meeting the circles at C and D respectively. G is the midpoint of CD. Line BG intersects circles (O) and (O1) at E and F, respectively. Show that EG = GF. 426 Chapter 6. Topics from Geometry F E G D C B A 1 O O X Figure 6-341 Constructive description ( (points A B C) (midpoint X A B) (on O1 (t X X A)) (inter O (l X O1) (b A C)) (inter D (l C A) (cir O1 A)) (midpoint G C D) (inter E (l B G) (cir O B)) (inter F (l B G) (cir O1 B)) (midpoint G F E) ) Example 6.342 (0.133, 6, 14) Three circles, centers A, B, C, have a point D in common and intersect two-by-two in the points A1, B1, C1. The common chord DC1 of the ﬁrst two circles meets the third in C2. Let A2, B2 be the analogous points on the other two circles. Prove that the segments A1A2, B1B2, C1C2 are twice as long as the altitudes of the triangle ABC. C B A D F 1 C 2 C Figure 6-342 Constructive description ( (points A B C D) (foot F C A B) (inter C1 (cir B D) (cir A D)) (inter C2 (l D C1) (cir C D)) ( C2C1 CF = 2) ) The eliminants C1C2 CF C2 = −(PCC1C−PCDC) PDC1D · DC1 CF PDC1D C1 = (4)·(PBDB·PBAD−PBAD·PABD+PADA·PABD) PABA DC1 CF C1 = (2)·S ABD S ABC PCC1C C1 = −(PCDC·PABA−2PBDB·PBAD−2PBCB·PBAD +4PBAD·PABD−2PADA·PABD−2PACA·PABD)/PABA S ABC=1 2(y1·x0) S ABD=1 2(y2·x0) PABD=(−2(x2−x0)·x0) PACA=2(y2 1+x2 1) PADA=2(y2 2+x2 2) PBAD=2(x2·x0) PBCB=2(y2 1+x2 1−2x1·x0+x2 0) PBDB=2(y2 2+x2 2−2x2·x0+x2 0) PABA=2(x2 0) PCDC=2(y2 2−2y2·y1+y2 1+x2 2−2x2·x1+x2 1) The machine proof 1 2(−C1C2 CF ) C2 = −(−DC1 CF ·PCC1C+ DC1 CF ·PCDC) (2)·PDC1D simplify = PCC1C−PCDC (2)·PDC1D · DC1 CF C1 = (−2PCDC ·PABA+2PBDB·PBAD+2PBCB·PBAD−4PBAD·PABD+2PADA·PABD+2PACA·PABD)·(2S ABD)·PABA (2)·(4PBDB·PBAD−4PBAD·PABD+4PADA·PABD)·S ABC·PABA 6.5 Circles 427 simplify = −(PCDC·PABA−PBDB·PBAD−PBCB·PBAD+2PBAD·PABD−PADA·PABD−PACA·PABD)·S ABD (2)·(PBDB·PBAD−PBAD·PABD+PADA·PABD)·S ABC coor = −(−8y2·y1·x2 0)·y2·x0·(2) (2)·(4y2 2·x2 0)·y1·x0·(2) simplify = 1 In the above proof, we have A = (0, 0), B = (x0, 0), C = (x1, y1), and D = (x2, y2). Example 6.343 (0.050, 3, 8) Show that an altitude of a triangle is the radical axis of the two circles having for diameters the medians issued from the other two vertices. B C A 1 B 1 C M N Figure 6-343 Constructive description ( (points A B C) (midpoint B1 A C) (midpoint C1 A B) (midpoint M B B1) (midpoint N C C1) (on-radical A M B N C) ) The machine proof −PBMB+PAMA −PCNC+PANA N = −PBMB+PAMA −1 2 PCC1C+ 1 2 PAC1A+ 1 2 PACA M = (2)·( 1 2 PBB1B−1 2 PAB1A−1 2 PABA) PCC1C−PAC1A−PACA C1 = PBB1B−PAB1A−PABA 1 2 PBCB−1 2 PACA−1 2 PABA B1 = (2)·( 1 2 PBCB−1 2 PACA−1 2 PABA) PBCB−PACA−PABA simplify = 1 The eliminants PANA N = −1 4(PCC1C−2PAC1A−2PACA) PCNC N = 1 4(PCC1C) PAMA M = −1 4(PBB1B−2PAB1A−2PABA) PBMB M = 1 4(PBB1B) PAC1A C1 = 1 4(PABA) PCC1C C1 = 1 4(2PBCB+2PACA−PABA) PAB1A B1 = 1 4(PACA) PBB1B B1 = 1 4(2PBCB−PACA+2PABA) Example 6.344 (0.800, 12, 18) Let A and B be the two common points of two circles (O) and (O1). Through B a line is drawn meeting the circles at C and D respectively. Show AC/AD = OA/O1A. X O 1 O A B C D Figure 6-344 Constructive description ( (points A B C) (midpoint X A B) (on O1 (t X X A)) (inter O (l X O1) (b A C)) (inter D (l C B) (cir O1 B)) (eq-product A C O1 A A D O A) ) 428 Chapter 6. Topics from Geometry Example 6.345 (0.783, 16, 25) From a point P on the line joining the two common points A and B of two circles (O) and (O1) two secants PCE and PFD are drawn to the circles respectively. Show that PC · PE = PF · PD. D F C E P B A 1 O O X Figure 6-345 Constructive description ( (points A C F X) (on Y (t X X A)) (inter O (l X Y) (b A C)) (inter O1 (l X Y) (b A F)) (lratio B X A −1) (on P (l A X)) (inter E (l P C) (cir O C)) (inter D (l P F) (cir O1 F)) (eq-product P C P E P D P F) ) Example 6.346 (1.283, 60, 67) If three chords drawn through a point of a circle are taken for diameters of three circles, these circles intersect, in pairs, in three new points, which are collinear. A B C O D 1 M 2 M 3 M E F G Figure 6-346 Constructive description ( (circle A B C D) (midpoint M1 A D) (midpoint M2 B D) (midpoint M3 C D) (inter E (cir M2 D) (cir M1 D)) (inter F (cir M3 D) (cir M1 D)) (inter G (cir M3 D) (cir M2 D)) (collinear E F G) ) Example 6.347 (0.733, 11, 38) If three circles pass through the same point of the circumcircle of the triangle of their centers, these circles intersect, in pairs, in three collinear points. A B C O P 1 A 1 B 1 C Figure 6-347 Constructive description ( (circle A B C P) (inter A1 (cir B P) (cir C P)) (inter B1 (cir C P) (cir A P)) (inter C1 (cir B P) (cir A P)) (collinear A1 B1 C1) ) Example 6.348 (1.083, 24, 66) If three circles having a point in common intersect in pairs in three collinear points, their common point is cyclic with their centers. 6.5 Circles 429 1 A 1 B P 1 C A B C Figure 6-348 Constructive description ( (points A1 B1 P) (on C1 (l B1 A1)) (midpoint M1 P B1) (on T1 (t M1 M1 P) 1) (midpoint M2 P C1) (on T2 (t M2 M2 P) 1) (midpoint M3 P A1) (on T3 (t M3 M3 P) 1) (inter A (l M1 T1) (l M2 T2)) (inter B (l M3 T3) (l M2 T2)) (inter C (l M1 T1) (l M3 T3)) (cocircle A B C P) ) Example 6.349 (0.333, 7, 15) Show that the radical axis of the two circles having for diame-ters the diagonals AC, BD of a trapezoid ABCD passes through the point of intersection E of the nonparallel sides BC, AD. A B C D E 1 N 2 N I Figure 6-349 Constructive description ( (points A B C) (on D (p C A B)) (inter E (l A D) (l B C)) (midpoint N1 A C) (midpoint N2 B D) (on-radical E N1 A N2 B) ) Example 6.350 (0.250, 8, 13) Given two circles (A), (B) intersecting in E, F, show that the chord E1F1 determined in (A) by the lines MEE1, MFF1 joining E, F to any point M of (B) is perpendicular to MB. 1 F 1 E M F E B A D Figure 6-350 Constructive description ( (points E F M) (circumcenter B E F M) (midpoint D E F) (lratio A D B r) (inter E1 (l M E) (cir A E)) (inter F1 (l M F) (cir A F)) (perpendicular E1 F1 M B) ) Example 6.351 (0.667, 4, 44) From the midpoint C of arc AB of a circle, two secants are drawn meeting line AB at F, G, and the circle at D and E. Show that F, D, E, and G are on the same circle. C O A M B D E F G Figure 6-451 Constructive description 430 Chapter 6. Topics from Geometry ( (circle A C D E) (circumcenter O A C D) (foot M A O C) (lratio B M A −1) (inter F (l A M) (l D C)) (inter G (l A M) (l C E)) (cocircle D E F G) ) Example 6.352 (1.067 39 18) From point P two tangent lines PA and PB of a circle are drawn. D is the middle point of segment AB. Through D a secant EF is drawn. Then ∠EPD = ∠FPD. A B E O D P F Figure 6-352 Constructive description ( (points A B E) (circumcenter O A B E) (midpoint D A B) (inter P (l O D) (t A A O)) (inter F (l E D) (cir O E)) (eqangle E P O O P F) ) Example 6.353 (0.100 4 13 ) Let D and E be two points on sides AB and AC of triangle ABC such that DE ∥BC. Show that the circumcircles of triangle ABC and ADE are tangent. A B C O D E N Figure 6-353 Constructive description ( (points A B C) (circumcenter O A B C) (on D (l A B)) (inter E (l A C) (p D B C)) (circumcenter N A D E) (S ANO = 0) ) 6.5.3 The Inversion Deﬁnition Suppose that we have given a circle whose center is O and the radius has the length r , 0. Let P and Q be any two points collinear with O such that OP · OQ = r2. Then P is said to be the inverse of Q with regard to the circle, and the transformation from P to Q is called an inversion. The point O is called the center of inversion, the given circle the circle of inversion, and its radius the radius of inversion. Example 6.354 (0.001, 2, 3) Two inverse points divide the corresponding diameter harmoni-cally. 6.5 Circles 431 B A O P Q Figure 6-354 Constructive description ( (points B A) (midpoint O A B) (lratio P O A r) (lratio Q O A 1 r ) (harmonic A B P Q) ) The machine proof (−AP BP)/( AQ BQ) Q = −BO AO ·r+1 −r+1 · −AP BP P = −(r−1)·( BO AO ·r−1) (r−1)·(−BO AO +r) simplify = BO AO ·r−1 BO AO −r O = (−1 2 r−1 2)·( 1 2 ) (−1 2 r−1 2)·( 1 2 ) simplify = 1 The eliminants AQ BQ Q = r−1 BO AO ·r−1 AP BP P = r−1 −( BO AO −r) BO AO O = −(1) Example 6.355 From a point P outside a given circle, center O, the tangents are drawn to the circle. Show that P is the inverse of the point of the intersection of OP with the chord of contact. A C B O P E Figure 6-355 Constructive description ((points A C) (on B (t C C A)) (midpoint O A B) (inter P (l A O) (t C C O)) (foot E C A O) (inversion O A P E) ) The machine proof (−OP AO)/(−AO OE) E = PAOC PAOA · −OP AO P = −PCOC·PAOC PAOA·(−PAOC) simplify = PCOC PAOA O = 1 2 PCBC−1 4 PABA+ 1 2 PACA 1 4 PABA B = PCBC+PACA PCBC+PACA simplify = 1 The eliminants AO OE E =−PAOA PAOC OP AO P = PCOC −PAOC PAOA O =1 4(PABA) PCOC O =1 4(2PCBC−PABA+2PACA) PABA B =PCBC+PACA Example 6.356 (0.433, 10, 11) Prove that two pairs of inverse points with respect to the same circle are cyclic, or collinear. 432 Chapter 6. Topics from Geometry O A B P Q R S Figure 6-356 Constructive description ( (points A B) (on O (b A B)) (lratio P O A r1) (lratio Q O A 1 r1 ) (lratio R O B r2) (lratio S O B 1 r2 ) (cocircle P Q R S ) ) The eliminants S PQS S =S BPQ r2 PRSR S = PORO·r2 2−PORO·r2+PBRB·r2−PBOB·r2+PBOB r2 2 S PRS S =S OPR·r2−S OPR+S BPR r2 S QRS S =S OQR·r2−S OQR+S BQR r2 S PQR R =S BPQ·r2 PBRB R =(r2−1)2·PBOB PORO R =PBOB·r2 2 S BPR R =(r2−1)·S BOP S OPR R =S BOP·r2 S BQR R =(r2−1)·S BOQ S OQR R =S BOQ·r2 S BPQ Q =−(S BOP·r1−S BOP−S ABP) r1 PPQP Q = POPO·r2 1−POPO·r1+PAPA·r1−PAOA·r1+PAOA r2 1 S BOQ Q =S ABO r1 S ABP P = −((r1−1)·S ABO) S BOP P =S ABO·r1 PAPA P =(r1−1)2·PAOA POPO P =PAOA·r2 1 PBOB O =PAOA The machine proof PPQP·(−S QRS )·S PRS PRSR·S PQS ·(−S PQR) S = PPQP·(−S OQR·r2+S OQR−S BQR)·(S OPR·r2−S OPR+S BPR)·r2·r2 2 (PORO·r2 2−PORO·r2+PBRB·r2−PBOB·r2+PBOB)·S BPQ·(−S PQR)·r2 2 simplify = PPQP·(S OQR·r2−S OQR+S BQR)·(S OPR·r2−S OPR+S BPR)·r2 (PORO·r2 2−PORO·r2+PBRB·r2−PBOB·r2+PBOB)·S BPQ·S PQR R = PPQP·(S BOQ·r2 2−S BOQ)·(S BOP·r2 2−S BOP)·r2 (PBOB·r4 2−2PBOB·r2 2+PBOB)·S BPQ·S BPQ·r2 simplify = PPQP·S BOQ·S BOP PBOB·S 2 BPQ Q = (POPO·r2 1−POPO·r1+PAPA·r1−PAOA·r1+PAOA)·S ABO·S BOP·r2 1 PBOB·(−S BOP·r1+S BOP+S ABP)2·r1·r2 1 simplify = (POPO·r2 1−POPO·r1+PAPA·r1−PAOA·r1+PAOA)·S ABO·S BOP PBOB·(S BOP·r1−S BOP−S ABP)2·r1 P = (PAOA·r4 1−2PAOA·r2 1+PAOA)·S ABO·S ABO·r1 PBOB·(S ABO·r2 1−S ABO)2·r1 simplify = PAOA PBOB O = PAOA PAOA simplify = 1 Example 6.357 (0.067, 6, 9) The inverse of a line not passing through the center of inversion is a circle through that point. 6.5 Circles 433 Constructive description ( (points O A) (lratio Q O A 1 r ) (tratio R Q O r1) (lratio P O A r) (midpoint U P O) (lratio G O R POAO PORO ) (eqdistance G U U O) ) The eliminants PUGU G = PRUR·POAO+POUO·PORO−POUO·POAO−PORO·POAO+P2 OAO/PORO POUO U =1 4(POPO) PRUR U =1 4(2PRPR−POPO+2PORO) POPO P =POAO·(r)2 PRPR P =PARA·r−PORO·r+PORO+POAO·r2−POAO·r PORO R =(r2 1+1)·POQO PARA R =PAQA+POQO·r2 1 POQO Q = POAO (r)2 PAQA Q =(r−1)2·POAO (r)2 The machine proof PUGU POUO = PRUR·PORO·POAO+POUO·P2 ORO−POUO·PORO·POAO−P2 ORO·POAO+PORO·P2 OAO POUO·P2 ORO simplify = PRUR·POAO+POUO·PORO−POUO·POAO−PORO·POAO+P2 OAO POUO·PORO U = 1 2 PRPR·POAO+ 1 4 POPO·PORO−1 2 POPO·POAO−1 2 PORO·POAO+P2 OAO ( 1 4 POPO)·PORO P = 2PARA·POAO·r+PORO·POAO·r2−2PORO·POAO·r−2P2 OAO·r+4P2 OAO POAO·r2·PORO simplify = 2PARA·r+PORO·r2−2PORO·r−2POAO·r+4POAO (r)2·PORO R = 2PAQA·r+POQO·r2 1·r2+POQO·r2−2POQO·r−2POAO·r+4POAO (r)2·(POQO·r2 1+POQO) simplify = 2PAQA·r+POQO·r2 1·r2+POQO·r2−2POQO·r−2POAO·r+4POAO (r)2·(r2 1+1)·POQO Q = (POAO·r2 1·r4+POAO·r4)·r2 (r)2·(r2 1+1)·POAO·(r2)2 simplify = 1 Example 6.358 (0.133, 5, 11) If the circle U passes trough two inverse points of circle O. Then the inverse of any point on circle U with regard to circle O is on circle U. In other words, the inverse of circle U with regard to circle O is itself. A O X P Q U E F Figure 6-358 Constructive description ( (points A O X) (lratio P O A r) (lratio Q O A 1 r ) (midpoint MU P Q) (tratio U MU P rU) (inter E (l P X) (cir U P)) (inter F (l E O) (cir U E)) (inversion O A E F) ) 434 Chapter 6. Topics from Geometry The machine proof PAOA PEOF F = PAOA·POEO −PUEU·POEO+POEO·POUO simplify = PAOA −(PUEU−POUO) E = PAOA −(PPUP−POUO) U = −PAOA PPMU P−POMU O MU = −PAOA −OP PQ 2 ·PPQP−OP PQ ·PPQP simplify = PAOA ( OP PQ +1)· OP PQ ·PPQP Q = PAOA·r2·( OP AO ·r+1)2 (−OP AO ·r)·(POPO·r2−POPO·r+PAPA·r−PAOA·r+PAOA) simplify = PAOA·r·( OP AO ·r+1)2 −OP AO ·(POPO·r2−POPO·r+PAPA·r−PAOA·r+PAOA) P = PAOA·r·(r2−1)2·(−1) −r·(PAOA·r4−2PAOA·r2+PAOA)·((−1))2 simplify = 1 The eliminants PEOF F =−(PUEU−POUO) PUEU E =PPUP POUO U =PPMU P·r2 U+POMU O PPUP U =(r2 U+1)·PPMU P PPQP Q = POPO·r2−POPO·r+PAPA ·r−PAOA·r+PAOA (r)2 POMU O MU = 1 4((2 OP PQ +1)2·PPQP) PPMU P MU = 1 4(PPQP) OP PQ Q = −r OP AO ·r+1 · OP AO PAPA P =(r−1)2·PAOA POPO P =PAOA·(r)2, OP AO P = −(r) Example 6.359 (10.817, 267, 29) The inverse of a circle not passing through the center of inversion is a circle. O A X P Q U E F I R G Figure 6-359 Constructive description ( (points O A X) (lratio P O A r1) (lratio Q O A r2) (midpoint U P Q) (lratio E O A 1 r1 ) (lratio F O A 1 r2 ) (midpoint I E F) (inter R (l P X) (cir U P)) (lratio G O R POAO PORO ) (eqdistance G I I E) ) Example 6.360 (0.033, 4, 24) The harmonic conjugate of a ﬁxed point with respect to a vari-able pair of points which lie on a given circle and are collinear with the ﬁxed point, de-scribes a straight line. This line is called the polar of the ﬁxed point with the circle, and the ﬁxed point is said to be the pole of the line. O A E F P Q M Figure 6-360 Constructive description ( (circle E F A) (circumcenter O E F A) (inter P (l O A) (l E F)) (lratio Q O A OA OP) (inter M (l E F) (t Q Q A)) (harmonic E F M P) ) 6.5 Circles 435 Example 6.361 (0.216, 5, 6) Let P and Q be inverse points with regard to circle OA. Then for any point C on circle O, we have CP · AQ = CQ · AP. A B C O P Q Figure 6-361 Constructive description ( (points A B C) (circumcenter O A B C) (lratio P O A r) (lratio Q O A 1 r ) (eq-product C P A Q C Q A P) ) Example 6.362 (0.400, 8, 16) If the circle (B) passes through the center A of the circle (A), and a diameter of (A) meets the common chord of the two circles in F and the circle (B) again in G, show that the points F, G are inverse for the circle (A). G F D I E C B A Figure 6-362 Constructive description ( (points D E C) (circumcenter A D E C) (midpoint I E C) (inter B (l A I) (b A E)) (inter F (l E C) (l A D)) (inter G (l A D) (cir B A)) (inversion A D G F) ) Example 6.363 (0.466, 6, 15) Show that the two lines joining any point of a circle to the ends of a given chord meet the diameter perpendicular to that chord in two inverse points. Q P D F I E A O Figure 6-363 Constructive description ( (points A E D) (circumcenter O A E D) (foot I E A O) (lratio F I E −1) (inter P (l O A) (l D E)) (inter Q (l O A) (l D F)) (inversion O A P Q) ) Example 6.364 (0.700, 8, 25) TP, TQ are the tangents at the extremities of a chord PQ of a circle. The tangent at any point R of the circle meets PQ in S ; prove that TR is the polar of S . P A O Q R T S D 1 S Figure 6-364 Constructive description ( (points P Q R) (circumcenter O P Q R) 436 Chapter 6. Topics from Geometry (on A (t P P O)) (inter T (l A P) (t Q Q O)) (inter S (l P Q) (t R R O)) (inter S 1 (l T R) (l O S )) (inversion O P S S 1) ) 6.5.4 Orthogonal Circles Deﬁnition. Two circles O1 and O2 with a common point A are said to be orthogonal if O1A⊥O2A. Example 6.365 (0.050, 4, 7) If two circles are orthogonal, any two diametrically opposite points of one circle are conjugate with respect to the other circle. Constructive description ( (points D E) (on O1 (b D E)) (lratio F O1 E −1) (on O (t D D O1)) (foot G F O E) (inversion O D G E) ) The eliminants PEOG G =PEOF PEOF O =PEDF+PDO1D· OD DO1 2−4S DO1F· OD DO1 +4S DEO1· OD DO1 PDOD O =PDO1D·( OD DO1 )2 S DO1F F =S DEO1 PEDF F =2PEDO1−PDED PEDO1= −1 2(PEO1E−PDO1D−PDED) PEO1E O1 = PDO1D O 1 O D E A F G Figure 6-365 The machine proof PDOD PEOG G = PDOD PEOF O = PDO1D· OD DO1 2 PEDF+PDO1D· OD DO1 2−4S DO1F· OD DO1 +4S DEO1 · OD DO1 F = PDO1D·( OD DO1 )2 2PEDO1+PDO1D· OD DO1 2 −PDED py = PDO1D·( OD DO1 )2·(2) −2PEO1E+2PDO1D· OD DO1 2+2PDO1D O1 = −PDO1D· OD DO1 2 −PDO1D· OD DO1 2 simplify = 1 6.5 Circles 437 Example 6.366 (0.016, 1, 3) Show that the two poles of the common chord of two orthogonal circles with respect to these circles coincide with the centers of the given circles. A R 1 O D B 2 O Figure 6-366 Constructive description ( (points A R) (on O1 (b A R)) (foot D A R O1) (lratio B D A −1) (inter O2 (l O1 D) (t A A O1)) (inversion O1 R D O2) ) The machine proof (−O1D RO1 )/(−RO1 O1O2) O2 = PAO1A PAO1R · −O1D RO1 D = −PAO1R·PAO1A PAO1R·(−PRO1R) simplify = PAO1A PRO1R O1 = PAO1A PAO1A simplify = 1 The eliminants RO1 O1O2 O2 = −PAO1R PAO1A O1D RO1 D = PAO1R −PRO1R, PRO1R O1 = PAO1A Example 6.367 (0.001, 1, 5) A circle orthogonal to two given circles has its center on the radical axis of the two circles. Constructive description: ( (points D1 D2) (on O (b D1 D2)) (on O1 (t D1 D1 O)) (on O2 (t D2 D2 O)) (on-radical O O1 D1 O2 D2) ) 2 D 1 D 2 O 1 O O Figure 6-367 The machine proof POO1O−PD1O1D1 POO2O−PD2O2D2 O2 = POO1O−PD1O1D1 PD2OD2 O1 = PD1OD1 PD2OD2 O = PD1OD1 PD1OD1 simplify = 1 The eliminants PD2O2D2 O2 = PD2OD2·( O2D2 D2O )2 POO2O O2 = ( O2D2 D2O 2 +1)·PD2OD2 PD1O1D1 O1 = PD1OD1·( O1D1 D1O )2 POO1O O1 = ( O1D1 D1O 2 +1)·PD1OD1 PD2OD2 O =PD1OD1 Example 6.368 (0.283, 8, 12) If two circles are orthogonal, any two points of one of them collinear with the center of the second circle are inverse for that second circle. F E M D B C A Figure 6-368 Constructive description ( (points M E F) (circumcenter B M E F) 438 Chapter 6. Topics from Geometry (inter A (l E F) (t M M B)) (inversion A M E F) ) Example 6.369 (0.350, 4, 18) The two lines joining the points of intersection of two orthogo-nal circles to a point on one of the circles meet the other circle in two diametrically opposite points. D C A F I E P Q Figure 6-369 Constructive description ( (points C E F) (circumcenter Q C E F) (lratio D Q C −1) (midpoint I E F) (inter P (l Q I) (t E E Q)) (inter A (l E C) (cir P E)) (inter A1 (l F D) (l C E)) ( CA EA = CA1 EA1 ) ) Example 6.370 (0.850, 11, 19) Show that in a triangle ABC the circles on AH and BC as diameters are orthogonal. B C A H M N I Figure 6-370 Constructive description ( (points A B C) (orthocenter H A B C) (midpoint M B C) (midpoint N A H) (inter I (l A B) (cir N A)) (perpendicular M I N I) ) Example 6.371 (1.183, 10, 12) The circle IBC is orthogonal to the circle on IbIc as diameter. B C I A b I c I O M Figure 6-371 Constructive description ( (points B C I) (incenter A I B C) (inter IB (t C C I) (l B I)) (inter IC (t B B I) (l C I)) (circumcenter O B C I) (midpoint M IB IC) (perpendicular M B O B) ) Example 6.372 (0.533, 16, 32) Show that given two perpendicular diameters of two orthog-onal circles, the lines joining an end of one of these diameters to the ends of the other pass through the points common to the two circles. 6.5 Circles 439 A B P D E F G I Figure 6-372 Constructive description ( (points P D I) (circumcenter A P D I) (lratio E A D −1) (midpoint X P I) (inter B (l A X) (t P P A)) (inter G (l I E) (cir B I)) (lratio F B G −1) (perpendicular F G D E) ) Example 6.373 (0.083, 4, 15) Show that if AB is a diameter and M any point of a circle, center O, the two circles AMO, BMO are orthogonal. O A B M I J P Figure 6-373 Constructive description ( (points A M) (on B (t M M A)) (midpoint O A B) (circumcenter I A M O) (circumcenter J B M O) (perpendicular I M M J) ) Example 6.374 (0.783, 11, 23) If the line joining the ends A, B of a diameter AB of a given circle (O) to a given point P meets (O) again in A1, B1, show that the circle PA1B1 is orthogonal to (O). 1 A 1 B A O B P 1 O Figure 6-374 Constructive description ( (points A1 B1 A) (circumcenter O A1 B1 A) (lratio B O A −1) (inter P (l A A1) (l B B1)) (circumcenter O1 A1 B1 P) (perpendicular O1 A1 A1 O) ) 6.5.5 The Simson Line Let D be a point on the circumscribed circle of triangle ABC. From D three perpendiculars are drawn to the three sides BC, AC, and AB of triangle ABC. Let E, F, and G be the three feet respectively. Then E,F and G are collinear. The line EFG is called the Simson line of the point D with respect to the triangle ABC, and D is called the pole of the simson line. For machine proofs of Simson’s theorem see Example 3.79 on page 144, Example 3.106 on page 164 and Example 5.55 on page 248. 440 Chapter 6. Topics from Geometry Example 6.375 (0.933, 3, 37) The Simson line bisects the line joining its pole to the ortho-center of the triangle. A B C O D F G H N Figure 6-375 Constructive description ( (circle A B C D) (orthocenter H A B C) (foot F D A C) (foot G D A B) (inter N (l G F) (l D H)) (midpoint N D H) ) Example 6.376 (0.050, 1, 8) Let D be a point on the circumcircle of triangle ABC. If line DA is parallel to BC, show that the Simson line D(ABC) is parallel to the circumradius OA. G F D O C B A Figure 6-376 Constructive description ( (points A B C) (circumcenter O A B C) (pratio PD A B C 1) (inter D (l A PD) (cir O A)) (foot F D A C) (foot G D A B) (parallel G F O A) ) The machine proof −S AOG −S AOF G = PBAD·S ABO (−S AOF)·PABA F = −PBAD·S ABO·PACA (−PCAD·S ACO)·PABA D = (2POAPD ·PBAPD)·S ABO·PACA·PAPDA (2POAPD·PCAPD )·S ACO·PABA·PAPDA simplify = PBAPD·S ABO·PACA PCAPD ·S ACO·PABA PD = PABC·S ABO·PACA PACB·S ACO·PABA·(−1) O = −PABC·PACB·PABA·PACA·(−32S ABC) PACB·PACA·PABC·PABA·(32S ABC) simplify = 1 The eliminants S AOG G =−PBAD·S ABO PABA S AOF F = −PCAD·S ACO PACA PCAD D = (2)·POAPD·PCAPD PAPDA PBAD D = (2)·POAPD·PBAPD PAPDA PCAPD PD = PACB PBAPD PD = −(PABC) S ACO O = PACA·PABC (−32)·S ABC S ABO O = PACB·PABA (32)·S ABC Example 6.377 (0.033, 1, 8) If E, F, G are the feet of the perpendiculars from a point D of the circumcircle of a triangle ABC upon its sides BC, CA, AB, prove that the triangle DFG, DBC are similar. G F D O C B A Figure 6-377 Constructive description ( (circle A B C D) (foot F D A C) (foot G D A B) (eq-product D B D F D C D G) ) 6.5 Circles 441 The machine proof PBDB·PDFD PCDC·PDGD G = PBDB·PDFD·PABA PCDC·(16S 2 ABD) F = PBDB·(16S 2 ACD)·PABA (16)·PCDC·S 2 ABD·PACA co−cir = (2BD 2)·(−CD·AD·AC)2·(2AB 2)·(2d)2 (2CD 2)·(−BD·AD·AB)2·(2AC 2)·(2d)2 simplify = 1 The eliminants PDGD G = (16)·S 2 ABD PABA PDFD F = (16)·S 2 ACD PACA PACA=2(AC 2) S ABD= BD·AD·AB (−2)·d PCDC=2(CD 2) PABA=2(AB 2) S ACD=CD·AD·AC (−2)·d PBDB=2(BD 2) Example 6.378 (0.050, 3, 5) Simson lines corresponding to pairs of diametrically opposite points on the circumcircle of a triangle meet a side of the triangle in two isotomic points. O A B D D G G C Figure 6-378 1 1 1 Constructive description ( (circle A B C D) (circumcenter O A B D) (lratio D1 O D −1) (foot G D A B) (foot G1 D1 A B) (midpoint C1 A B) (midpoint C1 G G1) ) The machine proof −GC1 G1C1 C1 = AG AB −1 2 −AG1 AB + 1 2 G1 = −( AG AB −1 2)·PABA PBAD1−1 2 PABA G = −(PBAD−1 2 PABA)·PABA (PBAD1−1 2 PABA)·PABA simplify = −(PBAD−1 2 PABA) PBAD1−1 2 PABA D1 = −(PBAD−1 2 PABA) 2PBAO−PBAD−1 2 PABA O = −(PBAD−1 2 PABA)·(2) (2)·(−PBAD+ 1 2 PABA) simplify = 1 The eliminants GC1 G1C1 C1 = AG AB −1 2 AG1 AB −1 2 AG1 AB G1 = PBAD1 PABA AG AB G = PBAD PABA PBAD1 D1 =2(PBAO−1 2 PBAD) PBAO O =1 2(PABA) Example 6.379 (1.817, 4, 35) If the perpendicular from a point D of the circumcircle (O) of a triangle ABC to the sides BC, CA, AB meet (O) again in the points N, M, L, the three lines AN, BM, CL are parallel to the simson of D for ABC. A B C O D F G L Figure 6-379 Constructive description ( (circle A B C D) (circumcenter O A B C) (foot F D A C) 442 Chapter 6. Topics from Geometry (foot G D A B) (inter L (l D G) (cir O D)) (parallel C L F G) ) Example 6.380 (4.283, 26, 25) Show that the simson of the point where an altitude cuts the circumcircle again passes through the foot of the altitude and is antiparallel to the corre-sponding side of the triangle with respect to the other two sides. G F E D H O C B A Figure 6-380 Constructive description ( (points A B C) (circumcenter O A B C) (foot G C A B) (inter D (l C G) (cir O C)) (foot E D A C) (foot F D B C) (inter G1 (l A B) (l E F)) ( AG BG = AG1 BG1 ) ) Example 6.381 (1.633, 4, 39) If the Simson line D(ABC) meets BC in E and the altitude from A in K, show that the line DK is parallel to EH, where H is the orthocenter of ABC. B C A O D E G H K Figure 6-381 Constructive description ( (circle A B C D) (foot E D B C) (foot G D A B) (orthocenter H A B C) (inter K (l E G) (l A H)) (parallel D K E H) ) Example 6.382 (1.483, 7, 44) Show that the symmetries, with respect to the sides of a triangle, of a point on its circumcircle lie on a line passing through the orthocenter of the triangle. H 1 F 1 G G F D O C B A Figure 6-382 Constructive description ( (circle A B C D) (orthocenter H A B C) (foot F D A C) (foot G D A B) (lratio G1 G D −1) (lratio F1 F D −1) (collinear F1 G1 H) 6.5 Circles 443 Example 6.383 (4.133, 4, 53) The four Simson lines of four points of a circle, each taken for the triangle formed by the remaining three points, are concurrent. A B C O D E F G H I J K Figure 6-383 Constructive description ( (circle A B C D) (circumcenter O A B C) (foot E D A B) (foot F D A C) (foot G C A B) (foot H C A D) (foot I A D B) (foot J A B C) (inter K (l H G) (l E F)) (inter M (l E F) (l I J)) ( EK FK = EM FM ) ) Example 6.384 (5.333, 58, 76) If two triangles are inscribed in the same circle and are sym-metrical with respect to the center of that circle, show that the two simsons of any point of the circle for these triangles are rectangular. 1 F 1 G F G 1 C 1 B 1 A D O C B A Figure 6-384 Constructive description ( (circle A B C D) (circumcenter O A B C) (lratio A1 O A −1) (lratio B1 O B −1) (lratio C1 O C −1) (foot G D A B) (foot F D A C) (foot G1 D A1 B1) (foot F1 D A1 C1) (perpendicular G1 F1 G F) ) Example 6.385 (1.683, 15, 48) Show that the Simson lines of the three points where the al-titudes of a triangle cut the circumcircle again form a triangle homothetic to the orthic triangle, and its circumcenter coincides with orthocenter of the orthic triangle. I 1 C 1 B 1 A 2 F 2 E 2 D 1 F 1 E 1 D G H F E D O C B A Figure 6-385 Constructive description ( (points A B C) (circumcenter O A B C) (orthocenter H A B C) (foot D A B C) (foot E B A C) (foot F C A B) (orthocenter G D E F) (lratio D1 D H −1) (lratio E1 E H −1) (lratio F1 F H −1) (foot D2 D1 A B) (foot E2 E1 A B) (foot F2 F1 A C) (inter A1 (l E E2) (l D D2)) (inter B1 (l D D2) (l F F2)) (inter C1 (l E E2) (l F F2)) (inter I (l A1 F) (l B1 E)) ( IB1 IE = IA1 IF ) ) 444 Chapter 6. Topics from Geometry Example 6.386 (0.633, 3, 40) The perpendiculars dropped upon the sides BC, CA, AB of the triangle from a point P on its circumcircle meet these sides in L, M, N and the circle in A1, B1, C1. The Simson line LMN meets B1C1, C1A1, A1B1 in L1, M1, N1. Prove that the lines AL1, BM1, CN1 are concurrent. I 1 N 1 M 1 L 1 C 1 B 1 A F E D N M L P C O X B A Figure 6-386 Constructive description ( (circle A B C P) (circumcenter O A B C) (midpoint X A B) (foot L P B C) (foot M P A C) (foot N P A B) (foot D O P L) (foot E O P M) (foot F O P N) (lratio A1 D P −1) (lratio B1 E P −1) (lratio C1 F P −1) (inter L1 (l B1 C1) (l M N)) (inter M1 (l C1 A1) (l M N)) (inter N1 (l A1 B1) (l M N)) (inter I (l B M1) (l A L1)) (inter J (l A L1) (l C N1)) ( AI L1I = AJ L1J ) ) Example 6.387 (0.467, 5, 38) The circumradius OP of the triangle ABC meets the sides of the triangle in the points A1, B1, C1. Show that the projections A2, B2, C2 of the points A1, B1, C1 upon the lines AP, BP, CP lie on the Simson line of P for ABC. F G 2 A 1 A P O C B A Figure6-387 Constructive description ( (circle A B C P) (circumcenter O A B C) (inter A1 (l B C) (l P O)) (foot A2 A1 A P) (foot G P A B) (foot F P A C) (inter F1 (l A C) (l G A2)) ( AF1 CF1 = AF CF ) ) Example 6.388 (0.950, 3, 37) Let L, M, N be the projections of the point P of the circumcircle of the triangle ABC upon the sides BC, CA, AB, and let the Simson line LMN meet the altitudes AD, BE, CF in the points L1, M1, N1. Show that the segments LM, L1M1 are equal to the projection of the sides upon the Simson line. A B C H O P L M N 1 L 1 M 1 N 1 A 1 B Figure 6-388 Constructive description ( (circle A B C P) (circumcenter O A B C) (orthocenter H A B C) (foot L P B C) (foot M P A C) (foot N P A B) (inter L1 (l A H) (l N M)) 6.5 Circles 445 (inter M1 (l B H) (l N M)) (inter N1 (l C H) (l N M)) (foot A1 A M N) (foot B1 B M N) ( L1M1 ML = 1) ) 6.5.6 The Pascal Conﬁguration For a machine proof of Pascal’s theorem, see Example 3.80 on page 145. Example 6.389 (The Converse of Pascal’s Theorem On a Circle) (0.933, 7, 44) S Q P F E D O C B A Figure 6-389 Constructive description ( (circle A E B D C) (circumcenter O A B C) (inter P (l A B) (l E D)) (lratio Q B C r) (inter S (l D C) (l P Q)) (inter F (l E Q) (l A S )) (cocircle F A B C) ) Example 6.390 (Pascal’s Theorem: The General Case) (0.666, 3, 23) Let A, B, C, D, F and E be six points with P = AB ∩DE, Q = BC ∩EF and S = CD ∩FA collinear. Then P1 = AC ∩DE, Q1 = BE ∩CF and S 1 = AB ∩FD are collinear. Constructive description ((points B A D E S ) (on C (l D S )) (inter P (l B A) (l D E)) (inter Q (l B C) (l S P)) (inter F (l E Q) (l A S )) (inter S 1 (l D F) (l B A)) (inter P1 (l A C) (l D E)) (inter Z2 (l B E) (l S 1 P1)) (inter Z1 (l C F) (l S 1 P1)) ( S1Z1 P1Z1 · P1Z2 S1Z2 = 1) ) The eliminants S1Z1 P1Z1 Z1 = S CFS1 S CFP1 P1Z2 S1Z2 Z2 = S BEP1 S BES1 S CFP1 P1 = −S DEC·S ACF S ADCE S BEP1 P1 = S AEC·S BDE S ADCE S BES1 S1 = −S BDF·S BAE S BDAF S CFS1 S1 = S DCF·S BAF S BDAF S BDF F =S AS Q·S BDE+S AES ·S BDQ S AES Q S ACF F =−S ASC·S AEQ S AES Q S BAF F =S AEQ·S BAS S AES Q S DCF F = S ES Q·S ADC S AES Q S BDQ Q = S BS P·S BDC S BSCP The eliminants S AS Q Q =S AS P·S BSC S BSCP S ES Q Q =S ES P·S BSC S BSCP S BS P P = −S BDE·S BAS S BDAE S AS P P = −S ADE·S BAS S BDAE S ES P P =S DES ·S BAE S BDAE S BDC C =S BDS · DC DS S ASC C =( DC DS −1)·S ADS S DEC C =S DES · DC DS S AEC C =S AES · DC DS +S ADE· DC DS −S ADE S ADC C =S ADS · DC DS S BSC C =( DC DS −1)·S BDS 446 Chapter 6. Topics from Geometry 1 Q 1 P 1 S B Q P S E D F C A Figure 6-390 The machine proof S1Z1 P1Z1 · P1Z2 S1Z2 Z1 = S CFS1 S CFP1 · P1Z2 S1Z2 Z2 = S CFS1·S BEP1 S CFP1 ·S BES1 P1 = S CFS1 ·(−S AEC·S BDE)·S ADCE (−S DEC·S ACF)·S BES1·(−S ADCE) simplify = −S CFS1 ·S AEC·S BDE S DEC·S ACF·S BES1 S 1 = −(−S DCF·S BAF)·S AEC·S BDE·S BDAF S DEC·S ACF·(−S BDF·S BAE)·(−S BDAF) simplify = S DCF·S BAF·S AEC·S BDE S DEC·S ACF·S BDF·S BAE F = S ES Q·S ADC·S AEQ·S BAS ·S AEC·S BDE·(−S AES Q)·S AES Q S DEC·(−S ASC·S AEQ)·(−S AS Q·S BDE−S AES ·S BDQ)·S BAE·(S AES Q)2 simplify = −S ES Q·S ADC·S BAS ·S AEC·S BDE S DEC·S ASC·(S AS Q·S BDE+S AES ·S BDQ)·S BAE Q = −(−S ES P·S BSC)·S ADC·S BAS ·S AEC·S BDE·S BSCP·(−S BSCP) S DEC·S ASC·(−S BSCP·S AS P·S BSC·S BDE−S BSCP·S AES ·S BS P·S BDC)·S BAE·(−S BSCP) simplify = −S ES P·S BSC·S ADC·S BAS ·S AEC·S BDE S DEC·S ASC·(S AS P·S BSC·S BDE+S AES ·S BS P·S BDC)·S BAE P = −(−S DES ·S BAE)·S BSC·S ADC·S BAS ·S AEC·S BDE·(S BDAE)2 S DEC·S ASC·(−S BDAE·S AES ·S BDC·S BDE·S BAS −S BDAE·S ADE·S BSC·S BDE·S BAS )·S BAE·(−S BDAE) simplify = S DES ·S BSC·S ADC·S AEC S DEC·S ASC·(S AES ·S BDC+S ADE·S BSC) C = S DES ·(S BDS · DC DS −S BDS )·S ADS · DC DS ·(S AES · DC DS +S ADE· DC DS −S ADE) S DES · DC DS ·(S ADS · DC DS −S ADS )·(S AES ·S BDS · DC DS +S ADE·S BDS · DC DS −S ADE·S BDS ) simplify = 1 Example 6.391 (Brianchon’s Theorem) (1.750, 2, 137) The dual of Pascal theorem. O A B C D E F 1 A 1 B 1 C 1 D 1 E 1 F I Figure 6-391 Constructive description ( (circle A B C D E F) (circumcenter O A B C) (on BT (t B B O)) (on AT (t A A O)) (on CT (t C C O)) (on DT (t D D O)) (on ET (t E E O)) (on FT (t F F O)) (inter A1 (l B BT) (l A AT)) (inter D1 (l E ET ) (l D DT )) (inter B1 (l C CT ) (l B BT)) (inter E1 (l F FT ) (l E ET )) (inter C1 (l D DT) (l C CT )) (inter F1 (l A AT) (l F FT )) (inter I (l B1 E1) (l A1 D1)) (inter J (l A1 D1) (l C1 F1)) ( A1I D1I = A1J D1J ) ) Example 6.392 (Kirkman’s theorem) (0.650, 2, 50) Given six points A, B, C, D, E, and F on a circle (or a conic), the three Pascal lines [BAECDF], [CDBFEA], [FECABD] are concurrent. There are 60 Kirkman points for one Pascal conﬁguration. 6.5 Circles 447 I Y X T S Q P F E D O C B A Figure 6-392 Constructive description ( (circle A B C D E F) (inter P (l C D) (l A B)) (inter Q (l D F) (l A E)) (inter S (l A C) (l B F)) (inter T (l B D) (l A E)) (inter X (l A B) (l E F)) (inter Y (l D F) (l A C)) (inter I (l S T) (l P Q)) (inter J (l P Q) (l X Y)) ( PI QI = PJ QJ ) ) Example 6.393 (Steiner’s theorem) (0.700, 2, 63) Given six points A, B, C, D, E, and F on a circle (or a conic), the three Pascal lines [ABEDCF], [CDAFEB], [EFCBAD] are con-current. There are 20 steiner points for one Pascal conﬁguration. I Y X T S Q P F E D O C B A Figure 6-393 Constructive description ( (circle A B C D E F) (inter P (l C D) (l A B)) (inter Q (l F A) (l D E)) (inter S (l B C) (l F A)) (inter T (l A D) (l B E)) (inter X (l A B) (l E F)) (inter Y (l C F) (l A D)) (inter I (l S T) (l P Q)) (inter J (l P Q) (l X Y)) ( PI QI = PJ QJ ) ) Example 6.394 (0.967, 2, 30) Given ﬁve points A0, A1, A2, A3 and A4, then points A0A1∩A2A3, A0A1 ∩A2A4, A0A2 ∩A1A3, A0A2 ∩A1A4, A0A3 ∩A1A2, A0A4 ∩A1A2 are on the same conic. (There are 60 such conics for ﬁve points.) S Q P 5 P 4 P 3 P 2 P 1 P 0 P 4 A 3 A 2 A 1 A 0 A Figure 6-394 Constructive description ( (points A0 A1 A2 A3 A4) (inter P0 (l A2 A3) (l A0 A1)) (inter P1 (l A2 A4) (l A0 A1)) (inter P2 (l A1 A3) (l A0 A2)) (inter P3 (l A1 A4) (l A0 A2)) (inter P4 (l A1 A2) (l A0 A3)) (inter P5 (l A1 A2) (l A0 A4)) (inter P (l P3 P4) (l P0 P1)) (inter Q (l P4 P5) (l P1 P2)) (inter S (l P5 P0) (l P2 P3)) (collinear P Q S ) ) 448 Chapter 6. Topics from Geometry Example 6.395 (4.383, 3, 84) Given six points A0, A1, A2, A3, A4 and A5 on one conic, then points A0A1 ∩A2A3, A0A1 ∩A4A5, A0A2 ∩A1A3, A0A3 ∩A1A2, A0A4 ∩A1A5, A0A5 ∩A1A4 are on the same conic. (There are 45 such conics for one Pascal conﬁguration.) Z Y X 5 P 4 P 3 P 2 P 1 P 0 P 5 A 4 A 3 A O 2 A 1 A 0 A Figure 6-395 Constructive description ( (circle A0 A1 A2 A3 A4 A5) (inter P0 (l A2 A3) (l A0 A1)) (inter P1 (l A4 A5) (l A0 A1)) (inter P2 (l A1 A3) (l A0 A2)) (inter P3 (l A1 A2) (l A0 A3)) (inter P4 (l A1 A5) (l A0 A4)) (inter P5 (l A1 A4) (l A0 A5)) (inter X (l P3 P4) (l P0 P1)) (inter Y (l P4 P5) (l P1 P2)) (inter Z (l P5 P0) (l P2 P3)) (inter Z1 (l P2 P3) (l X Y)) ( P2Z P3Z = P2Z1 P3Z1 ) ) Example 6.396 (4.566, 4, 75) Given six points A0, A1, A2, A3, A4 and A5 on one conic, then points A0A1 ∩A2A3, A0A1 ∩A4A5, A0A2 ∩A1A4, A0A3 ∩A1A5, A0A4 ∩A1A2, A0A5 ∩A1A3 are on the same conic. There are 90 such conics for one Pascal conﬁguration. Z Y X 5 P 4 P 3 P 2 P 1 P 0 P 5 A 4 A 3 A O 2 A 1 A 0 A Figure 6-396 Constructive description ( (circle A0 A1 A2 A3 A4 A5) (inter P0 (l A2 A3) (l A0 A1)) (inter P1 (l A4 A5) (l A0 A1)) (inter P2 (l A1 A4) (l A0 A2)) (inter P3 (l A1 A5) (l A0 A3)) (inter P4 (l A1 A2) (l A0 A4)) (inter P5 (l A1 A3) (l A0 A5)) (inter Y (l P4 P5) (l P1 P2)) (inter Z (l P5 P0) (l P2 P3)) (inter X (l P3 P4) (l P0 P1)) (inter W (l P0 P1) (l Y Z)) ( P1X P0X = P1W P0W ) ) Example 6.397 6 (0.866, 2, 55) Given six points A0, A1, A2, A3, A4 and A5 on one conic, then points A0A1 ∩A2A3, A0A1 ∩A4A5, A0A2 ∩A3A4, A0A3 ∩A2A5, A1A4 ∩A2A5, A1A5 ∩A3A4 are on the same conic. There are 60 such conics for one Pascal conﬁguration. 6Example 6.397 was a new theorem found by Wu in 1980 . Examples 6.395 and 6.396 were found by us. 6.5 Circles 449 0 A 1 A O 2 A 3 A 4 A 5 A 0 P 1 P 2 P 3 P 4 P 5 P X Y Z Figure 6-397 Constructive description ( (circle A0 A1 A2 A3 A4 A5) (inter P0 (l A2 A3) (l A0 A1)) (inter P1 (l A4 A5) (l A0 A1)) (inter P2 (l A3 A4) (l A0 A2)) (inter P3 (l A2 A5) (l A0 A3)) (inter P4 (l A2 A5) (l A1 A4)) (inter P5 (l A3 A4) (l A1 A5)) (inter X (l P4 P2) (l A0 A1)) (inter Y (l A3 A4) (l P1 P3)) (inter Z (l P5 P0) (l A2 A5)) (inter W (l P5 P0) (l X Y)) ( P5Z P0Z = P5W P0W ) ) 6.5.7 Cantor’s Theorems Example 6.398 (0.083, 3, 8) The perpendiculars from the midpoints of the sides of a triangle to the tangent lines of the circumcircle at the third vertex of the triangle are concurrent, and this concurrent point is the center of the nine point circle of the triangle A B C O L M N Figure 6-398 Constructive description ( (points A B C) (circumcenter O A B C) (midpoint L B C) (midpoint M A C) (inter N (p L O A) (p M O B)) (eqdistance N L N M) ) The machine proof PLNL PMNM N = S 2 BLOM·PAOA·S 2 ABO S 2 ALOM·PBOB·S 2 ABO simplify = S 2 BLOM·PAOA S 2 ALOM·PBOB M = (−S BOL−1 2S BCO+ 1 2 S ABO)2·PAOA (−S AOL−1 2 S ACO)2·PBOB L = (−1 2 S ABO)2·PAOA (−1 2 S ABO)2·PBOB simplify = PAOA PBOB O = PBCB·PACA·PABA·(64S 2 ABC) PBCB·PACA·PABA·(64S 2 ABC) simplify = 1 The eliminants PMNM N = S 2 ALOM·PBOB S 2 ABO PLNL N = S 2 BLOM·PAOA S 2 ABO S ALOM M = −(S AOL+ 1 2 S ACO) S BLOM M = −(S BOL+ 1 2 S BCO−1 2 S ABO) S AOL L = −1 2(S ACO+S ABO) S BOL L = −1 2(S BCO) PBOB O = PBCB·PACA·PABA (64)·S 2 ABC PAOA O = PBCB·PACA·PABA (64)·S 2 ABC Example 6.399 (0.017, 5, 5) Let A, B, C, D be four points on a circle O. The perpendiculars from the centroids of the four triangles ABC, ABD, ACD, and BCD to the tangent lines of 450 Chapter 6. Topics from Geometry circle O at points D, C, B, A are concurrent. Constructive description ( (circle A B C D) (circumcenter O A B C) (centroid M A B C) (centroid N A B D) (centroid L B C D) (inter P (p M O D) (p N O C)) (parallel P L O A) ) The eliminants S AOP P =S CMON·S ADO+S CDO·S AOM S CDO S AOL L = −1 3(S ADO+S ACO+S ABO) S CMON N = −(S COM+ 1 3 S CDO−1 3 S BCO−1 3 S ACO) S AOM M = −1 3(S ACO+S ABO) S COM M = 1 3(S BCO+S ACO) A B C O D M N L P Figure 6-399 The machine proof −S AOP −S AOL P = S CMON·S ADO+S CDO·S AOM (−S AOL)·(−S CDO) L = (S CMON·S ADO+S CDO·S AOM)·(3) (−S ADO−S ACO−S ABO)·S CDO N = (−3)·(−3S COM·S ADO+3S CDO·S AOM−S CDO·S ADO+S BCO·S ADO+S ADO·S ACO) (S ADO+S ACO+S ABO)·S CDO·(3) M = (3)·(3S CDO·S ADO+3S CDO·S ACO+3S CDO·S ABO) (S ADO+S ACO+S ABO)·S CDO·(3)2 simplify = 1 Example 6.400 (1.900, 42, 81) Let A, B, C, D, E be ﬁve points on a circle O. The perpendic-ulars from the centroids of the triangles whose vertices are from A, B, C, D, E to the lines joining the remaining two points are concurrent. A B C O D E 1 A 1 B 1 C 2 A 2 B 2 C N Figure 6-400 Constructive description ( (circle A B C D E) (circumcenter O A B C) (centroid A1 B C D) (centroid B1 A C D) (centroid C1 A B D) (midpoint A2 A E) (midpoint B2 B E) (midpoint C2 C E) (inter N (p A1 A2 O) (p B1 B2 O)) (perpendicular N C1 C E) ) For more results related to Cantor’s theorems, see Example 3.82 on page 147, Example 5.50 on page 245, Example 5.51 on page 245, and Example 5.56 on page 248. 6.6 A Summary We have given 400 machine proved geometry theorems in Sections 6.2–6.5 and 78 ma-chine solved geometry problems in Part I of the book. Thus totally there are 478 machine solved geometry problems in this book, including 280 proofs produced automatically by a computer program. 6.6. A Summary 451 In order to access the overall performance of the algorithm/program, we will ﬁrst list the machine computation times and proof lengths of the examples in Part I. No. page time maxt lems 2.35 73 0.017 1 3 2.36 74 0.117 2 9 2.37 75 0.067 1 7 2.41 78 0.050 1 5 2.42 80 0.133 1 14 2.46 83 0.067 2 4 2.47 84 0.033 1 3 2.52 86 0.083 4 5 2.53 87 0.250 6 8 2.54 88 0.300 9 13 2.55 89 0.300 10 7 2.56 91 0.167 2 17 2.58 92 0.517 17 18 2.59 93 1.033 15 18 2.62 94 0.050 1 4 2.65 96 0.117 1 10 2.66 98 0.117 1 10 3.36 120 0.001 1 2 3.40 123 0.050 3 3 3.41 124 0.017 1 4 3.42 124 0.750 3 15 3.43 125 0.083 2 10 3.44 126 0.033 1 7 3.45 127 0.250 6 16 3.51 130 0.067 1 6 3.52 131 0.050 3 6 3.53 132 1086.8 3125 65 3.68 139 0.050 2 2 3.69 140 0.067 2 5 3.70 140 0.033 1 5 3.71 141 0.033 3 3 3.79 144 0.033 1 12 3.80 145 0.083 1 14 3.81 146 0.050 1 14 3.82 147 0.067 4 5 3.102 156 1.050 48 15 3.105 162 0.867 5 8 3.106 164 0.117 2 7 3.107 164 0.033 2 4 No. page time maxt lems 3.108 165 0.200 5 8 3.109 165 0.350 4 7 3.110 166 0.050 3 7 3.111 167 1.383 4 11 3.112 167 1.100 5 10 4.23 180 0.033 1 4 4.24 180 0.017 1 2 4.25 180 0.033 2 7 4.40 189 0.067 2 6 4.41 190 0.883 9 10 4.47 192 0.083 3 7 4.53 194 0.017 1 4 4.54 195 0.033 1 3 4.61 198 0.083 4 8 4.62 198 0.133 3 8 4.63 199 0.083 2 5 4.64 200 0.067 3 6 4.65 201 0.083 3 6 4.87 210 0.100 4 7 4.88 211 0.117 5 6 4.89 211 0.067 3 6 4.90 212 0.133 4 6 4.91 213 0.167 4 10 4.92 215 99.200 140 78 5.38 237 0.083 4 8 5.46 243 0.017 2 4 5.47 243 0.100 3 7 5.48 244 0.117 4 5 5.49 244 0.083 5 5 5.55 248 0.017 1 9 5.56 248 0.717 4 5 5.57 250 0.083 5 7 5.58 252 0.083 6 5 5.59 252 0.117 9 6 5.61 253 0.067 4 3 5.62 254 0.133 5 3 5.63 255 0.017 3 2 5.64 255 0.050 3 3 5.65 256 0.267 7 9 452 Chapter 6. Topics from Geometry Table 1. Statistics for the Examples in Part I We use a triple (time,maxt,lems) to measure how diﬃcult a machine proof is: 1. time is the time needed to complete the machine proof. The program is imple-mented on a NexT Turbo workstation (25 MIPS) using AKCL (Austin-Kyoto Com-mon Lisp). 2. maxt is the number of terms of the maximal polynomial occurring in the machine proof. Thus maxt measures the amount of computation needed in the proof. 3. lems is the number of elimination lemmas used to eliminate points from geometry quantities. In other words, lems is the number of deduction steps in the proof. The following table contains some statistics for the timings and proof lengths of the 478 machine solved problems in this book. Proving Time Proof Length Deduction Step Time (secs) % of Thm. Maxterm % of Thm. Lemmas % of Thm. t ≤0.1 45.3% m = 1 16.9% l ≤3 7.1% t ≤0.5 68.8% m ≤2 33.0% l ≤5 16.7% t ≤1 85.5% m ≤5 66.9% l ≤10 42.6% t ≤5 97.45% m ≤10 81.7% l ≤20 73.2% t ≤10 98.9% m ≤100 98.7% l ≤50 95.1% t < 1087 100% m ≤3125 100% l ≤137 100% Table 2. Statistics for the 478 Theorems Remark. 1. We can see that our program is very fast and can produce short proofs for many diﬃcult geometry theorems. Eighty-ﬁve percent of the 478 proofs can be completed within one second, and the average maximal term and deduction step for the 478 examples are 14.877and 17.35 steps respectively. 2. If we set a standard that a machine proof is readable if one of the following condi-tions holds (1) the maximal term in the proof is less than or equal to 5; (2) the deduction step of the proof is less than or equal to 10; or (3) the maximal term in the proof is less than or equal to 10, and the deduction step is less than or equal to 20, 7If not considering Morley’s theorem (see 3 of the this remark), the average maximal term is 8.37. 6.6. A Summary 453 then 76.9 percent (or 368) of the proofs of the 478 theorems produced by our prover are readable. The machine proofs for 59.2% or 283 of the 478 theorems are actually presented in this book. 3. In spite of this success, there are still some geometry theorems for which the method/program performs badly. For instance, for Morley’s theorem (on page 132), we have time= 1086.8, maxt = 3125, and lems = 65. There are two main factors in the description of the geometry statements that aﬀect the machine proof: the number of points in the statement and the type of constructions needed to describe it. Generally speaking, the number of points in a geometry statement is ﬁxed and reﬂects the diﬃculty level of the statement in nature. On the other hand, the diﬃculty related to the type of con-structions is somehow artiﬁcial. According to our experience, the geometry relations can be listed in ascending order of diﬃculties as follows: collinear, parallel, ratios, perpendicular, circles, angles. Morley’s theorem involves information mainly about angles, and most of the other “diﬃcult geometry theorems” involve perpendiculars, circles, or angles. We have two ideas for further improvement of the method/program. First, we can put elimination results for more constructions into the program instead of dividing these constructions into other simple constructions. Second, we can use new geometry quantities to produce short and readable proofs. In our method, we mainly use areas and Pythagoras diﬀerences, which deal perfectly with geometry statements about collinear and parallel, but do not always work well for a geometry statement about perpendicular, circles, and angles. For instance, to express the fact that the sum of two angles is equal to another angle, we have to use a complicated equation of areas and Pythagoras diﬀerences. Besides area and Pythagoras diﬀerence, we also discussed how to use other geometry quantities such as the vector, the complex number, and the full-angle, to produce short and readable proofs. The approach based on full-angles presented in Section 3.8 is quite promising. 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Hou, A Note on Wu Wen-Ts¨ un’s Non-degenerate Conditions, IC/91/160, International Center for Theoretic Physics. + List of Symbols △ABC 5 triangle ABC ▽ABC 5 the area of triangle ABC (ABCD) 270 the cross-ratio ) ∠ABCD 37 the oriented angle from AB to CD ∠[AB, CD] 44 the full-angle from AB to CD [− − → AB, − − → CD] 231 the exterior product ⟨− − → AB, − − → CD⟩ 230 the inner product [x, y] 226 the exterior product ⟨x, y⟩ 223 the inner product − → A 229 the vector from the origin to A − − → AB 229 the vector from A to B AB 3 the directed line segment from A to B f AB 39 the oriented chord c AB 142 the cochord of f AB AB ∩CD 9 the intersection of AB and CD AB ∥CD 17 AB is parallel to CD AB⊥CD 30 AB is perpendicular to CD CH 61 the Hilbert intersection point statements in the plane CL 109 the linear constructive statements in the plane CABC 27 the co-area of triangle ABC O(ABCD) 271 the cross-ratio for ﬁve points PABC 28 the Pythagoras diﬀerence of triangle ABC for point B PABCD 30 the Pythagoras diﬀerence of quadrilateral ABCD S ABC 6 the signed area of triangle ABC S ABCD 8 the signed area of quadrilateral ABCD SH 182 the Hilbert intersection point statements in the space VABCD 171 the signed volume of tetrahedron ABCD VABCDE 173 the signed volume of polyhedron A-BCD-E K[x] 150 the ring of polynomials class(P) 150 the class of polynomial P init(P) 150 the initial of polynomial P ld(P) 150 the leading degree of polynomial P lv(P) 150 the leading variable of polynomial P 469 470 List of Symbols prem(P, Q) 151 the pseudo division of P for Q a A = b 72 point A is eliminated from a a simplify = b 72 a is simpliﬁed a py = b 135 Pythagoras diﬀerence is expanded a cons = b 123 constant is substituted a herron = b 135 using Herron-Qin’s formula a area−co = b 68 using the area coordinates a 2lines = b 79 using the two-line conﬁguration a co−cir = b 144 using the co-circle theorem a volume−co = b 199 using the volume coordinates Index a machine proof, 72 a proof is at the ﬁrst level or lemma level, 71 a proof is at the second level or proposition level, 72 a proof is at the third level or axiom level, 72 AFFINE, 70 aﬃne geometry, 83 aﬃne geometry of dimension three, 178 aﬃne geometry., 82 aﬃne plane geometry , 82 aﬃne space, 229 algebraic approach, 81 algebraic extension, 153 algorithm PLANE, 119 algorithm Solid, 210 algorithm: VECTOR, 240 ALINE, 129 angle, 51, 132 anticenter, 396 anticomplementary triangle, 292 ARATIO, 136, 182 area coordinate system, 133 area coordinates, 68, 118 Artin, 81 associated ﬁeld, 81 auxiliary parallelograms, 58 basis, 222 BLINE, 110 BPLANE, 184 Brahmagupta’s formula, 40 Brocard points, 362 Butterﬂy theorem, 42 butterﬂy theorem, 42, 405 butterﬂy theorem for quadrilaterals, 405 Cantor’s First Theorem, 245 Cantor’s Second Theorem, 245 Cantor’s theorem, 147, 248, 449 Cartesian product, 81 Cayley-Menger formula, 207 center of inversion, 430 CENTROID, 138 centroid, 245, 365 centroid of Tetrahedra, 195 centroid theorem, 13, 243, 285 centroid theorem for tetrahedra, 192, 234 Ceva’s Theorem, 10, 11 Ceva’s theorem, 11, 62, 94, 95, 266 ceva’s theorem, 61 Ceva’s Theorem for Skew Quadrilaterals, 194 cevian, 266 cevian triangle, 266 CIR, 148, 186 CIRC, 164 CIRCLE, 144 circle of inversion, 430 CIRCUMCENTER, 138 circumcenter, 308 circumcircle, 308 class, 150 co-angle inequality, 24 co-angle Theorem, 25 co-angle theorem, 21 co-angle triangles, 21 co-area, 27 co-circle theorem, 40 co-circle theorem for Pythagoras diﬀerence, 43 co-face theorem, 174 co-side theorem, 9, 57, 263 471 472 List of Symbols co-side triangles, 8 co-trihedral theorem, 179 co-vertex theorem, 173 cochord, 142 COCIRCLE, 130 COLL, 63, 183 COLLINEAR, 114 collinear, 4 Collins, 53 complete quadrilateral, 372 complex number, 249 conﬁguration, 90 CONG, 113 congruent, 226 conjugate rays, 271 consistency of proofs, 52 CONSTANT, 123 construction, 60, 110, 181 constructive conﬁguration, 184 constructive geometry statement, 9 constructive geometry statements, 109 constructive statement, 182 coordinates, 222 COPL, 183 coplanar, 171 cosine law, 36 CPLANE, 185 cross ratio, 270 cross-ratio, 271 cyclic quadrilateral, 395 derive unknown formula, 86 Desargues’ Axiom, 19, 82 Desargues’ axiom, 83 Desargues’ theorem, 276 dimension, 172 dimension axioms, 55 Dimension of vector space, 222 directed line segment, 3 directed lines, 3 distance formula in area coordinate system, 134 DPLANE, 185 Echols’ ﬁrst theorem, 252 Echols’ second theorem, 252 Echols’ theorem in general form, 253 elementary geometry, 53 eliminate area ratios, 190 eliminate co-circle points, 144 eliminate free points from the areas, 67 eliminate length ratios, 193 eliminate points from Pythagoras diﬀerence, 114 eliminate points from areas, 65 eliminate points from length ratios, 66 eliminate volumes, 188 elimination, 10 EQ-PRODUCT, 114 EQANGLE, 130 EQDISTANCE, 114 equilateral triangle, 127, 251 Erdos’ inequality, 26 Euclid, 81 Euclid’s parallel axiom, 58 Euclidean plane, 3 Euclidean space of dimension three, 225 Euler line, 326 Euler points, 329 Euler triangle, 329 Euler’s theorem, 141 exscribed circles, 333 extension, 152 exterior product, 226, 231 Fano plane, 91 Feuerbach’s theorem, 329 ﬁeld, 221 ﬁeld associated with, 81, 83 ﬁnal remainder, 151 ﬁnite geometries, 85 ﬂat full-angle, 45 FOOT, 112 FOOT2LINE, 182, 233 FOOT2PLANE, 185, 233 full-angle, 44, 128, 157 List of Symbols 473 Gauss line, 375 Gauss point, 375 Gauss-line theorem, 74 Gelernter, 52 general butterﬂy theorem, 41, 146 generally false, 155 generic point, 153 geometric approach, 81 geometric information base, 159 geometric objects, 110 geometric quantities, 109, 233 geometric quantity, 60, 181 geometry knowledge base, 160 geometry quantity, 128 Gergonne point, 341 Gr¨ unbaum, 94 Hanson, 52 HARMONIC, 114, 123 harmonic pencil, 271 harmonic sequence, 5, 122, 271, 372 Herron-Qin formula in the Minkowskian ge-ometry, 246 Herron-Qin’s Formula, 108 Herron-Qin’s formula, 37, 109, 135, 231 Herron-Qin’s formula for quadrilateral, 108 Herron-Qin’s formula for quadrilaterals, 38, 39 Herron-Qin’s Formula for Tetrahedra, 206 Herron-Qin’s formula for tetrahedron, 207 Hilbert, 59, 81 Hilbert intersection point statement, 61 Hilbert’s intersection point statements, 182 Hilbert’s Mechanization theorem, 59 homothetic, 296 homothetic center, 296 hyperbolic trigonometric, 247 INCENTER, 138 incenter theorem, 333 incidence, 82 initial, 150 inner product, 223, 230 inscribed angle theorem, 46, 131 inscribed circle, 333 INTER, 110 intercept triangles, 347 International Mathematical Olympia, 421 International Mathematical Olympiad, 124, 200, 213, 336 international mathematical olympiad, 25 inverse, 430 INVERSION, 123 inversion, 123, 430 irreducible, 152 isogonal conjugate point, 346 isogonal conjugates, 306 isotomic, 265 isotropic, 224 isotropic line, 229 Lagrange identity, 228 leading degree, 150 leading variable, 150 Lemoine axis, 315 Lemoine Point, 346 Lesening’s theorem, 276 light cone, 224 LINE, 110 line equation in the area coordinate system, 134 linear constructive geometry statements, 111 linear geometry quantity, 115 Loveland, 52 LRATIO, 122 medial triangle, 292 median, 285 Menelaus’ Theorem, 87 Menelaus’ theorem, 11, 12, 73, 263 Menelaus’ theorem for skew quadrilaterals, 180 Meta theorem, 241 metric vector space, 223 MIDPOINT, 114, 122 Minkowskian plane geometry, 246 474 List of Symbols Minkowskian space, 225 Miquel Point, 164 Monge’s theorem, 215 Morley’s theorem, 132, 453 MRATIO, 122 Nagel point, 342 ndg condition, 61, 111 NE-SQUARE, 255 NE-TRIANGLE, 252 negative Euclidean space, 226 negative Minkowskian space, 226 Nehring’s theorem, 278 nine point circle, 49, 329 nine-point circle theorem, 329 non-degenerate conditions, 52 nondegenerate condition, 62 nonsingular, 223 number of points, 453 OLINE, 184 ON, 110 oriented angle, 36, 37 oriented chord, 39, 141 oriented quadrilateral, 7 oriented triangle, 6, 55 orthic triangle, 300 ORTHOCENTER, 138 Orthocenter theorem, 32 Orthocenter theorem for tetrahedron, 204 Orthocenter theorem in Minkowskian Geom-etry, 247 Orthocenter-dual, 32 orthocentric, 302 orthocentric quadrilateral, 302 orthodiagonal quadrilateral, 401 orthogonal circles, 436 Pappus point theorem, 275 Pappus’ line, 275 Pappus’ Theorem, 92, 274 Pappus’ theorem, 14, 80 Pappus-dual, 275 PARA, 63 paradox, 52 PARALLEL, 114 parallel, 17, 175, 229, 232 parallel translation, 106 parallelogram, 17, 19, 20, 51, 58 PARTIO, 111 Pascal’s conﬁguration, 59 Pascal’s theorem, 41, 445 Pascal’s theorem on a circle, 145 Pascalian Axiom, 82 Pascalian axiom, 19, 83, 84 PE-SQUARE, 255 PE-TRIANGLE, 252 pedal triangle, 354 PERP, 113, 183 PERPENDICULAR, 114 perpendicular, 223 PLANE, 184 plane, 172 PLINE, 110 POINT, 110 polar, 434 pole, 434, 439 polygons, 12 polygrams, 94 position coordinates, 4, 56 position ratio, 56 position ratios, 4 PPLANE, 184 PRATIO, 181 predicate form, 63, 113, 182 predicates, 63 PRLL, 183 PRLP, 183 prover, 52 pseudo division], 151 pseudo remainder, 151 Ptolemy’s theorem, 40, 142 pure constructive, 241 pure point of intersection theorem, 59 Pythagoras diﬀerence, 28, 103 List of Symbols 475 Pythagoras diﬀerence theorem, 28 Pythagoras diﬀerence for an oriented quadri-lateral, 30 Pythagorean theorem, 24, 28, 225, 230 quadratic geometry quantity, 116 ratio constructions, 122 ratio of the directed segments, 55 ratio of two parallel line segments, 17, 58 rectangular coordinate system, 225 reduced, 150 reducible, 152 regular n-polygons, 253 remainder formula, 151, 153 right full-angle, 45 segment arithmetic, 83 Seidenberg, 53 Shephard, 94 signed area, 55 signed area of an oriented quadrilateral, 55 signed area of an oriented triangle, 6 signed area of triangle, 231 similar, 22 Simson line, 439 Simson’s theorem, 43, 144, 164 Simson’s theorem in Minkowskian geome-try, 248 sine law, 34 skew area coordinate system, 133 skew volume coordinate system, 215 solid metric geometry, 229 SPHERE, 186 square, 224 square distance, 230 squares, 123 SRATIO, 233 Steiner’s theorem, 189 Steiner-Lehmus’ theorem, 26 structure Ω, 82 subspace, 222 successive pseudo division, 151 symmedian, 346 symmedian point, 346 symmetric matrix, 223 SYMMETRY, 122 TANGENT, 114 tangent function, 128 tangential triangle, 317 Tarski, 53 the parameter set, 151 the radius of inversion, 430 the volume method, 198 Theorem of Pratt-Wu, 370 theorem of three perpendiculars, 204 theory of geometry, 81 TLINE, 110 TPLANE, 184 transcendental over, 152 translation, 176 transversal, 94 transversals for polygrams, 95, 99 TRATIO, 111 triangular form, 151 trilinear polar, 358 triple scalar product, 228 trisector, 132 tritangent circles, 333 two-line conﬁguration, 79 type of constructions, 453 unordered geometry, 53 vector, 229 vector space, 221 vectors, 222 visual angle theorem, 35 volume, 232 volume coordinates, 196 Wu, 53, 150 Zhang, 2, 81
14519	https://scholars.huji.ac.il/sites/default/files/ronniekosloff/files/jp072121s.pdf	Mechanism of Thermal Unimolecular Decomposition of TNT (2,4,6-Trinitrotoluene): A DFT Study Revital Cohen,† Yehuda Zeiri,‡ Elhanan Wurzberg,§ and Ronnie Kosloff,\| Chemical Research Support Unit, Department of Chemistry, Weizmann Institute of Science, RehoVot, 76100, Israel, Chemistry Department, NRCN, P.O. Box 9001, Beer-SheVa 84190, Israel, Rafael Intelligence and Security Directorate, P.O. Box 2250, Haifa, 31021, Israel, and Fritz Haber Center for Molecular Dynamics, Hebrew UniVersity, Jerusalem 91904, Israel ReceiVed: March 16, 2007; In Final Form: June 15, 2007 The widespread and long-term use of TNT has led to extensive study of its thermal and explosive properties. Although much research on the thermolysis of TNT and polynitro organic compounds has been undertaken, the kinetics and mechanism of the initiation and propagation reactions and their dependence on the temperature and pressure are unclear. Here, we report a comprehensive computational DFT investigation of the unimolecular adiabatic (thermal) decomposition of TNT. On the basis of previous experimental observations, we have postulated three possible pathways for TNT decomposition, keeping the aromatic ring intact, and calculated them at room temperature (298 K), 800, 900, 1500, 1700, and 2000 K and at the detonation temperature of 3500 K. Our calculations suggest that at relatively low temperatures, reaction of the methyl substituent on the ring (C-H R attack), leading to the formation of 2,4-dinitro-anthranil, is both kinetically and thermodynamically the most favorable pathway, while homolysis of the C-NO2 bond is endergonic and kinetically less favorable. At ∼1250-1500 K, the situation changes, and the C-NO2 homolysis pathway dominates TNT decomposition. Rearrangement of the NO2 moiety to ONO followed by O-NO homolysis is a thermodynamically more favorable pathway than the C-NO2 homolysis pathway at room temperature and is the most exergonic pathway at high temperatures; however, at all temperatures, the C-NO2 f C-ONO rearrangement-homolysis pathway is kinetically unfavorable as compared to the other two pathways. The computational temperature analysis we have performed sheds light on the pathway that might lead to a TNT explosion and on the temperature in which it becomes exergonic. The results appear to correlate closely with the experimentally derived shock wave detonation time (100-200 fs) for which only the C-NO2 homolysis pathway is kinetically accessible. Introduction 2,4,6-Trinitrotoluene (TNT) has been available as a pure material since 1870. At present, TNT is a major military explosive despite the development of more potent alternatives. TNT is cheap and safe, it is easy to prepare, and it has a low hygroscopicity, relatively low sensitivity to impact and friction, good thermal stability, and relatively high power during detonation. The widespread and long-term use of TNT has led to extensive study of its thermal and explosive properties.1-9 Considerable evidence exists that the thermal decomposition processes of TNT and other explosives are related to their sensitivity to impact and shock energy.1 Consequently, thermal decomposition chemistry is fundamentally important in the explosives field,10,11 and the determination of the kinetics and mechanism of the thermal decomposition of TNT remains a fundamental aspect of its characterization. There is substantial evidence that one or more of the following initiation steps take place during the thermolysis of polynitro compounds: (1) C-NO2 homolysis, (2) isomerization of the nitro (NO2) group to the nitrite (ONO), and (3) reactions of the nonenergetic substituent on the ring (CH3 in the case of TNT).1,3 The relative importance of these initiation steps depends on the reaction conditions (temperature and pressure). Currently, it is believed that at lower temperatures (<800-900 °C), the initiation chemistry of TNT is dominated by oxidation reactions of the methyl group and that at elevated temperatures (>800-900 °C), C-NO2 homolysis dominates the initiation process.2 However, there is no direct evidence for the latter occurrence but rather circumstantial indications for the formation of an aromatic radical.9 NO2(g) was rarely detected as a decomposition product; however, large quantities of NO(g) have been ob-served.12,13 Knowledge about aromatic -NO2 to -ONO rear-rangement is deficient, although nitro-nitrite isomerization has been shown to take place in nitrobenzene derivatives containing donor substituents.3,14 Decomposition of the nitrite is a plausible source of the large amount of NO(g) that is generated by many polynitrobenzene derivatives upon rapid thermal decomposi-tion.1,3,14 Unlike nitro aliphatic compounds, where the nitrite isomer of the respective nitro compound has been isolated, no aryl nitrite compound has been isolated to date. This might suggest that if and when formed, aromatic nitrite rapidly decomposes to form NO(g) and an oxygen radical. Indirect evidence for the formation of the latter exists but is relatively scarce.3,9,14 Corresponding author. E-mail: ronnie@fh.huji.ac.il; fax: 972-2-6513742. † Weizmann Institute of Science. ‡ NRCN. § Rafael Intelligence and Security Directorate. \| Hebrew University. 11074 J. Phys. Chem. A 2007, 111, 11074-11083 10.1021/jp072121s CCC: $37.00 © 2007 American Chemical Society Published on Web 10/04/2007 Although much research on the thermolysis of TNT and polynitro organic compounds has been undertaken, the kinetics and mechanism of the initiation and propagation reactions and their dependence on the temperature and pressure are unclear. To understand the explosive nature of TNT, the kinetics and mechanism of its thermal decomposition need to be examined in detail to clarify its detonation mechanism. Very few theoreti-cal studies have been reported for TNT or other nitro aromatic compounds.8,15-17 We have previously reported a computational study on the thermal decomposition leading to detonation of triacetonetriperoxide (TATP).10,11 Calculations related to TNT decomposition are limited to small model systems8,15-17 and to relatively low levels of theory.8,15 We are not aware of any comprehensive thermochemical DFT analysis of TNT decom-position. It is precarious to equate the findings of unimolecular chemistry in the gas phase directly to the reactions in the condensed phase. Nevertheless, we suggest that the first steps of the reaction schemes induced by heat, light, and ionizing radiation in different phases of nitro aromatic compounds are similar. We therefore exclude further oxidation by atmospheric oxygen. The rationale is a separation of time scales between the reactions that take place on the time scale of detonation wave propagation and subsequent secondary reactions that further degrade the parent molecule. Computational investigation of the mechanisms that are involved in the thermal decomposi-tion of a single TNT molecule will therefore shed light on the chemical processes that are involved in its detonation. Here, we report a comprehensive computational DFT inves-tigation of the unimolecular adiabatic (thermal) decomposition of TNT. It is conceivable that during TNT detonation, a large portion of it decomposes to water, CO, and CO2 as final products supplying the main energy release. These products are formed in a second stage of reactions that occur after the detonation shock wave passes. In our investigation, we concentrated on the processes that led to the explosion rather than on the processes that took place during and after it. Moreover, such reactions may involve atmospheric species as well as nearby radicals and first stage detonation products. Exploring these reaction mechanisms is beyond the scope of this paper. It has been shown that aromatic ring fission does not take place until most of the attached substituents are removed.1,4 We believe that ring fission is a process that most probably takes place during TNT detonation and not the process that leads to it as the C-C bond dissociation energy (BDE) (∼100-150 kcal/ mol), especially in aromatic rings, is much larger than the C-N (∼70 kcal/mol), O-N (∼50 kcal/mol), or C-H (∼100 kcal/ mol) BDEs; the dissociation of these bonds occurs at the initial stages (and usually the rate determining state (RDS)) of the decomposition pathways considered here (vide infra).18 Thus, we concentrate our investigation on the kinetics and mechanism of the processes that take place before the aromatic ring is dissociated. In our investigation, TNT decomposition is assumed to be unimolecular. We base our assumption on the experimental evidence that the detonation rate of TNT is ∼6900 m/s and that the size of the TNT unit cell in the bulk is roughly 10-20 Å.19-21 Accordingly, the time it takes the shock wave generated by the initial detonation to cross approximately one unit cell is roughly 150-300 fs. This time scale is probably too short to allow bimolecular reactions (followed, i.e., by radical formation) that will cause the explosion. Such reactions were shown to take place during millisecond time scales.2 Thus, detonation of TNT is a result of a unimolecular decomposition rather than of intermolecular radical chain propagation reactions.22 It is inferred that the detonation of TNT and similar molecules is caused by a rapid release of work accompanied by an abrupt increase in the pressure due to a fast increase in the number of decomposi-tion products. Following this inference, we investigated three decomposition routes that led to detonation;23 their initial steps were previously proposed to have contributed to TNT decom-position initiation.1 Two of our proposed mechanisms result in the evolution of four gas molecules out of one bulk molecule. All the investigated routes are described in Scheme 1. In the first route, C-NO2 homolytic cleavage takes place followed by subsequent C-NO2 homolysis steps forming eventually three NO2(g) molecules and an aromatic triradical intermediate. In the second route, C-H attack on the nitro substituent takes place to form 2,4-dinitro-anthranil (DNAn) and water, and in the third route, nitro-nitrite (C-NO2 f C-ONO) rearrangement takes place followed by O-NO homolytic cleavage and the release of NO(g) for each of the three NO2 substituents.23 There is indirect experimental evidence for the initial step of the first route (C-NO2 homolysis)1,9,13 and for the second route2,6,7 to occur in the thermal decomposition of TNT, while for the third route, the only indirect evidence found so far is the release of large quantities of NO(g) in a few experiments SCHEME 1 Mechanism of Thermal Unimolecular Decomposition of TNT J. Phys. Chem. A, Vol. 111, No. 43, 2007 11075 and the formation of a phenolic radical.1,12,13 We have calculated the free energy pathways for these three routes using the Gaussian 0324 program package in the uB3LYP25,26/cc-pVDZ27 level of theory. We took into account the thermal corrections to the relative energies of the intermediate and transition states both at room temperature (298 K) and at elevated temperatures, including the estimated detonation temperature of 3500 K generated during TNT explosion. On the basis of the temperature analysis, we attempted to estimate the most favorable pathway that leads to TNT detonation and the temperature range for which it becomes dominant. The temperature analysis was carried out to obtain general guidance for the difference between the thermal decomposition at room temperature and the thermal decomposition at higher temperatures. The analysis is aimed at qualitative identification of the process that takes place during the initiation stage eventually leading to detonation. Computational Methods All calculations were carried out using DFT as implemented in the Gaussian 03 program.24 The Becke three-parameter hybrid density functional method with the Lee-Yang-Parr correlation functional approximation (B3LYP)25,26 was employed together with the Dunning correlation-consistent polarized valence double-ú (cc-pVDZ) basis set.27 This level of theory was shown to be adequate for calculating the detonation mechanism of similar sized systems.10 The structures that had an open shell character were located using the unrestricted uB3LYP method wave function. This spin-unrestricted approach has been demonstrated to describe cor-rectly diradical intermediates as well as transition states when multireference problems are present.28 Geometry optimizations for minima were carried out using the standard Berny algo-rithm29,30 in redundant internal coordinates up to the neighbor-hood of the solution and, then, if not converged, continued the optimization using analytical second derivatives.31 Optimizations for transition states were carried out with an initial guess for the transition state being generated from manual manipulation of the geometry using MOLDEN.32 In cases where this approach failed to converge, we used analytical second derivatives at every step. Zero-point and RRHO (rigid rotor-harmonic oscillator) thermal corrections (to obtain ∆S and ∆G values) were obtained from the unscaled computed frequencies both for room tem-perature (298 K) and for higher temperatures (800, 900, 1500, 1700, 2000, and 3500 K). Results and Discussion Thermal Decomposition of TNT at Ambient Temperature C-NO2 Homolysis. The C-NO2 bond is usually the weakest bond of nitro aromatic compounds. It was experimentally suggested that C-NO2 homolysis is the dominant initial reaction that takes place in the decomposition channel of 3- and 4-nitrotoluene and also occurs for 2-nitrotoluene.1 In the gas phase, it was proposed that this reaction competes with two other reactions (NO2 isomerization and C-H R attack, see following description).33 Nevertheless, C-NO2 homolysis is most probably the favored decomposition channel of nitro aromatic compounds in the gas phase at high temperatures.1 We have calculated a pathway that is initiated by C-NO2 homolysis followed by subsequent C-NO2 homolysis of the two remaining substituents of TNT since C-NO2 is the weakest bond of TNT, and it has been shown that aromatic ring fission does not take place until most of the attached substituents are removed.1,4 This is the most straightforward way to generate four gas molecules out of one. The enthalpy and free energy pathway for the homolytic cleavage of all three nitro substituents in TNT is presented in Figure 1. The increasing contribution of entropy to the ∆G energy is indicated by the increasing energy differences between the ∆G and the ∆H pathways as more NO2 molecules are being eliminated. According to our calculations during the first step of C-NO2 homolytic cleavage in TNT, the ortho NO2 is eliminated, and an aromatic radical and NO2(g) are formed with ∆G ) 43.8 kcal/mol (∆H ) 58 kcal/mol). This value is consistent with the activation energy value of 40.9 kcal/mol that was assigned for NO2 removal in an early radical producing step obtained by EPR studies.34 It also resembles the value obtained by semiem-pirical calculations of C-NO2 homolysis.15 The ∆G value for initial cleavage of the para-NO2 substituent was found to be 8 kcal/mol higher in energy than for o-NO2; thus, it is less likely to be cleaved initially. The pathway continues with the homolysis of the two remaining C-NO2 bonds, which are the weakest bonds in TNT. The energy invested in the cleavage of the second o-C-NO2 bond was found to be ∆G ) 53.1 kcal/mol (∆H ) 66.5 kcal/ mol), 3.2 kcal/mol lower in energy than the energy required to cleave the p-C-NO2 bond; thus, it is conceivable that o-NO2 will be eliminated preferentially. The homolysis energy of the third p-C-NO2 bond was calculated as ∆G ) 60.4 kcal/mol (∆H ) 74 kcal/mol). We could not locate the transition state for C-NO2 homolysis. However, it should be noted that normally radical recombination, which is the microscopic Figure 1. Enthalpy and free energy plots for C-NO2 homolysis pathway as calculated at the uB3LYP/cc-pVDZ level of theory. 11076 J. Phys. Chem. A, Vol. 111, No. 43, 2007 Cohen et al. reversible reaction of homolysis, takes place without an activa-tion barrier.35 In this case, the observed activation energy for a bond dissociation process is equal to the bond dissociation energy (BDEC-NO2).36 Accordingly, the barrier height for C-NO2 homolysis is expected to be equal or similar to its BDE (43.8 kcal/mol for the first NO2, 53.1 kcal/mol for the second, and 60.4 kcal/mol for the third), as reported for N-NO2 homolysis.35,37,38 The closed shell singlet spin derivative of o,o-2NO2 was found to be 18.5 kcal/mol energetically more stable than the o,o-NO2 diradical (triplet); thus, it could be formed as an intermediate (instead of the o,o-2NO2 diradical) before the third C-NO2 homolysis takes place. Figure 1 shows that at room temperature, the calculated C-NO2 homolysis pathway of TNT is both kinetically and thermodynamically unfavorable since every step in the pathway is endergonic. However, the calculated ∆S associated with each C-NO2 homolysis step is positive, and its values are between 44 and 48 eu, which are relatively large. Since the entropy of C-NO2 homolysis is positive, its contribution to the ∆G value as the temperature increases will rise according to ∆G ) ∆H - T∆S. As a result, at elevated temperatures, the overall ∆G for the pathway will decrease, and the pathway will be less endergonic. C-H R Attack To Form 2,4-DNAn. NO2(g) was observed from decomposition of nitro aromatic compounds in the gas phase but was rarely detected in the decomposition of these molecules in the bulk state. This is in contrast to aliphatic nitro compounds in which NO2(g) is always detected during their decomposition.1 These observations suggest that nitro aromatic compounds may have alternate decomposition channels whose activation energies lie below that of C-NO2 homolysis, especially when other substituents are present on the ring. There is considerable evidence for the formation of anthranil and its derivatives in the slow thermal decomposition of nitro aromatic compounds.1,2,4,6,7,33,39 The mechanism of DNAn formation was studied in several nitro aromatic compounds, and it was suggested that intramolecular hydrogen transfer from the methyl substituent to the NO2 moiety is the RDS of the reaction.15,17,39 We have calculated the enthalpy and free energy pathway for the formation of 2,4-DNAn from TNT decomposition. The pathway is presented in Figure 2. Since the entropy contribution to the C-H R attack pathway is negligible (until the last water elimination step), the free energy values are very similar to the enthalpy values along the whole pathway. In the first step of the C-H R attack mechanism, the hydrogen atom is transferred from the toluene CH3 group to the o-nitro substituent in a concerted (one step) mechanism (tautomeriza-tion). The calculated barrier height for the tautomerization is 37.7 kcal/mol, and the transition state that leads to the formation of tautomer 1 is an ordered six-membered ring with the methyl hydrogen positioned between the methyl carbon and the NO2 oxygen in distances of 1.483 and 1.155 Å, respectively (Figure 3). In the second step, there is a rotation around the CdN bond, generating the more stable (by 3.6 kcal/mol) tautomer 1a with the oxygen atom positioned in close proximity to the methylene group. The next step is ring closure by an attack of the HONO oxygen on the methylene carbon to form the five-membered ring of the hydro-DNAn with a calculated ∆Gq ) 17.6 kcal/ mol for the transition. The last step, which makes the overall pathway exergonic and thus thermodynamically favorable, is the elimination of a water molecule by hydroxyl attack on the methylene hydrogen and formation of the rigid heterocyclic DNAn, which is 29.1 kcal/mol more stable than TNT. The barrier height for water elimination is 39.4 kcal/mol, which according to our calculations is also the RDS of the reaction. This is in contrast to previous reports claiming that the hydrogen transfer step is the RDS of the reaction.40 It should be noted, however, that previous observations were based on a primary H/D isotope effect, suggesting that C-H bond scission is involved in the RDS.40 Elimination of a water molecule also involves C-H bond scission, and thus, our calculations are supported by experimental evidence. The C-H R attack pathway is exergonic at room temperature, unlike the endergonic C-NO2 homolysis pathway, and it is also kinetically more favorable (as the RDS is lower in energy than the initial C-NO2 homolysis step). Thus, at mild temperatures, Figure 2. Enthalpy and free energy plots for the C-H R attack pathway as calculated at the uB3LYP/cc-pVDZ level of theory. Figure 3. TS for H-transfer in the C-H R attack pathway. Mechanism of Thermal Unimolecular Decomposition of TNT J. Phys. Chem. A, Vol. 111, No. 43, 2007 11077 it is more likely to take place than the C-NO2 homolysis pathway. However, as the temperature increases, the ordered transition states for hydrogen transfer (∆Sq ) -5.4 eu), ring formation (∆Sq ) -4.6 eu), and water elimination (∆Sq ) -5.7 eu) processes are expected to increase in energy due to the increased contribution of the entropy term (T∆S) to ∆G. Thus, at elevated temperatures, the C-H R attack pathway might become kinetically unfavorable. C-NO2 f C-ONO Rearrangement-Homolysis. Experi-mental evidence suggest that C-NO2 homolysis in the gas phase, is a major early thermal decomposition reaction of nitro aromatic compounds at high temperatures; however, the role of this channel as compared to C-NO2 isomerization is difficult to assess in the condensed phase. NO2(g) was rarely detected as a decomposition product in experimental studies of the thermal decomposition of TNT.1 However, the evolution of large quantities of NO(g) has been reported in a few studies.12,13 Nitro-nitrite isomerization is known to take place in aliphatic nitro compounds41 and has been shown to take place in nitrobenzene derivatives containing electron donating substit-uents.3,14 Decomposition of the nitrite is a plausible source of the large amount of NO(g) generated upon rapid thermal decomposition of nitro aromatic compounds. In the condensed phase, or under increased pressure, isomerization could be important because of TNT’s low volume of activation. We have calculated the pathway for nitro-nitrite isomeriza-tion and decomposition of the nitrite as a possible pathway for the decomposition of TNT and the generation of three molecules of NO(g). The pathway is described in Figure 4. Figure 4 describes the C-NO2 f C-ONO rearrangement pathway and the C-NO2 f C-ONO rearrangement-homolysis pathway. Since entropy contributions to the C-NO2 f C-ONO rearrangement pathway are negligible, the free energy values are very similar to the enthalpy values along the pathway. However, in the C-NO2 f C-ONO rearrangement-homolysis pathway, the increasing contribution of entropy to the ∆G energy is indicated by the increasing energy differences between the ∆G and the ∆H pathways as more NO molecules are being eliminated. In the first step of both pathways, the o-NO2 substituent isomerizes to its nitrite (ONO) derivative. This step is mildly exergonic (∆G ) -9.2 kcal/mol); however, the barrier height (∆Gq ) 54.9 kcal/mol) for the rearrangement at room temper-ature is relatively high (which makes this step the RDS of the pathway), suggesting that this step is kinetically unfavorable at room temperature. It should be noted that the experimentally estimated activation energy for C-NO2 f C-ONO rearrange-ment in nitro aromatic compounds is 56 kcal/mol, which closely resembles our calculated barrier for the transition. The transition state (TS) structure for C-NO2 f C-ONO rearrangement is an ordered TS in which both the O and the N atoms are positioned in close proximity to the aromatic carbon (Figure 5). The next step is O-NO homolysis. The energy required to cleave the O-NO bond is only 6.7 kcal/mol, and the products O and NO(g) are slightly more stable than TNT, which makes the formation of O and NO(g) thermodynamically favorable. We have not found the TS for O-NO homolysis; however, we expect this reaction to be barrierless as the microscopic reversible reaction (the recombination of radicals) is mostly regarded to be barrierless. If a barrier exists for O-NO homolysis, it is expected to be significantly lower than for the previous rearrangement step. Thus, most probably as soon as the nitrite derivative of TNT is formed, O-NO cleavage will take place. This postulation is also supported by the fact that no aromatic nitrite was isolated to date; however, the formation of a phenolic radical and NO(g) was observed in a few experiments.1 The pathway continues with two sequences of NO2 isomer-ization and O-NO homolysis to form eventually 3O and 3NO-(g). The overall process at 298 K of C-NO2 f C-ONO rearrangement followed by O-NO homolysis is roughly isoen-ergetic with TNT. However, the intermediates O and 2O as well as their rearranged derivatives 2ONO and 3ONO are somewhat Figure 4. Enthalpy and free energy plots for the C-NO2 f C-ONO rearrangement and C-NO2 f C-ONO rearrangement-homolysis pathways as calculated at the uB3LYP/cc-pVDZ level of theory. Figure 5. TS for NO2 to ONO rearrangement. 11078 J. Phys. Chem. A, Vol. 111, No. 43, 2007 Cohen et al. more stable than TNT. We have also calculated the pathway for C-NO2 f C-ONO rearrangement without subsequent O-NO homolysis for all three substituents and found it to be slightly more exergonic than the rearrangement-homolysis pathway. However, since the barrier for C-NO2 f C-ONO isomerization is significantly higher than the one for O-NO homolysis, the rearrangement pathway is kinetically unfavorable. The sequential C-NO2 f C-ONO rearrangement-homolysis pathway leads to the formation of four molecules out of one: three of them are of nitric oxide (NO(g)), which is a very stable gas molecule. The associated entropy gain for each O-NO homolysis step is calculated as ∆S ) ∼40 eu, and this relatively large entropy gain is expected to contribute significantly to the ∆G value at elevated temperatures, which might cause this pathway to become thermodynamically favorable. In addition, the formation of four gas molecules from one molecule in the bulk phase can boost the pressure abruptly, which might lead to a TNT explosion. However, the RDS of this pathway is not O-NO homolysis but rather a NO2 rearrangement that has a negative entropy. At elevated temperatures, the negative entropy term will increase the free energy for this step, causing it to be less kinetically favorable; thus, as the temperature increases, the C-NO2 f C-ONO rearrangement-homolysis pathway is expected to become exergonic but kinetically unfavorable. The result of O-NO homolysis is a stable oxygen radical molecule and NO(g). The relative stability of the products of O-NO homolysis can be understood by the exceptional stability of the NO(g) molecule. In addition, while TNT is a highly electron deficient compound due to its three NO2 substituents, the TNT oxygen radical derivative is more electron-rich. Thus, release of -one to two NO molecules from TNT stabilizes it by increasing the electron density on the ring. Unlike in the C-NO2 homolysis pathway, here the singlet equivalent of 2O (not shown) is higher in energy by 26 kcal/ mol than the triplet diradical (2O), and thus, it is unlikely that it will be formed. In this pathway, as in the C-NO2 homolysis pathway, the ortho substituents are easier to cleave than the para substituents as the para cleaved NO (p-O, not shown) and p,o-2O (not shown) are 7 and 6.5 kcal/mol higher in energy than o-O and o,o-2O, respectively. On the basis of room temperature analysis, it appears that the most favorable pathway, both kinetically and thermody-namically, is the C-H R attack, which has both the lowest kinetic RDS barrier and is the most exergonic pathway.42 The C-NO2 homolysis pathway is endergonic and, thus, will not occur. The C-NO2 f C-ONO rearrangement-homolysis pathway is isoenergetic with TNT; however, its RDS barrier is larger than the C-H R attack and, thus, will be slower than C-H R and less favorable. Thermal Decomposition of TNT at Elevated Tempera-tures. The situation is expected to change at elevated temper-atures in which the entropy contributes significantly to ∆G. Since ∆G ) ∆H- T∆S, as T increases, the contribution of the second term, T∆S, becomes greater. In this case, pathways that involve a significant increase in entropy, such as C-NO2 homolysis and O-NO homolysis, are expected to become more energetically favorable. Pathways that involve a decreased entropy of products and in TS, such as C-H R attack, are expected to be less favorable. The C-NO2 f C-ONO rearrangement-homolysis pathway involves both reduction in the entropy of the C-NO2 f C-ONO rearrangement TS and an increase in the entropy of the products of O-NO homolysis. Although it is expected to be more thermodynamically favorable, it will be less kinetically favorable due to the constrained geometry of the rearrangement transition states. We have conducted a thermochemical analysis of the calculated processes at 3500 K, which is the estimated temperature of the shock wave generated during TNT detonation, and at two additional intermediate temperatures to verify our assumptions. Using the freqchk option of the Gaussian 03 program, we obtained thermal corrections to the Gibbs free energy at 900, 1700, and 3500 K (based on the RRHO approximation). The results of this analysis are presented in Figures 6-8. Since the enthalpy (∆H) dependence on temperature is negligible, the figures only describe the free energy pathways (∆G). As expected, at elevated temperatures, all pathways become more exergonic. The C-NO2 f C-ONO pathway is the most exergonic, and the C-H R attack is the least exergonic one. However, kinetically, the situation is different. Figure 6 shows that for the C-H R attack pathway, although the overall path becomes more exergonic with increasing the temperature, its TS significantly increases in energy (i.e., from 37.7 kcal/mol at room temperature to 60.8 kcal/mol at 3500 K), which makes this pathway the most kinetically unfavorable one at high Figure 6. Free energy plots at 298, 900, 1700, and 3500 K for the C-H R attack pathway as calculated at the B3LYP/cc-pVDZ level of theory. Mechanism of Thermal Unimolecular Decomposition of TNT J. Phys. Chem. A, Vol. 111, No. 43, 2007 11079 temperatures with respect to the other two pathways. In contrast, the C-NO2 homolysis pathway, although less exergonic than the C-NO2 f C-ONO pathway, becomes the most kinetically favorable pathway to take place at elevated temperatures as the barrier to cleave the C-NO2 bond diminishes with increasing the temperature. It should be noted, however, that at 900 K, this pathway is still endergonic and thus is not expected to generate significant heat and to lead to a detonation at this temperature. At 1700 K, it becomes moderately exergonic and kinetically favorable; thus, most probably it is the major decomposition pathway of TNT at this temperature. The C-NO2 f C-ONO rearrangement-homolysis pathway becomes the most exergonic pathway at elevated temperatures. However, the first rearrangement step, which is also the RDS of this pathway, is entropically unfavorable; therefore, this pathway becomes less kinetically favorable than the C-NO2 homolysis. Comparing the C-NO2 f C-ONO rearrangement-homoly-sis and the C-H R attack pathways, it should be noted that although the RDS step of the former is entropically unfavorable, its activation energy rises only slightly with temperature, whereas in the C-H R attack pathway, the activation energy rise with the temperature is more significant. Accordingly, the C-NO2 f C-ONO rearrangement-homolysis pathway be-comes kinetically more favorable than the C-H R attack pathway at temperatures as high as 3500 K (but less favorable at temperatures lower than ∼2000 K). However, at such elevated temperatures, the C-NO2 homolysis pathway is dominant; thus, the contribution of either the C-H R attack pathway or the C-NO2 f C-ONO pathway to TNT decomposition at high temperatures is negligible. Our results at elevated temperatures are supported by experimental evidence showing that at moderate temperatures, the C-H R attack pathway dominates the thermal decomposition Figure 7. Free energy plots at 298, 900, 1700, and 3500 K for the C-NO2 homolysis pathway as calculated at the B3LYP/cc-pVDZ level of theory. Figure 8. Free energy plots at 298, 900, 1700, and 3500 K for the C-NO2 f C-ONO rearrangement-homolysis pathway as calculated at the B3LYP/cc-pVDZ level of theory. 11080 J. Phys. Chem. A, Vol. 111, No. 43, 2007 Cohen et al. of TNT, and at elevated temperatures, C-NO2 homolysis is the most important pathway. According to our calculations, the most exergonic reaction taking place at 3500 K is the C-NO2 f C-ONO rearrangement-homolysis. However, it is less favorable kinetically than C-NO2 homolysis at this temperature and is expected to be slower. We have conducted a kinetic analysis of the relative rates of these two reactions based on the Eyring equation where k is the rate, kB is Boltzmann’s constant, h is Planck’s constant, ∆Gq is the RDS free energy barrier, R is the gas constant, T is the temperature in Kelvin, and t1/2 is the half-life time in seconds. Implementation of these equations in the present case (where T ) 3500 K) yields a reactant (TNT) half-life time of 50 ps for the C-NO2 f C-ONO pathway (∆Gq ) 60 kcal/ mol) and 10 fs for the C-NO2 pathway (∆Gq ∼0). Since the estimated reaction time during detonation is ∼150 fs (assuming a shock wave velocity of 6900 m/s during detonation and a typical TNT unit cell size of ∼10 Å), it appears that although the C-NO2 f C-ONO pathway is considerably more exer-gonic than the C-NO2 pathway and therefore is expected to generate more heat, its relative contribution to detonation is minor since it is 3.5 orders of magnitude slower than the C-NO2 homolysis reaction. The situation might be different under extreme pressure, where the C-NO2 f C-ONO rearrangement might become favorable with respect to C-NO2 homolysis. If rearrangement takes place to form three ONO derivatives of TNT before C-NO2 or C-ONO homolysis takes place, then O-NO homolysis (∆G298O-NOcleavage ) 6.7 kcal/mol) will be favored over C-ONO homolysis (∆G298C-ONOcleavage ) 53 kcal/mol). To estimate the temperature range in which the relative importance of the different pathways changes from C-H attack to C-NO2 homolysis, we have constructed a graph of ∆Gq versus T for all three pathways (Figure 9). The ∆Gq value for each pathway is taken as the RDS ∆Gq value for the specific temperature. In the case of C-NO2 homolysis, the pathway was separated into three individual pathways: first, second, and third C-NO2 eliminations, and for each elimination step, the ∆Gq value was taken as the ∆G for C-NO2 cleavage (since the homolysis process was considered to be barrierless). The value for which the ∆G versus T line crossed the X axis (0 kcal/mol) is the temperature in which the pathway becomes exergonic. Since for NO2 elimination pathways the ∆Gq values plotted are the ∆G values for C-NO2 homolysis, and this event is endergonic at low temperatures, only when ∆G reaches 0.0 kcal/ mol does the homolysis pathway become exergonic. Accord-ingly, Figure 9 shows that the temperatures for which C-NO2 homolysis becomes exergonic are ∼1250 K for cleaving one NO2, ∼1400 K for two NO2 groups, and ∼1500 K for elimination of all three NO2 groups. Below 1250 K, even if C-NO2 homolysis takes place (since it has a lower activation energy than the other pathways), it will not generate heat since it is endergonic and might even inhibit the detonation. Thus, at temperatures below 1250 K, the most favorable pathway is the C-H attack since it is exergonic and requires a lower energy than the C-NO2 f C-ONO rearrangement-homolysis path-way. The temperature in which the C-NO2 f C-ONO rearrangement-homolysis pathway becomes more favorable than the C-H attack pathway is ∼1800 K. At this temperature, the C-NO2 homolysis pathway is already the main decomposi-tion pathway. Thus, before C-NO2 homolysis begins, the C-H attack pathway is expected to be a major decomposition pathway, whereas C-NO2 f C-ONO rearrangement-ho-molysis is expected to have a minor contribution. To determine which pathway takes place at the estimated detonation time of the TNT detonation shock wave (100-200 fs), we constructed an Arrhenius plot (ln k vs 1/T) (Figure 10). The values of ln k were calculated using the Eyring equation (eq 1) in which the ∆Gq value for each pathway is assumed to be the RDS ∆Gq value at the specified temperature. For C-NO2 homolysis, the decomposition pathway was again separated into three individual paths: first, second, and third C-NO2 elimina-tions. For each elimination of an NO2 group, the ∆Gq value is assumed to be given by the ∆G value for C-NO2 cleavage. The ln k values for detonation with t1/2 in the range of 100-200 fs are 29.6-28.9, respectively (eq 2). The pathways that Figure 9. Activation free energy vs temperature for all computed TNT decomposition pathways. Inset illustrates the temperature in which the C-NO2 f C-ONO rearrangement pathway becomes more kinetically favorable than the CH attack pathway. k ) kBT h e-∆Gq/RT (1) t1/2 ) ln 2 k (2) Mechanism of Thermal Unimolecular Decomposition of TNT J. Phys. Chem. A, Vol. 111, No. 43, 2007 11081 cross the window of these ln k values represent the pathways that are kinetically accessible during TNT detonation in specific temperatures. According to Figure 10, only three pathways out of the five examined can “cross the window” of ln k values of 28.0 and 30.0, in the temperature range of 200-4000 K: first, NO2 elimination at ∼1200 K; second, NO2 elimination at ∼1400 K; and thirdm NO2 elimination at ∼1500 K. This suggests that for the estimated detonation time scale, only the C-NO2 homolysis pathway is kinetically accessible. The C-H attack and the C-NO2 f C-ONO rearrangement-homolysis pathways are too slow to generate a shock wave with a detonation velocity of 6900 m/s and thus are kinetically inaccessible. The results here suggest that the temperature range in which the C-NO2 homolysis occurs consistently with this time scale is 1200-1500 K. At this temperature range, it was shown that C-NO2 homolysis becomes exergonic and thus favorable with respect to the other two possible decomposition pathways. Thus, even though TNT decomposition can take place at lower temperatures than 1200 K, it most probably will not lead to detonation before C-NO2 homolysis becomes exer-gonic.43 In conclusion, we conducted a computational study of the unimolecular thermal decomposition of TNT to elucidate its explosive nature. On the basis of previous experimental observations, we have postulated three possible pathways for TNT decomposition, keeping the aromatic ring intact, and calculated them at room temperature (298 K), 800, 900, 1500, 1700, 2000, and at the detonation temperature of 3500 K (which is the estimated temperature generated during TNT explosion). Our calculations suggest that at relatively low temperatures, the C-H R attack, leading to the formation of 2,4-DNAn, is both kinetically and thermodynamically the most favorable pathway, while the C-NO2 homolysis pathway is endergonic and kinetically less favorable. At ∼1250-1500 K, the situation changes, and the C-NO2 homolysis pathway dominates TNT decomposition. C-H R attack becomes both kinetically unfa-vorable and thermodynamically less favorable as compared to the C-NO2 homolysis pathway. The C-NO2 f C-ONO rearrangement-homolysis pathway is thermodynamically more favorable than C-NO2 homolysis at room temperature and is the most exergonic pathway at high temperatures; however, at all temperatures, the C-NO2 f C-ONO rearrangement-homolysis pathway is kinetically unfavorable as compared to the other two pathways and thus is slower. As a result, its contribution to the TNT decomposition is expected to be minor. The computational temperature analysis shed light on the pathway that leads to TNT explosion and on the temperature when the decomposition becomes exergonic. The results appear to correlate closely with the experimental time scale (100-200 fs) derived from the shock wave velocity for which only the C-NO2 homolysis pathway is kinetically accessible. We are currently performing atomistic scale molecular dynamic simula-tions11,44,45 for initial chemical events in the thermal decomposi-tion of TNT; we have preliminary results that support our DFT calculations. Acknowledgment. This research was supported by the MAFAT Israel Defense Research Agency. The Fritz Haber Research Center is supported by the Minerva Gesellschaft fu ¨r die Forschung, GmbH Mu ¨nchen, FRG. Supporting Information Available: Full Gaussian 03 citation (ref 24) and XYZ coordinates of all calculated com-pounds. This material is available free of charge via the Internet at References and Notes (1) Brill, T. B.; James, K. J. Chem. ReV. 1993, 93, 2667. (2) Brill, T. B.; James, K. J. J. Phys. Chem. 1993, 97, 8759. (3) Brill, T. B.; James, K. J.; Chawla, R.; Nicol, G.; Shukla, A.; Futrell, I. H. J. Phys. Org. Chem. 1999, 12, 819. (4) Singh, G.; Kapoor, I. P. S.; Mannan, S. M.; Tiwari, S. K. J. Hazard. Mater. 1999, 68, 155. (5) Long, G. T.; Brems, B. A.; Wight, C. A. Thermochim. Acta 2002, 388, 175. (6) Dacons, J. C.; Adolph, H. G.; Kamlet, M. J. J. Phys. Chem. 1970, 74, 3035. (7) Rogers, R. N. Anal. Chem. 1967, 39, 730. (8) Dewar, M. J. S.; Ritchie, J. P.; Alster, J. J. Org. Chem. 1985, 50, 1031. Figure 10. Arrhenius plot (ln k vs 1/T) for TNT decomposition pathways. Ln k values are calculated from the Eyring equation (eq 1) in which the ∆Gq for each pathway is assumed to be the RDS at the specified temperature. 11082 J. Phys. Chem. A, Vol. 111, No. 43, 2007 Cohen et al. (9) Gonzalez, A. C.; Larson, C. W.; Mcmillen, D. F.; Golden, D. M. J. Phys. Chem. 1985, 89, 4809. (10) Dubnikova, F.; Kosloff, R.; Almog, J.; Zeiri, Y.; Boese, R.; Itzhaky, H.; Alt, A.; Keinan, E. J. Am. Chem. Soc. 2005, 127, 1146. (11) van Duin, A. C. T.; Zeiri, Y.; Dubnikova, F.; Kosloff, R.; Goddard, W. A. J. Am. Chem. Soc. 2005, 127, 11053. (12) Hand, C. W.; Merritt, C.; Dipietro, C. J. Org. Chem. 1977, 42, 841. (13) Fields, E. K.; Meyerson, S. J. Am. Chem. Soc. 1967, 89, 3224. (14) Brill, T. B.; James, K. J. In Molecular Processes in Nitroaromatic ExplosiVes: Proof and Internal Control of the Nitro-Nitrite Rearrangement; 11th Detonation Symposium; Office of Naval Research: Arlington, VA, 1998; p 917. (15) Turner, A. G.; Davis, L. P. J. Am. Chem. Soc. 1984, 106, 5447. (16) Kaufman, J. J.; Hariharan, P. C.; Roszak, S.; Vanhemert, M. J. Comput. Chem. 1987, 8, 736. (17) Cox, J. R.; Hillier, I. H. Chem. Phys. 1988, 124, 39. (18) (19) Carper, W. R.; Davis, L. P.; Extine, M. W. J. Phys. Chem. A 1982, 86, 459. (20) Golovina, N. I.; Titkov, A. N.; Raevskii, A. V.; Atovmyan, L. O. J. Solid State Chem. 1994, 113, 229. (21) Vrcelj, R. M.; Sherwood, J. N.; Kennedy, A. R.; Gallagher, H. G.; Gelbrich, T. Cryst. Growth Des. 2003, 3, 1027. (22) Although intermolecular reaction chemistry may not exist, the neighboring molecules create a cage effect that may alter the unimolecular chemistry conclusions. However, study of such an effect is beyond the scope of this paper. (23) Additional routes resulting from combining the mentioned pathways in different stages or ring fission after formation of radical species on the aromatic ring might be accessible pathways that have not been studied here. However, it should be noted that the RDS of each studied route is in most cases the initiation step; thus, the steps preceding it are less likely to contribute to the understanding of the kinetic processes leading to TNT detonation, which is the aim of this study. (24) Frisch, M. J.; et al. Gaussian 03, revision C.02; Gaussian, Inc.: Pittsburgh, PA, 2004. (25) Becke, A. D. J. Chem. Phys. 1993, 98, 5648. (26) Stevens, P. J.; Devlin, F. J.; Chabalowski, C. F.; Frisch, M. J. J. Phys. Chem. 1994, 98, 11623. (27) Dunning, T. H., Jr. J. Chem. Phys. 1989, 90, 1007. (28) Cremer, D.; Kraka, E.; Szalay, P. G. Chem. Phys. Lett. 1998, 292, 97. (29) Peng, C.; Ayala, P. Y.; Schlegel, H. B.; Frisch, M. J. J. Comput. Chem. 1996, 17, 49. (30) Schlegel, H. B. J. Comput. Chem. 1982, 3, 214. (31) Stratmann, R. E.; Burant, J. C.; Scuseria, G. E.; Frisch, M. J. J. Chem. Phys. 1997, 106, 10175. (32) Schaftenaar, G.; Noordik, J. H. J. Comput.-Aided Mol. Des. 2000, 14, 123. (33) He, Y. Z.; Cui, J. P.; Mallard, W. G.; Tsang, W. J. Am. Chem. Soc. 1988, 110, 3754. (34) Guidry, R. M.; Davis, L. P. Thermochim. Acta 1979, 32, 1. (35) Chakraborty, D.; Muller, R. P.; Dasgupta, S.; Goddard, W. A., III. J. Phys. Chem. A 2001, 105, 1302. (36) It should be noted, however, that while the enthalpy of radical recombination/homolysis reactions may be estimated well by the calculations of reaction energetics, often there are entropic effects making this more suspect for free energy barriers. The required analysis to address this issue is beyond the scope of this paper. (37) Wu, C. J.; Fried, L. E. J. Phys. Chem. A 1997, 101, 8675. (38) Chakraborty, D.; Muller, R. P.; Dasgupta, S.; Goddard, W. A. J. Phys. Chem. A 2000, 104, 2261. (39) Willadse, P; Zerner, B.; MacDonal, C. J. Org. Chem. 1973, 38, 3411. (40) Shackelford, S. A.; Beckmann, J. W.; Wilkes, J. S. J. Org. Chem. 1977, 42, 4201. (41) Gray, P.; Rathbone, P.; Williams, A. J. Chem. Soc. 1960, 3932. (42) It should be noted that the calculated HOMO-LUMO gap for TNT is ∼4.3 eV, which corresponds to ∼116 kcal/mol. Accordingly, it is unlikely that TNT decomposition is promoted by excitation as the excitation energy is higher than the energy required for thermal decomposition of TNT in each of the calculated pathways. (43) It should be noted that all calculations were performed at atmospheric pressure. It is conceivable that under conditions of higher pressure, the temperature in which detonation begins to take place should be lower. (44) Strachan, A.; Kober, E. M.; van Duin, A. C. T.; Oxgaard, J.; Goddard, W. A., III. J. Chem. Phys. 2005, 122, 54502. (45) Strachan, A.; van Duin, A. C. T.; Chakraborty, D.; Dasgupta, S.; Goddard, W. A. Phys. ReV. Lett. 2003, 91, 98301. Mechanism of Thermal Unimolecular Decomposition of TNT J. Phys. Chem. A, Vol. 111, No. 43, 2007 11083
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14522	http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1200745.html	SOLUTION: A court consists of 3 judges. Two of them, independently of each other, make correct decisions with a probability of p. The third judge agrees with the first two decisions when the SOLUTION: A court consists of 3 judges. Two of them, independently of each other, make correct decisions with a probability of p. The third judge agrees with the first two decisions when the Algebra->Probability-and-statistics -> SOLUTION: A court consists of 3 judges. Two of them, independently of each other, make correct decisions with a probability of p. The third judge agrees with the first two decisions when theLog On Algebra: Probability and statistics SectionSolvers SolversLessons LessonsAnswers archive Answers Discover more math Mathematics Algebra Inc Math Click here to see ALL problems on Probability-and-statistics Question 1200745: A court consists of 3 judges. Two of them, independently of each other, make correct decisions with a probability of p. The third judge agrees with the first two decisions when they match. In case of different opinions of the first two judges, the third judge decides on his own and makes a mistake with a probability of q. What is the probability that the court will not make a mistake if the verdict is decided by a majority vote? ([p, q] = [0.5, 0.43]) Found 2 solutions by math_tutor2020, ikleyn: Answer bymath_tutor2020(3817) (Show Source): You can put this solution on YOUR website! Answer: 0.535 This value is exact without any rounding done to it. =================================================================================================================== Explanation: Let's give the judges code names: {J1,J2,J3} We'll have them be presented in the order that was mentioned in the instructions. J1 and J2 make their decisions independently of each other. The probability they reach a correct decision is p = 0.5 for each of those two judges. If {J1,J2} reach the same outcome, then J3 rules the same way. If {J1,J2} reach differing outcomes, then J3 thinks for her/himself to reach their own decision (making a mistake with probability q=0.43). For this situation J3 makes the correct decision with probability 1-q=1-0.43=0.57 Here is a way to represent a two-way table of outcomes for the first two judges. J2 right J2 wrong J1 right J1 wrong For example, the upper left corner has both J1,J2 give the correct ruling. Here is a non-tabular format to represent those four outcomes: J1 right, J2 right J1 right, J2 wrong J1 wrong, J2 right J1 wrong, J2 wrong Let's list those four outcomes in a new table. We'll have them as the 4 rows. The 2 columns will be the outcomes for J3. J3 right J3 wrong J1 right,J2 right A J1 right,J2 wrong B C J1 wrong,J2 right D E J1 wrong,J2 wrong F I then added letters A through F to represent the various possible scenarios. Scenario A is when all three judges make the correct decision. The cell next to A is blank because J3 doesn't make the wrong decision when her/his two other colleagues make the correct decision (remember J3 mimics the other two judges if they rule the same way). This also explains why we have a blank cell in the bottom left corner. Scenario B is when J1 gets it right, J2 gets it wrong, J3 gets it right. Scenario C is when J1 gets it right, J2 gets it wrong, J3 gets it wrong. And so on. The letters A, B, and D correspond to situations where the court gets the correct outcome. This is because there are at least 2 judges that get the correct ruling. A = all 3 judges get it right B = J1 and J3 get it right D = J2 and J3 get it right All other situations involve at least 2 judges getting the wrong verdict; hence the wrong final outcome. To find the probability that the court gets the correct decision, we need to calculate the following: P(A) P(B) P(D) ----------------------------- P(A) = P(J1 right,J2 right) P(A) = P(J1 right)P(J2 right) P(A) = pp P(A) = p^2 P(A) = 0.5^2 P(A) = 0.25 This works because J1,J2 are independent of each other. J3 does not alter the outcome here, so we ignore the probability associated with this judge. P(B) = P(J1 right, J2 wrong, J3 right) P(B) = P(J1 right)P(J2 wrong)P(J3 right) P(B) = p(1-p)(1-q) P(B) = 0.5(1-0.5)(1-0.43) P(B) = 0.1425 P(D) = P(J1 wrong, J2 right, J3 right) P(D) = P(J1 wrong)P(J2 right)P(J3 right) P(D) = (1-p)p(1-q) P(D) = (1-0.5)0.5(1-0.43) P(D) = 0.1425 ----------------------------------------------------------- Conclusion: P(A) = 0.25 P(B) = 0.1425 P(D) = 0.1425 P(A)+P(B)+P(D) = 0.25+0.1425+0.1425 = 0.535 represents the probability the court gets the right decision. The court gets it right exactly 53.5% of the time. This is around 50%, so it's about as good as a coin toss in my opinion. Those don't seem like good odds. Answer byikleyn(52786) (Show Source): You can put this solution on YOUR website! . A court consists of 3 judges. Two of them, independently of each other, make correct decisions with a probability of p. The third judge agrees with the first two decisions when they match. In case of different opinions of the first two judges, the third judge decides on his own and makes a mistake with a probability of q. What is the probability that the court will not make a mistake if the verdict is decided by a majority vote? ([p, q] = [0.5, 0.43]) ~~~~~~~~~~~~~~~~~~~~~~~~~ Here I provide another solution for this given problem. It has a different form from the solution by the other tutor, but produces the same answer. I wrote it simply to represent another way to organize and express the thoughts. We do not know what cases the judges consider and what judgment/decisions they make. We only know that the decisions can be Right or Wrong . Therefore, based on given information, we only can make a table of probabilities for all possible situations. So, I made this table: it is below. 1 2 3 verdict decided Include (+) Individual by a majority not include (-) probabilities for each possible court decision -------------------------------------------------------------------------------------------- 1 R R R ---> R + 0.50.51 2 W W W ---> W - 3 R W R ---> R + 0.50.50.57 4 R W W ---> W - 5 W R R ---> R + 0.50.50.57 6 W R W ---> W - The leftmost column is for the numbers of lines. The digits 1, 2 and 3 in the horizontal upper line represent the judges. "R" represents right decision; "W" represents wrong decision. The symbols in the table below "1", "2" and "3" symbolize the decisions (R for right, W for wrong). The symbols in the column named "verdict decided by a majority" represent the logical consequence of the decisions made in columns 1, 2, and 3. It is how the court makes its final decision, based on individual decisions of the judges. The arrows ( ---> ) show the logical implications ("verdict decided by a majority"). Notice that in the table I listed ALL LOGICALLY POSSIBLE situations. There are NO other possible situations that would be consistent with the problem. According to the problem, the question is about the probability of the final court's decision to be right. So, in column "include or not include" I write "+" for right decisions to include them into the final count or "-" for wrong decisions to NOT include them into the final count. In the rightmost column, I calculated the probabilities for each possible RIGHT decision of the curt. From this consideration, the final probability of the right verdict of the court is 0.50.50.57 +0.50.50.57 = 0.535. ANSWER Solved. This my solution is written as a description of an ALGORITHM calculating the desired probability based on given input data.
14523	https://www.youtube.com/watch?v=Y-ebdlRb3HE	Buffers and Henderson-Hasselbalch \| Chemistry \| Khan Academy ailabhcmus 462 subscribers 3 likes Description 835 views Posted: 21 Nov 2016 Transcript: let's say I have some weak acid I'll call it ha ha a is a placeholder for uh really a a whole set of of elements that I could put there could be Florine it could be an ammonia molecule if you add an H it becomes ammonium so this isn't any particular element I'm talking about this is just kind of a general way of writing a an acid and let's say it's an equilibrium it's an equilibrium with of course and you've seen this multiple times a proton and all of this is in an aquous solution aquous between this proton jumping off of this and its conjugate base a minus and we could have also written a a base equilibrium where we say the conjugate base could disassociate or it could essentially grab a hydrogen from the water and create o and we've done that multiple times but that's not the point of this video so let's just think a little bit about what would happen to this equilibrium if we were to stress it in some way and you can already imagine that I'm about to touch on laler principle which essentially just says that look if you stress an equilibrium in any way the equilibrium moves in such a way to relieve that stress so let's say that the stress that I apply to this system let's say that the stress that I apply is I'm going to add I'm going to add let me do a different color color I'm going to add some strong base that's too dark I'm going to add some na o and we know this is a strong base when you put it in an aquous solution the sodium part just kind of dissociates but the the more important thing you have all this o in the solution which wants to grab hydrogens away so when you add this o to the solution what's going to happen for every mole that you add so for every not even just mole for every molecule you add out of this into the solution it's going to eat up a molecule of hydrogen right so for example if you had one mole of hydrogen molecules in your solution and you added one mole of sodium hydroxide to your solution right when you do that all of this is going to react with all of that and it's and the O's are going to react with the H's so just so you be and form water and they'll both just kind of disappear into the solution they didn't disappear they all turned into water and so all of this hydrogen will go away or at least the hydrogen that was initially there that one mole of hydrogens will disappear so what will what should happen to this reaction well we know this is an equilibrium reaction so as these hydrogens disappear as these hydrogens disappear because this is an equilibrium reaction or because this is a weak base more of this is going to be converted or the more of this is going to be converted into these two products to kind of make up for for that loss of hydrogen and you can even play with it on the mat so this is going to so this hydrogen goes down initially and you know all of this equilibrium then it starts getting to equilibrium very fast but this is going to go down this is going to go up and then this is going to go down less because sure when you put the sodium hydroxide there it just ate up all of the hydrogens but then you have this you can kind of view as this this this spare hydrogen capacity here to produce hydrogens and when these disappear this weak base will disassociate more the equilibrium will move more in this direction and then you'll have so immediately this will eat all of that but then when the equilibrium moves in that direction some of the a lot of the hydrogen will be replaced so if you think about what's happening if I just threw if I just threw this sodium hydroxide in water so if I just did NaOH and an aquous solution so that's just throwing it in water that disassociates completely into the sodium sodium cation and hydroxide annion and so you all of a sudden you would immediately increase the the quantity of O's by essentially the amount of the number of moles of sodium hydroxide you're adding and you would immediately increase the pH right remember when you increase the amount of O you would decrease the P right and that's just because it's the negative law l or so if you increase o You're decreasing p and you're increasing pH and just to think o is you're you're making it more basic more basic and a high pH is also very basic if you have a mole of this you end up with a pH of 14 and if you had a strong acid not a strong base you would end up with a pH of zero and and you know hopefully you're getting a little bit familiar with that concept right now but if it if it confuses you just play around with the logs a little bit and you'll eventually get it but just to get back to the point if I if if you just did this in water you immediately get a super high pH because the O concentration goes through the roof but if you do it here if you apply the the sodium hydroxide to this solution the solution that contains a weak acid and its conjugate base the weak acid and its conjugate base what happens sure it immediately reacts with all of this hydrogen and eats it all up but then you have this extra Supply here that just keeps providing more and more hydrogens and it'll make up a lot of the loss so essentially the the stress won't be as bad and over here you you you dramatically increase increase the pH when you just throw it on water here you're going to increase the pH by a lot lot lot less and in future videos we'll actually do the math of how much less it's increasing the pH but the way you could think about is this this is kind of a shock absorber for pH it even though you threw the strong base onto this into this solution it didn't increase the pH as much as you would have expected and you can make it the other way if I just wrote this exact same reaction is a basic reaction and remember this is this the same thing so if I just wrote this as a minus I just wrote as conjugate base is in equilibrium with the conjugate base grabbing some water from the surrounding aquous solution everything we're dealing with right now is in an aquous solution and of course that water that it grabbed from is now going to be an O remember these are just equivalent reactions here I'm writing it as an acidic reaction here I'm writing it as a basic reaction but they're equivalent now if you were to add a strong a a strong acid to this solution what would happen so if I were to throw hydrogen chloride into this well hydrogen chloride if you just throw it into straight up water without this solution it would completely disassociate into a bunch of hydrogens and a bunch of chlorine anion chlorine anion and it would immediately make it very acidic you would get you'd get to a very low PH if you had a mole of this if if your concentration was one molar then this will go to a pH of zero but what happens if you add hydrochloric acid to this solution right here this one that has this weak base and its conjugate weak acid well all that all of these hydrogen protons that disassociate from the hydrochloric acid are all going to react with these O's you have here and they're just going to cancel each other out they're just going to merge with these and turn into water and become part of the aquous solution so this the O's are going to go down initially but then you have this reserve of weak base here and we let talia's principle tells us look if we have a stressor that is decreasing our overall concentration of O then the reaction is going to move in the direction that relieves that stress so the reaction is going to go in that direction so you're going to have more of our weak base turning into a weak acid and producing more o so the so the pH won't go down as much as you would expect if you just threw this in water this is going to lower the ph but then you have more o that could be produced as this guy grabs more and more hydrogens from the water so the way to think about is it's kind of like a like a cushion or a spring on in terms of what a strong acid or B could do to the solution and that's why it's called a buffer a buffer buffer because it provides a cushion on acidity if you add a strong base to water you immediately increase its pH or you decrease its acidity dramatically but if you add a strong base to a buffer because of the of latalia principle essentially you're not going to affect the pH as much same thing if you add an acid to that same buffer it's not going to affect the pH as much as you would have expected if you had thrown that acid in water because the the equilibrium reaction can always kind of refill the amount of O that you lost in the if you're adding acid or it can refill the amount of hydrogen you lost if you're adding a base and that's why it's called a buffer it provides a cushion so it gives some stability to the solution's ph and so the actual definition really is just a solution of the definition of a buffer is just a solution of a weak acid in equilibrium with its conjugate weak base that's what a buffer is and it's called a buffer because it provides you this kind of cushion of pH it's it's kind of a a stress absorber or a shock absorber for the acidity of a solution now with that said let's explore a little bit the math of a buffer which is really just the math of a of a weak acid so if we rewrite the equation again so h a is in equilibrium everything's in an aquous solution with hydrogen and its conjugate base we know that there's an equilibrium constant for this we've done many videos on that the equilibrium constant here is equal to the concentration of our hydrogen proton times the con the concentration of our conjugate base when I say concentration I'm talking marity moles per liter divided by the concentration of our weak acid now let's take the log or let's take the negative log of both sides of this equ actually let me do something a little let's solve for hydrogen concentration because what I want to do is I want to figure out a formula and we'll call it the Henderson Hasselback formula which which a lot of books want you to memorize which I don't think you should I think you should always just be able to go from the this kind of basic assumption and get to it but let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula so if we want to solve for hydrogen we can multiply both sides by the reciprocal of this right here and you get hydrogen concentration and I'm flipping well let me just do Ka times I'm multiplying both sides times the reciprocal of that so times a the concentration of our weak acid divided by the concentration of our weak base is is equal to our concentration of our hydrogen fair enough now let's take the negative log of both sides so the negative log do that negative log of all of that stuff of your acidic equilibrium constant times times make sure let me see what green was I using times h a our weak acid divided by our weak base is equal to the negative log of our hydrogen concentration which is just our pH right negative log of hydrogen concentration is that's the definition of pH and I'll write the p and the H in different colors so you know that P is just means negative log minus log that's all a base 10 let's see if we can simplify this anymore so our our logarithmic propert properties we know that when you take the when you when you take the log of something and you multiply it that's the same thing as taking the log of this plus the log of that so this can be simplified to minus log of our Ka minus minus the log minus the log of our weak acid concentration divided by its conjugate base concentration is equal to the pH now this is just the pka of our weak acid which is just a negative log of of its equilibrium constant so this is just the pka and the minus log of ha over a if you what we can do is we could take make this a plus and just take this to the minus one power right that's just another logarithm property and you can review the logarithm videos if that confused you and this this to the minus one power just means invert this so we could say plus the logarithm of our conjugate base concentration divided by the weak acid concentration is equal to the ph and this right here this is called the Henderson Hasselback equation Henderson Hasselback and I really encourage you not to memorize it because if you do attempt to memorize it you're and within a few hours you're going to forget whether this was a plus over here you're going to forget this and you're going to forget whether you put the a minus or the ha on the numerator the denominator and if you forget that it's fatal the better thing is to just start from your base assumptions and if you weren't and trust me it took me a couple minutes to do it but if you just do it really fast on paper and you don't have to talk it through the way I did it'll take you no time at all to come to this equation it's much better than memorizing it and you you won't forget it when you're 30 years old but what's useful about this well it immediately relates pH to our PKA and this is a constant right for an equilibrium plus the log of the ratios between the a the acid and the conjugate base so if I the more conjugate base I have and the less acid I have the more my pH is going to increase right if this goes up and this is going down my pH is going to increase which makes sense because I have more base in the solution and if I have the inverse of that my
14524	https://www.academia.edu/17963288/Mathematical_Methods_in_the_Physical_Sciences_3e	(PDF) Mathematical Methods in the Physical Sciences, 3e close Welcome to Academia Sign up to get access to over 50 million papers By clicking Continue, you agree to our Terms of Use and Privacy Policy Continue with GoogleContinue with AppleContinue with Facebook email Continue with Email arrow_back Continue with Email Sign up or log in to continue. Email Next arrow_forward arrow_back Welcome to Academia Sign up to continue. By clicking Sign Up, you agree to our Terms of Use and Privacy Policy First name Last name Password Sign Up arrow_forward arrow_back Hi, Log in to continue. Password Reset password Log in arrow_forward arrow_back Password reset Check your email for your reset link. Your link was sent to Done arrow_back Facebook login is no longer available Reset your password to access your account: Reset Password Please hold while we log you in Academia.edu no longer supports Internet Explorer. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds toupgrade your browser. Log In Sign Up Log In Sign Up more About Press Papers Terms Privacy Copyright We're Hiring! Help Center less Outline keyboard_arrow_down Title All Topics Mathematics Mathematical Methods in the Physical Sciences, 3e 859 pages download Download Free PDF Download Free PDF Mathematical Methods in the Physical Sciences, 3e Adrian Lee description See full PDF download Download PDF format_quote Cite close Cite this paper bookmark Save to Library share Share close Sign up for access to the world's latest research Sign up for free arrow_forward check Get notified about relevant papers check Save papers to use in your research check Join the discussion with peers check Track your impact See full PDF download Download PDF Loading Preview MATHEMATICAL METHODS IN THE PHYSICAL SCIENCES Third Edition MARY L. BOAS DePaul University MATHEMATICAL METHODS IN THE PHYSICAL SCIENCES Sorry, preview is currently unavailable. You can download the paper by clicking the button above. Related papers MATHEMATICAL METHODS FOR PHYSICISTS SIXTH EDITION Rugi Baam download Download free PDFView PDF chevron_right Solucionario CAlculo Impares Martina Martínez Carballeira download Download free PDFView PDF chevron_right Alan Jeffrey Advanced Engineering Mathe Sumit N O N R O U T I N E Rajput download Download free PDFView PDF chevron_right Related topics Mathematicsadd Follow close Welcome to Academia Sign up to get access to over 50 million papers By clicking Continue, you agree to our Terms of Use and Privacy Policy Continue with GoogleContinue with AppleContinue with Facebook email Continue with Email arrow_back Continue with Email Sign up or log in to continue. Email Next arrow_forward arrow_back Welcome to Academia Sign up to continue. By clicking Sign Up, you agree to our Terms of Use and Privacy Policy First name Last name Password Sign Up arrow_forward arrow_back Hi, Log in to continue. Password Reset password Log in arrow_forward arrow_back Password reset Check your email for your reset link. Your link was sent to Done arrow_back Facebook login is no longer available Reset your password to access your account: Reset Password Please hold while we log you in Explore Papers Topics Features Mentions Analytics PDF Packages Advanced Search Search Alerts Journals Academia.edu Journals My submissions Reviewer Hub Why publish with us Testimonials Company About Careers Press Help Center Terms Privacy Copyright Content Policy 580 California St., Suite 400 San Francisco, CA, 94104 © 2025 Academia. All rights reserved
14525	https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence?srsltid=AfmBOorEuIFYqRimF1Mq4zWZGldJUxO4WDOeEhY6fort6iXrqdq7dFAg	Art of Problem Solving Arithmetic sequence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Arithmetic sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arithmetic sequence In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence. For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies. More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if . Contents [hide] 1 Properties 2 Sum 3 Problems 3.1 Introductory problems 3.2 Intermediate problems 4 See Also Properties Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, . A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to . Proof: Let the sequence have first term and common difference . Then using the above result, as desired. Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences. Sum An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value. The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to . Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required. The second is that if an arithmetic series has first term , common difference , and terms, it has value . Proof: The final term has value . Then by the above formula, the series has value This completes the proof. Problems Here are some problems with solutions that utilize arithmetic sequences and series. Introductory problems 2005 AMC 10A Problem 17 2006 AMC 10A Problem 19 2012 AIME I Problems/Problem 2 2004 AMC 10B Problems/Problem 10 2006 AMC 10A, Problem 9 2006 AMC 12A, Problem 12 Intermediate problems 2003 AIME I, Problem 2 Find the roots of the polynomial , given that the roots form an arithmetic progression. See Also Geometric sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
14526	https://pmc.ncbi.nlm.nih.gov/articles/PMC4427793/	Cytoskeletal dynamics: A view from the membrane - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice J Cell Biol . 2015 May 11;209(3):329–337. doi: 10.1083/jcb.201502062 Search in PMC Search in PubMed View in NLM Catalog Add to search Cytoskeletal dynamics: A view from the membrane Magdalena Bezanilla Magdalena Bezanilla 1 Department of Biology, University of Massachusetts Amherst, Amherst, MA 01003 Find articles by Magdalena Bezanilla 1,,✉, Amy S Gladfelter Amy S Gladfelter 2 Department of Biological Sciences, Dartmouth College, Hanover, NH 03755 Find articles by Amy S Gladfelter 2,, David R Kovar David R Kovar 3 Department of Molecular Genetics and Cell Biology and 4 Department of Biochemistry and Molecular Biology, The University of Chicago, Chicago, IL 60637 Find articles by David R Kovar 3,4,, Wei-Lih Lee Wei-Lih Lee 1 Department of Biology, University of Massachusetts Amherst, Amherst, MA 01003 Find articles by Wei-Lih Lee 1, Author information Article notes Copyright and License information 1 Department of Biology, University of Massachusetts Amherst, Amherst, MA 01003 2 Department of Biological Sciences, Dartmouth College, Hanover, NH 03755 3 Department of Molecular Genetics and Cell Biology and 4 Department of Biochemistry and Molecular Biology, The University of Chicago, Chicago, IL 60637 ✉ Correspondence to Magdalena Bezanilla: bezanilla@bio.umass.edu All authors contributed equally to this paper. ✉ Corresponding author. Received 2015 Feb 17; Accepted 2015 Apr 21. © 2015 Bezanilla et al. This article is distributed under the terms of an Attribution–Noncommercial–Share Alike–No Mirror Sites license for the first six months after the publication date (see After six months it is available under a Creative Commons License (Attribution–Noncommercial–Share Alike 3.0 Unported license, as described at PMC Copyright notice PMCID: PMC4427793 PMID: 25963816 Abstract Many aspects of cytoskeletal assembly and dynamics can be recapitulated in vitro; yet, how the cytoskeleton integrates signals in vivo across cellular membranes is far less understood. Recent work has demonstrated that the membrane alone, or through membrane-associated proteins, can effect dynamic changes to the cytoskeleton, thereby impacting cell physiology. Having identified mechanistic links between membranes and the actin, microtubule, and septin cytoskeletons, these studies highlight the membrane’s central role in coordinating these cytoskeletal systems to carry out essential processes, such as endocytosis, spindle positioning, and cellular compartmentalization. The cytoskeleton underlies many aspects of cell physiology, including mitosis, cell division, volume control, cell stiffness, cell polarity, and extracellular matrix patterning. These events in turn impact development and tissue differentiation. The cytoskeleton receives, integrates, and transmits both intracellular and extracellular signaling cues. Most of these cues have to signal through a lipid bilayer before reaching the cytoskeleton. Thus, membrane–cytoskeleton interactions are central to deciphering how cytoskeletal remodeling is integrated throughout cells and tissues. Although signaling occurs across both the plasma and intracellular membranes, in this review we focus on the interplay between the cytoskeleton and the plasma membrane, which is predominantly composed of phospholipids (for a detailed review of plasma membrane lipid composition and localization, see Suetsugu et al., 2014). Common to eukaryotic cytoskeletal networks is the fact that they are formed from proteins with the inherent ability to self-assemble into long polymers. These polymers exist in a dynamic equilibrium with a monomeric pool, resulting in constant turnover in the cell. The ensemble of regulatory proteins, which regulates these dynamics, acts as the interface between cellular signaling and cytoskeletal remodeling. Not surprisingly then, many regulators of the cytoskeleton interact with membranes. However, it is still mostly unclear how these interactions work to regulate cytoskeletal dynamics and pattern specific subcellular networks in vivo. The cytoskeletal networks composed of actin, microtubules, and septins integrate various signals received at the membrane, and facilitate distinct functions in response. Actin has long been known to be intimately associated with membranes, and two major forms of actin regulation have been linked to the plasma membrane: (1) modulation of the actin monomer pool by phosphoinositides; and (2) modulation of actin assembly factors by membrane-associated small GTPases, by membrane-associated proteins, and by direct binding of assembly factors to the membrane. Also at the membrane, the actin-rich cortex interfaces with the microtubule cytoskeleton to coordinate intracellular events. Recent work has revealed mechanistic insights into this coordination with respect to spindle orientation, a critical event in development. To organize intracellular events, the membrane is compartmentalized, and this appears to be partially mediated by septins. We discuss recent studies that are beginning to mechanistically probe these membrane-associated cytoskeletal networks. Membrane regulation of actin dynamics Cells simultaneously assemble, maintain, and disassemble different F-actin networks within a common cytoplasm; each are tailored to facilitate a particular fundamental process such as motility, polarization, division, or endocytosis (Chhabra and Higgs, 2007; Blanchoin et al., 2014). F-actin networks with specified organization and dynamics are produced through the coordinated action of different overlapping sets of diverse actin-binding proteins with an array of complementary properties that include actin monomer (G-actin) binding, assembly, end capping, bundling, and severing/disassembling (Blanchoin et al., 2014). F-actin network assembly, organization, and dynamics are therefore controlled by the spatial and temporal regulation of the activity of actin-binding proteins. The association of these actin-binding proteins with the membrane is multifaceted. In some cases, actin-binding proteins are modulated by binding directly to phosphoinositide lipids. In other cases, membrane-associated proteins modify the activity of actin-binding proteins. Subsets of actin-binding proteins are even integral membrane proteins. Phosphoinositide lipids associate with diverse types of actin-binding proteins, and either inhibit or stimulate their activity (for review see Saarikangas et al., 2010). The actin nucleation promotion factors, WAVE and WASP, facilitate actin polymerization via the Arp2/3 complex upon binding PI(4,5)P 2. In contrast, actin-capping protein, the F-actin–severing protein ADF/Cofilin, and the G-actin–binding protein profilin are all inhibited by binding PI(4,5)P 2 (Saarikangas et al., 2010). Regulation of actin-binding proteins by association with and/or release from phosphoinositide lipids is an exciting possibility that could help explain the self-organization of diverse F-actin networks. However, the importance of phosphoinositide lipid regulation of most actin binding proteins has not been validated in vivo. Membrane regulation of profilin Cells maintain a reserve of up to hundreds of micromolar of unassembled G-actin monomers, which is available for rapid polymerization upon activation of assembly factors and/or production of free actin filament ends (Pollard et al., 2000). Despite the effective critical concentration for actin assembly being only 0.1 µM, a higher concentration of unassembled actin is maintained in part by G-actin–binding proteins that prevent its de novo assembly. Profilin is the primary evolutionarily conserved small G-actin–binding protein (Carlsson et al., 1977), which prevents actin filament assembly by inhibiting the formation of actin dimer and/or trimer nuclei (Jockusch et al., 2007). Actin monomers bound by profilin can only be added to actin filaments that are assembled by actin assembly factors such as Arp2/3 complex, formin, and Ena/VASP (Dominguez, 2009). Profilin-bound actin was assumed to be equally incorporated into F-actin networks assembled by different nucleation factors. However, by simultaneously binding to G-actin and continuous stretches of proline residues that are found on specific actin assembly factors such as formin and Ena/VASP (Ferron et al., 2007), profilin significantly increases the elongation rate of formin-assembled filaments (Romero et al., 2004; Kovar et al., 2006). Conversely, profilin inhibits Arp2/3 complex–nucleated branch formation by competing with the nucleation-promoting factor WASP for G-actin (Suarez et al., 2015). As a result, profilin facilitates formin- and Ena/VASP-mediated actin assembly over assembly by the Arp2/3 complex (Rotty et al., 2015; Suarez et al., 2015). It is therefore likely that the spatial and temporal regulation of profilin helps govern the type of F-actin network assembled, as profilin activity determines whether G-actin is incorporated into networks generated by one actin assembly factor over another (Fig. 1 A). Figure 1. Open in a new tab Regulation of actin assembly by membrane lipids. (A) Membrane phosphoinositides such as PI(4,5)P 2 and PI(3,4,5)P 3 might control the spatial and temporal assembly of diverse actin filament networks by regulating profilin activity. Profilin bound to PI(4,5)P 2 cannot associate with actin, which potentially could establish a pool of free actin monomers that might favor the nucleation of branched actin filaments by the Arp2/3 complex, which is activated by binding to the WASP V-CA domain (left). Alternatively, phosphorylated phospholipase C (PLC) releases profilin by hydrolyzing PI(4,5)P 2, which could facilitate a pool of actin bound to profilin that might favor the elongation of unbranched actin filaments by formin (right) or Ena/VASP (not depicted). (B) Small activated GTPases of the Rho superfamily insert into the membrane via a covalent lipid modification. These GTPases recruit and activate a nucleation-promoting factor such as WASP/WAVE that further modulates Arp2/3 complex activity. F-BAR proteins interact with WASP and either activate or inhibit actin polymerization activity. These activities lead to diverse functions, as indicated in the text boxes. (C) Small activated GTPases of the Rho superfamily directly bind to and recruit formins to the membrane, where they activate actin polymerization. F-BAR proteins can further modulate actin dynamics by either activating or inhibiting formin activity at the membrane to drive processes such as membrane protrusion and cytokinesis. In eukaryotes, such as plants, that lack formins with obvious Rho-binding domains, many formins bind directly to the membrane via an N-terminal PTEN domain (dark blue) that binds to PI(3,5)P 2, driving polarized growth, or via an N-terminal transmembrane domain (red). Question marks designate hypothetical membrane-associated proteins that negatively or positively regulate formin-mediated actin polymerization. Diverse profilins also bind to membrane phosphoinositides such as PI(3,4,5)P 3 and PI(4,5)P 2, which inhibits profilin’s interactions with G-actin and proline-rich stretches (Lassing and Lindberg, 1985, 1988; Lu et al., 1996; Lambrechts et al., 2002; Moens and Bagatolli, 2007). Multiple hydrophobic regions of profilin, including the actin- and proline-rich–binding regions, have been implicated in binding to phosphoinositides (Jockusch et al., 2007). Association of profilin with membrane phosphoinositides has been proposed to regulate the temporal and spatial levels of profilin-actin by two possible mechanisms (Fig. 1 A). One possibility is that external signal-mediated phosphorylation of phospholipase C hydrolyzes PI(4,5)P 2, releasing membrane-bound profilin to presumably facilitate actin assembly by formin and Ena/VASP (Goldschmidt-Clermont et al., 1991). Second, sequestration of profilin to membrane regions with high concentrations of PI(4,5)P 2 could increase the level of free G-actin, unbound to profilin, that might preferentially incorporate into branched actin filament networks generated by the Arp2/3 complex. Despite the proposal of these general hypotheses nearly 25 years ago (Goldschmidt-Clermont et al., 1991), there is unfortunately little in vivo evidence that phosphoinositide regulation of profilin occurs (Saarikangas et al., 2010). However, most higher eukaryotes express multiple profilin isoforms that associate with the particular ligands with significantly different affinities, such as actin- or proline-rich ligands like formin, which could tailor them for different cellular roles (Jockusch et al., 2007). Therefore, regulation by phosphoinositides would theoretically be a convenient way for individual profilin isoforms to facilitate self-organization of diverse actin filament networks by favoring particular actin assembly factors at discrete cellular locations (Neidt et al., 2009; Mouneimne et al., 2012; Ding and Roy, 2013). Further work is required to explore this exciting possibility. Membrane regulation of actin assembly factors Mechanistic insights for the role of the membrane are emerging in the case of the regulation of actin assembly factors. The most well-documented example of this is modulation of actin polymerization by small GTPases of the Rho superfamily. Most actin assembly factors are inherently inactive, but can be activated at the right time and place by small GTPase signaling cascades (Chesarone and Goode, 2009; Campellone and Welch, 2010). When activated, these small GTPases dock on the membrane due to exposure of a covalent lipid modification that intercalates into the membrane. Many actin assembly factors have GTPase-binding domains; binding to the active GTPase induces a conformational change, usually relieving an auto-inhibited state (Fig. 1, B and C). In the case of Arp2/3 complex, the SCAR/WAVE complex interacts with active GTPases and in turn activates the Arp2/3 complex, which generates filaments. Recently, new insights have emerged with respect to control of actin assembly at specific membrane sites. The WAVE complex was found to interact with a sequence motif found on a large number of diverse membrane proteins, ranging from channels to cell adhesion molecules. Binding occurs on a conserved face of the WAVE complex, which when mutated in flies leads to defects in the organization of the actin cytoskeleton (Chen et al., 2014). Future work is needed to sort out the signaling networks connected to this diverse set of membrane proteins and the specific physiological signals leading to activation of Arp2/3 complex-mediated actin polymerization. While the details of specific membrane recruitment are still being sorted out, it is clear that small GTPases bind to and activate the SCAR/WAVE complex, which in turn activates the Arp2/3 complex. However, another actin assembly factor, the formins, are not always fully activated by binding small GTPases (Seth et al., 2006; Maiti et al., 2012). In fact, many formins have other mechanisms to bind to the membrane (for review see Cvrčková, 2013). For instance, in plants, formins do not have obvious GTPase-binding domains, and in fact, class I formins are integral membrane proteins themselves. Thus, regulation of these molecules at the membrane is likely mediated by interactions with proteins or specific lipids at the membrane (Fig. 1 C). In support of this, moss class II formins contain a PTEN domain that mediates binding to PI(3,5)P 2 (van Gisbergen et al., 2012). Recruitment to PI(3,5)P 2-rich membrane domains and the ability to rapidly elongate actin filaments is essential for formin function during polarized growth (Vidali et al., 2009; van Gisbergen et al., 2012). However, examination of formin molecules at the cell cortex demonstrated that only a fraction of these molecules generate actin filaments (van Gisbergen et al., 2012). Thus, additional molecules associated with PI(3,5)P 2 at the membrane likely modulate the activity of this formin (Fig. 1 C). Whether there is a common family of molecules in eukaryotes that regulates membrane activity of actin assembly factors is unclear. However, a possible candidate class of membrane-associated molecules is the Bin-Amphiphysin-Rvs (BAR) domain–containing proteins (Aspenström, 2009; Suetsugu et al., 2010; Cvrčková, 2013). The positively charged BAR domains, which are found on many different proteins (Suetsugu et al., 2010), form α-helical coiled-coils that fold up into a crescent shape. These domains do not have high specificity for a particular lipid, but rather through their structure can sense or participate in membrane bending (Suetsugu et al., 2010, 2014). In yeast and animals, a family of proteins with an extended BAR domain, known as F-BAR proteins, are essential scaffolds upon which cytoskeletal proteins can assemble in order to generate specific subcellular structures and functions (Roberts-Galbraith and Gould, 2010). During endocytosis, nucleation-promoting factors for the Arp2/3 complex are recruited to the membrane by interacting with F-BAR proteins. F-BAR proteins not only recruit nucleation-promoting factors, but also modify their activity (Kamioka et al., 2004; Itoh et al., 2005; Tsujita et al., 2006; Takano et al., 2008; Henne et al., 2010; Roberts-Galbraith and Gould, 2010; Wu et al., 2010). In budding yeast, two F-BAR proteins oppositely regulate Las17, a homologue of the WASP actin nucleation–promoting factor (Fig. 1 B). Early in endocytosis, Syp1 recruits WASP but maintains it in an inactive state (Rodal et al., 2003; Sun et al., 2006; Boettner et al., 2009; Feliciano and Di Pietro, 2012). Upon vesicle maturation, Bzz1 activates WASP activity (Sun et al., 2006), thereby inducing a burst of actin polymerization mediated by the ARP2/3 complex that promotes internalization of endocytic vesicles. Further physiological support for this model has come from studies in neurons (Dharmalingam et al., 2009) and animal cells (Tsujita et al., 2006). F-BAR proteins also recruit formins to membranes. In fission yeast, the F-BAR proteins Cdc15 and Imp2 help recruit the essential cytokinesis formin Cdc12 to the division site (Chang et al., 1997; Carnahan and Gould, 2003; Ren et al., 2015). Similarly, the budding yeast Cdc15 homologue Hof1p acts redundantly with Rvs167 (a BAR domain–containing protein also containing a C-terminal SH3) to promote formation of the contractile actin ring (Nkosi et al., 2013). Although F-BAR proteins have clearly defined roles in recruiting formins, several recent studies have revealed how F-BAR proteins directly modulate formin activity. In mammals, the F-BAR protein srGAP2 binds to and directly inhibits the actin-severing activity of the formin FMNL1, which is mediated by its formin homology (FH) 1 domain (Mason et al., 2011). During Drosophila melanogaster embryogenesis, the F-BAR protein Cip4 binds to the formin Dia’s FH1 domain and inhibits the ability of Dia to promote actin assembly. Cip4 is a known activator of the WASP–WAVE–Arp2/3 complex pathway. Thus, while Cip4 activates Arp2/3 complex activity, it can simultaneously inhibit Dia activity (Yan et al., 2013). More recently, it was demonstrated in budding yeast that the SH3 domain of the F-BAR protein Hof1p dampens the actin nucleation activity of the formin Bnr1p without displacing Bnr1p from the actin filament end (Fig. 1 C; Graziano et al., 2014). These studies suggest that F-BAR proteins may have a conserved role in regulating diverse sets of actin nucleation factors at the membrane. Thus, understanding how BAR domain–containing proteins interact with and regulate specific subsets of actin regulators may help to decipher the distinct F-actin domains at the cell cortex. Additionally, since BAR domain–containing proteins are found widely throughout eukaryotes (Ren et al., 2006), it is possible that these molecules may have been an early link between membranes and actin modulation that, with various elaborations, evolved differently in distinct lineages. Connecting actin and microtubules to the membrane enables cortical force generation The cell cortex in animal cells plays a fundamental role in cell division, migration, and polarization (Kunda et al., 2008; Pollard and Cooper, 2009; Stewart et al., 2011; Abu Shah and Keren, 2014). The cortex integrates external stimuli—from extracellular matrix and neighboring cells—and transmits them into the cell to effect cytoskeletal changes crucial for development. A key component of the cortex is the thin F-actin shell underneath the cell membrane that is crucial for providing cortical stiffness and is a key determinant of cell shape (Pollard and Cooper, 2009; Guo et al., 2013). Perturbations in cortical F-actin architecture can alter the physical properties of the cortex, thereby affecting cell stiffness and strength. A recent study demonstrates that the bulk of the actin cortex is nucleated by the formin mDia1 and Arp2/3 complex (Bovellan et al., 2014), which suggests that fine-tuning of F-actin cortical structure and mechanics may be mediated by adjusting the relative contribution of each actin assembly factor. Several studies (for reviews see Basu and Chang, 2007; Akhshi et al., 2014) show that changes in microtubule stability also positively and negatively regulate cortical F-actin structures, including formation of lamellipodia and stress fibers. Here we focus on the converse: regulation of microtubule function by the actin-rich cortex. An excellent example of this regulation is how these two elements set the orientation of the mitotic spindle, which determines the plane of cell division, thereby impacting cell fate and tissue organization. It has been known for quite some time that, during cell division, an intact cortical F-actin meshwork and an intact astral microtubule array are required for spindle orientation (O’Connell and Wang, 2000; Théry et al., 2005; Toyoshima and Nishida, 2007; Fink et al., 2011; Luxenburg et al., 2011; Castanon et al., 2013). However, how the F-actin cortex is involved in this process, and how the membrane supports the underlying cytoskeletal organization to bring about spindle alignment toward a specialized cortical domain, remains unclear in many cellular systems. The prevailing notion is that the F-actin network provides a platform for a cortical anchor, or a complex of anchoring proteins, that could either mediate attachment (i.e., tethering) of astral microtubules or recruit force generators such as motor proteins that exert pulling forces on the microtubules emanating from the spindle. In this notion, the plus ends of astral microtubules would engage with these cortical platforms through so-called +TIPs (plus tip tracking proteins), including adenomatous polyposis coli protein (APC), CLASP, CLIP170, LIS1, dynactin, and dynein (Coquelle et al., 2002; Rogers et al., 2002; Reilein and Nelson, 2005; Siller and Doe, 2008; Ruiz-Saenz et al., 2013). Data to support this idea has been found in several organisms, including Caenorhabditis elegans zygotes (Couwenbergs et al., 2007; Nguyen-Ngoc et al., 2007), Drosophila neuroblasts (Siller et al., 2006), and cultured human cells (Kiyomitsu and Cheeseman, 2012). These studies have identified an evolutionarily conserved ternary complex composed of Gαi, the α subunit of heterotrimeric G-protein; LGN, a leucine-glycine-asparagine repeat protein; and NuMA, a nuclear mitotic apparatus protein; as the cortical anchoring complex that recruits dynein as the force generator for spindle orientation. NuMA interacts with LGN (Du and Macara, 2004; Bowman et al., 2006; Siller et al., 2006), which in turn binds to the myristoylated Gαi that is directly attached to the membrane. NuMA can also bind the membrane directly through a C-terminal PIP-binding domain in a manner independent of LGN and Gαi (Zheng et al., 2014). Intriguingly, when the F-actin meshwork was disrupted, NuMA and Gαi dissociate from the cell cortex (Luxenburg et al., 2011; Machicoane et al., 2014; Zheng et al., 2014), signifying that their membrane association is weak. These observations raise interesting questions about the physical nature of the anchoring platform, and suggest that additional mechanisms may be required to attach anchoring proteins to the F-actin meshwork or to stabilize them at the cortex. Recent work has shown that the actin-binding proteins ezrin/radixin/moesin (ERM) are probably the missing puzzle pieces at the cell cortex mediating spindle orientation (Solinet et al., 2013; Machicoane et al., 2014). ERMs help organize the F-actin meshwork, bridging it to the cell membrane, and this may be necessary for establishing and maintaining the Gαi-LGN-NuMA cortical platform. ERMs, when activated by Ste20-like (SLK) kinase (Machicoane et al., 2014), adopt an open conformation that binds F-actin and the plasma membrane. An N-terminal FERM domain, which binds PI(4,5)P 2 directly (Fievet et al., 2004; Roch et al., 2010; Roubinet et al., 2011), mediates interaction with the membrane. Interestingly, the FERM domain also binds to and stabilizes microtubules (Solinet et al., 2013), possibly via interaction with CLASP family of +TIPs (Ruiz-Saenz et al., 2013), which suggests that ERMs may function as microtubule-tethering factors. However, evidence suggests that they do more than just tethering microtubules. Depletion of ERMs or inhibition of ERM activation leads to loss of cortical rigidity, mislocalization of LGN and NuMA, and abnormal spindle rocking behavior (Carreno et al., 2008; Machicoane et al., 2014). It is interesting to speculate that ERMs may be required to increase membrane rigidity by pinning the F-actin meshwork to the plasma membrane. As proposed (Zheng et al., 2014), this rigidity may enable the cortical platform to counteract astral microtubule–mediated and dynein-generated pulling forces on the cortical anchors. It is noteworthy that the budding yeast version of the dynein cortical anchor, Num1, interacts with the plasma membrane directly via a BAR-like domain and a PH domain (Farkasovsky and Küntzel, 1995; Tang et al., 2009, 2012; Klecker et al., 2013; Lackner et al., 2013). In budding yeast, actin is dispensable for maintenance of the Num1 cortical platform (Heil-Chapdelaine et al., 2000a) or to support dynein-dependent spindle movements (Heil-Chapdelaine et al., 2000b), as membrane rigidity is provided by turgor pressure and the cell wall. During animal development, it is conceivable that stabilization of membrane rigidity, as exemplified by ERMs, may represent a general mechanism for modulating pulling forces on astral microtubules (Fig. 2). It is therefore tempting to speculate whether the recently characterized human cortical actin–associated protein, MISP, which has a role in astral microtubule stability and spindle orientation (Zhu et al., 2013), would orchestrate actin cytoskeleton communication with the cell membrane and the astral microtubules in a similar manner. Deciphering how actin-dependent membrane rigidity is controlled locally at specific regions of the cell cortex will surely constitute a major challenge to unraveling the mechanisms governing spatial and temporal regulation of oriented cell division. Figure 2. Open in a new tab Regulation of microtubule tethering by actin-dependent membrane rigidity. ERM increases membrane rigidity to support Gαi-LGN-NuMA–dependent anchoring and pulling of astral microtubules by cytoplasmic dynein. Activated ERMs in an open conformation may link F-actin to the cell membrane. Membrane association of the Gαi-LGN-NuMA complex mediated by the lipid anchor on Gαi and the PIP-binding domain on NuMA are presumably weak. Stiffening of the membrane (indicated by straight phospholipid tails) or yet unidentified interactions with F-actin or ERMs may further stabilize the Gαi-LGN-NuMA platform to prevent anchorage detachment. Septins: links between polymer assembly and membrane function An additional layer of membrane compartmentalization is provided by septins. Septins are a component of the cytoskeleton that directly bind to membranes in order to polymerize and in turn help organize cell membranes. Knowing how membranes specify septin assemblies at a particular place and time is essential to understand the mechanistic role of septins in cytokinesis and beyond. Septins were first observed at the plasma membrane in budding yeast (Byers and Goetsch, 1976; Rodal et al., 2005; Ong et al., 2014). Early work found that human septins exhibit a preference for PI(4,5)P 2 and proposed that a conserved polybasic sequence in septins links them to phospholipids (Zhang et al., 1999). More recently, recombinant budding yeast septins were assembled on lipid monolayers containing high levels (10–50%) of PI(4,5)P 2 (Bertin et al., 2010). Interestingly, the presence of the lipids could promote filament formation even with septin proteins that were otherwise defective for polymerization, which suggests that membranes can facilitate filament assembly. The first dynamic look at septin assembly with reconstituted septin proteins supported lipid bilayers with low levels of PI(4,5)P 2, and single-molecule total internal reflection fluorescence (TIRF) imaging found that septin filaments elongate through diffusion in two dimensions and annealing (Bridges et al., 2014). The notion that polymerization occurs at the membrane is supported by the finding that cytosolic pools of septins in diverse fungi and mammals consist of minimal heteromeric rods (or heteroligomers) but not filaments (Sellin et al., 2011; Bridges et al., 2014). Thus, membranes are intimately involved in septin filament formation (Fig. 3 A). Figure 3. Open in a new tab Septins assemble on and organize membranes. (A) Membrane association of septin heteroligomers promotes assembly of septin filaments. (B) Septins may play a role in promoting or sensing membrane curvature. Posttranslational modifications (indicated by black stars) are implicated in this function. (C) By providing a diffusion barrier, septins play a role in compartmentalizing the membrane. Septins are frequently found in areas where membranes are highly curved, such as the mother-bud neck in yeast and the bases of dendritic spines and primary cilia (Fares et al., 1995; Xie et al., 2007; Hu et al., 2010). This raises the possibility that septins sense and/or generate curvature, which is supported by the finding that septin filaments can tubulate phospholipid liposomes in vitro (Tanaka-Takiguchi et al., 2009; Fig. 3 B). In vivo, septins are recruited to curved blebs of membrane that are pulled back toward the cell center, which suggests that there may be a capacity for them to recognize specific curvatures (Tanaka-Takiguchi et al., 2009; Gilden and Krummel, 2010). Recent work has shown that septins can promote the formation of curved and ordered bundles of F-actin at the highly curved membranes of furrow canals during embryo cellularization, which suggests that septins and actin may collaborate for curvature sensing (Mavrakis et al., 2014). In mycelia of filamentous fungal systems, there are septin regulatory kinases that are only required for straight septin filaments without impacting septins that assemble at the curved surfaces, which suggests that posttranslational modifications could influence curvature preference or sensing (DeMay et al., 2009; Fig. 3 B). Finally, there has been substantial interest in the role of septins as diffusional barriers, and work from yeast to human cilia has suggested the possibility that septins can functionally compartmentalize membranes (Takizawa et al., 2000; Barral et al., 2000; Hu et al., 2010; Fig. 3 C). Despite the first observations of a barrier function over a decade ago, the mechanism by which septin compartmentalizes membranes has proven to be highly elusive. The first clues as to a molecular basis for the ER-based barrier have come from several recent studies. Yeast genetics uncovered a link between sphingolipid domains and septin-based ER barriers, and a second study identified a role for one specific septin, Shs1, in these barriers (Chao et al., 2014; Clay et al., 2014). Finally, a critical functional role for septins in membrane compartmentalization came from a screen looking at regulators of calcium influx in cultured mammalian cells (Sharma et al., 2013). This study showed that septins are required for establishing PIP 2-rich microdomains at sites of ER–plasma membrane contacts. These functional studies, along with the development of reconstitution methods for probing the barrier properties in artificial lipid membranes, should pave the way for understanding how septins influence membrane diffusion. But it is clear that a reciprocal relationship between certain membrane domains and septins underlies their organization and function. Conclusions As more mechanistic connections emerge between the membrane and the cytoskeleton, it is becoming clear that a new generation of tools is needed. In particular, being able to track the dynamics and localization of specific lipid species, as well as physical methods to measure membrane rigidity in living cells, is critical. Additionally, most studies have been performed in individual cells, but not in the context of developing tissues or varied extracellular environments. Thus, how mechanical strains on the membrane translate into cytoskeletal reorganization ultimately effecting cell physiology and development constitutes the next generation of questions in cytoskeletal dynamics. Acknowledgments The authors thank the Marine Biological Laboratories for providing a collaborative research environment. The Marine Biological Laboratories also provided research support in the form of competitive research fellowships (The Nikon Award for summer investigation to M. Bezanilla; The Laura and Arthur Colwin Endowed Summer Research Fellowship to W.-L. Lee). This work was supported in part by the David and Lucille Packard Foundation (to M. Bezanilla), the National Science Foundation (MCB-1330171 to M. Bezanilla; MCB-507511 to A.S. Gladfelter), and the National Institutes of Health/National Institute of General Medical Sciences (GM076094 to W.-L. Lee; GM079265 to D.R. Kovar). The authors declare no competing financial interests. 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14527	https://wayground.com/admin/quiz/6799e73c6f345a3c979b228a/identify-real-and-complex-parts	Enter code Login/Signup Identify Real and Complex Parts 12th Grade 9 Qs Try it as a student Similar activities COMPLEX NUMBER University 10 Qs Finding Zeros of the Polynomial Function 11th Grade - University 11 Qs Finding Zeros of a Polynomial Function 11th Grade - University 11 Qs Complex Numbers Unit 6 9th - 12th Grade 12 Qs QUADRATIC EQUATIONS 9th - 12th Grade 10 Qs quadratic formula review 11th - 12th Grade 11 Qs Powers of i and Operations with Complex Numbers 8th - 12th Grade 11 Qs Exploring Complex Numbers and Imaginary Parts 12th Grade 10 Qs View more activities Identify Real and Complex Parts Quiz • Mathematics • 12th Grade • Hard Edit Worksheet Share Save Preview Use this activity Standards-aligned Anthony Clark Used 1+ times FREE Resource 9 questions Show all answers 1. FILL IN THE BLANK QUESTION 1 min • 1 pt Preview Edit In z= 4-3i, the real number part is _______. Tags CCSS.HSN.CN.A.1 2. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit Write the expression in the form of a+bi, where a and b are real numbers. 3 ⋅ √-25 15 -15 15i -15i Tags CCSS.HSN.CN.A.1 3. FILL IN THE BLANK QUESTION 1 min • 1 pt Preview Edit The real number part is __. Tags CCSS.HSN.CN.A.1 4. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit Irrational Number Real Number Imaginary Number Complex Number Tags CCSS.HSN.CN.A.1 5. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit The imaginary part of 5 is... 5 -5 0 None of the above Tags CCSS.HSN.CN.A.1 6. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit X and Y axis represent ______ Respectively Real and Real parts Real and Imaginary parts Imaginary parts Imaginary and Real parts Tags CCSS.HSN.CN.B.4 7. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit We're off to a good start learning about operations on imaginary numbers! Now let's try something complex (meaning a number with a real and imaginary part) Identify the real part in this complex number 7+3i 7 3 i 3i Answer explanation In the complex number 7+3i, the real part is the coefficient of the real component, which is 7. The imaginary part is represented by 3i, where 3 is the coefficient of the imaginary unit i. Tags CCSS.HSN.CN.A.1 8. MULTIPLE CHOICE QUESTION 1 min • 1 pt Preview Edit What is the sum of 6+3i and 7+12i 18+10i 13+15i 15+13i 10+18i Answer explanation To add the complex numbers 6+3i and 7+12i, combine the real parts (6+7) and the imaginary parts (3i+12i). This gives 13 for the real part and 15i for the imaginary part, resulting in 13+15i, which is the correct answer. Tags CCSS.HSN.CN.A.2 9. MULTIPLE CHOICE QUESTION 1 min • 12 pts Preview Edit What is the real portion of the number (3+5i) 3 5 -3 -5 Tags CCSS.HSN.CN.A.1 Similar Resources on Wayground View all 11 questions Cubic With Imaginary Roots Quiz • 11th Grade - University 14 questions Illustrative Math Complex Numbers Quiz • 11th Grade - University 12 questions Polar Complex Numbers Quiz • 12th Grade 14 questions Categorizing Real and Complex Numbers Quiz • 11th Grade - University 10 questions Quadratic Formula Imaginary Solutions Quiz • 10th Grade - University 11 questions Polynomial Functions and Complex Zeros Quiz • 11th Grade - University 13 questions Enhanced Algebra Data Unit Quiz • 10th - 12th Grade 8 questions Math 2 Unit 1 Quadratic Solutions Review Quiz • 9th - 12th Grade Popular Resources on Wayground 10 questions Video Games Quiz • 6th - 12th Grade 10 questions Lab Safety Procedures and Guidelines Interactive video • 6th - 10th Grade 25 questions Multiplication Facts Quiz • 5th Grade 10 questions UPDATED FOREST Kindness 9-22 Lesson • 9th - 12th Grade 22 questions Adding Integers Quiz • 6th Grade 15 questions Subtracting Integers Quiz • 7th Grade 20 questions US Constitution Quiz Quiz • 11th Grade 10 questions Exploring Digital Citizenship Essentials Interactive video • 6th - 10th Grade Discover more resources for Mathematics 15 questions ACT Math Practice Test Quiz • 9th - 12th Grade 20 questions Multi-Step Equations and Variables on Both Sides Quiz • 9th - 12th Grade 10 questions Angle Relationships with Parallel Lines and a Transversal Quiz • 9th - 12th Grade 12 questions Slope from Two Points Quiz • 9th - 12th Grade 10 questions Intro to Parallel and Perpendicular Slopes Quiz • 9th - 12th Grade 20 questions Domain and Range from a graph Quiz • 9th - 12th Grade 27 questions Week 5 Student Practice Set (Term 1) Quiz • 9th - 12th Grade 20 questions Solving Absolute Value Equations Quiz • 11th - 12th Grade © 2025 Quizizz Inc. (DBA Wayground)
14528	https://ocw.mit.edu/courses/res-6-012-introduction-to-probability-spring-2018/fc79badd15189319b93a55ac547956f5_MITRES_6_012S18_L06AS.pdf	LECTURE 6: Variance; Conditioning on an event; Multiple random variables • Variance and its properties -Variance of the Bernoulli and uniform PMFs • Conditioning a r.v. on an event -Conditional PMF, mean, variance -Total expectation theorem • Geometric PM F -Memorylessness -Mean value • Multiple random variables -Joint and marginal PMFs -Expected value rule -Linearity of expectations • The mean of the binomial PMF 1 Variance -a measure of the spread of a PMF • Random variable X, with mean JL = E[X] I • Distance from the mean: X -JL l' I I X • Average distance from the mean? )(-1' E[X-r] , E[xj -r "fA -1" =0 • Definition of variance: var(X) = E[(X - JL)2] =;:! 0 • Calculation, using the expected value rule, E[g(X)] = L9(x)px(x) x Standard deviation: ax = j var(X). 2 Properties of the variance 'vo. t (?>-'1 X ) var(aX + b) = a2var(X) = r-"ill/C\z.(~) • Notation: p = E[X] • Let Y = X + b -y .: £ " r+ b = )6 v 0 ~ (If) v(l((Y) > £[('(0)1)'3 ~ E[ (x+JI - (t'+ jI)i-j > E[('X-r)l} VQr(X) • Let Y = aX ".-;. £ ['1J -= Q. f" liar (Of) =f [(a. X _ ctr)'J=f [ Q 1 (.x -r)'] ;:. Q 1.. E[Or -rlJ ~ctt Vo!(... ) A useful formula: var(X) = E [X2] - (E[X])2 Vctt(X) -= E[eX -t')~J = f[ Xi - 2r K + f %.1 ~G[xtJ -2.t'~"[xJ +,.,t ::.6[x lJ -([;;[xJ)~ 3 Variance of the Bernoulli • • 1, w.p. p • X= 0, w.p.l - p o 'I,. 1 = (1- F)t.l' + (0- p)'-. (I-f) x -;: p_2p9.+/ + l,t-r '" p_pt.;p(l_p) var(X) = E[X2] - (E[X])2 ::; Ii [:>rJ _ U: ['X])'1 = P_P2 :: P(I - p) 4 1 Variance of the uniform n+l • • • • • • .'... ("",')(2"" t-') (, o 1 n x - (~r VIlAIC) ~r;f'x·] -(I;["J)'.''"' , ... :: -'. 11>("-'+9.) Ii Px ( ) x 1 .... .. ,,' b " + 1 • , • • • • x 5 Conditional PMF and expectation, given an event • Condition on an event A ' use conditional probabilities OSS"' ...... e PX (X) = P (X = x ) PXIA(X) = P(X = x I A) 1? (Ah 0 I>x(X) = 1 x E[X) = L XPX (x ) E[X I A) = L XPX IA(X) x x E[g(X») = L g(x )px(x) E [g(X) I A) = L g(x )PXIA(X) x x • 6 Example of conditioning • Let A = { X > 2} x 1 L:1 2 =--:. 3_ 4 ;..." E[X] = 2..? 1 2 3 4 x E[X I A] = 3 , var(X) = -(6-q)(b-Q.'~) 11 7 Total expectation theorem Al n B PCB) = PCAl) PCB I Al) + ... + PCAn) PCB I An) • B'" ~X =?::.5 8 Total expectation theorem '-----_-' Al n {x=X} A2 n {x=X} A3 n {X=X} P (A,) ~-"..:.!.'-- E [X I A d P-"-(A ,", 2) ,-- E [X I A2[ • P ( A 3) E[X I A3[ P CB) = peAl) P CB I AI) + ... + P(An) P CB I An) B'" ~X =?::.5 PX(x) = peAl) PX IA, (x) +... + P(An ) pXIAn(X) for o..ffl ')C. :7 -;;c p)«"') :: f(A.) ""'5""'):. 0 (~) -1-' •• z L· 'xlA, II ,?c I off,..j E[;>r)A,] E [X l = peAl) E [X I All +... + P(An ) E [X I Anl 9 Total expectation example 1 (A,):: 3' I i (A:d : :0 px(x) 2/9 2/9 2/9 E[x:I.ti,].=1 ' /9 ' /9 ' /9 E [X lAd = t 2. +. .• ':f 0'--' 2 :-------; 6---,7 S ,---O: x , I , 3> • I A, 10 Conditioning a geometric random variable • X: number of independent coin tosses until first head ; P(H) = p px(k) = (1 - p)k-lp, k = 1,2, ... Memorylessness: Px(k) Number of remaining coin tosses, conditioned on Tails in the first toss, is Geometric, with parameter p 123 4 5 6 7 8 9 k Conditioned on X> 1, X-I is geometric with para mater p x ! I H 11 Conditioning a geometric random variable • X: number of independent coin tosses until first head ; P(H) = p px(k) = (1 - p)k-lp, k = 1,2, ... Memorylessness: Px(k) Number of remaining coin tosses, conditioned on Tails in the first toss, is Geometric, with parameter p 123 4 5 6 7 8 9 k Conditioned on X> 1, X-I is geometric with parameter p -.. -12 Conditioning a geometric random variable • X: number of independent coin tosses until first head ; P(H) = p px(k) = (1 - p)k-lp, k = 1,2, ... Memorylessness: Px(k) Number of remaining coin tosses, conditioned on Tails in the first toss, is Geometric, with parameter p 123 4 5 6 7 8 9 k Conditioned on X> n , X - n is geometric with parameter p -'" -13 The mean of the geometric pX(k) 00 00 E[X] = L kpx(k) = L k( l _ p)k-lp k= l k=l E[X] = ~ 5 6 7 8 9 k p . fl, 1(= I £[xJ = I ~E[x-IJ ::'\ + p.£[X-I/X.:J] t (l-r)t;[X-'/)(>J] = I .. 0 + (I - p) 1:[x] 14 Multiple random variables and joint PMFs " , .... Qr~"'tlQ )? ... f. X: Px Q P(X = Y) = ~ Joint PMF: Px y(x, y) = P (X = x and Y = y) , Y : PY :to y 4 1120 2120 21 0 2 11< ('f) '-~ ~ ~9. 0 I x f., (~) ~ I • +- .3 I 2 3 4 L2>X,Y(x,y) = 1 x y Px(x) = L PX,Y(x , y) y py(y) = L PX,Y(x, y) I x ... P.C;) 20 lio ~o 15 More than two random variables PXYZ(x, y,z) =P(X =x and Y =y and Z = z) , , LL2>X,Y,Z(X, y,z) = 1 x y z PX(x) = LLPX,Y,Z(x,y,z) ~ y z , ,,-, , PX,y(x, y) = LPX,Y,Z(x, y, z) z • - 16 Functions of multiple random variables Z = g(X, Y) PMF: PZ ( z ) = P(Z = z ) = p (g(X ,Y) = ;) -= 2 PI<) '( ("'-", 1) (x, 'I): 't(?:.J1)=~ Expected value rule: E(g(X, Y)] = 2:2:og(x,y)PX,Y(x,y) x y ! . 17 Linearity of expectations E[aX +b] = aE[X] +b E[n 'fJ " E [ ~ (x./ '01 E[X+ Y] = (tc"',.7):= 7<:+/'.) E[X] +E[Y] = :2" ~ (7<:+7) ~)'f (;I:,y) "" 7 :: Z;}~ t y('k:,I') t 'Z ~ Y f, ("'d) 0: 1) I l<,Y I , % /' ".-.. ; =J':Cf)«;I:) + L7f (1} = E[:X] +-E['f] y x /' 18 Linearity of expectations E[aX + b) = aE[X) + b E[X + Y) = E[X) + E[Y) E[Xl + ... + X n) = E[Xl) + ... + E[Xn) 19 The mean of the binomial n • X: binomial with parameters n, p E[X] = L kC)pk(l _ p)n- k number of successes in n independent trials k = Q L J k E[X] = np /p X i = 1 if ith trial is a success; (indicator variable) X i = 0 otherwise -- 1 - P X = Xl + ... +Xn E[X] ~ £[1<,] + ••• "--""'--' I" 20 MIT OpenCourseWare Resource: Introduction to Probability John Tsitsiklis and Patrick Jaillet The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit:
14529	https://pmc.ncbi.nlm.nih.gov/articles/PMC4712787/	Photoreceptors at a glance - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice J Cell Sci . 2015 Nov 15;128(22):4039–4045. doi: 10.1242/jcs.175687 Search in PMC Search in PubMed View in NLM Catalog Add to search Photoreceptors at a glance Robert S Molday Robert S Molday 1 Department of Biochemistry and Molecular Biology, Centre for Macular Research, University of British Columbia, Vancouver, British Columbia, Canada, V6T 1Z3 2 Department of Ophthalmology and Visual Sciences, Centre for Macular Research, University of British Columbia, Vancouver, British Columbia, Canada, V5Z 3N9 Find articles by Robert S Molday 1,2,, Orson L Moritz Orson L Moritz 2 Department of Ophthalmology and Visual Sciences, Centre for Macular Research, University of British Columbia, Vancouver, British Columbia, Canada, V5Z 3N9 Find articles by Orson L Moritz 2 Author information Copyright and License information 1 Department of Biochemistry and Molecular Biology, Centre for Macular Research, University of British Columbia, Vancouver, British Columbia, Canada, V6T 1Z3 2 Department of Ophthalmology and Visual Sciences, Centre for Macular Research, University of British Columbia, Vancouver, British Columbia, Canada, V5Z 3N9 Author for correspondence (molday@interchange.ubc.ca) © 2015. Published by The Company of Biologists Ltd PMC Copyright notice PMCID: PMC4712787 PMID: 26574505 ABSTRACT Retinal photoreceptor cells contain a specialized outer segment (OS) compartment that functions in the capture of light and its conversion into electrical signals in a process known as phototransduction. In rods, photoisomerization of 11-cis to all-trans retinal within rhodopsin triggers a biochemical cascade culminating in the closure of cGMP-gated channels and hyperpolarization of the cell. Biochemical reactions return the cell to its ‘dark state’ and the visual cycle converts all-trans retinal back to 11-cis retinal for rhodopsin regeneration. OS are continuously renewed, with aged membrane removed at the distal end by phagocytosis and new membrane added at the proximal end through OS disk morphogenesis linked to protein trafficking. The molecular basis for disk morphogenesis remains to be defined in detail although several models have been proposed, and molecular mechanisms underlying protein trafficking are under active investigation. The aim of this Cell Science at a Glance article and the accompanying poster is to highlight our current understanding of photoreceptor structure, phototransduction, the visual cycle, OS renewal, protein trafficking and retinal degenerative diseases. KEY WORDS: Disk morphogenesis, Photoreceptors, Phototransduction, Protein trafficking, Retinal degenerative diseases, Visual cycle Summary: Rod and cone photoreceptors contain a unique light-sensitive outer segment compartment that has a central role in phototransduction, the visual cycle, membrane renewal, protein trafficking and many retinal degenerative diseases. Introduction Rod and cone photoreceptors are specialized neurons that function in the initial step of vision. These light-sensitive cells lie at the back of the retina adjacent to the retinal pigment epithelium (RPE), a cell layer that is vital for the survival of photoreceptors. Rod cells are highly sensitive to light and operate under dim lighting conditions. Cone cells function under ambient and bright lighting conditions, exhibit rapid responses to variations in light intensity, and are responsible for color vision and high visual acuity. The human retina contains 120 million rod cells and 6 million cone cells, with the latter concentrated in the central or macula region of the retina. Open in a new tab Both rods and cones are highly compartmentalized in structure and function. They consist of five principal regions: outer segment (OS), connecting cilium (CC), inner segment (IS), nuclear region and synaptic region (see poster). The OS functions in the capture of light and its conversion into electrical signals in a process known as phototransduction. The CC connects the OS with the IS, allowing for the trafficking of specific proteins to the OS. The IS contains the metabolic and biosynthetic machinery of the cell including the mitochondria, endoplasmic reticulum, Golgi complex, lysosomes and other subcellular organelles. The nuclear region is continuous with the inner segment and houses the nucleus. The photoreceptor finally terminates in the synaptic region, which consists of synaptic vesicles and a ribbon synapse for transmission of the neurotransmitter glutamate from photoreceptors to bipolar cells and other secondary neurons. The OS of rod photoreceptors has been the focus of numerous molecular, cellular, biochemical and physiological studies owing to its unique structure, accessibility, ease of isolation, and importance in the visual response, membrane renewal and retinal diseases. In this Cell Science at a Glance article and poster, we provide an overview of our current understanding of the molecular and cellular organization of photoreceptors with emphasis on the mechanisms underlying phototransduction, the visual cycle, OS structure and morphogenesis, and protein trafficking. We also highlight key proteins that have been associated with retinal degenerative diseases. Structural organization and protein composition of the rod OS The OS of the rod cell is a cylindrical structure consisting of a plasma membrane (PM) that encloses an ordered stack of over 1000 closely spaced disks. The length and diameter of the OS varies for different vertebrates. Mammalian rod OS typically have a length of 20–30 µm and a diameter of 1.2–2.0 µm (Gilliam et al., 2012; Nickell et al., 2007). Each disk is a closed structure and consists of two flattened membranes that are circumscribed by a hairpin rim region (see poster). The photopigment rhodopsin, which comprises >85% of the disk membrane protein, is arranged in the form of higher-order oligomers in the disk lamellae at a density of >25,000 rhodopsin molecules/µm 2 (Fotiadis et al., 2003; Gunkel et al., 2015; Liebman and Entine, 1974). The disk rim contains a number of specialized membrane proteins, the Per2–Rom1 complex consisting of the two tetraspanin membrane proteins Per2 (also known as PRPH2) and Rom1 that generate the highly curved disk rim (Clarke et al., 2000; Goldberg and Molday, 1996; Kevany et al., 2013; Khattree et al., 2013), the retinal-specific ATP-binding cassette transporter ABCA4 that facilitates the clearance of retinoids from disk membranes (Illing et al., 1997; Molday et al., 2009; Quazi and Molday, 2014; Sun et al., 1999; Weng et al., 1999), and members of the guanylate cyclase family GC-1 and GC-2 (also known as GUCY2D and GUCY2F respectively) that catalyze the synthesis of cyclic guanosine monophosphate (cGMP) from GTP (Karan et al., 2010, 2011; Nemet et al., 2015; Yang et al., 1995) (see poster). Per2 is most crucial for OS biogenesis and structure because deficiency in Per2, as found in rds mice, results in the absence of rod and cone OS (Arikawa et al., 1992; Connell et al., 1991; Sanyal and Jansen, 1981; Travis et al., 1989), and mutations in the gene encoding Per2 cause autosomal-dominant retinitis pigmentosa (ADRP) and macular dystrophy in humans (Farrar et al., 1991; Kajiwara et al., 1994, 1991; Wells et al., 1993). Proteomic analysis and biochemical studies have further shown that disk membranes also contain R9AP (also known as RGS9BP), which anchors the GAP complex RGS9 and the long splice isoform of the type 5 G protein β subunit (Gβ5) (RGS9–Gβ5) to the disk membrane, the ATP8A2–CDC50A (also known as TMEM30A) complex that functions as a phospholipid flippase, and progressive rod-cone degeneration protein (PRCD), a small protein of unknown function that is associated with progressive rod–cone degeneration (Coleman et al., 2009; Kwok et al., 2008; Skiba et al., 2013). Disks also contain membrane-associated proteins, including the trimeric G-protein transducin, phosphodiesterase (PDE6), which is composed of two large catalytic subunits (PDE6A and PDE6B) and two regulatory subunits (PDE6C), and retinol dehydrogenase 8 (RDH8). Glutamic-acid-rich proteins (GARPs) associate with the Per2 homotetramers and Per2–Rom1 heterotetramers at the rim region of disk membranes (Körschen et al., 1999; Poetsch et al., 2001) (see poster). The PM of rod OS contains substantial amounts of rhodopsin (Molday and Molday, 1987; Nemet et al., 2015) as well as the cyclic-nucleotide-gated (CNG) channel that consists of three CNGA1 and one CNGB1 subunits, with the latter harboring an N-terminal GARP domain, and the Na+–Ca2+–K+ exchanger (NCKX1, also known as SLC24A1) (Colville and Molday, 1996; Cook et al., 1989; Kaupp and Seifert, 2002; Reid et al., 1990; Zhong et al., 2002). The CNG channel forms a complex with NCKX1 in the PM (Bauer and Drechsler, 1992; Molday and Molday, 1998) (see poster). Thin filaments have been observed by using electron microscopy (EM) between disks and between the disk and PM (Roof and Heuser, 1982). Protein-protein interactions between the CNGB1 subunit of the CNG channel in the PM and Per2–Rom1 complex in the rim region of the disks are important in maintaining the organization of the rod OS and might represent the thin filaments observed by EM (Gilliam et al., 2012; Poetsch et al., 2001; Ritter et al., 2011; Zhang et al., 2009) (see poster). The proximity of the CNG channel–NCKX1 complex to other key phototransduction proteins such as PDE6, GCs, GCAP (guanylate-cyclase-activating protein 1) has been proposed to enhance the efficiency of phototransduction (Körschen et al., 1999; Nemet et al., 2015). Recent studies have also shown that a subset of CNG channels interact with the erythrocyte membrane protein band 4.1 (also known as EPB41), although the importance of this interaction remains to be investigated (Cheng and Molday, 2013). Cone photoreceptors have a similar structural organization, although the disks are continuous with the PM and the OS most often is of conical shape (Mustafi et al., 2009). The genes encoding some proteins, such as Per2, Rom1, GC-1, ABCA4, RDH8, RGS9, Gβ5, R9AP (i.e. RGS9–Gβ5–R9AP) and PRCD, are expressed in both rods and cones, whereas others, including those encoding opsins, CNG channels, NCKXs, PDEs and transducins, are encoded by homologous genes expressed in either rods or cones. Phototransduction Phototransduction in rods is initiated when light isomerizes the 11-cis-retinal chromophore of rhodopsin to its all-trans isomer that induces a conformational change (Arshavsky and Burns, 2012; Lamb and Pugh, 2006; Luo et al., 2008; Palczewski, 2014). The activated form of rhodopsin R (also known as metarhodopsin II) catalyzes the exchange of GDP for GTP on the α-subunit of the trimeric G-protein transducin (comprising subunits α, β and γ; see poster). The α-subunit of transducin with its bound GTP dissociates from the βγ subunits and activates PDE6, a complex of one PDE6A, one PDE6B and two PDE6G subunits, leading to the hydrolysis of cGMP. Reduction in intracellular cGMP results in the closure of CNG channels within the PM and cessation of Na+ and Ca 2+ influx, hyperpolarization of the rod cell and inhibition of glutamate release at the photoreceptor synapse. The closure of CNG channels also results in a decrease in intracellular Ca 2+ levels to below 50 nM as NCKX1 continues to efflux Ca 2+ from the OS. Quantitative studies indicate that photoisomerization of a single rhodopsin molecule results in the activation of 16 transducin proteins in mouse rods and of 60 in frog rods (Arshavsky and Burns, 2014). Further amplification is realized through PDE6-catalyzed hydrolysis of 2000 and 72,000 cGMP molecules in mouse and frog, respectively. The photoreceptor cell is returned to its dark state through a series of biochemical reactions (Lamb and Pugh, 2006; Pugh and Lamb, 1993) (see poster). Rhodopsin is phosphorylated by G-protein-coupled receptor kinase (GRK1) and inactivated following the binding of arrestin (Burns et al., 2006; Chen et al., 2012; Gurevich et al., 2011; Wilden et al., 1986). PDE6 is returned to its dark inactive state through the hydrolysis of GTP on the α-subunit of transducin, a reaction that is facilitated by the GTPase-activating protein (GAP) RGS9 (Arshavsky and Wensel, 2013). Cyclic GMP levels are re-established following the activation of GC through the Ca 2+ sensors guanylate-cyclase-activating proteins (GCAPs) (Baehr and Palczewski, 2007). When cGMP levels rise, CNG channels open and return the photoreceptor to its dark, partially depolarized state. The ubiquitous Ca 2+ sensor calmodulin (CaM) modulates the sensitivity of the channel for cGMP (Hsu and Molday, 1993). Phototransduction in cones occurs through a similar mechanism. Finally, for the regeneration of rhodopsin, all-trans retinal has to be converted back to 11-cis retinal. This occurs through a series of biochemical reactions known as the visual or retinoid cycle, which take place in both rod OS and RPE cells (Kiser et al., 2014; Saari, 2012). Mutations within most of the phototransduction proteins have been associated with retinal diseases. Mutations in the genes encoding rhodopsin, CNGA1, CNGB1, PDE6A or PD6B cause retinitis pigmentosa (RP), whereas mutations in the genes encoding arrestin, rhodopsin kinase or NCKX1 cause congenital stationary night blindness (CSNB). Furthermore, mutations in the genes encoding cone CNG channel subunits CNGA3 and CNGB3, cone PDEH or cone transducin (GNAT2) give rise to achromatopsia, and mutations in the genes encoding in GC-1 have been linked to Leber congenital amaurosis (LCA) and cone–rod dystrophy (CRD) (see Box 1 and table within the poster). Box 1. Inherited retinal diseases Inherited retinal diseases are a clinically and genetically heterogeneous group of disorders that constitute a main cause of blindness in the world (Bramall et al., 2010; Veleri et al., 2015). These disorders are typically characterized by the progressive loss in vision resulting from mutations of genes encoding proteins that are essential for photoreceptor development, function or survival. Over 238 disease-linked genes have now been identified ( The two principal types of retinal degenerative disease (RDD) are retinitis pigmentosa (RP) and macular degeneration (MD). RP, with a prevalence of 1 in 3500 people is typically characterized by the initial loss in night and peripheral vision due to the degeneration of rods, followed by loss in cone-mediated central vision that often leads to total blindness. RP can be inherited as an autosomal-dominant (AD) RP, autosomal-recessive (AR) RP or X-linked (XL) RP trait with ADRP accounting for 30-40% of the cases, ARRP for 50-60% and XLRP for 5-15% (Hartong et al., 2006). RP can be associated with other disorders, such as hearing loss (Ushers syndrome) and cognitive impairment, polydactylism, hypogenitalism, obesity and renal disease (Bardet-Biedl syndrome). MD is typically associated with loss in central vision with variable preservation of peripheral vision. Inherited forms of MD, often called macular dystrophies, are divided into subgroups on the basis of their clinical characteristics. Examples include Stargardt macular degeneration, Best disease, Doyne honeycomb retinal dystrophy, Sorsby fundus dystrophy, Bull's eye maculopathy, and X-linked retinoschisis. Age-related macular degeneration (AMD) is a leading cause of vision loss in the elderly. Although not considered an inherited RDD, genetic variants that encode complement factors and other proteins are known to increase one’s risk of acquiring AMD (Fritsche et al., 2014). Other clinically defined inherited RDDs include cone dystrophy (CD) characterized by cone degeneration, cone-rod dystrophy (CRD) associated with cone degeneration followed by rod degeneration, and Leber congenital amaurosis (LCA), an early-onset RDD characterized by severe loss of vision at birth or within the first year of life (den Hollander et al., 2008). Congenital stationary night blindness (CSNB) is a group of nonprogressive retinal disorders characterized by impaired night vision and associated with loss in rod or both rod and cone function. Achromatopsia (ACHM) is a nonprogressive cone disorder associated with partial or complete loss in color vision. Inherited retinal diseases have been linked to mutations in proteins that play crucial roles in processes such as phototransduction; the visual cycle; outer segment (OS) structure and morphogenesis; connecting cilium structure and transport, as well as in cellular functions that include protein trafficking; protein folding and post-translational modification; protein trafficking; RNA splicing and transcription; nucleotide, carbohydrate and lipid metabolism; extracellular matrix structure; ion transport; synaptic structure and neurotransmission; and development. The molecular and cellular mechanisms by which mutations in specific genes cause photoreceptor cell death are currently under extensive investigation. Visual or retinoid cycle In the conventional visual cycle, all-trans retinal released from rhodopsin following photoexcitation is reduced to all-trans retinol by RDH8 in disks (see poster). All-trans retinol is shuttled to RPE cells by the interphotoreceptor retinoid-binding protein (IRBP) where it is first converted to its retinyl esters by lecithin retinol acyl transferase (LRAT), before being isomerized to 11-cis-retinol by RPE65, oxidized to 11-cis-retinal by RDH5 and other RDHs, and delivered back to photoreceptors by IRBP for the regeneration of rhodopsin. However, a substantial fraction of all-trans retinal that is released from rhodopsin reversibly reacts with phosphatidylethanolamine (PE) to form the Schiff base adduct N-retinylidene-PE (N-ret-PE). This retinoid compound can become trapped on the luminal leaflet of disk membranes. ABCA4 actively transports or flips N-ret-PE to the cytoplasmic leaflet of disk membranes (Molday et al., 2009; Quazi et al., 2012) (see poster). All-trans retinal produced through the reversible dissociation of N-ret-PE is then reduced by RDH8 as part of the visual cycle. Retinal can diffuse from photoreceptor OS to the IS and RPE cells. Other RDH isozymes, including RDH12 and RDH10, protect these cellular compartments against retinal toxicity. Recent studies have shown that ABCA4 also plays a crucial role in the removal of excess 11-cis retinal that is not needed for the regeneration of rhodopsin (Boyer et al., 2012). ABCA4 can flip the 11-cis isomer of N-ret-PE from the luminal to the cytoplasmic leaflet of disks (Quazi and Molday, 2014). This transport function, coupled with chemical isomerization to all-trans-N-ret-PE, enables all-trans retinal to be reduced to all-trans retinol by RDH8 for entry into the visual cycle. This ensures that none of the 11-cis and all-trans retinal accumulate in disks. If these compounds are not efficiently removed, they can form toxic bisretinoid compounds, which accumulate in RPE cells upon OS phagocytosis. High levels of bisretinoids within lipofuscin deposits are found in individuals with Stargardt macular degeneration linked to mutations in the gene encoding ABCA4 as well as in Abca4-knockout mice (Allikmets et al., 1997; Mata et al., 2000; Molday and Zhang, 2010; Sparrow et al., 2012). In addition to the conventional visual cycle, cone photoreceptors use a modified visual cycle in which 11-cis retinal is resynthesized from all-trans retinol through a series of reactions that take place in Müller cells and cones (Mata et al., 2002). Most proteins that function in the visual cycle have been associated with retinal degenerative diseases. Mutations in the genes encoding LRAT and RPE65 have been linked to LCA, mutations in the gene encoding IRBP are associated with RP, and mutations in gene encoding RDH5 cause a rare form of CSNB termed fundus albipunctatus (FA) (Travis et al., 2007). OS – membrane turnover and relationship to non-motile cilia Rod and cone OS are structurally homologous to non-motile cilia. The CC, the only physical connection between OS and IS, is structurally equivalent to the ciliary transition zone (Gilliam et al., 2012). Passing through this 0.3-µm connection is an axoneme with a 9+0 arrangement of tubulin doublets that is anchored via the basal body to the ciliary rootlet, a structure that spans the length of the IS. The CC has been imaged in three dimensions by using cryo-EM (Gilliam et al., 2012). Most components of the CC are always present in other non-motile cilia, although unique components, such as RPGRIP and a splice variant of RPGR (Hong et al., 2001), are present. Profuse bi-directional trafficking of soluble and transmembrane proteins occurs through the CC. Owing to the high volume of transport (Besharse et al., 1977), the OS serves as a default destination for membrane proteins that lack localization information, for instance, due to mutations (Agbaga et al., 2014; Baker et al., 2008; Tam et al., 2000), with the relative degree of OS localization dependent on the rate of disk membrane synthesis (Pearring et al., 2013). OS are rapidly renewed to ensure maximum photosensitivity, which requires 10 days in mice, rats and Xenopus laevis (Besharse et al., 1977; Young, 1967), but 6 weeks in Rana pipiens. Radiolabeling shows that disk synthesis occurs at the base of the OS (Young and Droz, 1968). Older disks are displaced distally and eventually shed in packets from the tip of the OS, where they are phagocytosed by the RPE (Kevany and Palczewski, 2010; Young and Bok, 1969). Because cone disks are not physically isolated, mixing of new and old components of the disk membrane occurs (Young, 1969); however, disk renewal, similarly, involves incorporation of new components as well as shedding and phagocytosis of a fraction of the OS membranes (Anderson et al., 1978; Hogan et al., 1974). Already several decades ago, it has been proposed that disks originate as evaginations of the PM, which then develop a specialized rim region and, eventually, seal off completely in rods (Steinberg et al., 1980) (see poster). This model is well-supported by several lines of evidence, including EM studies (Besharse et al., 1977; Steinberg et al., 1980), incorporation of membrane-impermeable dyes, such as Lucifer yellow, into basal disks (Matsumoto and Besharse, 1985) and the presence of open contiguous disks in cones, which evolutionarily precede rods (Lamb et al., 2007). In contrast, more recently Sung and colleagues have proposed that rod disks are never continuous with the PM (Chuang et al., 2007) and originate from fusion of transport vesicles that transit the CC (Chuang et al., 2015). However, this model of disk synthesis, the evidence for which is limited to the output of a single laboratory, has been discounted by two recent findings. David Williams (Jules Stein Eye Institute, UCLA, CA) and co-workers imaged nascent disks in three dimensions by electron tomography, demonstrating both the presence of open disks and the source of enclosed disk profiles in standard electron micrographs (David Williams, personal communication). Ding et al. (2015) demonstrated that nascent disks are accessible to membrane-impermeable tannic acid, even in cases where 2D electron micrographs indicate enclosure by a plasma membrane, and that even short delays in fixation can generate vesicular structures. Moreover, structural studies indicate that vesicles of the dimensions observed could not pass through the basal body (Jin et al., 2010). At least two membrane proteins are exclusively found at the site of, and are likely to be involved in, disk synthesis within rods and cones: prominin-1, which is also associated with membrane evaginations in other cell types (Han et al., 2012; Maw et al., 2000), and pcdh-21 (also known as CDHR1), a photoreceptor-specific protocadherin (Rattner et al., 2001). These proteins form a complex of unknown function (Yang et al., 2008) that also includes the extracellular soluble protein eyes shut (EYS) (Nie et al., 2012). Protein trafficking to the OS Trafficking of several proteins between OS and IS has been examined in some detail, including that of rhodopsin, Per2, arrestin, transducin, guanylate cyclase and phosphodiesterase (Pearring et al., 2013). Rhodopsin, the most abundant protein in rod OS, is a transmembrane protein with a C-terminal ciliary targeting signal (Tam et al., 2000) that is also present in cone opsins. The large unidirectional flow of rhodopsin to the OS makes it a cargo of interest for ciliary trafficking studies (Wang and Deretic, 2015). Rhodopsin trafficking involves crossing an uncharacterized diffusion barrier that separates OS and IS PM components (Jin et al., 2010). Vesicles transporting rhodopsin fuse with the IS PM at a specialized convolution at the base of the CC, termed the periciliary ridge complex (Papermaster et al., 1985; Peters et al., 1983). This structure is hypertrophied in frog rods, but analogous structures are present in mammalian photoreceptors and other primary cilia. Small G-proteins are associated with membranes that contain newly synthesized rhodopsin (Deretic et al., 1995); these have also been implicated in rhodopsin trafficking following both in vivo and in vitro investigations that demonstrated an inhibition of trafficking by dominant-negative Rab8 (Moritz et al., 2001) and a direct interaction of Arf4 and Rab11 with the ciliary targeting signal (Deretic et al., 2005; Reish et al., 2014). Rab8 is also implicated through association of the Rab8 effector rabin8 with the multiple-protein complex comprised of seven Bardet–Biedl syndrome (BBS) proteins, the BBSome, which constitutes a coat complex coat that is involved in ciliary trafficking (Nachury et al., 2007), as well as the presence of the Rab8 GEF RPGR in the CC (Murga-Zamalloa et al., 2010). Other factors implicated in rhodopsin trafficking include the t-SNARE syntaxin-3 (Mazelova et al., 2009b), the Rab11 effector FIP3 (also known as IKBKG), the ArfGAP ASAP1 (Mazelova et al., 2009a), and cytoplasmic dynein (Tai et al., 1999). There is conflicting evidence as to whether intraflagellar transport IFT – which is mediated by kinesin-2 motor proteins – is involved, and kinesin-2 might be more crucial for cone opsin transport (Avasthi et al., 2009; Bhowmick et al., 2009; Insinna and Besharse, 2008; Jiang et al., 2015a,b; Keady et al., 2011; Marszalek et al., 2000; Trivedi et al., 2012). A distinct localization signal has also been identified in Per2 (Salinas et al., 2013; Tam et al., 2004), which is transported by a pathway that bypasses the Golgi complex (Fariss et al., 1997; Tian et al., 2014). Several proteins require cofactors for OS trafficking, including GC – which requires RD3 (Azadi et al., 2010), cone opsin (11-cis retinal) (Zhang et al., 2008) and phosphodiesterase (UNC119) (Zhang et al., 2011). The soluble proteins arrestin and transducin exhibit light-dependent trafficking (Peterson et al., 2003; Sokolov et al., 2002; Whelan and McGinnis, 1988). In response to light, arrestin migrates to rod OS, whereas transducin translocates to IS. Retrograde transport of transducin is linked to saturation of phosphodiesterase and does not occur in cones unless they are genetically modified to express rod opsin (Lobanova et al., 2010). Arrestin transport was originally thought to be caused by its binding to phosphorylated rhodopsin. However, it is more likely to be an active transport mechanism that ensures adequate quenching of phototransduction, possibly triggered by a phospholipase C cascade (Orisme et al., 2010), as rhodopsin phosphorylation is not required (Calvert et al., 2006; Mendez et al., 2003; Strissel et al., 2006). Tubulin has been proposed to bind with low affinity to arrestin in the IS (Nair et al., 2005). Conclusions and perspectives Considerable progress has been made in the characterization of photoreceptor cells and their role in the initial step of the visual process. Most of the rod and cone proteins that play crucial roles in OS structure, phototransduction and the visual cycle have been identified and characterized at molecular and cellular levels. The renewal of rod and cone OS has also been elucidated in detail at cellular level. However, the molecular mechanisms responsible for OS phagocytosis by RPE and disk morphogenesis require more-detailed studies. Additional studies are also needed to define the molecular and cellular basis for protein trafficking and sorting within photoreceptors as many of the proposed mechanisms are controversial and lacking in detail. Finally, further studies are needed to elucidate the molecular and cellular mechanisms responsible for inherited retinal degenerative diseases, and the biochemical pathways important for photoreceptor survival and photoreceptor cell death. Information from these studies is crucial for the development of rational therapeutic approaches to slow or prevent vision loss in individuals who suffer from various retinal diseases. Footnotes Competing interests The authors declare no competing or financial interests. Funding Research in the laboratory of R.S.M. is funded by the National Institutes of Health [grant number EY 02422], the Canadian Institutes for Health Research [grant numbers MOP-106667; CIHR RMF-92101], Foundation Fighting Blindness and the Macula Vision Research Foundation. Research in the laboratory of O.L.M. is funded by the Canadian Institutes for Health Research [grant number MOP-64400], Natural Sciences and Engineering Research Council of Canada [grant number RGPIN-2015-04326], and the Foundation Fighting Blindness. R.S.M. is a Canada Research Chair in Vision and Macular Degeneration. Deposited in PMC for release after 12 months. Cell science at a glance A high-resolution version of the poster is available for downloading in the online version of this article at jcs.biologists.org. Individual poster panels are available as JPEG files at References Agbaga M.-P., Tam B. M., Wong J. S., Yang L. L., Anderson R. E. and Moritz O. L. (2014). 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14530	https://www.wtamu.edu/~cbaird/all_homework_solutions/Jackson_4_5_Homework_Solution.pdf	Jackson 4.5 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: A localized charge density ρ(x , y , z) is placed in an external electrostatic field described by a potential Φ(0)( x , y , z) . The external potential varies slowly in space over a region where the charge density is different from zero. (a) From first principles calculate the total force acting on the charge distribution as an expansion in multipole moments times derivatives of the electric field, up to and including the quadrupole moments. Show that the force is F=q E (0)+(∇[p⋅E (0)])0+(∇[ 1 6∑ j, k Q jk ∂E j (0) ∂xk (x)])0 +... Compare this to the expansion (4.24) of the energy W. Note that (4.24) is a number – it is not a function of x that can be differentiated! What is its connection to F? (b) Repeat the calculation of part a for the total torque. For simplicity, evaluate only one Cartesian component of the torque, say N1. Show that this component is: N 1=[p×E (0)(0)]1+1 3[ ∂ ∂x3(∑ j Q2 j E j (0))−∂ ∂x2(∑ j Q 3 j E j (0))]0 +... SOLUTION: (a) The force in general is: F=∫ρ(x)E (0)(x)d 3x Let us keep track of the components. F=∑ i ̂ xi∫ρ(x)Ei (0)(x)d 3x Expand the external electric field in a Taylor series and only keep the focus on the first few terms because it varies slowly over space: Ei (0)(x)=[Ei (0)(x')+∑ j x j ∂ ∂x j' Ei (0)(x')+1 2∑ j, k x j xk ∂ ∂x j' ∂ ∂xk ' Ei (0)(x ')+...]x'=0 Note we have labeled the gradients as being done with respect to primed variables to tell them apart from the integration variables. Substitute the expansion into the force equation: F=[E (0)(x')q+∑ i ∑ j ̂ xi∫ρ(x) x jd 3x ∂ ∂x j' Ei (0)+1 2∑ i ∑ j , k ̂ xi∫ρ(x) x j xk d 3x ∂ ∂x j' ∂ ∂xk ' Ei+...]x'=0 Note that ∇×E=0 means that ∂Ei ∂x j =∂E j ∂xi . We use this on the dipole terms and the quadrupole terms: F=[E (0)(x')q+∑ i ∑ j ̂ xi∫ρ(x) x jd 3x ∂ ∂xi' E j (0)+1 2∑ i ∑ j ,k ̂ xi∫ρ(x) x j xk d 3x ∂ ∂xi' ∂ ∂xk ' E j (0)+...]x'=0 The electric fields do not depend on the unprimed variables and come out of the integrals, which was the point of the Taylor series expansion. After a little manipulation, we recognize the integrals that are left as the dipole moment and quadrupole moments: F=[E (0)(x')q+∑ i ̂ xi ∂ ∂xi' p⋅E (0)+1 6∑ i ̂ xi ∂ ∂xi' (∑ j ,k Q j k ∂ ∂xk ' E j (0)+∑ j ∂ ∂x j' E j (0)∫ρ(x)r 2d 3x)+...]x'=0 Next note that the external field is created by charges outside our volume of interest, so: ∇⋅E(0)=0 or ∑ i ∂Ei (0) ∂xi =0 This is the exact factor found in the last term shown, dropping that entire term out. Replacing the index notation with vector notation, we finally have: F=q E (0)(0)+(∇[p⋅E (0)])0+(∇[ 1 6∑ j ,k Q jk ∂E j (0) ∂x k (x)])0 +... Here we have switched primed variables to unprimed to match Jackson, and because the originally unprimed variables (the integration variables) are neatly tucked away now in the multipole moments. If we write the first term in terms of the potential, we can factor out the gradient operator: F=−∇[qΦ (0)−p⋅E (0)−1 6∑ j, k Q jk ∂E j (0) ∂xk (x)+...]0 Let us compare this to the expansion (Jackson 4.24) of the energy W: W =qΦ(0)−p⋅E(0)−1 6∑ i ∑ j Q ij ∂E j ∂xi (0)+... We see that using both equations, we recover the familiar expression: W =−∫F⋅d x(0) (b) The electrostatic torque on a charge distribution ρ as a result of the external field E(0) is: N=∫x×(ρ(x)E (0)(x))d 3x Let us look at one component, say component 1: N 1=∫ρ(x)( x2E3 (0)−x3E2 (0))d 3x Expand the electric field in a Taylor series: E2 (0)(x)=[E2 (0)(x')+∑ j x j ∂ ∂x j' E2 (0)(x')+...]x'=0 and E3 (0)(x)=[E3 (0)(x')+∑ j x j ∂ ∂x j' E3 (0)(x')+...]x'=0 and insert these into the torque equation to find: N 1=[E3 (0)(x ')∫ρ(x) x2d 3x−E2 (0)(x')∫ρ(x) x3d 3x +∑ j ∂ ∂x j ' E3 (0)∫ρ(x) x2 x j d 3x−∑ j ∂ ∂x j' E2 (0)∫ρ(x) x3x j d 3x ]x'=0+... We have moved the electric field components out of the integrals because they do not depend on the unprimed integration variables. This was the point of the Taylor expansion. Just as was done in the previous section, the non-diverging nature of the external electric field means that there is a piece equal to zero that we can add to get the last terms to look like the quadrupole moments. We end up with: N 1=[ E3 (0)(x ') p2−E2 (0)(x ') p3+1 3∑ j ∂ ∂x j' E3 (0)Q2 j−1 3∑ j ∂ ∂x j' E2 (0)Q 3 j]x'=0+... Again use the relation: ∂Ei ∂x j =∂E j ∂xi N 1=[ E3 (0)(x ') p2−E2 (0)(x ') p3+1 3 ∂ ∂x3'∑ j E j (0)Q2 j−1 3 ∂ ∂x2'∑ j E j (0)Q3 j]x'=0+... N 1=[p×E (0)(0)]1+1 3[ ∂ ∂x3(∑ j Q2 j E j (0))−∂ ∂x2(∑ j Q3 jE j (0))]0 +...
14531	https://www.neurology.org/doi/10.1212/WN9.0000000000000017	Comparison of the Diagnostic Performance of the Central Vein Sign and CSF Oligoclonal Bands Supporting the Diagnosis of Multiple Sclerosis \| Neurology Open Access Skip to main content Skip to main contentAAN.com AAN Publications Author Center About the Journals Press Releases 0 Cart Search Sign inSUBSCRIBE Quick Search anywhere Enter search term Quick search in Citations Journal Year Volume Issue Page Searching: Anywhere AnywhereCitation Advanced SearchSearch Trending Terms: Cognitive Disorders/Dementia Cerebrovascular Disease/Stroke Multiple Sclerosis Health Disparities navigate the sidebar menu Sign inSUBSCRIBE Quick Search anywhere Enter search term Journals Neurology® Neurology® Clinical Practice Neurology® Education Neurology® Genetics Neurology® Neuroimmunology & Neuroinflammation Neurology® Open Access Articles Latest Research Latest Non-Research Current Issue Past Issues Editorial Board Masthead All Neurology Journals Editorial Boards Topic Collections Submit a manuscript Author Center E-alerts Research Articles All Research Articles Null Hypothesis Non-Research Articles All Non-Research Articles Research Methods in Neurology Open Review Resident & Fellow Neurology® Podcast Letters to the Editor Practice Current Continuing Medical Education (CME) Topic Collections Multimedia Blogs Infographics Neurology® Video Journal Club Video Summaries About About the Journals Advertise Contact Us Editorial Boards Ethics Policies For Authors Author Center Submit Manuscript Sign Up for E-Alerts Press Releases Terms of Service Privacy Policy A peer-reviewed journal for research in neurology and the clinical neurosciences Articles Latest Research Latest Non-Research Current Issue Past Issues Editorial Board Masthead All Neurology Journals Editorial Boards Topic Collections Submit a manuscript Author Center E-alerts Reference #1 Research Article May 16, 2025 Open Access Comparison of the Diagnostic Performance of the Central Vein Sign and CSF Oligoclonal Bands Supporting the Diagnosis of Multiple Sclerosis VIEW EDITORIAL Christopher Martin Allen Margareta A.Clarke Hari V.Pai, Marija Cauchi, Jonathan Hawken, Zin M.Htet, Kimberley Allen-Philbey … Show All …, Bader Mohamed Deborah Fitzsimmons, Roshan Das Nair, Paul Morgan Christopher Partlett, Rob A.Dineen Klaus Schmierer Emma Clare Tallantyre, and Nikos Evangelou FewerAuthors Info & Affiliations June 2025 issue 1 (2) Letters to the Editor Track CitationsAdd to favorites full text Contents Abstract Introduction Methods Results Discussion Glossary Acknowledgment Study Funding Disclosure Publication History Footnote Supplementary Materials References Information & Authors Metrics & Citations View Options References Figures Tables Media Share Abstract Background and Objectives The central vein sign (CVS) describes the presence of venules within multiple sclerosis (MS) brain lesions, visible on T2-weighted MRI. In the upcoming revision of the MS diagnostic criteria, the simplified “rule of 6” (i.e., finding 6 lesions with a central venule) can support the diagnosis of MS as an alternative to lumbar puncture (LP). We evaluated whether a T2-weighted MRI scan is more sensitive than oligoclonal bands (OCBs) for diagnosing MS at presentation with a typical clinically isolated syndrome (CIS). We also compared the tolerability of LP and the additional MRI. Methods Participants requiring an LP to meet the 2017 McDonald diagnostic criteria for MS were enrolled in this multicenter, prospective, diagnostic superiority study from 3 UK neuroscience centers. A six-minute T2-weighted sequence was used to assess the CVS using 2 definitions: a 40% threshold of all eligible lesions and the rule of 6. These were compared with OCBs, using the clinical diagnosis at 18 months as the reference standard. Results Of 113 participants, 99 (mean age: 38, female: 73%) have completed all study activities: 80 were diagnosed with MS, 10 remained CIS, 8 had alternative diagnoses, and 1 remained without a diagnosis. No significant difference in diagnostic sensitivity was detected between 40% CVS threshold (90% [CI 81%–96%]) and OCB testing (84% [CI 74%–91%]) (p = 0.332). The rule of 6 had a sensitivity of 91% (CI 83%–96%). Side effects were reported by 75% following LP compared with 9% following MRI. All participants preferred their MRI scan over their LP. Discussion CVS and OCB testing is equally sensitive in supporting the diagnosis of MS in cases of typical CIS. CVS assessed using the 40% threshold, and the simpler rule of 6 produces equivalent diagnostic performance. Compared with OCB testing, CVS testing seems safer and better tolerated by patients. Further studies are needed to evaluate CVS specificity, particularly outside of typical CIS cases, as studied here. Classification of Evidence This study provides Class IV evidence that CSF OCBs and the CVS are equally sensitive in supporting a diagnosis of MS in patients presenting with CIS. Introduction Making a diagnosis of multiple sclerosis (MS) can be challenging due to other conditions that mimic the symptoms, examination findings, and investigation results seen in MS. Diagnostic uncertainty can therefore arise, and patients frequently wait months, sometimes years, before the diagnosis is confirmed and treatment can start.1 Diagnostic delays can affect disease outcomes because early diagnosis and treatment are important in preventing irreversible long-term disability.2,3 However, the main trade-off against early diagnosis is the risk of misdiagnosis because there is currently no definitive diagnostic test for MS.1 CSF oligoclonal bands (OCBs) unmatched in serum support the diagnosis of MS in patients with a typical clinically isolated syndrome (CIS) using the 2017 McDonald diagnostic criteria.4 This has led to an increase in diagnostic sensitivity compared with previous iterations of the criteria.5 Although the majority of people with established MS have unpaired OCB (88%–99%), their sensitivity is lower in newly presenting patients (55%–68%).6,7 OCBs are also not specific for MS, being present in other inflammatory or infectious conditions.8 Lumbar puncture (LP), required to collect CSF, is often painful and may cause iatrogenic morbidity, most commonly post-LP headaches.9,10 As a result, LPs are associated with additional health care costs such as hospitalization for monitoring, an anesthetist performing a blood patch, and time off work.10,11 The currently proposed modified diagnostic criteria introduce the central vein sign (CVS) as an imaging biomarker that can support the diagnosis of MS.12 The CVS is the presence of a vein or venule in the center of each lesion.13 Research from the past 2 decades, initially using ultra-high-field (7T) and subsequently lower field (1.5 and 3T), high-resolution T2-weighted (T2), and susceptibility-weighted (SWI) MRI, has shown that the presence of the CVS in white matter lesions is highly specific to MS, able to differentiate it from other neuroinflammatory diseases.14-22 Initial studies used a threshold of 40% of lesions with a central vein,21 and later, the simplified “rule of 6” was introduced which requires finding 6 lesions with a central vein to differentiate MS from non-MS.22 The rule of 6 has now been incorporated into the latest proposed diagnostic criteria; patients with a typical history, evidence of dissemination in space, and 6 lesions with a central vein will be eligible to be diagnosed with MS. Alternatively, in the absence of evidence of dissemination in space, those with evidence of dissemination in time and 6 lesions with a central vein or unmatched OCB can be diagnosed with MS.12 We aimed to explore the diagnostic utility of the CVS in comparison with OCB in a cohort of patients with typical CIS to determine whether LP could be replaced by an MRI scan. The primary research question of this prospective, diagnostic superiority study was is CVS testing with T2 MRI more sensitive than OCB testing for the diagnosis of MS in patients presenting with typical CIS? The secondary objectives addressed the following1: what is the specificity of the 2 diagnostic tests in patients with CIS?2 What is the sensitivity and specificity of the rule of 6?3 What are the side effects and tolerability of LP and MRI in the diagnostic pathway of MS? Methods Standard Protocol Approvals, Registrations, and Patient Consents DECISIve (DiagnosE using the CVS, clinical trials reference: NCT05533905) was a prospective, multicenter, pragmatic, single-group, rater-blinded, diagnostic accuracy study. The study was approved by a research ethics committee (19/LO/1499), and written informed consent was obtained from all participants before enrolment in the study. The study is reported in accordance with the STARD guidelines.23 Study Design and Participants Eligible participants aged 18–65 presented with a typical CIS4 for diagnostic evaluation of MS at 3 UK neuroscience centers (Queen's Medical Centre, University Hospital of Wales, and The Royal London Hospital) in which LP was undertaken to assess whether patients met the revised 2017 McDonald diagnostic criteria for MS.4 Exclusion criteria included participants who were unable to communicate in English, unable or unwilling to provide informed consent, and/or those who already fulfilled the diagnosis of MS. Consecutive patients who met the eligibility criteria were approached after their consultation with a neurologist, and those who consented to participate were enrolled before their LP, which was undertaken as per standard of care at each study site. The investigations were to take place as soon as possible after enrolment into the study, and the order of the LP and MRI was not fixed. Any additional clinical investigations, such as blood tests for aquaporin-4 or myelin oligodendrocyte glycoprotein antibodies, or visual evoked potentials, were performed at the discretion of the clinical team. MRI Protocol Acquisition The study MRI was performed either as a separate research test or added to a clinical MRI before the injection of a gadolinium contrast agent (if applicable). The following 2 sequences were acquired on a 3T scanner (Philips Achieva in Nottingham, Siemens Magnetom Prisma in Cardiff, and Siemens Magnetom Verio in London): three-dimensional (3D) T2 gradient echo, sagittal acquisition, 0.6 × 0.6 × 0.6 mm voxel size, 230 x 230 x 180 mm field of view, effective echo time 25 ms, repetition time of 55 ms, parallel imaging factor 2, 10-degree flip angle, echo planar imaging factor or multiecho options if available, scan duration of 6 minutes or less, 3D fluid-attenuated inversion recovery (FLAIR), sagittal acquisition to match 3D T2 location, 1 × 1 × 1mm voxel size, 230 × 230 × 180 mm field of view, manufacturer specific optimized acquisition settings, parallel imaging factor of 2, fat-saturation prepulse, and a scan duration of around 6 minutes. Image Analysis The treating neurologist did not view the T2 sequence or attempt to interpret the images, and they were not reported by a local radiologist. The MRI data acquired at each site were anonymized and transferred for blinded central review by 3 independent, blinded raters: M.A. Clarke, C.M. Allen, and H.V. Pai, neurology trainees at the time of the study, read the literature related to the CVS and were trained by R. Dineen, a professor of neuroradiology, who in addition to his research experience, assesses the CVS for clinical purposes. Training involved theoretical instruction on the imaging marker, followed by hands-on image interpretation with expert guidance. Proficiency was assessed through independent image interpretation and comparison with the RD readings. M.A. Clarke is a postdoctoral MS researcher who has been researching the CVS for 9 years. The images were split between the 3 raters and assessed independently. A proportion of scans was assessed by all raters to calculate interrater agreement. Each FLAIR scan was assessed in 3D Slicer Version 424, and all distinct lesions, which measured at least 3 mm in 1 plane, were manually marked with fiducial coordinates. The fiducial coordinates were then used to identify lesions on the corresponding T2 scan, which used for the assessment of the CVS. The North American Imaging in MS (NAIMS) Cooperative radiologic definition of a central vein was used (Table 1).13 Example CVS-positive lesions are shown in Figure 1. The CVS was positive when ≥40% of eligible lesions had a visible central vein. The cutoff value was selected before opening enrolment, following a recent prospective assessment of the CVS in cases of diagnostic uncertainty, which used the same MRI sequence and field strength as DECISIve.25 We also tested the performance of the rule of 6, which was met either if ≥6 CVS positive lesions were detected or if there were fewer than 6 lesions in total, ≥50% of eligible lesions displayed the CVS.22 Open in Viewer Table 1 NAIMS Cooperative CVS Criteria \| Central vein eligibility criteria \| Central vein exclusion criteria \| --- \| \| Assessed on T2 \| Assessed on FLAIR \| \| Appears as a thin hypointense line or small hypointense dot \| Infratentorial lesion location \| \| Can be visualized in at least 2 perpendicular MRI planes, and appears as a thin line in at least 1 plane \| Lesion merges with another lesion (confluent lesions) \| \| Has a small apparent diameter (<2 mm) \| Lesion is < 3 mm in diameter in any plane \| \| Is positioned centrally in the lesion \| Assessed on T2 \| \| Runs partially or entirely through the lesion \| Lesion is poorly visible (owing to motion or other MRI-related artifacts) \| \| \| Lesion has multiple distinct veins \| Expand Table Abbreviations: CVS = central vein sign; FLAIR = fluid-attenuated inversion recovery; NAIMS = North American Imaging in Multiple Sclerosis. Open in Viewer Figure 1 Example of Multiple Sclerosis Lesions With a Central Vein Sagittal fluid-attenuated inversion recovery (FLAIR) of MS lesions (marked by orange arrows) and the corresponding slice of the T2 image showing the lesions each have a central vein. Tolerability Analysis All DECISIve participants were invited to provide retrospective feedback on their experience of the LP and MRI scan once the 2 tests had been performed but before the clinical diagnoses were known. This was collected using a five-point Likert scale rating the overall experience (“1—very poor” to “5—excellent”). Participants were asked if either had caused any immediate or delayed problems. To further explore participants’ experiences, interviews took place using maximum variation sampling. The interviews were conducted by a person with MS, following training from the DECISIve study team. Clinical Follow-Up Usual clinical follow-up provided the source of study follow-up assessment data. At 18 months, electronic and physical health records were accessed by the local research team. Index Tests and Reference Standard The analysis sample included participants who underwent both investigations of interest (CSF OCB by LP and CVS assessment on T2 MRI). The index tests were CSF OCB unmatched in serum, a positive CVS (40% or more lesions with a central vein detected on T2 MRI) and a positive rule of six. The reference standard was clinical diagnosis 18 months after recruitment. Since a positive CSF result can fulfil the 2017 McDonald criteria without further evidence of clinical or radiologic activity (thus inherently biasing the comparison between the CVS and OCB), we also tested the performance of both tests against a clinical diagnosis based on new lesions and/or a second relapse (previously termed clinically definite MS). Statistical Analysis Sensitivity, specificity, predictive values, and likelihood ratios of all tests were calculated, including 95% CIs. We used MedCalc v23.05 to calculate receiver operating characteristic (ROC) for all the index tests. If a participant attempted a test but a definitive test result was not available, it was assumed to be negative. The sensitivity of the tests was compared using the McNemar test for paired proportions. Interrater agreement was assessed by comparing the percentage agreement between the blinded raters in diagnoses according to the CVS, using the 40% threshold and the rule of 6, in a sample of 5 randomly selected patients. Data Availability Anonymized data not published within this article will be made available on reasonable request from any qualified investigator. Results Participant Demographic and Clinical Characteristics A total of 113 participants were recruited over 30 months (7 November 2019 until 6 May 2022) across 3 participating sites (Figure 2). Fourteen participants withdrew from the study leading to a cohort of n = 99 whose demographic and clinical characteristics are given in Table 2. An alternative diagnosis was reached in 8 participants: these included migraine (n = 2), non-MS inflammatory disorder (n = 1), idiopathic transverse myelitis (n = 1), chronic small vessel ischemic disease (n = 1), ischemic optic neuropathy (n = 1), fibromyalgia (n = 1), and radiologically isolated syndrome (RIS) (n = 1). One patient remained without a clinical diagnosis. The median interval between the LP and T2 MRI tests was 12 weeks (IQR 3–29). The LP was performed before the research MRI in 74 participants, the MRI was performed before the LP in 15 participants, and 10 had the LP and the research MRI on the same day. Open in Viewer Figure 2 DECISIve Study Flow Diagram Open in Viewer Table 2 Baseline Characteristics of Participants Included in the Study \| Baseline characteristics \| Diagnosis of MS (N = 80) \| Diagnosis of CIS (N = 10) \| Alternative diagnosis (N = 9) \| Total (N = 99) \| --- --- \| Mean age at enrolment, y (SD) \| 38 (12) \| 40 (13) \| 39 (12) \| 38 (12) \| \| Sex \| \| \| \| \| \| Female, n (%) \| 60 (75) \| 8 (80) \| 6 (67) \| 74 (75) \| \| Male, n (%) \| 20 (25) \| 2 (20) \| 3 (33) \| 25 (25) \| \| Ethnicity \| \| \| \| \| \| White, n (%) \| 68 (85) \| 8 (80) \| 8 (89) \| 84 (85) \| \| Asian, n (%) \| 6 (8) \| 0 \| 1 (11) \| 7 (7) \| \| Black, n (%) \| 3 (4) \| 0 \| 0 \| 3 (3) \| \| Mixed, n (%) \| 1 (1) \| 1 (10) \| 0 \| 2 (2) \| \| Other, n (%) \| 2 (3) \| 0 \| 0 \| 2 (2) \| \| Not provided, n (%) \| 0 \| 1 (10) \| 0 \| 1 (1) \| \| Past or current tobacco use, n (%) \| 36 (46) \| 7 (70) \| 1 (11) \| 44 (45) \| \| Hypertension, n (%) \| 5 (6) \| 1 (10) \| 4 (44) \| 10 (10) \| \| Diabetes, n (%) \| 4 (5) \| 1 (10) \| 1 (11) \| 6 (6) \| \| Other medical comorbidities, n (%) \| 2 (3) \| 4 (40) \| 4 (44) \| 31 (31) \| \| Family history of MS (%) \| 8 (10) \| 0 \| 0 \| 8 (8) \| \| Mode of presentation \| \| \| \| \| \| GP, n (%) \| 15 (19) \| 4 (40) \| 4 (40) \| 23 (23) \| \| Emergency admission, n (%) \| 18 (23) \| 3 (30) \| 1 (11) \| 22 (22) \| \| Ophthalmology, n (%) \| 21(26) \| 1 (10) \| 3 (30) \| 25 (25) \| \| Other, n (%) \| 26 (33) \| 2 (20) \| 1 (10) \| 29 (29) \| \| Suspected additional relapse(s), n (%) \| 18 (23) \| 1 (10) \| 2 (22) \| 21 (21) \| \| Clinical brain MRI before enrolment, n (%) \| 71 (89) \| 8 (80) \| 9 (100) \| 88 (89) \| \| Number of lesions \| \| \| \| \| \| None, n (%) \| 0 \| 2 (25) \| 0 \| 2 (2) \| \| 1, n (%) \| 2 (3) \| 1 (13) \| 1 (11) \| 4 (5) \| \| 2–3, n (%) \| 7 (10) \| 1 (13) \| 2 (22) \| 10 (11) \| \| 4–9, n (%) \| 30 (42) \| 2 (25) \| 3 (33) \| 35 (40) \| \| 10+, n (%) \| 27 (38) \| 1 (13) \| 1 (11) \| 29 (33) \| \| Unknown, n (%) \| 5 (7) \| 1 (13) \| 2 (22) \| 8 (9) \| \| Clinical spine MRI before enrolment, n (%) \| 36 (45) \| 7 (70) \| 4 (44) \| 47 (47) \| \| Number of lesions \| \| \| \| \| \| None, n (%) \| 9 (25) \| 3 (43) \| 3 (75) \| 15 (32) \| \| 1, n (%) \| 8 (22) \| 2 (29) \| 0 \| 11 (23) \| \| 2–3, n (%) \| 16 (44) \| 1 (14) \| 1 (25) \| 18 (38) \| \| 4–9, n (%) \| 2 (6) \| 0 \| 0 \| 2 (4) \| \| 10+, n (%) \| 0 \| 0 \| 0 \| 0 \| \| Unknown, n (%) \| 1 (3) \| 1 (14) \| 0 \| 2 (4) \| \| Time from symptom onset to enrolment, wk (IQR) \| 18 (9–42) \| 19 (14–38) \| 34 (13–45) \| 19 (10–43) \| \| Site of enrolment \| \| \| \| \| \| Nottingham, n (%) \| 48 (60) \| 3 (30) \| 3 (33) \| 54 (55) \| \| Cardiff, n (%) \| 24 (30) \| 7 (70) \| 4 (44) \| 35 (35) \| \| London, n (%) \| 8 (10) \| 0 (0) \| 2 (22) \| 10 (10) \| Expand Table Abbreviations: CIS = clinically isolated syndrome; GP = general practitioner; MS = multiple sclerosis. Diagnostic Superiority, Sensitivity, and Specificity Analyses The comparison of sensitivities of the CVS and OCB testing in the MS vs non-MS group (excluding RIS and CIS) did not show superiority of the CVS assessment, as given in Table 3. Sensitivity of the CVS using the 40% threshold was 90% (CI 81%–96%), and OCB sensitivity was 84% (CI 74%–91%), McNemar test for paired proportions p = 0.332. The sensitivity of CVS assessment using the rule of 6 was 91% (CI 83%–96%). There were 6 discordant results for CVS assessment using the 40% threshold, and the rule of 6; 5 were people in the MS group: 3 participants did not reach the 40% threshold (results ranged from 32% to 38% of lesions with a central vein) while fulfilling the rule of 6 and 2 fulfilled the 40% threshold (40% and 44% of eligible lesions with a central vein) without fulfilling the rule of 6. One person with RIS also fulfilled the rule of 6 without satisfying the 40% threshold (21% of eligible lesions with a central vein). Open in Viewer Table 3 Comparison of the Sensitivity of the Central Vein Sign and Oligoclonal Band Testing \| \| \| CVS correctly diagnosed MS \| Total, (%) \| --- \| \| Yes \| No \| \| OCB correctly diagnosed MS \| Yes \| 61 \| 6 \| 67/80 (84) \| \| \| No \| 11 \| 2 \| 13/78 (17) \| \| Total, (%) \| 72/80 (90) \| 8/80 (10) \| 80 \| Abbreviations: CVS = central vein sign; MS = multiple sclerosis; OCB = unmatched oligoclonal bands. p value from the McNemar test: p = 0.332. OCB achieved 100% (CI 50%–100%) specificity, while specificity for CVS using the 40% threshold and the rule of 6 were both 57% (CI 18%–90%) (Table 4). Three patients with alternative conditions had a positive CVS according to the 40% threshold and the rule of 6. The diagnoses were fibromyalgia (50% of eligible lesions with a central vein), ischemic optic neuropathy (60% of eligible lesions with a central vein), and cerebrovascular disease (50% of eligible lesions with a central vein). The diagnostic accuracy of OCB, CVS using the 40% threshold, and CVS rule of 6 were 85% (CI 76%–92%), 87% (CI 79%–94%), and 89% (CI 80%–94%), respectively. Open in Viewer Table 4 Diagnostic Accuracy Measures for the Central Vein Sign and Oligoclonal Band Testing \| Test performance \| CVS \| Rule of 6 \| OCB \| --- --- \| \| Estimate (95% CI) \| Estimate (95% CI) \| Estimate (95% CI) \| \| Sensitivity \| 90.0 (81.2–95.6) \| 91.3 (82.8–96.4) \| 83.8 (73.8–91.1) \| \| Specificity \| 57.1 (18.4–90.1) \| 57.1 (18.4–90.1) \| 100.0 (59.0–100.0) \| \| Predictive values for MS \| \| \| \| \| Positive predictive value \| 96.0 (91.1–98.3) \| 96.1 (91.2–98.3) \| 100.0 (94.6–100.0) \| \| Negative predictive value \| 33.3 (16.6–55.6) \| 36.4 (18.0–59.8) \| 35.0 (24.7–47.0) \| \| Likelihood ratios for MS \| \| \| \| \| Positive likelihood ratio \| 2.1 (0.9–5.0) \| 2.13 (0.9–5.0) \| — \| \| Negative likelihood ratio \| 0.2 (0.1–0.4) \| 0.2 (0.1–0.4) \| 0.2 (0.1–0.3) \| Expand Table Abbreviations: CVS = central vein sign; MS = multiple sclerosis; OCB = unmatched oligoclonal bands. Information regarding new MRI lesions and/or clinical relapses in the 18 months of follow-up was available for 76 participants. When comparing the diagnostic sensitivity of the CVS 40% threshold and OCB testing against clinically definite MS (evidence of new lesions and/or relapses), the CVS sensitivity was 96% (CI 86%–99%) and OCB sensitivity was 79% (CI 65%–90%),the McNemar test for paired proportions p = 0.039. The sensitivity of the rule of 6 was 94% (CI 83%–99%). A cross tabulation of all the index tests (CSF OCB, CVS using the 40% threshold, and rule of 6), the reference standard (clinical diagnosis at 18-month follow-up), and the alternative reference standard (clinically defined MS at 18 months) are presented in eTables 1 and 2. A ROC curve of all the index tests is shown in eFigure 1. Interrater Agreement The three-way interrater agreement was 100% meaning all raters independently agreed on the CVS result using both the 40% threshold and rule of 6. The 5 assessed cases included 4 participants who had a central vein in more than 40% of eligible lesions; all of them were given a clinical diagnosis of MS. One participant did not reach the 40% threshold, and their clinical diagnosis remains unknown. This participant also had negative OCB. Tolerability Analysis For the MRI scan, the mean Likert tolerability score was 4.4 (good), SD 0.7, and for LP, it was 3.4 (fair), SD 1.2 (comparison with the Wilcoxon signed-rank test Z = −4.4, p< 0.001). Immediate or delayed problems were reported by 74 participants following their LP and 9 following their MRI scan. Headache was reported by 33% and back pain by 27%, which were the commonest problems following the LP, which necessitated up to a fortnight off work or usual caring responsibilities in 15% of study participants. The MRI scans occasionally caused brief dizziness or claustrophobia, but there were no reports of time off work. Seventeen participants took part in interviews (9 female, mean age 45 years [SD 13 years]). All expressed a preference for their MRI scan over their LP. The most striking difference between the 2 tests, at any time point, was the burden experienced during the LP. Many participants also reported considerable anxiety before their LP caused by sharing of negative accounts through social networks or online. Classification of Evidence This study provides Class IV evidence that CSF oligoclonal bands and the central vein sign are equally sensitive in supporting a diagnosis of MS in patients presenting with CIS. Discussion Diagnostic LP, known to be associated with iatrogenic morbidity, has been used by many centers aiming to expedite the diagnosis of MS as per the 2017 modified diagnostic criteria.4 Our results demonstrate that the CVS, including the rule of 6, has equivalent sensitivity to OCB testing using a prospective, multicenter study of participants presenting with typical CIS. Our study also included a tolerability analysis, which demonstrated higher tolerability of the additional MRI compared with LP. Also in our study, T2 scans were performed as a separate research intervention; however, a T2 sequence (or a similar sequence such as SWI) that takes minutes to acquire can easily be added on to standard MRI brain protocols when MS is suspected. This includes different scanner vendors and field strengths.18,20,26 Although the image analysis for this study was performed offline by trained image raters rather than radiologists using clinical PACS reporting systems, we do not foresee any major barriers to clinical translation of this approach. The level of agreement between raters in our study shows that with training and experience, it is possible to achieve consistent results when assessing the CVS. Radiologists and neurologists can also readily interpret the simplified rule of 6 without having to assess all lesions. One of the fundamental issues with diagnostic accuracy studies in MS is the lack of a diagnostic gold standard to use against any proposed test. Except in rare cases, biopsy is not a feasible approach, so relying on clinical diagnoses in our study led to 2 different diagnostic standards being used, determined by OCB status. Only OCB negative cases required proof of further radiologic or clinical disease activity. OCB showed higher specificity than CVS assessment in this study, perhaps due to this bias. In addition, clinical follow-up of only 18 months limited the accuracy of the final clinical diagnosis. The most likely outcome is that some participants whose final study diagnosis remains CIS will go on to be given a diagnosis of MS in the next 5 years, when evidence of further radiologic or clinical disease activity is detected. However, study duration could not be increased without affecting the timeliness of the study findings to influence clinical care and going beyond the acceptable duration of the study's funder. We also applied a different reference standard of clinically definite MS to a subset where this information was available: those with evidence of new MRI lesions and/or relapses during the follow-up period. Although this comparison remained biased as those with OCB were often commenced on disease-modifying treatment following their clinical diagnosis, which suppresses this activity. Using this standard, we were able to demonstrate a significantly higher diagnostic sensitivity of the CVS compared with OCB testing, but OCB specificity remained higher than CVS assessment. This study was conducted in experienced MS centers, and only 8 patients were given an alternative diagnosis at the end of the follow-up period. This inherently underpowered our study to reliably assess specificity due to the study only recruiting those with typical CIS. Further studies are needed to evaluate CVS specificity, particularly outside of typical CIS cases, as studied here. Recently, Toljan et al. compared OCB with CVS reporting similar sensitivity of the rule of 6 and OCB (71% vs 75%, respectively) with increased specificity of the CVS compared with OCB (86% vs 76%).27 We are encouraged by the result showing high specificity of the CVS in a study which intentionally enrolled MS and non-MS cases. In another study by the same group,28 the rule of 6 showed 65% sensitivity and 98% specificity in correctly classifying patients with MS. We believe that differences in the target population (people with typical CIS requiring only a positive LP to be diagnosed with MS in our study, and people with T2 lesions and suspicion of MS in the 2 other studies) led to the differences in reported sensitivity and specificity. CVS assessment requires NAIMS CVS eligible supratentorial lesions. Typical CIS presentations include optic neuritis and partial transverse myelitis, and patients may have no lesions eligible for CVS interpretation at their first presentation. The range assessed in DECISIve was 0–37. Although there is no evidence of a difference in performance between the CVS 40% threshold and the rule of 6 in DECISIve, combining high-quality diagnostic studies would create a larger data set to assess the reliability of CVS interpretation with high and low lesion counts. The proposed changes to the McDonald criteria specify that a positive LP or finding 6 lesions with a central vein can confirm a diagnosis of MS. With increasing recognition of MS misdiagnosis,29 the desire to reach a diagnosis earlier (even in RIS), and our findings of a unanimous preference for an additional MRI scan over an LP, we encourage the widespread adoption of susceptibility-based imaging for MS diagnosis. CVS assessment should be particularly helpful in older patients or those presenting with vascular risk factors. Our study was affected by the coronavirus pandemic. Initially, both diagnostic tests were planned to take place within 8 weeks of each other, to minimize the risk of bias. However, due to lack of MRI scanner capacity and recurrent periods of lockdowns, the CVS assessment took place at a systematically later timepoint for some while LPs continued as usual as they were classed as routine clinical care. Considering that the CVS presence is not related to timing of assessment,30 we do not think this affected our results. However, there were participants who withdrew consent because of the pandemic, which lowered the final power of the primary analysis. This multicenter study supports the use of CVS at centers that have a 3T MRI scanner and radiologists reporting through the rule of 6. Some patients may still require diagnostic LP, for example, to rule out viral infection in cases of partial transverse myelitis. However, we expect that the overall number of patients requiring diagnostic LP will reduce, thereby reducing the burden for patients and costs of health care services. Building on these findings, a future economic analysis can formally assess the costs and consequence of implementing the CVS. DECISIve has shown that the sensitivity of the CVS is comparable with testing for OCB at first presentation with typical CIS in a pragmatic prospective multicenter study which aimed to replicate its performance in routine clinical practice. CVS with a threshold of 40% and the rule of 6 produced equivalent diagnostic performance, suggesting this could easily be implemented in clinical practice in centers with a 3T MRI scanner. Glossary CIS clinically isolated syndrome CVS central vein sign FLAIR fluid-attenuated inversion recovery LP lumbar puncture MS multiple sclerosis NAIMS North American Imaging in Multiple Sclerosis OCB oligoclonal band ROC receiver operating characteristic SWI susceptibility-weighted Acknowledgment The authors thank all the study participants and our study research team, who helped recruit and scan patients for this study. Footnote Editorial Rethinking Multiple Sclerosis Diagnosis: Can the Central Vein Sign Replace Lumbar Puncture? Page e000021 Supplementary Materials eFigure_1.pdf Download 122.05 KB eTable_1.pdf Download 48.03 KB eTable_2.pdf Download 57.24 KB References 1. Solomon AJ, Arrambide G, Brownlee WJ, et al. Differential diagnosis of suspected multiple sclerosis: an updated consensus approach. Lancet Neurol. 2023;22(8):750-768. 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The work cannot be changed in any way or used commercially without permission from the journal. Publication History Received: January 7, 2025 Accepted: March 27, 2025 Published online: May 16, 2025 Published in issue: June 2025 Permissions Request permissions for this article. Request permissions Disclosure C.M. Allen has received grant income from the National Institute for Health and Care Research (NIHR). M.A. Clarke is a 2023 ECTRIMS postdoctoral research fellow. R. das Nair has received research grants from the UK NIHR, the UKMS Society, and Progressive MS Alliance; and has received speakers' bureau fees from Biogen, Novartis, and Merck. P.S. Morgan has received honoraria and travel funding to speak at educational meetings from Janssen; and research support from the UK National Institute of Health Research and General Electric Healthcare. R.A. Dineen has received research support from the UK National Institute of Health Research. K. Schmierer is a member of the MAGNIFY-MS steering committee and the MS Global Advisory Network (Merck KGaA); is Chief Investigator of ChariotMS, supported by the National Institute of Health Research EME programme, the MS Society of Great Britain and Northern Ireland, the National MS Society (US), Barts Charity, and Merck KGaA; has received further research support from Biogen, Roche, Novartis, and Sandoz; has received speaking honoraria from and/or served in an advisory role for Biogen, Merck, Novartis, Roche, Sanofi-Genzyme, and T.G. Therapeutics; and has received remuneration for teaching activities from Medscape and MS Academy. E.C. Tallantyre has received honoraria for consulting work from Biogen, Janssen, Merck, Novartis, and Roche; and has received travel grants to attend or speak at educational meetings from Biogen, Merck, Roche, Takeda, and Novartis. N. Evangelou has served as a member of advisory boards for Biogen, Merck, Novartis, and Roche; and has received grant income from the United Kingdom Multiple Sclerosis Society, the Medical Research Council, the Patient-Centered Outcomes Research Institute, and the National Institute for Health Research. All other authors have no relevant disclosures. Go to Neurology.org/OA for full disclosures. Study Funding This study was funded by NIHR Research for Patient Benefit (PB-PG-0418-20044). Authors Affiliations & Disclosures Expand All Christopher Martin Allen Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Margareta A.Clarke Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Hari V.Pai Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Marija Cauchi Division of Psychological Medicine and Clinical Neurosciences, Cardiff University; Department of Neurosciences, Mater Dei Hospital, Malta; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Jonathan Hawken Division of Psychological Medicine and Clinical Neuroscience, School of Medicine, Cardiff University; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Zin M.Htet Division of Psychological Medicine and Clinical Neuroscience, School of Medicine, Cardiff University; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Kimberley Allen-Philbey The Blizard Institute, Centre for Neuroscience, Surgery and Trauma, Queen Mary University of London; Clinical Board Medicine (Neuroscience), Barts Health NHS Trust; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Bader Mohamed The Blizard Institute, Centre for Neuroscience, Surgery and Trauma, Queen Mary University of London; Clinical Board Medicine (Neuroscience), Barts Health NHS Trust; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Deborah Fitzsimmons Swansea Centre for Health Economics, Swansea University; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Roshan Das Nair Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; Institute of Mental Health, Nottinghamshire Healthcare NHS Foundation Trust; Health Division, SINTEF, Trondheim, Norway; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Paul Morgan Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; NIHR Nottingham Biomedical Research Centre, University of Nottingham; and Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Christopher Partlett Nottingham Clinical Trials Unit, School of Medicine, Faculty of Medicine and Health Sciences, University of Nottingham. Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Rob A.Dineen Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; NIHR Nottingham Biomedical Research Centre, University of Nottingham; and Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Klaus Schmierer The Blizard Institute, Centre for Neuroscience, Surgery and Trauma, Queen Mary University of London; Clinical Board Medicine (Neuroscience), Barts Health NHS Trust; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Emma Clare Tallantyre Division of Psychological Medicine and Clinical Neuroscience, School of Medicine, Cardiff University; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Nikos Evangelou Mental Health and Clinical Neurosciences Academic Unit, School of Medicine, University of Nottingham; Disclosure Financial Disclosure: 1. NONE Research Support: 1. NONE Stock, Stock Options & Royalties: 1. NONE Legal Proceedings: 1. NONE View all articles by this author Notes Correspondence Dr. Evangelou nikos.evangelou@nottingham.ac.uk The Article Processing Charge was funded by the authors. Submitted and externally peer reviewed. The handling editor was Amy Kunchok, MBBS, MMed, FRACP, PhD. These authors contributed equally to this work as joint first authors. Author Contributions C.M. Allen: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data; study concept or design; analysis or interpretation of data. M.A. Clarke: drafting/revision of the manuscript for content, including medical writing for content; analysis or interpretation of data. H.V. Pai: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data; analysis or interpretation of data. M. Cauchi: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data. J. Hawken: major role in the acquisition of data. Z.M. Htet: major role in the acquisition of data. K. Allen-Philbey: major role in the acquisition of data. B. Mohamed: major role in the acquisition of data. D. Fitzsimmons: drafting/revision of the manuscript for content, including medical writing for content; study concept or design. R. Das Nair: drafting/revision of the manuscript for content, including medical writing for content; study concept or design. P. Morgan: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data. C. Partlett: drafting/revision of the manuscript for content, including medical writing for content; study concept or design; analysis or interpretation of data. R.A. Dineen: drafting/revision of the manuscript for content, including medical writing for content; study concept or design; analysis or interpretation of data. K. Schmierer: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data. E.C. Tallantyre: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data; analysis or interpretation of data. N. Evangelou: drafting/revision of the manuscript for content, including medical writing for content; major role in the acquisition of data; study concept or design; analysis or interpretation of data. Metrics & Citations Metrics Citations Metrics Article Metrics Accesses Citations No data available. 2,776 0 Total First 90 Days Total number of accesses for the first 90 days after content publication Citation information is sourced from Crossref Cited-by service. See more details Picked up by 13 news outlets Blogged by 2 Posted by 10 X users Referenced by 1 Bluesky users 4 readers on Mendeley Citations Download Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Select your manager software from the list below and click Download. Format [x] Direct Import Cited By View Options View options PDF and All Supplements Download PDF and Supplementary Material Download is in progress Full Text View Full Text Figures Open all in viewer Figure 1 Example of Multiple Sclerosis Lesions With a Central Vein Sagittal fluid-attenuated inversion recovery (FLAIR) of MS lesions (marked by orange arrows) and the corresponding slice of the T2 image showing the lesions each have a central vein. Go to FigureOpen in Viewer Figure 2 DECISIve Study Flow Diagram Go to FigureOpen in Viewer Tables Open all in viewer Table 1 NAIMS Cooperative CVS Criteria Go to TableOpen in Viewer Table 2 Baseline Characteristics of Participants Included in the Study Go to TableOpen in Viewer Table 3 Comparison of the Sensitivity of the Central Vein Sign and Oligoclonal Band Testing Go to TableOpen in Viewer Table 4 Diagnostic Accuracy Measures for the Central Vein Sign and Oligoclonal Band Testing Go to TableOpen in Viewer Media Share Share Share article link Copy Link Copied! Copying failed. Share FacebookX (formerly Twitter)LinkedInemail References References 1. Solomon AJ, Arrambide G, Brownlee WJ, et al. Differential diagnosis of suspected multiple sclerosis: an updated consensus approach. Lancet Neurol. 2023;22(8):750-768. Crossref Google Scholar a [...] is confirmed and treatment can start. b [...] no definitive diagnostic test for MS. c [...] objectives addressed the following 2. Comi G, Radaelli M, Soelberg Sørensen P. Evolving concepts in the treatment of relapsing multiple sclerosis. Lancet. 2017;389(10076):1347-1356. Crossref Google Scholar a [...] irreversible long-term disability. b [...] 2 diagnostic tests in patients with CIS? 3. Fernández Ó. Is there a change of paradigm towards more effective treatment early in the course of apparent high-risk MS? Mult Scler Relat Disord. 2017;17:75-83. Crossref Google Scholar a [...] irreversible long-term disability. b [...] and specificity of the rule of 6? 4. Thompson AJ, Banwell BL, Barkhof F, et al. Diagnosis of multiple sclerosis: 2017 revisions of the McDonald criteria. Lancet Neurol. 2018;17(2):162-173. Crossref PubMed Google Scholar a [...] the 2017 McDonald diagnostic criteria. b [...] aged 18–65 presented with a typical CIS c [...] 2017 McDonald diagnostic criteria for MS. d [...] per the 2017 modified diagnostic criteria. 5. Filippi M, Preziosa P, Meani A, et al.; MAGNIMS Study Group. Performance of the 2017 and 2010 revised McDonald criteria in predicting MS diagnosis after a clinically isolated syndrome: a MAGNIMS study. Neurology. 2022;98(1):e1-e14. Go to Citation Crossref PubMed Google Scholar 6. Schwenkenbecher P, Sarikidi A, Wurster U, et al. McDonald criteria 2010 and 2005 compared: persistence of high oligoclonal band prevalence despite almost doubled diagnostic sensitivity. Int J Mol Sci. 2016;17(9):1592. Go to Citation Crossref Google Scholar 7. Dobson R, Ramagopalan S, Davis A, Giovannoni G. Cerebrospinal fluid oligoclonal bands in multiple sclerosis and clinically isolated syndromes: a meta-analysis of prevalence, prognosis and effect of latitude. J Neurol Neurosurg Psychiatry. 2013;84(8):909-914. Go to Citation Crossref PubMed Google Scholar 8. Freedman MS, Thompson EJ, Deisenhammer F, et al. Recommended standard of cerebrospinal fluid analysis in the diagnosis of multiple sclerosis: a consensus statement. Arch Neurol. 2005;62(6):865-870. 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Google Scholar a [...] that can support the diagnosis of MS. b [...] or unmatched OCB can be diagnosed with MS. 13. Sati P, Oh J, Constable RT, et al.; NAIMS Cooperative. The central vein sign and its clinical evaluation for the diagnosis of multiple sclerosis: a consensus statement from the North American Imaging in Multiple Sclerosis Cooperative. Nat Rev Neurol. 2016;12(12):714-722. Crossref PubMed Google Scholar a [...] or venule in the center of each lesion. b [...] Table 1). 14. Tallantyre EC, Brookes MJ, Dixon JE, Morgan PS, Evangelou N, Morris PG. Demonstrating the perivascular distribution of Ms lesions in vivo with 7-tesla MRI. Neurology. 2008;70(22):2076-2078. Go to Citation Crossref PubMed Google Scholar 15. Mistry N, Dixon J, Tallantyre E, et al. Central veins in brain lesions visualized with high-field magnetic resonance imaging: a pathologically specific diagnostic biomarker for inflammatory demyelination in the brain. JAMA Neurol. 2013;70(5):623-628. Go to Citation Crossref Google Scholar 16. Tallantyre EC, Morgan PS, Dixon JE, et al. A comparison of 3T and 7T in the detection of small parenchymal veins within MS lesions. Invest Radiol. 2009;44(9):491-494. Go to Citation Crossref Google Scholar 17. Tan IL, Van Schijndel RA, Pouwels PJW, et al. MR venography of multiple sclerosis. AJNR Am J neuroradiology 2000;21(6):1039-1042. Go to Citation Google Scholar 18. Clarke MA, Pareto D, Pessini-Ferreira L, et al. Value of 3T susceptibility-weighted imaging in the diagnosis of multiple sclerosis. AJNR Am J neuroradiology. 2020;41(6):1001-1008. Crossref Google Scholar a [...] it from other neuroinflammatory diseases. b [...] scanner vendors and field strengths. 19. Hammond KE, Metcalf M, Carvajal L, et al. Quantitative in vivo magnetic resonance imaging of multiple sclerosis at 7 Tesla with sensitivity to iron. Ann Neurol. 2008;64(6):707-713. Go to Citation Crossref Google Scholar 20. Samaraweera APR, Clarke MA, Whitehead A, et al. The central vein sign in multiple sclerosis lesions is present irrespective of the T2 sequence at 3 T. J Neuroimaging. 2017;27(1):114-121. Crossref Google Scholar a [...] it from other neuroinflammatory diseases. b [...] scanner vendors and field strengths. 21. Tallantyre EC, Dixon JE, Donaldson I, et al. Ultra-high-field imaging distinguishes MS lesions from asymptomatic white matter lesions. Neurology. 2011;76(6):534-539. Crossref PubMed Google Scholar a [...] it from other neuroinflammatory diseases. b [...] of 40% of lesions with a central vein, 22. Mistry N, Abdel-Fahim R, Samaraweera A, et al. Imaging central veins in brain lesions with 3-T T2-weighted magnetic resonance imaging differentiates multiple sclerosis from microangiopathic brain lesions. Mult Scler. 2016;22(10):1289-1296. Crossref PubMed Google Scholar a [...] it from other neuroinflammatory diseases. b [...] vein to differentiate MS from non-MS. c [...] ≥50% of eligible lesions displayed the CVS. 23. Bossuyt PM, Reitsma JB, Bruns DE, et al.; STARD Group. Stard 2015: an updated list of essential items for reporting diagnostic accuracy studies. BMJ. 2015;351:h5527. Go to Citation Crossref PubMed Google Scholar 24. Fedorov A, Beichel R, Kalpathy-Cramer J, et al. 3D slicer as an image computing platform for the quantitative imaging network. Magn Reson Imaging. 2012;30(9):1323-1341. Go to Citation Crossref PubMed Google Scholar 25. Clarke MA, Samaraweera APR, Falah Y, et al. Single Test to ARrive at Multiple Sclerosis (STAR-MS) diagnosis: a prospective pilot study assessing the accuracy of the central vein sign in predicting multiple sclerosis in cases of diagnostic uncertainty. Mult Scler. 2020;26(4):433-441. Go to Citation Crossref PubMed Google Scholar 26. Maggi P, Absinta M, Grammatico M, et al. Central vein sign differentiates Multiple Sclerosis from central nervous system inflammatory vasculopathies. Ann Neurol. 2018;83(2):283-294. 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Go to Citation Crossref PubMed Google Scholar Cover image: The ExTINGUISHtrial study timeline of inebilizumab as a treatment for NMDAR encephalitis. Stylized by Kaitlyn Aman Ramm, Digital Multimedia/Graphics Coordinator. See page e000007 Volume 1 \| Number 2 June 2025 advertisement Open Review Pilot Project Explore peer reviews, author responses, and editor comments on selected articles. Learn More Letters to the Editor The Letters section represents an opportunity for ongoing author debate and post-publication peer review. View our submission guidelines for Letters to the Editor before submitting your comment. View the Letters for this article Submit a Letter for this article Recommended Neurology® Articles 19 Sep 2007 Do oligoclonal bands add information to MRI in first attacks of multiple sclerosis? M. Tintoré, A. Rovira, J. Río, C. Tur, R. Pelayo, C. Nos, N. Téllez, H. Perkal, M. Comabella, J. Sastre-Garriga, and [...] X. Montalban +7 authors Neurology® Articles 14 Jan 2003 New diagnostic criteria for multiple sclerosis: Application in first demyelinating episode M. Tintoré, A. Rovira, J. Río, C. Nos, E. Grivé, J. Sastre-Garriga, I. Pericot, E. Sánchez, M. Comabella, and [...] X. Montalban +6 authors Neurology® articles 1 May 1991 Magnetic resonance imaging of the head in the diagnosis of multiple sclerosis: A prospective 2‐year follow‐up with comparison of clinical evaluation, evoked potentials, oligoclonal banding, and CT K. H. Lee, S. A. Hashimoto, J. P. Hooge, L. F. Kastrukoff, J.J.F. Oger, D. K.B. Li, and [...] D. W. Paty +3 authors Neurology® Editorial 29 Oct 2021 Multiple Sclerosis Diagnostic Criteria: Moving Ahead or Walking in Place? Gary R. Cutterand [...] Marcus W. Koch +0 authors Neurology® Editorial 7 Mar 2018 Central vein sign in multiple sclerosis: Ready for front and center? Aaron E. Millerand [...] Peter A. Calabresi +0 authors Topics Class IVDemyelinating Disease (CNS)Diagnostic test assessmentHealth Services ResearchImagingLevel of Evidence ClassificationMRIMultiple sclerosis View full text\|Download PDF Figures Tables Others Close figure viewer Back to article Figure title goes here Change zoom level Go to figure location within the article Download figure Toggle share panel Toggle share panel Share Toggle information panel Toggle information panel All figures All tables All others View all material View all material View all material xrefBack.goTo xrefBack.goTo Request permissions Expand All Collapse Expand Table Show all references SHOW ALL BOOKS Authors Info & Affiliations Now Reading: Comparison of the Diagnostic Performance of the Central Vein Sign and CSF Oligoclonal Bands Supporting the Diagnosis of Multiple Sclerosis Track Citations Add to favorites Share ###### PREVIOUS ARTICLE Cerebral Microbleeds in Parkinson Disease Previous###### NEXT ARTICLE Natural Language Processing Measures of Spontaneous Spoken Recall in Patients With Temporal Lobe Epilepsy Next Skip slideshow ArticlesArticles Latest Articles Current Issue Past Issues AboutAbout About the Journals Advertise Browse Journals Contact Us Editorial Board Ethics Policies Authors and ReviewersAuthors and Reviewers Author Center Information for Reviewers Submit Manuscript SubscribersSubscribers Sign Up for eAlerts Subscribe Follow Us AAN.COM CONTINUUM BRAIN & LIFE NEUROLOGY TODAY Back to top © 2025 American Academy of Neurology Manage Cookie Preferences Advertise Privacy Policy Your California Privacy Choices Your Privacy To give you the best possible experience we use cookies and similar technologies. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Algorithm for fair distribution of numbers into two sets Ask Question Asked 16 years ago Modified12 years, 9 months ago Viewed 4k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Given a set of n numbers (1 <= n <= 100) where each number is an integer between 1 and 450,we need to distribute those set of numbers into two sets A and B, such that the following two cases hold true: The total numbers in each set differ by at most 1. The sum of all the numbers in A is as nearly equal as possible to the sum of all the numbers in B i.e. the distribution should be fair. Can someone please suggest an efficient algorithm for solving the above problem ? Thank You. algorithm math Share Share a link to this question Copy linkCC BY-SA 2.5 Improve this question Follow Follow this question to receive notifications edited Dec 19, 2012 at 15:05 Bill the Lizard 407k 213 213 gold badges 577 577 silver badges 892 892 bronze badges asked Oct 2, 2009 at 5:17 Ralph Ralph 1 Do you get the set of n numbers up front? or you don't really know ahead of time?Calyth –Calyth 2009-10-02 13:05:21 +00:00 Commented Oct 2, 2009 at 13:05 Add a comment\| 13 Answers 13 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. Since the numbers are small it is not NP-complete. To solve it you can use dynamic programming: Make a three-dimensional table of booleans where true at t[s, n, i] means that the sum s can be reached with a subset of n elements below index i. To compute the value for t[s, n, i] check t[s, n, i-1] and t[s - a[i], n-1, i-1]. Then look through the table at second index n/2 to find the best solution. Edit: You actually don't need the complete table at once. You can make a two dimensional table t_i[s, n] for each index i and compute the table for i from the table for i-1, so you only need two of these two-dimensional tables, which saves a lot of memory. (Thanks to Martin Hock.) Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Oct 7, 2009 at 6:26 answered Oct 2, 2009 at 9:10 starbluestarblue 57k 14 14 gold badges 101 101 silver badges 153 153 bronze badges 6 Comments Add a comment Olexiy OlexiyOver a year ago For a more detailed description of DP, see the TopCoder tutorial on DP (topcoder.com/tc?module=Static&d1=tutorials&d2=dynProg). 2009-10-02T10:20:29.027Z+00:00 1 Reply Copy link fortran fortranOver a year ago A problem is NP-complete regardless the size of the input in this case! :-p Another question is if this instance of the problem is feasible given its size 2009-10-02T10:38:12.053Z+00:00 3 Reply Copy link Olexiy OlexiyOver a year ago The constraint for input values (each of them <= 450) makes the problem solvable in a polynomical time. The problem can be stupidly brute-forced in O(N^450) (which is clearly polynomical). A quick N^3 solution is described in another comment. 2009-10-02T10:39:02.28Z+00:00 0 Reply Copy link ShreevatsaR ShreevatsaROver a year ago @fortran: NO! NP-completeness (or even notions like "polynomial time") are defined in terms of the input size. The DP algorithm described here is perfectly correct; its running time is a polynomial function of the numbers in the input (here 1 to 100). The only reason this is not a polynomial-time algorithm for the general Number Partitioning Problem is that there, the "input size" is the number of bits to describe the input, so the log of the numbers in the input. 2009-10-02T15:11:28.693Z+00:00 3 Reply Copy link Martin Hock Martin HockOver a year ago You can also do it with a 2D table. t[s, k] true means that the sum s can be reached with some subset of k elements. Your array should be of size [S, ceil(n/2)] where S is the sum of all elements and n is the number of elements. Then iterate through the elements i ... i[n]. For each element i[x], scan through the table and if t[s, n] is true, set t[s+i[x], n+1] true (unless this would fall off the table). Also, set t[i[x], 1] true. Finally, scan through the last column (all x for [x, ceil(n/2)]) for the optimal value. 2009-10-07T02:33:40.82Z+00:00 1 Reply Copy link Add a comment\|Show 1 more comment This answer is useful 3 Save this answer. Show activity on this post. This is a constrained version of the Number Partioning Problem. Usually the goal is to find any 2 disjoint subsets that minimize the difference of the sums. Your problem is constrained in the sense you only consider 1 possiblity: 2 sets of size N/2 (or 1 set of N/2 and one set of N/2+1 if the total number if uneven). This dramatically reduces the search space, but I can't thnk of a good algorithm at the moment, I'll think about it. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 5:51 FalainaFalaina 6,695 31 31 silver badges 31 31 bronze badges 3 Comments Add a comment Jackson JacksonOver a year ago My calculation shows that there are 1.00891344545564e+29 ways to partition 100 numbers into 2 sets of 50 (the worst case) so any brute force approach is still going to fail! Still as your link says number partitioning is seen as the easiest of the NP complete problems! 2009-10-02T09:06:59.013Z+00:00 0 Reply Copy link ShreevatsaR ShreevatsaROver a year ago The Number Partitioning Problem (or the knapsack problem, or the subset-sum problem) is NP-complete only when you define the input size as the "length" of the input, i.e., the number of bits needed to describe the numbers. But there are easy dynamic-programming algorithms that run in time polynomial in the numbers themselves (here 1 to 100); so the fact that it is NP-complete shouldn't be a deterrent. (In fact, these problems undergo a "phase transition" when the sizes of the numbers go from polynomial to exponential in n... but 100 is a very small number, so that's of no concern here.) 2009-10-02T15:21:43.773Z+00:00 2 Reply Copy link ShreevatsaR ShreevatsaROver a year ago In short: constrained versions of NP-hard problems aren't necessarily hard. Far from it. :-) 2009-10-02T15:37:16.74Z+00:00 0 Reply Copy link Add a comment This answer is useful 2 Save this answer. Show activity on this post. If the numbers are sequential then you just alternate assigning them between A and B. I suspect they are not, in which case... Assign the largest unassigned number to the group with the lowest sum unless the difference in size of the the groups is less than or equal to count of unassigned numbers (in which case assign all of the remaining numbers to smaller group). It won't find the best solution in all cases, but its close and simple. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 5:24 RichHRichH 6,148 2 2 gold badges 40 40 silver badges 62 62 bronze badges 2 Comments Add a comment C Pirate C PirateOver a year ago Unfortunately, alternation doesn't quite meet the specified goals. Imagine the sequence 1, 2, 3: assigning numbers alternately would give you A = { 1, 3 } and B = { 2 }, but a more fair distribution would be A = { 1, 2 } and B = { 3 } 2009-10-02T05:32:06.967Z+00:00 1 Reply Copy link RichH RichHOver a year ago I did say it wouldn't find the best solution, but it is a really simple way to find a solution that is close. 2009-10-04T19:40:56.623Z+00:00 0 Reply Copy link This answer is useful 2 Save this answer. Show activity on this post. Never mind, I thought the numbers were sequential. This looks kind of like the Knapsack Problem, which is NP hard. The numbers are sequential? Put the largest number in A Put the next largest number in B Put the next largest number in B Put the next largest number in A Repeat step 1 until all the numbers are assigned. Proof: After every multiple of 4 numbers has been assigned, A and B both contain the same number of items, and the sum of the items in each group are the same because (n) + (n - 3) == (n - 1) + (n - 2) In the last iteration we are at Step 1 above and we have either 0, 1 1, 2 [1,2], or 3 [1,2,3] numbers remaining. In case 0, we are done and the groups are equal in count and weight. In case 1, we assign the number 1 to group A. Group A has one more item and one more weight. This is as fair as we can get in this situation. In case 2, we assign the number 2 to group A and the number 1 to group B. Now the groups have the same number of items and group A has one extra weight. Again, this is as fair as we can get. In case 3, assign the number 3 to group A, and assign numbers 2 and 1 to group B. Now the groups have the same weight (3 == 2 + 1) and group B has one extra item. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Oct 2, 2009 at 5:35 answered Oct 2, 2009 at 5:29 mobmob 119k 18 18 gold badges 159 159 silver badges 291 291 bronze badges 2 Comments Add a comment Olexiy OlexiyOver a year ago Though the knapsack problem is definitely NP hard, this problem (with constraints on input) can be solved in polynomic time. 2009-10-02T10:35:46.203Z+00:00 0 Reply Copy link Patrice Bernassola Patrice BernassolaOver a year ago This does not work in all cases try with 10, 15, 20, 25, 30, 45, 75. The result is A=75, 25, 20 and B = 45, 30, 15, 10 Sum in A = 120, Sum in B = 100 whereas it is possible to have 110 in each 2009-10-02T11:16:18.95Z+00:00 1 Reply Copy link This answer is useful 2 Save this answer. Show activity on this post. First, find a solution to the problem without the first constraint (i.e. - making sums as close as possible). This problem can be solved using DP approach (you can read more about DP here, and the first problem - about coins - is very similar to yours). Once you can solve it, you can add one more state to your DP - the number of persons selected to the subset already. This gives you a N^3 algorithm. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Oct 2, 2009 at 10:30 answered Oct 2, 2009 at 10:16 OlexiyOlexiy 1,084 1 1 gold badge 6 6 silver badges 12 12 bronze badges 3 Comments Add a comment ShreevatsaR ShreevatsaROver a year ago This is correct, but a minor point: the algorithm is (N^2K) where K is the maximum size of a subset (here 100N), not necessarily N^3 in general. 2009-10-02T15:40:23.083Z+00:00 0 Reply Copy link Olexiy OlexiyOver a year ago 450N. Yes, sure, but without that constraint the problem is NP-complete 2009-10-02T17:11:39.243Z+00:00 0 Reply Copy link ShreevatsaR ShreevatsaROver a year ago Yeah, I know. Although 450 is a "constant", it is really more informative to treat it as a parameter and say that the algorithm is O(N^3K), rather than O(N^3). :-) 2009-10-03T21:07:42.54Z+00:00 0 Reply Copy link Add a comment This answer is useful 2 Save this answer. Show activity on this post. I have an algorithm for you. It is using a lot of recursive and iterative concepts. Assuming you have n number Xn with 1 <= n <= 100 and 1 <= Xn <= 450. If n < 3 then distribute numbers and stop algorithm, If n > 2 then sort your list of number in ascending order, Compute the total sum S of all numbers, Then divide the previous total S by (n - n%2)/2 and obtain the A value, Now we will create couple of numbers which addition will be as near as possible as A. Get the first number and find a second number in order to obtain a sum S1 as near as possible than A. Put S1 in a new list of number and keep in memory how the sum was computed in order to have the base numbers later. Execute 5. until numbers in the list is < 2. Then put the remaining numbers to the sum list and restart algorithm to point 1. with new list. Example: Assuming: n = 7 and numbers are 10, 75, 30, 45, 25, 15, 20 Pass 1: Since n > 2 so sort the list : 10, 15, 20, 25, 30, 45, 75 Sum S = 220 A = 220 / ((7-1)/2) = 73 Couples: 10 & 75 => 85 15 & 45 => 60 20 & 30 => 50 Remaining numbers are < 2 so add 25 in the sum list : 85(10,75), 60(15,45), 50(20,30), 25(25) Pass 2: n = 4 and numbers are 85, 60, 50, 25 List count is > 2 so sort list : 25(25), 50(20,30), 60(15,45), 85(10,75) Sum S is still the same (S=220) but A must be recompute : A = 220 / ((4-0)/2) = 110 Couples: 25 & 85 => 110 50 & 60 => 110 The Sum list is : 110(25(25),85(10,75)), 110(50(20,30),60(15,45)) Pass 3: n = 2 and numbers are 110, 110 n < 3 so distribute numbers: A = 25, 10, 75 B = 20, 30, 15, 45 This works on each scenario I have tested. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Oct 2, 2009 at 10:46 answered Oct 2, 2009 at 10:37 Patrice BernassolaPatrice Bernassola 14.5k 6 6 gold badges 51 51 silver badges 61 61 bronze badges 2 Comments Add a comment ShreevatsaR ShreevatsaROver a year ago This is a good attempt, but it cannot possibly work for all inputs -- because if this were correct, it would be possible to extend it and get a polynomial-time algorithm for the NP-complete number partitioning problem. This is bound to fail, but I cannot think of a counterexample right now. Do you have some code for this? Try with larger numbers (e.g. with a set of 10 numbers close to 450 that has exactly one good partition). 2009-10-03T20:55:00.257Z+00:00 0 Reply Copy link ShreevatsaR ShreevatsaROver a year ago Actually, the same counterexample that I gave for another solution works: take (19, 20, 21, 29, 31). The sum is 120, so your algorithm will try to find two pairs close to 120/((5-1)/2)=60 each: (19,31) and (20,29). It will then proceed with the list 21(21), 49(20,29), 50(19,31). Next it tries to find pairs with sum 120/((3-1)/2)=120: (21,50(19,31)). So it ends with partitions 71(21,19,31) and 49(20,29), but the optimal partition is 60(29,31) and 60(19,20,21). Your algorithm actually works for this example if the "pairing" is done from the largest number first, but that cannot be generalized :) 2009-10-03T23:13:36.39Z+00:00 0 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. your requirement in #2 needs clarification, because: "The sum of all the numbers in A is as nearly equal as possible to the sum of all the numbers in B" is clear, but then your statement "the distribution should be fair" makes everything unclear. What does 'fair' exactly mean? Does the process need a random element in it? Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 5:29 9b5b9b5b 1,536 11 11 silver badges 20 20 bronze badges 1 Comment Add a comment Ralph RalphOver a year ago Thanks for your reply Jon. By fair, i mean the same thing, i.e. the sum of numbers in each set should be as nearly equal as possible, and the total numbers in each set differ by atmost 1. 2009-10-02T05:35:46.507Z+00:00 0 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. @ShreevatsaR notes that the algorithm below is known as the greedy algorithm. It does not do very well with certain inputs (I tried 10 different sets of randomly generated sets of inputs of size 100 and in all cases, the sums were very close which led me to think sorting the input was enough for the success of this algorithm). See also "The Easiest Hard Problem", American Scientist, March-April 2002, recommended by ShreevatsaR. ``` !/usr/bin/perl use strict; use warnings; use List::Util qw( sum ); my @numbers = generate_list(); print "@numbers\n\n"; my (@A, @B); my $N = @numbers; while ( @numbers ) { my $n = pop @numbers; printf "Step: %d\n", $N - @numbers; { no warnings 'uninitialized'; if ( sum(@A) < sum(@B) ) { push @A, $n; } else { push @B, $n; } printf "A: %s\n\tsum: %d\n\tnum elements: %d\n", "@A", sum(@A), scalar @A; printf "B: %s\n\tsum: %d\n\tnum elements: %d\n\n", "@B", sum(@B), scalar @B; } } sub generate_list { grep { rand > 0.8 } 1 .. 450 } ``` Note that generate\_list returns a list in ascending order. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited May 23, 2017 at 12:19 CommunityBot 1 1 1 silver badge answered Oct 2, 2009 at 5:34 Sinan ÜnürSinan Ünür 118k 15 15 gold badges 201 201 silver badges 347 347 bronze badges 12 Comments Add a comment Olexiy OlexiyOver a year ago This algorithm doesn't guarantee the second constraint as well - for example, for the list {1, 1, 3, 3, 4} it will produce smth like {1, 3, 4}-{1, 3} instead of the correct answer {1, 1, 4} - {3, 3} 2009-10-02T10:27:46.983Z+00:00 0 Reply Copy link Sinan Ünür Sinan ÜnürOver a year ago @Olexiy: You should try running the code before downvoting. You seem to have missed the fact that I am poping from the main list. 2009-10-02T10:50:15.063Z+00:00 0 Reply Copy link Olexiy OlexiyOver a year ago I missed the fact that you pop from the end of the list - try {4, 3, 3, 1, 1} now 2009-10-02T14:29:01.007Z+00:00 0 Reply Copy link Sinan Ünür Sinan ÜnürOver a year ago @Olexiy How is it not clear that I am assuming that the input is sorted? 2009-10-02T14:59:23.117Z+00:00 0 Reply Copy link ShreevatsaR ShreevatsaROver a year ago Use {2, 2, 2, 3, 3}. The optimal solution is to split into two sets {2, 2, 2} and {3, 3} both equal to 6, but your code (which I ran) produces {3, 2} and {3, 2, 2}, which are 5 and 7. Your idea is the differencing heuristic for the number-partitioning problem, but although it works for many cases, it is known that it can be very far from optimal for bad cases. 2009-10-02T15:31:15.893Z+00:00 0 Reply Copy link Add a comment\|Show 7 more comments This answer is useful 0 Save this answer. Show activity on this post. I assume the numbers are not sequential, and you can't re-balance? Because of constraint 1, you're going to need to switch buckets every other insertion, always. So every time you're not forced to pick a bucket, pick a logical bucket (where adding the number would make the sum closer to the other bucket). If this bucket isn't the same one as your previous bucket, you get another turn where you're not forced. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 5:31 psychotikpsychotik 39.2k 35 35 gold badges 104 104 silver badges 136 136 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Any dual knapsack algorithm will do (regardless of distribution of numbers). Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 5:33 Esteban ArayaEsteban Araya 29.8k 26 26 gold badges 112 112 silver badges 141 141 bronze badges 1 Comment Add a comment fortran fortranOver a year ago An example or references would be great! :-) 2009-10-02T10:23:41.44Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. Simulated Annealing can quite quickly find better and better answers. You could keep 1. true while improving the nearness of 2. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 10:58 Stephen DenneStephen Denne 37.2k 10 10 gold badges 49 49 silver badges 61 61 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. If you need the perfect answer then you have to generate and loop through all of the possible sets of answers. If a pretty good answer is all you need then a technique like simulated annealing is the way to go. Heres some C code that uses a very primitive cooling schedule to find an answer. ``` include include define MAXPAR 50 define MAXTRIES 10000000 int data1[] = {192,130,446,328,40,174,218,31,59,234,26,365,253,11,198,98, 279,6,276,72,219,15,192,289,289,191,244,62,443,431,363,10 } ; int data2[] = { 1,2,3,4,5,6,7,8,9 } ; // What does the set sum to int sumSet ( int data[], int len ) { int result = 0 ; for ( int i=0; i < len; ++i ) result += data[i] ; return result ; } // Print out a set void printSet ( int data[], int len ) { for ( int i=0; i < len; ++i ) printf ( "%d ", data[i] ) ; printf ( " Sums to %d\n", sumSet ( data,len ) ) ; } // Partition the values using simulated annealing void partition ( int data[], size_t len ) { int set1[MAXPAR] = {0} ; // Parttition 1 int set2[MAXPAR] = {0} ; // Parttition 2 int set1Pos, set2Pos, dataPos, set1Len, set2Len ; // Data about the partitions int minDiff ; // The best solution found so far int sum1, sum2, diff ; int tries = MAXTRIES ; // Don't loop for ever set1Len = set2Len = -1 ; dataPos = 0 ; // Initialize the two partitions while ( dataPos < len ) { set1[++set1Len] = data[dataPos++] ; if ( dataPos < len ) set2[++set2Len] = data[dataPos++] ; } // Very primitive simulated annealing solution sum1 = sumSet ( set1, set1Len ) ; sum2 = sumSet ( set2, set2Len ) ; diff = sum1 - sum2 ; // The initial difference - we want to minimize this minDiff = sum1 + sum2 ; printf ( "Initial diff is %d\n", diff ) ; // Loop until a solution is found or all are tries are exhausted while ( diff != 0 && tries > 0 ) { // Look for swaps that improves the difference int newDiff, newSum1, newSum2 ; set1Pos = rand() % set1Len ; set2Pos = rand() % set2Len ; newSum1 = sum1 - set1[set1Pos] + set2[set2Pos] ; newSum2 = sum2 + set1[set1Pos] - set2[set2Pos] ; newDiff = newSum1 - newSum2 ; if ( abs ( newDiff ) < abs ( diff ) \|\| // Is this a better solution? tries/100 > rand() % MAXTRIES ) // Or shall we just swap anyway - chance of swap decreases as tries reduces { int tmp = set1[set1Pos] ; set1[set1Pos] = set2[set2Pos] ; set2[set2Pos] = tmp ; diff = newDiff ; sum1 = newSum1 ; sum2 = newSum2 ; // Print it out if its the best we have seen so far if ( abs ( diff ) < abs ( minDiff ) ) { minDiff = diff ; printSet ( set1, set1Len ) ; printSet ( set2, set2Len ) ; printf ( "diff of %d\n\n", abs ( diff ) ) ; } } --tries ; } printf ( "done\n" ) ; } int main ( int argc, char argv ) { // Change this to init rand from the clock say if you don't want the same // results repoduced evert time! srand ( 12345 ) ; partition ( data1, 31 ) ; partition ( data2, 9 ) ; return 0; } ``` Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Oct 2, 2009 at 11:43 JacksonJackson 5,657 2 2 gold badges 31 31 silver badges 51 51 bronze badges 1 Comment Add a comment user140327 user140327Over a year ago Hahaha, solve NP problem with PERL! I've seen everything now! ;-) 2009-10-02T15:14:13.507Z+00:00 0 Reply Copy link This answer is useful -1 Save this answer. Show activity on this post. I would give genetic algorithms a try, as this seems a very nice problem to apply them. The codification is just a binary string of length N, meaning 0 being in the first group and 1 in the second group. Give a negative fitness when the number of elements in each group differs, and a positive fitness when the sums are similar... Something like: fitness(gen) = (sum(gen)-n/2))^2 + (sum(values[i](-1)gen[i] for i in 0..n))^2 (And minimize the fitness) Of course this can give you a sub-optimal answer, but for large real world problems it's usually enough. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Oct 2, 2009 at 10:37 answered Oct 2, 2009 at 10:32 fortranfortran 76.5k 27 27 gold badges 142 142 silver badges 179 179 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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14533	https://ohe.mn.gov/sites/default/files/2025-05/2022%20Cumulative%20Student%20Debt%20Annual%20Report.pdf	Cumulative Median Student Loan Debt in Minnesota, 2021-2022 Academic Year August 2023 Author Nicole Whelan Financial Aid Research Analyst Tel: 651-259-3967 Nicole.Whelan@state.mn.us Minnesota Office of Higher Education 1450 Energy Park Drive, Suite 350 Saint Paul, MN 55108-5227 Tel: 651.642.0567 or 800.657.3866 TTY Relay: 800.627.3529 Fax: 651.642.0675 Email: info.ohe@state.mn.us About the Minnesota Office of Higher Education The Minnesota Office of Higher Education is a cabinet-level state agency providing students with financial aid programs and information to help them gain access to postsecondary education. The agency also serves as the state’s clearinghouse for data, research and analysis on postsecondary enrollment, financial aid, finance and trends. The Minnesota State Grant Program is the largest financial aid program administered by the Office of Higher Education, awarding more than $235 million in need-based grants to Minnesota residents attending accredited institutions in Minnesota. The agency oversees tuition reciprocity programs, a student loan program, Minnesota’s 529 College Savings Plan, licensing and early college awareness programs for youth. Maxwell Herrera Research Intern Maxwell.Herrera@state.mn.us Contents Executive Summary 1 Introduction 3 Borrowing by Award Type, All Institutions 4 Table 1: Statewide Cumulative Median Debt, 2018-2022 5 Figure 1: Average Cumulative Median Debt by Award Type, 2018-2022 7 Figure 2: Percent of Graduates with Loans by Award Type, 2018-2022 8 Figure 3: Average Cumulative Median Debt by Award Type and Sector, 2022 9 Figure 4: Percent of Graduates with Loans by Award Type and Sector, 2022 10 Table 2: Cumulative Median Debt by Sector, 2022 11 Figure 5: Percent Change in Average Cumulative Median Debt, 2012-2022 12 Sub-Baccalaureate Certificates 13 Figure 6: Average Cumulative Median Debt by Year and Sector, Sub-Baccalaureate Certificate Recipients 14 Figure 7: Percent of Graduates with Loans by Year and Sector, Sub-Baccalaureate Certificate Recipients 14 Table 3: Average Cumulative Median Debt for Minnesota Sub-Baccalaureate Certificate Recipients, 2018-2022 15 Associate Degrees 16 Figure 8: Average Cumulative Median Debt by Year and Sector, Associate Degree Recipients 16 Figure 9: Percent of Graduates with Loans by Year and Sector, Associate Degree Recipients 17 Table 4: Average Cumulative Median Debt for Minnesota Associate Degree Recipients, 2018-2022 17 Bachelor’s Degrees 18 Figure 10: Average Cumulative Median Debt of by Year and Sector, Bachelor’s Degree Recipients 18 Figure 11: Percent of Graduates with Loans by Year and Sector, Bachelor’s Degree Recipients 19 Table 5: Average Cumulative Median Debt for Minnesota Bachelor’s Degree Recipients, 2018-2022 19 Master’s Degrees 20 Figure 12: Average Cumulative Median Debt by Year and Sector, Master’s Degree Recipients 20 Figure 13: Percent of Graduates with Loans by Year and Sector, Master’s Degree Recipients 21 Table 6: Average Cumulative Median Debt for Minnesota Master’s Degree Recipients, 2018-2022 21 Graduate Certificates 22 Figure 14: Average Cumulative Median Debt by Year and Sector, Master’s Degree Recipients 23 Figure 15: Percent of Graduates with Loans by Year and Sector, Master’s Degree Recipients 23 Table 7: Average Cumulative Median Debt for Minnesota Graduate Certificate Recipients, 2019-2022 24 Doctoral Degrees 25 Figure 14: Average Cumulative Median Debt of by Year and Sector, Doctoral Degree Recipients 25 Figure 15: Percent of Graduates with Loans by Year and Sector, Doctoral Degree Recipients 26 Table 8: Average Cumulative Median Debt for Minnesota Doctoral Degree Recipients, 2018-2022 26 First Professional Degrees 27 Figure 16: Average Cumulative Median Debt of by Year and Sector, First Professional Degree Recipients 27 Figure 17: Percent of Graduates with Loans by Year and Sector, First Professional Degree Recipients 28 Table 9: Average Cumulative Median Debt for Minnesota First Professional Degree Recipients, 2018-2022 28 Conclusion 29 Appendix A – Cumulative Median Debt by Institution 30 Average Cumulative Median Debt of Graduates, All Sectors, 2022 30 Graduates with Sub-Baccalaureate Certificates, Minnesota State Colleges, 2022 30 Graduates with Sub-Baccalaureate Certificates, Minnesota State Universities, 2022 31 Graduates with Sub-Baccalaureate Certificates, University of Minnesota System, 2022 31 Graduates with Sub-Baccalaureate Certificates, Private Non-Profit Institutions, 2022 32 Graduates with Sub-Baccalaureate Certificates, Private For-Profit Institutions, 2022 32 Graduates with Associate Degrees, Minnesota State Colleges, 2022 33 Graduates with Associate Degrees, Minnesota State Universities, 2022 34 Graduates with Associate Degrees, Private Non-Profit Institutions, 2022 34 Graduates with Associate Degrees, Private For-Profit Institutions, 2022 34 Graduates with Bachelor’s Degrees, Minnesota State Universities, 2022 35 Graduates with Bachelor’s Degrees, University of Minnesota System, 2022 35 Graduates with Bachelor’s Degrees, Private Non-Profit Institutions, 2022 35 Graduates with Bachelor’s Degrees, Private For-Profit Institutions, 2022 36 Graduates with Master’s Degrees, Minnesota State Universities, 2022 36 Graduates with Master’s Degrees, University of Minnesota System, 2022 37 Graduates with Master’s Degrees, Private Non-Profit Institutions, 2022 37 Graduates with Graduate Certificates, Minnesota State Universities, 2022 37 Graduates with Graduate Certificates, Private Non-Profit, 2022 38 Graduates with Doctoral Degrees, Minnesota State Universities, 2022 38 Graduates with Doctoral Degrees, University of Minnesota System, 2022 39 Graduates with Doctoral Degrees, Private Non-Profit Institutions, 2022 39 Graduates with First Professional Degrees, University of Minnesota System, 2022 39 Graduates with First Professional Degrees, Private Non-Profit Institutions, 2022 39 Appendix B – Information on Loan Repayment 40 Cumulative Median Student Loan Debt, 2022 1 Executive Summary Student loan debt continues to play a prominent role in the national conversation around college affordability and cost. Students who graduate with higher student loan debt does lead to higher payments, and in turn, that does make financial freedom difficult for students. Because the impact of student loan debt on graduates is a critical area of concern for policymakers, the Office of Higher Education has reported on Cumulative Median Student Loan debt for graduates since 2012. The data presented in this report represent the results of the 2021-2022 Cumulative Median Debt survey, which is sent annually to postsecondary institutions eligible to participate in the Minnesota State Grant program for that academic year. The data collected for this report represents the median cumulative student loan debt by type of certificate and degree type for students who graduated with cumulative student loan debt from these institutions between July 1, 2021 and June 30, 2022. In this report, cumulative median student loan debt (“cumulative median debt”) refers to the median amount of debt incurred by a student while attending a specific institution and pursuing a specific postsecondary credential. The debt figures include debt from federal, state, and private sources that is known to the institutions. Debt figures do not include any debt incurred from previous institutions, and students who graduate with no debt are not included in the median. In addition, debt from students who do not complete a degree is not included, which is critical because even while federal student loans have been placed in pandemic-related forbearance, default rates are considerably higher for those who did not complete a degree1. When aggregated across multiple institutions, the measure reported is the average cumulative median debt. The Office thanks institutions that took the time to compile data for this report. Institution-level data can be found in the appendix. In 2022, the borrowing patterns for students graduating with debt varied by program: • Sub-baccalaureate certificates: The percentage of graduates from these programs taking out student loans in 2022 (40%) decreased by 3 percentage points from 2021 (43%). Average cumulative median debt increased by 1% from $9, 912 in 2021 to $10,022 in 2022 across all sectors. • Associate degrees: Average cumulative median debt increased over the year, from $15,215 in 2021 to $16,107 in 2022 across all sectors, an increase of 6% from the previous year. The percentage of graduates from these programs taking out student loans in 2022 (49%) declined by 3 percentage points from 2021 (52%). 1 Wong, Nancy. “College Scorecard Data Show High Rates of Non-Repayment on Federal Student Loans and Signs of Forbearance Abuse.” Retrieved September 1,2021, from Cumulative Median Student Loan Debt, 2022 2 • Bachelor’s degrees: Average cumulative median debt slightly increased by 1% from $23,858 in 2021 to $24,062 in 2022 across all sectors. The percentage of graduates from these programs taking out student loans remained steady in 2022 (63%) compared to 2021 (63%). • Master’s degrees: Average cumulative median debt increased by 5% from $35,110 in 2021 to $36,833 in 2022 across all sectors. The percentage of graduates from these programs who took out student loans in 2022 (52%) decreased by 2 percentage points from 2021 (54%). • Graduate certificates: Average cumulative median debt decreased by 9% from $25,844 in 2021 to $23,622 in 2022 across all sectors. The percentage of graduates from these programs who took out student loans in 2022 (30%) decreased by 9 percentage points from 2021 (39%). • Doctoral degrees: Average cumulative median debt increased by 12% from $63,879 in 2021 to $71,497 in 2022 across all sectors. The percentage of graduates from these programs who took out student loans in 2022 (34%) increased by 2 percentage points from 2021 (34%). • First professional degrees: Average cumulative median debt decreased by 9% from $155,074 in 2021 to $140,835 in 2022 across all sectors. The percentage of graduates from these programs who took out student loans in 2022 (75%) decreased by 3 percentage points from 2021 (78%). Since the Office began collecting data in 2012, the average cumulative median debt of graduates has decreased by 11% or more for bachelor’s degrees, associate degrees, and sub-baccalaureate certificates, while the average cumulative debt of graduate degree programs has increased between 9% to 30%. Across most award types, the percent of graduates who borrow has also decreased across this time span. Cumulative Median Student Loan Debt, 2022 3 Introduction On average, loans made up just under one-third of Minnesota undergraduate students’ financial aid in 2021, for a total of roughly $890 million2. In addition to grants, savings, and income earned from working while in school, loans are a critical means of financing a postsecondary education for most students. This report provides data on loans incurred by students grouped by award type, and includes information on debt accumulation for both undergraduate and graduate students. The data presented in this report represent the results of the 2021-2022 Cumulative Median Debt survey. The Office sends the survey annually to postsecondary institutions eligible to participate in the Minnesota State Grant and SELF Loan programs for that Fiscal Year. Data collected detail the median cumulative student loan debt by type of award for students who graduated with student loan debt from these institutions between July 1, 2021 and June 30, 2022.3 In this report, cumulative median student loan debt (“cumulative median debt”) refers to the median amount of debt incurred by a student while attending a specific institution and pursuing a specific postsecondary credential. When aggregated across multiple institutions, the measure reported is the average cumulative median debt.4 Cumulative median debt is only reported for students that completed a postsecondary credential at the given institution, and for students completing two or more awards, the median debt for the highest award conferred is used. The debt figures include debt from all federal, state, institutional, and private sources that is known to the institutions. Debt incurred from previous institutions is not included. In addition, students with no debt are not included in the median, and where fewer than 10 students graduated with debt, median cumulative debt is suppressed. Data is not inflation-adjusted. The Office thanks institutions that took the time to compile data for this report. Institution-level data can be found in the appendix. 2 Financial Aid Awarded to Undergraduates Attending Minnesota Institutions. Retrieved August 31, 2022, from 3 This report provides data on the weighted average of median debt amounts across institutions. The median is the value for which half of borrowers borrowed more and half borrowed less. Many reports on student debt report average debt at the campus level, rather than median debt. However, a small number of borrowers who borrow large amounts can result in higher average debt, even though the larger amounts are not typical for most borrowers. When aggregating across multiple institutions, the average cumulative median debt is used. This is the weighted (by number of borrowing graduates) average of the cumulative median debt across each institution. 4 Each year institutions report median cumulative debt amount by award type to OHE. For each award type, OHE calculates an average median cumulative debt. To generate this measure for each award type, OHE weights each institution’s median cumulative debt amount by the number of graduates that took out student loans for that award at the institution. These weighted median cumulative debt amounts by award type are then averaged across all institutions. Data was not collected for Walden University and Capella University. Cumulative Median Student Loan Debt, 2022 4 Borrowing by Award Type, All Institutions Between 2018 and 2022, the cumulative median debt of students for those completing all award types slightly increased or decreased from 2019 to 2022. In comparison, as shown in Figure 1, the largest amount of average total difference between these years was amongst students who received a First Professional degree. In 2022, the borrowing amounts for students graduating with debt varied by program: • Sub-baccalaureate certificates: In 2022, the average cumulative median debt for students graduating with debt from these programs was $10,022. • Associate degrees: In 2022, the average cumulative median debt for students graduating with debt from these programs was $16,107. • Bachelor’s degrees: In 2022, the average cumulative median debt for students graduating with debt from these programs was $24,062. • Master’s degrees: In 2022, the average cumulative median debt for students graduating with debt from these programs was $36,833. • Graduate certificates: In 2022, the average cumulative median debt for students graduating with debt from these programs was $23,622. • Doctoral degrees: In 2022, the average cumulative median debt for students graduating with debt from these programs was $71,497. • First professional degrees5: In 2022, the average cumulative median debt for students graduating with debt from these programs was $140,835. Results showed a lower percentage of students in sub-baccalaureate (40%), associate (49%), master’s (52%), graduate certificate (30%), and doctoral programs (36%) take out loans, compared to students in bachelor’s (63%) and first professional degree programs (75%) (Figure 2). Overall, the average cumulative median debt for students in doctoral and graduated certificate programs saw greater year-to-year fluctuation between 2018 and 2022 than other award types. This variation may be due to a smaller sample size. There are fewer students enrolled in these programs and fewer programs overall compared to the availability of master’s, bachelor’s, associate, and sub-baccalaureate programs (see Table 1 for student counts by year). This pattern of increased borrowing along with a decrease in students may be reflective of the limited non-loan options for financial aid for 5 Awards in the First Professional Degree award category include law, medicine, and other specialized professional degrees. Cumulative Median Student Loan Debt, 2022 5 Masters and Doctoral programs. These additional costs may prevent pursuant of other degrees as student preferences change under these circumstances. Table 1: Statewide Cumulative Median Debt, 2018-2022 Award Type Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt Sub-Baccalaureate Certificate 2018 10,336 4,915 48% $10,009 2019 10,511 4,817 46% $10,286 2020 9,587 4,326 45% $10,112 2021 9,459 4,105 43% $9,912 2022 9,242 3,687 40% $10,022 Associate Degree 2018 16,901 10,092 60% $16,297 2019 16,702 9,401 56% $15,562 2020 15,583 8,451 54% $15,175 2021 16,894 8,858 52% $15,215 2022 16,336 8,078 49% $16,107 Bachelor's Degree 2018 32,027 21,664 68% $25,453 2019 30,696 20,050 65% $24,641 2020 30,273 19,684 65% $24,793 2021 30,703 19,373 63% $23,858 2022 30,426 19,072 63% $24,062 Master's Degree 2018 11,371 5,768 51% $36,074 2019 10,548 5,730 54% $36,996 2020 10,219 5,583 55% $37,336 2021 10,153 5,517 54% $35,110 2022 10,062 5,246 52% $36,833 Graduate Certificate 2019 199 107 54% $28,416 2020 210 120 57% $31,922 2021 904 343 38% $25,844 2022 1,209 357 30% $23,622 Doctoral Degree 2018 1,110 405 36% $59,390 2019 1,219 408 33% $63,977 2020 1,162 428 37% $56,622 2021 1,250 422 34% $63,879 2022 1,435 512 36% $71,497 First Professional Degree 2018 1,355 1,109 82% $147,367 2019 1,357 1,087 80% $143,912 2020 1,419 1,039 73% $145,674 2021 1,542 1,197 78% $155,074 2022 1,646 1,241 75% $140,835 Cumulative Median Student Loan Debt, 2022 6 In addition to variances in borrowing by degree type, student loan borrowing patterns varied by sector in 2022 (Figure 3). • For sub-baccalaureate degrees, students at University of Minnesota system had the highest average cumulative median debt ($21,113). • For associate degrees, students at private for-profit institutions had the highest average cumulative median debt ($21,629). • For bachelor’s degrees, students at private for-profit institutions had the highest average cumulative median debt ($39,324). • For master’s degrees, students at the University of Minnesota system had the highest average cumulative median debt available ($40,232). • For graduate certificates, students at private-non-profit had the highest average cumulative median debt available ($24,048). • For doctoral degrees, students at private-non-profit had the highest average cumulative median debt ($73,906). • For first professional degrees, students at the University of Minnesota system had the highest average cumulative median debt ($166,935). Cumulative Median Student Loan Debt, 2022 7 Figure 1: Average Cumulative Median Debt by Award Type, 2018-2022 $10,009 $10,286 $10,112 $9,912 $10,022 $16,297 $15,562 $15,175 $15,215 $16,107 $25,453 $24,641 $24,793 $23,858 $24,062 $36,074 $36,996 $37,336 $35,110 $36,833 $28,416 $31,922 $25,844 $23,622 $59,390 $63,977 $56,622 $63,879 $71,497 $147,367 $143,912 $145,674 $155,074 $140,835 2018 2019 2020 2021 2022 2018 2019 2020 2021 2022 2018 2019 2020 2021 2022 2018 2019 2020 2021 2022 2019 2020 2021 2022 2018 2019 2020 2021 2022 2018 2019 2020 2021 2022 Sub-Baccalaureate Certificate Associate Degree Bachelor's Degree Master's Degree Graduate Certificate Doctoral Degree First Professional Degree Cumulative Median Student Loan Debt, 2022 8 Figure 2: Percent of Graduates with Loans by Award Type, 2018-2022 Cumulative Median Student Loan Debt, 2022 9 Figure 3: Average Cumulative Median Debt by Award Type and Sector, 2022 As shown in Figure 4, for sub-baccalaureate certificate (80%), associate degree (88%), and bachelor’s degree (85%) students from private for-profit institutions had the highest rates of borrowing across all sectors. For master’s degree programs, the highest rate of borrowing among graduates was at private non-profit institutions (59%). For doctoral (56%) programs, the highest rate of borrowing among Cumulative Median Student Loan Debt, 2022 10 graduates was at Minnesota State Universities. For first professional degree programs (76%), the highest rate of borrowing among graduates were private non-profit institutions. Figure 4: Percent of Graduates with Loans by Award Type and Sector, 2022 Cumulative Median Student Loan Debt, 2022 11 Table 2: Cumulative Median Debt by Sector, 2022 Award Type Sector Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt Sub-Baccalaureate Certificate Minnesota State Colleges 7,512 2,581 34% $10,159 Minnesota State Universities 37 20 54% $15,023 Private For-Profit 1225 974 80% $9,158 Private Non-Profit 235 99 42% $12,494 University of Minnesota 233 13 6% $21,113 Associate Degree Minnesota State Colleges 13,619 5,939 44% $14,510 Minnesota State Universities 284 200 70% $18,709 Private For-Profit 1,546 1,361 88% $21,629 Private Non-Profit 887 578 65% $18,616 Bachelor's Degree Minnesota State Universities 9,494 6,098 64% $22,544 Private For-Profit 156 133 85% $39,234 Private Non-Profit 10,606 7,187 68% $26,281 University of Minnesota 10,170 5,634 56% $22,519 Graduate Certificate Minnesota State Universities 256 113 44% $23,593 Private Non-Profit 749 216 29% $24,048 University of Minnesota 204 28 14% 20,454 Master's Degree Minnesota State Universities 2,038 1,095 54% $28,747 Private Non-Profit 4,910 2,899 59% $38,420 University of Minnesota 3,114 1,252 40% $40,232 Doctoral Degree Minnesota State Universities 136 76 56% $63,048 Private Non-Profit 375 197 53% $73,906 University of Minnesota 924 239 26% $72,199 First Professional Degree Private Non-Profit 803 613 76% $114,097 University of Minnesota 843 628 74% $166,935 Data collection on cumulative median debt began in 2012. Figure 5 reflects the overall percent change in cumulative median debt from 2012 to 2022. For students in sub-baccalaureate certificate (-11%), associate degree (-13%), and bachelor’s degree (-12%) programs, cumulative median debt declined Cumulative Median Student Loan Debt, 2022 12 during this period. For students in master’s degree (+15%), doctoral degree (+27%), and first professional degree programs (+9%), cumulative median debt increased during this period. Associate degree recipients saw the largest decline in median debt, while doctoral degree recipients saw the largest increase in median debt (Figure 5). Figure 5: Percent Change in Average Cumulative Median Debt, 2012-2022 Cumulative Median Student Loan Debt, 2022 13 Sub-Baccalaureate Certificates The U.S. Department of Education sets both annual and lifetime maximum borrowing limits of federal loans for undergraduate dependent students that are lower than for independent students. Dependent students are allowed to borrow $5,500 in their first year, followed by $6,500 in their second year, and $7,500 in their third and fourth years for a maximum of $31,000 over their undergraduate career. Limits for independent students are $9,500 in their first year, $10,500 in their second year, and $12,500 in their third and fourth years for a maximum of $57,500 over their undergraduate career. The limits are the same for students seeking a sub-baccalaureate certificate, associate degree, or a bachelor's degree. Students may choose to take out additional private loans on top of these limits. Students receiving sub-baccalaureate certificates in Minnesota primarily attended public two-year institutions. Certificates at the sub-baccalaureate level encompass programs in fields such as construction, health care, cosmetology, and many other fields. These programs generally take between nine and fifteen months to complete. Among sub-baccalaureate certificate recipients, average cumulative median debt increased by 1% from $9,912 in 2021 to $10,022 in 2022 across all sectors. The percentage of graduates from these programs taking out student loans decreased from 2021 (43%) to 2022 (40%). This trend is not consistent across all sectors. In 2022, certificate recipients at: • Minnesota State Colleges borrowed $10,159 (approximately 0% change from $10,111 in 2021), with 34% of graduates borrowing (down from 39% in 2021). • Minnesota State Universities borrowed $15,023 (down 39% from $24,477 in 2021), with 54% of graduates borrowing (up from 53% in 2021). • The University of Minnesota system borrowed $21,113 (up 30% from $16,257 in 2021), with 6% of graduates borrowing (approximately the same 6% in 2021). • Private non-profit institutions borrowed $12,494 (down 1% from $12,583 in 2021), with 42% of graduates borrowing (down from 45% in 2021). • Private for-profit institutions borrowed $9,158 (up 7% from $8,535 in 2021), with 80% of graduates borrowing (up from 77% in 2021). Cumulative Median Student Loan Debt, 2022 14 Figure 6: Average Cumulative Median Debt by Year and Sector, Sub-Baccalaureate Certificate Recipients Figure 7: Percent of Graduates with Loans by Year and Sector, Sub-Baccalaureate Certificate Recipients 81% 80% 80% Cumulative Median Student Loan Debt, 2022 15 Table 3: Average Cumulative Median Debt for Minnesota Sub-Baccalaureate Certificate Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 10,336 4,915 48% $10,009 2019 10,484 4,813 46% $10,286 2020 9,587 4,326 45% $10,112 2021 9,459 4,105 43% $9,912 2022 9,242 3,687 40% $10,022 Minnesota State Colleges 2018 8,099 3,492 43% $10,140 2019 8,512 3,468 41% $10,507 2020 7,667 3,152 41% $10,391 2021 7,590 2,994 39% $10,111 2022 7,512 2,581 34% $10,159 Minnesota State Universities 2020 31 17 55% $24,681 2021 47 25 53% $24,477 2022 37 20 54% $15,023 University of Minnesota 2018 296 16 5% $21,523 2019 259 18 7% $16,162 2020 328 26 8% $23,752 2021 340 19 6% $16,257 2022 233 13 6% $21,113 Private For-Profit 2018 1,742 1,316 76% $9,293 2019 1,511 1,218 81% $9,382 2020 1,277 1,023 80% $8,765 2021 1,268 971 77% $8,535 2022 1,225 974 80% $9,158 Private Non-Profit 2017 627 74 12% $12,633 2018 199 91 46% $13,310 2019 229 113 49% $12,308 2020 284 108 38% $9,161 2021 214 96 45% $12,583 2022 235 99 42% $12,494 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 16 Associate Degrees Students receiving associate degrees in Minnesota primarily attended public two-year institutions. Among associate degree recipients, average cumulative median debt increased by 6% from $15,215 in 2021 to $16,107 in 2022 across all sectors. Additionally, the percentage of graduates from these programs taking out student loans in 2022 (49%) declined by 3 percentage points from 2021 (52%). This trend is not consistent across all sectors. In 2022, associate degree recipients at: • Minnesota State Colleges borrowed $14,510 (approximately no change from $14,560 in 2021), with 44% of graduates borrowing (down from 47% in 2021). • Minnesota State Universities borrowed $18,709 (down nine percentage points from $18,750 in 2021), with 70% of graduates borrowing (down from 86% in 2021). • Private non-profit institutions borrowed $18,616 (up 31% from $14,264 in 2021), with 65% of graduates borrowing (up from 61% in 2021). • Private for-profit institutions borrowed $21,629 (up 15% from $18,750 in 2021), with 88% of graduates borrowing (up from 86% in 2021). Figure 8: Average Cumulative Median Debt by Year and Sector, Associate Degree Recipients Cumulative Median Student Loan Debt, 2022 17 Figure 9: Percent of Graduates with Loans by Year and Sector, Associate Degree Recipients Table 4: Average Cumulative Median Debt for Minnesota Associate Degree Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 16,901 10,092 60% $16,297 2019 16,702 9,401 56% $15,562 2020 15,583 8,451 54% $15,175 2021 16,894 8,858 52% $15,215 2022 16,336 8,078 49% $16,107 Minnesota State Colleges 2018 14,436 8,026 56% $15,647 2019 14,232 7,353 52% $15,040 2020 13,303 6,602 50% $14,325 2021 13,386 6,313 47% $14,560 2022 13,619 5,939 44% $14,510 Minnesota State Universities 2018 365 253 69% $15,406 2019 304 192 63% $17,537 2020 271 185 68% $17,262 2021 1,589 1,373 86% $18,750 2022 284 200 70% $18,709 Private For-Profit 2018 1,464 1,329 91% $19,815 2019 1,429 1,304 91% $19,381 2020 1,555 1,350 87% $18,410 2021 1,589 1,373 86% $18,750 2022 1,546 1,361 88% $21,629 Private Non-Profit 2018 636 484 76% $17,875 2019 737 552 75% $12,816 2020 454 314 69% $17,896 2021 1,594 966 61% $14,264 2022 887 578 65% $18,616 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 18 Bachelor’s Degrees Students receiving bachelor’s degrees in Minnesota primarily attended public four-year institutions or private non-profit four-year institutions. The majority of students in these programs were enrolled full time for both fall and spring terms. Among bachelor’s degree recipients, average cumulative median debt increased by 1%, from $23,858 in 2021 to $24,062 in 2022, across all sectors. Additionally, the percentage of graduates from these programs taking out student loans in 2022 (63%) remained statistically the same from 2021 (63%). This trend is not consistent across all sectors. In 2022, bachelor’s degree recipients at: • Minnesota State Universities borrowed $22,544 (down 2% from $22,946 in 2021), with 64% of graduates borrowing (down from 65% in 2021). • The University of Minnesota system borrowed $22,519 (no statistical change from $22,593 in 2021), with 56% of graduates borrowing (slightly down from 57% in 2021). • Private non-profit institutions borrowed $26,281 (up 2% from $25,754 in 2021), with 68% of graduates borrowing (up from 67% in 2021). • Private for-profit institutions borrowed $39,324 (up 5% from $37,517 in 2021), with 85% of graduates borrowing (down from 91% in 2021). Figure 10: Average Cumulative Median Debt of by Year and Sector, Bachelor’s Degree Recipients Cumulative Median Student Loan Debt, 2022 19 Figure 11: Percent of Graduates with Loans by Year and Sector, Bachelor’s Degree Recipients Table 5: Average Cumulative Median Debt for Minnesota Bachelor’s Degree Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 32,027 21,664 68% $25,453 2019 30,696 20,050 65% $24,641 2020 30,273 19,684 65% $24,793 2021 30,703 19,373 63% $23,858 2022 30,426 19,072 63% $24,062 Minnesota State Universities 2018 10,736 7,658 71% $24,784 2019 10,500 7,166 68% $24,117 2020 10,152 6,883 68% $24,021 2021 9,979 6,514 65% $22,946 2022 9,494 6,098 64% $22,544 University of Minnesota 2018 10,377 6,239 60% $23,545 2019 10,554 6,223 59% $23,227 2020 10,644 6,251 59% $23,634 2021 11,023 6,312 57% $22,593 2022 10,170 5,654 56% $22,519 Private For-Profit 2018 182 162 89% $33,037 2019 181 156 86% $26,937 2020 191 166 87% $29,164 2021 141 129 91% $37,517 2022 156 133 85% $39,324 Private Non-Profit 2018 10,732 7,605 71% $27,530 2019 9,461 6,505 69% $26,516 2020 9,286 6,384 69% $26,647 2021 9,560 6,418 67% $25,754 2022 10,606 7,187 68% $26,281 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 20 Master’s Degrees Students receiving master’s degrees in Minnesota primarily attended private non-profit four-year institutions or the University of Minnesota system. Most students in these programs were enrolled full time. Among master’s degree recipients, average cumulative median debt increased by 4.9% from $35,110 in 2021 to $36,833 in 2022 across all sectors. Additionally, the percentage of graduates from these programs taking out student loans in 2022 (52%) decreased by 2 percentage points from 2021 (54%). This trend is not consistent across all sectors. In 2022, master’s degree recipients at: • Minnesota State Universities borrowed $28,747 (decrease of approximately 12.2% from $32,742 in 2021), with 54% of graduates borrowing (down from 59% in 2021). • The University of Minnesota system borrowed $40,232 (an increase of about 1.7% from $39,552 in 2021), with 40% of graduates borrowing (down from 44% in 2021). • Private non-profit institutions borrowed $38,420 (up about 13.1% from $33,980 in 2021), with 59% of graduates borrowing (similarly 59% in 2021). • No private for-profit institutions reported graduates with master’s degrees in 2022. Because the number of students graduating with loans from these programs at private for-profit institutions was limited in 2020, cumulative median debt information is suppressed for that year. In 2020, 67% of total graduates borrowed loans. Figure 12: Average Cumulative Median Debt by Year and Sector, Master’s Degree Recipients Cumulative Median Student Loan Debt, 2022 21 Figure 13: Percent of Graduates with Loans by Year and Sector, Master’s Degree Recipients Table 6: Average Cumulative Median Debt for Minnesota Master’s Degree Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 11,371 5,768 51% $36,074 2019 10,548 5,730 54% $36,996 2020 10,219 5,583 55% $37,336 2021 10,153 5,517 54% $35,110 2022 10,062 5,246 52% $36,833 Minnesota State Universities 2018 1,902 1,093 57% $31,033 2019 1,892 1,102 58% $32,914 2020 2,046 1,175 57% $34,985 2021 1,984 1,175 59% $32,742 2022 2,038 1,095 54% $28,747 University of Minnesota 2018 3,536 1,573 44% $39,164 2019 3,600 1,602 45% $39,822 2020 3,304 1,436 43% $39,922 2021 3,156 1,380 44% $39,552 2022 3,114 1,252 40% $40,232 Private For-Profit 2017 66 58 88% $52,204 2018 5 4 80% Suppressed 2019 7 4 57% Suppressed 2020 12 8 67% Suppressed Private Non-Profit 2018 5,928 3,098 52% $36,220 2019 5,049 3,022 60% $36,942 2020 4,857 2,964 61% $36,972 2021 5,013 2,962 59% $33,980 2022 4,910 2,899 59% $38,420 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 22 Graduate Certificates Beginning in 2019, the Office of Higher Education includes information on cumulative debt of graduates from graduate certificate programs. Graduate certificate programs are short-term programs conferring a post-baccalaureate certificate. Examples such programs as include Elections Administration or Business Administration certification. In 2019 and 2020, information on graduates for these programs was reported only for Minnesota State Universities. In 2021, data is reported for Minnesota State Universities, as well as several private non-profit institutions. For 2022, the University of Minnesota system has also reported data for the benefit of this report. Among graduate certificate recipients, average cumulative median debt decreased by 9% from $25,844 in 2021 to $23,622 in 2022 across all sectors. Additionally, the percentage of graduates from these programs who took out student loans in 2022 (30%) decreased by 8 percentage points from 2021 (38%). These numbers are greatly impacted by the increase in institutions reporting graduates with this type of award in 2022. In 2022, graduate certificate recipients at: • Minnesota State Universities borrowed $32,538 (up 2% from $31,922 in 2020), with 60% of graduates borrowing (up from 57% in 2020). • Private non-profit institutions borrowed $20,579, with 29% of graduates borrowing (not reported in prior years). • The University of Minnesota system borrowed $20,454, with 14% of graduate borrowing loans, with no prior years to report or compare too. Cumulative Median Student Loan Debt, 2022 23 Figure 14: Average Cumulative Median Debt by Year and Sector, Master’s Degree Recipients Figure 15: Percent of Graduates with Loans by Year and Sector, Master’s Degree Recipients Cumulative Median Student Loan Debt, 2022 24 Table 7: Average Cumulative Median Debt for Minnesota Graduate Certificate Recipients, 2019-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2019 199 107 54% $28,416 2020 210 120 57% $31,922 2021 904 343 38% $25,844 2022 1,209 357 30% $23,622 Minnesota State Universities 2019 199 107 54% $28,416 2020 210 120 57% $31,922 2021 250 151 60% $32,538 2022 256 113 44% $23,593 Private Non-Profit 2021 654 192 29% $20,579 2022 749 216 29% $24,048 University of Minnesota 2022 204 28 14% $20,454 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 25 Doctoral Degrees Among doctoral degree recipients, average cumulative median debt increased by 12% from $63,879 in 2021 to $71,497 in 2022 across all sectors. Additionally, the percentage of graduates from these programs who took out student loans in 2022 (36%) increased by 2 percentage points from 2021 (34%). This trend is not consistent across all sectors. In 2022, doctoral degree recipients at: • Minnesota State Universities borrowed $63,048 (up 23% from $51,223 in 2021), with 56% of graduates borrowing (up from 55% in 2021). • The University of Minnesota system borrowed $72,199 (up 2% from $71,061 in 2021), with 26% of graduates borrowing (up from 25% in 2021). • Private non-profit institutions borrowed $73,906 (up 29% from $57,123 in 2021), with 53% of graduates borrowing (down from 60% in 2021). • No private for-profit institutions reported graduates with doctoral degrees in 2022. Because the number of students graduating with loans from these programs at private for-profit institutions was less than 10 in 2022, cumulative median debt information is suppressed for that year. Figure 14: Average Cumulative Median Debt of by Year and Sector, Doctoral Degree Recipients Cumulative Median Student Loan Debt, 2022 26 Figure 15: Percent of Graduates with Loans by Year and Sector, Doctoral Degree Recipients Table 8: Average Cumulative Median Debt for Minnesota Doctoral Degree Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 1,110 405 36% $59,390 2019 1,219 408 33% $63,977 2020 1,162 428 37% $56,622 2021 1,250 422 34% $63,879 2022 1,435 512 36% $71,497 Minnesota State Universities 2018 28 21 75% $50,376 2019 44 22 50% $67,848 2020 91 54 59% $54,605 2021 97 53 55% $51,223 2022 136 76 56% $63,048 University of Minnesota 2018 876 264 30% $58,565 2019 979 258 26% $58,507 2020 839 230 27% $56,101 2021 917 227 25% $71,061 2022 924 239 26% $72,199 Private For-Profit 2017 33 30 91% $198,316 2018 5 0 0% Suppressed 2020 12 0 0% Suppressed Private Non-Profit 2018 201 120 60% $62,783 2019 196 128 65% $74,340 2020 220 144 65% $58,211 2021 236 142 60% $57,123 2022 375 197 53% $73,906 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 27 First Professional Degrees Among first professional degree recipients, average cumulative median debt decreased by 9% from $155,074 in 2021 to $130,835 in 2022 across all sectors. Additionally, the percentage of graduates from these programs who took out student loans in 2022 (75%) decreased by 3 percentage points from 2021 (78%). This trend varies from sector to sector. In 2022, first professional degree recipients at: • The University of Minnesota system borrowed $166,935 (down 5% from $176,549 in 2021), with 74% of graduates borrowing (down 5 percentage points from 2021). • Private non-profit institutions borrowed $114,097 (down 10% from $127,116 in 2021), with 76% of graduates borrowing (up from 75% in 2021). • No private for-profit institutions reported data for either 2021 or 2022, with the last recorded information being from 2017. Due to the number of students graduating with loans from these programs at private for-profit institutions was less than 10 in 2022, cumulative median debt information is suppressed for that year, and thereby excluded from this report. Figure 16: Average Cumulative Median Debt of by Year and Sector, First Professional Degree Recipients Cumulative Median Student Loan Debt, 2022 28 Figure 17: Percent of Graduates with Loans by Year and Sector, First Professional Degree Recipients Table 9: Average Cumulative Median Debt for Minnesota First Professional Degree Recipients, 2018-2022 Sector Year Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt All Minnesota Total 2018 1,355 1,109 82% $147,367 2019 1,357 1,087 80% $143,912 2020 1,419 1,039 73% $145,674 2021 1,542 1,197 78% $155,074 2022 1,646 1,241 75% $140,835 University of Minnesota 2018 799 671 84% $169,967 2019 769 614 80% $173,069 2020 781 619 79% $172,692 2021 853 677 79% $176,549 2022 843 628 74% $166,935 Private Non-Profit 2018 556 438 70% $112,745 2019 588 473 80% $106,064 2020 638 420 66% $105,855 2021 689 520 75% $127,116 2022 803 613 76% $114,097 Note: Sectors with no graduates for the award type are not included. Cumulative Median Student Loan Debt, 2022 29 Conclusion Overall, average cumulative median debt decreased across most award levels in recent years, with some award types seeing a slight increase in borrowing during the 2021-2022 academic year. For most undergraduate degree types (associate degrees and bachelor’s degrees), this is reflective of a larger annual trend of decreasing debt amounts from past years. These findings stand in stark contrast to the changing rates of cumulative debt for students seeking graduate level degrees. Graduates receiving master’s degrees saw an increase in average cumulative median debt from 2021 to 2022, reversing a decrease in median debt reported for the 2020-2021 academic year. The largest overall average increase in debt was for students who received doctoral degrees in 2022; while first professional degrees seekers had a significant decrease in average debt from last year’s total when compared to other award types. These shifts contrast with the steady increases in median debt rates observed in previous years. Across all undergraduate award types, the percent of graduates borrowing has decreased from 2018 to 2022, with the percent of graduates borrowing for graduate award types has increased during the same time period. Cumulative Median Student Loan Debt, 2022 30 Appendix A – Cumulative Median Debt by Institution Average Cumulative Median Debt of Graduates, All Sectors, 2022 Award Type Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Average Cumulative Median Debt Sub-Baccalaureate Certificate 9,242 3,687 40% $10,022 Associate Degree 16,336 8,078 52% $16,107 Bachelor's Degree 30,426 19,072 63% $24,062 Graduate Certificates 1,209 357 30% $23,622 Master’s Degree 10,062 5,246 52% $36,833 Doctoral Degree 1,435 512 36% $71,497 First Professional Degree 1,646 1,241 75% $140,835 Graduates with Sub-Baccalaureate Certificates, Minnesota State Colleges, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Alexandria Technical & Community College 306 102 33% $8,470 Anoka Technical College 634 188 30% $8,638 Anoka-Ramsey Community College 44 13 30% $11,905 Central Lakes College 258 100 39% $9,500 Century College 546 166 30% $9,500 Dakota County Technical College 346 121 35% $9,500 Fond du Lac Tribal & Community College 20 11 55% $10,500 Hennepin Technical College 608 163 27% $9,500 Hibbing Community College 79 34 43% $9,500 Inver Hills Community College 240 31 13% $7,000 Itasca Community College 50 12 24% $9,925 Lake Superior College 271 104 38% $9,926 Mesabi Range College 153 63 41% $8,750 Minneapolis Community & Technical College 306 102 33% $12,050 Minnesota State College Southeast 200 100 50% $9,500 Minnesota State Community and Technical College 278 155 56% $10,250 Minnesota West Community & Technical College 409 92 22% $8,875 Normandale Community College 130 24 18% $12,000 North Hennepin Community College 342 130 38% $15,164 Northland Community & Technical College 333 143 43% $10,500 Northwest Technical College 77 43 56% $11,000 Pine Technical & Community College 139 51 37% $9,500 Cumulative Median Student Loan Debt, 2022 31 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Rainy River Community College 1 0 0% $0 Ridgewater College 199 73 37% $11,077 Riverland Community College 198 82 41% $12,000 Rochester Community and Technical College 193 72 37% $10,500 Saint Paul College 697 205 29% $9,500 South Central College 178 58 33% $10,875 St. Cloud Technical & Community College 246 124 50% $10,974 Vermilion Community College 31 19 61% $10,843 Sector Average 7,512 2,581 34% $10,159 Graduates with Sub-Baccalaureate Certificates, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 2 1 50% Suppressed Metropolitan State University 13 7 54% Suppressed Minnesota State University Moorhead 4 2 50% Suppressed Minnesota State University, Mankato 1 1 100% Suppressed Southwest Minnesota State University 1 1 100% Suppressed St. Cloud State University 16 8 50% Suppressed Winona State University 0 0 0% $0 Sector Average 37 20 54% $15,023 Graduates with Sub-Baccalaureate Certificates, University of Minnesota System, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt University of Minnesota Crookston 11 2 18% Suppressed University of Minnesota Duluth 31 1 3% Suppressed University of Minnesota Twin Cities 191 10 5% $25,124 Sector Average 233 13 6% $21,113 Cumulative Median Student Loan Debt, 2022 32 Graduates with Sub-Baccalaureate Certificates, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bethel University 27 6 22% Suppressed The College of St. Scholastica 1 0 0% $0 Concordia University – St. Paul 5 1 20% Suppressed Crown College 6 0 0% $0 Dunwoody College of Technology 28 17 61% $5,500 Herzing University 50 44 88% $13,119 Martin Luther College 57 9 16% Suppressed Northwestern Health Sciences University 14 6 43% Suppressed Oak Hills Christian College 6 0 0% $0 St. Catherine University 12 11 92% $24,000 University of St. Thomas 29 5 17% Suppressed Sector Average 235 99 42% $12,494 Graduates with Sub-Baccalaureate Certificates, Private For-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Academy College 11 1 9% Suppressed Avalon School of Cosmetology 17 4 24% Suppressed Aveda Arts and Sciences Institute Minneapolis 451 342 76% $6,333 Empire Beauty School 113 102 90% $3,085 Hastings Beauty School Inc. 9 9 100% Suppressed Minnesota School of Barbering 19 0 0% $0 Minnesota School of Cosmetology 163 138 85% $11,569 Nova Academy of Cosmetology 98 72 73% $10,000 PCI Academy 344 306 89% $12,865 Rasmussen University 11 1 9% Suppressed Sector Average 1,225 974 80% $9,158 Cumulative Median Student Loan Debt, 2022 33 Graduates with Associate Degrees, Minnesota State Colleges, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Alexandria Technical & Community College 327 163 50% $12,000 Anoka Technical College 181 90 50% $19,102 Anoka-Ramsey Community College 1047 401 38% $13,000 Central Lakes College 491 178 36% $12,006 Century College 1,125 497 44% $15,250 Dakota County Technical College 416 186 45% $12,000 Fond du Lac Tribal & Community College 111 64 58% $15,791 Hennepin Technical College 492 201 41% $14,031 Hibbing Community College 158 64 41% $14,740 Inver Hills Community College 481 184 38% $13,000 Itasca Community College 163 52 32% $10,500 Lake Superior College 653 343 53% $16,397 Mesabi Range College 124 35 28% $13,091 Minneapolis Community & Technical College 905 398 44% $20,000 Minnesota State College Southeast 161 82 51% $13,790 Minnesota State Community and Technical College 709 379 53% $13,900 Minnesota West Community & Technical College 237 117 49% $13,522 Normandale Community College 1,299 409 31% $13,000 North Hennepin Community College 590 225 38% $15,750 Northland Community & Technical College 365 224 61% $15,750 Northwest Technical College 114 84 74% $14,150 Pine Technical & Community College 126 50 40% $13,000 Rainy River Community College 52 13 25% $10,740 Ridgewater College 539 275 51% $13,000 Riverland Community College 346 141 41% $16,000 Rochester Community and Technical College 673 314 47% $14,976 Saint Paul College 553 211 38% $14,125 South Central College 366 159 43% $13,034 St. Cloud Technical & Community College 696 329 47% $14,400 Vermilion Community College 119 71 60% $12,881 Sector Average 13,619 5,939 44% $14,510 Cumulative Median Student Loan Debt, 2022 34 Graduates with Associate Degrees, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 57 46 81% $14,564 Minnesota State University Moorhead 0 0 0% $0 Minnesota State University, Mankato 5 3 60% Suppressed Southwest Minnesota State University 53 38 72% $18,274 St. Cloud State University 10 6 60% Suppressed Winona State University 106 60 57% $18,826 Sector Average 284 200 70% $18,709 Graduates with Associate Degrees, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bethel University 29 7 24% Suppressed Concordia University-St. Paul 22 18 82% $17,375 Crown College 18 7 39% Suppressed Dunwoody College of Technology 260 160 62% $10,500 Herzing University 37 35 95% $16,116 North Central University 31 15 48% $24,250 Northwestern Health Sciences University 59 41 69% $24,250 Oak Hills Christian College 6 4 67% Suppressed St. Catherine University 354 285 81% $22,515 University of Northwestern - St. Paul 5 2 40% Suppressed University of St. Thomas 50 4 8% Suppressed White Earth Tribal & Community College 16 0 0% $0 Sector Average 887 578 65% $18,616 Graduates with Associate Degrees, Private For-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Academy College 3 3 100% Suppressed Rasmussen University 1,543 1,358 88% $21,409 Sector Average 1,546 1,361 88% $21,629 Cumulative Median Student Loan Debt, 2022 35 Graduates with Bachelor’s Degrees, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 929 637 69% $21,700 Metropolitan State University 1,701 951 56% $17,500 Minnesota State University Moorhead 959 680 71% $23,525 Minnesota State University, Mankato 2,542 1,654 65% $22,702 Southwest Minnesota State University 427 283 66% $20,965 St. Cloud State University 1,528 913 60% $24,688 Winona State University 1,408 980 70% $25,502 Sector Average 9494 6098 64% $22,544 Graduates with Bachelor’s Degrees, University of Minnesota System, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt University of Minnesota Crookston 394 272 69% $21,318 University of Minnesota Duluth 1,914 1,320 69% $26,719 University of Minnesota Morris 280 177 63% $19,792 University of Minnesota Rochester 143 89 62% $22,265 University of Minnesota Twin Cities 7,439 3,796 51% $21,278 Sector Average 10,170 5,654 56% $22,519 Graduates with Bachelor’s Degrees, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Augsburg University 528 344 65% $26,125 Bethany Lutheran College 136 90 66% $27,000 Bethel University 631 434 69% $25,500 Carleton College 484 181 37% $18,029 College of Saint Benedict 349 247 71% $28,000 The College of St. Scholastica 696 495 71% $28,770 Concordia College 345 250 72% $31,260 Concordia University-St. Paul 708 483 68% $21,800 Crown College 146 112 77% $27,504 Dunwoody College of Technology 82 57 70% $12,500 Gustavus Adolphus College 367 296 81% $26,521 Hamline University 405 325 80% $26,722 Cumulative Median Student Loan Debt, 2022 36 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Herzing University 156 138 88% $37,547 Macalester College 522 302 58% $20,091 Martin Luther College 153 104 68% $22,050 North Central University 118 100 85% $27,000 Northwestern Health Sciences University 788 639 81% $25,000 Oak Hills Christian College 22 17 77% $19,291 Saint John's University 10 6 60% Suppressed Saint Mary's University of Minnesota 352 258 73% $35,933 St. Catherine University 363 243 67% $27,000 St. Olaf College 562 411 73% $26,748 University of Northwestern - St. Paul 660 380 58% $27,000 University of St. Thomas 473 360 76% $24,737 Sector Average 10,606 7,187 68% $26,281 Graduates with Bachelor’s Degrees, Private For-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Academy College 20 18 90% $132,469 Rasmussen College 136 115 85% $24,745 Sector Average 156 133 85% $39,324 Graduates with Master’s Degrees, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 88 39 44% $27,582 Metropolitan State University 252 153 61% $41,000 Minnesota State University Moorhead 277 167 60% $24,177 Minnesota State University, Mankato 602 319 53% $26,076 Southwest Minnesota State University 139 72 52% $16,897 St. Cloud State University 489 235 48% $29,269 Winona State University 191 110 58% $33,446 Sector Average 2,038 1,095 54% $28,747 Cumulative Median Student Loan Debt, 2022 37 Graduates with Master’s Degrees, University of Minnesota System, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt University of Minnesota Duluth 215 100 47% $36,372 University of Minnesota Twin Cities 2,899 1,152 40% $40,568 Sector Average 3114 1252 40% $40,232 Graduates with Master’s Degrees, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Adler Graduate School 78 62 79% $49,394 Augsburg University 290 193 67% $50,916 Bethel University 328 189 58% $44,675 Concordia College 12 8 67% Suppressed Concordia University-St. Paul 297 185 62% $57,151 Crown College 798 462 58% $24,400 Hamline University 120 74 62% $49,550 Herzing University 352 188 53% $31,679 Martin Luther College 4 4 100% Suppressed Minneapolis College of Art and Design 26 13 50% $18,009 North Central University 33 19 58% $31,000 Northwestern Health Sciences University 74 34 46% $25,550 Saint John's University 26 18 69% $90,339 Saint Mary's University of Minnesota 25 3 12% Suppressed St. Catherine University 1,074 671 62% $34,166 University of Northwestern - St. Paul 403 297 74% $48,863 University of St. Thomas 46 18 39% $27,589 Sector Average 4,910 2,899 59% $38,420 Graduates with Graduate Certificates, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 0 0 0% $0 Metropolitan State University 19 9 47% Suppressed Minnesota State University Moorhead 14 6 43% Suppressed Minnesota State University, Mankato 61 24 39% $21,380 Cumulative Median Student Loan Debt, 2022 38 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Southwest Minnesota State University 18 6 33% Suppressed St. Cloud State University 98 39 40% $20,695 Winona State University 46 29 63% $30,750 Sector Average 256 113 44% $23,593 Graduates with Graduate Certificates, Private Non-Profit, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Adler Graduate School 4 2 50% Suppressed Bemidji State University 44 10 23% $7,500 Bethel University 124 70 56% $23,712 Concordia University-St. Paul 5 2 40% Suppressed Crown College 1 0 0% $0 Hamline University 159 29 18% $18,798 Saint Mary's University of Minnesota 161 88 55% $27,290 St. Catherine University 46 13 28% $23,280 University of Northwestern - St. Paul 3 0 0% $0 University of St. Thomas 202 2 1% Suppressed Sector Average 749 216 29% $24,048 Graduates with Doctoral Degrees, Minnesota State Universities, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Bemidji State University 0 0 0% $0 Metropolitan State University 19 13 68% $71,000 Minnesota State University Moorhead 14 5 36% Suppressed Minnesota State University, Mankato 38 22 58% $62,411 Southwest Minnesota State University 0 0 0% $0 St. Cloud State University 13 5 38% Suppressed Winona State University 52 31 60% $61,500 Sector Average 136 76 56% $63,048 Cumulative Median Student Loan Debt, 2022 39 Graduates with Doctoral Degrees, University of Minnesota System, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt University of Minnesota Duluth 1 1 100% Suppressed University of Minnesota Twin Cities 923 238 26% $72,460 Sector Average 924 239 26% $72,199 Graduates with Doctoral Degrees, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Augsburg University 24 22 92% $79,877 Bethel University 36 23 64% $52,636 The College of St. Scholastica 233 105 45% $84,182 Concordia University-St. Paul 14 9 64% Suppressed Hamline University 5 4 80% Suppressed Saint Mary's University of Minnesota 24 7 29% Suppressed St. Catherine University 39 27 69% $67,535 University of St. Thomas 24 22 92% $79,877 Sector Average 375 197 53% $73,906 Graduates with First Professional Degrees, University of Minnesota System, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt University of Minnesota Twin Cities 843 628 74% $166,935 Sector Average 843 628 74% $166,935 Graduates with First Professional Degrees, Private Non-Profit Institutions, 2022 Institution Award Recipients Number of Graduates with Loans Percent of Graduates with Loans Cumulative Median Debt Concordia University-St. Paul 31 27 87% $58,053 Mitchell Hamline School of Law 344 252 73% $121,347 Northwestern Health Sciences University 166 132 80% $187,356 Saint Mary’s University of Minnesota 26 22 85% $85,235 St. Catherine University 73 52 71% $56,859 University of St. Thomas 163 128 79% $64,315 Sector Average 803 613 76% $114,097 Cumulative Median Student Loan Debt, 2022 40 Appendix B – Information on Loan Repayment Depending on the types and amounts of their loans, graduates may have several repayment options. For each award type in this report, available repayment options will be presented. Several factors determine their repayment amount, including: • Total amount borrowed • Interest rate(s) • Interest accruing while in school or deferment • Loan forgiveness options • Loan term(s) • Graduates’ income Potential repayment options for graduates include: • Standard repayment plan: monthly payments made for up to 10 years. • Graduated repayment: monthly payments made for up to 10 years (payments start low and increase every two years). • Pay As You Earn: monthly payments are limited to 10% of discretionary income; outstanding balances are forgiven after 20 years. • Save on a Valuable Education (Formerly Revised Pay as You Earn): monthly payments are based on your discretionary income—the difference between your adjusted gross income (AGI) and the amount as defined by the U.S. Department of Health and Human Services Poverty Guideline6 amount for your family size which increases the income exemption from 150% to 225% of the poverty line. • Income-based repayment: monthly payments are generally equal to 15% of current discretionary income (10% if one is borrowing federal loans for the first time); outstanding balances are forgiven after 20-25 years. • Income-contingent repayment: monthly payments are limited to 20% of current discretionary income; outstanding balances are forgiven after 25 years. • Extended Fixed repayment: monthly payments made for up to 25 years. • Extended Graduated repayment: monthly payments made for up to 25 years (payments start low and increase every two years). Estimates use the annual median wage for an award holder in that degree type two years after graduating during the 2019-2020 academic year and the statewide average of median cumulative debt for that degree type7. Estimates use an estimated graduated monthly repayment amount across eligible repayment plans, and are based on current interest rates for federal unsubsidized loans for undergraduate and graduate degrees. All income-based information assumes 5% annual wage growth. It is important to remember that these estimates do not include debt incurred at other institutions or for 6 7 Median wage data by award type is not available for degrees above a bachelor’s degree. Loan data for all graduate degree types uses average median wages for all graduate certificates and degrees. Source: Cumulative Median Student Loan Debt, 2022 41 other degrees. Moreover, these estimates do not include any PSLF (Public Service Loan Forgiveness) options for graduates. Loan Repayment Choices for Minnesota Sub-Baccalaureate Certificate Recipients (Annual Income of $35,4888) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $10,022 5.50% $3,030 $13,052 120 $109 4% to 2% Graduated $10,022 5.50% $3,826 $13,848 120 $62 to $186 2% to 4% Pay As you Earn $10,022 5.50% $2,675 $12,697 240 or less $95 to $105 3% to 2% Save on a Valuable Education (Formerly REPAYE) $10,022 5.50% $5,327 $15,349 240 or less $22 to $186 1% to 4% Income-Based Repayment $10,022 5.50% $2,675 $12,697 120 $95 to $105 3% to 2% Income-Contingent Repayment $10,022 5.50% $5,514 $15,536 300 or less $70 to $86 2% to 2% Source: U.S. Department of Education Loan Repayment Choices for Minnesota Associate Degree Recipients (Annual Income of $38,4119) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $16,107 5.50% $4,869 $20,976 120 $175 5% to 3% Graduated $16,107 5.50% $6,149 $22,256 120 $99 to $298 3% to 6% Pay As you Earn $16,107 5.50% $5,468 $21,575 240 or less $138 to $175 4% to 3% Save on a Valuable Education (Formerly REPAYE) $16,107 5.50% $9,871 $25,978 240 or less $47 to $276 1% to 5% Income-Based Repayment $16,107 5.50% $5,468 $21,575 120 $138 to $175 4% to 3% Income-Contingent Repayment $16,107 5.50% $8,248 $24,355 300 or less $118 to $142 4% to 3% Source: U.S. Department of Education 8 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). 9 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). Cumulative Median Student Loan Debt, 2022 42 Loan Repayment Choices for Minnesota Bachelor’s Degree Recipients (Annual Income of $45,76710) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $24,062 5.50% $7,274 $31,336 120 $259 7% to 4% Graduated $24,062 5.50% $9,186 $33,248 120 $149 to $446 4% to 7% Pay As You Earn $24,062 5.50% $8,482 $32,544 240 or less $199 to $261 5% to 4% Save on a Valuable Education (Formerly REPAYE) $24,062 5.50% $13,835 $37,897 240 or less $108 to $375 3% to 6% Income-Based Repayment $24,062 5.50% $8,482 $32,544 300 or less $199 to $261 5% to 4% Income-Contingent Repayment $24,062 5.50% $10,795 $34,857 300 or less $193 to $227 5% to 4% Source: U.S. Department of Education Loan Repayment Choices for Minnesota Master’s Degree Recipients (Annual Income of $69,50211) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $36,833 7.05% $14,600 $51,433 120 $429 7% to 5% Graduated $36,833 7.05% $18,662 $55,495 120 $248 to $744 4% to 8% Extended Fixed Repayment $36,833 7.05% $41,618 $78,451 300 $262 5% to 3% Extended Graduated Repayment $36,833 7.05% $48,005 $84,838 300 $216 to $368 4% to 4% Pay As You Earn $36,833 7.05% $15,039 $51,872 240 or less $397 to $429 7% to 5% Save on a Valuable Education (Formerly REPAYE) $36,833 7.05% $17,752 $54,585 240 or less $306 to $598 5% to 6% Income-Based Repayment $36,833 7.05% $15,039 $51,872 300 or less $397 to $429 7% to 5% Income-Contingent Repayment $36,833 7.05% $17,511 $54,344 300 or less $380 to $422 7% to 4% Source: U.S. Department of Education 10 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). 11 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). Cumulative Median Student Loan Debt, 2022 43 Loan Repayment Choices for Minnesota Graduate Certificate Recipients (Annual Income of $69,50212) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $25,844 7.05% $22,711 $48,555 120 $294 3% to 2% Graduated $25,844 7.05% $27,157 $53,001 120 $153 to $304 3% to 3% Income-Contingent Repayment $25,844 7.05% $12,432 $38,276 300 $268 to $297 5% to 3% Save on a Valuable Education (Formerly REPAYE) $25,844 7.05% $8,174 $34,018 300 $306 to $493 5% to 5% Source: U.S. Department of Education Loan Repayment Choices for Minnesota Doctoral Degree Recipients (Annual Income of $69,50213) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $71,497 7.05% $28,341 $99,838 120 $727 14% to 9% Graduated $71,497 7.05% $36,226 $107,723 120 $481 to $1444 8% to 15% Extended Fixed Repayment $71,497 7.05% $80,786 $152,283 300 $433 9% to 5% Extended Graduated Repayment $71,497 7.05% $93,198 $164,695 300 $420 to $715 7% to 8% Source: U.S. Department of Education Loan Repayment Choices for Minnesota First Professional Degree Recipients (Annual Income of $69,50214) Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Standard $155,074 7.05% $61,471 $216,545 120 $1,764 31% to 19% Graduated $155,074 7.05% $78,573 $233,647 120 $1044 to $3131 18% to 33% Extended Fixed Repayment $155,074 7.05% $175,221 $330,295 300 $1,051 19% to 12% 12 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). 13 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). 14 Most recent available annual median wage; Graduate Employment Outcomes Tool, Minnesota Department of Employment and Economic Development (DEED). Cumulative Median Student Loan Debt, 2022 44 Repayment Plan Original Borrowed Amount Current Interest Rate (unsubsidized loans) Interest Paid Total Repayment Amount Repayment Term (In Months) Monthly Payment Percent of Monthly Income Extended Graduated Repayment $155,074 7.05% $202,143 $357,217 300 $911 to $1550 16% to 16% Source: U.S. Department of Education 2022
14534	https://math.stackexchange.com/questions/4049465/absolute-value-inequality-implication	real analysis - Absolute Value Inequality Implication - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Absolute Value Inequality Implication Ask Question Asked 4 years, 6 months ago Modified4 years, 6 months ago Viewed 92 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. From my knowledge of the properties of absolute value inequalities, I'm failing to see why: \|f(x)−A\|<1→\|f(x)\|<\|A\|+1\|f(x)−A\|<1→\|f(x)\|<\|A\|+1 The only approach I know is to split up the two possible cases: (1):f(x)−A>−1(1):f(x)−A>−1 and (2):f(x)−A<1(2):f(x)−A<1 So f(x)<A+1 f(x)<A+1, but how does this imply \|f(x)\|<\|A\|+1\|f(x)\|<\|A\|+1 ? real-analysis analysis inequality absolute-value Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Mar 5, 2021 at 0:11 JKMJKM 449 3 3 silver badges 10 10 bronze badges 2 Triangle inequality: A+1≤\|A+1\|≤\|A\|+1 A+1≤\|A+1\|≤\|A\|+1 DanLewis3264 –DanLewis3264 2021-03-05 00:15:12 +00:00 Commented Mar 5, 2021 at 0:15 1 @bounceback Yes, but your comment doesn't address the issue that f(x)f(x) may be negative, so f(x)≤B f(x)≤B doesn't then imply \|f(x)\|≤B\|f(x)\|≤B.Clement C. –Clement C. 2021-03-05 00:20:46 +00:00 Commented Mar 5, 2021 at 0:20 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. The triangle inequality says \|a+b\|≤\|a\|+\|b\|\|a+b\|≤\|a\|+\|b\|. From this, setting a=(x−y)a=(x−y) and b=y b=y, we get \|x\|≤\|x−y\|+\|y\|\|x\|≤\|x−y\|+\|y\| from which we deduce \|x\|−\|y\|≤\|x−y\|.\|x\|−\|y\|≤\|x−y\|. This inequality gives you the result you want: \|f(x)\|−\|A\|≤\|f(x)−A\|<1\|f(x)\|−\|A\|≤\|f(x)−A\|<1 so from \|f(x)\|−\|A\|<1\|f(x)\|−\|A\|<1 you immediatly get \|f(x)\|<\|A\|+1.\|f(x)\|<\|A\|+1. In fact, the following holds: try to verify it: ∣∣\|x\|−\|y\|∣∣≤\|x±y\|≤\|x\|+\|y\|.\|\|x\|−\|y\|\|≤\|x±y\|≤\|x\|+\|y\|. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 5, 2021 at 0:16 Arturo MagidinArturo Magidin 419k 60 60 gold badges 864 864 silver badges 1.2k 1.2k bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. More generally: This is a consequence of the so-called reverse triangle inequality, itself a consequence of the triangle inequality: \|\|x\|−\|y\|\|≤\|x−y\|,x,y∈R(†)(†)\|\|x\|−\|y\|\|≤\|x−y\|,x,y∈R Apply it to f(x)f(x) and A A to get \|\|f(x)\|−\|A\|\|≤\|f(x)−A\|\|\|f(x)\|−\|A\|\|≤\|f(x)−A\| which implies \|f(x)\|≤\|f(A)\|+\|f(x)−A\|\|f(x)\|≤\|f(A)\|+\|f(x)−A\|. _The reverse triangle inequality, very useful, is quite immediate given the triangle inequality: from \|x\|=\|x−y+y\|≤\|x−y\|+\|y\|\|x\|=\|x−y+y\|≤\|x−y\|+\|y\| and \|y\|=\|y−x+x\|≤\|x−y\|+\|x\|\|y\|=\|y−x+x\|≤\|x−y\|+\|x\| we get \|\|x\|−\|y\|\|=max(\|x\|−\|y\|,\|y\|−\|x\|)≤\|x−y\|\|\|x\|−\|y\|\|=max(\|x\|−\|y\|,\|y\|−\|x\|)≤\|x−y\|._ Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 5, 2021 at 0:22 answered Mar 5, 2021 at 0:17 Clement C.Clement C. 68.5k 8 8 gold badges 82 82 silver badges 171 171 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis analysis inequality absolute-value See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 8Absolute value problem \|x−y\|=\|y−x\|\|x−y\|=\|y−x\| 2Prove triangle inequality using the properties of absolute value 3How to solve inequalities with more than one absolute value expression 1absolute value inequality with complex number 1Absolute value inequality with variable on both sides 1Absolute Value Inequality Case Study 0Absolute value inside absolute value inequality (from both sides) 6What does it mean to have an absolute value equal an absolute value? 0Inequality with 4 absolute values 8How to solve double absolute value inequality? 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14535	https://teflconcourse.com/training/inservice/lexicogrammar/polysemy	ELT Concourse: polysemy ELT Concourse teacher training ELT Concourse home A-Z site index Teacher training index Teacher development For teachers For trainers For managers For learners About language Language questions Other areas Academic English Business English Entering ELT Courses index Basic ELT course TKT The Bridge Language analysis Training to train Transcription Glossaries Articles Polysemy and Homonymy Mr Stiggins, with several most indubitable symptoms of having quite as much pine-apple rum-and-water about him as he could comfortably accommodate, took his hat, and his leave. Charles Dickens, The Pickwick Papers, Chapter XXVII He carried a strobe light and the responsibility for the lives of his men. Tim O'Brien, The Things They Carried You are free to execute your laws, and your citizens, as you see fit. Star Trek: The Next Generation The farmers in the valley grew potatoes, peanuts, and bored Kevin Flynn Yet time and her aunt moved slowly – and her patience and her ideas were nearly worn out before the tête-à-tête was over. Jane Austen: Pride and Prejudice All the citations above are examples of a literary device called zeugma (/ˈzjuːɡ.mə/). Their effect lies in the fact that a single verb or adjective is applied to two objects, one of which is appropriate, the other not. There are six verbs above which operate like this in these examples (take, carry, execute, grow, move and wear out) and they all exhibit polysemy: they have more than one meaning in English. If it weren't for polysemy, zeugma creation would not be possible. By some estimates, 40% of all words in English are polysemous with more, sometimes many more, than one meaning. What is important here, as we shall see, is that the various meanings of polysemous words are connected in some way (if obscurely) and in this polysemy differs from homonymy. Homonymy also refers to words having more than one meaning but in this case the meanings are wholly unconnected. Zeugma is sometimes referred to as syllepsis but on this site the latter term is reserved for concord problems in which one element can only agree with some of another element in a sentence as in, for example: John and Mary will each bring his / her / their favourite dessert in which it is unclear which possessive determiner should be applied. Homonymy, Polysemy, Monosemy and Word family Homonymyrefers to words which look and sound the same but are unrelated and have different, unrelated meanings. They are usually distinguished from homographs (words which look the same but are differently pronounced, e.g., lead [the metal] and lead [the verb]) and homophones (words which look different but are pronounced the same, e.g., rights, rites, writes, no, know etc.). Homonyms are usually derived differently, i.e., they have different etymologies, but that is not always the case. Other examples are: bank (a commercial enterprise concerned with money) and bank (the side of a river) May (the fifth month of the year) and may (a modal auxiliary verb) peep (make a bird-like noise) and peep (look cautiously) air (atmosphere) and air (a tune) rose (the past form of rise) and rose (a flower) mean (signify) and mean (unpleasant) and mean (primitive) and mean (average)Polysemyrefers to the fact that a word or phrase can have a variety of related meanings. For example, He was booked into the hotel vs. He was booked by the referee. Both these meanings of book imply some kind of registration (in a book of some kind) so they are clearly related but have different definitions. A phrase such as on the rocks exhibits the same phenomenon, referring to a drink with ice cubes or to a shipwreck. Polysemes are often formed by conversion in which the form of a word remains unchanged but its word class alters. Other examples are: pen (a writing implement) and pen (the action of writing) dish (a kind of plate) and dish (a meal) frame (a surrounding for a picture) and frame (falsely incriminate) cap (the top of a pen or bottle) and cap (a type of headgear) drive (control a vehicle) and drive (a short private roadway) satellite (an orbiting space craft) and satellite (a country or state dependent on a more powerful one) Polysemy is pronounced /pɒ.ˈlɪ.siː.mɪ/, incidentally, and the adjective is polysemous, pronounced /pɒ.ˈlɪ.siː.məs/.Monosemyrefers to a word having only one possible meaning in a language and which cannot lead to ambiguity. For example, the word orrery has no meaning in English other than a clockwork model of the solar system (Riemer, 2010:161).Word familyrefers to a single word (or lemma) which takes different grammatical forms while retaining its core meaning. Examples are: nation, nations, national, nationalise etc. s moke, smoked, smoking, smokes, smoker, smoky etc .(An alternative definition of word families is one used by teachers of children and refers to words which contain similar constituents so, for example: nest, best, lest, chest, crest etc. would qualify as a word family. That is not a useful definition of the term for our purposes.) Homonymy and polysemy occur in all natural languages. Artificial languages avoid both on the principle that, in an ideal world, any word should have a single unambiguous meaning. You may have noticed that the world is not ideal. And here's the snag: the problem of distinguishing between homonymy and polysemy is, in principle, insoluble. Lyons, cited in Laufer, in Schmitt and McCarthy (1997:152) This is why we are dealing with homonymy and polysemy in the same guide. Semantic development Over time, in all languages, words change their meanings. Sometimes a word's meaning is extended to other meanings and sometimes it is altered completely. In the first case, we arrive at a word which is polysemous: the primary meaning is retained but other related meanings are grafted on to the word. In the second case, we have what is called polysemy splitting in which the word may retain its primary meaning but a secondary meaning is now so different that we are justified in calling the resulting word a homonym, having the same form as another word but a wholly different meaning. There are two processes at work: Radiation In this process, a word retains its primary meaning but other meanings are developed from it by analogy or metaphor. Our example is the word head. The primary meaning of the word is the uppermost or foremost part of an animal's body and that meaning is retained in the language, of course. However, radiating from the primary meaning are all sorts of other meanings, all of which are related to the primary meaning, like this: With a little effort, the derived, radiating meanings can be inferred from the primary meaning even when word class changes by conversion from the noun to a verb. When radiation occurs, the meanings (in the outer circle) derived from the primary one (in the centre) are not related to each other but retain a relationship only to the central word. 2. Concatenation In this process, there is a chain of meanings, each related to the one before it but developing until the original meaning and the derived meanings are so different that they qualify as homonyms rather than related meanings of the same word. Our example is stew. The oldest meaning referred to a hot room or steam bath, from which the idea of slowly cooking developed and from there, the idea of a meal so cooked of a mix of vegetables and / or meat is a natural step. Metaphorically, the verb can be used to mean leave people without help to face the consequences of their actions, like this: We cannot infer the meaning of Don't bother to help. Let him stew. without making a real leap of imagination and the meaning of a nice, warming vegetable stew is not derivable from a hot room without a good deal of inspired guesswork, if it is derivable at all. The polysemes have split into two or more derivationally related terms which are no longer semantically related so they are homonyms. The primary meaning of hot room is now lost to Modern English but in some cases of concatenation, may be retained. A modern example is the word mouse which originally, of course, applied to a small furry rodent but whose meaning has now been extended to refer to the computer peripheral because of a presumed similarity of appearance. Concatenation often works in two ways: 1. narrowing: a word may have its meaning narrowed so, for example, the word book originally meant any piece of written material inscribed on any material but the meaning was narrowed to only a certain type of bound paper pages. 2. broadening: a word may have its meaning extended so, for example, the word bird was originally used only to apply to young birds but has now been broadened to include all avians. It is often possible, with a good etymological dictionary, to trace a word's meaning back to its primary source. However, as we explain below, etymology is not a good guide to meaning. So what's the problem? Primarily it is Lyons' problem, cited above, that it is in principle impossible to distinguish between homonymy and polysemy. A good example is the word horn. The word derives from Proto-Indo-European and there are cognate words in other Germanic languages (the word is the same in German, Swedish, Danish, Norwegian and Old Frisian and almost the same in Modern Frisian, Dutch and the extinct Gothic). The musical instrument was originally made from the horns of animals and the modern meaning is derived from that. Arguably, therefore, the word meaning the outgrowth on the heads of some animals and a musical instrument usually made of brass is an example of monosemy but many would happily suggest that the meanings have diverged to the extent that they are now homonyms and not even polysemes. Almost everyone would consider that the uses of the word to mean telephone or the way vehicles have of warning others are so far removed from the original meaning as to constitute separate lexemes. Most dictionaries will see them that way. Here are some more examples. Would you classify the following examples in black as a word: having only one meaning (monosemy) having two different but related meanings (polysemy) having two completely unrelated meanings (homonymy) 1 I keep the car in the garage vs.I keep a dog and two cats 2 I kept the change vs.I kept him waiting 3 My only grandmother is my father's mother vs.My only grandmother is my mother's mother 4 The car took off down the street vs.The plane took off on time 5 I've smoked too many fags today vs.This work is a real fag 6 The chalk quarry is now disused vs.They chased their quarry for miles 7 I got the bus to work vs.I got $10 for the work 8 His grave is in the churchyard vs.That was a grave mistake 9 I took a short holiday vs.I took a shower 10 The fruit punch was delicious vs.The boxer's punch was vicious 11 That's a vital difference vs.He has no vital signs 12 Why was he tired? vs.Why did he leave? 13 How did she seem? vs.How did you find out? 14 It's a current affairs programme vs.The current runs quickly under the bridge 15 She has fair hair vs.The referee wasn't fair 16 Will he fence in the Olympics next year? vs.I fenced in the garden 17 I read while he watched television vs.While this is interesting, it doesn't help much Click here for some notes when you have made your decisions. keep I keep the car in the garage vs. I keep a dog and two cats There are authorities who would argue strongly that keep is monosemous and that all the senses can be derived from looking at its fundamental meaning. Others would argue that in order to define the word keep in these examples, we need different definitions for each. In the first, we have a word meaning something like store but in the second, the sense is of look after / have at home. If you agree with that, for you the word is polysemous because the two senses are distinct but related to each other. 2. _keep I kept the change vs. I kept him waiting_ In these examples, the situation is slightly different because the verb appears alone in the first and in the phrase keep[someone]waiting in the second. Some would argue that the meanings are not related because in the first, the verb means retain and in the second it means something like had in, e.g.: I had him wait or oblige as in: I obliged him to wait so we have an example of homonymy. In fact, the verb in English is derived obscurely but may come from Old English cepan[seize or hold and observe]. The supporters of the homonymy position would argue that in the 1000 or so years the word has been around, its meanings have diverged and are now not related. 3. _grandmother My only grandmother is my father's mother vs. My only grandmother is my mother's mother_ Most people would aver that this is an example of monosemy because a single definition (parent's parent [female]) would suffice in both cases. Strictly, this is polysemy because the family relationship is different. In some languages, there are different words for the different types of relationship. Another example is the word cousin in English which can apply to a male or female relation and to the children of an aunt or an uncle. Some languages have different forms for each kind of relationship and many distinguish between male and female cousins. 4. _take off The car took off down the street vs. The plane took off on time_ This is a similar case to the examples in 2. Some would argue that the terms are the same in both cases, meaning leave quickly from or on a surface and that we can derive the meaning of either from understanding the meaning of the other. Others would argue, probably unsuccessfully, that the terms are homonyms (and not related in meaning). We can argue, probably more successfully, that we have an example of polysemy because the meanings clearly are related but slightly different. 5. _fag I've smoked too many fags today vs. This work is a real fag_ This is a clearer case of homonymy. The words are unrelated. The word means (in BrE slang) a cigarette and is first attested in 1888. The sense of hard and boring work (never plural) comes from BrE public school slang for a younger pupil forced to work for older boys. In AmE, the word has been used since 1922 to mean a homosexual man. That's a third example of homonymy. 6. _quarry The chalk quarry is now disused vs. They chased their quarry for miles_ This is an example of two different words which look and sound the same but have unrelated meanings: they are homonyms. The words derive from different roots: 14th century French and Latin respectively. 7. _get I got the bus to work vs. I got $10 for the work_ The verb get is notoriously polysemous and most dictionaries will list around 30 different meanings (without adding in particles, prepositions or adverbs). It's also a very good example of the blurred lines between the three phenomena. For example, is the word get the same in _I get the idea I get the money I get wet I get letters_ etc.? Do we have real homonyms in the case of It's getting dark vs. I'm getting a cold? 8. _grave His grave is in the churchyard vs. That was a grave mistake_ There are many who see a semantic connection between grave meaning serious or important and grave meaning a hole in the ground to receive a corpse. That would make this a case of polysemy. However, the examples are, in fact, arguably homonyms because the derivations are quite different. We have 'arguably' in that sentence because etymology is a poor guide to a word's meaning. Much may change over time. The adjective grave derives ultimately from the Latin gravis, meaning heavy and that meaning is extended to mean serious in Modern English. Hence. we have a grave expression, a grave problem etc. The noun grave derives from the Old English græf meaning a ditch and the word has been reduced to the meaning of a hole in the ground to receive a dead body. There is also a verb in Old English, grafen which meant dig and from that the verb also came to mean carve or chisel. In Modern English this only survives in the adjective graven for an image or the verb engrave. The words look and sound the same but have unconnected meanings and sources. 9. _take I took a short holiday vs. I took a shower_ This is not so clear because the two shades of meaning of take here are so closely related as to mean the same thing. We can substitute have in both cases so we could argue that there is no suggestion of either homonymy or polysemy. Others might disagree. 10. _punch The fruit punch was delicious vs. The boxer's punch was vicious_ Clear homonymy. The first word probably derives from Hindi, the second from Old French. They are wholly unrelated and the similarity is coincidental. People without real-time access to the etymology of the language may, of course, see some kind of connection between the alcohol content of a punch and the effects of a punch in the head. 11. _vital That's a vital difference vs. He has no vital signs_ Most people will classify this as an example of monosemy. In both cases the word refers to life (or its absence) but the first meaning is metaphorically derived from the second and can be understood with little difficulty once one is aware of the meaning of vital as manifesting life and can then extend it to mean essential to life or just plain essential. 12. _why Why was he tired? vs. Why did he leave? Most people would consider a simply question word like _why can only have one meaning but a moment's thought reveals a problem. The problem lies with another polysemous word: reason. That word can refer to purpose (future) or to cause (past). So we can distinguish: 1. Why was he tired? as an enquiry concerning the cause for a condition and means How come? It can be rephrased as: What previous action or event caused his tiredness? This is clearer with an example such as Why is the girl crying? which clearly enquires about the cause of an action. 2. Why did he go to London? which concerns the purpose of an action looking forward and means What for? It cannot be rephrased as What previous action or event caused him to go to London? because that implies looking back. It can, however, be rephrased as What was his future purpose in going to London? Again, this is made clearer by an example such as Why is the girl opening the box? because that is an enquiry concerning the girl's purpose, not the cause of her action. _how How did she seem? vs. How did you find out?_ The situation is slightly less clear with the word how. It usually refers to manner in some way (so, in that sense is not polysemous) but it evinces four sorts of answers (and that implies four sorts of meaning): 1. an adjectival modifier such as quite excited, a bit down, unhappy etc. in answer to, e.g.: How did she appear? 2. an adverbial modifier such as hurriedly, with great care, in a rush etc. in answer to, e.g.: How did she drive? 3. a clausal response, such as she heard from Mary, John explained it etc. in answer to, e.g.: How did she find out? 4. an adverbial measure of extent, such as a lot, very, not much, hours and hours, somewhat etc. in answer to, e.g.: _How unhappy was she? How long was the journey_ _current It's a current affairs programme vs. The current runs quickly under the bridge_ This example should be compared to the example of grave, above. The assumption might be that these are homonyms because the two meanings of the adjective and the noun do not seem to be connected. However, this is probably better seen as a case of polysemy. Both meanings ultimately derive from the Old French corant (Modern French courant) meaning running or flowing. In the second example, then, the meaning of a moving body of water is clear. In the first example, the derivation is identical but by the middle of the 15th century, the word had acquired its meaning of present, up to date or in progress. So, we get, e.g., the current situation, the current argument etc. It is also, incidentally, the root of the noun currency meaning money used in the country. The modern meaning with reference to affairs of politics and the state is from the 17th century and arrived at by extension. The word is polysemous but some might argue that the meanings have diverged sufficiently to treat them as homonyms. 15. _fair She has fair hair vs. The referee wasn't fair_ This example, too, should be compared the last one and to the example of grave. A modern assumption would be that fair meaning light coloured and fair meaning just or equitable (with its antonym unfair) are clearly so far apart in meaning that the words are homonyms. Not so, in truth, because the word has only one derivation from the Old English fæger which meant pleasing, beautiful, clear and even morally good. All the modern meanings of the word in English are extensions of the original meaning: When it means light coloured and applied to hair or eyes, the meaning reflects 13th century notions of what is beautiful. When it means according to justice, the word also means pleasing and is of 14th century origin. When it means reasonable as in, for example, a fair chance, the meaning is an extension of the idea of just or unbiased. (The word's use to mean a market or celebration day is, however, a real homonym example because that meaning comes from the Latin feriae meaning religious holidays.) 16. _fence Will he fence in the Olympics next year? vs. I fenced in the garden_ In modern English fence (verb: fight with swords) and fence (verb: make a barrier or noun: barrier) may be considered homonyms because the meanings are no longer connected. Both words, however, derive ultimately from the same source (the Latin defensus meaning defend). 17. _while I read while he watched television vs. While this is interesting, it doesn't help much_ In both the examples, the word is a subordinating conjunction but in the first case it is temporal (meaning, roughly, during the time that or when) and in the second it is concessive (meaning although). The connection can be seen because they both carry the meaning of states or events occurring simultaneously but for teaching purposes, they are better treated as homonyms although they are more like polysemes. Occasionally, a lexeme will betray whether it is polysemous of homonymous in its grammar, word class and colligational characteristics. For example, in _The tree has blown down The concept is not yet full blown He received a nasty blow on the head_ are all, on the face of things, varieties of the word blow but the words merely look similar and it's easy to see that blow is a verb (and irregular) concerned with moving air. It is cognate with the German word blähen. blown is an adjective related to being in full bloom (as of a flower), The verb bloom from which it originates is cognate with the German verb blühen. blow is a noun meaning a strike and may be cognate with the Middle Dutch blouwen, beat, but its origin is obscure. ### The appeal to etymology and word origins The etymological fallacy In this section we have occasionally and with some reluctance appealed to words' etymologies to decide if they are examples of homonymy or polysemy. This is an area beloved of the language pedant, almost always the ignorant language pedant, who sees a word's historical meaning as somehow its 'real' meaning. A word's real meaning is the way in which it is used now by native speakers of the language. It may be the case, for example, that the word toilet, originally spelled toylet, meant when it was imported from French, a cloth that tailors used to bundle up other pieces of cloth. It then moved its meaning by the process of concatenation explained above, to a cloth spread on a dressing table and over the centuries has shifted its meaning again to a porcelain bowl with a flush or the room in which you will find one. Those who insist that a word's original meaning is a guide to how it should be used need to tell us at which point in history we should freeze a word's meaning to stop the decline of the language (something, incidentally, which is not happening). Another good example is the word egregious, an adjective imported in the 1530s. It comes from a Latin phrase ex grege which means separated from or rising above the flock and originally meant distinguished or excellent. However, by the late 16th century the word came to mean its opposite in the modern sense of notably bad, even evil. This was, it has been speculated, because the word was used ironically so frequently that people came to believe it meant the opposite of what was intended. It is a neat example of how language changes over time (and not much time in this case) and of how etymology is not a good guide to meaning. The etymology of words is, indeed, an interesting (for some) window into the past and also tells the serious linguist something about how language changes under cultural pressures. That a word's etymology is a guide to its current meaning or how it should be used is arrant nonsense. That is the etymological fallacy. Syntactical homonymy There is a full guide on this site to this area, linked below, so only a few examples of this phenomenon will be given here to get the flavour. The expression refers to the fact that many function words can perform different, but often semantically related, grammatical functions. The same considerations concerning whether the use of a function word in a different word class or with a different meaning is homonymy or polysemy apply here, too. Here are some examples: since This troublesome word has a number of related meanings (and some distinctly different ones) so may be used as an example of both polysemy and homonymy but syntactically rather than semantically based. 1. It functions as a temporal preposition and refers to the following time span up until another event intervenes or until the present. For example: _She has worked here since October They had been in their jobs since the beginning of the year but left in October They have waited since the first meeting for an answer_ 2. The word is also a good example of syntactical homonymy because it can also function as: 1. an adverb _I saw her when she came to visit her mother, but not since He started on a low salary but has since been promoted and now earns well_ 2. a temporal subordinating conjunction: _He had lived there since he came to London but moved when he retired I have been at university since I was 18_ 3. a causal resultative subordinating conjunction _I had a drink in the bar, since I had an hour to kill before my train Since it was Sunday, I stayed in bed till noon_ yet Take these two examples: _He is tired and getting old. Yet he works a six-day week. I was tired yet happy with my efforts_ In the first case, yet is an adverbial acting as a conjunct referring anaphorically to the first sentence. It could be replaced with a more familiar conjunct performing the same function, such as however. In the second case, it is a conjunction meaning something like but. Whether you consider that the two meanings are close enough to be an example of polysemy is moot but, because the word occupies two separate word classes, this is better described as an example of syntactical homonymy. 3. for Is a good example of syntactical homonymy but also of plain polysemy. 1. In the first case, we can see that it can be both a preposition and a conjunction. As a preposition in, e.g.: I did the work for the money and as a coordinating conjunction (albeit slightly old-fashioned and formal) in, e.g.: I said nothing for I knew they wouldn't listen to me 2. In terms of polysemy, the preposition for has at least 14 different but variously related meanings. The full list of 14 is available here (new tab). Again, how closely you feel the meanings are related will determine whether you teach the area as an example of polysemy or separate the meanings and teach it as homonymy. The meanings include these five : Intended to be given: There's a letter for you Having the purpose: The house is not for sale Amount of time or distance: _He spoke for hours We drove for miles_ On the occasion of: I bought it him for his birthday In support of: I won't vote for the President again For more, as well as a consideration of gradience, a related concept, see the guide. Metaphor: obscure and transparent meaning Obscure and transparent meanings are relative terms not absolute categories. For example: To derive the meaning of coin a new word from the central meaning of coin [metal currency] is quite difficult but the relationship exists, albeit obscurely. The two meanings of coin are actually an example of polysemy but, for teaching purposes, probably best considered and taught as homonyms of coin. On the other hand, to derive the meaning of I banked the money from knowledge of the noun bank (the business) is a relatively easy thing to do because the relationship is much more transparent. Providing one is given a bit of pictorial help, it is not too hard to figure out the alternative meaning of carrot as allurement. There is an argument that alternative meanings of a word which are easily and obviously just a metaphorical extension of the base meaning do not constitute examples of polysemy. Prepositions are good examples of simple metaphorical extensions of meaning which are not too difficult to work out providing one knows the usual, locative meaning of the preposition. Examples are: _to agree among ourselves to have people working under you to meet at six to be in danger to be beyond help_ and so on. All these prepositional uses are metaphorical extensions of the locative (place) meaning of the preposition, and therefore examples of polysemy, but the meaning is usually transparent. It is often observed that prepositions of time closely parallel prepositions of place so, for example: We met in the summer and We met in the theatre both refer to longer times and larger places than: We met at six o'clock and We met at the ticket office (However, the system is by no means consistent because words such as: since, during, till are only used (as prepositions) for time, and beneath, behind, out of are reserved (as prepositions) for spatial relations.) ### Compositionality let the cat out of the bag The term compositionality is usually attributed to Gottlieb Frege who asserted that the meaning of a phrase or clause may be inferred from its constituents so, for example, we can disambiguate the homonym word bear in these phrases: _the fat bear bear a hand_ by looking for the connections between phrase, word class and word meaning. It is, however, not always so easy when fixed phrases are concerned and it is clear that the expression let the cat out of the bag cannot be unpacked by knowing the meaning of all the words which make up the idiom. Polysemous words in compounds present some difficulties. For example, in: doorman the use of door is not to mean an object but an entrance to something (and may not be a door at all). In: hot potato we may not be referring to a potato at all but to a situation which needs careful handling. Affixes may also be polysemous so, for example, in: The situation is hopeless the suffix, -less, clearly means without but in: She's a hopeless golfer the suffix does not carry the meaning of without and the word means something like very poor. In: unfortunate the prefix, un-, means not but in: untie the prefix implies reversing an action. In: hopeful the suffix means showing the characteristics of the stem noun, hope, but in spoonful it means the amount contained by the stem noun. Teaching in this area ### Learner difficulties Homonymy is common to all languages but, of course, is also different in all languages. For example, in English, the word bank may refer to a place to keep money or the side of a river or take the meaning of the verb rely. Similarly, in German, arm may mean poor or an upper limb, in French, pic may mean peak or woodpecker and louer means both hire and praise although the words are unrelated, having different origins (Riemer, 2010:162). There are no predictable cross-language patterns so we have to be alert to the fact that, e.g., since will only be recognised as a time preposition (since last weekend) and not as a subordinating conjunction (He paid since he had the money). 2. Polysemy is likewise not a predictable pattern. The word head in English may mean boss, lead, foam on beer, tip, top etc. but each of these meanings (all of which are related and so polysemous) would require a different word altogether in most languages. In German, the word Leiter is polysemous and has the meaning of leader but also, by metaphorical extension, ladder. Translation is perilous. 3. Verbs present particular difficulties: 1. Those sometimes described as delexicalised, such as make, do, pay, get etc., are unpredictable because patterns of meaning and their concepts will not translate, with the meaning often reliant on the noun with which they collocate. Most such verbs are polysemous insofar as they take their meaning from the noun with which they appear. They made friends and They made the beds are clearly different but related meanings of the verb make and, for teaching purposes, it is often wise to treat them as homonyms rather than polysemes. A wiser approach is not to treat them separately at all but to teach the lexical chunk in which they appear. 2. A verb like paint, in the sense of apply liquid colour to a surface, is, on the face of it, uncomplicated and monosemous. However, there is clearly a cline from the uncomplicated sense to the polysemous uses in the distinction between _paint the house paint a picture of the house_ and the metaphorical extension of, e.g.: he painted a depressing picture of the future of the country. Armed only with the base meaning, a learner may be excused for being confused by the existence of the polysemes. 3. Transitivity also presents difficulties and verbs may alter in meaning depending on whether they are used with or without an object. For example: _She left at 6 She left him her house They managed without much money They managed the compan_ y In addition, when used transitively or intransitively, the nature of the subject will often alter, for example: _I broke the glass The glass broke She ended the activity The activity ended_ and so on. 4. Even verbs which are used transitively may alter meaning in quite subtle ways. For example: The police dusted the window for fingerprints i.e., added dust He dusted the room i.e., removed dust Additionally, She smoked a cigarette and She smoked the cheese are clearly different but related meanings of the verb smoke. And _He ate dinner The acid ate the metal_ are different but connected meanings of the verb eat. 5. The verb mean itself is notorious in this respect. Jackendoff (2012:33 et seq.) identifies that the verb is used at least seven different ways: 1. can be translated as: Fisch in German means fish in English 2. is defined as: Smog means fog and smoke combined 3. is explained as: The flashing light means the hard drive is working 4. is causally linked to: Catching that train means getting up very early 5. intend: I mean to finish this today 6. impact: The fall in the value of the pound means that imports will be more expensive 7. emotional impact: Your help meant a great deal to us Connotation (i.e., the emotional meaning we attach to words) is also unpredictable. For example, pig may have an alternative meaning in English of glutton but that is more to do with cultural and historical issues than with the polysemous nature of the word. ### Presenting the data to our learners Whether we consider that words have a single meaning, two (or more) related meanings or two (or more) wholly unrelated meanings matters. Many very common words in English are polysemous and to deny our learners the data they need to be able to deploy the words accurately would seem perverse. Consider the word play as an example in these sentences: Arsenal are playing tonight. I'm playing football tomorrow. I'm playing John in the final. Who plays the hero? She plays the piano beautifully. Which orchestra is playing? He's playing you for a fool. Which play are you going to see? If you have doubts about the polysemous nature of the word, try the so-called 'and / and-so' test, constructing sentences such as _I am playing darts and John Moscow Dynamo are playing tonight and so is the Royal Philharmonic He played chess and the violin He played a trick on me and a waiter on stage_ If you create a zeugma, that's evidence of polysemy. How would you present this data to learners? Think for a moment and then click here to reveal an idea. Learners, given the example sentences (i.e., the data) can be encouraged to work out the polysemes for themselves and produce a chart like this if they find it helpful. This is a fairly simple example but you can do this at many levels with many lexemes. Try it with, e.g., office or reserve etc. to see what's meant. You may end up with diagrams such as: If you are feeling strong, you can try this with extremely polysemous words in English such as get, set, make, do, but you would be well advised to break the area down into sense units. ### Context and co-text The fact that polysemy is so common, underlines the need to make sure our learners have the data they need to understand lexis. For example, one could present the meaning of wake up as: =to stop sleeping: she woke up at 8 o'clock feeling good That would effectively provide your learners with the base meaning but provide none of the sense of any polysemes. Alternatively, you could use the idea above but extend it with: =to become more active: their first goal made us wake up and play better = become aware of problems: The President told the people it was time to wake up to the danger This doesn't have to be done graphically, of course. It is often just as effective to present alternative meanings (dealing with homonymy or polysemy) in words. For example: Task: Match the meaning of set to the example by drawing a line to connect them. to fix at a place or time (transitive verb)Are we all set? Can we go now? to go down(intransitive verb)he has a full set of Dickens' novels to put(transitive verb)set the clock for 6: we have to leave early to become solid (intransitive verb)there's only a set number of hours we can work without a break a collection(count noun)set it on the table by the door ready(as an ungradable adjective)the sun set slowly into the sea fixed (as an ungradable adjective)the glue has set now so it's safe to move it Focusing learners on word class is helpful. Arguably, by the way, only the notion of set as a collection and set meaning go down are true homonyms of set. The others are polysemes. Some would even see the sense of a collection as derivable from the idea of a fixed number (so polysemy) but it is unlikely that most learners would draw that out for themselves. Whether most learners would make the connection between _: a set number the glue has set_ and set an alarm clock which are all linked to a meaning of fix, without help is also open to question but they can be led there. Noticing metaphorical extensions of meaning or similarities of meaning is an important skill to foster. Unless we understand the nature of homonymy and polysemy, of course, we can't do that. Related guides semanticsfor more on the meaning of meaning syntactical homonymy and gradiencefor the guide to how function words may slip between word classes and also represent different communicative functions within word classes synonymyfor a guide to synonymy and five other related ideas: metonymy, synecdoche, simile, metaphor, hyponymy idiomaticityfor more on (non-)compositionality and transparency ambiguityfor a guide which considers polysemy as a source of ambiguity (and quite a lot more) contextfor more on what it affects and sources of context for teaching References: Jackendoff, R, 2012, A User's Guide to Thought and Meaning, Oxford: Oxford University Press Riemer, N, 2010, Introducing Semantics, New York: Cambridge University Press Schmitt, N and McCarthy, M, 1997, Vocabulary - Description, Acquisition and Pedagogy, Cambridge: Cambridge University Press Contact \| FAQs \| Copyright notice \| ELT Concourse charter \| Disclaimer and Privacy statement\| Search ELT Concourse
14536	https://www.gauthmath.com/solution/K21xrFAO3pM/Find-all-real-values-of-m-for-which-the-inequality-m-x-2-4-x-3-m-1-0-is-satisfie	Question Solution Paso 1: Juan afirma que no considera apropiado que los adolescentes vayan solos de vacaciones, y Pedro le responde acusándolo de querer encerrar a los adolescentes en casa sin vida social. Esta es una falacia conocida como "Hombre de paja", donde se distorsiona el argumento original de Juan para atacar una versión más débil y fácil de refutar. †Answer: Respuesta: La conversación entre Juan y Pedro muestra una falacia de Hombre de paja. Paso 2: En este caso, se está atacando la credibilidad de una persona en base a su conocimiento limitado en un área específica. Es una falacia de "Ataque personal" o "Ad Hominem", donde se desvía la atención del argumento principal hacia la persona que lo presenta. †Answer: Respuesta: La afirmación de atacar a alguien por su falta de conocimiento en un área específica es una falacia de Ataque personal. Paso 3: En este caso, se está utilizando la falacia de "Apelar a la autoridad", donde se argumenta que el tabaco produce cáncer porque el padre de la persona, que arregla tubos de escape de vehículos, lo afirmó. La afirmación se basa en la autoridad del padre en un área no relacionada directamente con la salud. †Answer: Respuesta: La afirmación sobre el tabaco y el cáncer es una falacia de Apelar a la autoridad. Paso 4: En este caso, se está utilizando la falacia conocida como "Apelar a la piedad", donde se intenta convencer a alguien de no terminar una relación apelando a la posible tristeza o sufrimiento que causaría al otro. †Answer: Respuesta: La afirmación sobre sufrir demasiado si terminan la relación es una falacia de Apelar a la piedad. Paso 5: Aquí se está utilizando una amenaza implícita para convencer a alguien, una falacia conocida como "Amenaza implícita". †Answer: Respuesta: La afirmación sobre quemarse en el infierno si no se cree es una falacia de Amenaza implícita. Paso 6: En este caso, se está cometiendo una falacia de "Falsa Causa", al asumir que la alta cantidad de puntajes altos en la PSU de la PUC se debe directamente a una mayor calidad de enseñanza. †Answer: Respuesta: La conclusión sobre la calidad de enseñanza de la PUC basada en los puntajes altos es una falacia de Falsa Causa. Paso 7: Esta afirmación presenta una falacia conocida como "Presunción de inocencia", donde se da por hecho que alguien es inocente solo porque no se ha demostrado su culpabilidad. †Answer: Respuesta: La afirmación sobre la inocencia basada en la falta de pruebas es una falacia de Presunción de inocencia. Paso 8: En este caso, se está utilizando la falacia de "Apelar a la tradición", al argumentar que una receta que se ha seguido por generaciones no puede fallar. †Answer: Respuesta: La afirmación sobre la receta que se ha seguido por generaciones es una falacia de Apelar a la tradición. Paso 9: Aquí se está cometiendo una falacia de " contact@gauthmath.com
14537	https://clinical-laboratory-diagnostics.com/k15.html	Hematology 15.1 Hematopoiesis Lothar Thomas Circulating blood cells play a multifactorial role in the maintenance of organ function. They react to stimuli and adapt their number and function to the requirements of the organism. The lifespan of the individual blood cells in the circulation is shown in Tab. 15.1-1 – Life span and daily turnover of blood cells in the circulation. Changes in the number and function of blood cells can result from diseases of the hematopoietic system or hematopoiesis is activated or compromised due to disorder of an organ or a systemic disease. 15.1.1 Clonal hematopoiesis of indeterminate potential (CHIP) CHIP is a hematologic precursor lesion that is defined by the presence of somatic mutations in peripheral blood cells but without evidence for the presence of leukemia or another hematologic neoplasm. 15.1.1.1 Blood count abnormalities in CHIP The majority of individuals with CHIP show a normal leukocyte morphology. While a slight elevation in the red cell size has been described in some studies. An increased red cell distribution width (RDW) is the most consistent blood count feature found in CHIP. Next generation sequencing based detection of clonal hematopoiesis has become a diagnostic standard in the hematologic workup of patients with suspected myeloid neoplasms. In contrast testing for CHIP is currently not recommended in the general population. In addition to the hematologic progression, CHIP has been associated with a plethora of non-hematologic diseases of which cardiovascular disorders are currently best understood and clinically established /34/. 15.1.2 Red blood cells Red blood cells are comprised of erythrocytes (99%) and reticulocytes (1%). They transport O2 from the lungs to the tissues and CO2 from the tissues to the lungs. The amount of O2 that can be carried per unit volume of blood is determined by the concentration of hemoglobin (Hb) in blood. About 1.34 to 1.39 mL O2 can be bound per gram of Hb. For osmotic reasons Hb is packed into red cells. They perform their function exclusively in the circulation. The human organism contains, on the average, a red blood cell mass of 2–3 liters, and the daily total circulating red blood cell mass is 3,000 liters. Every day some 2.0 × 1011 senescent erythrocytes (approximately 1% of all red blood cells) are removed from the circulation in adults and retained in the spleen. Spleen macrophages degrade the cells, and the components, especially iron, are reused by the bone marrow for the production of new red cells. Every erythrocyte enters the circulation as reticulocyte. Within 1–2 hours of acute onset of hypoxia circulating erythropoietin levels begin to increase and reticulocyte counts can be doubled one to two days following the hypoxia /1/. Red blood cell membrane The red blood cell (RBC) membrane is composed of a lipid bilayer in which approximately 20 major proteins and at least 850 minor ones are embedded /2/. The lipid bilayer acts as a barrier for the retention of cations and anions within the red cells, while it allows water molecules to pass through freely. The maintenance of high intracellular K+ and low intracellular Na+ compared with the corresponding ion concentrations in the plasma involves a passive outward movement of K+, which is pumped back by the action of an ATP dependent Na+/K+ pump in exchange for Na+. A further component of the RBC membrane is the cytoskeleton. This protein network laminates the inner surface of the membrane and contains the proteins spectrin and actin which are connected to each other in two protein complexes; ankrin and protein 4.1 complex. The red cell membrane is attached to the intracellular cytoskeleton by protein-protein and lipid-protein interactions that confer the erythrocyte shape, stability and deformability. During the life span the RBC is forced to cross the pores of splenic sinusoids thounds of times and has an ongoing relationship with the spleen that contributes to remodeling during the first week of its life. In addition the spleen plays the primary role in the removal of aged erythrocytes. Dynamics of red cells RBCs undergo a rapid reduction in volume and hemoglobin (Hb) in the few days after release from the bone marrow. This rapid phase is followed by a much longer period of slower reduction during which the volume and Hb are co regulate. The correlation between volume and Hb content increases as the cells mature from an initial correlation coefficient of approximately 0.40 in the reticulocyte population to approximately 0.85 in the full population /3/. Erythroblast adaption to iron During RBC maturation abundant synthesis of Hb is required. Since iron is essential for Hb synthesis, erythropoiesis consumes a major part of the body iron. The transcription factor Bach 1 is involved in the responses of erythroblasts to iron status. In Iron deficiency Bach 1 depresses the genes of globin synthesis in erythroblasts to balance the levels of heme and globin /4/. The erythroid regulator erythroferrone facilitates iron acquisition e.g., in hemorrhage-induced anemia. Erythroferrone is produced by erythroid precursors in the marrow and the spleen and mediates hepcidin suppression during increased erythropoietic activity stimulated by erythropoietin. Thus erythroferrone increases iron availability by suppressing hepcidin /5/. 15.1.3 Thrombocytes (platelets) Approximately 2 × 1010 thrombocytes per liter of blood are released from the bone marrow daily. The plasma membrane of the platelets contains receptors such as the glycoprotein GPIIb/IIIa receptor and the von Willebrand receptor (GPIb/V/IX). Thrombocytes repair vascular injuries and prevent excessive bleeding. 15.1.4 Leukocytes The function of leukocytes is carried out extra vascularly. Therefore the blood merely serves as a transport way which the leukocyte uses to move from one place to the other /1/. Polymorphonuclear neutrophil granulocytes (PMN) The PMN are mature neutrophil granulocytes and have a segmented nucleus with the segments separated by a filamentous strand. The PMN pool in blood is divided in two compartments of approximately equal size: the circulating pool and cells which adhere to the walls of the small vessels (marginal pool). Upon blood sampling cells of the circulating pool, but not the marginal pool, are determined reliably with the leukocyte count. There is, however, a constant exchange of cells between the two pools. In the tissues the PMN are capable of phagocytosis. Eosinophils have the same morphologic character and maturation sequence as neutrophils. They are filled with orange-colored acidophilic granules that contain basic proteins, eosinophilic peroxidase and other substances. IL-5 mobilizes eosinophils, induces chemotaxis and enhances the production of reactive oxygen radicals. Following a short circulation time in the blood, they participate in inflammatory tissue reactions, stimulated by the immune system, and they are also involved in defense mechanisms against helminth infections. Basophils have the same morphologic character and maturation sequence as neutrophils. Basophils lack mobility and phagocytosis. Their granules are filled with angiotonic substances involved in blood vessel contraction, such as histamine and serotonin. Lymphocytes vary in size from slightly larger than erythrocytes to as large or larger than monocytes. Lymphocytes are broadly classified in T cells, B cells and NK cells and are concerned with innate and adaptive immunity (see Section 21.1.4 – Innate immune response). Their sub populations are determined flowcytometrically. For monocytes, as for PMNs, the blood serves as a street from the bone marrow to the tissues. In blood the monocytes are distributed into a circulating and a marginal pool. Following stimulation, which usually takes place in the tissues, they are transformed into metabolically active macrophages. Their primary functions are the phagocytosis of aging erythrocytes and of microorganisms and the release of inflammatory cytokines for the regulation of the cellular and humoral immune defense. Cell distribution in the bone marrow The hematopoietic cells in the bone marrow are comprised of up to 60% granulocytes (mainly neutrophils), 20% erythropoietic precursors, and 15% lymphocytes, plasma cells, monocytes and megakaryocytes. 15.1.5 Hematopoietic system The hematopoietic system provides the blood with cells, maintains the steady state level of circulating blood cells, and responds to acute challenges. There is a constant turnover of hematopoietic cells as they reach their “use-by date” and are removed from the system /6/. Over a period of 7 years, an adult forms a quantity of blood cells that corresponds to his body weight. The hematopoietic system is organized in a hierarchical manner (Fig. 15.1-1 – Hierarchy of the hematopoietic system with myelopoiesis and lymphopoiesis). Pluripotent stem cells exist primarily in the bone marrow of adults and generate the numerous lineages found in blood. The differentiation of stem cells into blood cell lineages is regulated via a group of hematopoietic growth factors and cytokines. Activation of receptors on the surface of hematopoietic cells by growth factors leads to signalling events which enable each cell to differentiate. Among all of the hematopoietic cells, stem cells have the greatest capacity for self-renewal. Committed progenitor cells are limited to one or two cell lineages and have a low capacity for replication. Growth factors are required for the survival and proliferation of hematopoietic cells at all stages of development /7/. Effective erythropoiesis is the result of the interaction of hematopoietic cells and their supporting stroma. More than 95% of hematopoiesis takes place in the bone marrow; this is the only location where erythropoiesis, granulopoiesis, lymphopoiesis, monopoiesis and megakaryopoiesis proceed simultaneously. Under certain conditions, hematopoiesis can also occur in the spleen. 15.1.5.1 Hematopoietic micro environment Bone marrow stroma provides the micro environment for the hematopoietic cells and influences their proliferation and differentiation. Components of the micro environment are /8/: Fibroblasts, macrophages, adipocytes and accessory cells such as T lymphocytes and monocytes Extracellular matrix (e.g., collagen, laminin, fibronectin, and proteoglycans). The cells of the micro environment affect hematopoiesis, in a positive and negative manner, by the following mechanisms: Direct cell-cell contact; thus, pluripotent stem cells require contact with stromal cells Secretion of proteins which serve to maintain the structure of the extracellular matrix The formation of soluble and cell-associated cytokines. Colony stimulating factors (CSF), interleukins (IL-1, IL-3, IL-6, IL-12), as well as inhibitors such as tumor necrosis factor-α, transforming growth factor-β and interferon-γ are produced. 15.1.5.2 Hematopoietic stem cell The maintenance of blood cell populations is achieved by the proliferation and differentiation of precursor cells located primarily in the bone marrow. The precursor cells are derived from a common hematopoietic stem cell population that is established during embryogenesis and functions for the lifetime of the organism. In addition to its differentiation potential the stem cell has the capacity to produce daughter cells with the same or very similar proliferative and developmental potential as the parental cell /9/. Stem cells are defined as self-renewing cells, capable of forming all blood cell lineages for at least 4 months following transplantation in a recipient mouse /10/. Stem cell division proceeds asymmetrically. One daughter cell retains the pluripotency of the mother cell, while the other is limited (committed) to develop into a cell of a hematopoietic lineage. In adults, stem cells are in a resting stage or of minimal activity. These stages are induced by the pleiotropic hormone transforming growth factor-β (TGF-β) /11/. The hematopoietic progenitor cells are named according to the colony type (colony-forming unit, CFU) that they form in vitro culture /12/. The pluripotent stem cell is called CFU-blast (Fig. 15.1-1 – Hierarchy of the hematopoietic system with myelopoiesis and lymphopoiesis). The CFU-blast can form cells of all hematopoietic cell lineages and comprises 0.05% of the bone marrow cells. The CFU-blast expresses CD34+, but no lineage specific antigen and is, consequently, defined as CD34+DR–lin–. The expression of the HLA antigen DR is one of the earliest differentiation markers and leads from the CFU-blast to the committed progenitor cells. From these, the CFU-GEMM can differentiate into one of the following cell lineages: G, granulocytes; E, erythrocytes; M, monocytes; and M, megakaryocytes /13/. The differentiation of the CFU-blast into a committed progenitor cell (lineage-bound progenitor cells) is subject to control by growth factors. The most important of these is the stem cell factor (SCF), which is produced by the stromal cells. The SCF is responsible in particular for the long term survival of non-dividing, pluripotent stem cells. Pluripotent stem cells are also present in the peripheral blood; the CD34+ cells account for 0.15% of the mononuclear blood cells. Their fraction is increased to 0.6% during a course of chemotherapy, and they are harvested for stem cell transfer by means of apheresis. 15.1.5.3 Hematopoietic growth factors Hematopoietic growth factors are required /7/: For the survival and proliferation of hematopoietic factors at all stages of cell development. Factors that affect multi potential cells are the steel factor, the Fms-like tyrosine kinase 3 (FLT3) ligand, the granulocyte-macrophage colony-stimulating factor (GM-CSF), and IL-2, IL-3 and IL-7. For the development of progenitor cells, that are committed to one or two cell lineages. Lineage-specific hematopoietic growth factors are erythropoietin and thrombopoietin. Stem cell factor The stem cell factor (SCF) is synthesized by the bone marrow stroma. SCF binds to c-kit, a tyrosine kinase receptor localized on stem cells and committed progenitor cells. SCF generates or activates these cells to become colony stimulating factors (CSF). In cell culture, SCF manifests effects that are synergistic with those of the CSF /14/. Lineage-specific hematopoietic growth factors The lineage-specific hematopoietic growth factors are synthesized in the kidneys (erythropoietin), endothelial cells, fibroblasts, macrophages (GCSF), bone marrow stromal cells, and the liver (thrombopoietin). The effects of the lineage-specific growth factors are mediated by receptors of the cytokine receptor super family. The receptors are transmembrane proteins with one or two extracellular binding domains and one intracellular domain, which activates kinases of the Janus kinase (JAK) family. JAKs mediate signal transduction to the cell nucleus via multiple intermediate steps /7/. FMS-like tyrosine kinase 3 gene The FMS-like tyrosine kinase 3 (FLT3) is a transmembrane ligand-activated receptor tyrosine kinase that is normally expressed by hematopoietic stem cells and early myeloid and lymphoid progenitor cells, and is involved in the proliferation, differentiation and apoptosis of hematopoietic cells through various signaling pathways. The FTL3 gene is mutated in 25-30% of patients with acute myeloid leukemia (AML) because of the poor prognosis associated with FTL3- internal tandem duplication mutated AML, allogeneic hematopoietic stem-cell transplantation is commonly performed in first complete remission /32/. 15.1.5.4 Embryonic and fetal hematopoiesis The blood cells are formed in the 3rd–6th week of gestation in the yolk sac, the para aortic splanchnopleura, and the ventral region of the embryo, which is responsible for the aorta-gonad-mesonephros (AGM) development. Soon after their initial development in yolk-sac blood islands, primitive erythroblasts enter the newly formed vasculature. The colonization of the rudimentary liver, which is the principle location of embryonic and fetal blood formation until week 22 of gestation, ensues from these tissues very early on. The colonization of the bone marrow and the spleen takes place at this time. During the second half of pregnancy, the bone marrow becomes the major site of blood cell formation and it retains this function life long. The spleen remains a lymph organ. The first erythropoietic blood cells are primitive erythroblasts in the yolk sac on days 16–19 post-conception. The hemoglobin in these cells contains embryonic globin. The cells also maintain their nucleus for a certain period of time in the circulation. The differentiation of primitive erythroblasts occurs in the circulation through the accumulation of increasing quantities of hemoglobin. From the 8th week of development onwards definitive red blood cells, formed by the liver, gradually replace the primitive erythroblasts. The change is believed to be based upon a dose effect of the EKLF gene /15/. The definitive red blood cells are smaller, they no longer have a nucleus, and the formation of globin is limited to fetal or adult hemoglobin. The synthesis of erythropoietic cells is dependent upon erythropoietin, which is a mitogen as well as a survival factor for erythroid progenitor cells /16/. During the fetal period, the hemoglobin concentration and the hematocrit increase, while the mean corpuscular volume (MCV) decreases. Hematopoietic stem cells continue to self-renew and differentiate into mature blood cell lineages throughout their lifetime. Until a sufficient number of hematopoietic stem cells are produced in the fetal liver to supply the adult-type (definitive) erythrocytes, embryo/fetus-specific erythrocytes, called primitive erythrocytes, which are produced in the extra-embryonic yolk sac tissue, play a pivotal role in transporting oxygen to embryonic and fetal tissue. A committed hematopoietic cell population serves as a common precursor for primitive erythrocytes and lympho-myeloid lineages. Primitive erythrocytes are large in size and enter into circulation in their nucleated form. The erliest hematopoietic progenitors are found in the CD41+CD45– cell stage /16/. It has long been thought that hematopoiesis occurs in two phases (primitive and definitive erythropoiesis) during development, but some findings include that there are intermediate cell populations /16/. Definitive hematopoiesis produces all lympho-myeloid lineages and adult-type erythropoiesis. The leukocytes are initially formed from para aortic splanchnopleure and aorta-gonad-mesonephros cells. The stem cells that have migrated from there are responsible for the formation of B and T lymphocytes, granulocytes and natural killer (NK) cells. These cells are detectable from the 15th week of gestation onwards, and the mean leukocyte count is 0.8 × 109/L. From the 21st week of gestation, the leucocyte count is above 1 × 109/L. 15.1.6 Erythropoiesis Erythropoiesis occurs in three stages: the primitive, the fetal definitive and the adult definitive stages. After the primitive stage, which takes place in the yolk sac, definitive erythropoiesis moves to the fetal liver and the spleen but is finally restricted to the bone marrow, as adult definitve red blood cells, for the rest of the live. After birth, the location of erythropoiesis gradually switches to spongy flat bones, such as ilium, sternum, ribs and cranium, the sites which adults rely on mostly for steady-state erythropoiesis /17/.Extramedullary erythropoiesis results /17/: In the development of erythroblastic islands in other organs and tissues, in particular the spleen and liver In the abundance of erythroid precursor cells co-expressing CD71+. Continuous red cell production is required for the maintenance of the erythrocyte and hemoglobin concentration. Following a period of about 10 days for maturation and differentiation in the bone marrow some 2 × 1011 reticulocytes are released into the blood stream, daily. During maturation of the red cell, a marked accumulation of hemoglobin, making up over 95% of the total red blood cell protein, occurs. The differentiation of erythropoiesis begins at the stage of the pluripotent hematological stem cell. All successive cells have lost the capability of renewing erythropoiesis; only cell division and differentiation remains possible. The development of erythropoietic cells in the bone marrow takes place in the proliferation pool at first and thereafter in the maturation pool (Fig. 15.1-2 – Erythropoiesis develops a proliferation pool and a maturation pool). Proliferation pool In this phase of erythropoiesis, the stem cell acquires the potential of surface receptor expression to acquire signals for proliferation and differentiation. The most immature form of the committed erythroid progenitors is the burst-forming unit erythroid (BFU-E). The BFU-E requires 14 days or more for the formation of a cluster of mature erythroblasts. The BFU-E stage is followed by the colony forming unit erythroid (CFU-E). Both progenitors constitute some 0.3% of the nucleated erythropoietic bone marrow cells. A time period of 7 days is required for the formation of a cluster of 8–64 mature erythroblasts from the CFU-E. Maturation pool CFU-E cells enter this pool and mature to reticulocytes which are released in the circulation and become erythrocytes (Fig. 15.1-2 – Erythropoiesis develops a proliferation pool and a maturation pool). The maturation phase begins with the first progenitor cell that is cytologically distinguishable in the bone marrow, the pronormoblast. It has basophilic cytoplasm, a nucleus with, as a rule, six nucleoli and a diameter of 15–20 μm. As the stage of maturation progresses, the cell diameter and density of the erythropoietin receptors decrease and cellular hemoglobin content increases. Following rejection of the nucleus by orthochromatic erythroblasts the reticulocyte, which remains with the RNA residues, is released into circulation, where it matures within 24–48 hours to an erythrocyte. Erythropoietin (EPO) is an important regulatory hormone of erythropoiesis. It prevents apoptosis of CFU-E cells and induces their clonal expansion. The CFU-E has the highest EPO receptor density of the erythroid progenitor cells. Under stimulation with EPO, CFU-E proceeds through division stages up to orthochromatic erythroblasts which are no longer capable of division. The normal red blood cell achieves a mean cellular hemoglobin concentration (MCHC) of up to 360 g/L. A further increase of 10–20 g/L can take place if the cell loses water or membrane, which occurs only relatively seldom. The hemoglobin concentration of the maturing red blood cell regulates the number of cell divisions that the cell is to undergo. The pronormoblast normally undergoes four divisions, resulting in a total of 16 erythrocytes. With every division the volume of the cells declines. For heme formation see Section 7.1 – Iron metabolism and disorders and Ref. /18/. Changes in erythrocyte volume and hemoglobin concentration can result from: Iron deficiency or disorders of globin synthesis (thalassemia). The hemoglobin content of the erythrocyte is reduced. Normally, after four cell divisions in the maturation pool, the erythroblast is saturated with hemoglobin and the cell nucleus receives a signal to refrain from further cell division. This is not the case in iron deficiency and an additional cell division occurs with a volume reduction of the red cell. Vitamin B12 or folic acid deficiency. In this case the cellular hemoglobin content, which signals the erythroblast nucleus to cease cell division and to implement the expulsion of the nucleus, occurs earlier, resulting in fewer divisions and a larger erythrocyte volume. Hereditary disorders which result in microcytic normochromic anemia (MCHC normal). Examples are the spherocyte and the keratocyte. Hereditary spherocytosis is a disease in which the cell membrane of mature erythrocytes is reduced. Cellular hemoglobin and cell volume are normal, but the cell diameter is decreased. Due to the reduced diameter, the hematocrit is lower than expected, the calculated MCHC is elevated and the erythrocyte is seemingly hyperchromic. Keratocytes that develop in vivo due to the effect of oxidative medication, have a reduction of the cell membrane and of hemoglobin. Erythrocyte hemoglobin precipitates with the development of Heinz bodies, a proportion of the erythrocyte contains less hemoglobin and looks pale in the blood smear. Senescent erythrocytes are removed from the blood stream through phagocytosis by the reticuloendothelial system of the spleen. Senescent erythrocytes show volume deformation and decrease and increase in cell density. Cell membrane changes lead to a loss of carbohydrates on the cell surface. Based on the volume (MCV) and the mean hemoglobin concentration (MCHC) of erythrocytes anemias are classified as: Normocytic and normochromic Macrocytic and hyperchromic Microcytic and hypochromic Microcytic and normochromic. The classification of anemia can never be: Normocytic and hyperchromic Macrocytic and hyperchromic. Regulation of erythroid differentiation Red cell production increases 5 to 7-fold after blood loss or hemolysis, but does not overshoot because it is tightly regulated /19/. The regulation involves proliferation, differentiation and survival of erythroid progenitor and precursor cells. Although the cells of the proliferation pool act in concert each of these processes can be regulated independently of another. Following blood loss or hemolysis, stem cell factor (SCF/kit-ligand) and glucocorticoids increase proliferation in the BFU-E to colony-forming unit-erythroid (CFU-E) stages. However, SCF and glucocorticoids have no influence on the survival and differentiation of these erythroid progenitors. EPO, in turn, supports the maturation of erythroid progenitor cells from CFU-E to basophil erythroblasts, but it influences neither their maturation nor their differentiation. Proliferation, differentiation and survival of erythroid progenitors are controlled through activation and repression of specific genetic programs. An important transcriptional regulator in erythropoiesis is LIM domain-only protein 2 (LMO2). This protein is regulated at both a post-transcriptional and post-translational level in erythroid progenitors and contributes to the precision with which erythrocyte production is controlled. Erythroid precursor cells co-expressing CD71+ and the erythroid lineage marker CD235a (glycophorin A) are cells of extramedullary erythropoiesis (EE). These cells are defined as CD71+ erythroid cells (CEC). CEC are mainly erythroid precursors expressing high levels of CD71, including reticulocytes but excluding erythrocytes. EE may be considered a normal physiological process during pregnancy and in developing newborns. Regarding of the underlying mechanism, EE results in the development of erythroblastic islands in other organs/tissues, in particular the spleen and liver. EE results in an abundance of erythroid precursors or CEC in the periphery. Erythroid precursors have immunosuppressive or immunomodulatory properties and their expression can, therefore, have an impact on the effector functions of various different immune cells /17/. 15.1.7 Granulopoiesis The phagocytic cell lineages of granulocytes and monocytes have a bipodent committed progenitor the colony forming unit granulocyte-monocyte (CFU-GM). Colony stimulating factors (CSFs) such as the granulocyte colony stimulating factor (G-CSF) and the granulocyte macrophage-CSF (GM-CSF) stimulate the proliferation, differentiation and maturation of neutrophil and eosinophil granulocytes and monocytes. The importance of GM-CSF in granulopoiesis is similar to that of erythropoietin in erythropoiesis. Neutrophile production and storage Following the maturation from the stem cell, there is a continuous maturation from the earliest recognizable precursor the myeloblast to the polymorphonuclear neutrophil. In the bone marrow neutrophils and their precursors mature in two pools: The mitotic or production pool that is divided into myeloblasts, promyelocytes, and myelocytes. Four to five cell divisions occur from myeloblast to myelocyte stages, leading to a 32-fold increase in the number of cells. Transition time from myeloblast to myelocyte is 3–9 days; in proliferative stress the rate of granulopoiesis can be elevated by a factor of 20. The post-mitotic maturation and storage pool which is divided in metamyelocytes, bands and segmented neutrophils. In this pool mitosis no longer occurs, but a steady process of maturation and storage of bands and segmented neutrophils is taking place. Upon demand these cells are released into the circulation. Under non-pathological conditions, only post mitotic cells are released. Normally the cells spend about 10 days in the storage and maturation pool. The pool contains about 15–20 times as many granulocytes as are in the blood. Formation of granules Granules start to form at the stage of neutrophil maturation marked by transition from myeloblast to promyelocyte. From here on, formation of granule protein continues even up to the stage of segmented cells. Granules are believed to be formed by aggregation of immature transport vesicles that bud off from the trans-Golgi network (TGN). In TGN sorting takes place between constitutively secreted proteins and proteins that are routed into the regulated secretory pathway (i.e., go to granules) /20/. Stimulation and activation of granulocytes The increase in granulocytes in the circulation can be triggered by the following mechanisms /1, 21/: Demarcation; administration of epinephrine or severe exercise will decrease the proportion of marginated granulocytes and therefore induce neutrophilia. Since demarcation only represents redistribution of existing intravascular pools of granulocytes, it does not increase numbers of cells available to control infectious organisms. Leukocyte egress; increased numbers of maturating band-form neutrophils in the circulation results from the release of a post-mitotic storage pool rather than an increased rate of production of theses cells. Under non-pathological circumstances, only post-mitotic cells are released. Responses to endotoxin and infection occur within minutes to hours. Shift to the left; when an increased ratio of bands to segments is measured in the blood, with or without an increase in more mature neutrophils such as metamyelocytes, it is indicative of accelerated marrow release, usually accompanied by a reduction in size of the storage pool. The increased rate of granulopoieses as in severe inflammation results in a significant increase in post-mitotic cells only after 2–3 days. Neutrophil count remains high; this may be the case in chronic infection. The neutrophil count is in a new steady state of balanced but increased input and output which is maintained by accelerating production. Expression of the Fc-receptor (CD64). In resting neutrophils the level of CD64 expression is relative low (about 1000 molecules/cell), following activation it can increase up to 5 to 10-fold /22/. 15.1.8 Monopoiesis In bone marrow, the principal cytologically detectable cell is the monoblast. Immature cells migrate into tissues and body cavities where they mature into so-called wandering macrophages (pulmonary-alveolar macrophages, macrophages lining sinuses, Kupffer cells in liver, Langerhans’ cells in epidemis) /23/. Monocytes circulate in the blood with a half life time of 1–3 days, and then migrate into the tissue. Monocyte populations in the blood are CD14+CD16– and CD14+CD16+. Cluster of differentiation 64 (CD64): The membrane gylycoprotein CD64 (FC-gamma receptor 1) is found only on macrophages. CD64 binds monomeric antibodies of IgG type with high affinity. Treatment of polymorphonuclear granulocytes with interferone (IFN) gamma can induce CD64 expression on these cells. Some authors /35/ recommend monitoring of neutrophil CD64 as a prognostic marker in septic shock. In these patients CD64 is a valuable marker indicating the course of septic shock. In the initial 48 hours after admission decrease of CD64 concentration can aid in predicting the prognosis of septic patients. 15.1.9 Megakaryopoesis Megakaryopoiesis is a complex, stepwise process that takes place in the bone marrow. Committed progenitor cells, the colony forming unit megakaryocyte (CFU-MK), undergo a number of lineage commitment decisions lead to the production of polyploid megakaryocytes. As of a certain stage CFU-MK cease to proliferate, and endomitosis begins /24/. At this stage they are megaloblasts or immature megakaryocytes. The process of endomitosis involves DNA replication without cell division, and a polyploid cell with a single frequently lobed nucleus develops. In the process of DNA replication the nucleus does not divide and cytoplasm remains intact /25/. Replication usually occurs three times so that the mature megakaryocyte has a ploidy of 16, which, however, is variable and can be as high as 256. Maturing megakaryocytes acquire the competence of forming thrombocytes. This occurs through /24/: Up-modulation of a vast array of cytoskeletal, membrane and granule regulatory proteins Acquisition of stores of ribosomes, α-granules and dense granules Differentiation into cells that are large, polyploid, and have a cytoplasm filled with a complex system of interconnected cytoplasmic membranes (demarcation membrane system; DMS). The demarcations marks preformed platelet regions, also termed pro platelets. In the bone marrow sinusoids, pro platelets are separated from the cytoplasm by shear forces. One megakaryocyte can release 1,000–5,000 platelets. The pro platelet formation is a terminal process, and once has been completed the residual megakaryocyte nucleus in engulfed by macrophages within the bone marrow. Megakaryopoiesis and formation of thrombocytes is regulated by thrombopoietin (TPO), but other cytokines such as IL-6 and IL-11 also play a role /26/. The latter make it possible for the megakaryocyte to adapt to regions that permit maturation and thrombocyte formation. The effect of TPO on megakaryocytes and thrombocytes is mediated by the c-Mpl (CD110) receptor; however, only a low number is present on the thrombocytes. TPO is synthesized in constant amounts by the liver and its concentration is regulated to a certain extent by binding to its thrombocyte receptor and thus by the number of thrombocytes as well. If the blood thrombocyte count is high, increased amounts of TPO bind to c-Mpl and only small quantities bind to megakaryocytes; the result is down-regulation of megakaryopoiesis. If, on the other hand, the thrombocyte count is low, then plasma TPO concentration will be high and megakaryopoiesis is stimulated. Von Willebrand factor receptor The receptor GPIb/V/IX is an important regulator of pro platelet formation. In immune thrombocytopenia, autoantibodies develop and inhibit the release of pro platelets. Bernard-Soulier syndrome: in this hereditary autosomal recessive macro thrombocytopenia, defined platelet regions that are marked by the demarcation membrane system are larger than normal /31/. 15.1.10 Splenectomy Thromboembolism and pulmonary hypertension constitute frequently occurring complications (Odds ratio 4.08) in patients with transfusion-dependent thalassemia and can be associated with splenectomy. Thalassemias are congenital autosomal recessive hemoglobinopathies. The patients are unable to produce enough hemoglobin to survive without red blood cell transfusions. The two primary types of the disease, alpha- and beta-thalassemia, are distinguished by decreased or absent synthesis of either alpha-globin chains or beta-globin chains of the hemoglobin molecule. Nowadays thalassemias are differentiated as follows /36/: Beta thalassemia minor: asymptomatic microcytic anemia. Thalassemia syndromes: they are classified depending on clinical severity and transfusion requirement. The patients need maintaining a pre-transfusion level of hemoglobin of 95–105 g/L which suppresses erythropoiesis. Despite advances in therapeutic care, splenectomy is a treatment option in some patients. Splenectomy is related to major long-term complications (thromboembolic, pulmonary hypertension, and infections). The spleen normally removes damaged red blood cells (RBCs). In the absence of spleen negatively charged RBCs and hyperactive thrombocytes contribute to the hypercoagulable state of thalassemia, as these cells stay in blood circulation and activate thrombin production. 15.1.11 Therapeutic measures that compromise hematopoiesis Refer to Tab. 15.1-4 – Therapeutic measures that compromise hematopoiesis. 15.1.12 Hematopoietic equilibrium In normal hematopoiesis there is an equilibrium between the cell mass for each cell lineage in the blood, and the tissues, and their formation in bone marrow. There are three important variations that disturb this equilibrium: Hypo proliferation of the marrow or of one or more cell lineages. In this case either the marrow is incapable of reacting adequately to the increased stimulatory effect of growth factors (intrinsic marrow disorder), or substances that are important for the formation and maturation of cells are lacking (e.g., iron deficiency). Ineffectiveness of one or more cell lineages. Hematopoiesis is hypo proliferative due to the inhibitory effect of inflammatory cytokines. This is the case in anemia of chronic disease. Bone marrow failure, defined as the inability of hematopoiesis to meet the physiologic demands for the production of a sufficient number of functional blood cells is classified based on presumed etiology in inherited, secondary or idiopathic. An approach to evaluating patients presenting for the presence of inherited bone marrow failure syndromes is shown in Ref. /30/. Hyper proliferation of the marrow or of one or more cell lineages. This results in an intensified hematopoietic response to the demand. The compensatory capacity of the marrow is often inadequate to fulfill the requirements and cytopenia results. An example is thrombocytopenia in disseminated intravascular coagulation. Primary disorders of hematopoiesis A disease of one or more hematopoietic cell lineages is present (e.g., in leukemia, thalassemia, or pancytopenia). The differential diagnosis of pancytopenia is shown in Table 15.1-3 – Differential diagnosis of pancytopenia. Secondary disorders of hematopoiesis Hematopoiesis is compromised because of a local organ disease (splenomegaly) or a systemic disease (inflammation). The consequence is diminished or flawed blood cell formation or a reactive hematopoietic response with augmented synthesis and overproduction of precursors. Disorders may be caused by: Important substances that are deficient (iron, folic acid, vitamin B12, vitamin B6) A reactive response of the marrow (granulocytosis, granulocytopenia, thrombocytosis, thrombocytopenia, anemia) in systemic disease of the organism A reactive response according to hypoxia (polycythemia at high altitudes of above 3,000 meters). Secondary disorders of hematopoiesis are often complex and require special hematological and biochemical tests in addition to a blood count for clarification. 15.1.13 Investigation of hematopoiesis For the investigation of hematopoiesis, basic tests are distinguished from functional tests. Basic tests are descriptive measures of the condition (e.g., Hb value or leukocyte count). Functional tests describe hematopoietic function in response to a compromising situation (e.g., reticulocytosis in blood loss or a left shift of granulopoiesis in sepsis). 15.1.14 Basic hematologic tests Basic hematological tests for the detection of hematopoietic disorders are performed using hematology analyzers: Complete blood count Partial or complete leukocyte differentiation Reticulocyte count. Some analyzers also measure, in addition, reticulocyte indices like reticulocyte Hb content (CHr, RetHe), reticulocyte RNA content and the proportion of hypochromic red cells (%HYPO). Blood smear. This test is required as a supplementary investigation in symptomatic patients and if the hematology analyzer provides a signal indicating the presence of atypical cells. Important hematological tests are shown in Tab. 15.1-2 – Hematological investigations for assessment of the hematopoieses. Flow cytometric measurement of blood cell surface markers using specific antibodies (immunophenotyping). 15.1.15 Dermatoses associated with hematological malignancy Cutaneous manifestations of hematologic malignancy supports the concept of molecular defects associating with specific clinical features /37/. Neutrophilic or eosinophilic dermatoses are classified according to their clinico-pathological picture into superficial/epidermal, dermal, and deep forms, with a forth category that encompasses mixed as well as syndromic neutrophilic dermatoses. Neutrophilic and eosinophilic dermatoses associated with hematological malignancy represent a heterogenous group of skin conditions, which may either parallel the course of the underlying hematological disorder or be independent from it. Proposed pathophysiology of dermatosis of hematological malignancy is that leukemic cells may promote systemic inflammation, resulting in the clinico-pathological picture of Sweet syndrome, a rare neutrophilic dermatosis. Nelson DA. The biology of myelopoiesis. Clin Lab Med 1990; 10: 649–59. Andolfo I, Russo R, Gambale A, Iolascon A. New insights on hereditary erythrocyte membrane defects. Haematologica 2016; 101: 1284–94. Higgins JM, Mahadevan L. Phyiological and pathological population dynamics of circulating human red blood cells. PNAS 2010; 107: 20587–92. Kobayashi M, Kato H, Hada H, Itoh-Nakadai A, Fujiwara T, Muto A, et al. Iron-heme-Bach1 axis is involved in erythroblast adaption to iron deficiency. Haematologica 2017; 102: 454–65. Kautz E, Jung G, Du X, Gabayan V, Chapman J, Nasoff M, et al. Erythroferrone contributes to hepcidin suppression and iron overload in a mouse model of β-thalassemia. Blood 2015; 126: 2031–7. Valent P, Büsche G, Theurl I, Uras IZ, Germing U, Stauder R, et al. Normal and pathological erythropoiesis in adults: from gene regulation to targeted treatment. Haematologica 2018; 103 (10) 1593–1603 Kaushansky K. Lineage-specific hepatopoietic growth factors. N Engl J Med 2006; 2006; 354: 2034–45. Mayani H, Guibert LJ, Janowska-Wieczorek A. Biology of the hemopoietic microenvironment. Eur J Haematol 1992; 49; 225–33. Graham GJ, Wright EG. Hemapoietic stem cells: their heterogeneity and regulation. Int J Exp Path 1997; 78: 197–218. Orlic D, Bodine D. What defines a pluripotent hematopoietic stem cell (PHSC): will the real PHSC please stand up? Blood 1994; 84: 3991–4. Fortunel NO, Hatzfeld A, Hatzfeld JA. Transforming growth factor-β: pleiotropic role in the regulation of hematopoiesis. Blood 2000; 96: 2022–36. Huss R. Applications of hematopoietic stem cells and gene transfer. Infusionsther Transfusionsmed 1996; 23: 147–60. Migliaccio AR, Vannucchi AM, Migliaccio G. Molecular control of erythroid differentiation. Int J Hematol 1996; 64: 1–29. Ashman LK. The biology of stem cell factor and its receptor c-kit. IJBCB 1999; 31: 1037–51. Isen J, Fraser ST, He Z, Zhang H, Baron MH. Dose-dependent regulation of primitive erythroid maturation and identity by the transcription factor Eklf. Blood 2010; 116: 3972–80. Yamane Toshiyuki. Cellular basis of embryonic hematopoiesis and its implications in prenatal erythropoiesis. Int J Mol Sci 2020; 21: 9346. doi: 10.3390/ijms21249346. Elahi S, Mashhouri S. Immunogical consequences of extramedullary erythropoiesis:immunoregulatory functions of CD71+ erythroid cells. Haematologica 2020; 105 (6): 1478–83. Chiabrando D, Mercurio S, Tolosano E. Heme and erythropoieses: more than a structural role. Haematologica 2014; 99: 973–83. Brandt SJ, Koury MJ. Regulation of LMO2 mRNA and protein expression in erythroid differentiation. Haematologica 2009; 94: 447–8. Borregard N, Cowland JB. Granules of the human neutrophilic polymorphonuclear leukocyte. Blood 1997; 89: 3503–21. Jagels MA, Hugly TE. Mechanisms and mediators of neutrophilic leukocytosis. Immunopharmacology 1994; 28: 1–18. Hoffmann JJML. Neutrophil CD64: a diagnostic mar- ker for infection and sepsis. Clin Chem Lab Med 2009; 47: 903–16. Ziegler-Heitbrock HWL. Definition of human blood monocytes. J Leukoc Biol 2000; 67: 603–6. Geddis AE. The regulation of proplatelet production. Haematologica 2009; 94: 756–9. Cazzola M. Molecular basis of thrombocytosis. Haematologica 2008; 93: 646–8. Kaushansky K. Thrombopoietin: understanding and manipulating platelet production. Annu Rev Med 1997; 48: 1–11. Gulati GL, Hyun BH. The automated CBC. Hematol Oncol Clin North Am 1994; 8: 593–603. Krause JR. The automated white blood cell differential. Hematol Oncol Clin North Am 1994; 8: 605–16. Israels LG, Israels ED. Mechanisms in hematology. Winnipeg 1996; The University of Manitoba. West AH, Churpek JE. Old and new tools in the clinical diagnosis of inherited bone marrow failure syndromes. Hematology 2017; 1: 79–87. Veninga A, De Simone I, Heemskerk JWM, ten Cate H, van der Meijden PEJ. Clonal hematopoietic mutations linked to platelet traits and the risk of thrombosis or bleeding. Haematologica 2020; 105 (8): 2020–31. Bazarbachi A, Bug G, Baron F, Brissot E, Ciceri F, Dalle IA, Döhner H, et al. Clinical practice recommendation on hematopoietic stem cell transplantation for acute myeloid leukemia patients with FTL3-internal tandem duplication: a position statement for the Acute Leukemia Working Party of the European Society for Blood and Marrow transplantation. Haematologica 2020; 105 (6): 1507–16. Leaf RK, Dhami RS,Chun NS,Ta R. Case 11-2022: An 80-year-old woman with pancytopenia. N Engl J Med 2022; 386 (15): 1453–61. Hoermann G. Clonal hematopoiesis of indeterminate potential: clinical relevance of an incidental finding in liquid profiling. J Lab Med 2022; 44 (4): 301–310. Pham HM, Nguyen DLM, Duong MC, Phan XT, Tran LT, Tan LT, et al. Neutrophil CD64 – a prognostic marker of sepsis in intensive care unit: a prospective cohort studx. Front Med 2023. doi: 103389/fmed.2023.1251221. Tsampika-Vasileia K. Splenectomy is significantly associated with thrombosis but not with pulmonary hypertension in patients with transfusion-dependent thalassemia: a meta-analysis of observational studies. Front Med 2023. doi: 10.3389/fmed.2023.1259785. Maronese CA, Derlino F, Moltrasio C, Cattaneo D, Lurio A, Marzano AV. Neutrophilic and eosinophilic dermatoses associated with hematological malignancy. Front Med 2023. doi: 3389/fmed.2023.1324258. Gonzalez-Lopez T, Jarque I, Siniscalchi C, DI Micco P. Editorial: Infectious diseases and hematology. Frontiers Med 2024. doi: 10.3389/fmed.2024.1385874. 15.2 Erythrocytes (cell count and indices) Lothar Thomas The red blood cell (RBC) count is a basic examination for the evaluation of erythropoietic disorders. RBCs are investigated in further detail by measuring the hemoglobin concentration, the mean cell volume (MCV)of the erythrocyte and the red cell distribution width (RDW). Based on the measured RBC count, the MCV, and the hemoglobin concentration, hematology analyzers calculate the following parameters: Hematocrit, also referred to as packed cell volume Mean cell hemoglobin (MCH) Mean cell hemoglobin concentration (MCHC). MCV, MCH, and MCHC are referred to as RBC indices and serve to describe RBC changes and to differentiate erythropoietic disorders. A sensitive marker is the determination of the proportion of hypochromic red blood cells (%HYPO). 15.2.1 Erythrocyte count The main significance of the RBC count is based on the measurement of erythrocyte indices such as the hematocrit, MCV and MCHC. 15.2.1.1 Indication In combination with erythrocyte indices: Differentiation of anemia Diagnosis of polycythemia and polyglobulia. 15.2.1.2 Method of determination Current major analytic methods are variations of the flow cytometry using electrical resistance or impedance measurements. The methods incorporate laser technology (flow cytometry) or cytochemical techniques or both /1/. 15.2.1.2.1 The automated complete blood count (CBC) Principle: analysis of the CBC begins with the automatic aspiration of a specific amount of well-mixed blood specimen. The specimen is automatically aliquoted, diluted with appropriate diluents with or without added red cell lysing agent, and channeled to specific counting chambers, baths or flow cells for cell counting and sizing and to a spectrophotometric cuvette for hemoglobin determination/2/. Light scattering technology A specific amount of diluted blood sample (cell suspension) is hydrodynamically focused in a flow cell, which is illuminated by a narrow beam of light. As a cell passes through the illuminated area of the flow cell, it scatters light. The light scattered by each cell is detected by a photo detector and converted to an electrical pulse, whose amplitude is proportional to the cell volume. The number of pulses generated is proportional to the number of cells passing through the illuminated area. The measurement of scattered light at two angles (2.5° to 3.5° and 5 to 15°) in the forward direction permits the simultaneous determination of the red cell and platelet count, the size and Hb content of red cells and the differentiation of leukocytes /2/. The red blood cell count (RBC) and the platelet (PLT) count are obtained from the number of pulses generated within a predetermined size range (PLT 0–20 fL, RBC 30–180 fL). The MCV and CHCM (comparable to MCHC) are derived from histograms of the red blood cell volume and the Hb concentration (Fig. 15.2-1 – Erythrogram). The HCT, MCH and MCHC are calculated by formulas. The white blood cell count (WBC) is obtained from another cell suspension in which red blood cells have been lysed and leukocytes are fixed and stained for enzymes like myeloperoxidase. This stained cell suspension is hydrodynamically focus in another flow cell, which is illuminated by a tungsten-halogen light source. As the cells pass through the illuminated area, the light scatter as well as the light absorption are measured. The data obtained are displayed in a cytogram along with leukocyte count and differentiation. See Fig. 15.2-2 – Histogram of red cell distribution width. Impedance technology A specific amount of diluted blood sample flows through a small aperture located between two sensing electrodes. As each cell passes through the aperture, a momentary increase in impedance is recorded in the form of an electrical pulse. The amplitude of the pulse is proportional to the cell volume. The number of pulses generated is proportional to the number of cells passing the aperture /2/. The RBC and the platelet count, although performed on the same cell suspension, are determined separately by analyzing the number of pulses of cell size > 36 fL (red cells) and 2–20 fL (platelets). The HCT, MCH and MCHC are calculated by formulas. For hemoglobin determination red cells are lysed in a specific lysing bath and measured by the spectrophotometric cyanmethemoglobin method. The leukocyte count and differential are based on the principle of cell counting and sizing by detection and measurement of changes in electrical resistance as cells pass through an aperture between two electrodes suspended in a conductive diluent. The different cell types are distinguished electronically by the pulses they generate. Using a specific reagent that results in differential shrinkage of leukocytes, these cell types may be differentiated and expressed on a histogram /1/. 15.2.1.3 Specimen EDTA blood: 1 mL Capillary blood (heparinized) from finger tip or heel. 15.2.1.4 Reference interval Refer to references /2, 3, 4/.and Tab. 15.2-1 – Erythrocyte reference intervals. 15.2.1.5 Clinical significance The RBC count as a single parameter is of little diagnostic value. An evaluation of the red cell mass of the body (i.e., the differentiation between erythrocytopenia, erythrocytosis or a normal RBC mass) can only be achieved in combination with the hematocrit. The reason for the minimal significance of the RBC count is that changes in the plasma volume are reflected by the RBC count (e.g., as observed during pregnancy or in the case of disorders of water and electrolyte balance). For instance, in strenuous physical activity, depending on the intensity and the duration of the exercise, the RBC count, the hematocrit, and the hemoglobin concentration rise by 10–30% as compared to the baseline values. The red cell mass of healthy individuals determined with radioactively labeled autologous erythrocytes is 30–36 mL/kg body weight and the plasma volume, determined with radioactively labeled albumin, 33–39 mL/kg body weight /6/. The data are not valid in cachectic or obese individuals. In routine diagnostic investigation, the erythrocyte number is useful essentially for validating the credibility of the Hb value and the hematocrit. Thus, according to a rule the following relation is valid in cases with a normal blood count and in normocytic normochromic anemia: Erythrocyte count (106/μL) × 3 = Hb value (g/dL) × 3 = hematocrit (%) The rule assumes that the erythrocyte hemoglobin content changes linearly with erythrocyte volume, but this is only the case in normal erythropoiesis. Discrepancies are the rule with pathological changes such as: Iron deficiency anemia. Hypochromic erythrocytes are deformed, consequently the cell volume and the hematocrit are underestimated in impedance measurement. Beta thalassemia. These patients generally have mild anemia, elevated red cell count, normal or slightly reduced hematocrit, and decreased MCV. Macrocytic anemia caused by folic acid or vitamin B12 deficiency, alcoholism and chronic liver disease Hereditary spherocytosis with elevated hemoglobin content in relation to the MCV of the erythrocyte Interference with hemoglobin determination due to hyperlipidemia Cold agglutinins. The erythrocyte number is underestimated because of red cell aggregation. Marked hemolysis of the sample. The erythrocyte number is inappropriately low in comparison to the hemoglobin concentration. 15.2.1.5.1 Anemia Acute anemia (e.g., due to acute hemorrhage) is hardly recognizable by a decrease in the RBC count or the hematocrit during the first 24 h because RBC and plasma are lost in an equal ratio. A decline in the RBC count does not occur until fluid shifts occur to correct the blood volume deficit created by the blood loss. In chronic anemia, the blood volume is usually normal (i.e., decline in red cell mass and increase in the plasma volume). The RBC count and the hematocrit are usually reduced. However, in the case of pronounced microcytosis (e.g., severe iron deficiency anemia or thalassemia) the RBC count may be within the reference interval because of increased compensatory red cell production. The number of RBC can be also elevated in hereditary spherocytosis. See Section 15.4 – Hematocrit. 15.2.1.5.2 Cancer-associated anemia Use of erythropoiesis-stimulating agents (ESAs) to manage anemia rises hemoglobin levels and reduces the need for red blood cell transfusions. ESA therapy is suspected of causing thrombosis The ESMO and the ASCO/ASH clinical practice guidelines recommend /27, 28/: ESMO: ESA therapy is recommended in patients with symptomatic anemia who receive chemotherapy and present with an Hb level < 10 g/dL ASCO/ASH: ESAs may be offered to patients with chemotherapy-associated anemia whose cancer treatment is not curative in intent and whose hemoglobin has declined below 10 g/dL. 15.2.1.5.3 Polycythemia Conditions with an elevated red blood cell count are termed polycythemia or erythrocytosis. Absolute polycythemia, which is based on an expansion of erythropoiesis in the bone marrow, is distinguished from relative polycythemia, which is contingent upon a reduction in plasma volume. Polycythemias can be congenital or acquired. In secondary polycythemia, (erythrocytosis) increased erythrocyte formation results from augmented erythropoietin synthesis in hypoxic patients. Congenital polycythemia is rare and in newborns it is associated with a hematocrit of above 0.65 and hyper viscosity syndrome. Polycythemia, which is not to be mistaken for polycythemia vera, is present if the red blood cell mass is greater 25% of the expected value. In addition to the erythrocyte number, the hemoglobin level and hematocrit are elevated. For detailed information on polycythemia see Section 15.4 – Hematocrit. Mild polycythemia, which results from an increase in red blood cell mass and a reduction in plasma volume are observed in smokers. 15.2.1.6 Comments and problems Anticoagulant EDTA is recommended at 1.5–1.8 mg/mL of blood /7/. Among the EDTA salts, K3EDTA is the least suitable since it causes: The strongest shrinkage of the RBCs as the EDTA concentration rises The biggest time dependent MCV decline during the time interval between blood collection and automated measurement of the complete blood count. K2EDTA appears to cause only mild shrinkage while Na2EDTA, in comparison, leads to mild swelling /7/. A rise in the EDTA concentration generally results in a decline in the MCV. Sampling If the blood is collected from an individual in a sitting position after a change from an upright position that lasted for at least 15 min., the RBC count is 5–10% higher than after at least 15 min. in a supine position /7/. A venous occlusion time of more than 2 min. results in a rise of the cell count by an average of 10%; this corresponds to a 1–2-fold increase above the maximally tolerable imprecision. Blood collection immediately after strenuous physical activity is associated with an increase of the RBC count by about 10%. All of these changes are due to hemoconcentration. In comparison to venous blood collection in skin puncture specimens in children the red blood cell count is increased by 6% and in adults by 1,8% /8/. Interfering factors Cold agglutinins: cold agglutinins of a high titer lead to the aggregation of erythrocytes if the sample is stored at room temperature. Measuring the automated complete blood count, the RBC count is falsely low and the MCV is too high. Consequently, the calculated hematocrit is too low and the MCH as well as the MCHC are too high /9/. White blood cells: these cells are counted as part of the red cell count. The proportion is negligible as long as the WBC count is normal. If the WBC count is high as in chronic myeloid or lymphoid leukemia, the WBC count must be subtracted from the RBC count. Platelets: large platelets (e.g., essential thrombo-cythemia) are also counted as part of the red cell count. Intraindividual variation Within day variation 4%, day-to-day variation 5.8%, month-to-month variation 5.0% /10/. Stability At room temperature (20 °C) and at 4–8 °C 3 days, at 37 °C 36 h; after that continuous decrease /11/. 15.2.2 MCV, MCH, MCHC, RDW, %HYPO The evaluation of RBC is accomplished by additional measuring or calculating the following indices /6/: MCV (mean cell volume) of red cells. The MCV is expressed in femtoliters (fL = 10–15 L), and is either directly measured by the hematology analyzer or calculated as follows: MCH (mean cellular hemoglobin) of the RBC. MCH is expressed as pg/RBC and is calculated by the hematology analyzer according to the following equation: MCHC (mean cellular hemoglobin concentration). MCHC is expressed as red cell hemoglobin (g/dL or g/L) and is calculated as follows: RDW (red cell distribution width). The distribution of the MCV in a sample is shown in the form of a graph (Fig. 15.2-1 – Erythrogram). Some hematology analyzers also generate an erythrogram (the RBC hemoglobin content is plotted against their volume) (Fig. 15.2-2 – Histogram of red cell distribution width (RDW)). %HYPO, the proportion of hypochromic RBCs with a hemoglobin concentration of less than 280 g/L, expressed in % of the total RBCs. Based on the MCV, anemia is classified as normocytic, microcytic or macrocytic; and based on the MCH , as normochromic and hypochromic. Hyperchromic anemia does not exist. 15.2.2.1 Indication Differentiation and monitoring of anemia. Early diagnosis of anemia (%HYPO). 15.2.2.2 Method of determination MCV (see also Section 15.2.1.2 – Method of determination) Impedance technology: when the red blood cell flows through the aperture of the sensing electrodes, a sudden increase in impedance is recorded in the form of an electrical pulse. The amplitude of the pulse is proportional to the cell volume. Light scattering technology/2/: a specific amount of diluted blood sample is hydrodynamically focused in a flow cell that is illuminated by a laser or mercury-halogen light. The light that is scattered by every red blood cell in the flow cell is converted to an electrical pulse whose amplitude is proportional to the cell volume. The measurement of scattered light at two angles in the forward direction permits simultaneous determination of the size and hemoglobin content of the red cell. Proportion of hypochromic red cells (%HYPO) Determination can only be performed by hematology analyzers that transform the red blood cell to a sphere. The hematology analyzer is capable of separately measuring red cell corpuscular volume, corpuscular hemoglobin concentration (CHCM) together with the MCV. The proportion of red blood cells with a hemoglobin concentration of less than 280 g/L is recorded and expressed as %HYPO. The %HYPO play a prominent role in the assessment of iron deficient erythropoiesis. In addition the analyzer determines the fraction of microcytic (% MICRO) and macrocytic (% MACRO) red blood cells /12/. 15.2.2.3 Specimen EDTA blood: 1 mL 15.2.2.4 Reference interval See references /2, 3, 5, 13, 14/ and Tab. 15.2-2 – Reference intervals of erythrocyte indices. 15.2.2.5 Clinical significance The erythrocyte indices MCV, MCH, MCHC, RDW are important tools for /15/: Classification of anemia Detection of latent anemia Etiological clarification of anemia. The MCV should be estimated along with the RDW. Thus, in microcytic anemia, an elevated RDW is indicative of iron deficiency anemia, while a normal RDW suggests thalassemia. 15.2.2.5.1 MCV Determination of MCV supports the diagnostically important classification of anemia into normocytic, microcytic and macrocytic forms. It must be noted that the MCV is an averaged value, and that small populations of microcytic or macrocytic erythrocytes cannot be recognized. MCV depends on the hemoglobin content and hydration of the erythrocyte (i.e., plasma osmolality). Markedly hypochromic erythrocytes are subject to more severe deformation than those with normal Hb content and, in consequence, the MCV is underestimated. Normal MCV The MCV is normal if: The majority of the RBCs have a cell size within the reference interval; this is associated with an RDW that does not exceed 15 fL Both large and small RBCs are present; this is associated with a RDW that is > 15 fL. Such a situation occurs, for example, in immune hemolytic anemias (IHA) or in a microangiopathic hemolytic anemia. In IHA the simultaneous presence of microspherocytes which are smaller than normal RBCs and reticulocytes as well as polychromatic RBCs which are larger than normal RBCs may cause a mean of the MCV to be within the reference interval. In an increased RDW, a peripheral blood smear should be examined which, in this situation, will demonstrate anisocytosis, polychromasia, and microcytosis. In disseminated intravascular coagulation, fragmentocytes that are counted as small RBCs and large polychromatic RBCs may be present but the MCV may still be within the reference interval. Decreased MCV Microcytosis may be caused in iron, copper and vitamin B6 deficiency as well as in hereditary conditions. The most common cause is iron deficiency. Because of iron deficiency, the erythropoiesis undergoes one or more cell divisions as normal and with each cell division the RBC size becomes smaller. The RDW is > 15 fL, the peripheral blood smear shows microcytosis and anisocytosis. An increase in the RDW is an early sign of iron deficiency anemia. Hereditary sideroblastic anemia is a rare microcytic anemia, commonly with a markedly decreased MCV of below 60 fL. It is often misdiagnosed as thalassemia, particularly thalassemia inter media. Increased MCV Macrocytosis occurs due to the following causes: In regenerative anemia (e.g. supplementation of deficiency-induced anemias) Smoking, liver cirrhosis Alcoholism; the mean increase is about 5 fL i.e., 5–10% above the mean value of healthy controls. A cutoff level ≤ 96 fL is considered to be a threshold of non chronic alcoholism. MCV normalization is to be anticipated 3–4 months after abstention from alcohol; a cutoff of ≤ 94 fL is taken as the threshold for the assessment of alcohol abstinence. Due to its long half-life time, MCV is unsuitable as a clinical control marker for alcohol withdrawal /16/. Chronic liver disease. Some 20% of patients with non-alcoholic liver disease have macrocytosis that cannot be accounted for by folic acid or vitamin B12 deficiency or reticulocytosis due to bleeding. Vitamin B12 or folate deficiency. These deficiencies impair DNA synthesis which reduces the rate of nuclear replication and mitosis, resulting in fewer cellular divisions during RBC development. The decrease in divisions causes RBCs to be larger than normal. The rate of hemoglobin synthesis is not impaired. A normal MCV does not rule out a vitamin B12 or folate deficiency /17/. A major reason is that some 20% of vitamin deficient patients are also iron-deficient. Reticulocytosis. Depending on the regenerative erythropoietic response, the volume of reticulocytes is 3–10% greater than that of erythrocytes. Since the MCV measurement includes reticulocytes, a rise in the MCV of erythrocytes is to be expected in reticulocytosis of about 15% or more. Myelodysplastic syndrome, hereditary stomatocytosis. 15.2.2.5.2 MCH Generally, the erythrocyte volume is provided with 95% of the maximum possible content of Hb and, consequently, MCH correlates with MCV in most types of anemia. Accordingly, microcytosis corresponds to hypochromia, normocytosis to normochromia. Normal MCH Normal MCH is typical of healthy individuals but is often also observed in acute hemolytic anemia and in anemia of chronic disease (e.g., infection, inflammation, chronic liver disease, and malignant tumors). Low MCH Low MCH indicates a decrease in the hemoglobin content of the red cells and is, for example, typical of iron, copper and vitamin B6 deficiency. High MCH Conditions with a high MCH also cause an increase in the MCV. This is mainly the case in macrocytic anemias, (e.g., due to folate and vitamin B12 deficiency) as well as in regenerative anemias (e.g., as observed during the supplementation of iron deficiency anemia). 15.2.2.5.3 MCHC The MCHC is an estimate of the hemoglobin concentration of the red blood cell. Because of the comparable behavior of cell size and hemoglobin content of the individual red cell, the MCHC remains constant in many hematopoietic diseases. Hemogram abnormality related to red blood cell count, Hb concentration, MVC or hematocrit leads to abnormal calculated red blood cell indices, especially MCHC. Normal MCHC In many types of anemia the MCHC is within the reference interval. Decreased MCHC A decreased MCHC may occur in deficiency-induced anemias (e.g., iron, copper and vitamin B6 deficiency). If falsely low hemoglobin concentrations or a falsely elevated hematocrit are measured, the MCHC is also decreased. Increased MCHC An increased MCHC is found in the presence of cold agglutinins at a high titer and in hereditary spherocytosis. RDW The RDW allows evaluation of whether the RBC of a patient are isocytic or anisocytic. Very high RDW values are measured in acute hemolytic anemias and are a sign of an underlying reticulocytosis. Tab. 15.2-3 – Classification of anemia based on MCV and RDW shows the classification of anemias, dependent on the RDW and the MCV. According to the results of the Third National Health and Nutrition Examination Survey 1988–1994 (NHANES III), RDW is an indicator of mortality risk in middle-aged individuals and in elderly. The mortality risk in individuals without carcinoma or cardiovascular disease, but with an RDW of over 14.05%, was twice as high as in those with an RDW of below 12.6% /19/. 15.2.2.5.4 %HYPO Determination of the proportion of hypochromic red cells is a more sensitive marker than erythrocyte indices for detection of hypochromic red cells and thus iron-restricted erythropoiesis. A decrease in erythrocyte hemoglobin content of up to 10% does not lead to a significant change in MCV, MCH or MCHC. The determination of %HYPO has a detection limit of 2% with acceptable precision. Incipient erythrocyte iron deficiency is, therefore, recognized earlier than by the determination of erythrocyte indices. This is also the case with monitoring of iron supplementation. A response is indicated with the decline in %HYPO after 2–3 weeks /20/. In individuals with low ferritin levels, the iron stores are empty. In this situation, a normal %HYPO indicates that iron-restricted erythropoiesis has not, as yet, occurred, and that the iron supply for erythropoiesis is still at a level that enables normal hemoglobin synthesis (latent iron deficiency, iron deficiency without anemia). Patients with chronic kidney disease should have a balanced iron metabolism, particularly under treatment with erythopoiesis stimulating agents (ESA). The European Best Practice Guidelines for the Management of Anemia in Patients with Chronic Renal Failure recommend a minimal ferritin concentration ≥ 100 μg/L (better 200–500 μg/L), %HYPO below 10 (better below 2.5) or, instead of %HYPO, transferrin saturation above 20% (better 30–40%) /21/. 15.2.2.5.5 %MICRO/%HYPO This ratio is a screening test for β-thalassemia if the microcytic red blood cell fraction is above 18%. A ratio above 0.90 with greater than 18% microcytic erythrocytes is indicator of β-thalassemia /22/. The diagnostic sensitivity of the ratio is acceptable and its specificity is moderate. 15.2.2.5.6 %MACRO If reticulocytosis is not present, the macrocytic red blood cell fraction (%MACRO) is an earlier indicator than MCV with regard to the following: Detection and assessment of the progression of vitamin B12 or folate deficiency anemia Assessment of alcohol abstinence by monitoring of red blood cell volume. The classification of anemia based on MCV, MCH and MCHC is shown in Tab. 15.2-4 – Classification of anemia based on MCV, MCH and MCHC. 15.2.2.6 Comments and problems MCV If the lower threshold of the hematology analyzer is set too high, the MCV is too high because small RBCs are no longer measured. If the upper threshold is too high, white blood cells are measured as well. In the presence of several RBC populations, the MCV represents only the arithmetic mean. Therefore, without knowledge of the RDW, microcytosis in particular may be overlooked. The MCV measured by hematology analyzers yields lower results than the manual method because in the procedure for determining packed cell volume by the micro hematocrit method, the hematocrit is measured too high because of the inclusion of a minimal amount of plasma. MCH In the case of pronounced hypertriglyceridemia and in leukocytosis of > 50 × 109/L, the hemoglobin concentration and thus also the MCH are determined at too high a level because of light absorption and light scattering. MCHC Since the inter individual variation of MCHC is small, this red cell index is well suited for plausibility control of hemogram. It is also suited for controlling the analytical reliability of the hematology analyzer used (e.g., by comparison of the variability of the daily mean from day to day, and for verifying the adjustment of the threshold values). Spurious increased MCHC induces an analytical alarm. Either it is just an artefact, or it refers to a true pathological sample. If it is a problem of the analytical reliability of the analyzer it needs prompt corrective action to ensure delivery of right results to the clinician e.g., red blood cell count (RBC), Hb, and MCV. Problems causing an increase of MCHC are /25/: decreased RBC (cold agglutination) RBC disease lipemic, hemolytic, and icteric plasma. leukocytosis spherocytosis electrolyte disorders. Hyperglycemia Due to the swelling of RBCs, blood glucose levels above 600 mg/dL (33.3 mmol/L) cause a rise in the MCV and the hematocrit as well as a decline in the MCHC /23/. Cold agglutinins See section 15.2.1.6. Cold agglutinins are found to be increased in /24/: Infection with Mycoplasma pneumoniae. In this case high-titer IgM class cold agglutinins are detectable The Epstein-Barr virus (EBV) infection (infectious mononucleosis). This refers to class IgM or IgG anti-i cold agglutinins. Malignant, lymphoproliferative B-cell diseases such as chronic lymphocytic leukemia or other malignant lymphomas. In these cases monoclonal immunoglobulins, particularly of the IgM class, are responsible for the cold agglutination of red blood cells. Warming of blood samples to 37 °C prior to measurement cancels the cold agglutination. Cold agglutination can be verified with a blood smear. Agglutination of the erythrocytes is confirmed at 250-fold magnification. RDW Intraindividual variation is within 24 hours 1.7%, day-to-day 5.9% and month-to-month 5.3% /10/. Stability /11/ Erythrocyte count: with room temperature (RT) and at 4 °C 72 h MCV: at 4 °C 3 days, at RT 12 h, at 37 °C 8 h MCH: at 4 °C and RT 3 days; at 37 °C 24 h MCHC: at 4 °C 3 days; at RT 7 h RDW: at 4 °C 3 days; at RT 7 h. Reticulocyte count In reticulocyte count 8.6% of cases exhibited interference. The main interferents of spuriously high reticulocyte count were caused by parasites, such as malaria, as well as autofluorescence due to drugs. The main interferents of spuriously low reticulocyte count were caused by red blood cell fragments. The presence of numerous red blood cell fragments in blood samples can cause spuriously low red blood cell count and high platelet count, thus affecting the reticulocyte count /26/. References Krause JR. The automated white blood cell differential. Hematol Oncol Clin North Am 1994; 8: 593–603. Gulati GL, Hyun BH. The automated CBC. Hematol Oncol Clin North Am 1994; 8: 605–616. Nebe T, Bentzien F, Bruegel M, Fiedler GM, Gutensohn K, Heimpel H, et al. Multizentrische Ermittlung von Referenzbereichen für Parameter des maschinellen Blutbildes. J Lab Med 2011; 35: 3–28. Saarinen UM, Siimes MA. Developmental changes in red blood cell counts and indices of infants after exclusion of iron deficiency by laboratory criteria and continuous iron supplementation. J Pediatr 1978; 92: 412–6. Taylor MRH, Holland CV, Spencer R, Jackson JF, O’Connors GI, Donnell JRO. Haematological reference ranges for school children. Clin Lab Haem 1997; 19: 1–15. Geaghan SM. Hematologic values and appearances in the healthy fetus, neonate and child. Clin Lab Med 1999; 19: 1–37. NCCLS. Procedure for determining packed cell volume by the microhematocrit method – second edition; approved standard. NCCLS Document H7–A2, Vol 13, No 9. Villanova: NCCLS, 1993. Schalk E, Scheinpflug K, Mohren M. Kapilläre Blutbildanalysen in der klinischen Praxis: eine sichere, zuverlässige und valide Methode. J Lab Med 2009; 33: 303–9. Wisser H. Störungen der Messgrößen des kleinen Blutbildes durch Antikörper. GIT Labor-Medizin 1995; 83–5. Costongs GMPJ, Janson PCW, Bas BM, et al. Short-term and long-term intra-individual variations and critical differences of haematological laboratory parameters. J Clin Chem Clin Biochem 1985; 23: 69–76. Empfehlungen der Arbeitsgruppe Präanalytik der DGKC. Stabilität der Messgröße in der Probenmatrix. DG Klinische Chemie Mitteilungen 1996; 27: 74–8. Buttarello M, Bulian P, Venudo A, et al. Laboratory evaluation of the Miles H3 automated reticulocyte counter. A comparative study with manual reference method and Sysmex R-1000. Arch Pathol Lab Med 1995; 119: 1141–8. Thomas C, Thomas L. Biochemical markers and hematologic indices in the diagnosis of functional iron deficiency. Clin Chem 2002; 48: 1066–76. Segel GB, Oski FA. Hematology of the newborn. In: Williams WJ, Beutler E, Erslev AJ, Lichtman MA (eds). Hematology. New York 1990; McGraw Hill: 476. Tyler RD, Cowell RL. Classification and diagnosis of anemia. Comp Haematol Int 1996; 6: 1–16. Peach HG, Bath NE, Farish S. Predictive value of MCV for hazardous drinking in the community. Clin Lab Haem 1997; 19: 85–7. Oosterhuis WP, Niessen RWLM, Bossuyt PMM, Sanders GTB, Sturk A. Diagnostik value of the mean corpuscular volume in the detection of vitamin B12 deficiency. Scand J Clin Lab Invest 2000; 60: 9–18. Anagnou J. Mikrozytose, Hypochromie und Erythrozytenindizes im Wandel. Dtsch Med Wschr 1985; 110: 657–8. Patel KV, Ferrucci L, Ershler WB, Longo DL, Guralnik JM. Red blood cell distribution width and the risk of death in middle-aged and older adults. Arch Intern Med 2009; 169: 515–23. McDougal IC. What is the most appropriate strategy to monitor functional iron deficiency in the dialysed patient on rHuEPO therapy? Merits of percentage hypochromic red cells as a marker of functional iron deficiency. Nephrol Dial Transplant 1998; 13: 847–9. European Best Practice Guidelines for the Management of Anemia in Patients with Chronic Renal Failure. Target guideline 6: assessing and optimizing iron stores. Nephrol Dial Transplant 1999; 14 (Suppl 5): 14–5. d’Onofrio G, Zini G, Ricerca BM, Mancini S, Mango G. Automated measurement of red blood cell microcytosis and hypochromia in iron deficiency and β thalassemia trait. Arch Pathol Lab Med 1992; 116: 84–9. van Duijnhoven HLP, Treskes M. Marked interference of hyperglycemia in measurements of mean (red) cell volume by Technicon H analyzers. Clin Chem 1996, 42: 76–80. Strobel SL, Panke TW, Bills GL. Cold erythrocyte agglutination and infectious mononucleosis. Laboratory Medicine 1993; 24: 219–21. Haddad YB, Faure C, Boubaya M, Arpin M, Cointe S, Frankel D, et al. Increased mean corpuscular hemoglobin concentration: artefact or pathological condition? Int Jnl Lab Hem 2017; 39. 32–41. Jiang H, Wang J, Wang K, Gu J, Chen J, Wang Z. Interferents of automated reticulocyte analysis integrated with relevant clinical cases. Clin Lab 2019: 65; 1251–9. Bohlius J, Bohlke K, Castelli R, Djulbegovic B, Lustberg MB, Martino M, et al. Management of cancer-associated anemia with erythropoiesis-stimulating agents: AsCO/ASH clinical practice guideline update. Blood Advancea 2019; 3 (8): 1197–1210. Aapro M, Beguin Y, Bokemeyer C, Jordan K, Herrstedt J., et al Management of anaemia and iron deficiency in patients with cancer: ESMO clinical practice guidelines. doi: 10.1093/annonc/mdx/7658. 15.3 Hemoglobin concentration Lothar Thomas The hemoglobin (Hb) concentration is a function of the number of red blood cells their MCH and the proportion of blood plasma. With a constant plasma volume, there is a direct relationship between blood Hb level and red blood cell mass. A decrease in an organisms’ Hb content is termed anemia. Anemia is present if the red blood cell mass of the body is reduced (normal 21–27 mL/kg BW in women and 24–32 mL/kg BW in men) /1/. Since the red blood cell mass can only be measured with radioactive methods the determination of the Hb value in a defined blood volume is a simple alternative. This is because the Hb concentration includes the product of the erythrocyte number and the Hb content of the red cell. Anemia is diagnosed from the Hb level if the blood volume is normal. This is the case in anemia that has been present for longer than 48 hours. The reason is that in this case, the decrease in red blood cell mass is compensated for by an increase of plasma volume. The blood volume is 90 mL/kg BW in newborns and in older children and in adults it is 80 mL/kg BW. 15.3.1 Indication Diagnosis, and monitoring in anemia, erythrocytosis, and polycythemia. 15.3.2 Method of determination Hemiglobin cyanide method /2/ Principle: in solution, Fe2+ of Hb is oxidized by potassium hexacyanoferrate [K4Fe(CN)6] to Fe3+. Hemiglobin (Hi) is formed and generates HiCN with cyanide ions (CN–), which are provided in the solution by potassium cyanide (Tab. 15.3-1 – Formation of hemiglobin cyanide from hemoglobin). HiCN has an absorption maximum at 540 nm, and the absorption of HiCN is proportional to Hb concentration. Hematology analyzers are calibrated with secondary HiCN standards that contain 500–800 mg/L HiCN. For HiCN methods employing a 250-fold dilution of blood samples, this provides an equivalent Hb concentration of 125–200 g/L. The HiCN method is the reference method. See also Section 15.2.1.2 – Method of determination. 15.3.3 Specimen EDTA blood (disodium or dipotassium EDTA): 1 mL Capillary blood (heparinized capillaries): 0.02–0.05 mL. 15.3.4 Reference interval See references /3, 4, 5, 6/ and Tab. 15.3-2 – Hemoglobin reference intervals. 15.3.5 Clinical significance The Hb concentration combined with the hematocrit and the RBC count is an important criterion for the diagnosis and differentiation of anemia, erythrocytosis, and polycythemia. Frequent causes of anemia are shown in Fig. 15.3-1 – Frequent causes of anemia. 15.3.5.1 Diagnosis of anemia The term anemia describes a decline of the Hb level, due to /7, 8/: An absolute reduction in the erythrocyte number (e.g., in anemia of chronic disease) A decrease in erythrocyte Hb content. This can be the case with a normal, slightly reduced or even increased cell count (e.g., in iron deficiency anemia, heterozygous β-thalassemia). An increase in plasma volume with a relative reduction of the erythrocyte number with normal or even increased red blood cell mass (e.g., during the last trimester of pregnancy). In this case, the condition is referred to as pseudoanemia. The diagnosis of anemia is an important aspect of the practice of hematology. To decide whether a patient is anemic compared on the basis of the population distribution of Hb values is problematic /9/. Proposed lower thresholds are shown in Tab. 15.3-3 – Proposed low hemoglobin thresholds for adults. In children, the threshold values for anemia can only be related as a function of age. The proposed thresholds of normal of the Centers for Disease Control (CDC) in the USA /10/ are shown in Tab. 15.3-4 – Proposed low hemoglobin thresholds for children . For certain groups of individuals such as smokers, or as a function of living conditions (living at high altitudes), adjustments in the Hb level are required. Refer to: Tab. 15.3-5 – Correction for anemia during an extended stay at high altitude Tab. 15.3-6 – Correction for anemia in smokers. 15.3.5.2 Extent of anemia Depending on the Hb level, the extent of anemia is classified as /11/: Mild; from the lower reference interval value to 100 g/L Moderate 100–80 g/L Severe 80–65 g/L Life-threatening; below 65 g/L. 15.3.5.3 Clinical symptoms of anemia Clinical symptoms are cold and pale skin, fatigue, palpitation, low endurance level, depression, disturbance of cognitive function, and a general reduction in quality of life. In severe anemia, if the Hb value is reduced by about 50%, cardiac decompensation with congestive heart failure may occur, since coronary blood flow will have reached its maximum. This situation is particularly critical in patients with pre-existing coronary heart disease. Mild proteinuria can also occur in severe anemia, due to a decrease in renal blood flow /12/. The severity of clinical symptoms is dependent upon the extent of anemia and other factors /12/: The physiological status of the patient. Healthy young individuals have fewer and milder symptoms than elderly individuals. Co morbidity. Thus, in individuals with multiple co morbidities or in bed-ridden patients, even only slight decreases in the Hb concentration lead to symptoms such as tiredness, falling upon standing up, claudication or angina pectoris-like symptoms. The pace of occurrence of anemia. Anemias that develop slowly are often first noticed in apparently healthy individuals in situations of physical or mental stress. 15.3.5.4 Prevalence of anemia In Europe and the USA, approximately 1% of adult males and 3–5% of adult females are anemic; in Africa, the numbers are 27% and 48%, and in southeast Asia they are 40% and 57%. The most common causes are deficient nutrition and malnutrition, as well as hookworm infestation /13/. Approximately half of all anemia is contingent upon iron deficiency; worldwide, some 500 million individuals are believed to suffer from iron deficiency anemia, while the number of individuals with iron deficiency but without anemia is 3 times as high. 15.3.5.5 Tolerance to anemia Healthy young individuals with normovolemia These individuals show no indication of a critical change in O2 supply down to a Hb value of 50 g/L. From ≤ 60 g/L, however, ECG changes and disturbances of cognitive function may occur. Hb values of 45–50 g/L are an absolute indication for substitution with banked blood. Patients with cardiovascular disease (CVD) Patients with stable CVD tolerate Hb values of 70–80 g/L without hypoxic damage. Values below 70 g/L increase morbidity and mortality. Surgical patients /14/ A preoperative Hb value ≤ 100 g/L is associated with increased peri operative mortality. This is not the case with a peri operative decline in Hb to ≤ 100 g/L in patients without CVD. The mortality risk is greatest if, in CVD patients, the intraoperative fall is ≥ 40 g/L. With regard to all patients, a peri operative fall in Hb down to 70 g/L is associated with increased morbidity but not with increased mortality. Nonetheless, each reduction of 10 g/L below 70 g/L increases the mortality risk 1.5-fold. Patients without CVD and a preoperative Hb value of 60–90 g/L and only minimal pre-operative blood loss have an odds ratio of 1.4 for mortality, in comparison with those with pre-operative Hb values > 120 g/L (see also Tab. 15.4-2 – Hematocrit reference intervals. Intensive care patients Ventilated intensive care patients with poly trauma and sepsis do not seem to benefit from transfusion to Hb values > 90 g/L. Only in the event of massive blood loss or diffuse bleeding diathesis does an Hb value > 100 g/L seem to contribute to the stabilization of blood coagulation. 15.3.5.6 Classification of anemia Anemia can be classified as follows: According to pathogenesis. However, this classification is problematic, since in many forms of anemia multiple mechanisms contribute to pathogenesis. Based on erythrocyte morphology into microcytic, normocytic and macrocytic forms; or based on red blood cell Hb concentration into hypochromic and normochromic forms. This classification has gained acceptance in the differential diagnosis of anemia. According to erythropoietic regenerativity as hypo-, normo- and hyper-regenerative anemia According to the form of progression into acute and chronic Into congenital and acquired anemia. Classification of anemia according to erythropoiesis regenerativity is shown in Tab. 15.3-7 – Classification of anemia according to erythropoietic activity of the bone marrow. Acute anemia Acute anemia (e.g., due to hemorrhage) can only be recognized with the hematocrit, the erythrocyte count and the Hb concentration following 24 hours, because no sufficient compensatory increase in plasma volume occurs during the initial hours. Chronic anemia In patients with chronic anemia the blood volume is normal, since plasma volume has increased according to the decrease in red blood cell mass. As a result there is a decrease in the erythrocyte count and a reduction of the hematocrit and Hb value. Relative anemia Relative anemia is a condition with normal red blood cell mass, but in which the total blood volume is increased due to an increased plasma volume. The cause is a regulatory change in the water and electrolyte balance (e.g., during pregnancy). In contrast to chronic anemia, in which serum total protein is usually found to be normal, total protein in relative anemia (pseudoanemia) is low or low-normal, with the exception of Waldenströms’ macroglobulinemia. Aplastic anemia Aplastic anemia refers to the failure of marrow to form blood, and hematopoietic failure is the end-organ effect of diverse pathophysiological mechanisms. Bone marrow cellularity is decreased, the blood count indicates pancytopenia /73/. 15.3.5.7 Differentiation of anemias Microcytic hypochromic anemia is the most common form of anemia. It is a very heterogenous group of diseases that may be either acquired or inherited. Microcytic hypochromia is often diagnosed in children, adolescents, women before menopause and in pregnant women (mostly due to iron deficiency) /64/. 15.3.5.7.1 Microcytic hypochromic anemias Microcytic hypochromic anemia can result from /64/: Iron deficiency A defect in globin genes (hemoglobinopathies or thalassemias) A defect in heme synthesis A defect in iron availability or in iron acquisition by the erythroid precursors. The microcytic anemias can be sideroblastic, a trait which reflects the implications of different gene abnormalities. The inherited microcytosis due to defects in heme synthesis are /64/: Sideroblastic anemias (refer to Tab. 15.3-9– Classification and differentiation of microcytic anemia) Erythropoietic protoporphyria (refer to Tab. 14.6-6) Congenital erythropoietic porphyria (refer to Tab. 14.6-6). The inherited microcytosis due to iron metabolism deficiency are /64/: Iron deficiency anemia (refer to Tab. 7.1-2) Ferroportin disease Hereditary atransferrinemia (refer to Tab. 7.1-9) Hereditary aceruloplasminemia (refer to Tab. 7.1-9). For anemia differentiation the following Tables are recommended: Tab. 15.3-8 – Blood cell status in certain groups of individuals and patients Tab. 15.3-9 – Classification and differentiation of microcytic anemia Tab. 15.3-10 – Hematological data in patients with β-thalassemia Tab. 15.3-11 – Classification and differentiation of normocytic anemias Tab. 15.3-12 – Causes of hemolytic anemias Tab. 15.3-13 – Laboratory findings in sickle cell anemia 15.3.5.7.2 Macrocytic anemias The interaction between folate and vitamin B12 is responsible for the megaloblastic anemia seen in both vitamin deficiencies. Dyssynchrony between the maturation of cytoplasm and that of nulei leads to macrocytosis, immature nuclei, and hyper segmentation in granulocytes in the peripheral blood. The dysplastic and hyper cellular bone marrow can be mistaken for signs of acute leukemia. The ineffective erythropoiesis results in intramedullary hemolysis and release of lactate dehydrogenase. Clinical and Laboratory findings of megaloblastic anemia are shown in: Tab. 15.3-14 – Classification and differentiation of macrocytic anemias Section 13.3 – Vitamin B12 deficiency Section 13.4 – Folate deficiency. 15.3.5.7.3 Which test to choose? Important laboratory tests for the differentiation and evaluation of the pathogenesis of anemias are: Hematocrit as a measure for assessing the red cell blood volume fraction Erythrocyte indices MCV, MCH and blood smear findings, for the identification of iron deficiency, vitamin B12 or folic acid deficiency or changes in erythrocyte shape Reticulocyte count or reticulocyte index (RI) as a measure of the effectiveness of erythropoiesis; based on the RI, anemias are classified into normo-, hypo- and hyper regenerative Reticulocyte indices such as the reticulocyte Hb content (CHr, Ret-He) for early diagnosis of iron-restricted erythropoiesis Ferritin as an indicator of iron stores Zinc protoporphyrin for diagnosis of iron-restricted erythropoiesis Transferrin saturation (TSAT) or soluble transferrin receptor (sTfR) for estimation of functional iron Soluble transferrin receptor (sTfR) in relation to hematocrit for the detection of intrinsic hypo proliferative erythropoiesis. See Section 7.4 – Soluble transferrin receptor. Erythropoietin (EPO) concentration relative to the hematocrit value, for assessing the adequate stimulation of erythropoiesis (see Fig. 15.10-1 – Expected range of EPO concentration as a function of the hematocrit value) EPO and assessment of observed/predicted (O/P) ratio; see also Section 7.4 – Soluble transferrin receptor. An O/P ratio of below 0.8 is indicative of inadequately low EPO stimulation and a value of above 1.2 suggests hyper proliferative erythropoiesis due to augmented EPO stimulation. C-reactive protein for the detection of anemia of inflammation and infection Haptoglobin as indicator of hemolytic anemia. In most cases of anemia erythropoiesis is compromised due to: Nutritional deficiencies (e.g., iron, folate, vitamin B12) Chronic disease (e.g., systemic inflammation, solid malignant tumor, leukemia, lymphoma) Organic disease (e.g., kidney, liver, thyroid or small intestine). A bone marrow investigation is usually not necessary: In cases of compromised erythropoiesis (except lymphoma and leukemia) In all cases of microcytic anemia (except for suspected sideroblastic anemia). 15.3.5.8 Red blood cell function in anemia The symptoms and severity of anemia depend on various factors, including the degree of anemia, the rapidity of onset, and the age and the physiologic status of the patient /12/. At rest, the amount of O2 required by the whole body ranges between 200 to 300 mL per minute. In health, the amount of O2 delivered to the whole body exceeds resting O2 requirements by a factor of 2–4-fold. In cardiac output of 5 liters per minute and oxygen saturation of 99% the delivery will be 1032 mL per minute. An isolated decrease in Hb to 100 g/L will result in an O2 delivery of 688 mL per minute and a Hb decrease to 50 g/L to an delivery of 342 mL per minute /15/. There are several mechanisms by which the body tries to counterbalance the effects of anemia /12/: Increased cardiac output with increase in the oxygen transport capacity, which is very effective, but metabolically expensive Increase of the respiratory rate Oxygen binding by hemoglobin Reduction of pH in capillary blood and tissues, which leads to better dissociation of O2 from Hb and to vasodilation. The plasma bicarbonate concentration is reduced due to hyperventilation and the compensatory renal bicarbonate excretion. Increased erythropoietin EPO secretion (see Tab. 15.10-1 – Reference intervals for erythropoietin); within 1–2 hours of acute onset of hypoxia (normobaric or hypobaric) circulating EPO levels begin to increase and erythropoiesis is stimulated. However, the erythropoietic activity is diminished dramatically when the arterial O2 saturation decreases below 60%. The blood is shunted from presumably non vital donor organs to oxygen-sensitive recipient organs. 15.3.5.9 Increase in the O2 transport capacity Stimulated erythropoiesis increases the number of red blood cells and the amount of Hb in the circulation. Any increase in the Hb level brings a proportional increase in O2 per volume of blood transported. The contribution of enhanced heart rate to the increase in cardiac output and the increase in the O2 transport capacity is variable. Hemodynamic compensation occurs solely via the increased stroke volume down to a fall in the Hb level to 75 g/L. If, additionally, extreme conditions are present, this compensation is limited by myocardial tissue. Under conditions of more intense performance requirements or a further decline in Hb, the shift in the oxyhemoglobin curve to the right is important. With utilization-related compensation the oxygen reserves are exploited in an extreme manner. This utilization is as high as 90% in most organs, corresponding to a decline in blood O2 saturation from normally 150 mL/L to 10 mL/L. The shift of the oxyhemoglobin curve to the right does not occur with acute loss of blood but only following a number of days due to enhanced formation of 2,3-DPG and its effect on Hb (Fig. 15.3-3 – Structure and function of hemoglobin). Due to an increased erythrocyte 2,3-DPG formation, there occurs a reduction in O2 affinity for Hb and, in consequence, a shift to the right of the Hb-O2 affinity curve, with increased O2 release in the tissues (see Fig. 15.4-4 – Effect of 2,3-diphosphoglycerate (2,3-DPG) on O2 saturation curve). In this way, half of the oxygen deficit caused by moderate anemia is compensated for. 15.3.5.10 Oxygen binding by hemoglobin The O2 transport by Hb depends on the concentration of Hb and the Hb-O2 affinity, which determines the ease of O2 loading onto or unloading from the Hb molecule /16/. The dependency of the oxygen saturation of Hb on the PO2 in the blood is described by the Hb-O2 affinity curve (Fig. 15.3-2 – O2 affinity curve). The curve is characterized by the P50, the PO2 at which Hb is saturated by 50% with O2. The physiological significance of a variable Hb-O2 affinity lies in the appropriate adjustment of the HbO2-binding in order to optimize both arterial O2 loading and peripheral O2 release. A shift to the left of the Hb-O2 affinity curve facilitates arterial oxygen loading a shift to the right indicates a low Hb-O2 affinity, which favours the oxygen release from Hb to the tissues. The Hb-O2 affinity curve is shifted /16/: To the right due to acidosis, a high CO2, high erythrocyte 2,3-DPG and an elevated body temperature To the left due to alkalosis, hypocapnia, a decreased 2,3-DPG, and a low temperature. Effect of pH /16/ The effect of pH on the Hb-O2 affinity is due to pH-dependent pK changes of ionizable amino acid residues of the Hb molecule. Acidosis stabilizes the deoxy form of the Hb molecule and decreases the Hb-O2 affinity (Bohr effect). Effect of CO2 /16/ The main effect of CO2 on the Hb-O2 affinity is due to its effect on cell pH. Another effect is that an increase in CO2 at constant pH shifts the Hb-O2 affinity curve towards the right through the reversible formation of carbamates from CO2 and N-terminal residues of alpha- and beta chains of the Hb molecule. 15.3.5.11 Oxygen delivery in anemia The erythrocytes are endowed with a content of 2,3-bisphosphoglycerate (2,3 BPG), that is nearly 103 times as high as the content in other body cells /16, 83/. The 2,3-BPG is synthesized in a side path of red cell glycolysis. Alkalosis stimulates glycolysis and increases 2,3-BPG in erythrocytes, whereas acidosis reduces the erythrocyte content of 2,3-BPG. The enzyme responsible for 2,3-BPG synthesis is 2,3-DPG mutase and for breakdown is 2,3-BPG phosphatase. A reduction of free 2,3-BPG by binding to Hb upon deoxygenation increases its rate of synthesis and elevates to total 2,3-BPG content in the erythrocyte. Generally, organic phosphates bind preferentially to deoxy-Hb with an affinity 100 times higher than to that to oxy-Hb. The binding stabilizes the deoxy form of Hb. 2,3 BPG binds specifically to deoxy-Hb and lowers the fractional oxygen saturation, and the oxygen-binding curve is shifted to the right. The function of 2,3-BPG in the binding and release of O2 to the Hb molecule is shown in Fig. 15.4-4 – Effect of 2,3-diphosphoglycerate (2,3-BPG) on O2 saturation. All patients with anemia have increased content of 2,3-BPG that enhances oxygen delivery (P02). In a person with anemia at any given oxygen tension the oxygen saturation is reduced. At a venous PO2 of 40 torr the oxygen saturation is about 80% and 20% of oxygen in the blood is unloaded. In cases of anemia elevate concentrations of 2,3 PBG in erythrocytes lower oxygen affinity of the blood allows a much higher fraction of oxygen to be unloaded. Example /83/: In an individual with a Hb concentration of 150 g/L the oxygen-carrying capacity of the blood is 20 mL per 100 mL of blood. In the arteries and the capillary bed about 20% of the oxygen (about 4 mL of blood) will be unloaded per 100 mL of blood. In anemic patients with a Hb concentration of 75 g/L the oxygen binding capacity is halved. In this patient 20% of the oxygen (about 2 mL) would be unloaded per 100 mL of blood. However, the patient’s erythrocytes have increased content of 2,3 BPG and lower affinity for oxygen, for this reason about 3 mL is released. By this way the deficit in red cells is compensated. 15.3.5.12 Findings in hypoxic anemia The following findings are indicative of hypoxic anemia: tachycardia, hypotension, O2 extraction > 50%, mixed venous O2 saturation < 50%, central venous O2 saturation < 60%, mixed venous PO2 < 32 mm Hg and lactic acidosis (lactate > 2 mmol/L). 15.3.5.13 Polycythemia and polyglobulia In Caucasians suspicion of polycythemia or polyglobulia is unlikely if in women the Hb value is < 165 g/L or the hematocrit is < 0.50 and in men the Hb value is < 180 g/L or the hematocrit is < 0.55 /1/. Refer to Chapter 15.4 – Hematocrit. 15.3.6 Comments and problems Anticoagulation For venous blood 1.5–2.2 mg EDTA (dipotassium or di sodium salt) per mL, the final EDTA concentration is 3.7 or 5.4 μmol/L /2/. Method of determination The use of quartz cuvettes for the spectrophotometric measurement is an interfering factor with respect to the hemiglobin cyanide method. The use of reagent blanks solves the problem. A further problem is the turbidity of samples. Ultrafiltration reduces turbidity so that the ratio of absorption required for the reference method A540/504 is ≥ 1.59 /17/. Lipemia and leukocytes Turbid blood leads to a rise in Hb of up to 30 g/L because the HiCN solution becomes turbid /2/. Leukocyte values > 100 × 109/L can have the same effect as lipemia /2/. Thrombocytes Values > 700 × 109/L cause the HiCN solution to become turbid and lead to erroneously high Hb values /2/. References Fairbanks VF, Klee GG, Wiseman GA, Hoyer JD, Teferi A, Petitt RM, Silverstein MN. Measurement of blood volume and red cell mass: reexamination of 51Cr and 125J methods. Blood Cells, Molecules and Diseases 1996; 22: 169–86. NCCLS. Reference and selected procedures for the quantitative determination of hemoglobin in blood – second edition; approved standard. NCCLS Document H15–A2, Vol 14 No 6. Villanova: NCCLS, 1994. 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Clinical and hematologic features of 300 patients affected by hereditary spherocytosis grouped according to the type of membrane protein defect. Haematologica 2008; 93: 1310–7. Harigae H, Furuyama K. Hereditary sideroblastic anemia: pathophysiology and gene mutations. Int J Hematol 2010; 92: 425–31. Fishbane S, Cohen DJ, Coyne DW, Djamali A, Sigh AK, Wish JB. Posttransplant anemia: the role of sirolimus. Kidney Int 2009; 76: 376–82. Mix TCH, Kazmi W, Khan S, et al. Anemia: a continuing problem following kidney transplantation. Am J Transplant 2003; 3: 1426–33. Taylor RD, Cowell RL. Classification and diagnosis of anemia. Comp Haematol Int 1996; 6: 1–6. Tabbara IA. Hemolytic anemias. Med Clin North Am 1992; 76: 649–68. Ataga KI. Hypercoagulability and thrombotic complications in hemolytic anemias. Haematologica 2009; 94: 1481–3. Firth PG, Head A. Sickle cell disease and anesthesia. Anesthesiology 2004; 101: 766–85. Dorn-Beineke A, Frietsch T. Sickle cell disease – pathophysiolpgy, clinical and diagnostic implications. Clin Chem Lab Med 2002; 40: 1075–84. Ware RE, Rees RC, Sarnaik SA, Iyer RV, Alvarez OA, Casella JF, et al. Renal function in infants with sickle cell anemia: baseline data from the Baby Hug Trial. J Pediatr 2010; 156: 66–70. Dobie G. Sickle cell disease and thromboembolism: New insights on the pathophysiology, diagnosis, and treatment. Clin Lab 2023; 69: 1335–48. Lee JE, Jang JH, Kim JS, Yoon SS, Lee JH, Kim YK, et al. Clinical signs and symptoms associated with increased risk for thrombosis in patients with paroxysmal nocturnal hemoglobinuria from a Korean Registry. Int J Hematol 2013; 97: 749–57. Scheiring J, Rosales A, Zimmerhackl LB. Today’s understanding of the haemolytic uremic syndrome. Eur J Pediatr 2010; 169: 7–13. Robson KJH, Weatherall DJ. Malarial anemia and STAT6. Haematologica 2009; 94: 157–9. Weiss G. Iron metabolism in the anemia of chronic disease. Biochem Biophys Acta 2009; 48: 57–63. Thomas C, Thomas L. Anemia of chronic disease: pathophysiology and laboratory diagnosis. Lab Hematol 2005; 11: 14–23. Aapro M, Österborg A, Gascon P, Ludwig H, Beguin Y. Prevalence and management of cancer-related anaemia, iron deficiency and the specific role of i.v. iron. Ann Oncol 2012; 23: 1954–62. Ludwig H, van Belle S, Barrett-Lee P, Birgegard G, Bokemeyer C, Gascon P, et al. The European Cancer anemia survey (ECAS): a large, multinational, prospective survey defining the prevalence, incidence, and treatment of anemia in cancer patients. Eur J Cancer 2004; 40: 2293–306. Caro JJ, Salas M, Ward A, Goss G. Anemia as an independent prognostic factor for survival in patients with cancer: a systematic quantitative review. Cancer 2001; 91: 2214–21. Bron D, Meuleman N, Mascaux C. Biological basis of anemia. Seminars Oncol 2001; 28 , Suppl 8: 1–6. Aapro M, Brguin Y, Bokemeyer C, Dicato M, Gascon P, Glaspy J, et al. Management of anemia and iron deficiency in patients with cancer: ESMO Clinical Practice Guidelines. Ann Oncol 2018. doi: 10.1093/annonc/mdx758. Sleijfer S, Lugtenburg PJ. Aplastic anemia: a review. Netherl J Med 2003; 61: 157–63. Coyle TE. Hematologic complications of human immunodeficiency virus infection and the acquired immunodeficiency syndrome. Med Clin North Am 1997; 81: 449–70. Iolascon A, Esposito M, Rosso R. Clinical aspects and pathogenesis of congenital dyserythropoietic anemias: from morphology to molecular approach. Haematologica 2012; 97: 1786–94. Heimpel H. Congenital dyserythropoietic anemias: epidemiology, clinical significance, and progress in understanding their pathogenesis. Ann Hematol 2004; 83: 613–21. Maitra P, Caughey M, Robinson L, Desal PC, Jones S, Nourale M, et al. Risk factors for mortality in adult patients with sickle cell disease: a meta-analysis of studie in North America and EUrope. Haematologica 2017; 102: 626–36. Stabler SP. Vitamin B12 deficiency. N Engl J Med 2013; 368: 149–60. Chadebech P, Loustau V, Janvier D, Languille L, Ripa J, Tamagne M, et al. Clinical severity in adult warm autoimmune hemolytic anemia and its relationship to antibody specificity. Haematologica 2018; 103: e35. Inamoto Y, Lee SJ. Late effects of blood and marrow transplantation. Haematologica 2017; 102: 614–25. Dierickx D, Habermann TM. Post-transplantation lymphoproliferative disorders in adults. N Engl J Med 2018; 378. 549–62. De Falco L, Sanchez M, Silvestri L, Kannengieser C, Muckenthaler MU, Iolascon A, et al. Iron refractory iron deficiency anemia. Haematologica 2013; 98: 845–53. Millot S, Delaby C, Moulouel B, Lefebvre T, Pilard N, Ducrot N, et al. Hemolytic anemia repressed hepcidin level without hepatocyte iron overload: lesson from Günther disease model. Haematologica 2017; 102: 260–70. Young NS. Aplastic anemia. N Engl J Med 2018; 379 (17): 1643–55. Petrelli F, Ghidini M, Ghidini A, Sgroi G, Vavassori I, Petro D, et al. Red cell transfusions and the survival in patients with cancer undergoing curative surgery: a systematic review and meta-analysis. Surgery Today 2021. doi: 10.1007/s00595-020-02192-3. Bienstock JL, Eke AC, Hueppchen NA. Postpartum hemorrhage. N Engl J Med 2021; 384 (17): 1635–45. Pawlak R, Berger J, Hines I. Iron status of vegetarian adults: a review of literature. Analytic Review 2018; 6: 486–97. Ambroszkiewicz J, Klemarczyk W, Mazur J, Gajewska J, Rowicka G, Strucinska M, Chelchowska M. Serum hepcidin and soluble transferrin receptor in the assessment of iron metabolism in children on a vegetarian diet. Biol Trace Elem Res 2017; 180: 182–90. Hillmen P, Szer J, Weitz I, Röth A, Höchsmann B, Panse J, et al. Pegcetacoplan versus Eculizumab in paroxysmal nocturnal hemoglobinuria. N Engl J Med 2021; 384: 1028–37. Serjeant G. Sickle cell disease: Thoughts for India from the Jamaican Cohort Study. Frontiers in Medicine 2021; 8: Article 745189. Kanter J, Walters MC, Krishnamurti L, Mapara JL, Kwiatkowski JL, Rifkin-Zenenberg S, et al. Biologic and clinical efficacy of LentiGlobin for sickle cell disease. N Engl J Med 2022; 386 (7): 617–28. Notaro R, Luzzatto L. Breakthrough hemolysis in pNH with proximal or terminal complement inhibition. N Engl J Med 2022; 387 (2): 160–6. Block J, Rashkova C, Castanon I, Zoghi S, Platon J, Ardy RC, et al. A systemic inflammation and normocytic anemia in DOCK11 deficiency. N Engl J Med 2023; 389 (6): 527–39. Bunn HF. Oxygen delivery in the treatment of anemia. N Engl J Med 2022; 387 (25): 2362–5. Wang H. Lin P. Association between sarcopenia and hemoglobin level: a systematic review and meta-analysis. doi: 10.3389/fmed 2024.1424227. 15.4 Hematocrit (HCT) Lothar Thomas The HCT, or packed cell volume (PCV) is a measure of the ratio of the volume occupied by the red cells to the volume of whole blood in a sample of capillary or venous blood. The ratio is measured after appropriate centrifugation and is expressed as a fraction (e.g., 0.42 and not 42%). The HCT is used, together with the red cell count, in calculating the MCV and, together with hemoglobin content, in calculating the MCHC /1/. 15.4.1 Indication Detection of anemia or polycythemia Estimating changes in hemodilution and hemoconcentration Reference for assessing erythropoietin formation in relation to the extent of anemia; see section 15.10. 15.4.2 Method of determination Micro hematocrit method Disposable type I, class B borosilicate glass capillary tubes, 75 mm in length, internal diameter 1.15 mm, are recommended /1/. Wall thickness should be 0.20 mm. Use of a special micro hematocrit centrifuge with rotor radius greater than 8 cm, reaching maximal speed within 30 seconds, capable of sustaining a relative centrifugal force (RCF) of (10,000 to 15,000) × g at the periphery for 5 min., without the rotor exceeding a temperature of 45 °C. Calculation of RCF: see Tab. 15.4-1 – Equations for HCT calculation. Hematology analyzer Principle: see Section 15.2.1.2 – Method of determination. In the most simple case, HCT is calculated as shown in the equation as product of RBC × MCV. The RBC is the red blood count. Some hematology analyzers determine the sum of electrical pulses and divide them by the number of pulses. Calculation is performed according to the equation in Tab. 15.4-1 – Equations for HCT calculation. The hematology analyzer results are adapted to the micro hematocrit method. 15.4.3 Specimen Whole blood (di sodium EDTA, tripotassium EDTA or heparin as anticoagulant): 1 mL Capillary blood (heparinized capillaries): 0.05 mL 15.4.4 Reference interval Refer to references /1, 2, 3, 4, 5, 6, 7, 8/ and Tab. 15.4-2 – Hematocrit reference intervals. 15.4.5 Clinical significance The HCT determination is a simple method for the detection of anemia, and erythrocytosis. It is useful, in addition, for identifying the extent of changes with hemodilution and hemoconcentration. The HCT is dependent upon: The red blood cell mass of the organism (17–32 mL/kg body weight in women and 20–36 mL/kg body weight in men) /9/ The mean cell volume of erythrocytes The plasma volume, the reference interval is 30–45 mL/kg body weight. 15.4.5.1 Decrease of HCT Besides a decrease of the Hb level, a decrease in HCT is a diagnostic criterion of anemia. Exceptions are patients: With hyper hydration (e.g., postoperative patients with minimal blood loss but overcompensated volume substitution). In such cases there is an increased plasma volume in the presence of still normal red blood cell mass, called pseudoanemia. With acute bleeding, in which the erythrocyte loss has not yet been compensated for by an increase in plasma volume, anemia is already present pathophysiologically, but is not yet indicated by a reduction of HCT or the Hb level. In healthy normovolemic adults, HCT can decline to 0.15–0.20 before abnormal regional myocardial blood flow distribution, indicated by a rise in cardiac lactate formation, develops. In the fetus and the newborn, anemia can be very severe before disturbances of cardiac function occur /10/. An elevated incidence of postoperative cardiac complications is found in patients with peripheral vascular disease in whom HCT is below 0.29 /11/. Cardiac disease is the most frequent cause of death in hemodialysis patients. An increase in HCT from below 0.30 to the range of 0.30–0.38 reduces the incidence of myocardial infarction by 30% within a time period of 30 months /12/. In patients with congestive heart failure and HCT ≤ 0.24, mortality and re-hospitalization risks are increased by 51% and 17%, respectively, in comparison to patients with HCT of 0.40–0.44. Although anemia is an independent risk factor, a patient’s co morbidities also play an important role /13/. HCT plays a major role in primary hemostasis by influencing blood viscosity and platelet adhesion. HCT values below 0.30 lead to a reduction in blood viscosity and platelet adhesion and the bleeding time may be prolonged /14/. It is that during continuous venovenous hemofiltration (CVVH) higher HCT values are accompanied by an increased hemostasis activation during CVVH. This is not the case in CVVH patients with HCT values between 0.30 and 0.35 /15/. Diseases and conditions with decreased HCT are shown in Tab. 15.4-3 – Diseases and conditions with decreased HCT. In patients with acute blood loss the number of blood donations is dependent on the actual HCT and the desired HCT in patients with anemia. A calculation is shown in Tab. 15.4-4 – Calculation of desired packed red cells in dependence of HCT. 15.4.5.2 Increase of HCT HCT elevations of over 0.48 in women and over 0.51 in men are due to: An absolute increase in the red cell mass, associated with erythrocytosis and polycythemia A plasma volume decrease (e.g., associated with exsiccosis). Clinically, there is an association between a high-normal or high HCT and increased blood viscosity, blood vessel disease, metabolic disease and thrombosis. According to: The Framingham Study /16/, Caucasians with a HCT over 0.50 had a 2–6 fold higher relative risk of stroke over the study period of 87 months than individuals with lower values The Puerto Heart Program /17/, individuals from the urban population with a HCT over 0.49 had a 2-fold higher risk of cardiovascular disease within 8 years than individuals with a HCT below 0.42 A British study /18/ the risk of non-insulin dependent diabetes mellitus increased significantly with increasing HCT levels. There was more than a fourfold elevation in relative risk of diabetes among men with a HCT of ≥ 0.48 relative to those with a HCT below 0.42, adjusted for age and body mass index. 15.4.5.3 Erythrocytosis Erythrocytosis is defined as an absolute increase in red blood cell mass and can be associated with an elevated HCT and hemoglobin level (Fig. 15.4-1 – Classification of erythrocytosis). An HCT > 0.51 in a male and > 0.48 in a female individual is above the normal threshold and thus elevated. Initial classification of erythrocytosis is on the basis of whether it is a primary process (also known as familial and congenital polycythemia) or a secondary process where the red cell production is driven by some other process /19/. Some clinicians use the term polycythemia interchangeably with erythrocytosis, the two are not synonymous. Polycythemia refers to an increased number of any hematopoietic cell in blood, be it erythrocytes, thrombocytes or leukocytes. A complicating matter is the term polycythemia vera, a type of chronic myeloid leukemia that only affects the erythroid lineage. 15.4.5.3.1 Primary erythrocytosis Primary erythrocytosis (also known as primary familial and congenital polycythemia, PFCP) is a pathology of erythroid progenitors, which display hypersensitivity to erythropoietin /19, 20/. PFCP is characterized by an isolated primary polycythemia in which an increased red cell mass is associated with subnormal erythropoietin levels. Refer to: Tab. 15.4-5 – Diseases and conditions with elevated HCT Tab. 15.4-6 – WHO criteria for polycythemia vera. A genetic abnormalites in concern to the EPO level are: Defects of the EPO signalling pathway (EPO receptor mutations). Refer to Fig. 15.4-2 – Homodimer erythropoietin receptor High oxygen affinity hemoglobins Bisphosphoglycerate mutase deficiency. Refer to Fig. 15.4-4 – Effect of 2,3 diphosphoglycerate on O2 saturation Mutations in genes in the oxygen sensing pathway. Refer to Fig. 15.4-3 – Increased synthesis of erythropoietin. 15.4.5.3.2 Secondary erythrocytosis Secondary erythrocytosis results from a mechanism other than intrinsic to the bone marrow and is driven by increased production of erythropoietin (EPO)(predominantly reactive) /19, 20/. The responsiveness of erythroid progenitors to circulating EPO is usually normal. Erythrocytosis is associated with inappropriately normal or raised EPO levels indicating a defect in the control of EPO synthesis by the oxygen-sensing pathway. The increased EPO secretion may represent either a physiologic response to tissue hypoxia, or a deregulation of the oxygen-dependent EPO synthesis. Refer to Tab. 15.4-5 – Diseases and conditions with elevated HCT. Etiologically, there are a number of different ways where a secondary erythrocytosis can result /25/: Central hypoxia driven process (e.g., reduced oxygen supply will lead to stimulation of EPO production and erythrocytosis) Local renal hypoxia driven process (e.g., a local hypoxia in the kidney leads to increased EPO production and erythrocytosis) Pathological EPO production (e.g., meningioma, cerebellar hemangioblastoma hepatocellular carcinoma, parathyroid adenomas) Exogenous EPO administration (EPO doping). 15.4.5.3.3 Idiopathic erythrocytosis The cause of erythrocytosis cannot be elucidated. These patients can be differentiated on the basis of their EPO levels /19/: One third have levels below the normal range and are likely display hypersensitivity to EPO Two thirds have inappropriately normal EPO levels for elevated HCT or increased EPO concentrations. This group are likely to have defects in the oxygen sensing pathway. 15.4.5.4 Decreased plasma volume Decreased plasma volume (hemoconcentration) in the presence of normal red blood cell mass leads to relative erythrocytosis. The decrease in plasma volume is due to disturbances of the water and electrolyte balance. These can be caused by /27/: Insufficient intake of fluids (e.g., small children, elderly individuals, seriously ill individuals) Insensible loss of water (e.g., perspiration) Diarrhea and vomiting Polyuria (e.g., due to disorders of the ADH thirst mechanism, abuse of diuretics, diabetes mellitus) Ingestion of medicines for the improvement of cardiac function Smoking; smokers may have relative, absolute or combined erythrocytosis Abuse of alcohol, tea, caffeine-containing and cola-containing drinks. 15.4.6 Comments and problems Blood collection Venous occlusion for too long a period (more than 2 min.) causes a significant rise in HCT /22/. Anticoagulant If EDTA concentrations are used that are higher than recommended (see Section 15.3 – Hemoglobin concentration), a falsely low HCT occurs because of a decline in the MCV /2/. Sample Arterial blood has an approximately 2% higher HCT than venous blood. Method of determination Because of trapped plasma, the values of the PCV are approximately 2% higher than the HCT determined by hematology analyzers. The differences are higher yet if abnormal RBCs are present (e.g., sickle cells, thalassemia, iron deficiency, spherocytes, macrocytes) /2/. In the presence of a high reticulocyte or WBC count, HCT determinations using hematology analyzers result in the calculation of falsely elevated values, because the higher cell volumes of reticulocytes and white blood cells enter into the calculation of the HCT. Falsely low values are determined in the case of in vitro hemolysis, autoagglutination, and microcytosis. Point of care analyzers that are used for bedside HCT determinations, measure the HCT by means of the conductivity of undiluted blood. In patients with an increased plasma osmolality, the HCT determination gives lower values /23/. Stability Up to 24 h at room temperature or 4–8 °C, if the HCT is determined by an automated hematology analyzer. Centrifugation within 6 h for the micro hematocrit method /1/. Leukemia The white leukocyte pellet must be ignored when the results of the capillary method are interpreted /1/. Intraindividual variation Within-one-day variation 4.6%, day-to-day variation 4.1%, month-to-month variation 3.4% /24/. References NCCLS. Procedure for determining packed cell volume by the microhematocrit method – second edition; approved standard. NCCLS Document H7–A2, Vol 13, No 9. Villanova: NCCLS, 1993. Fairbanks VF, Tefferi AY. Normal ranges for packed cell volume and hemoglobin concentration in adults: relevance to apparent polycythemia. Eur J Hematol 2000; 65: 285–96. Kujala UM. Hemoglobin and packed-cell volume in endurance athlets prior to rhEPO. Int J Sports Med 2000; 21: 228. Geaghan SM. Hematologic values and appearances in the healthy fetus, neonate and child. Clin Lab Med 1999; 19: 1–37. Mordechai S, Merlob P, Reisner SH. Neonatal polycythemia: I. Early diagnosis and incidence relating to time of sampling. Pediatrics 1984; 73: 7–10. Shohat M, Reisner SH, Mimounie F, Merlob P. Neonatal polycythemia: II. Definition related to time of sampling. Pediatrics 1984; 73: 11–3. Segel GB, Oski FA. Hematology of the newborn. In: Williams WJ, Beutler E, Erslev AJ, Lichtman MA. Hematology, 4th ed. New York: McGraw-Hill, 1990: 102. Taylor MRH, Holland CV, Spencer R, Jackson JF, O’Connors GI, O’Donnell JRO. Haematological reference ranges for schoolchildren. Clin Lab Haem 1997; 19: 1–15. Fairbanks VF, Klee GG, Wiseman GA, Hoyer JD, Tefferi A, Petitt RM, Silverstein MN. Measurement of blood volume and red cell mass: reexamination of 51Cr and 125J methods. Blood Cells, Molecules and Diseases 1996; 22: 169–86. Alverson DC. The physiologic impact of anemia in the neonate. In: Bifano EM, Ehrenkranz RA (eds). Clinics in perinatology 1995; 22: 609–25. Leone BJ, Spahn DR. Anemia, hemodilution, and oxygen delivery. Anest Analg 1992; 75: 651–3. Besarab A, Bolton K, Brown JK, et al. The effects of normal as compared with low hematocrit values in patients with cardiac disease who are receiving hemodialysis and Epoetin. N Engl J Med 1999; 339: 584–90. Kosiborod M, Curtis JP, Wang Y, Smith GL, Masoudi FA, Foody JM, et al. Anemia and outcomes in patients with heart failure. Arch Intern Med 2005; 165: 2237–44. Boneu F, Fernandez F. The role of the hematocrit in bleeding. Transfus Med Rev 1987; 1: 182. Stefanidis I, Heintz B, Frank D, Mertens PR, Kierdorf HP. Influence of hematocrit on hemostasis in continuous venovenous hemofiltration during acute renal failure. Kidney Int 1999; 56 Suppl 72: 51–5. Heyman A, Karp HR, Heyden S, et al. Cerebrovascular disease in the biracial population of Evans county, Georgia. Arch Intern Med 1971; 128: 949–51. Soriie PD, Garcia-Palmieri MR, Lstas R, Havlik RJ. Hematocrit and the risk of coronary heart disease: the Puerto Rico Heart Program. Am Heart J 1981; 101: 456–61. Wannamethee SG, Perry IY, Shaper AG. Hematocrit and risk of NIDDM. Diabetes 1996; 45: 576–9. McMullin MF. The classification and diagnosis of erythrocytosis. Int Jnl Lab Hem 2008; 30: 447–59. Cao M, Olsen RJ, Zu Y. Polycythemia vera. Arch Pathol Lab Med 2006; 130: 1126–32. Landaw SA. Polycythemia vera and other polycythemic states. Clin Lab Med 1990; 10: 857–71. Junge B, Hoffmeister H, Feddersen HM, Röcker L. Standardisierung der Blutentnahme. Dtsch Med Wschr 1978; 103: 260–5. Stott AW, Hortin GL, Wilhite TR, Miller SB, Smith CH, Landt M. Analytical artifacts in hematocrit measurements by whole-blood chemistry analyzers. Clin Chem 1995; 41: 306–11. Costongs GMPJ, Janson PCW, Bas BM, et al. Short-term and long-term intraindividual variations and critical differences of haematological laboratory parameters. J Clin Chem Clin Biochem 1985; 23: 69–76. Balloch AJ, Cauchi MN. Reference ranges for haematology parameters in pregnancy derived from patient populations. Clin Lab Haemat 1993; 15: 7–14. Selby GB, Eichner ER. Hematocrit and performance: the effect of endurance training on blood volume: Sem Hematol 1994; 31: 122–7. Vincent JL, Sakr Y, Creteur J. Anemia in the intensive care unit. Can J Anesth 2003; 50: S53–S55). Sachs V. Hämotherapie – ein Leitfaden. Ärztl Lab 1984; 30: 204–20. Musallam KM, Tamim HM, Richards T, Spahn DR, Rosendaal FR, Habbal A, et al. Preoperative anemia and postoperative outcomes in non-cardiac surgery: a retrospective cohort study. Lancet 2011; 378: 1396–1407. Bernard AC, Davenport DL, Chang PK, Vaughan TB, Zwischenberger JB. Intraoperative transfusion of 1 U to 2 U packed red blood cells is associated with increased 30-day mortality, surgical site-infection, pneumonia, and sepsis in general surgery patients. J Am Coll Surg 2009; 208: 931–7. Cao M, Olsen RJ, Zu Y. Polycythemia vera. Arch Pathol Lab Med 2006; 130: 1126–32. Spivak JL. Myeloproliferative neoplasms. N Engl J Med 2017; 376: 2168–81. Gordeuk VR, Key NS, Prchal JT. Re-evaluation of hematocrit as a determinant of thrombotic risk in erythrocytosis Haematologica 2019; 104 (4): 653–8. Pasquier F, Marty C, Balligand T, Verdier F, Grosjean S, Gryshkova V, et al. New pathogenic mechanisms induced by germline erythropoietin receptor mutations in primary erythrocytosis. Haematologica 201(, 103. 675–86. Drew JH, Guaran RL, Grauer S, et al. Cord whole blood viscosity: measurement, definition, incidence and clinical features. J Pediatr Child Health 1991; 27: 363–5. Gaston RS, Julian BA, Curtis JJ. Posttransplant erythrocytosis: an enigma revisited. Am J Kidney Dis 1994; 24: 1–11. Fallert-Müller A. Lexikon der Biochemie. Heidelberg 1999; Spektrum Akademischer Verlag: 430. Camps C, Petousi N, Bento C, Cario H, Copley RR, McMullin MF, et al. Gene panel sequencing improves the diagnostic work-up of patients with idiopathic erythrocytosis and identifies new mutations. Haematologica 2016, 101: 1306–18. Drevon L, Maslah N. Anemia and hemodilution: analysis and a single center cohort based on 2,858 red cell mass measurements. Haematologica 2020; 106 (4): 1166–71. Borani O, Varettoni M, Riccaboni G. Erythrocytosis in congenital heart defects: hints for diagnosis and therapy from a clinical case. Front Med 2024. doi: 103389/fmed.2024.1419092. 15.5 Dyshemoglobins and cell free hemoglobin Lothar Thomas Increased concentrations of dyshemoglobins such as methemoglobin (hemiglobin), carboxyhemoglobin and sulfhemoglobin limit the oxygen-carrying capacity of arterial blood. These hemoglobin (Hb) fractions are not detected by blood gas analysis when measuring blood oxygen status and therefore can cause misinterpretation when the functional oxygen saturation is used for patient evaluation /1/. 15.5.1 Methemoglobin (metHb), hemiglobin (Hi) Methemoglobin is an oxidized form of Hb. Many oxidant chemicals and drugs are capable of inducing methemoglobinemia and can also cause hemolysis. 15.5.1.1 Indication Cyanosis and suspicion in: Toxic methemoglobinemia Hereditary methemoglobinemia Decrease in arterial oxygen saturation which, from the clinical point of view, primarily cannot be accounted for. 15.5.1.2 Method of determination Principle: metHb has a spectral absorbance curve with a characteristic maximum at 630 nm, in weak acid solution. For increase of the analytical specificity a measurement of absorption difference is made. MetHb reacts with cyanide ions provided by KCN to form hemiglobin cyanide which has no absorption at 630 nm /2/. 15.5.1.3 Specimen EDTA blood, heparinized blood: 1 mL For the determination, hemolysate is used (1 part of whole blood + 5 parts of distilled water). 15.5.1.4 Reference interval MetHb: 0.2–1.0% /3/ 15.5.1.5 Clinical significance In vivo, ferrous (Fe2+) Hb is continuously oxidized to ferric (Fe3+) Hb and hence metHb is produced. The reduction of metHb to Hb is catalyzed by the NADH-dependent methemoglobin reductase. To a small extent, a nonenzymatic reduction of metHb is also possible by ascorbic acid or reduced glutathione. The presence of methemoglobinemia reduces the oxygen-binding capacity, since Fe3+ is no longer able to reversibly bind oxygen. Methemoglobinemia is said to be present if the proportion of oxidized iron in hemoglobin exceeds 1%. MetHb inducers act by oxidizing Fe2+ to Fe3+ hemoglobin, resulting in the formation of metHb. Methemoglobinemia should be considered if a patient’s arterial oxygen saturation measured with pulse oximetry, is as low as about 85%. 15.5.1.5.1 Hereditary methemoglobinemia The term describes an autosomal recessive inherited condition in which there is a deficiency in NADH-dependent methemoglobin reductase; its activity is below 20%. The metHb proportion in blood is 8–40%; finding of a chocolate brown blood color. Heterozygous metHb reductase deficiency is not associated with methemoglobinemia /4/. Neonates and individuals with congenital NADH-dependent metHb reductase activity or glucose-6-phosphate dehydrogenase deficiency have an impaired ability to regenerate normal Hb and are therefore more likely to accumulate metHb after oxidant exposure. Thus, the administration of drugs or nitrate-containing water, which is converted to nitrite in the gastrointestinal tract, can cause methemoglobinemia. The clinical manifestations of congenital methemoglobinemia are cyanosis and neurological disturbances. 15.5.1.5.2 Toxic methemoglobinemia Activity of NADH-dependent met-Hb reductase is normal in toxic methemoglobinemia. Methemoglobinemia occurs /3, 5/: In occupational risk groups especially chemical and munitions workers Dependent upon metHb inducers (drugs, chemicals) which directly convert Hb to metHb or indirectly promote conversion via the formation of oxygen radicals in the circulation. An important environmental source of methemoglobinemia in infants is nitrate-contaminated well water. Refer to: Tab. 15.5-1 – Agents capable of inducing methemoglobinemia Tab. 15.5-2 – Correlation of clinical symptoms with increasing proportions of metHb. Treatment involves administration of oxygen and infusion of methylene blue for the reduction of Fe3+ to Fe2+. 1 mg of methylene blue/kg body weight is administered in a 10 minute infusion. Mild methemoglobinemia (below 20%) will resolve spontaneously without intervention. 15.5.1.6 Comments and problems Stability In intact RBCs, metHb is stable only for about 5 h because it is rapidly converted back to Hb. However, if the blood sample is diluted with 5 parts of distilled water and the erythrocytes are hemolyzed, stability is guaranteed for at least 45 h. Fluoride should not be used, as its interaction with metHb leads to erroneously low total Hb and metHb values /1/. Interference factors Hypertriglyceridemia and hyperbilirubinemia interfere with the metHb determination. Using multi-wavelength photometry, interference by triglycerides or by bilirubin does not occur until the concentration reaches levels of about 1000 mg/dL (11.4 mmol/L) or about 10 mg/dL (171 μmol/L), respectively /6/. 15.5.2 Carboxyhemoglobin (COHb) Carbon monoxide (C0) is a colorless, odorless, tasteless and nonirritating gas produced by incomplete combustion of any carbon-containing material. Sources of exposures include smoke inhalations in fires, automobile exhaust fumes, faulty or poorly ventilated charcoal, kerosene or gas stoves, cigarette smoke and methyl chloride /5/. Silent killer is the most common nickname for CO. 15.5.2.1 Indication Non-specific signs such as flu-like symptoms, headache, syncope, seizures occurring for the first time, and a history of carbon monoxide exposure. 15.5.2.2 Method of determination CO measurement methods /9/: CO monitor: the volume of CO is measured in the end tidal breath in parts per million (ppm) and is correlated to blood Hb concentration COHb measurement: Hb is measured spectrophoto-metrically at multiple wavelengths. Different correction fractures are used in order to eliminate interferences (e.g., due to elevated bilirubin or high triglyceride levels). A triple wavelength measurement is performed (e.g., at 415, 380 and 450 nm /17/ or at 578, 562 and 598 nm) total blood carbon monoxide (TBCO): The measurement of TBCO allows the measurement of total amount of CO present in the blood sample at the time of sampling and includes the CO bound to Hb and the amount of free CO present in the blood sample. A disadvantage of TBCO measurement is that it is analyzed using GC-MS/MS only HPLC combined with absorption spectrophotometry: the method is implemented in two steps. At first, protein-bound and free Hb are separated by HPLC. The free Hb fraction is diluted with acetic acid, H2O2 and tetramethylbenzidine. A blue dye is formed and measured spectrophotometrically at 600 nm. 15.5.2.3 Specimen Whole blood (EDTA, oxalate, heparin): 5 mL The sample tube needs to be filled in such a manner that, if at all possible, only a small volume of air is above the blood. 15.5.2.4 Reference interval \| COHb /8/ \| Non-smokers \| ≤ 3,0% \| \| Smokers \| ≤ 10,0% \| Recommended values 15.5.2.5 Clinical significance Environmental CO exposure is typically below 0.001% (10 ppm), but may be higher in urban areas. The quantity of CO absorbed by the body is dependent upon the respiratory minute volume, duration of exposure and CO and O2 concentrations in the environment. After cooking with a gas stove indoor CO concentration reach 100 ppm. A cigarette smoker is exposed to 400–500 ppm of CO and automobile exhaust may contain 10,000 ppm CO. A 4-hour exposure to 70 ppm results in a COHb content of 10%, and following an exposure of the same duration to 350 ppm CO may lead COHb level of 40% /9/. In the USA, the Occupational Safety and Health Administration permissible CO exposure in workers is 50 ppm averaged over an 8-hour work day /9/. The most frequent sources of CO poisoning include exposures to smoke inhalations in fires, automobile exhaust fumes, faulty or poorly ventilated charcoal, kerosene or gas stoves and cigarette smoke. In Germany, the maximal allowable CO concentration is set at 30 ppm (33 mg/m3) /8/. The pathological effect of CO is due to a decrease in O2 transport capacity of the blood and a reduction in tissue O2 extraction and a direct CO toxicity at the cellular level. At low levels endogenously CO functions as a neurotransmitter (see Section 19.2 – Oxidative stress) and modulates inflammation, apoptosis, and cell proliferation /10/. CO affinity for Hb is more than 240 times that of O2 and shifts the oxyhemoglobin dissociation curve to the left (see Fig. 15.4-4 – Effect of 2,3-diphosphoglycerate on O2 saturation curve). CO is preferentially bound in the lungs and causes a blockade of O2 diffusion in the lungs and muscles, since CO dissociates less rapidly from Hb than O2. The presence of COHb always signifies diminished O2 supply to the tissues, which is not detected by measuring the Hb value and which is very much more marked than an equivalent reduction in Hb /10/. Clinical symptoms are shown in Tab. 15.5-3 – Clinical symptoms in dependence of the COHb concentration. 15.5.2.5.1 COHb in healthy individuals Serum COHb levels in nonsmokers would be expected to have levels of less than 1–3% from endogenous production and background environmental exposure. Increases in non-smokers to some 3% are mainly associated with passive smoking, rather than environmental pollution. CO content in tobacco smoke is 4% and most smokers have, depending on the number of cigarettes they smoke daily, a blood COHb content of 3–8%. Certain professional groups (e.g., non-smoking fire fighters) have COHb values that are 1–2% higher than those in non-smokers. The elimination half-life time of COHb is approximately 4 hours with the breathing of room air and 1 hour with the breathing of 100% O2 /9/. 15.5.2.5.2 Constant mild CO elevations At a constant CO fraction of 25 ppm, the blood COHb content will be 3.5%, at 50 ppm it increases to 6–8% /9/. Even short-lasting low CO elevations in the inhaled, air can lead to COHb values of 2–6% and can trigger angina pectoris-like symptoms and arrhythmia in patients with atherosclerosis, under stress /11/. 15.5.2.5.3 Pregnancy and smoking CO crosses the placenta readily and small quantities of COHb in maternal blood have a considerable effect on the fetal O2 supply. Fetal Hb has high O2 affinity (P50, 19.4 mm Hg) in comparison with adult Hb (P50, 26.3 mm Hg at sea level), and this favors O2 uptake from the hypoxic maternal uterine blood. At the nadir of the HbO2 dissociation curve, fetal O2 saturation is only 75–80%. Therefore the fetus is vulnerable even to small fluctuations in O2 saturation /10/. If a pregnant woman smokes, for example, one pack of cigarettes daily, the COHb fraction may be 6% or more. In consequence, the maternal P50 is decreased from 26 to 23 mmHg and the O2 partial pressure in the uterine blood declines from 38 to 32 mmHg, which leads to a reduction of diffusion gradients toward the placenta. As a result, the umbilical cord blood O2 partial pressure drops from 28 to 22 mmHg, fetal arterial O2 saturation decreases from 75% to 58% and the fetus goes into a state of O2 hypoxia /12/. 15.5.2.5.4 CO poisoning The clinical symptoms of acute CO poisoning are headache, nausea, confusion, stupor and coma. In mild poisoning with CO values up to 25%, the patient represents the physician with flue-like symptoms /9/. The symptoms in chronic CO poisoning are headache, nausea light-headedness, cerebellar dysfunction and mood disorders. Nonetheless there is, in addition, considerable impairment of intellectual function, with lapses of concentration and cognitive disturbances. Even 3 years after the patient is removed from the environment, more than 40% of patients still manifest neurological problems. CO-associated symptomatology is often not taken into consideration so that the corresponding examinations are not performed. The clinical severity of CO poisoning often correlates poorly with the blood COHb level and is, rather, determined by the duration of poisoning. Therefore, a patient who attains a high COHb level after a brief, high-level exposure may not manifest any clinical symptomatology, whereas a patient who attains the same COHb level after a prolonged lower-level exposure may be significantly symptomatic /9/. Five severity grades of CO poisoning are established and, accordingly, a COHb fraction of over 50% is considered to be potentially fatal (Tab. 15.5-3 – Clinical symptoms in dependence of the COHb concentration). The cherry red appearance of blood and tissues is an unreliable sign and is only found with the most severe CO poisonings. In forensic medicine, a COHb content exceeding 50% is indicative of CO poisoning as the primary cause of death. Values of 10–50% indicate that smoke was inhaled and that CO may be a factor related to the cause of death, but that the victim was still alive when the fire began. Pathophysiologically, the poor correlation between CO poisoning and clinical effects is explained on the basis of the combination of Hypoxia due to formation of COHb and a direct toxic effect of soluble CO in blood at the cellular level. The effects of CO are not confined to the period immediately after exposure /9/. Persistent or delayed neurological symptoms with loss of memory, confusion, ataxia, seizures, urinary and bowel incontinence, disorientation and psychiatric symptoms develop after a latency period of 2–40 days /9/. 15.5.2.5.5 Increased endogenous CO accumulation Endogenous CO accumulates during degradation of Hb and myoglobin as well as of enzymes containing heme structures, such as peroxidase, catalase or cytochrome C. The endogenous formation of CO is responsible for physiological concentrations of COHb in the blood. Elevated endogenously dependent COHb concentrations occur in conditions of severe hemolysis or myolysis. In COPD patients, COHb values are higher in stage IV than in stages II or III /13/. Values of COHb as high as 12% are measured in neonatal jaundice. In sickle cell anemia, endogenous CO leads to a worsening of the situation: CO and, therefore, COHb, accumulate increasingly due to the shortened lifespan of erythrocytes COHb increases the affinity of other binding sites for O2, an effect originally described by Haldane. This shifts the oxygen dissociation curve to the left makes it less S-shaped and more hyperbolic (See Fig. 15.3-2 – The O2 dissociation curves in dependence of PO2 and the red cell content of 2,3-DPG). In sickle cell disease this shift in the dissociation curve would have the effect of reducing the level of deoxyhemoglobin and thus the tendency for HbS polymerization and sickling (see also Tab. 15.3-9 – Classification and differentiation of microcytic anemia) /14/. 15.5.2.6 Comments and problems Sample Blood sampling should be performed with hermetic test tubes with screw caps. For laboratory storage, there should be as little free airspace as possible on top of the anticoagulated blood /15/. Method of determination COHb measurement: Despite being a direct biomarker of CO exposure often low COHb concentrations are measured in patients with CO exposure suspicion or with symptoms that are normally associated to higher COHb concentrations. CO can be measured in breath through a CO monitor which measures the volume of CO in the end tidal breath in parts per million (ppm) and is correlated to blood COHb values. Total blood carbon monoxide (TBCO). The poisoning diagnosis is only confined to the effects caused by direct CO toxicity e.g., increased oxidative stress, impaired mitochondrial function, inflammation of the brain. Gas chromatography: this methode is the gold standard and capable of detecting low concentrations below 2.5% COHb. For this reason, tests in which hemolysis-dependent COHb accumulation is to be measured should only be performed using gas chromatography. CO spectrometric results correlate with those of gas chromatography as of COHb values of ≥ 2.5%/9/, but good comparability is only ensured with values of over 5% /14/. Gas chromatography has a limit of detection of less than 0.1% COHb /15/. Interference factors Hypertriglyceridemia leads to erroneously high values with spectrophotometric determinations, because a constant absorption factor is added. Methemoglobinemia also causes falsely high COHb values with the use of this method /6, 15/. Stability Whole blood should be stored refrigerated or deep frozen and exposure of samples to strong light must be avoided. The total sample CO content can remain constant over weeks to years at 3 °C or deep frozen /15/. 15.5.3 Free hemoglobin Cell free hemoglobin (plasma free heme, plasmaprotein bound heme and cell free hemoglobin) are mediators of tissue injury in hemolytic disease. Cell free hemoglobin is a potent scavenger of nitric oxide (NO), activates inflammation and undergoes redox cycling reactions that cause oxidative stress. Cell free heme also stimulates oxidative stress, activates TLR4, and the inflammasome leading to exacerbated inflammation-mediated tissue injury. TLR4 is a transmembrane protein, member of the toll-like receptor family, which belongs to the pattern recognition receptor family. Cell free hemoglobin and total Hb will co-exist and need to be distinguished from each other when quantifying /16/. 15.5.3.1 Indication Hemolytic disease Extracorporeal membrane oxygenation support Cardiopulmonary bypass Transfusion toxicity Dialysis Sickle cell disease 15.5.3.2 Method of determination Spectrometric method Principle: Hb is measured at multiple wavelengths and different correction factures are used in order to eliminate interferences (e.g., due to elevated bilirubin or high triglyceride levels). A triple wavelength measurement, is performed (e.g., at 415, 380 and 450 nm /17/ or at 578, 562 and 598 nm). Immunonephelometric quantification Rabbit antiserum raised against human hemoglobin A with no cross reactivity against myoglobin is used /18/. HPLC combined with absorption spectrophotometry Principle: the method is implemented in two steps. At first, protein-bound and free Hb are separated by HPLC. The free Hb fraction is diluted with acetic acid, H2O2 and tetramethylbenzidine. A blue dye is formed and measured spectrophotometrically at 600 nm. Inspection of plasma Hemolysis is visible as free Hb at concentrations ≥ 300 mg/L /19/. 15.5.3.3 Specimen Heparin plasma, citrate plasma: 1 mL 15.5.3.4 Reference interval \| Free hemoglobin: \| Plasma < 100 μg/L /20/ \| 15.5.3.5 Clinical significance Normal catabolism of hemoglobin (Hb) takes place through extravascular hemolysis. Tissue macrophages engulf senescent erythrocytes or take up Hb of ruptured erythocytes. Within macrophages degradation of Hb and heme, catalyzed by hemoxygenase, leads to the formation of amino acids, iron, CO and biliverdin, then converted to bilirubin /20/. In intravascular hemolysis free Hb dissociates in αβ-dimers or follows spontaneous oxidative reactions. The clearance of αβ-dimers or oxidative products takes place after they are complexed by the plasmaproteins haptoglobin and hemopexin. Haptoglobin reacts with αβ-dimers and neutralizes its reactive ferric part whereas hemopexin binds free heme. The complexed haptoglobin (Hp) and hemopexin (Hx) are taken up by hepatocytes and splenocytes (receptor CD163 for Hb-Hp complexes and CD91 for heme-Hx complexes). Part of the dissociated αβ-dimers (32 kD) cross the wall of the vessels and reach the extravascular spaces. Hb dimers freely circulate in the blood stream /20/: In overwhelming scavenging capacity of haptoglobin In patients with haptoglobin phenotype Hp 1-1. Phenotype Hp 2-2 individuals with complexes of free Hb and Hp 2-2 exhibit higher functional affinity for the scavenger receptor CD163 on macrophages as other phenotypes. There is a marked relationship between free Hb concentration in plasma and phenotypes of haptoglobin. Individuals with Hp 1-1 have the highest concentration free Hb, followed by phenotype Hp 2-1. The concentration of free Hb of phenotype Hp 2-2 is 61% than in those with the Hp 1-1 phenotype. In Hp 1-1 and Hp 2-1 individuals the concentration of free Hb is also positively correlated with the Haptoglobin concentration /21/. Diseases and pathophysiologic aspects of consequences of free hemoglobin are shown in Tab. 15.5-4 – Hemolysis and plasma free hemoglobin. In addition to the normal degradation of erythrocytes, daily hemolysis of 1% of the red blood cell mass, representing some 3 g of Hb, leads to complete disappearance of haptoglobin from plasma and to detection of free Hb /21/. A rise in free Hb is a more sensitive indicator of intravascular hemolysis than an increase in LD. Thus, with LD activity of 165 U/L, only hemolysis that leads to a free Hb concentration of 800 μg/L causes a rise in LD of 58% and thus a value that is above the upper reference interval value /22/. Severe hemolysis always manifests increased LD and if the rate of hemolysis is > 5%, representing a daily release of Hb of ≥ 15 g, non-conjugated bilirubin is also increased. In vivo and in vitro hemolysis can be distinguished by determination of potassium and haptoglobin. A hemolytic sample with elevated potassium and normal haptoglobin value raises suspicion of the presence of in vitro hemolysis. Refer to Tab. 15.5-4 – Hemolysis and plasma free hemoglobin. 15.5.3.6 Comments and problems Anticoagulant EDTA must not be used as anticoagulant. Free Hb is some 20-fold higher in EDTA plasma than in heparin plasma. Serum should not be investigated, since Hb is released from the erythrocytes during coagulation. In absence of intravascular hemolysis, the serum concentration of free Hb can be 50–100 mg/L /18/. Method of determination COHb measurement: Despite being a direct biomarker of CO exposure often low COHb levels are measured in patients with CO exposure suspicion or with symptoms that are normally associated to higher COHb levels. These discrepancies can be due to previously administered oxygen that reduced the COHb levels prior to hospitak admission.When using COHb as biomarker the effects caused by direct CO toxicity (e.g., increased oxidative stress, impaired mitochondrial function inflammation in brain tissue) are not taken into account. Carbon monoxide pulse oximetry can not be a screening tool for CO poisoning in the emergency department due to its low sensitivity. This method of determination can have a role in the prehospital setting as a tool to quickly identify CO poisoned patients /29/. While bilirubin and hyperlipidemia strongly interfere with the spectrophotometric methods according to reference /17/, they hardly do so /23/ with the method described in reference /24/. Only hyperlipidemia visible to the naked eye causes interferences in the immunonephelometric method /23/. Reference range /20/ Reference ranges are dependent on the method of determination. Old data from healthy individuals found free Hb concentrations below 50 mg/L. New studies agree to consider as pathologic cut-off value of 100 mg/L. References Lim SF, Tan IK. Quantitative determination of methemoglobin and carboxyhemoglobin by co-oximetry, and effect of anticoagulants. Ann Clin Biochem 1999; 36: 774–76. Taulier A, Levillain P, Lemonnier A. Determining methemoglobin in blood by zero crossing, point first, derivative spectrophotometry. Clin Chem 1987; 33: 1767–70. Beutler E. Methemoglobinemia and sulfhemoglobinemia. In: Williams WJ, Beutler E, Erslev AJ, Lichtman MA. Hematology, 4th ed. New York: McGraw-Hill, 1990: 743. Wolak E, Byerly FL, Mason T, Cairns BA. Methemo- globinemia in critically ill burned patients. Am J Crit Care 2005; 14: 104–8. Olson KR (ed). Poisoning & Drug Overdose. Norwalk 1990, Appleton and Lange. Carl B, Kaehler H, Schlegel R. Einfache photometrische Bestimmung von Carboxyhämoglobin und Methämoglobin durch Mehrwellenlängenmessung. Lab Med 1990; 14: 299–305. Brown CM, Levy SA, Susann PW. Methemoglobinemia: life-threatening complication of endoscopy premedication. Am J Gastroent 1994; 89: 1108–9. Deutsche Forschungsgemeinschaft. Photometrische Bestimmung von Carboxy-Hämoglobin (COHb) im Blut. Mitteilung VIII der Senatskommission für klinisch-toxikologische Analytik. Weinheim: VCH, 1988: 1–104. Oliverio S. Current challenges in carbon monoxide poisoning diagnosis from an analytical perspective. 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Absorbance and redox based approaches for measuring free heme and free hemoglobin in biological matrices. Redox biology 2016; 9: 167–77. Harboe M. A method for determination of hemoglobin in plasma by near-ultraviolet spectrophotometry. Scand J Clin Lab Invest 1959; 11: 66–70. Lammers M, Gressner AM. Immunonephelometric quantification of free hemoglobin. J Clin Chem Clin Biochem 1987; 25: 363–7. Guder W, Fonseca-Wollheim F. Heil W, et al. Die hämolytische, ikterische und lipämische Probe. Empfehlungen zur Erkennung und Vermeidung klinisch relevanter Störungen. DG Klinische Chemie-Mitteilungen 1999; 30: 167–77. Dufour N, Radjou A, Thuong M. Hemolysis and plasma free hemoglobin during extracorporeal membrane oxygenation support: from clinical implications to laboratory details. ASAIO Journal 2020; 66: 239–46. Na Na, Ouyang J, Taes YEC, Delanghe J. Serum free hemoglobin concentrations in healthy individuals are related to haptoglobin type. Clin Chem 2005; 51 (9): 1754–5. Podlasek SJ, McPherson RA. New lactate dehydrogenase-IgM complexes. Laboratory Medicine 1989; 20: 617–9. Kahn SE, Watkins BF, Bermes EW. An evaluation of a spectrophotometric scanning technique for measurement of plasma hemoglobin. Ann Clin Lab Sci 1981; 11: 126–31. Bednar R, Bayer PM. Freies Hämoglobin im Plasma. Vergleich zweier spektralphotometrischer Methoden. Bilirubin als Störfaktor. Lab Med 1994; 18: 196–9. Rother RP, Bell L, Hillmen P, Gladwin MT. The clinical sequelae of intravascular hemolysis and extracellular plasma hemoglobin: a novel mechanism of human disease. JAMA 2005; 293: 1653–62. Chambers SD, Ceccio SL, Annich GA, Bartlett RH. Extreme negative pressure does not cause erythrocyte damage in flowing blood. ASAIO J. 1999; 45: 431–5. Adamzik M, Hamburger T, Petrat F, Peters J, de Groot H, Hartmann M. Free hemoglobin concentration in severe sepsis: methods of measurement and predictin of outcome. Critical Care 2012; 16: R125 431–5. Gladwin MT, Kanias T, Kim-Shapiro DB. Hemolysis and cell-free hemoglobin drive an intrinsic mechanism for huma disease. J Clin Invest 2012; 122 (4): 1205–8. Ramponi G. The diagnostic accuracy of carbon monoxide pulse oximetry in adults with suspected acute carbon monoxide poisoning: a systematic review. Front Med 2023. doi: 10.3389/fmed.2023.1250845. 15.6 Reticulocyte count and reticulocyte indices Lothar Thomas The reticulocyte analysis of peripheral blood samples is an indicator of the erythropoietic activity of the bone marrow. Automated reticulocyte analysis involves the measurement of the reticulocyte count and most often of reticulocyte indices such as The cell volume (MCVr) The cellular concentration (CHCMr) and content of hemoglobin (CHr, RetHe) The reticulocyte maturity pattern. 15.6.1 The reticulocyte The reticulocyte is an immature red blood cell (RBC) from which the nucleus has been extruded. Reticulocytes are transitional RBC between nucleated RBC and the mature RBC. A reticulocyte is a RBC which, when stained with a supravital stain, contains precipitable ribonucleic acid (RNA). The RNA content is still high enough to be identified using optical fluorescent methods in the flow cytometer or panoptically under the microscope. To be identified with microscopic techniques as a reticulocyte, the cell must contain two or more blue-colored RNA particles that are visible without requiring fine focus adjustment on the individual cell to confirm their presence The granules should be distant from the cell margin to avoid confusion with Heinz bodies /1/. The following groups are distinguished morphologically /2/: Group 0; normoblasts and megaloblasts, both contain a nucleus and a dense perinuclear reticulum Group 1; reticulocytes with a reticulum in the form of dense clumps Group 2; reticulocytes with an annular reticulum Group 3; reticulocytes with a diffusely scattered reticulum Group 4; reticulocytes with a reticulum in the form of scattered granules and fragments. At this point the final maturation process stage has been reached. The reticulocyte gradually loses its granules and fragments and becomes a mature erythrocyte. In the blood of healthy individuals, only a small fraction of reticulocytes belongs to groups 1 and 2, approximately 30% to group 3 and over 60% to group 4 /3/. In hyper-regenerative erythropoiesis, there occurs an increase of group 1 and 2 reticulocytes and the blood smear shows polychromasia of the erythrocytes /4/. 15.6.1.1 Reticulocyte maturation During the normal maturation of RBC in the bone marrow, there is a gradual condensation of the nuclear chromatin with a reduction of both nuclear and cellular size. When the nucleus becomes pyknotic, it is extruded. In parallel, the synthesis of Hb increases. The heme portion of Hb is synthesized in the mitochondria, and the globin chains in the polyribosomes. In erythroblasts the globin chains in RNA-containing ribosomes as well as mitochondrial heme are formed. Coupling of both components takes place in the mitochondria /5/. The RNA containing poly ribosomes can remain in the non nucleated RBCs for up to 4 days. During this period of time there is a continuous decrease in poly ribosomes and in Hb formation. About 25% of Hb is synthesized in the reticulocyte or the equivalently polychromatophilic anucleate stage of erythroid development. These cells have a greater volume than mature erythrocytes. Since Hb synthesis is not yet complete, the macrocytic erythrocytes will stain poly chromatically with Wrights-Giesma stain. The appearance of polychromatophilic macrocytes in the blood reflects the premature release of reticulocytes from the bone marrow and indicates enhanced erythropoiesis /4/. The primary reticulocyte maturation (3 days) occurs in the bone marrow. The reticulocyte leaves the bone marrow and completes its final maturity in the peripheral blood within one day. With the loss of protein-synthesizing poly ribosomes there is cessation of Hb synthesis and transformation from reticulocyte to mature erythrocyte /5/. Following pyknosis of the nucleus, reticulocytes can be stained with supravital dyes such as brilliant cresyl blue or new methylene blue. The polyribosomal RNA is stained. RNA can also be displayed with fluorescent dyes such as thiazole orange, acridine orange or pyronin Y /6/. 15.6.1.2 Disorders of reticulocyte maturation In acute blood loss with a decline in the Hb level to below 80 g/L, the bone marrow is stressed with resulting hyper regenerative erythropoiesis. This leads to the following changes in maturation: Reticulocytes no longer reside for 3 days but rather for only 1.5 days in the bone marrow and their final maturation is achieved in the blood. They reside in the circulation 1.7–3 for instead of 0.8–1.2 days /7/. The result is a rise in the reticulocyte count, with a shift of the reticulocytes towards maturation groups 1–3. Stress reticulocytes appear in the peripheral circulation. These are reticulocytes with a large volume and high RNA content (macro reticulocytes). Erythrocytes that are formed from these stress reticulocytes have a shortened life span /8/. 15.6.2 Reticulocyte count and derived indices The absolute reticulocyte count, the reticulocyte index and the reticulocyte production index are indicators of erythropoietic activity and permit conclusions regarding the effectiveness of erythropoiesis. 15.6.2.1 Indication Assessment of bone marrow erythropoietic activity following a diagnosis of anemia Distinguishing hemolytic or posthemorrhagic anemia (hyper regenerative) from anemia of chronic disease (hypo regenerative) Monitoring therapy (megaloblastic anemia, iron deficiency anemia) Checking early regeneration after marrow or stem cell transplant Monitoring therapy with erythropoiesis stimulating agents (ESA). 15.6.2.2 Method of determination The following can be determined: Reticulocyte count Reticulocyte index (hematocrit correction) Reticulocyte production index (shift correction). 15.6.2.2.1 Reticulocyte count Microscopic counting Staining of unfixed cells with vital stains produces a net-like precipitate, the granulofilamentous material, in the immature red blood cell. This is achieved by mixing whole blood with a vital stain (brilliant cresyl blue or new methylene blue) in a 1 : 1 ratio, followed by the preparation of blood smears on several slides. After air-drying, a microscopic examination is performed. Among 1000 red blood cells, all cells that contain a bluish, thread-like or granular precipitate are counted /9/. Automated reticulocyte counting The automated systems have in common the rapid analysis of whole blood suspensions in flow-through systems, with the red cell population interrogated on a cell-by-cell basis by laser light. The reticulocytes are stained in a separate procedure. Some analyzers utilize a detection principle based upon light absorption or scatter caused by the reagent-RNA precipitate within the reticulocyte. Others employ fluorescent reticulocyte reagents /10/. Fluorescence-activated cytometry: a fluorescent dye (e.g. thiazole orange, acridine orange, pyronin Y) binds to the RNA of the reticulocyte. The extent of the fluorescence emission that is produced by the excitation of the bound dye is directly proportional to the RNA content of the reticulocyte and inversely proportional to its maturity stage /6/. Flow cytometry: the detection of reticulocytes is based on the principle that precipitation and staining, due to the use of fixation and staining reagents, causes light to be absorbed or scattered. The stains employed include, for example, methylene blue or oxazine /10/. The reticulocyte count is either expressed as a percentage (number of reticulocytes/100 red blood cells) or as absolute cell count (109/L) /9/. 15.6.2.2.2 Reticulocyte index; RI (hematocrit correction) The reticulocyte count can be increased in relation to erythrocyte count either because more reticulocytes are in the circulation, or there are fewer erythrocytes /9/. Therefore, the observed reticulocyte count may be corrected to a normal hematocrit of 0.45 (45%) according to the equation 1 in Tab. 15.6-1 – Hematocrit correction and shift correction. 15.6.2.2.3 Reticulocyte production index; RPI (shift correction) Counts corrected for hematocrit (HCT) are not perfect indices of production, because the reticulocyte count can also be altered by premature release of cells from the marrow (shift) /9/. If shift cells (polychromatophilic macrocytes) are detected in the Wright-stained smear, an empirical correction must be applied for RBC maturation. The maturation time of the reticulocyte is taken as: 1 day at HCT of 0.45 (45%) 1.5 days at a HCT of 0.35 (35%) 2 days at a HCT of 0.25 (25%) 3 days at a HCT of 0.15 (15%) The calculation of the RPI is shown in Tab. 15.6-1 – Hematocrit correction and shift correction. If the patient has a corrected reticulocyte count (RI) of 10% and a HCT of 25% the RPI is 10/2 = 5. In this case a shift (RPI) > 3 is considered to represent an adequate erythropoietic response while an RPI < 2 is inadequate. The expected lower limits of the RPI for an adequate erythropoietic response in dependence of the HCT are shown in Tab. 15.6-2 – Expected lower RPI limits and reticulocyte count as function of HCT. 15.6.2.3 Specimen EDTA blood: 1 mL 15.6.2.4 Reference interval Refer to Tab. 15.6-3 – Reticulocyte reference intervals. 15.6.2.5 Clinical significance The reticulocyte count is an important indicator of bone marrow erythropoietic activity and of a significant reduction of erythrocytes. 15.6.2.5.1 Absolute reticulocyte count The reticulocyte count per volume is a measure of the bone marrow’s effectivity in forming mature RBC. This holds true for normo regenerative erythropoiesis (steady state), as well as for hyper- and hypo regenerative erythropoiesis /15/. 15.6.2.5.2 Relative reticulocyte count Expression of reticulocytes as percentage of red cells is a measure of the erythrocyte life span. It allows for an estimation of the shortening of the erythrocyte life span in chronic anemia in the steady state. The higher the relative reticulocyte count (%), the shorter the life span. Often, inexperienced physicians do not know that an apparently elevated percentage reticulocyte count may not be elevated when correction is made for a low erythrocyte count /15/. 15.6.2.5.3 Reticulocyte index (RI) In the steady state, after HCT correction an increased percentage of reticulocytes can be an indicator of a shortened erythrocyte life span. The higher the RI, the shorter the life span. 15.6.2.5.4 Reticulocyte production index (RPI) An elevated reticulocyte count can reflect /15/: A hyper regenerative erythropoiesis because of a shortening of the erythrocyte life span A premature release of red cells from the marrow. The RPI reflects the increase or reduction in erythropoiesis. Under ESA stimulation, an increase in reactivity of up to 8-fold is possible. Hypo reactivity is present if an increase in reactivity is below 2. Erythropoietin-stimulated reactivity increases as the HCT falls. In hyper regenerative erythropoiesis, the reticulocyte maturation time in bone marrow is shortened, proportional to the decline in HCT and the residence time in peripheral blood is prolonged. The HCT dependent prolonged residence time in blood is corrected with calculation of the RPI /1/. Without correction of the prolonged residence time, with increasing anemia erythrocyte production will be overestimated due to the absolute reticulocyte count and the erythrocyte life span will be underestimated due to the percentage of reticulocyte count. The expected RPI in dependence of HCT and reticulocyte count that can be achieved in intact marrow function are shown in Tab. 15.6-2 – Expected lower RPI limits and reticulocyte count as function of HCT /16/. With an HCT of 0.35, an RPI ≥ 2 is indicative of regenerative, an RPI ≥ 3 points to hyper regenerative erythropoiesis. An RPI < 2 denotes hypo regenerative erythropoiesis (e.g. anemia associated with chronic disease due to inflammation, infection, malignant disease or intrinsic hypoplasia of erythropoiesis). 15.6.2.5.5 Reticulocytosis With a HCT below 0.30 it is necessary to determine the RPI since, otherwise, reticulocytosis will be diagnosed too frequently /15/. Reticulocyte determination is particularly relevant in normocytic anemia. In macrocytic anemia, reticulocytosis is indicative of untreated folate or vitamin B12 deficiency /17/. In microcytic anemia, the reticulocyte count should only be determined if ferritin level or the transferrin saturation do not convey an unequivocal statement. Refer to: Fig. 15.6-1 – Differentiation of normocytic anemia by reticulocyte count Fig. 15.6-2 – Differentiation of macrocytic anemia by reticulocyte count Fig. 15.6-3 – Differentiation of microcytic anemia by reticulocyte count Tab. 15.6-4 – Diseases and conditions with reticulocytosis. 15.6.2.5.6 Reticulocytopenia Reticulocytopenia is a sign of hypo regenerative erythropoiesis and occurs in: Deficiency anemia (e.g. iron, copper or vitamin B6, vitamin B12 and folate deficiency) Anemia associated with chronic disease (infection, chronic inflammation), malignancy, chronic liver disease) caused by inflammatory cytokine-induced hypo proliferation of erythropoiesis /18/ Chronic renal insufficiency. Erythropoietic proliferation is diminished due to inappropriately low erythropoietin secretion /19/. Myelodysplastic syndrome (MDS). Anemia associated with MDS is hypo proliferative, with peripheral reticulocytopenia in spite of hyper cellular marrow and up to 90% ineffective erythropoiesis /20/. Concentration of soluble transferrin receptor is often elevated. Congenital dyserythropoietic anemia. Three types are distinguished. They manifest moderate anemia with Hb values around 90 g/L and a low to normal reticulocyte count /21/. 15.6.2.6 Comments and problems Method of determination Microscopic method: the microscopic analysis of the blood smear is imprecise, with intra- and inter laboratory coefficients of 25% and 25–50%, respectively, if the reticulocyte count is determined on the basis of 1000 red cells /9/. Howell-Jolly bodies, Heinz bodies and malaria parasites are also stained with the smear /9/. Hematology analyzer: counting accuracy is higher since some 10,000 cells are counted. The inter laboratory coefficient of variation for reticulocytosis > 2.5% is 24%, being approximately half as large as with microscopic counting with the same reticulocyte count /26/. Potential sources of error associated with flow cytometry are caused by Howell-Jolly bodies, nucleated red blood cells, sickle cells, giant thrombocytes, cold agglutinins, parasites (malaria, babesiosis), and platelet clumps. Cytometric methods distinguish systematically from one another based on the use of different dyes. Stability Strong depending on the staining technique and the method of determination. At 20 °C, decline after 24 h, at 4– 8 °C possible for up to 72 h and longer /27/. 15.6.3 Retikulocyte maturity index and immature reticulocyte fraction The Reticulocyte maturity index (RMI) and Immature Reticulocyte Fraction (IRF) are indices for the RNA content of reticulocytes. The quantification is based on the determination of the RNA content in the reticulocyte. Immature reticulocytes have a higher RNA content than mature forms. A rise in the number of reticulocytes in the blood leads to an increase in RMI and IRF. 15.6.3.1 Indication Assessment of erythropoietic activity: In severe anemia (e.g., hemolytic anemia) Following bone marrow transplantation and after chemotherapy. 15.6.3.2 Method of determination Reticulocyte maturity index (RMI) Hematology analyzers for reticulocyte counting utilize reagents that bind specifically and rapidly to nucleic acids, such as ethidium bromide, auramine O (Sysmex), CD4K530 (Abbott), oxazine 750 (Siemens), or new methylene blue (Beckman Coulter). The light energy of a laser is absorbed or scattered. Quantitative flow cytometric fluorescence measurements with the reagents like auramine O or ethidium bromide are also used. The reticulocyte population is generally classified, according to the corresponding threshold value that has been set, into 3 maturation stages: low fluorescence reticulocytes (LFR), medium fluorescence reticulocytes (MFR), and high fluorescence reticulocytes (HFR) /28/ (Fig. 15.6-4 – Differentiation of reticulocytes in maturation stages). Healthy individuals normally have reticulocytes of the LFR fraction. Immature Reticulocyte Fraction (IRF) The IRF is calculated as follows: IRF (%) = HFR (%) + MFR (%) 15.6.3.3 Specimen EDTA blood: 1 mL 15.6.3.4 Reference interval Refer to Tab. 15.6-5 – RMI and IRF reference intervals. 15.6.3.5 Clinical significance The maturation stage of reticulocytes in peripheral blood is mainly dependent upon: The severity of anemia The iron, vitamin B12 and folic acid status Stimulation with erythropoietin The presence of systemic inflammation. The RMI or IRF do not manifest a clear relationship to production of erythrocytes or to a reduction of their life span. However, under non-steady state conditions, they provide an early indication of incipient regeneration or suppression of erythropoiesis or its responsiveness to ESA therapy /15/. The diagnostic value of the RMI in combination with the reticulocyte count for clarification of anemia is shown in Tab. 15.6-6 – Diagnostic significance of reticulocyte count and reticulocyte maturity index /32/. 15.6.3.5.1 Acute hemorrhage In association with a decreased Hb, the reticulocyte maturation time in bone marrow is shortened and increasing numbers of immature reticulocytes pass into the blood. RMI and IRF increase due to elevations in MFR and HFR fractions /10/. Both of these indices increase within 5–8 hours of severe acute loss of blood, while a rise in reticulocytes only becomes significant after 24–48 hours. 15.6.3.5.2 Renal anemia, pernicious anemia, myelodysplastic syndrome Reticulocyte maturation is delayed and patients with these diseases have a reduced reticulocyte count in association with elevated RMI and IRF /10/. 15.6.3.5.3 Bone marrow transplantation The RMI is the earliest clinical indicator of engraftment following bone marrow transplantation /30/. If, on the 21st post-transplant day, the reticulocyte count reaches a value of 15 × 109/L and HFR a value of 0.5 × 109/L, this reflects to a degree of 100% the functioning of the transplant /32/. In contrast to an increase in polymorphonuclear neutrophils and the reticulocyte count, however, the RMI and IRF manifest no obvious advantages /31/. 15.6.3.6 Comments and problems There are various definitions of immature reticulocytes. In some studies, only the HFR fraction is considered, while in others the sum of HFR and MFR. For determination of the IFR, the HFR and MFR fraction are taken into consideration /31/. Method of determination The methods used by different hematology analyzers are not comparable, since no standard reticulocyte preparations are available and the manufacturers each have their own different calibration methods. Since the HFR fraction normally is only a small proportion of the reticulocytes, approximately 500 reticulocytes are found when 50,000 red blood cells are counted and only a few of these fall within the HFR group. Considerable disturbances with false positive HFR results can occur due to large blood platelets, leukocytes, erythroblasts, erythrocytes with malaria parasites, or Howell-Jolly bodies. Typically chronic lymphocytic leukemia may give spuriously RMI values /31/. 15.6.4 Reticulocyte indices Reticulocyte analysis has been extended from the simple counting of reticulocytes to precise measurements of cellular indices such as volume, hemoglobin concentration, and content /33/. 15.6.5 Mean cell volume of reticulocytes (MCVr) During maturation of erythropoietic progenitor cells in the bone marrow, their volume decreases continuously. Marked decreases occur: In bone marrow from stage 0 (orthochromatic erythroblast) to stage 1 (reticulocyte with reticulum appearing in the form of a dense clump), due to loss of the cell nucleus In blood during transition from stage 4 (reticulocyte with reticulum appearing as a few scattered granules or fragments) to the mature erythrocyte. The circulating reticulocyte has a diameter of 8.5 μm some 1–1.5 μm larger than the erythrocyte. The reticulocyte volume is on average 20% larger than that of the mature erythrocyte. Based on median mature erythrocyte volume (MCV) of 88 fL, median reticulocyte volume (MCVr) is around 106 fL. 15.6.5.1 Indication Suspected stress erythropoiesis following acute bleeding, deficient oxygen supply, over stimulation of bone marrow through ESA-therapy Assessment of therapeutic responsiveness in deficiency anemia (iron, folic acid, vitamin B12). 15.6.5.2 Method of determination Principle: laser-based technology utilized in the Advia 120 hematology analyzer allows simultaneous measurement of volume and Hb concentration of both erythrocytes and reticulocytes. From the measurements of cell volume and Hb concentration, the Hb content of every cell is determined. EDTA blood is diluted with reticulocyte reagent. The reagent contains sodium dodecyl sulfate, which forces red blood cells into a spherical shape. The oxacine staining method used for reticulocyte analysis measures staining intensity and the indices such as MCVr, and Hb concentration (CHCMr). The mean Hb content of reticulocytes (CHr) is calculated from the product of the volume times Hb concentration of each reticulocyte /31/. With Sysmex hematology analyzers, reticulocyte MCV is derived from forward scattering of fluorescence-labeled reticulocytes. 15.6.5.3 Specimen EDTA blood: 1 mL 15.6.5.4 Reference interval \| Adults /34/: 92–120 fL \| 15.6.5.5 Clinical significance A decrease in MCVr is typical in iron deficiency, while a rise may indicate folic acid and vitamin B12 deficiency anemia. Macro reticulocytes are formed in erythropoiesis associated with stress situations (e.g., in acute hypoxia). 15.6.5.5.1 Stress erythropoiesis The MCV of macro reticulocytes is ≥ 27% larger than that of erythrocytes /4/. In comparison with erythrocytes, MCV can be increased by a factor of 3. Macro reticulocytes are found under stress situations and are also designated as stress reticulocytes. In the blood smear they can be detectable as polychromatophilic erythrocytes. Stress reticulocytes are found: 5–8 h after severe acute bleeding During a stay at high altitudes As a response to successful treatment of iron deficiency anemia in the early phase (first 2–3 days) As a response to ESA therapy (ESA, erythropoiesis-stimulating agents). In individuals with normal iron status and without acute phase response, administration of ESA leads to a rise in the reticulocyte count and MCVr, however the reticulocyte Hb concentration (CHCMr) decreases. If functional iron deficiency develops, no increase in MCVr occurs and small reticulocytes are released in the blood. 15.6.5.5.2 Hemolytic anemia In acute and massive hemolytic anemia (e.g., autoimmune hemolysis) increased production of erythropoietin stimulates erythropoiesis and development of iron-restrictive erythropoiesis. The result is an inappropriately low reticulocyte count. The MCVr is reduced and an inversion of the MCVr/MCV ratio, which is normally greater than 1.0, occurs /35/. It has been shown in phlebotomy investigations that this situation occurs some 2 days following a reduction in transferrin saturation. The same results are observed in over stimulation with ESA. 15.6.5.5.3 Folic acid and vitamin B12 deficiency anemia Macrocytosis of erythrocytes and reticulocytes is present in both of these forms of anemia, and the MCVr/MCV ratio is greater than 1.0. With treatment of anemia, an inversion of the ratio with reticulocytes being smaller than erythrocytes is seen as a response to vitamin B12 therapy for megaloblastic anemia. The cause is a decrease in MCV which occurs sooner in the reticulocyte than in the erythrocyte. In a study /35/ the produced reticulocytes after 17 days treatment had an MCVr of 108.8 fL, while the MCV of erythrocytes was still 109.8 fL, because most of the circulating erythrocytes had been formed prior to the administration of vitamin B12. 15.6.6 Reticulocyte hemoglobin content (CHr or RetHe) Reticulocytes have a higher fluid content, a 1–3 pg higher Hb content, and up to some 20% greater volume than erythrocytes. The results is that reticulocytes are more hypochromic than erythrocytes. 15.6.6.1 Indication Evaluation the iron demand of erythropoiesis. Assessment the therapeutic response of iron deficiency anemia within several days after the start of treatment Monitoring of iron supply for erythropoiesis under ESA therapy. 15.6.6.2 Method of determination On Advia 120, the Hb concentration of individual reticulocytes (CHCMr) is measured and the Hb content is calculated according to the equation CHr = MCVr × CHCMr /31/. In addition, the Hb content of the reticulocyte fraction (RFHb) can be determined according to the equation RFHb = Reticulocyte number × CHr. On Sysmex NE-2100, the Ret-He is determined. The light intensity of the forward scattered fluorescence-labeled reticulocyte correlates with its Hb content. Results are expressed in pg /36/. 15.6.6.3 Specimen EDTA blood: 1 mL 15.6.6.4 Reference interval Tab. 15.6-7 – Reference interval for reticulocyte hemoglobin. 15.6.6.5 Clinical significance CHr and Ret-He are markers for determining the actual iron demand of erythropoiesis. A value below 28 pg indicates iron-restricted erythropoiesis. When iron demand starts, CHr and Ret-He decrease within 48–72 h. Other markers such as %HYPO or biochemical markers of iron metabolism indicate changes following 10–20 days at the earliest, while erythrocyte MCV and MCH do so only after 2 to 3 months. 15.6.6.5.1 Iron deficiency The detection of iron deficiency before microcytic hypochromic anemia develops is important to avoid systemic complications of the iron deficiency. Classical biochemical parameters for early diagnosis of iron deficiency are serum ferritin, the transferrin saturation and the soluble transferrin receptor. In children, in youth during the growth spurt, endurance athletes, women of menstrual age, and repeat blood donors, the ferritin value can be critical as an indicator of iron deficiency, since these individuals already have low values in spite of a still adequate supply of iron /41/. In healthy individuals treated with ESA, it could be shown that decreases in CHr and Ret-He are early indicators of iron deficiency /42/. As early as 3–5 days following the start of ESA therapy, a significant decline in CHr or Ret-He was measured. In the diagnostic investigation of iron deficiency in children, CHr was of higher predictive value than biochemical markers of iron deficiency /43/. The fraction of hypochromic reticulocytes (CHCMr below 270 g/L), which is normally less than 25%, is also an early indicator of iron-restricted erythropoiesis. As iron deficiency progresses, CHr and Ret-He decrease and the hypochromic reticulocyte fraction increases. 15.6.6.5.2 Functional iron deficiency Functional iron deficiency (ID) is a state of iron-poor erythropoiesis in which there is insufficient mobilization of iron from the stores in the presence of increased demands. Anemia with functional ID develops during increased erythropoiesis mediated either by endogenous erythropoietin responses to anemia, or by therapy with erythropoiesis stimulating agents (ESAs). In functional ID imbalance between the surging iron requirements of the stimulated erythroid marrow and iron availability is the pathophysiology. Iron is sequestered in macrophages of the reticulo-endothelial system. The iron stores are normal or increased (ferritin above 100–299 μg/L) in combination with transferrin saturation less than 20%. Thus, in functional iron deficiency there is a situation in which the iron demand of erythropoiesis, due to overstimulated erythropoiesis, exceeds the rate of iron release from the stores and the transferrin iron transporting capacity. In consequence, hypochromic reticulocytes and erythrocytes are released from the bone marrow. The reticulocyte Hb content (CHr, RetHe) is decreased /44/. In hemodialysis patients who are on ESA therapy, monitoring of CHr or RetHe is a better indicator of iron demand than biochemical markers of iron metabolism (ferritin, transferrin saturation). CHr and RetHe below 28 pg are indicative of functional ID /41/: In tumor patients, due to bleeding, hemolysis and cytostatic therapy In anemia of chronic disease (ACD). IL-6 triggered by inflammation induces hepatocyte synthesis of hepcidin, which leads to augmented iron retention in the macrophages and reduced intestinal iron absorption. In this way, iron turnover is reduced, resulting in functional ID, which is developed in 5–10% of ACD patients. Reticulocytes and erythrocytes have a reduced Hb content, but their MCV value is usually sub-normal. In dialysis patients, the threshold value of functional ID is ≤ 29 pg /45/. Functional ID is also present if CHr or RetHe are smaller than the erythrocyte MCH value (CH inversion) /37/. 15.6.6.5.3 Diagnostic diagram Development of iron deficiency and functional ID as a function of the iron supply can be identified and monitored with the aid of the diagnostic diagram shown in Fig. 15.6-5 – Diagnostic diagram for the assessment, monitoring and therapy of the iron status. In the diagram the ferritin index (sTfR/log10 ferritin) serves as indicator of the iron supply for erythropoiesis, since it correlates well with stainable iron of the bone marrow. The CHr or RetHe are indicators of the iron demand of erythropoiesis /41/. In a study /46/ the differentiation of anemia was not possible in 32% of the cases with classical serial determination of hematological and iron metabolism markers, but with the application of the diagram this percentage decreased to only 14%. In a treatment study of tumor patients, the concept of the diagram and the therapeutic recommendations for the different forms of anemia proved to be of value /47/. 15.6.6.5.4 Sickle cell anemia In sickle cell anemia, reticulocyte Hb concentration (CHCMr) is elevated and, as a rule, greater than 380 g/L. CHCMr decreases within the first two weeks of treatment with hydroxyurea, due to improved hydration of the sickled cells; in consequence, the tendency of the cells to undergo sickling is attenuated /48/. Compared with healthy individuals, patients with sickle cell anemia have a 2–3-fold higher Hb contained in the reticulocyte compartment (retHb) as normal controls. The Hb in the red cell compartment (rbcHb) is decreased. In a study /49/ the retHb values in controls were 1.76 ± 0.69 g/L and in SS4α 6.5 ± 4.2 g/L. The rbcHb/retHb ratio is an important indicator of sickle cell disease. In healthy individuals, as well as in iron deficiency anemia, the ratio is ≥ 50, in sickle cell anemia it is ≤ 50. RbcHb is the remainder obtained by the subtraction of retHb from total Hb. 15.6.6.6 Comments and problems Method of determination CHr and Ret-He are to be determined with the hematology analyzers Siemens Advia and Sysmex analyzers, but CHCMr only with Advia. All data, therefore, relate to the use of these analyzers. Stability Upon storage of the blood samples for 60 minutes, MCVr increases and CHCMr decreases, but CHr and Ret-He are stable for at least 24 hours /31/. References NCCLS. Methods for reticulocyte counting (flow cytometry and supravital dyes). Approved guideline. NCCLS Document H 44-A, Vol 17, No 15. Villanova: NCCLS, 1997. Heilmeyer L, Westhäuser R. Reifungsstudien an überlebenden Retikulozyten in vitro und ihre Bedeutung für die Schätzung der täglichen Hämoglobinbildung in vivo. Zschr Klin Med 1932; 121: 361–655. Löwenstein LM. The mammalian reticulocyte. Int Rev Cytol 1959; 8: 135–74. Perrotta AL. The polychromatophilic erythrocyte. Am J Clin Pathol 1972; 57; 471–7. Pappas AA, Owens RB, Flick JT. Reticulocyte counting by flow cytometry. Ann Clin Lab Sci 1992; 22: 125–32. Houwen B. Reticulocyte maturation. Blood Cells 1992; 18: 167–86. Hillman RS. Characteristics of marrow production and reticulocyte maturation in normal man in response to anemia. J Clin Invest 1969; 48: 443–53. Come SE, Shohet SB, Robinson SH. Surface remodeling vs whole cell hemolysis of reticulocytes produced with erythroid stimulation or iron deficiency anemia. Blood 1974; 44: 817–30. NCCLS. Method for reticulocyte counting. Proposed standard. Order Code HP 16-P, Vol 5, No 10. Villanova NCCLS, 1985. Davis BH, Bigelow NC. Automated reticulocyte analysis. Hematol Oncol Clin North Am 1994; 8: 617–30. Herkner KR, Bock A, Mulh A, et al. Reticulocyte counting and maturity grading in pediatrics using the Sysmex R-1000 automated reticulocyte analyzer. Sysmex J Int 1992; 2: 4–9. Geaghan SM. Hematologic values and appearances in the healthy fetus, neonate and child. Clin Lab Med 1999; 19: 1–37. Butarello M, Bulian P, Farina G, Temporin V, Toffolo L, Trabuio E, et al. Flow cytometric reticulocyte counting. Parallel evaluation of five automated analyzers: an NCCLS-ICSH approach. Am J Clin Pathol 2001; 115: 100–11. Nebe T, Bentzien F, Bruegel M, Fiedler GM, Gutensohn K, Heimpel H, et al. Multizentrische Ermittlung von Referenzbereichen für Parameter des maschinellen Blutbildes. J Lab Med 2011; 35: 3–28. Nebe CT, Diem H, Heimpel H. Aktuelle Aspekte zur Bestimmung der Retikulozytenzahl. J Lab Med 2010; 34: 295–304. Heimpel H, Diem H, Nebe CT. Die Bestimmung der Retikulozytenzahl: Eine alte Methode gewinnt neue Bedeutung. Med Klin 2010; 105: 538–43. Thomas L, Messinger M. Bedeutung der Retikulozytenbestimmung zur Differenzierung und Behandlungskontrolle der Anämie. Lab Med 1994; 18: 305–10. Means Jr RT, Krantz SB. Progress in understanding the pathogenesis of the anemia of chronic disease. Blood 1992; 80: 1639–47. European Best Practice Guidelines for the Management of Anemia in Patients with Chronic Renal Failure. Nephrol Dial Transplant 1999; 14, Suppl 5. Bowen D, Williams K, Phillips I, Cavill I. Cytometric analysis and maturation characteristics of reticulocytes from myelodysplasic patients. Clin Lab Haem 1996; 18: 155–60. Delaunay J, Iolascon A. The congenital dyserythropoietic anemias. Baillaire’s Clinical Haematology 1999; 12: 691–705. Pekrun M, Lakomek M, Eber S, Konrad H, Schröter W. Diagnostik des Pyruvatkinasemangels bei erhöhter Retikulozytenzahl. Dtsch Med Wschr 1995; 120: 1620–4. Jelkmann W, Lundby C. Blood doping and its detection. Blood 2011; 118: 2395–404. Koepke JF, Koepke JA. Reticulocytes. Clin Lab Haematol 1986; 8: 169–79. Aulesa C, Ortola J, Olive T, Ortega JJ. Temporary increase in the reticulocytes after transfusion of RBC concentrates in the BMT patients. Sysmex J Intern 1994; 4: 96–8. Davis BH, Bigelow NC, Koepke JA, Borowitz MJ, Houwen B, et al. Flow cytometric reticulocyte analysis. Multiinstitutional interlaboratory correlation study. Am J Clin Pathol 1994; 102: 468–72. Arbeitsgruppe Präanalytik der DGKC. Stellungnahme zu den Empfehlungen Stabilität der Messgröße in der Probenmatrix. DG Klinische Chemie Mitteilungen 1996; 27: 27–9. Davis BH, DiCorato M, Bigelow NC, et al. Proposal for standardization of flow cytometric reticulocyte maturity index (RMI) measurements. Cytometry 1993; 14: 318–26. Van den Bossche J, Devreese K, Malfait R, van de Vyvere M, Wauters A, Neels H, et al. Reference intervals for complete blood count determined on different hematology analyzers: ABX Pentra 120 Retic, Coulter Gen S, Sysmex SE 9500, Abbott Cell Dyn 4000 and Bayer Advia 120. Clin Chem Lab Med 2002; 40: 69–73. Kanold J, Bezou M, Coulet M, et al. Evaluation of erythropoietic/hematopoietic reconstitution after BMT by highly fluorescent reticulocyte counts compares favorably with traditional peripheral blood cell counting. Bone Marrow Transplant 1993; 11: 313–8. Brugnara C. Reticulocyte cellular indices: a new approach in the diagnosis of anemias and monitoring of erythropoietic function. Crit Rev Clin Lab Sci 2000; 37: 93–130. Davies S, Cavill I, Bentley N, et al. Evaluation of erythropoiesis after bone marrow transplantation. Br J Haematol 1992; 81: 12–7. Brugnara C, Hipp MJ, Irving PJ, Lathrop H, Lee PA, Minchello EA, Winkelman J. Automated reticulocyte counting and measurement of reticulocyte cellular indices. Evaluation of the Miles H3 blood analyzer. Am J Clin Pathol 1994; 102: 623–32. Buttarello M, Bulian P, Venudo A, et al. Laboratory evaluation of the Miles H3 automated reticulocyte counter. A comparative study with manual reference method and Sysmex R-1000. Arch Pathol Lab Med 1995; 119: 1141–8. d’Onofrio G, Chirillo R, Zini G, et al. Simultaneous measurement of reticulocyte and red blood cell indices in healthy subjects and patients with microcytic and macrocytic anemia. Blood 1995; 85: 818–23. Thomas L, Franck S, Messinger M, Linssen J, Thome M, Thomas C. Reticulocyte hemoglobin measurement– comparison of two methods in the diagnosis of iron-restricted erythropoiesis. Clin Chem Lab Med 2005; 43: 1193–1202. Thomas C, Thomas L. Biochemical markers and hematologic indices in the diagnosis of functional iron deficiency. Clin Chem 2002; 1066–76. Brugnara C, Zelmanovic D, Sorette M, Ballas SK, Platt O. Reticulocyte hemoglobin. An integrated parameter for evaluation of erythropoietic activity. Am J Clin Pathol 1997; 108: 133–42. Ervasti M, Kotisaari S, Sankilampi U, Heinonen S, Punnonen K. The relationship between red blood cell and reticulocyte indices and serum markers of iron status in the cord blood of newborns. Clin Chem Lab Med 2007; 45: 1000–3. Ullrich C, Wu A, Rieber S, Wingerter S, Brugnara C, Shapiro D, Bernstein H. Screening healthy infants for iron deficiency using reticulocyte hemoglobin content. JAMA 2005; 294: 924–30. Thomas L, Thomas C. Anämie bei Eisenmangel und Störungen im Eisenstoffwechsel. Dtsch Med Wschr 2002; 127: 1591–4. Brugnara C, Laufer MR, Friedman AJ, Bridges K, Platt O. Reticulocyte hemoglobin content (CHr): early indicator of iron deficiency and response to therapy (letter). Blood 1994; 83: 3100. Brugnara C, Zurakowski D, DiCanzio J, Boyd T, Platt O. Reticulocyte hemoglobin content to diagnose iron deficiency in children. JAMA 1999; 281: 2225–30. Cavill I. Functional iron deficiency. Blood 1993; 82: 1377. Fishbane S, Shapiro W, Dutka P, Valenzuela OF, Faubert J. A randomized trial of iron deficiency testing strategies in hemodialysis patients. Kidney Intern 2001; 60: 2406–11. Leers MPG, Keuren JFE, Oosterhuis WP. The value of the Thomas-plot in the diagnostic workup of anemic patients referred by general practitioners. Int J Haematol 2010; 32: 572–81. Steinmetz HT, Tsamaloukas A, Schmitz S, Wiegand R, Rohrberg R, Eggert J, et al. A new concept for the differential diagnosid and therapy of anemia in cancer patients. Support Care Cancer 2011; 19: 261–9. Bridges KR, Barabino GD, Brugnara C, et al. A multiparameter analysis of sickle erythrocytes in patients undergoing hydrooxyurea therapy. Blood 1996; 88: 4701–10. Brugnara C, Zelmanovic D, Sorette M, Ballas SK, Platt O. Reticulocyte hemoglobin. Am J Clin Pathol 1997; 108: 133–41. 15.7 Hemoglobinopathies Elisabeth Kohne Human hemoglobin consists of the main component HbA and several minor components of which only HbA2 is genetically determined, whereas the so-called modified hemoglobins HbA1a, HbA1b and HbA1c are only formed secondarily within the erythrocyte under the influence of exogenous factors. During the fetal period the embryonic hemoglobins Hb Gower 1, Hb Gower 2 and Hb Portland are synthesized first being replaced by HbF from the 5–7th gestational week. At term newborns have 80% HbF and 20% HbA. During the first 6 months of life a gradual shift in hemoglobin takes place until the adult pattern is reached after the 12th month. The common structural principle of all hemoglobins is their composition of four heme molecules and the protein portions (globin) each consisting of two pairs of identical polypeptide chains (Tab. 15.7-1 – Normal human hemoglobins) /1, 2, 3/. The umbrella term “hemoglobinopathy” includes all genetic hemoglobin disorders. These are divided into two main groups as follows: Thalassemia syndromes Structural hemoglobin variants (abnormal hemoglobins). Both are caused by mutations and/or deletions in the α- or β-genes. When gene defects cause hemoglobin synthesis disorders, this gives rise to thalassemia. Hemoglobin structure in these cases is normal. When they change in the hemoglobin structure, this gives rise to abnormal hemoglobin. There are also mixed forms that combine features of both groups (e.g., β0/β+-thalassemias, HbSC disease and HbE α-thalassemias). The common features of the pathophysiology and various disease patterns are limited, and as a result so are the possibilities for summarizing them /2, 4/. Epidemiology of hemoglobinopathies With about 7% of the worldwide population being carriers, hemoglobinopathies are the most common monogenic diseases and one of the world’s major health problems. They were originally found mainly in the Mediterranean area and large parts of Asia and Africa. Global migration has spread them from those areas all over the world. In many parts of Europe today, hemoglobin defects are classified as endemic diseases (Tab. 15.7-2 – Prevalence of hemoglobinopathy gene carriers in the world’s population) /4, 5/. Germany is one of the countries in which hemoglobinopathies have increased in recent years and represent a relevant healthcare problem in German medicine. Most frequently occurring are thalassemias and sickle cell disease /4, 5, 6/. No epidemiological prevalence studies exist, but based upon calculations using currently available figures, it is assumed that genetic carriers of hemoglobinopathy make up 1% of the present German population /4/. An overview /5/of the most important hemoglobin diseases occurring in this country, listed according to prevalence, is found in Tab. 15.7-3 – The most important hemoglobinopathies in Germany. Clinical laboratory investigations The type and extent of laboratory investigations depend on the questions to be answered in each case based on the clinical and hematologic data and the history (ethnic background). The presence of a hemoglobinopathy is ruled out if all hematologic parameters are normal (exception: asymptomatic anomalies). If on the other hand hemoglobin analysis reveals normal values in the setting of very specific hematologic findings further laboratory investigations may have to be performed. Peculiarities A large variety of defects due to the varied ethnic composition of the population is typical for countries such as the USA or Germany, which have a significant proportion of immigrants. The clinical presentations are very different, varying from mild hypochromia and microcytosis with or without anemia to complex clinical presentations. These peculiarities represent a challenge for the clinical laboratory in regard both to the methods of analysis and the knowledge required for the interpretation of the findings. In doubtful cases consultation with a specialized laboratory may be necessary. 15.7.1 Indication Microcytic hypochromic anemia after iron deficiency has been ruled out Chronic hemolytic anemia Vascular obliteration crises of unclear etiology in patients from areas in which HbS and/or HbC is widespread Drug-induced anemia Erythrocytosis and/or cyanosis caused by hematological factors Hydrops fetalis of unclear etiology Prevention (testing of family members, diagnosis of partners for genetic counseling) Prenatal diagnosis. Generalized hemoglobin electrophoresis for all cases of anemia cannot be justified economically, particularly in those with no background of migration. Indication for DNA analysis Within the framework of thalassemia diagnostic investigation: Genetic testing of β-thalassemia major Molecular diagnosis of β-thalassemia inter media Combination of different forms of thalassemia or of thalassemias with structural Hb anomalies Suspected silent beta thalassemia trait carriers Diagnostic investigation of α-thalassemia Issues related to genetics and preventive medicine. Diagnostic investigation of Hb structure anomalies Identification of rare anomalies For clarification in cases of pre-existing suspicion, but lacking electrophoretic or chromatographic separation In the presence of issues related to genetics and preventive medicine Forms of different hemoglobinopathies combined with one another or with thalassemia. DNA analyses should and may be implemented only in cases where the issues cannot be resolved with the use of conventional hemoglobin analyses. 15.7.2 Method of determination Laboratory examinations consist of the routine hematologic evaluation, cytologic tests, and hemolysate analyses including hemoglobin electrophoresis using different buffer and pH systems and chromatographic procedures. The stepwise approach shown in Tab. 15.7-4 – Program for the laboratory diagnostic investigation of hemoglobinopathies has proven itself for practical use. The help of a specialized laboratory may be necessary since, particularly for economic reasons, even large laboratories have to contain the expenditures related to personnel and equipment in this special, originally purely hematological, field. The plan that is depicted in Fig. 15.7-1 – Clinical laboratory diagnosis of thalassemia syndromes and hemoglobinopathies has proven itself as a practical procedure for hemoglobinopathy analysis. 15.7.2.1 Basic hematological diagnosis In suspected hemoglobinopathy a complete red blood cell count, including reticulocyte count, is a requirement. The RDW (red cell distribution width) value also provides important evidence which is usually elevated as a measure of anisocytosis in iron deficiency, while in thalassemia minor normal RDW values are usually observed (Fig. 15.7-2 – Use of RDW values in diagnostic investigation of thalassemia minor). The assessment of the blood smear elicits diagnostically valuable information because the thalassemias, as well as most structural hemoglobin anomalies, express characteristic changes in erythrocyte morphology /2/. 15.7.2.2 Clinical chemistry tests These include biomarkers of iron status (ferritin, transferrin saturation) and the hemolysis parameters haptoglobin, LD, bilirubin, and the Coombs test. 15.7.2.3 Hemoglobin analyses Hb electrophoresis Routine electrophoresis is performed on cellulose acetate membranes with an alkaline Tris-EDTA borate buffer (pH 8.5) (micro zone electrophoresis). Fig. 15.7-3 – Separation schema for normal and abnormal hemoglobin on micro zone electrophoresis). Under these circumstances abnormal hemoglobins are only separated from normal ones if they have a difference in electrical charge. The relative concentrations of the separated bands can be determined by densitometry, or the fractions eluted and measured. For HbA2 both procedures have inherent errors. Using acid agarose gel electrophoresis with a maleic acid buffer at pH 6.1, as an additional technique, hemoglobins may be separated which migrate together on alkaline electrophoresis /2, 7, 8, 9, 10, 11, 12/. A well established procedure for qualitative and quantitative evaluations (HbA2) is starch block electrophoresis. However, it is worthwhile only for laboratories with a high demand on account of its complexity. Depending on the experience in the laboratory isoelectric focusing may also be used. HPLC HPLC (cation-anion exchange systems) is the optimal method for separating normal and abnormal hemoglobins. HPLC is also the method of choice for quantitative analysis of all separable hemoglobin fractions /2, 7, 8, 10, 11/. 15.7.2.4 HbF determination and HbF cell detection Alkali denaturation is the classical method for quantification of HbF. A cyan hemoglobin solution is produced using a hemolysate followed by the addition of NaOH. During this process HbA2 is denatured. Subsequent precipitation using ammonium sulfate leaves HbF remaining in solution where it may be measured photometrically /2/. The acid elution method serves to detect HbF cells in peripheral blood smear preparations. HbA is eluted from erythrocytes by a citric acid-phosphate buffer at pH 3.2 while HbF remains in the cells and is stained. In the case of a negative HbF cell result the process of alkali denaturation is superfluous. 15.7.2.5 HbS solubility test The solubility test is used to differentiate between HbS and anomalous hemoglobins with identical migration patterns during electrophoresis, such as HbD and HbG. In a hemolysate to which dithionite has been added for removal of oxygen, HbS is the only hemoglobin to precipitate causing marked turbidity of the reaction mixture /2/. 15.7.2.6 Identification of abnormal hemoglobins The common abnormal hemoglobins HbS, HbC, HbE and HbD represent more than 90% of all structural hemoglobin anomalies seen in daily laboratory practice /2, 7, 8, 10, 11/. As a rule, these hemoglobins can be directly diagnosed electrophoretically or according to their chromatographic properties with, as necessary, chemical test procedures such as the solubility test. DNA analyses are employed for the identification of rare pathological hemoglobin variants. Molecular biologic testing should be restricted to those hemoglobin anomalies that cannot be clarified with conventional methods. 15.7.2.7 DNA analyses The molecular biologic procedures are listed in Tab. 15.7-5 – Basic information on molecular biologic methodology /13/. 15.7.3 Specimen Erythrocytes from blood with anticoagulants are required. EDTA blood (in coated sample containers) is best and is suitable for simultaneous determination of the blood cell count, RBC morphology and erythrocyte enzymes. EDTA blood may be stored for a few days without significant deterioration, the investigation of α-thalassemia being an exception. For routine examinations 5 mL of blood are usually adequate, larger volumes are needed in the case of severe anemia. 15.7.4 Reference interval Refer to Tab. 15.7-6 – Reference intervals for hemoglobins. 15.7.5 Clinical significance Basic forms of hemoglobinopathies are shown in Tab. 15.7-7 – Basic forms of hemoglobinopathies. Generally significant criteria are the age of the patient (e.g. for assessing an HbF value) the ethnic background, the family history, and the clinical hematologic findings. Hemoglobin analysis includes: Assessment of quantitative changes of normal hemoglobins such as the increase in HbA2 and/or HbF typical of thalassemias Exclusion or confirmation of an abnormal variant and its identification and quantification In each case the question has to be addressed of whether the abnormal hemoglobin is responsible for the clinical symptoms or is simply a random finding without pathologic significance. 15.7.5.1 Thalassemia syndromes This term includes all thalassemic hemoglobin synthesis disorders. Thalassemias are one of the most common single-gene disorders in the world. The presence of mutations in the globin genes cause the reduction or absence of globin chains and results in thalassemia. The mutations occur in the alpha, beta, gamma, and delta genes, and as a result, the synthesis of hemoglobin is disturbed. Different genotypes in thalassemia cause various phenotypes. The phenotypes vary from a thalassemia minor to thalassemia major with mild or severe anemia, respectively. The couples who are the carriers of thalassemia play an important role in the birth of infants with thalassemia major so that the chance of the birth of α-thalassemia major infant in a thalassemia couple is 25%. In autosomal recessive conditions alpha- and beta-thalassemias have the greatest significance. Heterozygous thalassemia carriers are not completely healthy: they always have symptoms that require differentiation with mild, iron-refractory, microcytic hypochromic anemia /10, 11, 14, 15, 16/. 15.7.5.1.1 α-thalassemias α-thalassemias are a common monogenic gene disorder caused by a defect in α-globin genes located on the short arms of chromosome 16. α-thalassemias are caused by an α-globin chain synthesis defect. At the molecular level, they result from partial (α+) or total (α0) deletions, or more rarely mutations, of one or more of the α-globin genes (αα/αα). They occur mainly in Africa, Arab nations, and more frequently, South-East Asia. All become manifest perinatally. There are four clinical pictures of α-thalassemia, according to the number of genes affected by loss of function (Tab. 15.7-8 – Compilation of the most important criteria of α-thalassemias): Clinically inapparent α-thalassemia minima (heterozygous α+-thalassemia, –α/αα). This can be identified on the basis of mild hypochromia with a barely reduction in hemoglobin concentration. α-thalassemia minor (heterozygous α0-thalassemia, ––/αα, or homozygous α+-thalassemia, –α/–α) with mild anemia, hypochromia and microcytosis HbH disease (compound heterozygous α+/α0-thalassemia with three inactive α-genes, ––/–α) moderate hypochromic hemolytic anemia with splenomegaly. Anemic crisis are caused by viral infections and oxidants (drugs). Complications include cardiac problems, gall stones, lower leg ulcers, and folic acid deficiency. Hemoglobin (Hb)-Bart’s hydrops fetalis (homozygous α0-thalassemia) is the most severe form of α-thalassemias with very serious hemolytic anemia already present in utero and marked by a lack of any α-globin chain synthesis (––/––), with hydrops and ascites. Hb Bart’s hydrops fetalis is the most severe form of α-thalassemias and is incompatible with life. If both husband and wife are carriers of both α-globin gene deletions, they have a 25% chance of having a child with Hb Bart’s hyrops fetalis in each pregnancy /18/. 15.7.5.1.2 β-thalassemias β-thalassemias are a prevalent anemic condition worldwide, particularly in Southeast Asia, arises due to genetic irregularities affecting globin chain synthesis. Mutations in the ß-globin genes lead to deficient or non-functional ß-globin chain production causing anemia and relate complications. A notable β-hemoglobin variant linked to a relative ß-thalassemia phenotype is hemoglobin E, resulting from GAG to AAG transition at codon 26 in the β-globin gene. The prevalence of β-thalassemia and HbE carriers in Thailand ranges from 3 to 9% and 10 to 53%, respectively. β-thalassemia carriers experience mild to asymptomatic anemia, whereas patients with β-thalassemia major encompassing homozygous β-thalassemia and thalassemia/HbE endure moderate to severe anemia necessitating treatments such as blood transfusions and iron chelation.Given the high prevalence of these carriers, β-thalassemia major cases are likely widespread, making it a substantial public health and socioeconomic concern. Accurate quantification of hemoglobin A2 is vital for diagnosing β-thalassemia carriers /19/. Refer to Tab. 15.7-9 – Compilation of the most important criteria of β-thalassemias. 15.7.5.1.3 β-thalassemia minor The starting point for the diagnosis of thalassemia minor (heterozygous β-thalassemia) is the complete blood count. The significant laboratory markers of β-thalassemia are increased HbA2 and/or increased HbF values. If the MCH is below 27 pg and the HbA2 above 3.5%, diagnosis of heterozygous β-thalassemia is made. The majority of β-thalassemia trait carriers have MCH reductions to 23–19 pg and HbA2 values of 4.0–6.0%; HbA2 increases to 6.5–8.0% can occur. In some 30% of the cases, HbF increase to 1–3%, occasionally to 3.0–15%, occurs simultaneously. To be noted are age-dependent higher HbF values in young children with β-thalassemia minor. The iron status (ferritin, transferrin saturation) is, as a rule, normal. Exceptions (i.e., iron deficiency in β-thalassemia minor) can occur in children and during pregnancy. Simultaneous iron deficiency can lead to the temporary underestimation of HbA2. If there is uncertainty, a reevaluation following correction of the iron deficiency is necessary. 15.7.5.1.4 β-thalassemia major The homozygous or mixed heterozygous β-thalassemia presents no earlier than the age of some three to five months. At the time of diagnosis the anemia is variable; hemoglobin values are usually below 80 g/L. Anemia is always hypochromic, with an MCH ≤ 22 pg and an MCV between 50 and 60 fL. Thalassemia-like erythrocyte morphology with significant poikilocytosis can be seen on the blood smear. The hemoglobin analysis shows variable fractions of HbA, HbF and HbA2. In general it can be assumed that with an HbF increase between 20 and 98%, along with a typical hematological values, thalassemia major or thalassemia inter media is present. The patient’s transfusion status must be considered during assessment. There is an increasingly frequent diagnostic problem that occurs in patients with treated thalassemia major, such as those receiving continuous transfusion therapy. As a result of treatment, the full clinical picture is no longer seen. The diagnosis can only be confirmed with DNA analysis. The issue concerns mainly young patients or young adults who come to Germany within the framework of a family reunion. 15.7.5.1.5 β-thalassemia inter media The mild homozygous or mixed heterozygous β-thalassemia refers, primarily, to a clinical diagnosis in patients with a hemoglobin pattern resembling that of thalassemia major, who stand out due to a minimal or no transfusion requirement. Diagnostic differentiation with regard to thalassemia major is made over a period of time by regular clinical hematological monitoring. If necessary, DNA analysis is performed. Thereby, either a high level of residual β-globin gene activity is demonstrated, or classical thalassemia major is found, albeit with additional influencing factors, particularly hereditary HbF persistence or α-thalassemia. 15.7.5.1.5.1 Diagnostic significance of HbF persistence Hereditary HbF persistence is a clinically harmless, frequently congenital proliferation of HbF. The specific diagnostic significance of elevated HbF values in β-thalassemia was mentioned. δβ-thalassemia is also characterized by high HbF values. In sickle cell disease, a high HbF value has a positive prognostic significance. Apart from hemoglobinopathies, HbF proliferation can occur as secondary phenomenon in many hematological diseases /2/. 15.7.5.2 Abnormal hemoglobins This group of autosomal dominant inherited hemoglobin disorders are caused by structural defects resulting from an altered amino acid sequence in the α- or β chains. Clinicians must distinguish between clinical harmless hemoglobin anomalies and those that cause illness. These latter are divided into the following four groups /2, 5, 10, 11, 17/: HbS; variants with a tendency to aggregate and with sickle cell formation (e.g., sickle syndromes) HbE; variants with abnormal hemoglobin synthesis (e.g., HbE) Unstable hemoglobins, e.g. Hb Köln; variants with a tendency to precipitate and with hemolysis (unstable hemoglobins e.g., Hb Köln) Variants with abnormal oxygen transportation and congenital polycythemia (e.g. Hb Johnstown), or with congenital cyanosis such as abnormal methemoglobins and HbM abnomalities (e.g. HbM Iwate). The forms in both of the two last groups cause serious illness when heterozygous. When homozygous, they are fatal. The main hemoglobin abnormalities are HbS, HbE and HbC. The large groups of rare hemoglobin abnormalities that occur in isolated cases all over the world should also be monitored. These are often accompanied by hemolysis, polycythemia, and/or cyanosis. Identifying these is an important part of differential diagnosis of hematological diseases where other efforts towards diagnosis have proved inconclusive. They come first in the diagnostic investigation of hemoglobin defects and comprise more than 90% of the anomalies. Less common in routine laboratory practice are HbD and HbG variants, HbO-arab, HbG and HbJ. All other hemoglobin defects (e.g., the unstable hemoglobins Hb Köln, Hb Zürich) are extremely rare /5/. They occur with the same frequency in Germans and in foreigners (i.e., only in individual persons or families) and are not included in standard hemoglobin diagnostic investigation. The clinical classification of pathological hemoglobin variants are shown in Tab. 15.7-10 – Clinical classification of the most important pathological hemoglobin variants. 15.7.5.2.1 HbS and sickle-cell disease The term sickle cell disease includes all manifestations of abnormal HbS levels (proportion of HbS > 50%). These include homozygous sickle-cell disease (HbSS) and a range of mixed heterozygous hemoglobinopathies (HbS/β-thalassemia, HbSC disease, and other combinations). According to the International Nomenclature the previously commonly-used term sickle-cell anemia should not be used, as the dominant aspects of the disease are vascular obliterations and the organ damage they cause, not anemia /1, 2, 17/. Cardinal symptoms and diagnostic criteria Symptoms begin before the age of 1 year, with chronic hemolytic anemia and developmental disorders. The main problems are pain crisis (sickle cell crisis) that can affect the back, extremities, thorax, abdomen, and the CNS in particular. Patients are also susceptible to infection. HbS is the most dangerous of all hemoglobinopathies. The sickle cells caused by a lack of oxygen lead to vascular obliterations, so infarctions with tissue death can occur in almost all organs (skin, liver, spleen, bone, kidney, retina, CNS). Chronic hemolytic anemia can usually be well tolerated. Aplastic crisis are seen with severe anemia following viral infections. Molecular genetic diagnostic investigation in HbS and sickle cell disease: HbS is mainly diagnosed with conventional Hb analysis. Special indications for DNA testing are combined forms of HbS with other anomalies, with β- or α-thalassemia, and issues related to prenatal diagnosis. HbS heterozygosity Heterozygous HbS gene carriers are not affected clinically or hematologically and have a normal blood count. The diagnosis is based on the detection of HbS in typical position on electrophoresis, the proportion of which, as determined by HPLC, is quantitatively lower than that of HbA and accounts for 35–40% of total hemoglobin. HbS values of below 30% are suspicious with regard to the presence of iron deficiency or an coexisting α-thalassemia. The MCH is decreased in both cases. HbS homozygosity The hemoglobin level is usually 60–90 g/L. Sickle cells and target cells are found in the blood smear. In HbSS (homozygous form) no normal HbA is found on hemoglobin electrophoresis. The HbF fraction is variable, usually in the range of 5 to 15%. Higher HbF fraction often occur. HbS-β-thalassemia The hematological presentation resembles that of sickle cell disease. Differentiating characteristics in relation to HbS homozygosity are found in microcytosis and hypochromia. Differentiation of the HbS-βo and HbS-β+ forms is accomplished, in the simplest case, by demonstrating an HbA fraction in the HbS-β+ combination, while in the HbS-β0 form no HbA is present. The diagnosis can be confirmed by means of an Hb analysis, or molecular genetically (Tab. 15.7-11 – Diagnostic characteristics of sickle cell β-thalassemia). Notes on HbS-β-thalassemia: An elevated HbA2 value in HbS heterozygosity is not an attribute indicative of HbS-β-thalassemia HbS-β-thalassemias are sickle cell diseases with variable clinical symptoms; they should not be interpreted as a type of thalassemia. 15.7.5.2.2 HbE and HbE disease HbE is a common Hb variant native to South-East Asia. Its disease pattern is similar to that of β-thalassemias. HbE is also unstable, which means that hemolysis can be caused by viral infections and medications. HbE is often combined with thalassemias, which may result in serious major-form hemoglobinopathies /1, 2, 10, 11/. HbE heterozygosity Mild, variable hypochromia (MCH 25 pg) and microcytosis are present. The HbE fraction is, as a rule, around 30–45%; the remainder is HbA. HbF is not elevated. At lower HbE concentration, the concomitant presence of iron deficiency or α-thalassemia must be considered. HbE homozygosity (HbE disease) Characteristic is hypochromia with microcytosis (MCH 20 pg, MCV 65 fL) in the presence of pronounced erythrocytosis. In addition, the presence of abundant target cells. The HbE fraction is approximately 95%; the remainder is HbF and HbA2. On electrophoresis, HbE migrates identical to HbO, HbC and HbA2. Molecular biologic or special electrophoretic, immunologic and chromatographic methods, particularly HPLC, permit the differentiation of these anomalies. HbE in combination with other anomalies The combination with β-thalassemia (HbE-β-Thal), leads to moderate to severe hypochromic and dyserythropoietic anemia, corresponding to thalassemia inter media or thalassemia major. In combination with α-thalassemia the HbE fraction, depending on the number of inactive α-globin genes, is significantly reduced, while hypochromia is more pronounced. Refer to: Tab. 15.7-12 – Characteristic features of the most important hemoglobin variants Tab. 15.7-7 – Basic forms of hemoglobinopathies. 15.7.5.2.3 HbC anomaly and HbC disease HbC homozygosity, or HbC disease, progress in a similar way to sickle-cell disease, but is less serious. Variable hemolytic anemia is the most dominant form. Heterozygous HbC gene carriers enjoy complete clinical health /1, 2, 10, 11/. HbC heterozygosity HbC trait carriers do not manifest anemia. Target cells are detectable in the blood smear. MCHC is elevated. On alkaline Hb electrophoresis, the HbC fraction migrates in a typical position, identical to HbA2, and in trait carriers the quantitative HbC fraction accounts for 30–40% of hemoglobin. Differentiation from hemoglobins with identical migration properties (i.e., HbO and HbE) is accomplished with acidic electrophoresis. HPLC is employed for quantitative analyses. HbC homozygosity (HbC disease) The blood count shows predominantly target cells (Tab. 15.7-12 – Characteristic features of the most important hemoglobin variants). The Hb level is 100–120 g/L. The MCHC value is above 350 g/L. Almost 100% HbC is seen on hemoglobin electrophoresis. HbF may be slightly increased. 15.7.6 Comments and problems Method of determination The initial emphasis during investigation of hemoglobin abnormalities is on excluding or confirming the presence of thalassemia or an abnormal hemoglobin. Usually electrophoresis, cytological and biochemical tests alone are sufficient for diagnosis of thalassemia. In certain cases structural studies may become necessary and this is usually undertaken in specifically designated laboratories. The same often holds true for other specialized procedures (e.g., methemoglobin) and spectral analyses, oxygen affinity and 2,3-DPG determinations. Pre analytical factors The laboratory should be informed about a preceding blood transfusion since interpretation of test results under these circumstances requires special experience. In doubtful cases investigations must be performed after degradation of the foreign erythrocytes. Analytical factors Lack of carefulness during analytical procedures leads to error. For instance, inadequate removal of serum proteins results in spurious bands on electrophoresis, most commonly seen anodal to HbA, but occasionally on the cathodal side. Similar problems arise from denatured products from old, contaminated, or hemolyzed blood samples. It is advisable to conduct parallel examinations of normal control samples and if possible of reference samples. The well known problem of identical speed of migration of the various hemoglobins must be addressed by using a combination of different methods. Stability Cooling of the specimen is not required unless extremely high temperatures are present; for some tests it may even be deleterious. It is recommended that the shortest possible transport time for the samples and optimal timing of the transport are arranged (e.g., on weekends when the laboratory may offer to process the incoming specimens). Samples are stable for 1 week at 8–12 °C for hemoglobin analysis but only for 24 h for the blood cell count and the peripheral blood smear. 15.7.7 Pathophysiology Common causes of hemoglobinopathies are mutations and/or deletions in the α or β-globin genes. If the genetic defects lead to disorders of Hb synthesis, thalassemia develops. In such cases the hemoglobin structure is normal. If genetic defects provoke changes in the Hb structure, abnormal hemoglobins are formed. Among the individual groups, there are many combination and interactive forms (Tab. 15.7-7 – Basic forms of hemoglobinopathies). Thalassemia syndromes /2, 14, 15/ Common criteria: the inheritance is autosomal recessive. Nomenclature and classification of thalassemias are based on the globin chain that is affected in each case by the synthesis disorder. A common characteristic is reduced synthesis of the affected chain type, with a loss of synthesis balance. The pathological mechanisms that result are responsible for the disease symptomatology: A decrease in hemoglobin synthesis causes anemia. Severe forms of anemia are provoked mainly by ineffective erythropoiesis and hemolysis A shortage of substrate leads to a reduced hemoglobinization of erythrocytes (i.e., to hypochromia and microcytosis). Heterozygous thalassemia carriers are not completely healthy; rather, they have symptomatology, with mild, iron refractory hypochromic microcytic anemia, that at any rate necessitates clarification. Homozygous or mixed (compound) heterozygous major forms are associated with severe hypochromic hemolytic anemia and complex diseases. α-thalassemia /14, 15/ α-thalassemia is caused by a synthesis disorder at the level of the α-globin chains. The molecular basis is partial (α+) or complete (α0) deletion, less frequently mutations of one or more of the four α-globin genes (αα/αα). As a function of the number of genes that are affected by the loss of activity, there are 4 α-thalassemia phenotypes that are already fully expressed perinatally (Tab. 15.7-8 – Compilation of the most important criteria of α-thalassemias): Clinically silent α-thalassemia minima (heterozygous α+ thalassemia; −α/αα), recognizable by mild hypochromia with barely measurable reduction in hemoglobin value α-thalassemia minor (heterozygous α0-thalassemia; −−/αα) or homozygous α+-thalassemia; (−α/−α) with mild anemia, hypochromia and microcytosis. In HbH disease (compound heterozygote α+/α0-thalassemia) with three inactive α-genes (−α/−−), the pathophysiology is provoked by deficient Hb synthesis and unstable excess of HbH, which is formed from β-chain tetrameres (β4). The result is an intermediate incidence of disease with hypochromic hemolytic anemia and splenomegaly. Anemia crises occur during viral infection and due to noxious oxidative agents (medication). Complications are cardiac problems, biliary calculi, lower leg ulcers and folic acid deficiency. The Hb Bart’s hydrops fetalis syndrome (homozygous α0-thalassemia; −−/−−) is an hemoglobin synthesis disorder that occurs as early as the intrauterine period. Here hemoglobin is made up mainly of the non-functioning γ-chain tetramer Hb Bart’s (γ4). This syndrome is, consequently, not compatible with life. The pale and generally edematous children fall ill already at the intrauterine stage, during the last third of the fetal life, of severely hemolytic anemia with hydrops and ascites; without therapy they usually die before or shortly after birth (Tab. 15.7-8 – Compilation of the most important criteria of α-thalassemias). β-thalassemia /2, 14/ The β-thalassemia syndromes are the result of β-globin chain synthesis disorders. Molecular causes are β-globin gene mutations. Hematological changes are not manifested before the 3rd–6th months of life. The considerable degree of phenotypic heterogeneity is the result of the large number of different mutations that can affect any stage of gene expression with different pathophysiological effects. Depending upon whether β-chain deficiency is partial or complete, β+- or β0-thalassemias are distinguished (Tab. 15.7-9 – Compilation of the most important criteria of β-thalassemias). Thalassemia minor: the deficient Hb synthesis causes the typical mildly-pronounced hypochromic microcytic anemia (i.e., Hb values are slightly decreased or lie within the lower reference interval). Thalassemia major: homozygous β-thalassemia is a serious disease (thalassemia major). Here, not enough β-chains are synthesized while the unaffected α-chains are produced in excess. The α-chain excess acts on erythroid precursors to cause their precocious demise, thereby resulting in highly ineffective erythropoiesis. The correlate of this pathological mechanism is incipient severe anemia, which begins in parallel to the hemoglobin switch and, if left untreated, leads to death within a few years. An iron utilization disorder and increased iron resorption are threatening complications thereof. In consequence, and as a result of the obligatory continuous transfusion therapy, hemosiderosis with multiple organ defects develops during the course of the disease. For secondary hemochromatosis, refer to Tab. 7.1-9 – Iron overload not due to disorders of the hepcidin-ferroportin axis. Thalassemia inter media: this thalassemia of moderate severity is a mild homozygous or mixed heterozygous thalassemia with varying need for transfusions. δβ-thalassemia: due to a gene deletion, the δ- and β-chains are decreased or not synthesized at all. Heterozygous forms have a typical thalassemia minor constellation, with a 5–10% increase in HbF, but rather decreased HbA2. Homozygous δβ-thalassemia causes an intermediate syndrome, because mean Hb values are maintained via the reactivation of γ-chain synthesis. Hb Lepore anomalies: Hb Lepore is an abnormal form of hemoglobin, the non-α-chain represents a fusion product of δ- and β-chain. Overall hemoglobin synthesis is significantly reduced. The clinical and hematological picture is similar to that seen in β-thalassemia. Hb Lepore homozygosity, or combined Hb Lepore/β-thalassemia, is similar to thalassemia major, while the heterozygous Hb Lepore constellation corresponds to thalassemia minor. Abnormal hemoglobins (structural hemoglobin variants) /1, 2, 4/ Common criteria: these types of autosomal dominant inherited hemoglobin disorders arise on the basis of flawed genetic codes as products of modified amino acid sequences or deletions in hemoglobin α- and β-chains. Hemoglobin anomalies that are clinically harmless or that cause clinical disease must be distinguished. The latter are classified, based on their pathophysiology, into five well-defined groups (Tab. 15.7-10 – Clinical classification of the most important pathological hemoglobin variants). Molecular biological principles: more than 90% of the over 1100 hemoglobin variants that have been discovered as yet are caused by point mutations or missense mutation in one of the globin genes. Furthermore, there are few variants with elongated or shortened globin chain due to mutations of the normal terminator codon or nonsense and frame shift mutations in the third exon of the β-globin gene. Fusion genes between adjacent genes result in fusion proteins (e.g., Hb Lepore) which is also related to β-thalassemia. HbS and sickle cell disease: the pathophysiology is caused by the sickle cell hemoglobin HbS, the structural defect of which provokes a tendency for aggregation of Hb molecules, especially under conditions of oxygen deprivation. In consequence, erythrocytes take on a crescent shape, lose their deformability and alter their rheological properties. The clinical picture is characterized by a variety of phenotypic manifestations, of which homozygous sickle cell disease and sickle cell β-thalassemia are the most severe. The clinical symptomatology that develops following the first half year of life is comprised of two marked complexes of symptoms: vascular obstruction with multiple tissue and organ damage and chronic hemolytic anemia. HbE anomaly and HbE disease: a characteristic feature is reduced hemoglobin production (i.e., HbE manifestation is similar to that occurring in β-thalassemias). Furthermore, HbE is slightly unstable so that hemolysis is triggered under the influence of oxidative substances. HbC anomaly and HbC disease: the pathophysiology of HbC is, like that of HbS, based on disturbed solubility with formation of intracellular crystals in erythrocytes and increased tendency for cell aggregation. Combinations of abnormal hemoglobins and thalassemias: these combinations form a special disease group. Combinations of β-thalassemia and abnormal β-hemoglobins are of the greatest practical relevance. In these cases, due to the interaction of different pathophysiological effects, characteristic biochemical constellations or clinical and hematological syndromes occur. Demonstration of a genetic predisposition to thalassemia must no lead to the interpretation of the clinical picture as a type of thalassemia, because clinically the thalassemia component plays no role. Hemoglobinopathies with disturbed oxygen transport function: within this group, three disorders are distinguished: a) Anomalies in permanent methemoglobin status (HbM anomalies) b) Anomalies with elevated oxygen affinity c) Anomalies with reduced oxygen affinity. Characteristics of the hemoglobinopathies mentioned under (b) are polyglobulia and/or cyanosis. The ( c ) group is associated with anemia and cyanosis. Important features of abnormal hemoglobins are shown in Tab. 15.7-12 – Characteristic features of the most important hemoglobin abnormalities. Unstable hemoglobins: certain amino acid substitutions provoke an instability of the hemoglobin structure, where the abnormal hemoglobin is denatured within the erythrocyte spontaneously or usually under the influence of oxidizing substances (medications) or also in viral infections. Different molecular structures of the over 150 various unstable hemoglobins lead to more or less different pathological mechanisms and diseases. The best known and most common anomaly is Hb Köln; other examples in the German population are Hb Tübingen, Hb Presbyterian, Hb Freiburg and Hb Zürich. The laboratory findings are chronic hemolytic anemia, erythrocytes containing Heinz bodies, and excretion of a brown pigment in the urine (mesobilifuscinuria). References Steinberg MH, Forget BG, Higgs DR, Nagel RL. Disorders of Hemoglobin: Genetics, pathophysiology and clinical management. Cambridge; University Press 2001. Kleihauer E (ed), unter Mitarbeit von Kohne E, Kulozik AE. Anomale Hämoglobine und Thalassämiesyndrome: Grundlagen und Klinik. Landsberg; ecomed Verlagsgesellschaft: 1996. Kulozik AE. Hämoglobinopathien. In: Ganten D, Ruckpaul K, eds. Monogen bedingte Erbkrankheiten. Berlin: Springer-Verlag 2000; 370–92. Kohne E. Hämoglobinopathien: Klinische Erscheinungsbilder, diagnostische und therapeutische Hinweise. Dtsch Ärztebl 2011; 108: 532–40. Kohne E, Kleihauer E. Hämoglobinopathien – eine Langzeitstudie über vier Jahrzehnte. Dtsch Ärztebl 2010; 107: 65–71. Kulozik AE. Hämoglobinopathien nehmen zu: Dtsch Ärztebl Int 2010; 107: 63–4. British Committeee for Standards in Haematology. Guideline. The laboratory diagnosis of haemoglobinopathies: Brit J Haematol 1998; 101: 783–92. Wajcman H, Préhu C, Bardakdjian-Michau J, Promé D, Riou J, Godart C, Mathis M, Hurtrel D, Galactéros F. Abnormal hemoglobins: laboratory methods. Hemoglobin 2001; 25: 169–81. Wild BJ, Bain BJ. Detection and quantitation of normal and variant haemoglobins: an analytical review. Ann Clin Biochem 2004; 41: 355–69. Kohne E. Diagnostik von Hämoglobinopathien: J Lab Med 2004; 28: 400–9. Taher AT, Musallam KM, Cappellini MD. β-thalassemias. N Engl J Med 2021; 384 (8): 727–43. Wild B, Bain BJ. Investigation of abnormal haemoglobins and thalassaemia. In: Dacie and Lewis. Practical Haematology. 10th Edition. Elsevier Churchill Livingstone 2006; 271–311 Grody W, Nakamura R, Kiechle F, Strom C. Molecular Diagnostics. Techniques and Applications for the Clinical Laobratory. Elsevier Company 2009. Weatherall DJ, Clegg JB. The Thalassaemia Syndromes: 4th Edition. Blackwell Science Ltd, Oxford 2001. Higgs DR, Weatherall DJ. The Alpha Thalassaemias: Cell Mol Life Sci 2009; 66: 1154–62. Kiani AA, Mohamadinejad M, Shokrgozar N, Abbasian S. Mutations in thalassemia carrier couples: The importance of prenatal diagnostic tests. Clin Lab 2022; 68: 1021–9. Dickerhoff R. Sichelzellkrankheit. In: Leitlinien Kinderheilkunde und Jugendmedizin. I 1 2006; 1–7. Überarbeitet 03/2010. Gan G, Li Y, Bai J, Jiang M, Zheng L, Li Y. Misdiagnosis of Hb Bart’s disease: prenatal screening and diagnosis of thalassemia in special population. Clin Chem Lab Med 2023; 61 (10): e210–e213. Satthakarn S, Panyasai K, Phasit A, Panyasai S. Reliability of hemoglobin A2 value as measured by the premier resolution system for screening of β-thalassemia. J Clin Chem Clin Biochem 2023. doi: 10.1515/cclm-2023-1006. 15.8 Erythrocyte enzymes Elisabeth Kohne Erythrocyte enzymes play a key role in the regulation of the intracellular metabolism of erythrocytes. Enzyme deficiencies lead to impaired energy supply and may cause hemolytic anemias by decreasing the erythrocyte life span. These conditions were referred to in the past as congenital non-spherocytic hemolytic anemias /1, 2, 3/. Most of the enzymatic deficiencies are not due to quantitative deficiencies of the enzyme protein but, in a similar way to the hemoglobinopathies, to defective enzyme proteins (enzymopathies). As a result of alterations in functional characteristics more rapid enzyme inactivation ensues equivalent to a reduction or lack in enzyme activity. Genetic causes of enzymopathies are mutations in the region of the coding genes. The inheritance of most enzyme defects is autosomal recessive (X linked). Heterozygous trait carriers manifest reduced enzyme activity but, as a rule, they are not sick. Enzymopathies caused by homozygous (or compound heterozygous) inheritance can present with diverse clinical symptomatology, but hemolytic anemia comprises the largest group. Other important manifestations are polycythemia (erythrocytosis) and, additionally, syndromes of developmental disorders and severe neurological deficits /3/. The majority of enzymopathies can be detected in neonates by means of laboratory diagnostic investigation. The most important erythrocyte enzyme defects are: Pyruvate kinase (PK) deficiency, the most common enzyme defect of glycolysis /4/ Glucose-6-phosphate dehydrogenase (G6PD) deficiency, the most common enzyme defect of the pentose phosphate cycle /2/. Further rare but clinically relevant enzymopathies are: Triose phosphate isomerase deficiency, which leads to neuromuscular symptomatology Hexokinase deficiency, which is associated with chronic hemolytic anemia. Glucose-6-phosphate isomerase deficiency, which causes severe hemolytic anemia 2,3-diphosphoglycerate mutase deficiency, associated with polyglobulia. Of the variety of erythrocyte enzyme defects, PK deficiency and G6PD deficiency are described in greater detail in the following. Regarding other or more rare enzymopathies, the reader is referred to the specialized literature /3/. Prevalence and distribution With more than 400 million genetic carriers, erythrocyte enzyme defects are among the most prevalent hereditary metabolic disorders. G6PD deficiency, the geographic distribution of which mostly coincides with malaria regions, ranks at the top of the list, because there is a selection advantage for carriers of the G6PD deficiency genetic endowment /2/. In Mediterranean countries, Southeast Asia and Africa, and in the Afro-American population, the frequency of the genetic G6PD deficiency is between 10–20%. In Central and Northern Europe this enzyme deficiency was originally rare, but the number of affected individuals has greatly increased due to immigration of a large number of persons from endemic regions /1, 2, 3/. The second most prevalent enzyme defect is PK deficiency, the origin of which is mainly Central and Northern Europe as well as North America. In the current German population, it is to be expected that at most 10% of all patients with congenital hemolytic anemia have an enzymopathy /1/. 15.8.1 Indication Chronic hemolytic anemia, the etiology of which remains uncertain following hematological pre-diagnostic investigation. 15.8.2 Method of determination Determination of enzyme activity For routine purposes commercial automated enzyme analysis systems are normally used /1, 2, 3/. The enzyme activities are measured spectrophotometrically in the hemolysate. The latter is incubated at 37 °C with a mixture of substrate and reagents designed to make the enzyme activity to be determined rate-limiting. The parameter to be determined spectrophotometrically is the change in concentration of the pyridine nucleotides NADH or NADPH. The enzyme activities are calculated by using standardized formulae. In some of the enzymopathies the functional deficit is associated with characteristic alterations in a wide array of physicochemically properties of the enzyme. In these settings the sole determination of enzyme activities no longer suffices for establishing the diagnosis; instead, additional parameters (e.g. enzyme kinetics, enzyme electrophoresis, pH optimum, and thermostability) have to be considered. DNA analyses G6PD: if the G6PD gene mutation in the patient under evaluation is known, a PCR-based DNA analysis can be performed. In the meantime, individual laboratories have begun to offer complete sequencing of the G6PD gene, with which the known G6PD mutations can be ruled out or confirmed /1, 2/. The indication is limited to the diagnostic confirmation of heterozygous enzyme deficiency carriers (e.g., the question of the carrier status or, in extreme cases, of prenatal diagnosis). PK: in patients of Central and Northern European origin, targeted PCR-restriction enzyme analysis can be performed due to the high prevalence of type 1529 G to A mutations. This method can also be utilized for prenatal diagnosis in afflicted families if a 1529 G to A PK defect is present in the parents. For molecular diagnosis of PK, only a few special laboratories that can confirm or rule out currently known PK variants exist in Europe /4, 5/. This also applies to other erythrocytic enzymopathies. 15.8.3 Specimen EDTA blood: 5 mL 15.8.4 Reference interval Refer to Tab. 15.8-1 – Reference intervals for erythrocyte enzymes. 15.8.5 Clinical significance The investigation of enzyme deficiency states is usually indicated by decreased or minimal residual enzyme activities. Making a defined diagnosis from the laboratory findings may pose difficulties on account of the large variety of erythrocyte enzymatic defects which may occur. It must be noted that: Reticulocytes and young erythrocytes have a higher level of enzyme activity than senescent cells. In consequence, during hemolytic crises of any etiology, a false normal finding may be obtained in a markedly red cell population. Particular attention must therefore be paid to relative activities of various enzymes and their cell distribution. Falsely normal enzyme activities may be measured after recent blood transfusion. Patients with hypochromic anemia have seemingly elevated enzyme activity if the hemoglobin level is taken as reference for enzyme activity. 15.8.5.1 Glucose-6-phosphate dehydrogenase deficiency A classification of abnormal enzyme findings according to the following categories is helpful /2/: Slightly enzyme deficiency Moderate enzyme deficiency Severe enzyme deficiency Most severe enzyme deficiency. Group 1 is insignificant from a clinical point of view. In group 2, hemolysis is triggered only by oxidative stress. In group 3, extremely severe hemolytic crises occur under oxidative stress. Group 4 is characterized by permanent hemolysis which in addition can worsen critically. Peculiarities resulting from the X-linked inheritance of G6PD deficiency must be taken into account. Depending on the genotypes involved which differ in males and females the following manifestations occur: Men can be hemizygously normal or hemizygously deficient Women may be homozygously normal, homozygously deficient or heterozygous. Heterozygous women have normal, intermediate, or very low G6PD activities according to the timing of X-chromosome inactivation (Lyon hypothesis). At the time of acute hemolysis the diagnosis of G6PD deficiency may be very difficult since only those cells with relatively high enzyme activity remain following the destruction of cells deficient in enzyme activity. A follow-up examination may be warranted. In a few cases of G6PD deficiency the life span of the erythrocytes is markedly shortened, even without exposure. The resulting chronic non-spherocytic hemolytic anemia is distinguished from the forms known to be associated with defects of glycolysis (e.g., PK deficiency) only insofar as, additionally, hemolytic crises can be triggered by noxious oxidative agents. In all cases that have been examined closely, enzyme variants with particularly unfavorable kinetic properties, poor stability and low activity were found. With regard to further details, see Tab. 15.8-2 – Findings and enzymatic characteristics associated with the most important erythrocyte enzymopathies. 15.8.5.2 Pyruvate kinase deficiency The level of enzyme activity by itself is an unreliable diagnostic parameter since there is no exact correlation between the enzyme activity and severity of the hemolytic anemia /4, 7, 8/. The following observations, based on experience, provide useful indications. The majority of PK mutants associated with severe hemolytic anemia have enzyme activities of less than 30% of normal. In mild clinical course the enzyme activity may be significantly higher or, more rarely lower than they. Reticulocytes must always be taken into consideration. In every case monitoring of enzyme activity and reticulocyte counts, are advisable. The types of clinical-hematological presentations and the enzyme patterns found in cases of G6PD and pyruvate kinase deficiency are summarized in Tab. 15.8-2– Findings and enzymatic characteristics associated with the most important erythrocyte enzymopathies. 15.8.6 Comments and problems Stability Storage of samples for 3–4 days at 4–6 °C possible. 15.8.7 Pathophysiology Glucose-6-phosphate dehydrogenase deficiency /1, 2, 6/ The coding sequence of the polypeptide chain comprising 515 amino acids of the G6PD wild-type GdB+ is localized in the nucleotides of exons 2–13 of the G6PD gene. The GdA variant of black Africans is based on mutation 376 A → G (structure; 126 Asn → Asp); the Mediterranean type GdB is caused by mutation 563 C → T (Structure; 188 Ser → Phe). Many other variants with different phenotypic expression have been found and it is considered to be a proven fact that the population of Southeast Asia is also affected to a great extent. The clinically most important enzyme defects are the Mediterranean type (Gd Mediterranean, almost inactive) or the abnormal variants of the black population type A+ (GdA+, moderate reduction in activity) or type A (GdA, residual activity 5–15%). In the absence of oxidative damage, the vast majority of the G6PD deficiency carriers have a normal Hb concentration and reticulocyte number. Hematological crises are triggered by a range of oxidative injuries, in particular infections, medications (Tab. 15.8-3 – Medication and chemicals that cause hemolysis in G6PD deficiency), fava beans, acidosis and other factors. Hydrogen peroxide or other free radicals oxidize reduced glutathione, which due to the enzyme deficiency cannot be reduced again. The resulting precipitation of hemoglobin in the form of Heinz bodies causes massive intravascular hemolysis. This can, following destruction of the aged cells with the minimal enzyme activity, be self-limiting. An exception is the Mediterranean type deficiency (e.g.; favism). Pyruvate kinase (PK) deficiency /4, 5, 7, 8/ PK deficiency is based on a variety of heterogeneous mutations that, in turn, generate very different functional enzyme properties. The different molecular PK forms are coded by two genes: The gene M or muscle type gene, detectable in leukocytes and in many tissues The gene L or liver type gene, which also controls erythrocyte PK and is termed R type. Only variants of the PK-L gene cause hemolytic anemia. It is assumed that as a result of reduced ATP synthesis the cell membrane is damaged. The problem especially concerns reticulocytes and young erythrocytes. Due to the loss of energy sources, the cell membrane integrity is disturbed, resulting in increased rigidity and shrinking of the erythrocyte. In some cases, irregularly shaped cells are observed. They are sequestered by the spleen and phagocytosed by the reticuloendothelial system. The spleen eliminates mainly young cells, because following splenectomy the reticulocyte number increases considerably. The anemia can be so mild that diagnosis is not made before adulthood. In severe forms, already a newborn can become severely ill, manifesting the whole clinical picture of neonatal hemolytic disease. Splenomegaly usually does not develop before the age of 4 to 6 months. Pallor and anemia are the rule in the first weeks of life. Later hemolytic crises, particularly when infection is present, alternate with periods in which anemia is only moderately severe. Other hereditary red cell enzymopathies are only of secondary significance in the pathogenesis of enzymopenic hemolytic anemia, and are only detected in individual cases (overview Ref. /1, 2/). References Jacobasch G. Hereditäre Membrandefekte und Enzymopathien roter Blutzellen. In: Ganten D, Ruckpaul K (eds). Handbuch der Molekularen Medizin. Berlin: Springer- Verlag, 2000; 6: 392–441. Beutler E. Glucose-6-phosphate dehydrogenase deficiency and other red cell enzyme abnormalities. In: Beutler E, Lichtman MA, Coller BS, Kipps TJ, Seligsohn U (eds). Williams Hematology 6th ed. New York: McGraw-Hill, 2001: 527–45. Prchal JT, Gregg XT. Red Cell Enzymes. Hematology Am Soc Hematol Educ Program 2005; 19–23. Zanella A, Bianchi P. Red cell pyruvate kinase deficiency: from genetics to clinical manifestations. Baillieres Best Pract Res Clin Haematol 2000; 13: 57–81. Pissard S, Max-Audit I, Skopinski L, Vasson A, Vivien Pascal, Bimet C, Goossens M, Galacteros F, Wajcman H. Pyruvate kinase deficiency in France: a 3-year study reveals 27 new mutations. J Haematology 2006; 133: 683–9. Mehta A, Mason PJ, Vulliamy TJ. Glucose-6-Phosphatdehydrogenase deficiency. Baillières Best Pract Res Clin Haematol 2000; 13: 21–38. Beutler E, Blume KG, Kaplan JC, Löhr GW, Ramot B, Valentine WN. International Commitee for Standardization in Haematology: Recommended methods for red-cell enzyme assays. Brit J Haemat 1977; 35: 331–40. Pekrun A, Schröter W. Erythrozytenenzymdefekte als Ursache angeborener hämolytischer Anämien. In: Huber H, Löffler H, Faber V, eds. Methoden der diagnostischen Hämatologie. Berlin: Springer, 1994. 15.9 Enzymopenic methemoglobinemia Elisabeth Kohne Enzymopenic methemoglobinemia is a group of rare diseases with autosomal recessive inheritance that are provoked causally by NADH-cytochrome B5 reductase deficiency /1, 2/. The general clinical term for all forms is congenital recessive methemoglobinemia. Cytochrome B5 reductase exists in two variants, a soluble erythrocyte form that is involved in metHb reduction in erythrocytes and a membrane-bound form in different somatic cell systems that is integrated into numerous metabolic processes. Accordingly, enzyme deficiencies can exert different effects and can be associated with two different diseases: Hereditary enzymopenic methemoglobinemia type I is limited to erythrocytes. Homozygous or mixed heterozygous patients become ill with methemoglobinemia, while heteroyzgotes remain normal. Hereditary enzymopenic methemoglobinemia type II is the generalized form. Apart from congenital methemoglobinemia, there are progressive neurological symptoms with very severe psychomotor disorders and death in early childhood. 15.9.1 Indication Differential diagnosis of methemoglobinemia Uncertain situation with regard to cyanosis Cyanosis does not improve under oxygenation In conditions of cyanosis with normal or slightly reduced (about 85%) arterial blood oxygen saturation. 15.9.2 Method of determination Firstly, the blood metHb concentration is measured (see Section 15.5 – Dyshemoglobins). Determination of metHb reductase (cytochrome B5 reductase) activity in erythrocyte hemolysate is performed spectrophotometrically /3/. To confirm the diagnosis, determine the hereditary status and classify the type of hereditary methemoglobinemia that has been demonstrated, an DNA analysis is required. 15.9.3 Reference interval \| Cytochrome B5 reductase activity in erythrocytes /3/ \| \| \| Neonates \| 9.61 ± 1,91 IU/g Hb \| \| Adults \| 19.2 ± 3.9 IU/g Hb \| 15.9.4 Clinical significance In patients with congenital persistent methemoglobinemia and enzymatically determined cytochrome B5 reductase deficiency, a diagnosis of enzymopenic methemoglobinemia is made. In heterozygosity, clinical symptoms are absent. With the demonstration of the underlying basic molecular defect, the disease type can be classified. The investigation, which includes parental genetic counseling, is a component of the diagnostic program. If necessary, prenatal diagnosis can be performed. 15.9.4.1 General symptomatology The patients attract attention due to cyanosis that begins at birth. Methemoglobin values in newborns may exceed 40%. In older children and adults, the values are usually around 10–25%, but values as high as 40% may occur. The intake of food with different vitamin C content is the explanation for seasonal fluctuations. Some patients develop moderate compensatory polyglobulia. 15.9.4.2 Enzymopenic methemoglobinemia type I Type I cytochrome B5 reductase deficiency is characterized by the fact that it is limited to erythrocytes. Homozygous or mixed heterozygous patients become ill with uncomplicated methemoglobinemia; heteroyzgous trait carriers are normal but they are, nonetheless, sensitive to oxidizing substances. 15.9.4.3 Enzymopenic methemoglobinemia type II Type II is the generalized and lethal form of cytochrome B5 reductase deficiency, which, apart from the methemoglobinemia, is associated with progressive neurological symptoms. Neurological changes include severe disorders of mental development, microcephaly, dwarfism, cataract formation, as well as seizures, opisthotonus and generalized hypertension. The defect affects not only erythrocytes but also microsomal cytochrome B5 reduction in the liver, brain, muscles, leukocytes, thrombocytes and fibroblasts. Severe disorders of lipid metabolism with a reduction in brain cerebrosides, elevation of palmitic acid and decrease in linoleic acid in fatty tissues and characteristic changes in phospholipids, phospho glycerides and cholesterol in the liver, kidneys, spleen, muscle and adrenal glands also occur. The disease usually causes death in early childhood. References Kleihauer E, Kohne E, Kulozik AE. Anomale Hämoglobine und Thalassämie-Syndrome: Grundlagen und Klinik. Landsberg; ecomed Verlagsgesellschaft, 1996. Percy MJ, Lappin TR. Recessive congenital methaemoglobinaemia: cytochrome b5 reductase deficiency. Brit J Haemat 2008; 141: 298–308. Beutler E. Red cell metabolism. A manual of biochemical methods. 3th ed. Grune and Stratton, Inc 1984. 15.10 Erythropoietin (EPO) Lothar Thomas The organism’s red blood cell mass is kept constant under physiological conditions to assure optimal oxygen supply to the tissues. Every day some 2 × 1011 erythrocytes are lost due to blood cell turnover and without replacement hemoglobin (Hb) values would drop by 1 g/L in 24 hours. Since both the Hb content and the life span of the erythrocyte are genetically determined by its volume, red blood cell mass can only be preserved by dynamic adjustment of erythropoiesis. This occurs through a sensitive homeostatic mechanism that links the production of red blood cells to the oxygen requirements of the tissues. This process is mediated by EPO, an interleukin that is synthesized in the kidneys. EPO maintains the red blood cell mass by promoting the survival, proliferation, and differentiation of erythrocytic progenitors. If oxygen supply is reduced, increased quantities of EPO are produced, resulting in hyper proliferation of erythropoiesis /1/. Inhibition of EPO synthesis leads to the development of hypo proliferative, normocytic, normochromic anemia. In patients with anemia increase in erythropoietic activity stimulated by endogenous or exogenous EPO facilitates compensatory iron acquisition during recovery from hemorrhage-induced anemia. The erythroid regulator erythroferrone (ERFE) is produced by erythroid precursors in the marrow and the spleen and acts directly on the liver to decrease hepcidin production, and thereby increase iron availability for new red blood cell synthesis /2/. Refer to: Fig. 15.10-1 – Compensation of O2 deficiency in patients with acute hemorrhage. 15.10.1 Indication Normocytic anemia of uncertain etiology In hypo regenerative erythropoiesis for the differentiation of inadequate EPO synthesis from intrinsic hypo proliferation of bone marrow (see also Section 7.4 – Soluble transferrin receptor (sTfR)) Differentiation of erythrocytosis (polycythemia), see also Section 15.4 – Hematocrit Suspicion and progression monitoring of para neoplastic EPO formation Prior to treatment of non-renal anemia with erythropoiesis-stimulating agents (ESAs) Recognition of a fetal emergency. 15.10.2 Method of determination Radioimmunoassay As tracer 125J-labeled human recombinant EPO is used; the primary antibody is generated in rabbits, the secondary antibody for the precipitation of the immune complex is a goat antibody. Immunometric assay EPO is detected with the use of two monoclonal antibodies that are directed against recombinant EPO. The secondary antibody is enzyme labeled or with a chemoluminescent label /3/. Reference for calibration is the WHO 2nd IRP 67/343. 15.10.3 Specimen Serum, plasma (heparin): 1 mL Additionally, EDTA blood should be collected so that a reference of the EPO level to the hematocrit or the Hb value can be made. 15.10.4 Reference interval Refer to References /1, 4/ and Tab. 15.10-1 – Reference intervals for erythropoietin. 15.10.5 Clinical significance Tissue hypoxia stimulates the synthesis of EPO, while normal tissue oxygen supply suppresses its formation. 15.10.5.1 Assessment of erythropoietic activity The EPO concentration must be assessed relative to the red blood cell mass, the indirect measure of which is the hematocrit (HCT) or the Hb concentration. In this way it can be established, in chronic anemia, whether a stimulation of erythropoiesis is adequate for a decrease in HCT or Hb. There is an inverse relationship between the Hb level or the HCT and the decadic logarithm of serum EPO concentration. This is the case only in anemia, but not within the reference interval of the Hb or HCT /7/. With a slight decline in the HCT to 0.38–0.35 or in Hb to 125–115 g/L, EPO begins to increase slightly in the reference interval. However, a marked rise in EPO occurs only as of a HCT≤ 0.30 or an Hb ≤ 100 g/L (Fig. 15.10-2 – Expected range of EPO concentration as a function of the hematocrit /5/. For the same degree of anemia, however, an adequate increase in EPO does not always occur; this depends, rather, on the cause of the anemia /6/. Thus, the most marked increases in EPO are seen in aplastic anemia, and the minimal rises are observed in anemia of chronic disease and in the final stages of chronic renal insufficiency; iron deficiency anemia lies in between (Fig. 15.10-3 – Relationship between HCT and EPO concentration). EPO levels are lower in fetuses than in adults, in spite of the very regenerative nature of erythropoiesis. The mean fetal EPO concentration is ≤ 5 U/L until the 37th week of gestation. Due to fetal stress during birth, the values increase 10-fold on the average and then decrease during the first week of life. At the age of 7–12 weeks, infants manifest physiological anemia with an Hb value of 90–110 g/L; EPO values are also low, with the exception of a slight increase in the Hb nadir /8/. The assessment of deficient EPO formation is based on the EPO level in comparison with reference patients (iron deficiency anemia, hemolytic anemia, thalassemia inter media) /7/ with comparable Hb values or HCT (Fig. 15.10–3 – Relationship between HCT and EPO concentration). With values in the range of diminished EPO formation, the following causes are to be considered: Neonatal anemia Anemia of inflammation (rheumatoid arthritis, chronic infection, AIDS, inflammatory intestinal disease, autoimmune disease, critically ill patients) Cancer-related anemia with or without chemotherapy (solid tumors, multiple myeloma, malignant lymphoma) /8/ Erythrocytosis (polycythemia), see Section 15.4 – Hematocrit. The behavior of serum EPO in different forms of anemia is shown in: Tab. 15.10-2 – Diseases and conditions associated with an adequate rise in EPO concentration Tab. 15.10-3 – Diseases and conditions with inadequately low increase in EPO concentration Tab. 15.10-4 – Diseases and conditions with a disproportionate rise in EPO concentration. 15.10.5.2 Chronic kidney disease According to the National Kidney Foundation in the USA, the term chronic kidney disease includes /10/, patients with chronically diminished kidney function, dialysis-dependent patients and those with a malfunctioning kidney transplant. In untreated cases, renal anemia leads to the following disorders: reduced oxygen availability to the tissues, increased cardiac output, enlargement of the heart, ventricular hypertrophy, angina pectoris, congestive heart failure, decreased mental alertness, reduced immune response, and menstrual disorders. In children, delayed growth and diminished intellectual capacity occur. Renal anemia is based on inadequate EPO formation due to reduction of functional kidney tissue. Additional factors are iron deficiency, loss of blood, acute and chronic inflammation, aluminum toxicity, and a reduction in erythrocyte life span. A clarification of chronic kidney disease-associated anemia should be performed if /10/: In premenopausal women and prepubertal female individuals the Hb value is < 110 g/L (HCT < 0.33) In men and in postmenopausal women, the Hb value is < 120 g/L (HCT < 0.36). Increasing the Hb value to 130–150 g/L by ESA therapy does not reduce the risk of cardiovascular events within 3 years /11/. Iron metabolism and ESA in end-stage renal disease, see Tab. 7.3-4 – Diseases and conditions associated with elevated serum ferritin concentrations. 15.10.5.3 Chronic inflammation In anemic patients with chronic inflammatory disease, the inflammatory decrease of EPO formation is the key cause of normocytic, normochromic anemia /12/. Rheumatoid arthritis These patients have, at times, Hb values < 80–90 g/L, but hardly need banked blood. In severely anemic patients with rheumatoid disease, anemia is linked to iron deficiency. ESA therapy is not recommended in general, but can be important in individual cases. AIDS Approximately two thirds of these patients have normocytic, normochromic anemia, which is aggravated under treatment with AZT. ESA therapy may be justified in symptomatic patients. Tumor-associated anemia /13/ Anemia is common in tumor patients (see also Tab. 15.3-11 – Classification and differentiation of normocytic anemia). The inflammatory inhibition of EPO formation is of particular relevance and depends on the release of inflammatory cytokines and hepcidin. Chemotherapy and radiotherapy amplify the anemia due to the inhibitory effect on the bone marrow and the reduction in EPO formation (Fig. 15.10-4 – Activation of the immune system in acute phase response). 15.10.5.4 Critically ill patients In the critically ill, anemia is multifactorial and corresponds to the anemia of chronic disease. It is contingent upon inadequate low EPO formation, hypo proliferation of erythropoiesis (hepcidin increase and functional iron deficiency), due to intrinsic factors, and a reduced erythrocyte life span /14/. 15.10.5.5 Therapy with erythropoiesis-stimulating agents (ESA) Erythropoiesis and ESA Approximately 90% of hemodialysis patients receive ESA therapy (7000–8000 units rHuEPO per week), and 70% receive iron substitution (300 mg per month) in addition. Thus the erythron is extended and fatty marrow is replaced through hematopoiesis. The stimulation leads mainly to proliferation of early erythroid cells of the colony forming unit erythroid (CFU-E) (Fig. 15.1-2 – Erythropoiesis is separated into a proliferation pool and a maturation pool). In contrast, the pool of late erythroid progenitor cells is increased in cases of chronic red blood cell requirement (e.g., in chronic hemolytic anemia). If the marrow is already hyperproliferative due to stimulation by endogenous EPO, ESA does not lead to a meaningful further increase in regenerativity. Thus, erythropoiesis enhanced 2.9-fold can only be increased to 3.6-fold with high doses of ESA /15/. The stimulatory effect of ESA is dependent upon: The extent of regenerative bone marrow (no intrinsic hypo proliferation) The iron supply of erythropoiesis The presence of an acute phase reaction (CRP level). Iron availability Due to enhanced proliferation of erythropoiesis, ESA stimulation leads to an increased requirement for iron and to the distribution of iron from the stores into the bone marrow. With normal iron reserves in healthy individuals, intrinsic EPO stimulation can only increase erythropoiesis by 3-fold above the normal rate, without the generation of hypochromic red blood cells. If erythropoiesis is overstimulated by the administration of ESA, iron requirement exceeds the supply; this condition is termed functional iron deficiency. In consequence, hypochromic red blood cells are formed. This situation occurs particularly in the early regeneration phase of erythropoiesis following administration of ESA. If the iron stores are repleted and if erythropoiesis is adequately stimulated with ESA, functional iron deficiency only occurs if the ferritin value is below 100 μg/L. Under ESA therapy, the ESA dose and iron supply must be coordinated with one another. Acute phase reaction (APR) Systemic inflammatory changes in the organism are termed APR. Besides functional iron deficiency, APR is one of the most important reasons for the different responsiveness of patients to ESA and for a elevated ESA requirement /15/. Patients with inflammation have elevated serum concentrations of IL-6, CRP, fibrinogen and ferritin. Transferrin and transferrin saturation are reduced. The cause of reduced bone marrow responsiveness in inflammation is an enhanced expression of IFN-γ, IL-6 and hepcidin /16/. In bone marrow, these mediators antagonize the anti-apoptotic effect of EPO on CFU-E and lead to ineffective erythropoiesis (Fig. 15.10-4 – Activation of the immune system in acute phase response). In addition, an increase in hepcidin leads to a disturbance in iron distribution due to inhibited iron release from macrophages and enterocytes via the ferroportin receptor (see also Section 7.6 – Hepcidin). The resulting functional iron deficiency intensifies the ESA resistance, leading to anemia. Patients with chronic kidney disease, with malignancy, with chronic inflammatory disease, as well as critically ill patients, often suffer from anemia which affects, to a significant degree, morbidity, mortality and the quality of life. The majority of these patients have normocytic, normochromic erythrocytes. ESA treatment is advantageous for these patients, particularly the use of banked blood is reduced. Additional parenteral administration of iron enhances the erythropoietic response to ESA. ESA treatment is expensive. It is therefore important to select patients who are likely to respond to ESA and monitor the erythropoietic response. Tests before starting ESA therapy and for therapeutic monitoring are listed in Tab. 15.10-5 – Tests for the assessment of erythropoiesis under ESA therapy. Treatment with ESA is not free of risks. Thus, investigations show that ESA therapy may be associated with /9/: An elevated risk of venous thrombosis Enhanced tumor progression and, possibly, a shortening of survival. ESA therapy in chronic renal insufficiency Anemia is a strong predictor of cardiovascular disease and mortality in patients with chronic kidney disease /10/. The correction of anemia with values of < 110 g/L with ESA therapy, to target values of 110–120 g/L, improves the condition of the patients. Optimal iron reserves are present with ferritin values of 200–500 μg/L and transferrin saturation (TSAT) of > 20% (Tab. 15.10-5 – Tests for the assessment of erythropoiesis under ESA therapy). According to the National Kidney Foundation of the USA, intravenous iron therapy should be implemented in patients with functional iron deficiency with a TSAT < 20% if the ferritin value is ≤ 800 μg/L. Refer also to Tab. 15.10-6 – Erythropoiesis stimulating agents. ESA therapy in tumor-associated anemia ESA therapy should only be initiated in patients with symptomatic anemia with an Hb level below 100 g/L and in patients with asymptomatic anemia with an Hb level below 80 g/L. The Hb target is a stable level of about 120 g/L /17/. Effectivity of ESA therapy The effectivity of ESA therapy depends on the cause of the anemia and the presence of inflammation. Patients with end stage renal disease have the best response rate, which is some 70%. Tumor patients have a response rate of 30–70%, but in those with myelodysplastic syndrome it is only 20%. A rise in the Hb value of 10 g/L within 4 weeks after initiation of therapy is considered to be a positive response. 15.10.6 Comments and problems Sampling Should be performed during the morning due to daily fluctuations of EPO concentrations. Maximal values are seen at midnight, the nadir occurs during the morning hours. Method of determination Values obtained with commercials assays are comparable /1/. The measurement of synthetic EPO preparations elicits very different results due to variability in structure and immunogenetics. Precision: depending on the test CV% is 7–24 in the range of 5–10 U/L. Only at 50 U/L is CV% below 10. Accuracy: when tested with standard 87/684, the accuracy of 6 assays was ± 25% and with one assay it was + 200% /18/. Reference interval Since the reference interval is very broad, and a significant increase beyond the upper value of the reference interval can only be detected with a HCT of less than 0.30 or a Hb of below 100 g/L, reference must be made to one of these two parameters. The reference curve relating the HCT or Hb to the EPO concentration can be generated based upon an investigation of cohorts that include patients with iron deficiency anemia, hemolytic anemia or thalassemia inter media /7/. EPO concentrations are not age dependent /19/. Half-life With normal Hb values, the half-life of endogenous EPO is 5.2 hours; in patients with anemia it is 1.5–2.9 hours. Stability In serum at least 2 weeks at room temperature /20/. 15.10.7 Pathophysiology In combination with other hematopoietic growth factors, EPO is a physiological regulator of erythropoiesis (Fig. 15.10-5 – Site of action of erythropoietin and other hematopoietic growth factors in erythropoiesis) /20/. Depending on the developmental stage of erythroid progenitor cells, EPO can be a mitogen, an apoptosis inhibitor, or a differentiating factor. Thus the CFU-E cells require contact with EPO molecules in order not to become apoptotic /21/. In healthy individuals, circulating EPO originates from neuronal fibroblasts near the proximal tubular cells. The synthesis of EPO is controlled at a transcriptional level due to dynamic changes in the oxygen tension. The production increases under hypoxic conditions (Fig. 15.10-6 – Erythropoietin feedback control loop) /1/. In minor amounts EPO is produced in the liver and brain. EPO exerts its influence by binding to a specific receptor, the EPO receptor (EPOR) on the cell surface of erythroid progenitor cells, resulting in erythroid proliferation, differentiation and inhibition of apoptosis (Fig. 15.10-7 – Homodimer erythropoietin receptor following showing phosphorylated tyrosines). The EPOR homodimerizes in the presence of EPO and auto phosphorylation of the Janus tyrosin kinase 2 (Jak2) occurs. Once JAK2 is activated, specific EPOR tyrosines are phosphorylated and form docking sites for adapter molecules such as Grb 2, the signal transducer and activator of transcription 5 (STAT5) and phosphatidylinositol-3-kinase (PI3-K). Activated STAT5 forms dimers and trans locates into the nucleus where it induces transcription of genes involved in proliferation and cell survival. PI3-K, via activated Akt, also induces the expression of several anti-apoptotic proteins, such as Bcl-2 and Bclx, to prolong cell survival by Janus kinase 2 (JAK 2) and the subsequent transfer by STAT 5 to the cell nucleus. Two groups of EPO receptor mutations are clinically relevant (see also Section 15.4 – Hematocrit (HCT)) /22/: Replacement of arginine with cysteine at position 129 of the extracellular portion of the EPOR with the consequence of spontaneous homodimerization of the EPOR in the absence of EPO. The continuous stimulation of the progenitor cells, with development of erythrocytosis, is the result. The modification of the intracellular C-terminal portion of the EPOR, with deficiency of the binding site for phosphatase SHP-1. Physiologically, this enzyme removes phosphate groups from the tyrosine residues of the C-terminus of the EPOR thereby inactivating signal transduction. The removal of the SHP-1 binding site increases the sensitivity of the cell for EPO, leading to erythrocytosis. EPO is a glycoprotein of MW 30–34 kDa. Its biological activity is dependent upon its tertiary structure, which is comprised of 4 α-helices with 2 long loops and 1 short loop. Endogenous and recombinant EPO have the same sequence of 165 amino acids, but they differ from one another with regard to their glycosylation. Endogenous EPO is more acidic than rHuEPO and they can be distinguished from one another through isoelectric focusing of a urine sample. Four oligosaccharide chains make up 35–40% of the molecular mass. Recombinant HuEPO that is formed in mammalian cell cultures (e.g. of the Chinese hamster) is fully glycosylated, but it is not glycosylated if synthesized in E. coli cultures /23/. The fetal liver is the main site of EPO production from the 20th week of gestation. Following birth, the renal interstitial cells increasingly take over the role of EPO formation, and in adults 85% of the daily EPO synthesis takes place in the kidneys. Under normal conditions, EPO synthesis is activated as a response to a reduction in the red blood cell mass (anemia) or to reduced O2 saturation of erythrocyte hemoglobin (hypoxemia) /24/. The result of hypoxic stimulation is increased formation of hypoxia-inducible factor-1 (HIF-1) in the kindey. HIF-1, as the most important EPO gene regulating factor, stimulates EPO synthesis. HIF-1 is a global regulator of cellular and systemic O2 homeostasis. HIF-1 also regulates vessel formation, promotes the survival of ischemic cells and plays a role in carcinogenesis. Hypoxia-driven EPO gene expression in peritubular fibroblasts results from an array consisting of the kidney inducible element(KIE) and a negative regulatory element (NRE) /23/. 15.10.7.1 Roxadustat Roxadustat an oral hypoxia-inducible factor prolylhydrolase inhibitor increases endogenous EPO concentration through the stabilization of hypoxia inducible factor (HIF; refer to section 15.4), which is achieved by inhibiting the activity of its pan-prolylhyroxylase domain. A study /54/ showed that roxadustat increased the hemoglobin concentration and reduced ferritin and transferrin saturation after patients undergoing peritoneal dialysis were switched from an Erythropoietin stimulating Agent (ESA). References Jelkmann W. Regulation of erythropoietin production. J Physiol 2011; 589: 1251–8. Kautz L, Jung G, Du X, Gabayan V, Chapman J, Nasoff M, et al. Erythroferrone contributes to hepcidin suppression and iron overload in a mouse model of β-thalassemia. Blood 2015; 126: 2031–7. 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A new concept for the differential diagnosis and therapy of anemia in cancer patients. Support Care Cancer 2011; 19: 261–9. Kidanewold A, Woldu B, Enawga B. Role of erythropoiesis stimulating agents in the treatment of anemia: a literature review. Clin Lab 2021; 67: 883–92. Kawai Y, Toya Y, Wakui H, Fujikawa T, Ueda E, Azushima K, et al. Comparison of the effects of weekly and biweekly intravenous CERA administration on erythropoiesis: a randomized controlled trial.J Clin Hypertens 2021; 00: 1–9. Eckardt KU, Agarwal R, Aswad A, Awad A, Block GA, Bacci NR, et al. Safety and efficacy of vadadustat for anemia in patients undergoing dialysis. N Engl J Med 2021; 384: 1601–12. Chertow GM, Pergola PE, Farag YMK, Agarwal R, Arnold S, Bako G, et al. Vadadustat in patients with anemia and non-dialysis-dependent CKD. N Engl J Med 2021; 384: 1589–600. Singh AK, Carroll K, Perkovic V, Solomon S, Jha V, Johansen KL, et al. Daprodustat for the treatment of anemia in patients undergoing dialysis. N Engl J Med 2021. doi: 10.1056/NEJMoa2113379. Hirai K, Kaneko S, Yanai K, Taisuke K, Kiyonori I Ookawara S. Effects of Roxadustat on anemia, iron metabolism, and lipid metabolism in patients with non-dialysis chronic disease. Front Med 2023; 22 February. doi: 10.3389/fmed.2023.1071342. 15.11 Thrombocyte count and thrombocyte indices Lothar Thomas Thrombocytes, also called platelets, are anucleic, granula rich cells, the functions are: Maintenance of hemostasis Initiation of tissue repair following vascular injury and in inflammation. In healthy individuals, the bone marrow releases 1× 1011 thrombocytes daily, approximately one half of the daily erythrocyte production. Megakaryopoiesis and thrombocyte formation are regulated by thrombopoietin. Thrombocytes arise from the fragmentation of megakaryocytes, a process in which the formation includes elongated strands of cytoplasm, also called prothrombocytes. Refer also Section 15.1 – Hematopoiesis) Thrombocytes in the blood are normally in the resting stage. Following physiological stimulation, they undergo a change in shape, with subsequent adhesion to a surface, followed by aggregation. The changes in shape that thereby occur can be observed as surface changes in vivo using physical methods, and with immunological methods by measuring the expression of receptors. In platelet aggregation or following marked stimulation, the platelets are degranulated and the granula membrane is transferred to the thrombocyte surface. In this way, neoantigens are expressed on the activated thrombocyte in the form of glycoproteins. These processes can be identified diagnostically in the laboratory with the use of immunofluorescence and flow cytometry. But flow cytometry alone is also capable of recognizing changes in platelet shape. Thus, changes in forward and side scattering of light reveal, respectively, changes in platelet volume and in granularity. In addition to this chapter refer also to Section 16.1-3 – Thrombocytes Chapter 17 – Thrombocyte diagnostics. 15.11.1 Indication Bleeding of uncertain etiology Exclusion of a bleeding tendency Monitoring during radiotherapy and under treatment with cytostatics and heparin In the presence of erythrocytosis In splenomegaly. Suspicion of bone marrow disease (myelophthisis, myeloproliferation). Suspicion of destruction, consumption (e.g., sepsis) or reactive proliferation of thrombocytes. 15.11.2 Thrombocyte count 15.11.2.1 Microscopic counting Principle: venous or capillary blood is diluted with a hypotonic solution containing a platelet aggregation inhibitor (1% ammonium oxalate solution). This is performed using a pipette into which blood is aspirated up to the 1 mark, and the diluting solution (1% ammonium oxalate solution) up to the 101 mark. For the purpose of hemolysis, the sealed pipette is rotated for 15 minutes. Counting is performed in the Neubauer chamber or the Thoma chamber. Before starting counting, the thrombocytes are to sediment in a chamber for 10 minutes /1/. The reference method is a hemocytometer count with the use of a phase contrast microscope /2/. 15.11.2.2 Hematology analyzer The determination can be performed using the impedance method, the optical method, or with immunoplatelet counting. One-dimensional and two-dimensional optical methods are distinguished. Some hematology analyzers measure according to the impedance and the optical method. Impedance method The platelets are counted using the electric resistance measurement (impedance) method. If a thrombocyte that migrates through a capillary interrupts the applied electrical circuit, an impulse is triggered. Thrombocytes and erythrocytes from the same cell suspension are determined. For thrombocytes the pulses are counted with a window of 2–20 fL and for red blood cells in one > 36 fL. The average size of all pulses in the thrombocyte histogram is expressed as the mean platelet volume (MPV) /3/. One-dimensional optical method A laser diode is used for generation of monochromatic light. The light is directed to the flow cell at an angle of 2–3°. According to the Mie scattering theory, the intensity of the monochromatic light that is scattered from a homogeneous particle depends on its volume and the difference in the refraction index between the particle and its surrounding medium. The thrombocyte count is determined by scattered light impulses that are generated, while the MPV is determined by the scattered light intensity /3/. Two-dimensional optical method The principle corresponds to that of the one-dimensional method. With the use of a detergent, however, a spherical structure is imposed on the platelets, and the scattered light is measured at two angles, a low angle of 2–3° and a high angle of 5–15°. With the two-dimensional light scattering technology, normal-sized platelets, large platelets (20–30 fL), red blood cell (RBC) fragments, erythrocyte ghosts, microcytes, and cellular debris can be differentiated /4/. Combination of impedance and optical methods Some hematology analyzers use both counting methods to distinguish between platelets and other particles, particularly with low platelet counts. With low thrombocyte count and in the presence of other particles, the impedance measurement manifests a bias toward higher values than optical measurement because other particles are also recognized /5/. Immunoplatelet counting Antibodies against CD41 (GPIIb), CD42 (GPIb) and CD 61 (GPIIIa) are used to identify platelets which are measured by flow cytometry. The thrombocyte count is calculated from the relationship between the detected fluorescence and the number of red blood cells (RBC ratio) that are identified with a blood cell analyzer by means of impedance counting. The platelet count is calculated by multiplying the RBC ratio with the number of red blood cells. The advantage of the RBC ratio is that the thrombocyte count is independent of dilution and pipetting errors /6/. An alternative method involves the addition of a fixed quantity of latex particles to the sample. The number of latex particles is determined by impedance counting and the platelet-dependent fluorescence is derived from that number /7/. 15.11.3 Thrombocyte indices Apart from the thrombocyte count, hematology analyzers determine or establish the mean platelet volume (MPV), the platelet crit (PCT), and the platelet distribution width (PDW). The Advia 120 determines, in addition, the mean platelet component concentration (MPC), the platelet component distribution width (PCDW), the mean platelet mass (MPM), and the platelet mass distribution width (PMDW). Changes in the MPC are a gauge of platelet activation. The broadening of the PDW indicates hyper regenerative thrombopoiesis. 15.11.3.1 Immature Platelet Fraction (IPF) Determination of the IPF (Immature Platelet Fraction) is useful for the clarification of thrombocytopenia. Procedures: the flow cytometric determination is performed (e.g., with the Sysmex XE-2100) using the fluorescent dyes polymethine and oxazine. Both dyes penetrate the thrombocyte cell membrane and label thrombocyte and erythrocyte RNA. The labeled cells are directed by a semiconductor diode laser beam and the resulting forward scattering (platelet volume) and fluorescence intensity (platelet RNA content) are recorded. The IPF and mature thrombocyte fraction are distinguished using a computerized algorithm. The IPF is expressed as a relative fraction of the total thrombocyte fraction. 15.11.4 Specimen EDTA blood: 1 mL Capillary blood (EDTA-coated capillaries): 0.02 mL 15.11.5 Reference interval In thrombocyte count sex and genetic background are not important for reference intervals. The effect of age is higher as of sex and genetic background. Reference intervals (2.5 to 97.5 percentiles) in population setting are as follows (x 109/L): Below 15 years 176–452; females 15–64 years 165–405; males 15–64 years 141–364; females > 64 years 140–379, males > 64 years 122–350 /22/. Reference intervals for platelet indices are published in Ref. /60/. The coefficient of variation (CV) from day to day is 6.7%, and from month to month 10.6%, respectively /21/. 15.11.6 Clinical significance The hemostatic system consists of coagulation factors, the vascular wall and thrombocytes. The circulating lifespan of thrombocytes that are released from the bone marrow is 7–10 days, until they are removed by macrophages of the reticuloendothelial system. In healthy individuals approximately one third of all thrombocytes are contained in the spleen, while the remainder is in the circulation. The thrombocyte pool in the spleen is readily available and there exists a free exchange of thrombocytes between the spleen and the circulation. To fulfill their hemostatic function thrombocytes must not only be fully functional, in addition their number must lie within certain limits. Clinicians consider a thrombocyte count of (100–400) × 109/L to be normal, and only occasionally do clinically normal individuals manifest values that are outside this range. With thrombocytopenia ≤ 10 × 109/L there is a danger of bleeding, and in thrombocytosis ≥ 450 × 109/L the risk of thrombotic events is increased. Since, in many diseases, determination of the complete blood count is performed with the use of hematology analyzers, identification of thrombocytopenia and thrombocytosis is far more common than the clinical cases with clinical hemostaseological symptomatology. 15.11.6.1 Thrombocytosis The terms thrombocythemia and thrombocytosis are used synonymously and are not defined unambiguously. Commonly, they are taken to mean a rise in the thrombocyte count to above 450 × 109/L. Individuals with values of (350–450) × 109/L should be monitored. The extent of thrombocytosis is classified arbitrarily according to the thrombocyte count as /12/: Mild with (450–700) × 109/L. Moderate with (700–900) × 109/L. Severe with more than 900 × 109/L. Based on their etiology, thrombocytosis is classified as: Hereditary or familial Clonal forms, which are associated with myeloproliferative or myelodysplastic disease Secondary forms: these are reactive forms of thrombocytosis. Clonal forms of thrombocytosis are termed primary thrombocytoses. Since thromboembolic events occur more frequently in primary thrombocytoses, the differentiation of primary and secondary forms is important. 15.11.6.1.1 Primary thrombocytosis Primary thrombocytoses are the consequence of myeloproliferative and myelodysplastic disorders or are genetically or are familial. Hereditary causes are activating mutations in the MPL gene, which codes for the megakaryocyte and thrombocyte receptors of the same name, or the TPHO gene, which codes for thrombopoietin. 15.11.6.1.2 Secondary thrombocytosis Either there is increased formation of thrombocytes due to a stimulus or thrombocytes are released into peripheral circulation from the splenic pool. Increased release from the splenic pool occurs (e.g., in physical effort, stress, or administration of catecholamines). Bone marrow thrombocytopoiesis is stimulated by the peripheral loss of thrombocytes (e.g., due to immunological factors, sepsis, blood loss, or oncogenes). Secondary thrombocytosis hardly ever triggers thrombosis. Following recovery, the thrombocyte count decreases once again. In the bone marrow, the megakaryocytes are increased and are seldom dysmorphic. The large thrombocytes that appear in peripheral circulation are round, their functionality is normal, and the platelets do not tend to aggregate spontaneously, as do the primary thrombocytes /12/. Approximately 88% of thrombocytoses over 500 × 109/L are secondary in nature and are usually caused by an inflammatory event /13/. Differentiation The differences between primary and secondary thrombocytosis in laboratory diagnostic investigation are listed in Tab. 15.11-2 – Differences between primary and secondary thrombocytosis. 15.11.6.2 Thrombocytopenia Thrombocytopenia refers to a decrease in the count to < 100 × 109/L. Threatening complications of bleeding do not occur, however /14/: In ambulatory patients with aplastic anemia and counts of ≥ 5 × 109/L Following minor surgical procedures in patients with fever > 38 °C or following thrombocyte transfusion due to recent grade 3 bleeding according to WHO, if the count was ≥ 10 × 109/L. An important point for such a decision is the accuracy of the thrombocyte count with values of this order of magnitude /5/. The benefits of the administration of banked thrombocytes in patients with a thrombocyte count ≥ 5 × 109/L without bleeding is not documented. Frequent clinical findings in thrombocytopenia are petechiae, purpura, mild to moderate mucosal bleeding, bilateral epistaxis, and gastrointestinal, pulmonary and urogenital bleeding. Symmetrical petechiae and purpura, which affect both the trunk and the extremities are characteristic. In routine clinical practice, thrombocytopenia is frequently associated etiologically with the ingestion of medication, chemotherapy, sepsis, disseminated intravascular coagulation, or massive blood transfusions. All cases of thrombocytopenia require laboratory confirmation by means of an additional method of counting, and the investigation of a blood smear. Etiologies of thrombocytopenia are: Reduced thrombocyte formation. Elevated thrombocyte turnover. Disturbed thrombocyte distribution or a dilution-associated increase of plasma volume. Pseudo thrombocytopenia. Findings for differentiation of etiologies are shown in Tab. 15.11-3 – Differentiation of the mechanism of thrombocytopenia. 15.11.6.2.1 Reduced thrombocyte formation Reduced formation is relatively rare (children < 5%; adults < 10%) the cause of thrombocytopenia. A distinction is made between /15/: Hereditary forms (e.g., Wiskott-Aldrich syndrome, Chediak-Higashi syndrome, thrombocytopenia-absent radius syndrome, Alport syndrome, Fechtner syndrome, Trousseau Syndrome, May-Hegglin anomaly, type IIB von Willebrand disease, Bernard-Soulier syndrome, gray platelet syndrome, platelet type von Willebrand syndrome, Mediterranean macro thrombocytopenia), and congenital thrombocytopenia (macro thrombocytes). Disease and therapy associated transient thrombocytopenia following chemotherapy for malignant hematological neoplasia or following bone marrow infiltration of solid tumors, following stem cell injury due to radiation or medication. The thrombocyte count should be ≥ 10 × 109/L; in necrotizing tumors ≥ 50 × 109/L. 15.11.6.2.2 Increased thrombocyte turnover Destruction of the circulating thrombocytes is the most frequent cause of thrombocytopenia. Two essential forms are distinguished: Immune thrombocytopenia (ITP). This refers to enhanced clearance of thrombocytes from circulation, due to thrombocyte-associated IgG and complement activation. ITPs are complications of HIV and hepatitis C viral infections and following the treatment of patients with Helicobacter pylori infection. Medications also cause ITP /16/. A selection is listed in Tab. 15.11-4 – Medication-associated autoimmune thrombocytopenia. Non-immune associated: disseminated intravascular coagulation (DIC), sepsis, hemolytic uremic syndrome, thrombotic thrombocytopenic purpura, intraoperative, following multiple blood transfusions. Acquired thrombocytopenia due to destruction or peripheral consumption are, in most cases, severe forms of the disease. The thrombocyte count is markedly reduced, the MPV and thrombocyte function are normal, but the lifespan of the thrombocytes is reduced. The bone marrow manifests hyper regenerative megakaryopoiesis. 15.11.6.2.3 Distribution and dilution associated thrombocytopenia Patients with splenomegaly sequester thrombocytes. Hypersplenism usually leads to mild thrombocytopenia of (50–100) × 109/L. The cause is the storage of up to 90% of the thrombocytes in the enlarged spleen. The lifespan of the thrombocytes is only slightly shortened, which shows that they are only sequestered and not destroyed. In patients with splenomegaly and liver cirrhosis, thrombocytopenia is due to the sequestration of the platelets in the spleen to a lesser extent than it is to decreased hepatic TPO formation /17/. Dilutional thrombocytopenia is caused by transfusion therapy in massive blood loss. 15.11.6.2.4 Thrombocytopenia and thrombocyte transfusion The following banked thrombocyte concentrates (TC) are available as thrombocyte substitutes /17/: Pooled units from 4–6 donors with (240–360) × 109 platelets in 200–350 mL of plasma or plasma replacement solution Apheresis TC from a single donor with (200–400) × 109 thrombocytes in 200–300 mL of plasma. A TC contains < 3 × 109 erythrocytes and < 1 × 106 leukocytes. The thrombocyte recovery rate in peripheral blood is only 60–70%, since the remainder is kept in the spleen. In sepsis, in disseminated intravascular coagulation, or in the presence of thrombocyte antibodies, the recovery rate is even lower. Fresh donor thrombocytes are detectable for 7–10 days in peripheral blood. The thrombocyte count should be determined prior to transfusion, as well as 1 and 20 hours post-transfusion. If a refractory condition is present, the increase following 1 hour is < 7,5 × 109/L and following 20 hours < 4,5 × 109/L. Recommendations for transfusion are provided in Tab. 15.11-5 – Recommendation for thrombocyte transfusion. Critically ill patients often have disorders of hemostasis. Thrombocytopenia < 50 × 109/L was observed in a multicenter study /18/ in 13.7% of the cases and 35.4% of the patients with severe thrombocytopenia died in intensive care. The administration of thrombocyte concentrates was very inconsistent. Approximately 40% of the transfusions were administered with a thrombocyte count > 50 × 109/L and 34% in spite of the fact that with this thrombocyte count on the day of the transfusion there was no significant bleeding. With a transfusion of, on the average, 1.7 units, the mean thrombocyte increase was: 18.5 × 109/L [interquartile range (2.0–35.5) × 109/L]. In patients with severe thrombocytopenia (platelet count 10–50) × 109/L witholding prophylactic platelet transfusion before central venous catheter (CVC) placement did not meet the predefined margin for noninferiority and resulted in more CVC-platelet bleeding events than prophylactic platelet transfusion /63/. 15.11.6.3 Diseases and conditions with thrombocytosis or thrombocytopenia Diseases and conditions with primary thrombocytosis are listed in Tab. 15.11-6 – Diseases and conditions with primary thrombocytosis Diseases and conditions with secondary thrombocytosis are shown in Tab. 15.11-7 – Diseases and conditions with secondary thrombocytosis. Thrombocytopenias of various etiologies are listed in Tab. 15.11-8 – Thrombocytopenia of various origins. Pregnancy related thrombozytopenias are listed in Tab. 15.11-9 – Pregnancy associated thrombocytopenia. 15.11.6.4 Mean platelet volume (MPV) The MPV can be used, in combination with the thrombocyte distribution width, to distinguish conditions of reduced formation from those with increased thrombocyte destruction. 15.11.6.4.1 Acute bleeding In the presence of a significant decrease in the thrombocyte count, the MPV is increased and the platelet distribution width (PDW) is broadened. 15.11.6.4.2 Disturbances of thrombocyte formation In aplastic anemia, megaloblastic anemia, chemotherapy of malignant tumors, acute leukemia, and systemic lupus erythematosus, the thrombocyte count and the MPV are decreased, and the PDW is broadened. With improvement of the clinical symptomatology, or following chemotherapy, a rise in the MPV occurs before the thrombocyte count. 15.11.6.4.3 Immune thrombocytopenia The MPV and the PDW are normal. 15.11.6.4.4 Hereditary thrombocytopenia All forms of thrombocytopenia with x-linked recessive inheritance have a low MPV with a left shift of the log-normal volume distribution (e.g., Wiskott-Aldrich syndrome). Hereditary thrombocytopenia with macro thrombocytes and elevated PDW are: Alport syndrome, May-Hegglin anomaly, Sebastian anomaly, type IIB von Willebrand disease, Bernard-Soulier syndrome, Mediterranean macro thrombocytopenia, and autosomal dominant thrombocytopenia /15/. 15.11.6.4.5 Reactive thrombocytosis Under reactive conditions with an elevated thrombocyte count due to release from the splenic pool, which is the case with infection, tumor, rheumatoid arthritis, pancreatitis, or following a surgical procedure, MPV and PDW are normal. In myeloproliferative syndrome, the MPV may be elevated and the PDW may be broadened. 15.11.6.4.6 Splenectomy The thrombocyte count and the MPV may be elevated, and the PDW may be broadened. 15.11.6.4.7 MPV, metabolism and ischemic heart disease In comparison with normal thrombocytes, those with an increased MPV have increased metabolic activity and an elevated potential for thrombosis. Increased MPV is associated with obesity, diabetes mellitus, and ischemic cardiovascular events. In one study /20/, patients with acute coronary syndrome had a higher MPV value and a lower thrombocyte count than patients with stable angina pectoris and healthy controls. Thus, the mean values in 60 individuals in each group were: Healthy controls: 257 × 109/L, MPV 9.1 fL Stable angina pectoris: 267 × 109/L, MPV 10.0 fL Acute coronary syndrome: 201 × 109/L, MPV 11.0 fL. 15.11.6.5 Immature Platelet Fraction (IPF) In thrombocytopenia it is important to differentiate whether the cause is failure or suppression of the bone marrow, increased peripheral platelet consumption, or peripheral platelet destruction. In the two latter cases, the bone marrow releases immature thrombocytes with high RNA content. These thrombocytes are analogous to the reticulocytes in erythropoiesis, and are also termed reticulated platelets. Reticulated platelets are identical to the IPF. The fraction of reticulated platelets, or the IPF, reflects the rate of thrombopoiesis. The IPF reference interval is 1.1–6.1%. A particularly high IPF is demonstrated in autoimmune thrombocytopenia (9.2–33.1%) and acute thrombocytopenic purpura (11.2–30.9%). With adequate therapy the IPF falls, and the thrombocyte count increases /19/. 15.11.7 Comments and problems Method of determination In thrombocytopenic patients, variations are seen between thrombocyte determinations performed on hematology analyzers with different counting methods. Causes are: The inability to distinguish between thrombocytes, fragmented erythrocytes and microcytes. In chronic lymphocytic leukemia, nuclear and cytoplasmic lymphocyte fragments may also be counted as thrombocytes Macro thrombocytes are excluded from the platelet count with impedance counting and one-dimensional optical methods. Pseudo thrombozytopenia, caused by platelet clumping, is often found in clinical routines. However pseudo thrombocytosis resulting from fragmentation of red blood cells is a very rare phenomenon. Lipemia Platelet count can be measured by fluorescence method, optical method, and impedance method. The fluorescence method can more accurately reflect the true platelet count of lipemia specimens compared with impedance method and optical method /61/. Immunoplatelet counting Likewise, immunoplatelet counting is not free of disadvantages. Thus /6/: Platelet aggregates, platelet complexes with leukocytes, and macro thrombocytes can lead to gating problems with flow cytometry Due to antibodies (CD 41, CD 42b, CD 61), thrombocytes are not counted in Bernard-Soulier syndrome and in Glanzmann’s thrombasthenia Thrombocyte autoantibodies in patients receiving therapy against glycoproteins (e.g., anti-GP IIb/IIIb; Reopro), can interfere with the determination. Repetition of thrombocyte count In thrombocytopenia of clinically uncertain etiology, the laboratory must repeat the determination using a different procedure. Reference interval Age, sex and genetic background modulate platelet count in healthy people. The effect of aging is much bigger than those of sex and ethnicity. Reference intervals (2.5th and 97.5th percentiles) estimated on the global population were as follows (× 109/L): all < 15 years 176–452; women 15–64 years 156–405; men 15–64 years 141–364; women > 64 years 140–379; men > 64 years 122–350 /22/. Pseudothrombocytosis Abnormally shaped erythrocytes in the form of spherocytes, microspherocytes, and cells with budding projections (asterisks) can be the reason of pseudothrombocytosis in patients with severe burn injuries. Heat-induced disruption of red cell membranes can generate smaller cells that may be incorrectly counted as platelets by automated blood counters /62/. Intraindividual variation CV over the course of the day: 6.7%; from day to day: 11.5%; from month to month: 10.6% /21/. Stability The thrombocyte count is stabile at room temperature for at least 24 hours; and with certain hematology analyzers, for up to 168 hours. References ICSH: International Committee for Standardization in Hematology. ICSH Expert Panel on Cytometrie. 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Studies on the immune response in heparin-induced thrombocytopenia. Blood 2009; 113: 4963–9. Steinbrecher O, Mitrovic M, Eischer S, Sinkovec H, Eichinger S, Kyrle PA. Clinical and laboratory characteristics of cyclic thrombocytopenia: an observational study. Haematologica 2020; 105: e198-e200. Bick RL, Strauss JF, Frenkel EP. Thrombosis and hemorrhage in oncology patients. Hematol Oncol Clin North Am 1996; 10: 875–907. Kaushansky K. The thrombocytopenia of cancer. Hematol Oncol Clin North Am 1996; 10: 431–55. Crowther MA, Burrows RF, Ginsberg J, Kelton JG. Thrombocytopenia in pregnancy: diagnosis, pathogenesis and management. Blood Reviews 1996; 10: 8–16. Druzin ML, Stier ES. Maternal platelet count at delivery in patients with idiopathic thrombocytopenic purpura, not related to perioperative complications. J Am Coll Surg 1994; 179: 264–6. McCrae KR, Samuels P, Schreiber AD. Pregnancy associated thrombocytopenia: pathogenesis and management. Blood 1992; 80: 2697–714. Mandelbrot L, Schlienger I, Bongain A, et al. Thrombocytopenia in pregnant women infected with human immunodeficiency virus: maternal and neonatal outcome. Am J Obstet Gynecol 1994; 171: 252–7. Hermann W, Risch L, Grebhardt C, Nydegger UE, Sakem B, Imperiali M, et al. Reference intervals for platelet indices in seniors and frequency of abnormal results in a population-based setting: a comparison between directly and indirectly estimated reference intervals. J Lab Med 2021; 45 (2): 125–9. Dou M, Cao Pj. Evaluation of platelet count by different methods in hyperlipidemia specimens. Clin Lab 2022; 68 (4): 755–62. Liapis K, Papadakis S. Pseudothrombocytosis. N Engl J Med 2023; 388: 746. van Baarle FLF, van de Weerdt EK, van der Velden WJFM, Ruiterkamp RA, Tuinman PR, Ypma PF, et al. Platelet transfusion before CVC placement in patients with thrombocytopenia. N Engl J Med 2023; 388 (21): 1956–65. 15.12 Leukocyte count Lothar Thomas Leukocytes are produced in the bone marrow and the lymphoid organs, they use the blood stream as their route of transport and exert their effects in the tissues. A distinction is made between the following forms: Neutrophils: the cells are divided into two pools of approximately the same size in the vascular system. Only the cells belonging to the circulating pool are captured with blood sampling, not those of the marginated pool, in which the cells adhere loosely to the vascular intima. There is a continuous exchange of cells between the circulating and the marginated pools. The lifespan of the neutrophils in the blood is 21 hours and in the tissues it is 4–5 days. Neutrophils are phagocytic cells, and they are in the front lines of the defense against infectious pathogens. See Section 19.7 – Polymorphonuclear neutrophil function). Eosinophils: the synthesis of their granular proteins begins with their transformation from eosinophilic myeloblasts to promyelocytes. Interleukin-5 is an important survival factor in the maturation of eosinophils and prevents apoptosis. Eosinophils react to immunological stimuli and are important effector cells in allergic and parasitic inflammatory events. They circulate in the blood for only a few hours before passing into the tissues. Basophils: like eosinophils, basophils differentiate from agranular progenitor cells, and leave the bone marrow as mature cells. In the tissues, they reside in the small blood vessels and the post-capillary venules, and following a stimulus they migrate into the interstitium. They play an important role in the early phase of an IgG-mediated allergic reaction. In contact allergies, basophils located in the skin release the content of their granules over a period of days. Monocytes: this cell population circulates in the blood with a transit time of 14 hours and, like the neutrophils, are distributed into two pools. Following their passage into the tissues, the cells can be activated and are transformed into metabolically active macrophages. The macrophage-supported immune defense ensues through the phagocytosis of inflammatory pathogens as well as the formation of pro-inflammatory cytokines and activation of the immune system. Lymphocytes: these mononuclear cells include B cells, T cells and natural killer (NK) cells. Morphologically, these cells are similar. Refer to Section 21 – Immune system. 15.12.1 Indication Determination of the leukocyte count, including the sub populations, is indicated in: Initial investigations within the framework of the automated blood count Suspicion of hematologic disease e.g., abnormal blood count, hemolysis, thrombosis, lymph node swelling, splenomegaly Inflammation, infection, tissue necrosis, toxic disorders of the hematopoietic system (medication, radiation) Fever, shock, breathing difficulties, abdominal pain, urogenital symptoms, headache, disturbance of consciousness Frequent bacterial infections Allergic disease and helminthiasis Assessment of the progression and treatment of the symptoms mentioned. 15.12.2 Method of determination Automated hematology analyzer with a broad number of parameters and a high throughput are used in clinical routines. 15.12.2.1 Manual counting method In a micro collection device 1 part blood and 10 parts dilution solution (acetic acid 1–3% or Türk’s solution) are mixed. Thereby, the erythrocytes are hemolyzed and the leukocyte nuclei become more refractive. The counting chamber (e.g., the Neubauer chamber) is filled with diluted sample, and 4 corner squares (4 mm2) are counted. Leukocyte count (109/L) = sum of all counted cells × 25 × 106. The coefficient of variation of the hemocytometer count is around 15% with low leukocyte counts, and improves to 6.5% with normal or increased numbers. The counting chamber method continues to be indispensable for verifying the accuracy of determinations with hematology analyzers. 15.12.2.2 Hematology analyzer The counting and differentiation of the leukocytes is performed optically, with a combination of impedance and optical methods, or of optical methods with a cytochemical reaction. Beforehand, however, the erythrocytes are destroyed with the use of a detergent /1, 2/. Volume conductivity light scattering technology The cell counting and three part differential is based on the principle of cell counting and sizing by detection and measurement of changes in electrical resistance as cells pass through an aperture between two electrodes suspended in a conductive diluent. If a leukocyte passes through the aperture, a momentary rise in impedance, in the form of an electric pulse, is recorded. The pulse amplitude is proportional to the cell volume. Pulses from cells with a volume of greater than 35 fL are recorded. The different leukocytes are distinguished electronically by the pulses they generate. Using a special reagent that results in different cell shrinkage of leukocytes, these cells are differentiated and expressed on a histogram (Fig. 15.12-1 – Hematogram of leukocyte differentiation using volume, conductivity and scatter technology). Flow cytometry cytochemical technology These systems perform the differentiation of leukocytes in two separate channels, which are termed the peroxidase and basophil/lobularity channels. The differentiation of the leukocytes, based upon their size and myeloperoxidase (MPO) content, is accomplished in the peroxidase channel. In the leukogram, the leukocytes are represented in a system of coordinates as clouds, whereas the MPO and the volume are plotted on the x and y axes, respectively Large leukocytes with high MPO content, such as neutrophils, are shown in the upper right part of the leukogram, while small MPO-free cells such as lymphocytes are found in the lower left section. The differentiation of the basophils is accomplished following the removal of the cytoplasm with acidic buffer, and the counting of the nuclei (nucleogram) in the basophil/lobularity channel. Refer to Fig. 15.12-2 – Hematogram of leukocyte differentiation with combined flow cytometry and cytochemistry methodology. 15.12.2.3 Blood smear See Section 15.13 for cell differentiation. 15.12.3 Specimen EDTA blood: 1 mL Hemocytometer count method, capillary blood: 0.1 mL 15.12.4 Reference interval Refer to references /3, 4, 5, 6, 7, 8/ and Tab. 15.12-1 – Leukocyte reference intervals. Reference intervals of lymphocyte subpopulations refer to Ref. /73/. 15.12.5 Clinical significance Stepwise diagnostics in hematology is important in patients with an abnormal white blood cell (WBC) count. 15.12.5.1 Classifying white blood cell disorders In patients with abnormal WBC count, review the CBC with differential and examine the peripheral blood smear e.g., atypical cells, neoplastic cells, parasites. Characterize the WBC count as normal, high or low according to the leukocyte differentials expressed in absolute numbers. White blood cell count Leukocyte counts of (4–10) × 109/L within the framework of a screening evaluation, are considered to be definitely normal; values of (2.5–4) × 109/L are borderline, and those below 2.5 × 109/L are classified as definitely pathological. Smokers may have values as high as 12 × 109/L, and in heavy smokers these may reach 15 × 109/L. In a study, non-smokers had mean values of 6.1 × 109/L, while those of smokers were 10% higher /7/. Changes in the WBC count are primarily due to a change in the number of polymorphonuclear neutrophils (PMN) or of lymphocytes. The relative PMN fraction in healthy individuals is 40–75% of the leukocyte count. Infections are the primary cause of leukocytosis. Thus, a typical acute infection is characterized by the following processes: neutrophilic combat phase, monocytic recovery phase, and eosinophilic healing phase. In chronic infection each of the 3 phases can be present, depending upon whether the acute (neutrophilia), subacute or remittent (monocytosis) or chronic (lymphocytosis) phase persists. Viral infections and certain bacterial infections, such as enteric fever, generally do not follow the depicted course of events. A low WBC count is often drug-induced, caused by bone marrow disease (e.g., neoplasia and leukemia), a hereditary disorder of development, immunosupressive diseases or sepsis. Peripheral blood smear When the WBC count is high the microscopic differential may evaluate: Granulocytosis, lymphocytosis, monocytosis? Left shift and toxic granulation? Eosinophilia, basophilia? Atypic cells, progenitors, blasts? Hairy cells? Red cell morphology? Nucleated red blood cells? Useful further information Initial questions to be answered are: Any clinical information? Infection (antibodies to Epstein-Barr virus)? Inflammation (C-reactive protein)? Thrombocytes? Hemoglobin level? LD, urea, uric acid (increased cell turnover) Haptoglobin (hemolysis)? 15.12.5.2 Neutrophile granulocytosis Polymorphic nuclear cells (PMN) The PMN progenitor cells in the bone marrow are differentiated into those that are capable or incapable of cell division. Cells that are capable of cell division are found in the mitotic pool, while those that are no longer capable of cell division undergo their post mitotic maturation in the storage pool. Myeloblasts, promyelocytes and myelocytes belong to the mitotic pool, while metamyelocytes, bands and mature granulocytes are part of the post mitotic pool. The newly formed and mature cells dwell in the pools for some 10 days, before they are released into the blood stream. The storage pool contains 15–20 times as many granulocytes as the peripheral blood /10/. The kinetics of the PMN is shown in Tab. 15.12-2 – Kinetics of the neutrophil granulocytes. During the first hours of life, the newborn has marked neutrophilia with peak values after 12 hours of over 10 × 109/L; as of the third day there is a continuous fall, with stable granulocyte values in the range of (2.0–7.0) × 109/L. There is a distinct left shift in comparison with children of 2 years of age. Healthy children with very low birth weight and without perinatal or neonatal complications manifest considerable variation in their leukocyte counts, and some 95% are rated as neutropenic according to the Manroe criteria (Fig. 15.12–3 – Neutrophil count in the newborn) /4/. Normally, only PMN are released into the blood. On the other hand, the bone marrow contains more banded than PMN. If the requirement for PMN cannot be met by the bone marrow, more banded neutrophils and metamyelocytes will be continuously released. Thus, if the fraction of banded neutrophils and, perhaps, that of the metamyelocytes as well, increases in the blood, that is a sign of augmented release of granulocytes from the marrow and a reduction of the storage pool. This course of events is also termed a left shift. The PMN determination in peripheral blood provides only limited information on the neutrophil mass, because the circulation contains only 5–10% of the body’s neutrophil pool, and only 2% of the PMN neutrophil life cycle is captured. Three processes can cause neutrophile granulocytosis /11/: A shift in the PMN from the vascular marginal pool into the circulating pool. This is the case with hard labor, psychological stress, catecholamine release and the administration of noradrenaline. A rise of maximally 2-fold above the initial value occurs. This course of events is also termed pseudo neutrophilia, since a true increase in the blood neutrophil count does not occur. The enhanced release of PMN from the storage pool. This form of leukocytosis (e.g., in response to endotoxin) lasts for only some minutes to a few hours. Increased granulopoiesis. This takes place in the maturation pool and, following stimulation of the bone marrow, at least 2–3 days are required before proliferation of the cells in the storage pool occurs. With proliferative stress (e.g., infection) granulopoiesis can be increased by a factor of 20. In this case, the PMN maturation time is shortened from 10 to 2 days. In conditions of high peripheral requirements and an empty storage pool, cells of the mitotic pool such as myelocytes and promyelocytes are also released directly into the peripheral blood (leukemoid reaction). If granulopoiesis does not respond sufficiently to the requirements of the tissues (e.g., with systemic infection) neutropenia may result; this occurs in sepsis, in some 20% of the cases. Fc receptor antigen CD64 Neutrophils express the Fc receptor antigen CD64. The antigen is expressed by myeloid stem cells, and is maintained up to the metamyelocyte stage. PMN and banded neutrophils have only approximately 1000 of these antigens on their cell membrane. Newborn and premature infants express a higher proportion of the antigen. Upon activation, PMN increase the number of antigens by 5–10-fold, and the measurement of neutrophil CD64 thereby permits a distinction between healthy and inflammation /12/. In systemic inflammation, as in cases of infection and sepsis, pro-inflammatory cytokines like IFN-γ , IL-1 and IL-6, as well as GCSF, activate the up-regulation of CD64 on the granulocytes. In the diagnostic investigation of a systemic infection, CD64 is believed to have a diagnostic sensitivity of 90% with a specificity of 90–100%; this in comparison with the leukocyte count which has a sensitivity of 60% with a specificity of 51% / /13/. Neutrophilia In school children and adults, neutrophilia is present if the number of PMN and their precursors is greater than 7.5 × 109/L. Elevated granulocytes are an indication of: Inflammation due to acute inflammation like infection and acute tissue necrosis. In chronic infection neutrophilia may be present due to enhanced granulopoiesis. The peripheral requirements for granulocytes is increased, but there is overcompensation of granulopoiesis with the development of an enlarged storage pool. Myeloid leukemia The ingestion of glucocorticoids. These lead to neutrophilia due to an increase in granulopoiesis, augmented migration from the marginal to the circulating vascular pool, and reduced migration of the granulocytes from the blood stream. The Monroe and the Rodwell criteria are utilized for the diagnostic investigation of neonatal sepsis (Tab. 15.12-3 – Criteria of neonatal sepsis). Diseases and conditions with neutrophilic granulocytosis are listed in Tab. 15.12-4 – Diseases and conditions with neutrophilia. 15.12.5.3 Neutropenia Neutropenia is defined as a decrease of the PMN including the band forms. The number is determined with a hematology analyzer, and if a decrease of below 0.5 × 109/L is indicated, a blood smear is counted as a control and the percentage of PMN and band forms is multiplied with the leukocyte number. Normally, in healthy adults and children over 5 years of age, the number of PMN is greater than 1.5 × 109/L. In persons of African origin and in other ethnic groups (e. g,. in Israel and Jordan) values as low as 1.0 × 109/L are also normal. These individuals also have a low leukocyte count. Thus, only 25.2% of male Caucasians aged 3–74 years have a leukocyte count of below 5.0 × 109/L, while for blacks the corresponding figure is 48.1%. In females, the corresponding figures are 27.1% and 42.6% /15/. Neutropenia is classified as /16/: Mild; the neutrophil count is 1.0–1.5 × 109/L Moderate; the neutrophil count is 0.5–1.0 × 109/L Severe; the neutrophil count is below 0.5 × 109/L. This classification predicts the risk of a pyogenic infection, if the neutropenia has been present for more than 2–3 months. Only patients with severe neutropenia are at risk for bacterial infections /10/. The infection is usually due to the patient’s own flora. The most common symptoms are gingivitis, ulcerations and oral thrush. Febrile neutropenia is present if /17/: The oral temperature, with a single measurement, is ≥ 38.3 °C, or ≥ 38 °C for at least 1 hour and The neutrophil count is < 0.5 × 109/L or < 1.0 × 109/L with a tendency to decline to ≤ 0.5 × 109/L. In severe neutropenia following chemotherapy for solid tumors, treatment with hematopoietic growth factors (G-CSF, GM-CSF) is indicated if the neutropenia has been present for longer than 10 days or if there is fever of > 38.1 °C. Causes of neutropenia The most frequent cause of neutropenia is a toxic disorder of granulopoiesis due to medication (Tab. 15.12-5 – Drugs that can cause neutropenia). Congenital neutropenia (Tab. 15.12-6 – Congenital neutropenia) is usually severe in nature. The causes of secondary neutropenia are diverse (Tab. 15.12-7 – Secondary neutropenia) are diverse. Infections with neutropenia are shown in Tab. 15.12-8 – Acute febrile infections with leukopenia and/or thrombocytopenia. 15.12.5.4 Lymphocytosis Lymphocytes Lymphocytes are formed in the bone marrow and in the secondary lymph organs such as the spleen and the lymph nodes. The lymphocytes of the blood represent only some 2% of the lymphocytes of the organism. These are found in the spleen, the lymph nodes and the organ-associated lymphatic system. In contrast to the granulocytes, the lymphocytes enter continuously the circulation and migrate in the fluid compartments before they return to the lymph nodes; this process is known as homing /18/. The lymphocyte count in the blood is constant only 2% of the lymphocytes circulate in the blood, and in any case for less than 1 hour /10/. Most of the lymphocytes in the blood are small, intermitotic cells in the resting stage. A small proportion of the cells is medium-sized, apparently originating from the small lymphocytes, and in an activated state. In addition, large lymphocytes with coarse eosinophilic granulation (large granular lymphocytes, LGL) occur during infection. Lymphocytes are immune cells which express, due to immunophenotyping, specific identifiable features on their surface. Based upon their functional activity in immune defense, and their surface features (CD classification), three classes of lymphocytes are distinguished: T cells that are released from the thymus gland, and are primarily responsible for cell-mediated immunity B cells, the origin of which is the bone marrow and the secondary lymph organs. They are the precursors of the immunoglobulin-forming plasma cells, and are responsible for the humoral immune response. Natural killer (NK) cells. They express no surface lymphocyte features, and are active within the framework of the non-specific immune defenses. The NK cells are also distinguishable morphologically since, in the blood smear, some of them are visible LGL’s. In the peripheral blood, 65–80% of the lymphocytes are T cells, 8–15% are B cells and 10% are NK cells. NK cells do not recirculate, like the other lymphocytes, from the blood to the lymph nodes and back. The lymphocyte count is subject to influencing factors to a considerable degree; it is higher in the late afternoon and in the evening than during the morning. Lymphocytosis caused by increased lymph flow occurs after brief physical exertion. Following longer and more marked exertion, and in cases of systemic infection such as sepsis, lymphopenia and eosinopenia develop along with a rise in PMN. Large granular lymphocytes (LGL) In response to pathogens that attack the organism, two different cell types react: cytotoxic T lymphocyte and natural killer (NK) cells /19/. These are morphologically identical, and they appear microscopically as large lymphocytes with azurophilic cytoplasmic granules. They are, however, functionally different, and they recognize antigens via T cell receptors and NK cell receptors. In healthy individuals, only a small fraction of the circulating lymphocytes are LGL’s. The majority of the LGL’s are cytotoxic T cells of phenotype CD3+ CD8+ CD4–, while the few NK cells are of phenotype CD3– CD16+. The polyclonal proliferation of LGL’s occurs in a reactive and transient manner within the framework of viral infection (EBV, Cytomegalo virus), autoimmune disease, malignancies and also, in some cases, splenectomy. The clonal proliferation of LGL’s is maintained over the long term, independently of whether or not the patient is symptomatic. The proliferation can affect T cells and NK cells, in spite of the fact that the WHO has designated NK proliferation as a separate disease, defined as chronic lymphoproliferative disease of the NK cells. NK cell lymphoproliferative disease is poorly understood. T cell LGL leukemia is a clonal proliferation of end-differentiated cytotoxic T cells, with a functional α/β+T cell receptor and a pattern CD8+ CD4– or, occasionally CD4+CD8–/+dim. Lymphocytosis In adults, lymphocytosis is present with a cell count of > 4.0 × 109/L. In children, the reference interval is age-dependent (Fig. 15.12-4 – Age-dependency of the lymphocyte count in children). Thus, the lower reference interval value at the age of 8 months is 4.5 × 109/L, and at the age of 18 years it is 1.0 × 109/L /20/. Disorders in association with lymphocytosis are: Viral infections such as infectious mononucleosis. Many of the lymphocytes are transformed by the Epstein-Barr virus and manifest a colorful image in the blood smear. The virus infects only B cells, not T cells. The lymphocyte count is (6–15) × 109/L. A similar picture, with transformed lymphocytes, is also seen with Cytomegalo virus and viral hepatitis. In these infections the lymphocyte count is only slightly elevated, if at all. Infections like toxoplasmosis, enteric fever, brucellosis or pertussis. In pertussis, the lymphocyte count increases over 20 × 109/L, while the lymphocytes are small cells of normal appearance /10/. Neoplasias of the lymphocytic system, such as in acute and chronic lymphatic leukemia and, occasionally, in non-Hodgkin lymphoma. Indicators of a lymphoproliferative disease are: If lymphocyte count exceeds 5 × 109/L If accompanied by > 5% smudge cells Atypic cells e.g., pro lymphocytes, mantle cells, Sezary cells Clinical signs of lymphoproliferation (lymphadenopathy, splenomegaly, B symptoms) Further procedure: immunophentyping of peripheral blood, nodal histopathology. 15.12.5.5 Lymphopenia Depending on the literature source, lymphopenia in adults is defined as a cell count of below 1.5 × 109/L, or below 1.0 × 109/L /20/. In children, the lower threshold is age-dependent (Fig. 15.12–4 – Age-dependency of the lymphocyte count in children). Diseases and conditions with lymphopenia are shown in Tab. 15.12–9 – Diseases and conditions with lymphopenia. Small children of less than 3 months of age with no previous underlying disease, who are admitted to hospital with acute symptoms on an emergency basis, often require life-saving measures if lymphopenia is present. Thus, out of 42 lymphopenic children, 26 had to be admitted to the pediatric intensive care unit; by way of comparison, this was only necessary for 1 out or 42 non-lymphopenic children /22/. 15.12.5.6 Monocytosis Monocytes Monocytes emerge from the bone marrow and are part of the inflammatory response and immune response to infection. Monocytes share a common stem cell with the granulocytes. During maturation they go their own separate ways, but in the bone marrow the morphology of the pro monocytes and the promyelocytes is almost identical, and their small population is integrated into that of the neutrophil progenitor cells during counting. They can only be differentiated with the use of esterase staining. The monocytes are released into the blood following two to three cell divisions. No storage pool analogous to that of the neutrophils exists. Following a transit time of 14 hours they leave the blood and migrate into the tissues. They mature in the tissues, and the nature of the maturation (i.e., their enzyme pattern) is a function of the tissue where the maturation process takes place. Thus, as alveolar macrophages of the lungs, they have a different protein make-up than the Kupffer stellate cell of the liver or the peritoneal macrophages. Fusion of macrophages can occur in the tissues, with the formation of large cells such as Langhans giant cells, which are found in granulomatous inflammation such as that occurring in tuberculosis. Like lymphocytes, macrophages can also divide in situations of increased need /10/. Important macrophage tasks are phagocytosis and the killing of microbes. The internalized pathogens are processed, and are presented to helper T cells in association with an HLA class II antigen (see Section 21.1.2 – Immune recognition). In inflammation, extravasation of the PMN initially precedes that of the monocytes at the affected site. Thus, the PMN granules release proteins that stimulate the expression of vascular β-integrin and formyl peptide receptors, to which activated monocytes bind. At the site of the inflammation, the PMN that are subject to apoptosis release lysophosphatidylcholine, which binds to the monocyte G2A receptors and attracts them to the affected site /23/. Monocytosis In adults and school children, an increase in the monocyte count to over 0.9 × 109/L is termed monocytosis. Newborns and small children have a higher upper reference interval value. Monocytes are found in infection, autoimmune disease, systemic hematological disease, with solid tumors and for various other reasons. In infectious diseases monocyte distribution width is increased. Diseases and conditions with monocytosis are shown in Tab. 15.12-10 – Diseases and conditions with monocytosis. Signs of neoplastic monocytosis: Immature Monocytes (promonocytes) and/or blasts Thrombocytopenia and/or anemia Organ infiltration (skin, lung, spleen) No conclusive medical explanation for monocytosis Bone marrow diagnostics (cytology, histopathology, cytogenetics) are important investigations. 15.12.5.7 Eosinophilia Eosinophils The eosinophil granulocyte arises from a pluripotent stem cell as a cell of the myeloid series, which differentiates under the influence of hematopoietic growth factors. In individuals with healthy bone marrow, the eosinophil fraction of the white blood cells is some 3%, of which one third are promyelocytes and myelocytes, 26% are metamyelocytes, and one third are mature eosinophils that dwell in the storage pool. The polyclonal synthesis of eosinophils takes place following stimulation with GM-CSF, IL-3 and IL-5. The mature eosinophil enters the blood after approximately 4 days, where it circulates for 6–18 hours before migrating into the tissues. The eosinophil stays for 2–5 days in the mucosal membranes of the respiratory and gastrointestinal tracts, as well as in the skin, before undergoing apoptosis: unless this is delayed by GM-CSF, IL-3 and IL-5. The daily turnover of the eosinophils is 2.2 × 108 cells/kg BW; the post mitotic storage pool contains (9–14) × 108 cells/kg BW /24/. Eosinophilia Eosinophilia in the blood entails a 100-fold more marked eosinophilia in the tissues. The tissue concentration can be high even if the eosinophil count in the blood is low. The blood eosinophil count manifests diurnal fluctuations. Highest values are measured during the evening, and the lowest values are found in the morning. Eosinophilia is present with a cell count of > 0,5 × 109/L. Eosinophilia is classified as: Mild, with up to 1.5 × 109/L Severe, with > 1,5 × 109/L. If eosinophilia is present, the following should be considered /25/: A reactive event such as an allergic/atopic disease, urticaria, parasitic infection, malignant disease or vascular collagen disease If the afore mentioned causes are ruled out and the hyper eosinophilia persists, the diagnosis of idiopathic hyper eosinophilic syndrome is likely, in particular if the hyper eosinophilia has been present for longer than 6 months and is above 1.5 × 109/L. The accumulation of eosinophils in the blood and the peripheral tissues can be due to /26/: A disorder of the myeloid cell lineage (primary eosinophilia). This can occur late in the eosinophil differentiation process, and can lead to the rare diagnosis of eosinophilic leukemia. If the disorder occurs early in the differentiation process, and if the eosinophils are a component of the malignant clone that also forms other myeloid cells, or even lymphoid cells, then the eosinophilia is an event within the framework of a myeloproliferative disease. The increased formation of eosinophil-stimulating cytokines by non-myeloid cells (secondary eosinophilia). The result is the polyclonal formation of eosinophils. Stimulated by inflammatory mediators like IL-5, eotaxin, platelet-activating factor, C5a and C3a, eosinophils migrate to the site of inflammation in the tissue and release from their granule proteins that are destructive to tissue, such as eosinophil cationic protein (ECP) and reactive oxygen radicals /27/. For the differential diagnosis of eosinophilia see Tab. 15.12-11 – Differential diagnosis of eosinophilia. 15.12.5.8 Basophilia Basophils Basophils make up some 0.3% of the circulating leukocytes, and are related functionally to the mast cells /28, 29/. Both cell types express the functionally active IgG receptor, and form the same effector molecules, such as histamine, lipid mediators, (leukotrienes, prostaglandins), serine proteases and interleukines (IL-4, IL-13, IL-6). Basophils leave the bone marrow as mature cells, and are characterized by the expression of FcεRI, CD49b, and the high affinity IL-3 receptor CD123. Their concentration in the bone marrow, the liver, the spleen, and the peripheral blood is low under basal conditions. Within the framework of the specific immune response, and in types of inflammation, they pass into the peripheral tissues and the lymph nodes. Mast cells accomplish their maturation in the peripheral tissues such as the skin, the small intestine and the peritoneal cavity. They are characterized by the expression of FcεRI and c-Kit. Under receptor-mediated stimulation (FcεRi), basophils and mast cells release effector molecules. Basophils accelerate the TH2 immune response, since they release IL-4 within minutes of activation. The activation occurs through the cross-linking of two surface-bound IgE molecules, or by means of a large number of substances in an antigen-independent manner. Basophilia Elevated basophils are commonly associated with immediate type hypersensitivity reactions. Total IgE is often also increased. There is, however, no correlation between the rise in IgE and the basophil count. Little increase (> 2–3%) may be indicative for a myoproliferative neoplasia. Basophilia can occur in: Allergic inflammation such as hypersensitivity to medication or food, erythroderma, urticaria, and rheumatoid arthritis Parasitic infection Stem cell disease like myeloid leukemia, myeloproliferative syndrome, M. Waldenstroem Diabetes mellitus, myxedema, estrogen-containing medication Infectious disease such as tuberculosis, chicken pox, influenza Post-splenectomy syndrome. 15.12.6 Comments and problems Method of determination The lysis of the erythrocytes is time-critical. If the lysis time is too short, or if relatively lysis-resistant red blood cells are present (e.g., reticulocytes or erythrocytes of newborns) then the leukocyte count will be overestimated. If the lysis time is too long, or if the leukocytes have previously been damaged (e.g., in chemotherapy or sepsis) because their volume is smaller than normal and they will not be counted. Giant thrombocytes may be counted along with the leukocytes. For the determination of the leukocyte count in the blood smear, see Section 15.13. Blood sampling Reliable values are obtained with EDTA blood; capillary blood sampling should only be performed as an exception. With capillary blood sampling, the first two drops of blood following puncture are to be discarded. In newborns and small children, the leukocyte count depends upon the blood sampling. Compared with capillary blood sampling from the heel, venous and arterial sampling yield only 82 ± 3,5% and 77 ± 5,3% of the leukocyte count, respectively. Blood sampling following violent screaming leads to a rise in capillary-sampled leukocytes to 146 ± 6,1% of the resting value /30/. IgM-mediated neutrophil agglutination A case that caused pseudo neutropenia has been described. The neutrophils were loaded with IgM antibodies /31/. Cryoglobulins At room temperature and with counting by means of hematology analyzers, cryoglobulins lead to pseudo leukocytosis. Within this temperature range protein crystals, the size of leukocytes, are formed, and these are captured with automated counted as leukocytes. The protein crystals disappear if the sample is warmed to 37 °C. Intraindividual variation of leukocyte counts The CV within 1 day is 19.9%, the day-to-day CV is 16.3%, and the month-to-month CV is 17.3% /32/. Diurnal variation of leukocyte counts The diurnal variation shows lower values during the morning and higher values in the late afternoon and at night; at 12 o’clock at night 9.5 × 109/L and 7 × 109/L between 8.00 and 11.00 o'clock /72/. Pseudoleukocytosis Pseudoleukocytosis may occur under several conditions: presence of nucleated red blood cells (NRC), cryoglobulins, cryofibrinogen, platelet clumps, microorganisms, lysis-resistant red blood cells, lipids. NRC are especially common in hemolytic anemia, massive blood loss, severe oxygen deficiency, acute and chronic leukemias, malignant tumor, osteomyelofibrosis, and invasive candidiasis /76/. Stability In EDTA blood at room temperature and refrigerated within 72 hours, dependent upon the hematology analyzer /33/: Total leukocytes: decrease of 0.6–5.1% Neutrophils: increase of 1.3–10.3%, with one analyzer, a decrease of 1.2% is reported Monocytes: with one analyzer, an increase of up to 31%, with another, a decrease of 28–78% Lymphocytes: on one analyzer at room temperature an increase of 5%, on another, a decrease of 3–14%. References Gulati GL, Bong HH. The automated CBC. Hematol Oncol Clin North Am 1994; 8: 593–603. Brugnara C, Kratz A. Automated hematology analyzers: State of the art. Clinics Review articles 2015; 35: No1. de Waele M, Foulon W, Renmans W, Segers E, Smet L, Jochmans K, van Camp B. Hematologic values and lymphocyte subsets in fetal blood. Am J Clin Pathol 1988; 89: 742–6. Geaghan SM. Hematologic values and appearances in the healthy fetus, neonate and child. Clin Lab Med 1999; 19: 1–37. Soldin SJ, Brugnara C, Hicks JM. Pediatric reference ranges. Washington; AACC Press 1999: 189–232. Taylor MRH, Holland CV, Spencer R, Jackson JF, O’Conners GI, O’Donnell JRO. Haematological reference ranges for school children. Clin Lab Haem 1997; 19: 1–15. Nebe T, Bentzien F, Bruegel M, Fiedler GM, Gutensohn K, Heimpel H, et al. Multizentrische Ermittlung von Referenzbereichen für Parameter des maschinellen Blutbildes. J Lab Med 2011; 35: 3–28. van den Bossche J, Devreese K, Malfait R, van de Vyvere M, Wauters A, Neels H, de Schouwer P. Reference intervals for a complete blood count determined on different automated hematology analysers: Abx Pentra 120 Retic, Coulter Gen-S, Sysmex SE 9500; Abbott Cell Dyn 4000 and Bayer Advia 120. Clin Chem Lab Med 2002; 40: 69–73. Manroe BL, Rosenfeld CR, Weinberg AG, Browne R. The differential leukocyte count in the assessment and outcome of early-onset neonatal group B streptococcal disease. J Pediat 1977; 91: 632–7. Boggs DR, Winkelstein A. White cell manual. Philadelphia; Davis Co 1983: 29–60. Jagels MA, Hugli TE. Mechanisms and mediators of neutrophilic leucocytosis. Immunopharmacology 1994; 28: 1–8. Hoffmann JJML. Neutrophil CD64: a diagnostic marker for infection and sepsis. Clin Chem Lab Med 2009; 47. 903–16. Davis BH, Olsen SH, Ahmad E, Bigelow NC. Neutrophil CD64 is an improved indicator of infection or sepsis in emergency department patients. Arch Pathol Lab Med 2006; 130: 654–61. Rodwell RL, Leslie AL, Tudehope DI. Early diagnosis of neonatal sepsis using a hematologic scoring system. J Pediat 1988; 112: 861–7. Haddy TB, Rana SR, Castro O. Benign ethnic neutropenia: What is a normal absolute neutrophil count? J Lab Clin Med 1999; 133: 15–22. Pizzo PA. Management of fever in patients with cancer and treatment-induced neutropenia. N Engl J Med 1993; 328: 1323–32. Hughes WT, Armstrong D, Bodey GP, et al. 1997 Guidelines for the use of antimicrobial agents in neutropenic patients with unexplained fever. Clin Infect Dis 1997; 25: 551–73. Westermann J, Pabst R. Distribution of lymphocyte subsets and natural killer cells in the human body. Clin Investig 1992; 70: 539–44. Zambello R, Semenzato G. Large granular lymphocyte disorders: new etiopathogenetic clues as a rationale for innovative therapeutic approaches. Haematologica 2009; 94: 1341–5. Schoentag RA, Cangiarella J. The nuances of lymphocytopenia. Clin Lab Med 1993; 13: 923–36. Altman PL, Dittmer DS, ed. Blood and other body fluids. Washington; Federation of American Societies for Experimental Biology, 1961: 125. Adamski JK, Arkwright PD, Will AM, Patel L. Transient lymphopenia in acutely unwell young infants. Arch Dis Child 2002; 86: 200–1. Soehnlein O, Lindbom L, Weber C. Mechanisms underlaying neutrophil-mediated monocyte recruitment. Blood 2009; 114: 4613–23. Giembycz MA, Lindsay MA. Pharmacology of the eosinophil. Pharmacological Reviews 1999; 51: 213–339. Gotlib J. Molecular classification and pathogenesis of eosinophilic disorders: 2005 update. Acta Haematol 2005; 114: 7–25. Roufosse F, Cogan E, Goldman M. Recent advances in pathogenesis and management of hypereosinophilic syndromes. Allergy 2004; 59 673–89. Staumont-Salle, Dombrowicz D, Capron M, Delaporte E. Eosinophils and urticaria. Crit Rev in Allergy & Immunology 2006; 30: 13–8. Ohnmacht C, Voehringer D. Basophil effector function and homeostasis during helminth infection. Blood 2009; 113: 2816–25. Mack M, Rosenkranz AR. Basophils and mast cells in renal injury. Kidney Int 2009; 76: 1142–7. Christensen RDD, Rothstein G. Pitfalls in the interpretation of leucocyte counts of newborn infants. Am J Clin Pathol 1979; 72: 608–11. Carr ME, Whitehead J, Carlson P, et al. Case report: immunoglobulin M-mediated temperature-dependent neutrophil agglutination as a cause of pseudoneutropenia. Am J Med Sci 1996; 311: 92–5. Costongs GMPJ, Janson PCW, Bas BM. Short-term and long-term intra-individual variations and critical differences of haematological laboratory parameters. J Clin Chem Clin Biochem 1985; 23: 69–76. Imeri F, Herklotz R, Risch L, Arbetsleitner C, Zerlauth M, Risch GM, et al. Stability of hematological analytes depends on the hematology analyser used: a stability study with Bayer Advia 120, Beckman Coulter LH 750 and Sysmex XE 2100. Clin Chim Acta 2008; 397: 68–71. Weimann A, Lun A, Lun S, Zimmermann M, Borges AC, Ziebig R, et al. Leukocyte, neutrophil, immature granulocyte counts and interleukin-6 are superior to procalcitonin, C-reactive protein and delta He for detection of mild inflammation: data from marathon runners producing mild systemic inflammation visible immediately after the run. J Lab Med 2010; 34: 53–9. Rochlitz CF, Herrmann R. Rauchen und Leukozytose. Dtsch Med Wschr 1990; 115: 756. Belo L, Santos-Silva A, Rocha S, Caslake M, Cooney J, Pereira-Leite L, Quintanilha A, Rebelo I. Eur J Obstet Gynecol Reprod Biol 2005; 123: 46–51. Masri-Iraqi H, Robenshtok E, Tzvetov G, Manistersky Y, Shimon I. Elevated white blood cell counts in Cushings disease: association with hypercortisolism. Pituitary 2013; OI 10.1007/s11102–013–05220 Madjid M, Awan I, Willerson JT, Casscells SW. Leucocyte count and coronary heart disease. J Am Coll Cardiol 2004; 44: 1945–56. Kunz E, Gunzer U, Haubitz I. Das Risiko der Splenektomie. Med Klin 1994; 89: 515–21. Dale DC, Link DC. The many causes of severe congenital neutropenia. N Engl J Med 2009; 360: 3–5. Boztug K, Appaswamy G, Ashikov A, Schäffer AA, Salzer U, Distelhorst J, et al. A syndrome with congenital neutropenia and mutations in G6PC3. N Engl J Med 2009; 360: 32–43. Dale DC, Bolyard AA, Aprikyan A. Cyclic neutropenia. Semin Hematol 2002; 39: 89–94. Kannourakis G. Glycogen storage disease. Semin Hematol 2002; 39: 103–6. Cham B, Bonilla MA, Winkelstein J. Neutropenia associated with primary immunodeficiency syndromes. Semin Hematol 2002; 39: 107–12. Juul SE, Calhoun DA, Christensen RD. Idiopathic neutropenia in very low birthweight infants. Acta Paediatr 1998; 87: 963–8. Boxer L, Dale DC. Neutropenia: Causes und consequences. Semin Hematol 2002; 39: 75–81. Palmblad JEW, von dem Borne AEG. Idiopathic, immune, infectious, and idiosyncratic neutropenias. Semin Hematol 2002; 39: 113–20. Bux J, Hofmann C, Kauth T, et al. Serological and clinical aspects of granulocyte antibodies leading to alloimune neonatal neutropenia. Transfus Med 1992; 2: 43–9. Conway LT, Clay ME, Kline WE, et al. Natural history of primary autoimmune neutropenia in infancy. Pediatrics 1987; 79: 728–33. Starkebaum G. Chronic neutropenia associated with autoimmune disease. Sem Hematol 2002; 39: 121–7. Nossent JC, Swaak AJ. Prevalence and significance of hematological abnormalities in patients with systemic lupus erythematodes. QJMed 1991; 80: 605–12. Curtis BR. Non-chemotherapy drug-induced neutropenia: key ponts to manage the challenges. Hematology 2017; 187–93. Kim S, Demetri GD. Chemotherapy and neutropenia. Hematol Oncol Clin North Am 1996; 10: 377–95. Schröder JO, Schwab UM, Schnabel A, et al. Häufung MTX-assoziierter Zytopenien bei älteren Patienten. Dtsch Ärztebl 1996; 93: B-257–8. Takata K, Kubota S, Fukata S, Kodo T, Nishiara E, Ito M, et al. Methamizole-induced agranulocytosis in patients with Grave’s disease is more frequent with an initial dose of 30 mg daily than with 15 mg daily. Thyroid 2009; 19: 559–63. Schattner A. Campylobacter jejuni and cytopenias. Am J Med 2013. doi: 10.1016/j.amjmed.2013.07.003. Friedman AD. Hematologic manifestation of viral infections. Pediatr Ann 1996; 25: 555–60. 58 Link H, Nietsch J, Kerkmann M, Ortner P. Adherence to granulocyte-colony stimulating factor (G-CSF) guidelines to reduce the incidence of febrile neutropenia after chemotherapy- a representative sample survey in Germany. Support Care Cancer 2016, 24: 367–76. De Jager CPC, van Wijk PTL, Mathorea RB, de Jongh-Leuvenik J, van der Poll T, Wever PC. Lymphocytopenia and neutrophil-lymphocyte count ratio predict bacteremia better than conventional infection markers in an emergency care unit. Critical Care 2010; 14: R192. Lichtman M. Monocyte and macrophage disorders: classification. In: Williams WJ, Beutler E, Erslev A, Lichtman MA (eds). Hematology. New York; McGraw-Hill 1990; 882–5. Germing U, Gattermann N. Mastering the multitude of monocytosis. Haematologica 2019. doi: 10.3324%2Fhaematol.2019.227546. Erhardt S, Burchard GD. Eosinophilie bei Tropenrückkehrern und Migranten. Dtsch Ärtzebl 2009; 105: 801–8. Reilly A, Becker J, Meyer J, Rackoff W. Hypereosinophilia. Med Pediatr Oncol 1992; 20: 23–9. Udwadia FE. Tropical eosinophilia: a review. Respiratory Medicine 1993; 87: 17–21. Wenzel SE. Eosinophils in asthma – closing the loop or opening the door. N Engl J Med 2009; 360: 1026–8. Bousquet J, Chanez P, Lacoste JY, et al. Eosinophilic inflammation in asthma. N Engl J Med 1990; 323: 1033–9. Cottin V, Cordier JF. Eosinophilic pneumonias. Allergy 2005; 60: 841–57. Roufosse F. Hypereosinophilic syndrome variants: diagnostic and therapeutic considerations. Haematologica 2009; 94: 188–93. Cantarini L, Volpi N, Carbotti P, Greco G, Agliano M, Bellisai F, et al. Eosinophilia-associated muscle disorders: an immunohistochemical study with tissue localisation of major basic protein in distinct clinicopthological forms. J Clin Pathol 2009; 62: 442–7. Stefanini M, Claustro JC, Motos RA, et al. Blood and bone marrow eosinophilia in maglinant tumors: role of blood and tissue eosinophil colony stimulating factors in two patients. Cancer 1991; 68: 543–8. Skokowa J. Circumventing mutation to nix neutropenia. N Engl J Med 2021; 384 (20): 1956-8. Torge A, Haeckel R, Özcürümez M, Krebs A, Junker R. Diurnal variation of leukocyte counts affects the indirect estimation of reference intervals. J Lab Med 2021; 45 (2): 121–4. Huenecke S, Behl M, Fadler C, Zimmermann SY, Bochennek K, Tramsen L, et al. Age Matched lymphocyte subpopulation reference values in childhood and adolescence: application of exponential regression analysis. Eur J Haematol 2008; 80: 532–9. Manieri E, Dondi A, Neri I, Lanari M. Drug rash with eosinophilia and systemic symptoms (DRESS) syndrome in childhood: a narrative review. Front Med 2023. doi: 10.3389/fmed.2023.1108345. Hung WK, Hung WH. Drug reaction with eosinophilia and systemic symptoms. N Engl J Med 2022; 387 (2): 167. Bensaoud H, Buttafuoco F, Lali S, Chevalier C. Early detection of peripheral candidiasis further to cytographic interferences in Sysmex XN-9000 hematology analyzer. Clin Chem Lab Med 2024: 62 (3): e78–e80. Xu X, Shen G. Eosinophils were significantly elevated in the chronic subdural hematoma fluids: Two case reports. Clin Lab 2024; 70: 2363–5. 15.13 Blood smear examination Lothar Thomas The blood smear examination is the cornerstone of the diagnosis of diseases of the hematopoietic system and of other organ or systemic disorders which compromise hematopoiesis. Despite automated hematology analyzers, flow cytometry, cytogenetics and molecular diagnostics. The efficacy of automated analyzers in generating reliable results and flagging the ones that need to be verified by additional means including a blood smear examination has been widely documented. Hematology analyzers are highly efficient because of the improved precision and turnaround time, but they can not eliminate the need for the blood smear examination in selected cases. The blood smear provides the investigator with insight into the structure of the blood cells, something that is not possible with automated counting and differentiation /1, 2, 3, 4/. Blood smear examinations serve two important objectives: It often provides the most useful information It serves as a quality control tool in verifying the results generated by automated hematology analyzers. The International Consensus Group for Hematology Review has elaborated 41 criteria for when an automated blood count should be supplemented with a blood smear. In a modification of the criteria, the blood smear fraction is, in many laboratories, 10% in ambulatory patients and 20% in in-patients /5/. 15.13.1 Indication Historically, leukocyte differentiation was perfomed microscopically. The most modern hematology laboratories use automated hematology analyzers. 15.13.1.1 Decision-oriented laboratory testing Decision-oriented laboratory testing using microscopic differentiation is still performed in some situations following automated CBC and WBC /5/. Red cell morphology Because many morphologic red cell shape changes cannot be recognized by hematology analyzers it is necessary to examine a blood smear in patients presenting with anemia /1/. Reasons for examining a blood smear even in the presence of a normal automated complete blood count include the following: Clinical information that the patient had a prior red cell abnormality and its current status is questioned A family history of a morphologic red cell abnormality Investigation whether the patient has a hypo splenic or post splenectomy blood picture Determination of the presence of parasites. Leukocyte morphology The evaluation of leukocytes consists of three parts /4/. Verification of the white blood cell count Differential leukocyte count Detection of morphologic abnormalities within a population of cells Thrombocyte morphology Thrombocyte evaluation consists of two parts /4/: Verification of the automated blood count Evaluation of the platelet morphology (verification of the automated mean platelet volume, identification of (e.g., aggregates, giant platelets their differentiation from erythrocyte fragments). None blood cells Other important cells are /4/: Endothelial cells, epithelial cells, tumor cells Organism; intracellular (plasmodium, babesia), extracellular (trypanosomas, micro filariae). 15.13.1.2 Clinical indication /2/ Anemia, unexplained jaundice, or both Sickle cell disease (dactylitis or sudden splenic enlargement and pallor in a young child or, in an older child or adult, limb, abdominal, or chest pain) Thrombocytopenia (e.g., petechiae or abnormal bruising) or neutropenia (e.g., unexpected or severe infection) Lymphoma or other lymphoproliferative disorder (lymphadenopathy, splenomegaly, enlargement of the thymus or other lymphoid organs, skin lesions suggestive of infiltration, bone pain, and systemic symptoms such as fever, sweating, itching, and weight loss) Myeloproliferative disease (splenomegaly, plethora, itching, or weight loss) Disseminated intravascular coagulation Acute or recent-onset of renal failure or unexplained renal enlargement, particularly in a child Retinal disorders, hemorrhages, exudates, signs of hyper viscosity, or optic atrophy Bacterial or parasitic disease that can be diagnosed from a blood smear Disseminated non hematopoietic cancer (weight loss, malaise, bone pain) General ill health, often with malaise and fever, suggesting infectious mononucleosis or other viral infection or inflammatory or malignant disease. 15.13.2 Sample preparation Anticoagulation K2or K3-EDTA or sodium citrate, but not heparin, are acceptable for venous or capillary blood sampling. Heparin is obsolete because of /3/: Developing thrombocyte clumps that interfere with the morphologic interpretation of the smear Heparin causes the development of a purple-blue hue on stained films. Sample processing Films should be prepared within 2 hours after blood collection. If this is not possible, the sample should be stored for less than 8 hours at 4 °C but storage at room temperature (22 °C) is also possible. Long-term storage should be at 4 °C. After prolonged storage blood samples should always be mixed by a minimum of 10 complete (180 degree) inversions /3/. Preparation of the film Approximately 5 μL of blood should be used to allow for a wedge blood film of appropriate thickness and of 2.5–4 cm length /3/. The blood drop should be placed about 1 cm from the end of the slide (opposite labeling end). After the blood drop has been placed on the slide, the spreader slide is moved slowly backward at about a 30 to 45 degree angle toward the blood drop (Fig. 15.13-1 – Blood smear method). The faster the blood is spread on the slide the thicker the film. The smear is thicker at the front, becomes thinner and has a short feathered edge towards the back (Fig. 15.13-2 – Relevant areas for the evaluation of the blood smear). The thickness and the length of the smear are influenced by /3/: The size of the blood drop and the HCT (the bigger the drop and the higher the HCT, the thicker the smear) The angle of the spreader (the greater the angle, the thicker and shorter the film) The speed of spreading (fast spreading produces thicker films). Drying of the film Air drying for 1 hour without forced air circulation is sufficient /3/. At higher relative humidity (≥ 70%), forced air drying is recommended. If slides cannot be stained immediately, fixation with methanol is necessary within 4 hours, but preferably less than one hour, after air-drying; otherwise the plasma causes gray-blue background effects. 15.13.2.1 Staining Blood films are typically stained by Romanowsky dyes. A Romanowsky stain is any stain containing oxidized methylene blue and/or its products of oxidation (azure B), and a halogenated fluorescein dye, usually eosin B or Y. The most commonly used procedures which ensure that all cell types in a blood film can be identified reliably include Wright-Giemsa stain and the Pappenheim stain. Wright-Giemsa stain The stain is composed of basic dyes (methylene blue and its derivatives) and an acidic dye (eosin) in a 2 : 1 ratio /6/. Eosin is termed acidic because it binds to cellular components such as hemoglobin, eosinophilic granule contents, and basic granular proteins. Methylene blue a member of the blue staining thiazin dyes (azure A, azure B, methylene violet) are termed basic dyes. They bind to acidic components, such as DNA, RNA, neutrophil granules and acidic cellular proteins. The use of basic and acidic mixtures of dyes leads to a panoptic (multicolored) staining of the cells. The best panoptic staining is obtained, according to Pappenheim’s specifications, with follow-up staining of the smear, initially with the May-Grünwald dye, and then with the Giemsa stain. Basic stains (blue) bind to acidic cell components such as DNA, RNA and acidic cytoplasmic and granular proteins. Eosin is an acidic dye (red) and it binds to basic cellular components like hemoglobin and basic proteins in the cytoplasm and in certain granules (eosinophil granules). For staining instructions according to May-Grünwald-Giemsa, Pappenheim, or Wright-Giemsa, see Ref. /3, 6, 7/. Pappenheim stain May-Grünwald’s eosin methylene blue and Giemsa’s azure methylene blue are used for staining. A lot of modifications are described /7/. Requirements for acceptable blood film Desirable qualities are: Minimum 2.5 cm in length terminating at least 1 cm from the end of the slide Gradual transition in thickness from the thick to thin area, ending in a squared or straight edge No artifact introduced by the technique Macroscopically, pale red and violet-blue in the thin and thicker sections of the smear, respectively. Color shifts in the staining process due to the predominance of alkaline (blue cast) or acidic (red cast) constituents are prevented with the use of buffered distilled water at pH 6.8–7.2 as the rinsing solution. Microscopically, the red blood cells should be pale red, and the leukocyte nuclei should be more purple than blue. The blood cells should be free of vacuoles and other artifacts, and the glass slides should not be contaminated with precipitates from dye residues /3/. 15.13.2.2 Cytochemical staining Cytochemical stains serve to demonstrate enzymes and other components of the blood cells. They are useful in the differentiation of immature hematopoietic cells and the differentiation of lymphocytes and are indicated: To differentiate between myeloid and lymphocytic leukemias To differentiate precursors of the granulocytic and the monocytic lineage To identify ALL subtypes To characterize cells in CLL and hairy cell leukemia To differentiate between reactive leukocytosis, leukemoid reaction and neoplastic myeloproliferative disease To detect enzymatic defects, especially in neutrophilic granulocytes (e.g., partial peroxidase defect in the case of the myelodysplastic syndrome) To differentiate between a rise in reactive and a rise in neutrophil granulocytes by means of the neutrophil alkaline phosphatase. NAP score 100 neutrophil leukocytes are scored from 0 to 4+ on the basis of the intensity of the precipitated dye in their cytoplasm. The sum of the cells in each category multiplied by its category factor yields the score (reference interval 13–130). Because of a time-dependent activity loss, the smears must be processed within 3 days. The determination of the NAP score is useful for differentiating CML (low score) from other myeloproliferative diseases and from leukemoid reactions. In polycythemia vera, the NAP score is elevated in approximately 70% of the cases and thus supports this diagnosis versus a reactive erythrocytosis. 15.13.2.3 Artifacts Artifacts can arise at all steps of the blood smear procedure. Anticoagulant With the use of EDTA blood for the smear, the following artifacts may arise: agglutination of leukocytes, satellitism (e.g., platelet coating of granulocytes) as well as agglutination of platelets. IgG autoantibodies may induce agglutination of blood cells in EDTA-containing blood but not in citrated blood. IgM autoantibodies are often EDTA independent. Some but not all autoantibodies are of the cold-reactive type. EDTA-associated artifacts include not only agglutination of granulocytes, but also of lymphocytes and circulating lymphoma cells /7/. Age of the sample Best results are obtained when films are prepared within 2 hours of blood collection because of degenerative artifacts that may manifest (e.g., vacuolization of granulocyte and monocyte cytoplasm, or fragmentation and lobulation of granulocytes). The lymphocytes take on an apoptotic appearance, so that only with difficulty they can be attributed to a viral infection or to a lymphoproliferative disease /3/. Preparation of the film When the film is prepared, certain cell types may be damaged especially in conditions with large numbers of atypical lymphocytes. In chronic lymphocytic leukemia ruptured cells may be produced or are present primarily (smudge cells or Gumprecht Kernschatten) /3/. In order to minimize the cellular changes that occur during the preparation of the smear, the addition of 1 drop of albumin (22%) to 5 drops of blood, followed by usual spreading, is recommended. Drying and fixation Optimal results are obtained if the fixation and the staining are performed directly after the air drying of the film. If it is not possible to stain the smears immediately, they should be fixed in methanol within 1 hour of air drying. If this does not occur, the background of the smear takes on a grey-blue appearance, due to changes that have occurred in the blood plasma. Air drying that proceeds too slowly causes the contraction of the cells, a reduction in the fine structure of the nucleus, and the formation of cytoplasmic vacuoles. The staining of a smear that is still moist leads to the irregular spread of the blood film and to poorly differentiable cell morphology /3/. If the water content is in excess of 3% in the methanol fixation solution, artifacts occur, such as decreasing crispness of cellular appearance and the development of artificial cytoplasmic vacuoles /3/. 15.13.3 Microscopy The systematic approach to macroscopic examination of the blood smear is to select an area near the feathered edge and away from the thick portion and looking for a zone where no more than 50% of the erythrocytes are in contact with one another /9/. Thin areas of the smear, in which the erythrocytes form discrete strips (feathered edge), should not be analyzed. As a rule, the optimal analysis site is found towards the last third of the smear (Fig. 15.13-2 – Relevant areas for the evaluation of the blood smear). The evaluation of the cells includes: The verification of the cell counts provided by the hematology analyzer The differentiation of the leukocytes The analysis of the morphology of the red blood cells and the thrombocytes Consideration of cell aggregates and satellite phenomena The search for toxic or degenerative cellular changes and infectious pathogens. The smear is initially scanned with a low magnification (100 to 250-fold), for a reliable quantitative estimate of the cell counts. After finding the best suited area for evaluation individual cells should be investigated at higher magnification /10/. 15.13.3.1 Verification of the cell counts Scanning the smear at low (100x) magnification a reliable quantitative estimate of the counts from the blood smear is attainable by the following calculations: For normal erythrocyte count 600/field For normal thrombocyte count 30–60/field; for a count of (400–1000) × 109/L, 75–125/field; for a count of (50–100) × 109/L, 25–30/field and with a count of (20–50) × 109/L, 10–15/field For normal leukocyte count, 1–2/field; for a leukocyte count of (20–50) × 109/L, 5–6/field; for a leukocyte count of (50–100) × 109/L, 10–15/field; for a leukocyte count above 100 × 109/L, 20–30/field. With the use of a 1000-fold magnification, only approximately one sixth of the cells is counted in each case. For the verification of the cell count, the mean value from 10 view fields should be used. 15.13.3.2 Leukocyte differentiation The differentiation of the leukocytes is performed at a 400 to 1000-fold magnification. Destroyed or pyknotic cells are not included in the count. If, however, their fraction is large, this should be noted in the laboratory report. Smears from patients with chronic lymphocytic leukemia often contain a large number of smudged cells. In these cases a repeat smear on a blood sample to which albumin has been added should be made. In the laboratory results the fraction of individual leukocytes, in relation to 100 counted leukocytes, should be expressed in percent. If other nucleated cells (normoblasts, megaloblasts, megakaryocytes) or cell nuclei are present, these should be counted and their fraction should be expressed per 100 leukocytes /4/. Gumprecht Kernschatten are counted as lymphocytes. Morphological changes in the leukocytes must be characterized and interpreted. Refer to: Tab. 15.13-1 – Artifacts and morphological changes of leukocytes in the blood smear Fig. 15.13-3 – Morphological granulocyte alterations Fig. 15.13-4 – Morphological lymphocyte alterations. 15.13.3.3 Platelet morphology Thrombocyte evaluation is usually performed under 100x oil immersion objective lens (1,000 × magnification) /4/. A blood smear from a healthy individual usually shows 1 thrombocyte per 10 to 30 erythrocytes, or 7 to 21 thrombocytes per field at 1000 × magnification. Platelet satellitosis may cause falsely low automated thrombocyte counts and is detected by microscopy. The platelet morphology is of importance; giant platelets are seen in hereditary disorders and thrombocytopenia in cases with increased thrombocyte consumption. Refer to: Tab. 15.11-3 – Differentiation of the mechanism of thrombocytopenia Fig. 15.13-5 – Morphological thrombocyte alterations Tab. 15.13-3 – Artifacts and morphological platelet changes in the blood smear. 15.13.3.4 Red blood cell morphology Four features of red cells should be evaluated on a peripheral blood film: size, shape color and inclusions /1/. Red cell size In the thicker areas the cell size is underestimated and in the thin areas it is overestimated. Red cell shape Abnormalities in red cell shape are described by the term poikilocytosis. Red cell shape abnormalities are shown in: Fig. 15.13-6 – Morphological erythrocyte alterations Tab. 15.13-2 – Artifacts and morphological red blood cell changes in the blood smear. 15.13.3.5 Blood smear in infection In addition to leukocytosis, a variety of morphologic changes in neutrophils may be seen in reactive states, which are collectively designated as toxic changes. The main changes are toxic granulation, Döhle bodies and vacuolization. Toxic granulation is present in two thirds of patients with sepsis, but lacks specificity for infectious states. In noninfectious reactive conditions toxic granulation may be also seen. Toxic vacuolization is a useful morphologic change during bacterial sepsis /11/. The monocytes manifest nonspecific morphological changes. The morphological changes of the lymphocytes in EBV infection and in cytomegaly are listed in Tab. 15.13-1 – Artifacts and morphological changes of leukocytes in the blood smear. The blood smear is suitable only in rare cases, or not at all, as the primary diagnostic method for the identification of infections. A summary is shown in Tab. 15.13-4 – Morphological blood smear findings in infectious disease. 15.13.3.6 Blood smear in hypo- and hypersplenism In hematology the terms hypersplenism and hyposplenism refer to changes in the blood cell count that are due to altered function of the spleen. Normal splenic function consists of storing reticulocytes for maturation that are released into the circulation, the removal of cell particles and nuclear fragments, and the final configuration of erythrocytes. Immature cells that are released by the bone marrow are eliminated. Aged or damaged erythrocytes are taken up and degraded by the spleen. The enhanced degradation of erythrocytes in a hemolytic reaction leads to splenomegaly. The spleen also maintains the thrombocyte count constant. In acute complete release of thrombocytes the increase in peripheral thrombocyte count is about 50 × 109/L. Hyposplenism This hematologic symptom may result from splenectomy, and from extensive immunological and lymphoproliferative diseases, such as hairy cell leukemia or amyloidosis of the spleen. Associated changes can be /12/: Transient or long-time increased thrombocytosis An abnormal blood smear: target cells, Howell-Jolly bodies, and red blood cells of irregular shape (spiculated cells). Additionally, nucleated red blood cells, immature granulocytes and megakaryocyte residues are seen in small numbers. Hypersplenism This term refers to a clinical symptom and not to a pathological or a histological entity. The spleen can be enlarged, or it may simply manifest hyper function. The latter results in cytopenia that can concern one or more hematopoietic cell lineages. The affected cell lineage manifests bone marrow proliferation, and can also express morphological changes or the interruption of maturation. Changes that may be associated with hypersplenism are /12/: Permanent thrombocytopenia Spherocytic cells, teardrop cells, and schistocytes in the blood smear. References Pierre RV. Red cell morphology and the peripheral blood film. Clin Lab Med 2002; 22: 25–61. Bain BJ. Diagnosis from the blood smear. N Engl J Med 2005; 353: 498–507. Houwen B. Blood film preparation and staining procedures. Clin Lab Med 2002; 22: 1–14. Gulati GL, Bong HH. Blood smear examination. Hematol Oncol Clin North Am 1994; 8: 631–50. Barnes PW, McFadden SL, Machin SJ, Simson E. The international consensus group for hematology review: suggested criteria for action following automated CBC and WBC differential analysis. Lab Hematol 2005; 11: 83–90. Woronzoff-Dashkoff KK. The Wright-Giemsa stain. Clin Lab Med 2002; 22: 15–23. Bucher U. Labormethoden in der Hämatologie. Bern; Huber 1988; 59–65, 87. Dalal BI, Brigden ML. Artifacts that may be present on a blood film. Clin Lab Med 2002; 22: 81–100. NCCLS. Reference leukocyte differential count (proportional) and evaluation of instrumental methods. Villanova 1992; NCCLS Document H20–A Vol 12 No 1. Nosanchuk J, Chang J, Bennett J. The analytical basis for the use of platelet estimates from peripheral blood smears. Am J Clin Pathol 1978; 69: 383–7. Kroft SH. Infectious diseases manifested in the peripheral blood. Clin Lab Med 2002; 22: 253–77. Constantino BT. Pelger-Huet anomaly – morphology, mechanism, and significance in the peripheral blood film. Labmedicine 2005; 30: 103–7. Moreira AMB, Vieira LM, Carvalho M. Acquired Pelger-Huet anomaly associated with ibuprofen therapy. Clin Chim Acta 2009; 409: 140–1. Blood images in hematology. Howell-Jolly body-like inclusions in neutrophils. Blood 2009; 114: 2860. Baumann H, Bettelheim P, Diem H, Gassmann W, Nebe T. Lymphozytenmorphologie im Blutausstrich – Vorstellung einer überarbeiteten Nomenklatur und Systematik. J Lab Med 2011; 35: 261–70. Mallah HS, Brown MR, Rossi TM, Block RC. Parenteral fish oil-associated Burr cell anemia. J Pediatr 2010; 156: 324–6. Glader E. Hemolytic anemias in children. Clin Lab Med 1999; 19: 87–111. Stiegler H, Fischer Y, Steiner S, et al. Sudden onset of EDTA-dependent pseudothrombocytopenia after therapy with the glycoprotein IIb/IIIa antagonist c7E3 Fab. Ann Hematol 2000; 79: 161–4. Bizarro N, Goldschmeding R, dem Borne AE. Platelet satellism is Fc gamma RIII (CD16) receptor mediated. Am J Clin Pathol 1995; 103: 740–4. Becker RC, Giuliani M, Savage RA, et al. Massive hemolysis in C. perfringens infections. J Surg Oncol 1987; 35: 13–8. Baumgarten BU, Röllinghoff M, Bogdan C. Ehrlichien. Dtsch Ärztebl 2000; 97: A2456–A2461. Weinberg GA. Laboratory diagnosis of Ehrlichosis and Babesiosis. Pediat Infect Dis 2001; 20: 435–7. Eberhard ML, Lammie PJ. Laboratory diagnosis of filariasis. Clin Lab Med 1991; 11: 977–1010. De Braekeleer E, Douet-Guilbert N, Rowe D, Brown N, Morel F, Berthou C, et al. ABL1 fusion genes in hematological malignancies: a review. Eur J Haematol 2011; 86: 361–71. Foa R, Chiaretti S. Philadelphia chromosome – positive acute lymphoblastic leukemia. N Engl J Med 2022; 386: 2399–2411. 15.14 Bone marrow examination Torsten Haferlach The laboratory diagnostic investigation of leukemia and lymphoma has evolved considerably. A large part of the progress that has been made is related, on the one hand, to the increasing understanding of the pathophysiological mechanisms of these diseases and, on the other hand, to the further development of laboratory techniques and instruments, including computer supported procedure for the analysis of the findings. Today, a prompt and comprehensive laboratory diagnostic investigation often leads to targeted therapy and includes best possible prognosis. At the same time the available methods should be utilized in the laboratory work flow in a targeted and cost effective way, in the sense of a step-wise diagnostic investigation. This presupposes strict algorithms with regard to sample collection and laboratory examinations. With this swift further development of diagnostic investigation, it is difficult for the requesting physician to always ask his laboratory the appropriate question. Furthermore, it is a difficult challenge for the laboratories to keep abreast of the swift methodological developments regarding primary diagnosis and, similarly, the demonstration of minimal residual disease, which is increasingly required. For these reasons, an up-to-date orientation for the routine diagnostic investigation of leukemia and lymphoma is provided herein to the referring physician and the investigating laboratory. 15.14.1 Bone marrow specimens Bone marrow histology and aspiration A bone marrow evaluation is generally indicated on the basis of the clinical findings, and only upon detection of changes in the blood cell count or its composition. If there is evidence of disease of the hematological system, or of lymphoma, an investigation of the bone marrow aspirate, often supplemented with a biopsy/histology, is necessary /1, 2, 3, 4/. Site of bone marrow aspiration The posterior iliac crest (spina iliaca posterior superior) is the preferred site of specimen collection. The puncture of the iliac crest at this site is a technically more simple and less painful procedure than sternal puncture, which should be performed only in exceptional cases and which is only suitable for aspiration; in fact, this procedure should be considered obsolete. Prior pelvic radiation, punctio sicca, or a patient who is too obese constitute such exceptions. But in the latter case as well, or in mechanically ventilated patients in supine position, pelvic puncture can, in fact, be performed at the spina iliaca anterior superior or the anterior iliac crest. Here, the os ilium at the spina iliaca posterior superior is at its widest (approximately 3 cm) so that, injuries to vital organs or to large vessels are largely ruled out if the correct technique is applied (Fig. 15.14-1 – Transverse section at the level of the biopsy site). Whenever possible, the procedure and its associated risks of complications are explained to the patient on the previous day. The puncture can also be performed on an out-patient basis on patients at risk of bleeding, but in such cases a sufficiently long follow-up period of observation must be assured. In special cases, the administration of banked thrombocytes must be considered prior to puncture, in order to achieve thrombocyte counts of above 20 × 109/L. Premedication with tranquilizers is only necessary for extremely sensitive patients, while in children the use of a short anesthesia may be recommended, following discussion with the anesthetist. Puncture is possible in the supine or the abdominal position; the author recommends the abdominal position since the patient is, thereby, stable and safe, even with the use of considerable exertion, and does not have to be stabilized further. Bone marrow biopsy In patients without leukemia, or with bone marrow aspirates containing very few cells, a biopsy for histology, in addition to the aspiration, is necessary. In myeloproliferative disease and chronic lymphocytic leukemia, as well as for the staging of lymphoma, bone marrow histology is always required. In order to obtain artifact-free tissue, the histology should be performed prior to aspiration. In the abdominal position, a minimum of 10 mL of local anesthetic is administered following thorough disinfection and the sterile covering of the skin to the periosteum of the spina iliaca posterior superior. The time required to achieve satisfactory anesthesia is at least 5 minutes. The original sternal puncture needle (to be used at the pelvis without the spacer) and a histology needle (e.g., the so-called Jamshidi needle) are to be used for the puncture (Fig. 15.14-2 – Jamshidi needle in comparison with a usual puncture needle). Disposable instruments are highly suitable. In children, a short anesthesia is, in some cases, to be recommended, following discussion with the anesthetist. Bone marrow histology An 8-gauge needle is preferred; in young patients with strong bones, the 11-gauge needle has proven successful. For particularly obese patients, a longer needle is available. The Jamshidi needle is placed on the middle of the posterior iliac crest with a mandrel and, following removal of the mandrel, is inserted through the bony cortex in the direction of the spina iliaca anterior, which is usually palpable. In this way, a biopsy core length of up to 4 cm is possible, allowing for a representative assessment of the bone marrow. Furthermore, a large biopsy permits easier rotation and better adhesion of the punched bone marrow cylinder in the canula. Once the cylinder has been obtained, imprint smears can be made if necessary and, depending upon the planned embedding method, the biopsy is fixed according to the instructions of each laboratory. If, in cases of punctio sicca and non leukemic disease, material for the cytogenetic analysis cannot be aspired, a punch cylinder can also be placed directly following sampling in a physiological saline solution containing heparin. The tube is then dispatched, together with the punch biopsy, to the genetics laboratory. In these cases, the markers of immunophenotyping can be performed with the histology of a second punch biopsy following paraffin embedding. Bone marrow aspiration The bone marrow aspiration at the iliac crest follows the sampling of the biopsy material. Usually, a Klima and Rosegger puncture needle without arresting disc is used. Here as well, disposable needles are preferred, due to the fact that they are sharper as well as for hygienic reasons. The bone marrow puncture is made through already existing skin incision, approximately 1 cm from the histology biopsy site and at an angle to the direction of the biopsy. Prior to aspiration the patient is informed that short-lived pain which, in any case, cannot be prevented with the careful use of a local anesthetic will be felt. Aspiration is made quickly and strongly with a full stroke of a 10 mL syringe. Thereby, 2–3 mL of bone marrow suffice for the first EDTA sample for further cytomorphological analysis, in order to avoid dilution with peripheral blood. A second and a third syringe, with heparin as anticoagulant, can then be withdrawn, up to 5 or 10 mL in each case. If all the samples are taken from a single site, the last aspirates may manifest a different cell composition than the first one, due to increasing dilution with blood. If the aspiration is unsatisfactory, the position of the puncture needle must be altered by rotating under suction, or by means of a repeated puncture. In punctio sicca the rotating of the needle in the bone while continuing to aspirate often does help to achieve a successful sampling. If the puncture is still unsuccessful the other side as well, or the anterior iliac crest, can be punctured following appropriate anesthesia. A sternal puncture (ultima ratio) is also possible, but in such a case a depth stop must definitely be used. In infants, the tibial crest below the attachment of the M. quadriceps can also be considered for puncture in exceptional cases. Following the application of an adhesive bandage, the puncture site is compressed by lying on a sand bag. The site is monitored at least 30 minutes later for bleeding. Sample distribution Aspirates for the following analyses are taken routinely, with the additives for the syringes in each case in brackets: Aspiration cytology: 0.5 mL EDTA ad 2–3 mL bone marrow aspirate (maximum) Immunophenotyping: 0.5 mL heparin or EDTA ad 5 mL bone marrow aspirate (maximum) Cytogenetics: 0.5 mL heparin ad 5–10 mL bone marrow aspirate (maximum) Molecular genetics: 0.5 mL heparin or EDTA ad 5 mL bone marrow aspirate (maximum). The aspiration cytology with EDTA always comes first, since potential heparin contamination from the cone of the syringe can cause serious staining artifacts in the Pappenheim staining. All syringes must be marked prior to puncture with the patient’s name, the material, and also with the type of anticoagulant, and then immediately sent to the corresponding laboratories. As a general rule, the smaller the sampling volume, the less the contamination with peripheral blood, and the more representative is the aspired bone marrow. If heparinized samples are used for immunocytolgy, EDTA smears that have not yet been stained are to be provided as well. Unlimited sample quality become increasingly important as the level of difficulty of the diagnostic issue increases (e.g., for the monitoring of minimal residual disease). This signifies that for cytomorphology, the smear should be made, if at all possible, within 3 hours of aspiration, and then sufficiently (for 30 minutes) air-dried, before it is packaged for dispatch. These also still allow for gut staining results over a number of days. The material for cytogenetics, immunophenotyping and molecular genetic analytic procedures should reach the investigating laboratory within 24 hours. In summary, the proposed algorithm for the sampling of bone marrow and its dispatch to the individual laboratories is presented in Tab. 15.14-1 – Methods for investigation of bone marrow in primary diagnosis or suspected diagnosis of leukemia and lymphoma. The biopsy cylinder should be dispatched within one day, in the fixing solution recommended by the laboratory. 15.14.2 Processing and analysis of blood smear and bone marrow aspirate Analysis of peripheral blood If peripheral blood is examined, the blood count, including thrombocytes, MCV and MCH, reticulocytes, and the differential, are indispensable basic tests. In hematological diseases in particular, interfering factors can be present in EDTA blood, and these interfere with the widely used automated measurements of the blood count, leading to erroneous results. In consequence, a few examples: Leukocytes of over 100 × 109/L manifest, in many instruments, falsely high hemoglobin levels (10–20 g/L too high) Fragmentocytes e.g., in disseminated intravascular coagulation or the splitting of cytoplasma from blasts, e.g. in AML M4 or M5 can lead to falsely high thrombocyte counts (up to 30 × 109/L too high) In all cases of reduced platelet counts as determined by automated counting, thrombocyte agglutinins or EDTA-induced pseudo thrombocytopenia should be excluded by the investigation of a blood smear In almost all cases, the differential count is compiled primarily by automated hematological systems. Attention should be paid to the fact that the instruments are focused on the differentiation of normal blood from pathological complete blood counts. In the presence of a pathological cell population, the analyzer documents a signal, which of necessity leads to a blood smear and microscopic analysis. Since hematology analyzers produce more or less reliable results, the microscopic analysis of peripheral blood is an absolute requirement in every case of suspicion of hematological disease or for its monitoring /6/. For preparation and microscopy of peripheral blood refer to Section 15.13 – Blood smear examination. Processing and assessment of the bone marrow aspirate The examination of the bone marrow is part and parcel of the basic diagnostic investigation of hematological disease. There are different opinions as to whether or not aspiration cytology alone is sufficient in acute leukemia, or whether, rather, supplementary biopsy is indicated. For the assessment of leukemic cells in acute leukemia, good quality blood and bone marrow smears, at a minimum, are most suitable. These should be sufficiently dried (more than 30 minutes) before dispatch and staining. The additional assessment of a bone marrow biopsy is certainly indicated in cases of an unsuccessful aspiration (punctio sicca); it allows for immunohistological analysis as well. In myeloproliferative disease, the bone marrow histology plays an important role at the time of diagnosis because, among other things, of the necessary evaluation of the degree of fibrosis. The diagnostic value is crucially dependent upon the subsequent processing of the collected material. The best smears for semi quantitative estimate of cell density are obtained when a cover slide is placed on the bone marrow fragments on the glass slide (following filtration or collected from a watch glass), and then the sample is smeared with the cover slide as a single stroking motion using moderate pressure to thin out clumps (Fig. 15.14-3 – Preparation of a smear of bone marrow fragments). The bone marrow preparations must then be air-dried for 30–60 minutes. The subsequent Pappenheim or May-Grünwald-Giemsa staining procedures represent the standard methods for blood and bone marrow imaging. Bone marrow smears that are suitable for assessment contain bone marrow fragments at the center of the smear, and mixed bone marrow and blood at the periphery. Following careful observation of the entire preparation at low magnification, the cellularity and different distribution patterns, e.g., chronic lymphatic leukemia or lymphoma, even with nodular infiltrates, or the detection of tumor cells can be assessed. Thereon, a single analysis of at least 200–500 cells from 2 representative sections of the smear is performed. The distribution of bone marrow cells from healthy individuals is shown in Tab. 15.14-2 – Percentage of bone marrow cells in healthy individuals The additional examination of at least two smears is performed according to the following criteria: Assessment of the cell density: a decrease can be a function of the sampling and the smearing procedures. A true decrease in cellularity may only be assumed if marrow fragments with fat and stromal cells are detectable. The age of the patient is to be taken into consideration; cellularity decreases physiologically with age. Assessment of the erythropoiesis to granulopoiesis ratio (EP:GP). The normal ratio is approximately 1 : 3 to 4. The cytological examination of the smear only allows for a relative quantitative assessment. Additionally, the distribution of the different stages of maturity, including in particular the percentage of blasts, as well as changes in the cytoplasm and cell nuclei and signs of dysplasia. The eosinophil, basophil and monocyte fractions are specified. The quantitative and qualitative assessment of the megakaryocyte number, and the distribution and fine structure of the lymphocytes and the plasma and reticular cells are necessary. Assessment of absolute cell density and possible heterogeneous cell distribution in medullary cavities, histological slices must be evaluated, likewise in an age-dependent manner. Iron storage in the reticular cells as well as the quantification of sideroblasts and ringed sideroblasts (stained with Berlin blue) can be better performed cytomorphologically than histologically. For the demonstration of blasts, or with regard to the question of myelodysplastic syndrome, the following cytochemical staining procedures are performed on blood and bone marrow smears for further morphological differentiation: The myeloperoxidase staining is obligatory, and, possibly, for histology, chloracetate esterase or Sudan Black as a hint regarding affiliation with the granulocyte lineage, as well as non-specific esterases. The latter provides an indication of affiliation with the monocyte lineage (Tab. 15.14-1 – Methods for investigation of bone marrow in primary diagnosis or suspected diagnosis of leukemia and lymphoma). The PAS staining as an indication of affiliation with the lymphatic and erythrocyte series in AML-M6, as well as acid phosphatase as a hint of the T cell nature of an acute lymphatic leukemia, are optional and, today, no longer necessary because of the use of immunophenotyping. With the obligatory use of immunophenotyping, fluorescence in situ hybridization (FISH) or molecular genetics (PCR), the following tests are not applied due to deficient sensitivity and specificity: Alkaline leukocyte-neutrophil phosphatase (ANP); previously in connection with the exclusion of CML (today BCR-ABL) Tartrate resistant acid phosphatase for the diagnosis of hairy cell leukemia (today CD103 as APAAP or immunophenotyping/immunohistology) Terminal deoxynucleotidyl transferase (TdT) on the smear in acute lymphocytic leukemia (today immunophenotyping). 15.14.3 Processing and assessment of bone marrow histology Bone marrow biopsy allows not only for an assessment of the individual cellular components of the bone marrow but also, for a true quantitative assessment of the cell content, as well as for an analysis of the topographic distribution of the cells and their iron content, and statements concerning the bone marrow stroma and osseous changes. The significance of the bone marrow biopsy in the diagnostic investigation of hematological disease is closely related to the particular clinical issue. Thus, bone marrow histology indispensable to the diagnosis in, e.g., primary myelofibrosis and in aplastic anemia. In other diseases, such as, e.g., acute leukemia, it can provide valuable supplementary information on cytomorphology. Paraffin embedding employed in the diagnosis of suspected leukemia/lymphoma allows a precise diagnostic expert appraisal of the tissue /1, 5, 7/. The following standard stains are used for all biopsies: Giemsa for cellular details PAS for cellular details and for the demonstration of memory cells Gomori silver for the display of filaments Berlin blue for evaluation of the iron content Naphthol-AS-D chloroacetate esterase reaction for the display of granulopoiesis and of mast cells. Immunohistochemistry is used, in addition, for the display and characterization of the cells. The following markers are routinely employed for the investigation of hematological disorders: Myeloperoxidase and CD15 for granulopoiesis CD61 for megakaryopoiesis Glycophorin A for erythropoiesis CD34 for progenitor cells. CD68 for cells the monocyte-macrophage system. CD117 for mast cells. Furthermore, immunohistological investigations are employed, particularly in the characterization of lymphatic aggregates, in hairy cell leukemia, in the sub typing of infiltrates of malignant lymphoma, e.g., CD19, CD20, CD3, CD103, CD138, and in the identification of metastases of non-marrow tumor cells, e.g., cytokeratin and hormone receptors. References Mufti GJ, Flandrin G, Schäfer HE, Sandberg AA, Kanfer EJ: An Atlas of Malignant Haematology. London; Verlag Martin Dunitz, 1996. Löffler H, Rastetter J, Haferlach T. Atlas der klinischen Hämatologie, 6. Aufl. Heidelberg; Springer, 2004. Theml H, Diem H, Haferlach T. Taschenatlas der Hämatologie, 5. Aufl. Stuttgart; Thieme, 2002. Haferlach T, Labordiagnostik bei Leukämien und Lymphomen, 2. Aufl.. Bremen; UNI-MED, 2007. Bain JB, Clark DM, Lampert IA, Koch S: Knochenmarkpathologie. Atlas und Lehrbuch. Blackwell; Berlin, 2000. Haferlach T, Haferlach C, Kern W, Labordiagnostik in der Hämatologie, Köln, Deutscher Ärzte-Verlag, 2011 Swerdlow SH, Campo E, Harris NL, Jaffe ES. WHO Classification of tumours of haematopoietic and lymphoid tissues. Lyon, International Agency for Research on Cancer (IARC), 2008. Löffler H, Haferlach T. Hämatologische Erkrankungen, Heidelberg, Springer, 2010. 15.15 Acute leukemias Torsten Haferlach In acute leukemia, the diagnostic possibilities have been extended in a fundamental manner. On the one hand, numerous methods of diagnostic investigation have been established and, on the other hand, insights from correlations of biological parameters and clinical course have led in many cases to diagnostic findings becoming basic prerequisites for the initiation of appropriate therapy. Furthermore, important prognostic biomarkers are provided by these diagnostic investigations. These reasons suffice, therefore, in order that the diagnostic investigation be carried out in a comprehensive and, at the same time, goal-oriented manner. The insights have also led to new classifications /1/. Thus, classifications according to morphology (FAB classification), cytogenetics (WHO classification to a certain extent) and, in addition, results from immunophenotyping (similar to the EGIL classification), are interconnected and must be taken into consideration /2, 3, 4, 5, 6/. This leads to potential algorithms in the diagnostic investigation of leukemia, not only for the primary investigations but also for the follow-up examinations for the monitoring of treatment (minimal residual disease, MRD). At the same time, much new data, especially from molecular methods such as, gene expression analysis and next-generation sequencing (NGS), are to be expected in the short term. These will, precisely in the field of hematology, supplement current diagnostic procedures in the short term and, in some cases, perhaps even completely replace them. Sample collection The parallel investigation of blood and bone marrow should be striven for in all cases of suspected acute leukemia. For the creation of smears for cytomorphology and cytochemistry (at least 6–8 smears should be prepared). If alkaline phosphatase anti-alkaline phosphatase (APAAP) or fluorescence in situ hybridization (FISH) are to be determined, EDTA or citrate must be used as anticoagulant. The material that has been rendered incoagulable in this manner can also be utilized, as required, for immunophenotyping and for molecular analysis (e.g., with the polymerase chain reaction). The last two methods mentioned, as well as FISH, can also be implemented, as an alternative, with heparin blood or bone marrow. Heparin must be used as the anticoagulant for cytogenetics, since only the vital cells that go into metaphase are available for the chromosome analysis. Thus, a current diagnostic investigation of leukemia requires, optimally, EDTA and heparin blood and bone marrow. In total, 5–20 mL of bone marrow should be collected. In addition, in acute leukemia, sampling for histology by means of an iliac crest puncture, must be taken into consideration as a function of the suspected diagnosis /7/. An orientation is provided in Tab. 15.15-1 – Methods for the diagnostic investigation of acute leukemia. Classification For the first diagnosis and classification of acute leukemia, the morphological criteria of the FAB classification, die EGIL classification and the WHO classification as well, are still employed in parallel today on account of their clinical relevance and feasibility /1, 2, 3, 4, 5, 6/. Above and beyond that, ALL is classified in an even more precise manner according to immunophenotyping and cytogenetics/molecular genetics. This renders the orientation complicated at the present time, but in the midterm it will contribute, through better understanding of the biology of individual leukemia subgroups, to a more specific diagnosis and, finally, to targeted treatment decisions as well. To make a diagnosis is, thus, not always to be considered the same as classification in the sense of the WHO document. 15.15.1 Diagnostic investigation of acute leukemia 15.15.1.1 Acute myeloid leukemia (AML) Depending upon the leukemia subtype, the methods and classification models are of special importance in the primary diagnostic investigation and in follow-up investigations. According to the FAB and WHO classifications, the bone marrow smear should be analyzed. The counting of 200–500 bone marrow cells is recommended. According to the FAB classification, the blast fraction should be greater than 30% of the nucleated cells of the bone marrow; WHO has set the limiting value at ≥ 20%, and this is generally considered to be valid. Furthermore, according to WHO, AML with specific genetic aberrations [i.e., AML with t(15;17) or t(8;21)or inv(16) or 11q23] are also ruled out with this limiting value. One also speaks of AML if the corresponding chromosomal changes are demonstrated, and the bone marrow blast fraction is under 20%. The cases are, however, very rare. The algorithm for the FAB classification, which is further helpful in daily clinical practice, and the FAB AML subgroups, are shown in: Fig. 15.15-1 – Previous FAB classification: bone marrow blast cell cutoff between AML and MDS Tab. 15.15-2 – Previous FAB classification of AML, can be performed as initial diagnostic step before WHO classification. For the diagnosis of ALL, it should initially be demonstrated purely cytomorphologically, as well as with cytochemistry (myeloperoxidase below 3%, nonspecific esterase negative) that AML L-M1–M6 is not present. Immunophenotyping is required for further goal-oriented classification; it separates, on the one hand, AML-M0 and AML-M7 according to FAB as well as bilineal and biphenotypic acute leukemia according to EGIL and WHO from the classical ALL of B and T lineage. These are then divided into additional subgroups based upon their marker profiles. See Section 15.15 – Acute leukemias Section 15.16 – Myelodysplastic syndrome. First Assessment of the bone marrow in AML and ALL The assessment includes: Cell content on the smear, small fragments present Description of the blast morphology according to size, nucleus-to-plasma ratio, cytoplasm color, inclusions, Auer rods, pseudo Chediak bodies Fraction of promyelocytes, myelocytes and metamyelocytes, rods, segmented, eosinophils, abnormal eosinophils, basophils, monocytes Fraction of nucleated cells of the erythrocyte series, by stage of maturation Fraction of mature lymphocytes, plasma cells If necessary, available tissue mast cells Number and form of the megakaryocytes. Extended definition of the blasts It has often been attempted to subdivide the myeloblasts further according to shape and, in particular, to the number of cytoplasmic granules. From the current standpoint it seems to make sense to proceed according to the following classification: Type I blasts: myeloblasts with immature cytoplasm, agranular, the nucleus may contain a few nucleoli Type II blasts: similar to type I, but with 20 (other classifications 15) azurophilic granules in the cytoplasm. The subtype that was previously differentiated as a type III blast, containing more than 15 or 20 granules, should be omitted from routine use. It should be remembered that the reproducibility of this blast classification is extremely limited, and is lacking in clinical or classificatory relevance. It can, therefore, be ignored in most cases in the classification of AML and MDS. Furthermore, according to morphology and in particular in the absence of cytochemistry as well as (if necessary) with the assistance of APAAP or immunophenotyping it should be possible to subdivide the blasts as follows: Atypical promyelocytes in AML-M3 and M3 variants with translocation (15;17) and the demonstration of PML-RARA Monoblasts and promonoblasts, especially following nonspecific esterase reaction in AML-M4, AML-M5a und AML-M5b Magakaryoblasts following measurement of CD41 or CD61 in AML-M7 with APAAP or immunophenotyping. Proerythroblasts are not added to the true blasts, since it is difficult to ascribe them to the normal or the malignant populations. Assessment of dysplasia Apart from the blast fraction – especially in AML – which makes possible a demarcation relative to myelodysplasia, the detection of dysplastic changes in the three cell lineages has also become important in AML, according to WHO 2008 (WHO classification, Tab. 15.15-3 – WHO classification of AML). The Goasguen and Bennett criteria are to be used for the assessment of dysplasia in AML according to WHO /1, 8/: Dysgranulopoiesis ≥ 50% of the segmented cells (at least 10) are agranular or hypogranular, or manifest pseudo-Pelger-Huet changes or peroxidase defect More than 100 cells should be assessed Dyserythropoiesis ≥ 50% of at least 25 nucleated cells of the erythropoietic series manifest one of the following morphological abnormalities: Karyorrhexis Megaloblastoid changes Multinuclearity Nuclear fragmentation Dysmegakaryopoiesis ≥ 50% of at least 6 megakaryocytes manifest one of the following abnormalities: Micromegakaryocytes Multiple individual nuclei Large mononuclear core form In the assessment of dysplasia, attention should be paid in particular to the fact that in AML, at least 50% of all cells of a given series have to manifest one or more of the above-mentioned changes. Only in such cases is the series considered to be dysplastic. This is in contrast to the assessment of the same dysplasia criteria in myelodysplastic syndromes, in which only 10% of the observed cells of each series has to fulfill these criteria, in order that the condition be considered to be dysplasia. In AML, dysplasia according to WHO 2008 is required, because with the demonstration of 2 or 3 dysplastic lines a patient could be allocated to the subgroup “AML with myelodysplastic-like changes.” Morphological classification of AML according to the FAB criteria The FAB classification is based upon the stage of maturity of the blasts, the cell lineage affiliation and the number of blasts, as well as the assessment of the cytochemistry, particularly myeloperoxidase (MPO) and the nonspecific esterase (NSE) reaction in the bone marrow smear. Even if they, in fact, should no longer be used in the times of the WHO 2008 classification, they continue to be of practical relevance from the clinical point of view (one cannot wait 7–10 days for a complete diagnosis according to the WHO criteria). The maturation of the blasts and their respective differentiation with regard to normal and to abnormal promyelocytes (AML-M3) can be assessed according to Tab. 15.15-4 – Differentiation of the blasts from abnormal promyelocytes and from normal promyelocytes. The further sub-classification of AML, based upon morphological properties according to FAB, is shown in Tab. 15.15-2 – Previous FAB classification of AML, can be performed as initial diagnostic step before WHO classification. Noteworthy morphological aspects with regard to the sub-entities AML-M0: Differential diagnostically, biphenotypic acute leukemia (EGIL classification), and also ALL, in particular with t(9;22), as well as AML-M5a, come into question. Immunophenotyping is required for the definitive establishment of AML-M0. AML-M1: Because of the limited volume of cytoplasm and the low degree of maturity one often sees, particularly in the peripheral blood, so-called pseudo-nucleoli, corresponding to the Golgi apparatus that appears in the nucleus. They should not be confused with giant nucleoli. There exists a certain correlation between these morphological findings and a mutation in the NPM1 gene. The maturation of granulopoiesis to the stage of the promyelocyte or beyond is below 10%. AML-M2: The maturation of granulopoiesis to the stage of the promyelocyte or beyond is above 10%. One finds a certain correlation between this FAB subtype with AML with translocation t(8;21). Of these cases of genetically defined AML, 90% manifest an M2 FAB subtype, and 10% an M1 subtype. The AML-M2 cases with t(8;21) often have type II blasts, or cells that are close to the promyelocyte (previously so-called type III blasts). According to WHO, however, AML is always present, even if the fraction of unambiguous blasts in the strict sense were below 20% (Tab. 15.15-3 – WHO classification of AML 2008)). This AML subtype often manifests long needle-like Auer rods, dysgranulopoiesis, and mild eosinophilia. AML-M3: Most prominent are abnormal promyelocytes, often with bundles of Auer rods, so-called faggot cells. The MPO is always more than 90% strongly positive. Frequent leukopenia in the peripheral blood. AML-M3 variant(v): the nuclei are bi-lobular, and can easily be confused with monocytes. Granules are hardly distinguishable and Auer rods occur less commonly than in M3. The MPO is strongly positive here as well, while the NSE is negative. Frequent leucocytosis in the peripheral blood. AML-M4: Composite of myeloid and monocytic blasts, with bone marrow MPO above 3% and NSE above 20%. The monocytic blast fraction lies, thereby, between 20% and 80%. AML-M4Eo: Blasts as in M4 and similar cytochemistry but, apart from that, unambiguous so-called pathological eosinophils with distinct dark granules. In contrast to normal eosinophils, these eosinophils are positive with chloracetate esterase. Proof for this sub-type is found in the demonstration of an inversion in chromosome 16 or the molecular correlate CBFB-MYH11. AML-M5a: Mostly a monomorphic blast population with relatively blue cytoplasm with a net-like internal structure. Very marked NSE in over 80% of the blasts. Differential diagnostically, M0 ALL (also Burkitt type) and dedifferentiated multiple myeloma must be taken into consideration. AML-M5b: With regard to the maturation of the mono blastic cells, this mature sub-type of mono blastic AML is easier to diagnose in the blood than in the bone marrrow. Here as well, the NSE is moderately to strongly positive in over 80% of the cases. This is more pronounced in the bone marrow than in the blood. In AML-M5a and AML-M5b, gingiva hyperplasia or skin infiltrates, caused by the blasts, are frequently present at the time of diagnosis. AML-M6: More than 50% of all nucleated cells belong to the erythrocyte series. Of the other, non-erythropoietic cells, at least 20–30% are blasts (according to FAB). Of more historical interest in the relevance of the strong PAS positivity in the immature cells erythrocytes. This can provide support for the M6 diagnosis, since normal erythropoiesis does not manifest this reaction, and PAS is not longer absolutely necessary in the diagnostic investigation of AML, particularly of AML-M6. AML-M7: With this sub-type, which is observed far more frequently in children than in adults, bone marrow fibrosis with punctio sicca often occurs. Even if the blasts have a certain specific morphology, with cytoplasmic evaginations, and look, to a certain extent, like megakaryocytes, the diagnosis cannot be made with certitude based solely on cytomorphological criteria. In suspected AML-M7, immunophenotyping or APAAP with at least CD41 or CD61 must be performed. WHO classification of acute myeloid leukemia (AML) Within the framework of a purely cytomorphological differentiation between AML and MDS, as well as with respect to bilineal acute leukemia and ALL, it has again initially been specified in the 2008 WHO classification of AML (Tab. 15.15-3 – WHO classification of AML 2008) that with greater than 20% blasts in the bone marrow, one is dealing with AML and thereby, the category of RAEB-T, which belongs to myelodysplasia, is omitted /1/. As the most important first step towards a biological classification of clinical symptomatology, the WHO has consolidated four genetically defined sub-types of AML with specific balanced trans locations into as a single group; this as a first stage in the sense of a hierarchy. For the four genetically defined groups named in Tab. 15.15-3 – WHO classification of AML (2008), the term AML is also valid, to a certain extent, if the number of bone marrow blasts is below 20%. The integration into MDS is, thereby, ruled out. As the second step in the WHO classification, a sub-group that was defined, inter alia, on the basis of morphological criteria was exposed – namely, AML with multilineal dysplasia. Thereby, the dysplastic changes are classified according to the criteria of Goasguen and Bennett /8/. The WHO considers multilineal to refer to the demonstration of dysplasia in 2 or 3 cell lineages in at least 50% of the analyzed bone marrow cells. In so doing, a far higher limiting value for AML than for MDS as evidence for dysplasia is established; for the latter, only 10% of the cells must display dysplastic properties. This group also includes patients who, at the same time, have only cytogenic changes, such as those found in MDS, as well as those who, in their previous medical history, had MDS or MDS/MPN overlap. This AML category, thereby defined in a miscellaneous manner according to WHO, must be further evaluated with regard to its autonomy in prospective clinical studies. Various analyses suggest that although, admittedly, the morphological differences can be captured and that they partially correlate with certain genetic sub-groups, an independent prognostic relevance remains to be demonstrated. At the third stage of the WHO hierarchy, subgroups are identified that can only be defined while taking into account the patient’s medical history; here, therapy-associated MDS and AML in particular are included. The further sub-classification of these AML groups according to the type of previous therapy (according to alkylating agents, topoisomerase II inhibitors, other medication or radiotherapy) is absolutely necessary for the description of the biology and the clinical course. Not only clinical differences but also, and in particular, cytogenetic and molecular genetic differences stand out in these sub-groups, the further identification and biological description of which, via this WHO proposal, is made possible in a meaningful manner. AML following previous MDS or MPS is also captured in this sub-group according to WHO. Only at the fourth stage of the WHO AML classification are the old, purely morphological or immunophenotypic sub-groups according to the FAB criteria shown referred to. Thereby, further sub-entities are added: Acute basophilic leukemia, acute pan myelosis with fibrosis, and myeloid sarcoma or chloroma. It remains to be seen whether these very rare AML sub-groups emerge more clearly with their compilation according to the WHO criteria. To a certain extent, they can be diagnosed only when the histological examination is taken into consideration. Biphenotypic leukemia, which is also now classified according to WHO, is dealt with in Section 15.15 – Acute leukemias. They should be defined on the basis of the immunophenotype, according to the criteria of the EGIL group /6/, and they are classified between acute lymphatic and acute myeloid leukemia. In this group, undifferentiated acute leukemia with bilineal properties and the genetic changes t(9;22) or t(4;11) is accounted for. An algorithm for the primary diagnosis of AML and follow up investigations is presented in Fig. 15.15-2 – Standard diagnostic investigation of acute lymphatic leukemia. Cytogenetics, FISH and molecular genetics in leukemia. In leukemia, characteristic chromosomal aberrations, which define autonomous entities with typical morphology and characteristic clinical course, are recognized. They are also increasingly linked with direct therapeutic consequences. Thus, in the new WHO classification of acute myeloid leukemia, specific chromosomal changes have been incorporated as decisive classification criteria. Apart from classical cytogenetics – the analysis of chromosomal changes – fluorescence in situ hybridization (FISH) and, particularly, molecular techniques such as the polymerase chain reaction (PCR), increasingly and independently play an additional very important role. This applies to the diagnostic investigation and classification of leukemia, as well as to the determination of the therapeutic response (minimal residual disease, MRD). With regard to cytogenetics and molecular genetics in leukemia, see the review literature /7, 9/. Refer to Fig. 15.15-1 – Methods for the diagnostic investigation of acute leukemia Tab. 15.15-5 – Laboratory findings in acute myeloid leukemia 15.15.1.2 Acute lymphocytic leukemia (ALL) Cytomorphology, cytochemistry (Pappenheim, myeloperoxidase, previously also PAS) and especially immunophenotyping, chromosome analysis and molecular genetics are responsible for the primary diagnostic investigation of ALL (Fig. 15.15-2 – Standard diagnostic investigation of acute lymphocytic leukemia). The therapy is essentially influenced by the findings from immunophenotyping and cytogenetics/ molecular genetics. Apart from the distinction between B and T cell lineages, many additional aspects are to be considered. Above and beyond that numerous chromosomal aberrations, which are of prognostic and therapeutic relevance, have been described. On the one hand ALL is classified based upon karyotype into so-called ploidy groups (i.e., according to the number of chromosomes) and on the other hand the classification is made according to structural aberrations. The most common translocation in adult ALL is that of t(9;22)(q34;q11). It is associated with an unfavorable prognosis and necessitates targeted therapy. With the introduction of (e.g., Imatinib) a specific tyrosine kinase inhibitor, the detection of the Philadelphia chromosome and its molecular correlate, the BCR-ABL rearrangement, has taken on a high level of therapeutic relevance. In a small proportion of ALL, so-called crytic BCR-ABL rearrangements, which are not detectable with chromosome analysis, are present. Therefore, screening with FISH or RT-PCR should be performed in all ALL B-lineage cases, in view of the important therapeutic consequences /7, 10/. Within the framework of scientific analyses, assessment of the therapeutic response is provided by real time PCR, either with leukemia-specific fusion transcripts or with patient-specific immunoglobulin or T-cell receptor rearrangements. Crucial progress in therapy control may be expected to emerge from these data. Initial results in children are very promising /11/. The classification of ALL according to purely morphological criteria, as was the case with the FAB classification (L1, L2, L3), no longer plays a role. If need be, a certain reminiscence is rendered by the fact that the subtype of mature B-ALL with t(8;14) usually corresponds to the so-called (FAB) L3 sub-type. The cells then usually manifest a very dark blue cytoplasm and a many vacuoles. However, in this case as well, the morphology alone cannot be considered to be sufficient, due to the immense therapeutic consequences. It absolutely must be supplemented by immunophenotyping and cytogenetics or molecular genetics. 15.15.2 Significance of the bone marrow histology in acute leukemia In contrast to the significance of the bone marrow histology in chronic leukemia and in the staging of lymphoma, its relevance in acute leukemia is, rather, secondary. Individual aspects speak, apart from the previously described methods of smear cytomorphology and cytochemistry, immunophenotyping and cytogenetics and molecular genetics, for the supplementary implementation of a bone marrow punch in: Punctio sicca and leukopenia or a lack of blasts in the peripheral blood Marked fibrosis (e.g., AML-M7) or packed marrow, which then similarly leads to aspiration punctio sicca Determination of angiogenesis is corresponding therapeutic studies with angiogenesis inhibitors Differential diagnosis of severe aplastic anemia and hypo cellular MDS, or in suspicion of bone marrow tumor cell infiltration Staging and suspected diagnosis of lymphoma, in order to be able to capture the bone marrow involvement with certitude. In this case the bone marrow punch, including immunohistology, and immunophenotyping from the bone marrow aspirate, are complementary but not always concordant, so that the most correct assertion can only be made with the use of both methods. Therefore, in the diagnosis of acute leukemia, it is imperative to consider a bone marrow biopsy in every individual case. Admittedly, this can certainly usually be omitted but in certain cases, where necessary, it should be performed at the same time as the other samplings, in order to avoid a second procedure. For further current literature on the diagnostic investigation of leukemia, see Ref. /12, 13/. References Swerdlow SH, Campo E, Poleri SA, Harris NL, Stein H, Siebert L, et al., The 2016 revision of the World health Organization classification of lymphoid neoplasms. Blood 2016; 127: 2375–90. Arber DA, Orazi A, Hasserjian R, Thiele J, Borowitz MJ, Le Beau M, et al. The 2016 revision of the World Health Organization (WHO) classification of myeloid neoplasms and acute leukemia. Blood 2016. doi: 10.1182/blood2016-03-643544. Bennett JM, Catovsky D, Daniel MT, et al. Proposed evised criteria for the classification of acute myeloid leukemia. A report of the French-American-British Cooperative Group. Ann Intern Med 1985; 103: 620–5. Bennett JM, Catovsky D, Daniel MT, et al. Criteria for the diagnosis of acute leukemia of megakaryocyte lineage (M7). Ann Intern Med 1985; 103: 460–2. Bennett JM, Catovsky D, Daniel MT, et al. Proposal for the recognition of minimally differentiated acute myeloid leukemia (AML-M0). Br J Haematol 1991; 78: 325–9. Béné M-C, Castoldi G, Knapp W, et al. Proposals for the immunological classification of acute leukemias. European Group for the Immu-nological Characterization of Leukemias (EGIL). Leukemia 1995; 9: 1783–6. Löffler H, Haferlach T, Schoch C. WHO-Klassifikation der akuten myeloischen Leukämien (AML) und der myelodysplastischen Syndrome (MDS) – Hämatologische Erkrankungen – Ein diagnostisches Handbuch. DMW 2002; 127: 447–5. Heidelberg; Springer; 2010. Goasguen JE, Matsuo T, Cox C, Bennett JM. Evaluation of the dysmyelopoiesis in 336 patients with de novo acute myeloid leukemia: Major importance of dysgranulopoiesis for remission and survival. Leukemia 1992; 6: 520–5. Haferlach T. (Hrsg.), Ludwig W-D, Haferlach T, Schoch C. Labordiagnostik bei Leukämien und Lymphomen. Classification of acute leukemias. Bremen, Unimed 2007. Schoch C, Schnittger S, Bursch S, et al. Comparison of chromosome banding analysis, interphase- and hypermetaphase-FISH, qualitative and quantitative PCR for diagnosis and for follow-up in chronic myeloid leukemia: A study on 350 cases. Leukemia 2002; 16: 53–9. van Dongen JJM, Seriu T, Panzer-Grünmayer ER, et al. Prognostic value of minimal residual disease in acute lymphoblastic leukaemia in childhood. Lancet 1998; 352: 1731–8. Löffler H, Rastetter J, Haferlach T. Atlas der klinischen Hämatologie. Heidelberg; Springer; 2004. Theml H, Diem H, Haferlach T. Taschenatlas der Hämatologiel. Stuttgart; Thieme, 2002. Journal article. A patient with hypokalemia and hypoxemia – what is the culprit? Clin Chem 2023; 69 (11): 1220–5. 15.16 Myelodysplastic syndrome Torsten Haferlach The myelodysplastic syndromes (MDS) are myeloid neoplasms characterized by clonal proliferation of hematopoietic stem cells, recurrent genetic abnormalities, myelodysplasia, ineffective hematopoiesis, peripheral-blood cytopenia, and high risk of evolution to acute myeloid leukemia /1/. The MDS arises within the context of genetic changes in hematopoietic stem cells. These induce ineffective hematopoiesis, often with cytopenia in the peripheral blood. Clinically, symptoms of anemia (ineffective erythropoiesis), increased susceptibility to infection (granulocytopenia), and hemorrhagic diathesis (thrombocytopenia) emerge. MDS are a group of diseases that develop in advanced age; the median age at the time of occurrence is 69 years. The annual incidence at over 70 years of age is 30 : 100,000. 15.16.1 Diagnostic investigation of MDS The diagnosis of MDS is currently made based upon cytomorphological test of bone marrow and peripheral blood. The objective of the diagnosis is the differentiation of MDS from other clonal myeloid diseases, such as AML, and also from paroxysmal nocturnal hemoglobinuria (PNH), from severe aplastic anemia, and from reactive and other benign changes which may be associated with dysplastic hematopoiesis. Apart from cytomorphology, cytogenetics, which not only confirms the presence of a clonal disease in the case of an aberrant karyotype, but also is of considerable prognostic value and is acknowledged in the new WHO classification, is also of central diagnostic value /1/. Up until now, however, it has been necessary to apply various morphological classifications to MDS. Thus, up until recently, the FAB classification, which is now being extended and, in a stepwise manner, replaced by the WHO classification, was considered to be valid /1, 2/. The WHO classification of MDS is based, in the first place, mainly upon cytomorphological criteria: (i)dysplasia of granulopoiesis and/or erythropoiesis and/or megakaryopoiesis, (ii) the demonstration of ring sideroblasts, (iii) the relative and absolute number of monocytes, and (iv) the fraction of myeloid blasts in the blood and/or the bone marrow. For proof of dysplasia in MDS, only 10% of the cells have to manifest the dysplasia criteria /3/ mentioned in: Section 15.15 – Acute leukemias Tab. 15.16-1 – WHO classification of myelodysplastic syndrome since 2008. In the current WHO proposal for the classification of MDS the RAEB-T category, with a bone marrow blast fraction of ≥ 20–30%, is assigned to AML (Tab. 15.16-2 – International Prognostic Scoring System: grouping). The RA and RARS entities are maintained, while the RAEB and CMML entities are further subdivided according to their bone marrow blast fraction. Furthermore, the category of refractory cytopenia with multi lineage dysplasia (RCDM), and a subgroup of unclassifiable MDS, have been introduced. It remains to be seen whether this further subdivision of MDS – which is, again, based upon purely morphological criteria – is improved and changed very predictably with molecular markers, and whether new clinical relevance will have to be defined and tested /2/. It is, however, certainly meaningful to list the 5q-syndrome entity as a distinct category, since it can be clearly differentiated, clinically and genetically, from all other MDS subgroups, and has a better prognosis as well as a more specific therapeutic option (e.g., lenalidomide). Here, however, additional insights from molecular genetics (e.g., TP53 mutations in patients with 5q- syndrome) increasingly play a role /3/. Within the framework of the cytomorphological diagnostic investigation, at least 200 (500, according to WHO) bone marrow cells and 20 megakaryocytes should be evaluated; in MDS, signs of dysplasia should be demonstrable in at least 10% of the cells. Pseudo- Pelger neutrophils, peroxidase defects of segmented neutrophils, ring sideroblasts and micro megakaryocytes, and the proliferation of bone marrow blasts to 5–19%. These morphological changes partially correlate with the presence of clonal cytogenetic and molecular genetic markers; they also exhibit, however, considerable examiner dependence. It remains to be seen whether this further division of MDS is clinically relevant. It is, however, certainly meaningful to list the 5q- syndrome entity as a distinct category, since it can be clearly differentiated, clinically and genetically, from all other MDS subgroups, and is associated with a more favorable prognosis. This does not hold true for the 5q- syndrome which, with a clear genetic marker, is characterized by bone marrow blasts of below 5% and, often, a normal or even elevated peripheral blood thrombocyte count. In the morphological area, therefore, the evaluation of neutrophil hypo granulation should not remain the sole diagnostic criterion. In general terms, it is often difficult to diagnose early stage refractory anemia (RA) with cytopenia and dysplasia with only one lineage, and the monitoring of disease progression at intervals of 2–3 months prior to making a final diagnosis of MDS. Similarly, the recently introduced entities of RN for pure neutropenia and RT for pure thrombocytopenia which, along with RA are low listed as RCUD (refractory cytopenia with one lineage dysplasia), should be assessed. With reference to the differentiation of hypoplastic MDS and aplastic anemia, it must be taken into consideration that signs of dysplasia in erythropoiesis also occur in the latter, and, therefore, in contrast to dysplasia signs in other lines and bone marrow blast cell proliferation, are of no diagnostic value. The demonstration of cytogenetic changes certainly speaks rather for the presence of MDS, than for aplastic anemia, but proves neither the one nor the other. PNH should also be included in the differential diagnosis. Histology, including immunohistology, is strongly recommended for the differentiation of these entities. Apart from cytomorphology and cytogenetics, multi parametric flow cytometry is increasingly finding its way into the diagnostic work-up in suspected MDS. Thus, for the different lineages of granulopoiesis, monocytopoiesis and erythropoiesis, this method features the possibility of qualitatively assessing signs of dysplasia in the form of aberrant antigen expression patterns, and of quantifying the blasts. The correlation with the cytomorphological findings is, consequently, high. In addition, valuable diagnostic information, of prognostic relevance, can be gathered in cases that are difficult to classify cytomorphologically. 15.16.2 Recognition of prognostic factors in the diagnosis of MDS In can generally be said that the prognosis in patients with MDS is reduced in comparison with the general population, particularly in younger patients and in cases of high-risk MDS. The currently most meaningful and widely used system for the estimation of prognosis is the International Prognostic Scoring System (IPSS) /4/, which is based on the biomarker of bone marrow blast cells, karyotype changes and number of cytopenias. Refer to: Tab. 15.16-3 – International Prognostic Scoring System: prognostic classification Tab. 15.16-4 – Characteristic cytogenetic alterations in MDS, in comparison with AML. The scoring system is based upon 816 patients with MDS, the large majority of which was not previously treated, so that the spontaneous course could be estimated. The biological insights into the fundamental principles and the clinical course of MDS, and the parallels to acute myeloid leukemia (AML) that are present with regard to many aspects have, at least in young and high-risk patients, lead to the use of AML-typical therapies and partially limit, thereby, the applicability of the IPSS score. New insights from cytogenetics have to be incorporated /5, 6/. The consideration of the requirement for transfusion for the prognosis estimation is proposed – WPSS /7/. The closeness of the correlation between the two diseases of MDS and AML is shown by the genetic aberrations (Tab. 15.16-4 – Characteristic cytogenetic alterations in MDS, in comparison with AML). The table shows that in the diagnostic investigation of MDS, while cytomorphology is of substantial and fundamental relevance, the question of prognosis and the distinct biological entity can certainly be answered with other parameters, particularly cytogenetic parameters /8/. In this context, the IPSS score has already pointed out the right direction. Thus, it holds true for MDS as well as for acute leukemia and chronic myeloproliferative syndromes, that only by means of a combination of cytomorphology /5, 9/ and cytogenetics, possibly informative at present, supplemented with immunophenotyping and molecular genetics, can the diagnosis be made. The WHO classification of 2008 /1/ will, therefore, have to be extended promptly in order to show, beyond the classification, its clinical and prognostic relevance in prospective studies. The therapy that is recommended following diagnosis is more clearly focused, away from an often only supportive, and at times even nihilistic approach, with substances having biologically different modes of action, such as lenalidomide or azacitidine; it can only be employed, in individual patients and in a targeted manner, along with the use of the laboratory methods mentioned in the foregoing. References Cazzola M. Myelodysplastic syndromes. N Engl J. Med 2020; 383: 1358–74. Bejar R, Stevenson K, Abdel-Wahab O, et al. Clinical effect of point mutations in myelodysplastic syndromes. N Engl.J Med 2011; 364: 2496–506. Jadersten M, Saft L, Smith A et al. TP53 mutations in low-risk myelodysplastic syndromes with del(5q) predict disease progression. J Clin Oncol 2011; 29: 1971–9. Greenberg P, Cox C, Le Beau MM et al. International Scoring System for evaluating prognosis in myelodysplastic syndromes. Blood 1997; 89: 2079–88. Haase D, Germing U, Schanz J et al. New insights into the prognostic impact of the karyotype in MDS and correlation with subtypes: evidence from a core dataset of 2124 patients. Blood 2007; 110: 4385–95. Schanz J, Steidl C, Fonatsch C et al. Coalesced multicentric analysis of 2,351 patients with myelodysplastic syndromes indicates an under-estimation of poor-risk cytogenetics of myelodysplastic syndromes in the international prognostic scoring system. J Clin Oncol 2011; 29: 1963–70. Malcovati L, Germing U, Kuendgen A et al. Time-dependent prognostic scoring system for predicting survival and leukemic evolution in myelodysplastic syndromes. J Clin. Oncol. 2007; 25 (23): 3503–3510. Haferlach C, Bacher U, Haferlach T et al. The inv(3)(q21q26)/t(3;3)(q21;q26) is frequently accompanied by alterations of the RUNX1, KRAS and NRAS and NF1 genes and mediates adverse prognosis both in MDS and in AML: a study in 39 cases of MDS or AML. Leu-kemia 2011; 25: 874–7. Haferlach T. The molecular pathology of myelodysplastic syndrome. Pathobiology 2018; 23: 1-6. 15.17 Myoproliferative neoplasms Torsten Haferlach Under the umbrella term of chronic myeloproliferative neoplasms (MNP), the following entities have, in the narrower sense, been consolidated up to the present time: Polycythemia vera (PV) Essential thrombocythemia (ET) Primary myelofibrosis (PMF) Chronic myeloid leukemia (CML). The new WHO classification of 2008 has added the following entities /1/: Chronic eosinophilic leukemia (not specified more precisely) Chronic neutrophilic leukemia and mastocytosis. MPN’s occur sporadically; they are thereby traced back to acquired and not congenital clonal genetic changes. In all cases, it is presumed that the starting cells are pluripotent hematopoietic stem cells. By means of the stepwise elucidation of a clinically, morphologically, cytogenetically and molecular genetically well-defined clinical picture, new possibilities have now become available, especially in the diagnostic investigation and therapy of CML and, since 2005, for JAK2-mutated MPN as well /2, 3/. All the more important, first of all, is, therefore, the differentiation within MPN and the definition of BCR-ABL1-positive CML, based upon morphological, clinical and, particularly, diagnostic laboratory findings. Refer to: Tab. 15.17-1 – Clinical characteristics and orienting markers in the differential diagnosis of MPN Tab. 15.17-2 – Methods at the time of diagnosis in chronic myeloproliferative neoplasia. However, overlapping traits, and phenotypes of the other MPN’s that to blend into one another, are also seen clinically since, in all of these, JAK2 mutations are observed. Furthermore, in the diagnosis and differential diagnosis of MPN, bone marrow histology is, in addition to the cytogenetic and molecular genetic diagnostic investigation of major importance (Fig. 15.17-1 – Algorithm for molecular genetic testing in myeloproliferative neoplasia). With the exception of ET, MPN manifests an almost complete displacement of the fatty bone marrow. Specifically, in the diagnostic investigation of MPN, the points listed in the Tables are to be taken into consideration: Tab. 15.17-1 – Clinical characteristics and orienting markers in the differential diagnosis of MPN. Tab. 15.17-2 – Methods at the time of diagnosis in chronic myeloproliferative neoplasia. 15.17.1 Chronic myeloid leukemia Within the MPN group, chronic myeloid leukemia (CML) is, cytogenetically, the best-studied form. The presence of a Philadephia translocation distinguishes CML unambiguously from all other myoproliferative diseases. As early as 1960 Nowell and Hungerford demonstrated, for the very first time, a tumor-specific chromosomal change in patients with CML /4/. At that time, they discovered a small marker chromosome which was later named Philadelphia chromosome. Following the introduction of banding techniques in cytogenetics, it was possible to identify Philadelphia chromosome as a shortened chromosome 22. It was shown that shortening of chromosome 22 occurs due to a translocation between the long arms of chromosomes 9 and 22: t(9;22)(q34; q11) /5/. The development of molecular genetic procedures permitted the identification of the gene involved: ABL1 on the long arm of chromosome 9 in chromosome band 9q34, and the so-called breakpoint cluster region (BCR) on the long arm of chromosome 22 in chromosome band 22q11. In this way, translocation t(9;22)(q34;q11) leads, at the molecular level, to the formation of two leukemia-specific hybrid genes: BCR-ABL1 on derivative chromosome 22, and ABL1-BCR, on derivative chromosome 9. The BCR-ABL1 gene codes for a chimeric protein, which manifests elevated tyrosine kinase activity in comparison with normal ABL ; this plays a determining role in the pathogenesis of CML /6/. Consequently, the diagnosis of CML is dependent upon the clinical picture and the blood and bone marrow findings, but also absolutely requires the demonstration of the pathognomonic BCR-ABL1 fusion transcript. This applies particularly since the introduction of therapy with tyrosine kinase inhibitors /7/. Patients with CML and a cytogenetically normal karyotype (ca. 5%) manifest, both with fluorescence in situ hybridization (FISH) as well as with RT-PCR, a BCR-ABL1 rearrangement. Using metaphase-FISH, the BCR-ABL1 fusion gene can be demonstrated either on chromosome 22 or less often on chromosome 9. This form of CML is designated Philadelphia-negative, BCR-ABL1-positive CML, but this is irrelevant clinically. However, it leads to the suggestion that chromosome analysis alone is not sufficient in the diagnosis of CML. Cytomorphology and histology Of all of the MPN forms, the most severe leukocytosis (up to 700 × 109/L) occurs in CML. There occurs, thereby, a left shift, through to the blast cells (usually below 5%), in both peripheral blood and bone marrow aspirate samples. The bone marrow is hyper cellular, and exhibits a massive increase in granulopoiesis in comparison with erythropoiesis (up to a ratio of 20: 1; normal is 3 : 1). In addition, eosinophilia and, in particular and relatively pathognomonically, basophilia are found. In all MPN, but in particular in CML, glycolipid-storing cells, so-called pseudo-Gaucher cells, and sea blue histiocytes, are found in many cases, due to increased cell turnover in the bone marrow. In CML, fibrosis of the bone marrow is seldom seen at the time of diagnosis. The diagnostic investigation of CML, at first diagnosis and during the disease course, should follow certain algorithms which can then govern treatment /8/. Cytomorphological testing of the blood and bone marrow, as well as bone marrow histology, are generally considered to be obligatory with regard to the primary diagnosis. In addition, chromosome analysis (optimally on bone marrow), and FISH and PCR analyses for BCR-ABL1. PCR can be performed quantitatively and serves, in cases of a favorable therapeutic response, as a sensitive marker of disease progression. It is recommended that, in the further course, evaluations take place at intervals of three months. Classical cytogenetic testing should also be performed again later on an annual basis. The reason is that under tyrosine kinase inhibitor treatment as well as, occasionally, under interferon therapy, previously observed Philadelphia-independent cytogenetic changes such as trisomy 8 or monosomy 7 occur /9/. Their relevance and influence on disease progression is unclear, but is still being further validated. Furthermore, TKI resistance due to mutations at the site of action of the substances has been observed. In suspected cases, mutation analysis should be performed, as this can make possible a modification of treatment. Following allogenic bone marrow transplantation, the renewed demonstration of the BCR-ABL1 fusion transcript by means of quantitative PCR, which is predictive of a relapse, is clinically relevant. CML phases in relation to the clinical course With the use of various therapeutic modalities (e.g., TKI, interferon and allogenic transplantation) the CML clinical course scenarios have changed in comparison with what they were previously. Patient survival has been substantially prolonged. Nonetheless, it remains meaningful, from the clinical-morphological point of view, to use the 3 phases of CML for orientation (Tab. 15.17-3 – WHO classification of CML based upon the morphological findings) /1/. This division has also been retained in the new WHO classification, and allows for clear patient assignment. It can never, however, be permitted to take on therapeutic consequences on its own. Rather, classical cytogenetics, FISH, PCR and quantitative PCR, as well as mutation analysis, are to be performed in order to make the correct diagnosis and, in particular, to guide treatment in CML patients /8/. 15.17.2 Polycythemia vera Polycythemia vera (PV) is one of the Philadelphia chromosome negative classical myeloproliferative neoplasms, a group of disorders that also encompasses primary myelofibrosis and essential thrombocythemia. PV is distinguished by an augmented erythrocyte mass resulting from autonomous myeloproliferation. The concentration of erythropoietin is depressed. Cytomorphology and histology In Polycythemia vera (PV), the blood count shows a substantial elevation of the hematocrit to over 50%, the hemoglobin level is 16–22 g/dL, and moderate leukocytosis and thrombocytosis are observed. Bone marrow cytology in PV shows hyper cellularity as well as uniform proliferation of all three cell lineages. On iron staining, iron reserve, which is incorporated into the numerous erythrocytes, is lacking. Histologically, one sees an elevated megakaryocyte number with giant forms and cluster formation, as well as augmented granulopoiesis and erythropoiesis with deficient storage iron, hyperplasia of the sinus system, and varying degrees of fibrosis as well as osteopenia. In PV, the bone marrow findings are normal in only 10% of the cases. The diagnostic criteria are listed in Tab. 15.17-4 – Diagnostic criteria of polycythemia vera. The diagnosis of PV may not, however, be made without molecular analyses of JAK2 V617F mutations (in 95% of PV cases); reactive polyglobulia can, thereby, be ruled out in almost all cases /1, 2, 3/ (Tab. 15.17-4 and Fig. 15.17-1 – Algorithm for molecular genetic testing in myeloproliferative neoplasia). Patients with pure polyglobulia occasionally manifest a JAK2 exon 12 mutation /10/. Differential diagnostically, however, hypoxia caused by cardiac or pulmonary factors, erythropoietin-producing tumors, elevated androgens or hyperplasia of the red cells due to nicotine abuse must still be ruled out in patients without JAK2 mutations. However, a bone marrow puncture is no longer obligatory for the diagnosis of PV, since the biomarkers can usually also be tested using peripheral blood. The implementation of cytogenetic testing in PV is recommended, according to the WHO classification. The incidence of chromosome aberrations increases with the persistence of the disease. It is higher in patients who are receiving myelosuppressive therapy. In these cases it is impossible to say, however, whether this reflects an influence of the therapy or, rather, is simply related to the fact that patients with progressive disease are more often treated with myelosuppresive agents. The transformation of the disease into MDS or AML is, likewise, associated with a higher rate of karyotype changes. Overall, it therefore seems that the demonstration of chromosome aberrations is associated with poor prognosis. 15.17.3 Primary myelofibrosis Cytomorphology and histology In primary myelofibrosis (PMF), the peripheral blood findings are uncharacteristic, usually anemia with reticulocytosis as well as functional splenectomy, Howell-Jolly bodies, and teardrop erythrocytes. In severe bone marrow fibrosis with extramedullary hematopoiesis, normoblasts also appear in the peripheral blood. The leukocyte and thrombocyte counts do not manifest unequivocal changes and, not uncommonly, numbers that are rather below normal or even increased are present at the time of diagnosis. Due to disturbed thrombocyte function, bleeding time is sometimes prolonged. In PMF, the bone marrow cytology is often not evaluable on account of the severe fibrosis and the consequent punctio sicca. If this is, nonetheless, required and imposed in addition to the histology, it can be attempted to make touch imprints of a biopsy punch. The histological picture shows fibrosis of varying severity (MF 0–3) with the optional occurrence of woven bone formation, inflammatory marrow changes with lymphocyte infiltrates, erythrocyte extravasates, plasmacytosis and inflammatory vascular changes with enlarged sinusoides, vessel wall sclerosis, and intrasinusoidal hematopoiesis. In addition, clusters of atypical megakaryocytes and megaloblastoid erythropoiesis are found. At diagnosis, blast cells are usually not increased. During the course of the disease, cytopenia increases, the extent of the splenomegaly may become very great, transformation to acute leukemia is observed. Cytogenetically, a deletion in the long arm of chromosome 20(20q-) is the most common chromosome aberration in patients with PMF. It is observed, like the deletion in the long arm of chromosome 13, in 6–7% of the patients. Further karyotype alterations that have been described include numerical changes in chromosomes 7, 8 and 9, as well as structural aberrations of 1q and 5q /1/. It is recommended that chromosome analysis be performed; in punctio sicca, a biopsy punch can also be processed in physiological saline containing heparin for further analysis. Apart from that, molecular markers should definitely be determined: in the first place, JAK2 and MPLW515 (Tab. 15.17-5 – Diagnosis criteria for primary myelofibrosis). Newer, targeted therapies have recently been made available /11/. 15.17.4 Essential thrombocythemia Cytomorphology and histology The cardinal symptom in essential thrombocythemia (ET) is marked thrombocytosis, which can reach as high as 5 × 1010/L. According to WHO, ≥ 450 × 109/L are required (Tab. 15.17-6 – Diagnostic criteria of essential thrombocythemia). In the smear, giant thrombocytes and thrombocyte aggregates are observed. Bleeding time can be normal or shortened but, due to disturbances of thrombocyte function, may also be prolonged. Potassium is released from the thrombocytes, leading to hyperkalemia. In the bone marrow, in the presence of normal granulopoiesis and erythropoiesis, ET manifests diagnosis-determining megakaryocyte clusters, which are often located around the central-intermediate sinus. The diagnosis of ET is, at times, a diagnosis of exclusion. It, as well, necessitates a step-wise algorithm of molecular diagnostic investigation (Fig. 15.17-1 – Algorithm for molecular genetic testing in myeloproliferative neoplasia). Only some 5% of ET patients have clonal karyotype abnormalities. In these cases, a gain of chromosome 9 is most frequently observed. The performance of a complete chromosome analysis seems, therefore, not to be a requirement; it is, nonetheless, meaningful within the framework of an initial diagnosis in suspected MPN. 15.17.5 Chronic eosinophilic leukemia, not otherwise specified According to the WHO classification of 2008, chronic eosinophilic leukemia (not otherwise specified) is also listed in the chapter on MPN. It is classified here because, in contrast, diseases with eosinophilia and evidence of molecular changes in the FIP1L1-PDGFRA, FGFR1, or PDGFRA and PDGFRB genes are allocated to a separate chapter /1, 12/. Thus, following exclusion of the molecular changes or even specific cytogenetic findings, this CEL, which cannot be classified in a more precise manner, remains as part of the MPN chapter. It is defined as eosinophilia in the blood (1.5 × 109/L), less than 20% blast cells, and no chromosome alterations. 15.17.6 Mastocytosis Mastocytosis is also listed in the MPN chapter of the WHO classification of 2008. The following are distinguished: cutaneous mastocytosis (CM), indolent systemic mastocytosis (ISM), systemic mastocytosis with an associated clonal hematologic non-mast cell lineage disease (SM-AHNMD), aggressive systemic mastocytosis (ASM), mast cell leukemia (MCL), mast cell sarcoma (MSC), and extra cutaneous mastocytoma. Further detail with regard to the complex diagnostic criteria for the individual subgroups is not addressed /13/. Among others, various dermatological investigations are required, as are laboratory biomarkers (e.g., measurement of serum tryptase), immunophenotyping, or histology and immunohistology. In addition, gene mutations are to be investigated such as the analysis of KIT (usually D816V) mutations. References Swerdlow SH, Campo E, Harris NL, Jaffe ES, Pileri SA, Stein H, Thiele J, Vardiman JW: WHO Classification of Tumours of Haematopoietic and Lymphoid Tissues. 4th. 2008. Lyon, International Agency for Research on Cancer (IARC). James C, Ugo V, Le Couedic JP et al. A unique clonal JAK2 mutation leading to constitutive signalling causes polycythaemia vera. Nature 2005; 434 (7037): 1144–1148. Kralovics R, Passamonti F, Buser AS et al. A gain-of-function mutation of JAK2 in myeloproliferative disorders. N Engl J Med 2005; 352 (17): 1779–1790. Nowell PC, Hungerford DA. A minute chromosome in human granulocytic leukemia. Science 1960; 132: 1497. Rowley JD. A new consistent chromosomal abnormality in chronic myelogenous leukaemia identified by quinacrine fluorescence and Giemsa staining. Nature 1973; 243: 290–3. Heisterkamp N, Stephenson JR, Groffen J, Hansen PF, de Klein A, Bartram CR u. Grosveld G. Localization of the c-abl oncogene adjacent to a translocation breakpoint in chronic myelocytic leukaemia. Nature 1983; 306: 239–42. Druker BJ, Talpaz M, Resta DJ, Peng B, Buchdunger E, Ford JM, et al. Efficacy and safety of a specific inhibitor of the BCR-ABL tyrosine kinase in chronic myeloid leukemia. N Engl J Med 2001; 344: 1031–7. Baccarani M, Cortes J, Pane F et al. Chronic myeloid leukemia: an update of concepts and management recommendations of European Leukemia Net. J Clin Oncol 2009; 27: 6041–51. Schoch C, Haferlach T, Kern W, et al. Occurrence of additional chromosome aberrations in chronic myeloid leukemia patients treated with imatinib mesylate. Leukemia 2003; 17: 461–3. Scott LM, Tong W, Levine RL et al. JAK2 exon 12 mutations in polycythemia vera and idiopathic erythrocytosis. N Engl J Med 2007; 356: 459–68. Verstovsek S, Kantarjian H, Mesa RA et al. Safety and efficacy of INCB018424, a JAK1 and JAK2 inhibitor, in myelofibrosis. N Engl J Med 2010; 363: 1117–27. Cools J, de Angelo DJ, Gotlib J, et al. A tyrosine kinase created by fusion of the PDGFRA and FIP1L1 genes as a therapeutic target of imatinib in idiopathic hypereosinophilic syndrome. N Engl J Med 2003; 348: 1201–14. Horny HP, Sotlar K, Valent P. Mastocytosis: state of the art. Pathobiology 2007; 74 (2): 121–132. 15.18 Immunophenotyping of acute leukemia and non-Hodgkin lymphoma Richard Schabath, Wolf-Dieter Ludwig The diagnosis and classification of acute leukemia (AL) and non-Hodgkin lymphoma (NHL) is oriented to the classification of the World Health Organization (WHO) for tumors of the hematopoietic and lymphatic tissue /1/. In this document, acute myeloid leukemia (AML) and acute lymphocytic leukemia (ALL) of precursor B and precursor T cells are differentiated from mature B, T and NK cell neoplasms. In the differential diagnostic differentiation of the subtypes of AL and NHL, morphological and histological features, immunophenotype, and genotype are taken into consideration. The analysis of these cell biological features of tumor cells, commonly implemented in parallel, has made the identification of new subtypes or entities, whose clinical picture and therapeutic influenceability are different, substantially easier. In addition, understanding of the mechanisms that are pathogenetically relevant to leukemia genesis and lymphoma genesis could be improved by means of the characterization of genetic changes in leukemia and lymphoma cells /2, 3, 4, 5/. Immunophenotyping in bone marrow and/or peripheral blood is an essential component in the initial diagnostic investigation and the course of AL and leukemic NHL. It is based upon the demonstration of different antigens using monoclonal antibodies (MAB), which are expressed by precursor and/or mature myelopoietic and lymphopoietic cells, often as a function of their stage of maturation /1, 4, 6, 7, 8/. The binding of MAB’s to membrane-bound or intracellular antigens can be analyzed using different methods. Thereby, the labeling of cell suspensions with immunofluorescence techniques and their analysis using multi parametric flow cytometry, as well as immunoenzymatic staining, are considered to be routine methods /8, 9, 10, 11, 12, 40/. The advantages of flow cytometric immunophenotyping are: Rapid analysis of the samples in spite of high cell counts (over 106 cells in a single sample) High sensitivity of detection Simultaneous analysis of various biomarkers such as 2–10 fluorescent and light scattering properties of the cells Exact quantification of the results Statistical data analysis. For recommendations, indications, standardization and quality assurance of immunophenotyping in AL and NHL by multiparametric flow cytometry and immunoenzymatic procedures, see Section 52.2 and references /4, 7, 9, 10, 11, 40/. 15.18.1 Immunological classification of acute leukemia The essential objectives of the immunophenotyping of acute leukemia are: To classify morphologically and cytochemically undifferentiated leukemia of B cell, T cell, NK cell and myeloid cell lineage, as well as to determine the stage of maturity of the leukemia cells To identify biologically and/or prognostically relevant subtypes and to diagnose them in a standardized, study-conform manner To identify the expression of proteins that are involved in the regulation of important cell biology functions, (e.g.; adhesion, proliferation, differentiation, and apoptosis) Detection of target structures for targeted therapy (e.g., with MAB’s against CD20, CD33 or CD52) To make possible the treatment of non-eliminated residual leukemia cells (minimal residual disease) by means of its identification. Fig. 15.18-1 – Flow diagram for immunophenotyping in acute leukemia illustrates the stepwise approach in the immunological classification of AL /4, 20/. Starting from the morphological diagnosis of AL, which is always based upon the light microscopic evaluation of pan-optically stained smears, an unequivocal diagnosis and definition of the subtype is made by means of lineage assignment of the leukemia blast cells and the determination of the immunological subtype and stage of maturity of the AL. Lineage assignment of the leukemia blast cells The lineage assignment of the leukemia blast cells occurs based upon the detection of membrane-bound or cytoplasmic antigens that are expressed by immature lymphoid or myeloid progenitor cells. Particularly relevant to the diagnostic investigation of AL are the cytoplasmic antigens that are already expressed in very immature cells in a lineage specific manner /13, 14/: CD13, CD33, myeloperoxidase (MPO) and lysozyme in myeloid cells CD19, cyCD22 and cyCD79a in B lymphoid progenitor cells cyCD3 in T lymphoid progenitor cells. These antigens, which can be detected immunoenzymatically or following fixation and permeabilization of leukemia cell suspensions as well as with flow cytometry /15/, are expressed to a certain extent, including membrane-bound, by mature leukemia or lymphoma cells (CD3, CD22, CD79a). Determination of the immunological subtype and stage of maturity The determination of the immunological subtype and stage of maturity of AL using MAB’s against antigens whose expression is limited to immature lympho-hematopoietic progenitor cells, or is associated with different stages of differentiation of lymphopoiesis or myelopoiesis. Based upon their expression pattern, the antigens that are important for immunophenotyping of AL are subdivided into: Lineage-specific traits: (i) MPO for the myeloid cell lineage, (ii) cy/m for immunoglobulins, (iii) CD22, CD79a for B and T cell receptors α/β and γ/δ, (iv) CD3 for the T cell line Panmyeloid antigens (e.g., CD13, CD33, CD65) Pan-B (e.g., CD19, CD22, CD79a) Pan-T antigens (e.g., CD3, CD2, CD5, CD7) Lineage-associated antigens (e.g., CD14 for monocytic, CD15 for granulocytic, CD235a (Glycophorin A) for erythrocytic and CD41 or CD61 for megakaryocytic cells) Progenitor cell-associated antigens, such as CD10, CD34, CD117, HLA-DR, terminal deoxynucleotidyl transferase (TdT). The analysis of the antigens listed in Fig. 15.18-1 – Flow diagram for immunophenotyping in acute leukemia allows for unambiguous lineage assignment and subtype definition in almost all cases /20/. Tab. 15.18-1 – Basic panel for the initial diagnosis in acute leukemia shows the basic panel for the initial diagnostic investigation of AL, recommended by the German competence network for acute and chronic leukemia. 15.18.1.1 Acute undifferentiated leukemia Morphological or cytochemical acute undifferentiated leukemia (AUL) in which, with immunophenotyping, not more than one antigen of myeloid or B or T lymphoid cell lineage is detectable, is only diagnosed very infrequently (less than 1% of all AL) /1, 16, 17/. AUL, also designated morphologically as stem cell leukemia, expresses progenitor cell-associated antigens (CD34, CD38, CD117, HLA-DR, TdT) exclusively. The definitive assignment of AUL, important for the therapeutic course of action, necessitates additional analyses such as, ultrastructural demonstration of MPO or platelet peroxidase /18, 19/ or of cytogenetic or molecular genetic abnormalities /1, 4/ which are characteristic of AML or ALL. These complex analyses are not, however, routinely performed within the framework of the initial leukemia diagnostic investigation. 15.18.1.2 Immunophenotyping of ALL The identification and assignment of ALL to B or T cell lineage is easily possible, based upon the cyCD3, cyCD22, cyCD79a antigens that are first expressed intra-cytoplasmatically in lymphoid progenitor cells /14, 20, 21/ as well as the membrane antigens CD7 and CD19. These antigens can be detected in immature precursor B or T cells, and in over 99% of the corresponding immunological subtypes of ALL. For the recognition of immature T-ALL subtypes, the simultaneous analysis of CD7 and cyCD3 is important, since some 10–20% of immature AML cells express CD7 /4/, while cyCD3 is considered to be a specific marker of the T cell series /13/. The precise characterization of the immunological subtype, or the determination of the stage of maturity of ALL, is accomplished by the analysis of additional B cell and T cell-associated antigens (Fig. 15.18-1 – Flow diagram for immunophenotyping in acute leukemia). The characterization allows differentiation of the following subtypes /4, 21/: B progenitor cell ALL B cell ALL/Burkitt’s lymphoma T progenitor cell ALL The characteristics of these subtypes are shown in Tab. 15.18-2 – Terminology, frequency and phenotype of the immunological ALL subtypes. In 5–40% of ALL patients, the blast cells co express myeloid antigens (my + ALL). This is associated particularly with immature subtypes (pro-B ALL, pro-T ALL or pre-T ALL) /26/. The differentiation of the subtypes is important for the definition of risk groups and assignment to the different therapy forms in the German multi-center ALL therapeutic trials in children (ALL-BFM) and adults (GMALL): Pro-B-ALL (high-risk therapy in the GMALL study) B-ALL (own protocol in the GMALL and ALL-BFM studies) T-progenitor cell ALL subtypes are treated variously (Tab. 15.18-2 – Terminology, frequency and phenotype of the immunological ALL subtypes). 15.18.1.3 Immunophenotyping of AML In AML, immunophenotyping has not achieved the status that it has attained in ALL. Within the framework of classification, immunophentyping is necessary mainly for the identification of: Minimally differentiated AML, subtype M0: MPO cytochemically negative, expression of pan myeloid antigens and/or demonstration of MPO using MAB’s in the absence of lineage-specific B cell or T cell characteristics /19/. Acute megakaryoblastic leukemia, subtype M7: membrane-bound, infrequently solely intracytoplasmic expression of CD41 and/or CD61 /18/. In 75–90% of all AML patients, leukemia cells express the antigens CD13, CD33 and CD65; however, all three pan myeloid markers are detectable in only approximately 55% of the patients, while at least one of these antigens is found in more than 98% of the cases /22/. MPO can be detected using MAB’s in just 90% of AML patients. In addition, CD117, the receptor for the stem cell factor (c-kit protooncogene), which is expressed by 1–4% of normal bone marrow cells and some 60–70% of AML cells, but only rarely by immature ALL cells, has turned out to be a valuable marker for the immunological characterization of AML /23/. By means of the simultaneous analysis of these antigens and the use of MAB’s against erythroids (e.g., glycophorin A) platelet-associated antigens such as CD41 and CD61, as well as MPO, close to 100% of AML cases can be identified with immunological cell markers, and differentiated from ALL. For the differentiation of AML and ALL, the demonstration of TdT is hardly relevant, since 10–40% of AML cases, depending upon the method used (immunofluorescence or immunocytochemistry), express the enzyme TdT, which was previously considered to be a lymphoid marker. It is possible to distinguish between immature myeloid and granulocytic differentiated leukemia cells by means of combined intracytoplasmic demonstration of MPO and lactoferrin (LF), where undifferentiated myeloid cells are MPO-positive and LF-negative, while granulocytic differentiated leukemia cells express both MPO and LF /14/. Due to the lack of availability of monoblast-specific MAB’s it is, as a rule, not possible to immunologically differentiate immature monoblastic leukemia (FAB M5a) from other subtypes of immature AML (e.g., FAB M0/M1). In AML, as in ALL, aberrant expression of blast cell lymphoid markers occurs. In 10–25% of the patients, co-expression of T-lymphoid antigens, especially CD4, CD7 and/or CD2, can be demonstrated, while B-lymphoid antigens including CD10 (below 10%) are only expressed infrequently /25/. Due to the heterogeneity of antigen expression in AML, with the exception of the FAB-M0 and FAB-M7 subtypes, it is not possible to establish, with certainty, a correlation of the immunophenotype with morphologically or cytochemically defined FAB subtypes, or with cytogenetically and molecular biologically defined WHO subgroups of AML /22, 24/. A correlation can only provide an initial clue with regard to the AML subtype, until the cytogenetic and molecular biology results are made available (Tab. 15.18-3 – Correlation of the FAB classification, cytogenetics and immunophenotype). 15.18.1.4 Mixed phenotype acute leukemia AL subtypes in which the pathological blast cell population simultaneously expresses myeloid and lymphoid antigens (aberrant antigen expression) are diagnosed with increasing frequency (up to 5% of all AL). These subtypes were initially classified as biphenotypic acute leukemia (BAL) /20, 36/. With the publication of the WHO classification of 2008, an internationally accepted nomenclature for this heterogeneous subgroup of AL now exists /1/. The mixed phenotype acute leukemia (MPAL) immunological classification score is based upon this classification /20/ and has replaced the EGIL recommendations (Tab. 15.18-4 – Score for the definition of mixed phenotype acute leukemia). However, with regard to MPAL prognosis and therapy, it is mainly the underlying cytogenetic or molecular biological aberration and not the immunophenotype that is critical /36, 37/. 15.18.1.5 Detection of residual leukemia cells in AL with immunophenotyping In AL, the demonstration of residual leukemia cells during and following the initial course of chemotherapy has acquired increasing relevance with regard to subsequent therapeutic planning and prognostic assessment of the disease. Up to now, the assessment of remission has been based upon morphological evaluation of the bone marrow. Due to the low sensitivity of morphology (detection limit 10–2 i.e., 1 leukemia cell per 100 normal cells) more sensitive methods for the detection of minimal residual disease, such as immunophenotyping (detection limit 10–3–10–5) and molecular genetic analyses (detection limit of the polymerase chain reaction 10–4–10–5) have been employed /7, 27, 28, 38/. Detection of residual leukemia cells with immunophenotyping is based on the expression of leukemia-associated immunophenotypes, which can be depicted using multiparametric flow cytometry in the great majority of AL cases. The following are considered to be important biomarkers with regard to the distinction between leukemic and normal progenitor cells: Aberrant or asynchronous antigen expression Lack of expression of differentiation antigens Altered antigen density on leukemia cells. Apart from the rapidity of the test, the possibility of quantifying the number of residual leukemia cells and of determining their vitality are considered to be advantages of immunophenotyping by multiparametric flow cytometry for the demonstration of minimal residual disease. An important prerequisite for the unambiguous identification of residual leukemia cells with immunophenotyping is the precise characterization of the leukemic blasts at the time of diagnosis by multiparametric flow cytometry, with which the expression of 3–10 antigens per cell can also be simultaneously assessed, along with light scattering properties. Single-color or dual-color analyses of leukemia cells are not sufficient for the characterization of leukemia-specific traits and should, therefore, no longer be used for the diagnosis of minimal residual disease. Examples of suitable antigen combinations for the diagnosis of minimal residual disease in AL patients, and data on the frequency of occurrence of various aberrant or asynchronous phenotypes in leukemia subtypes, are shown in Tab. 15.18-5 – Antigen combinations for the diagnosis of minimal residual disease in patients with acute lymphatic leukemia. Based upon evidence for these combinations of antigens, it is possible to identify residual leukemia cells in a sample containing 10,000 normal hematopoietic progenitor cells (detection limit 10–4). In ALL, the clinical significance of immunophenotyping and molecular biological demonstration of residual leukemia cells at different time points during the course of treatment could be demonstrated in numerous studies and in some cases in prospective, multicenter studies /27, 38/. In AML as well, sufficient clinical data now exist to justify the routine use of immunophenotyping for the diagnosis of MRD /28/. However, for the detection of minimal residual disease there is considerable competition between flow cytometry and longer established molecular biology methods (e.g.,quantitative PCR). 15.18.1.6 Immunophenotyping of NHL Immunophenotyping also plays a decisive role in the diagnostic investigation of leukemic NHL. Its objectives, apart from the differentiation of mature lymphoid neoplasms from AL, are: Assignment of the malignant cells to B, T or NK cell lineages Proof or exclusion of clonality of malignant B cells using kappa or lambda light chain restriction Differentiation of mature lymphoid neoplasms, especially mature T cell neoplasms, from reactive conditions (e.g., EBV or Cytomegalovirus infections or toxoplasmosis) Monitoring of responsiveness to therapy (chemotherapy, monoclonal antibodies) by means of the early identification of residual leukemia or lymphoma cells (minimal residual disease). The assignment of mature B, T and NK cell neoplasms to the corresponding cell lineage or to a certain maturity stage, based upon a panel of different MAB’s, the composition of which is influenced to a considerable extent by the question under consideration (e.g., initial diagnostic investigation or demonstration of minimal residual disease) /7/. The immunophenotype expression profiles of normal B and T cells and of mature B, T and NK cell neoplasms that are depicted in: Tab. 15.18-6 – Immunological marker profile of normal B cells and chronic lymphoproliferative diseases of the B cell type Tab. 15.18-7 – Immunological marker profile of normal T cells and T cell type lymphoproliferative diseases are not applicable in all cases. There is, at times, overlapping of immunophenotyping with various subtypes, which allows for definitive assignment only along with additional evaluation findings, such as clinical, morphological, histological and genotype data /1, 7, 11, 29, 30, 31/. Scores have been proposed which, based upon a characteristic expression profile of membrane-bound antigens, significantly improve the precision of the differentiation between B-CLL and other mature B cell neoplasms, particularly for the differentiation of chronic lymphocytic leukemia of B cell lineage from other mature B cell neoplasms. In 90–95% of the cases, typical B-CLL can be distinguished from other mature B cell neoplasms by means of analysis of five antigens which are expressed on leukemia cells in B-CLL either not at all (CD 79b, FMC7), weakly (membrane-bound immunoglobulins), or strongly (CD5, CD23) /31/. In this regard, an additional useful marker is CD200, which is usually expressed by B-CLL cells, while mantel cell lymphoma which, as CD5-positive NHL can be problematic with regard to differentiation from B-CLL, is CD200 negative /39/. Studies of mutations in the variable portions of the immunoglobulin heavy chain gene (IgVH) have led to the recognition of two different cell biological subtypes of B-CLL which, with regard to clinical course, immunophenotype, and cytogenetic and molecular genetic findings, are clearly different /32/. Membrane-bound (CD38) and intracytoplasmic (ZAP-70) antigens are expressed differently by these two subtypes and are therefore increasingly used as diagnostic and prognostic biomarkers. CD38 or ZAP-70 positivity in B-CLL cells usually correlates with lack of mutation of the IgVH gene and an unfavorable clinical course /33, 34/. In mature B cell neoplasms as well, immunophenotyping by multiparametric flow cytometry is extremely sensitive of residual leukemia cells (detection limit 10–4) /35/. In site of increasingly precise information regarding the immunophenotype of mature B, T and NK cell neoplasms, as well as further methodical development, unambiguous assignment to the entities defined in the WHO classification is, based upon bone marrow or peripheral blood immunophenotyping, often not possible in leukemic mature lymphoid neoplasia /1/. 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The reliability and specificity of c-kit for the diagnosis of acute myeloid leukemias and undifferentiated leukemias. Blood 1998; 92: 596–99. Stasi R, Taylor CG, Venditti A, Del Poeta G, Aronica G, Bastianelli C, Simone MD, et al. Contribution of immunophenotype and genotypic analyses to the diagnosis of acute leukemia. Ann Hematol 1995; 71: 13–27. Drexler HG, Thiel E, Ludwig WD. Review of the incidence and clinical relevance of myeloid antigen-positive acute lymphoblastic leukemia. Leukemia 1991; 5: 637–45. Drexler HG, Thiel E, Ludwig WD. Acute myeloid leukemias expressing lymphoid-associated antigens: Diagnostic incidence and prognostic significance. Leukemia 1993; 7: 489–98. Campana D, Coustan-Smith E. Advances in the immunological monitoring of childhood acute lymphoblastic leukaemia. Best Practice & Research Clinical Haematology 2002; 15: 1–19. Kern W, Haferlach C, Haferlach T, Schnittger S. Monitoring of minimal residual disease in acute myeloid leukemia. 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Gerr H, Zimmermann M, Schrappe M, Dworzak M, Ludwig WD, Bradtke J, et al. Acute leukaemias of ambiguous lineage in children: characterization, prognosis and therapy recommendations. Br J Haematol 2010; 149: 84–92. Ratei R, Basso G, Dworzak M, Gaipa G, Veltroni M, Rhein P, et al. AIEOP-BFM-FCM-MRD-Study Group. Monitoring treatment response of childhood precursor B-cell acute lymphoblastic leukemia in the AIEOP-BFM-ALL 2000 protocol with multiparameter flow cytometry: predictive impact of early blast reduction on the remission status after induction. Leukemia. 2009; 23: 528–34. Palumbo GA, Parrinello N, Fargione G, Cardillo K, Chiarenza A, Berretta S, et al. CD200 expression may help in differential diagnosis between mantle cell lymphoma and B-cell chronic lymphocytic leukemia. Leukemia Research 2009; 33: 1212–6. Béné MC, Nebe T, Bettelheim P, Buldini B, Bumbea H, Kern W, et al. Immunophenotyping of acute leukemia and lymphoproliferative disorders: a consensus proposal of the European Leukemia Net Work Package 10. Leukemia 2011; 25: 567–74. 15.18.2 Myelodysplastic syndrome (MDS) MDS is a heterogene myeloid clonal disease originating from hematopoietic stem cells, characterized by ineffective hematopoiesis, refractory hemocytopenia, and high risk of transformation to acute myeloid leukemia. Table 15.1-1 Life span and daily turnover of blood cells in the circulation /1/ \| Blood cell \| Life span \| Turnover/ 24 hours \| \| Erythrocyte \| 120 days \| 2.0 × 1011 \| \| Reticulocyte \| 24 hours \| 2.0 × 1011 \| \| PMN1) \| 21 hours \| 1.0 × 1011 \| \| Eosinophil \| 6–18 hours \| \| \| Basophil \| 8 hours \| \| \| Monocyte \| 14 hours \| 8.4 × 109 \| \| Thrombocyte \| 10 days \| 1.0 × 1011 \| 1) Life span in tissue 4-5 days, PMN polymorphonuclear neutrophil granulocyte Table 15.1-2 Hematological tests for assessment of the hematopoieses /27, 28/ \| Comment \| \| Hemoglobin concentration (Hb value) The measurement of the hemoglobin concentration is the most reliable (precision, validity) and best standardized hematology test. The Hb value is an effective parameter for assessing the organism’s erythrocyte mass and thus also of the oxygen-transporting capacity of the blood, although it is only an indirect measure of red blood cell mass. \| \| Erythrocyte count The erythrocyte count, also known as the red blood cell count (RBC), is an indicator of the red blood cell mass of the organism. It is not, however, a good diagnostic parameter for the identification of a decrease in cell mass in anemia or an increase caused by polycythemia, since changes in the erythrocyte volume are not taken into consideration. \| \| MCV The mean corpuscular volume (MCV) is a red cell index of the size of the peripheral red blood cells. Hematology analyzers measure the MCV directly. However, since all red blood cells (including reticulocytes and, possibly, normoblasts) are measured, the MCV cannot automatically be equated with normocytosis, microcytosis or macrocytosis, without knowledge of the erythrocyte distribution width. \| \| MCH The mean corpuscular Hb content of the red blood cells (mean corpuscular hemoglobin, MCH) is a red cell index that expresses, in pg, the mean Hb content of all of the red blood cells. MCH is a measure that remains constant over the life span of the erythrocyte. The MCH determines the volume of the erythrocyte during maturation of the progenitor cells. \| \| %HYPO The Hb concentration in individual red blood cells is measured, and the proportion with a concentration of below 280 g/L is expressed as percentage of the total number. The %HYPO is a direct measure of the iron demand of erythropoiesis and can only be determined with certain hematology analyzers. \| \| MCHC The mean cellular hemoglobin concentration (MCHC) is a red cell index which expresses the Hb concentration of the circulating red blood cells in g/L. Changes in the MCHC are indicative of a disturbance in the relationship between the Hb content of the erythrocytes and their volume. Normochromic and hypochromic anemia are differentiated based upon the MCHC value. \| \| Hematocrit (HCT) The HCT, also termed packed red blood cell volume (PCV), is the product of the erythrocyte number × MCV. It is a redundant parameter in the diagnostic investigation of anemia. \| \| RDW The red blood cell distribution width (RDW) is calculated from the distribution histogram as the standard deviation or coefficient of variation of the MCV. In microcytosis, an elevated RDW is indicative of iron deficiency, and a normal RDW suggests the presence of heterozygous β-thalassemia. \| \| Reticulocyte count The reticulocyte count is an indicator of the erythropoietic effectiveness of the bone marrow. Erythropoiesis is differentiated into normo-, hypo- and hyper regenerative forms. \| \| Reticulocyte RNA Reticulocyte RNA content is an indicator of the extent of the stimulation of erythropoiesis. A large fraction of reticulocytes with high RNA content indicates intensive stimulation (e.g., following presence in high altitude after two days or after administration of erythropoiesis-stimulating agents). \| \| Reticulocyte hemoglobin content Mean reticulocyte Hb content (CHr, RetHe) is an early and dynamic indicator for the iron demand of erythropoiesis. In combination with the determination of %HYPO and MCH, a diagnostic pattern is available which differentiates short term iron demand of erythropoiesis (CHr decrease within 72 hours), middle term iron demand (%HYPO increase within 2 to 4 weeks) and longer term iron demand (MCH decrease within 3 to 6 months). The CHr and RetHe are effective markers for the diagnosis of iron-deficient erythropoiesis, a state where the iron demand of erythropoiesis is greater than its supply, independent of the stored iron content. \| \| Differentiated leukocyte count The differentiated leukocyte count, determined with a hematology analyzer in combination with the blood count, has gained acceptance based upon its acceptable precision and short analysis time. There are, nonetheless, limitations, and automated differentiation should be supplemented with the blood smear examination in symptomatic patients and if one or more blood cell count parameters are flagged by the hematology analyzer. \| \| Blood smear There are two relevant indications for a microscopic blood smear investigation: Verification of the hematology analyzer blood cell count, if this seems to be implausible The obtaining of further diagnostic information from blood cell morphology, particularly evidence for blood cells that are normally only detectable in the bone marrow. \| \| Thrombocyte count Indicator of the organism’s thrombocyte mass. \| \| MPV The mean platelet volume (MPV) is an indicator of stimulation of thrombopoiesis. With a low platelet count, an elevated MPV is indicative of peripheral platelet consumption, for (e.g., due to disseminated intravascular coagulation) while a reduced MPV is, rather, suggestive of an intrinsic disorder of megakaryopoiesis. \| Table 15.1-3 Differential diagnosis of pancytopenia /33/ \| Clinical and laboratory findings \| \| Aplastic Anemia Aplastic anemia is characterized by hypocellular bone marrow with an absence of myeloid forms. The term refers to an empty bone marrow. Congenital and acquired forms of aplastic anemia exist. Acquired forms are caused by toxins, infections or autoimmune diseases. \| \| Myelodysplastic Syndrome (MDS) MDS is a primary myeloid neoplasm with a risk of transformation to acute leukemia. Clonal proliferation of hematopoietic stem cells causes dysplasia of the bone marrow and peripheral blood cytopenias. For further information refer to Section 15.16 – Myelodysplastic syndrome. \| \| Systemic Diseases Increased concentration of tumor necrosis factor and interleukin-6 in infections and inflammations can cause suppression of the bone marrow that would cause reduced production of blood cells. \| \| Toxins, Medications Toxins and medications with myelosuppressive properties can cause low blood counts. Examples are illicit drugs like levamisole in cocaine, and excessive alcohol consumption. \| \| Nutritional Disorders Vitamin B12 and folic acid are required for DNA synthesis. In deficiencies arrest in the synthesis phase (s phase) of blood cell cycle occurs. Patients present with pancytopenia, magaloblastosis and elevated concentration of serum latate dehydrogenase. \| \| Sequestration and peripheral Destruction of Blood cells Peripheral destruction or sequestration of blood cells (sequestration in splenomegalia) can lead to pancytopenia. In most cases destruction of erythrocytes is predominant, immune thrombocytopenia and autoimmune thrombocytopenia can cause low thrombocyte and neutrophil counts, respectively. \| \| Myelophthisis Myelophthisis is the infiltration of bone marrow by non hematopoietic cells, resulting in disruption of normal hematopoiesis or the development of bone marrow fibrosis. Causes include solid tumors with metastasis to the bone marrow, granulomatous and lipid storage disorders. \| \| Acute MyeLoid Leukemia Patients with acute myeloid leukemia present with a pathologic cellular blood count: normochromic anemia, platelet count greatly diminished, and leukocyte count below 4,000/μl in 20–30% of cases. For further information refer to Section 15.15 – Acute leukemias. \| Table 15.1-4 Therapeutic measures that compromize hematopoiesis /38/ \| Clinical and laboratory findings \| \| CYtomegalovirus (CMV)-INFECTION Example: CMV infection is a common complication in patients undergoing hematopoietic stem cell transplantation. CMV infection can cause graft failure due to inhibition of the myelopoiesis. \| \| nosocomial infection Example: Nosocomial infections with multi-drug resistant organisms in intensive care units provide leukemoid neutrophilia and anemia. \| \| Voriconazole treatment Example: Voriconazole is used to treat progressive and potent life-threatening infections in immunocompromised patients. The adverse drug reactions related to voriconazole are varied. In some patients voriconazole can result in myelodysplastic syndrome-like adverse reactions. \| \| NEutropenia Example: Patients with neutropenia are at high risk of bacterial infections. Prolonged neutropenia increases the risk of fungal infections. \| Table 15.2-1 Erythrocyte reference intervals \| Adults /3/ \| ♀ 4.1–5.4 \| \| ♂ 4.4–5.9 \| \| Children /4, 5/ \| \| Fetus /6/ \| \| \| 1. day \| 4.3–6.3 \| 15. gestation wk \| 1.9–3.0 \| \| 0.5 month \| 3.9–5.9 \| 16. wk \| 2.2–3.2 \| \| 1 months \| 3.3–5.3 \| 17. wk \| 2.3–3.2 \| \| 2 months \| 3.1–4.3 \| 18.–21. wk \| 2.6–3.6 \| \| 4 months \| 3.5–5.1 \| 22.–25. wk \| 2.4–3.8 \| \| 6 months \| 3.9–5.5 \| 26.–29. wk \| 2.7–4.3 \| \| 9–12 months \| 4.0–5.3 \| > 30. gestation wk \| 2.5–5.1 \| \| 1.5–3.0 yrs \| 3.7–5.3 \| \| \| \| 4–9 yrs \| 3.9–5.1 \| \| \| \| 10–12 yrs \| 4.1–5.2 \| \| \| \| 13–16 yrs \| ♀ 4.0–5.0 \| \| ♂ 4.3–5.6 \| Values expressed in 106/μL or 1012/L, Values are 2.5th and 97.5th percentiles Table 15.2-2 Reference intervals of erythrocyte indices \| MCV (fL) \| MCH (pg) \| MCHC (g/L) \| RDW (%) \| \| Adults /2/ \| \| \| \| \| 80–96 \| 28–33 \| 330–360 \| < 15 \| \| %HYPO /13/ \| 1–5% \| \| \| \| Age \| MCV (fL) \| MCH (pg/cell) \| MCHC (g/L) \| \| Children /5, 14/ \| \| \| \| \| Umbilical cord \| 101–125 \| 33–41 \| 310–350 \| \| 1 day \| 98–122 \| 33–41 \| 310–350 \| \| 2–6 days \| 94–135 \| 29–41 \| 240–360 \| \| 14–23 days \| 84–128 \| 26–38 \| 260–340 \| \| 24–37 days \| 82–126 \| 26–38 \| 250–340 \| \| 40–50 days \| 81–125 \| 25–37 \| 260–340 \| \| 2.0–2.5 months \| 81–121 \| 24–36 \| 260–340 \| \| 3.0–3.5 months \| 77–113 \| 23–36 \| 260–340 \| \| 5–7 months \| 73–109 \| 21–33 \| 260–340 \| \| 8–10 months \| 74–106 \| 21–33 \| 280–320 \| \| 11–13.5 months \| 74–102 \| 23–31 \| 280–320 \| \| 1.5–3.0 yrs \| 73–101 \| 23–31 \| 260–340 \| \| 4–12 yrs \| 77–89 \| 25–31 \| 320–360 \| \| 13–16 yrs \| 79–92 \| 26–32 \| 320–360 \| \| Children /3/ \| MCV (fL) \| \| \| \| Fetus \| \| \| \| \| Week 15 \| 127–159 \| \| \| \| Week 16 \| 119–167 \| \| \| \| Week 17 \| 121–153 \| \| \| \| Weeks 18–21 \| 119–143 \| \| \| \| Weeks 22–25 \| 109–141 \| \| \| \| Weeks 26–29 \| 103–134 \| \| \| \| Week ≥ 30 \| 97–132 \| \| \| Values are 2.5 th and 97.5 th% confidence interval with non parametric distribution Values expressed as x ± 2s Values are instrument-dependent Conversion: 1 × 10–15 L = 1 fL; pg × 0.062 = fmol; g/L × 0.062 = mmol/L Table 15.2-3 Classification of anemia based on MCV and RDW /18/ \| Microcytic isocytic \| \| Microcytic anisocytic \| \| Normocytic isocytic \| \| Normocytic anisocytic \| \| Macrocytic isocytic \| \| Macrocytic anisocytic \| \| \| MCV \| RDW \| MCV \| RDW \| MCV \| RDW \| MCV \| RDW \| MCV \| RDW \| MCV \| RDW \| \| Decrease \| Normal \| Decrease \| Elevated \| Normal \| Normal \| Normal \| Elevated \| Elevated \| Normal \| Elevated \| Elevated \| \| β-thalassemia minor \| \| Iron deficiency anemia \| \| Anemia of chronic disease \| \| Osteomyelofibrosis \| \| Aplastic anemia \| \| Pernicious anemia \| \| Table 15.2-4 Classification of anemia based on MCV, MCH and MCHC /15/ \| Erythrocyte indices \| Clinical and laboratory findings \| \| MCV normal MCH normal MCHC normal \| Normochromic normocytic anemia: Non regenerative anemia (e.g., chronic renal disease, chronic inflammatory disease, systemic infection, chronic liver disease, malignant tumor, endocrine disease, maldigestion, malabsorption) \| \| MCV normal MCH elevated MCHC elevated \| Apparently hyperchromic anemia due to pre-analytic or analytic interference: Intravascular hemolysis, in vitro hemolysis Hyperlipidemia, causes erroneously high hemoglobin concentration Heinz bodies in toxic hemolytic anemia, unstable hemoglobin, enzymopathy Laboratory errors due to erroneously low hematocrit or erroneously high hemoglobin measurement. \| \| MCV normal MCH reduced MCHC normal \| Incipient iron deficiency anemia. The RDW is usually above 15%, the %HYPO above 5% and the reticulocyte Hb content (Ret-He, CHr) is below 28 pg. See Section 15.6 – Reticulocyte count and reticulocyte indices. \| \| MCV reduced MCH reduced MCHC reduced \| Most common form of anemia. In 80–90% of the cases in Northern and Central Europe, classical iron deficiency anemia or anemia of chronic disease with iron-restricted erythropoiesis is present; in some 5% of the cases, β-thalassemia is the cause. Hereditary sideroblastic anemia is a rare cause. \| \| MCV reduced MCH normal MCHC elevated \| Severe hereditary spherocytosis, a hemolytic disease with an increase of the erythrocyte number, the hemoglobin concentration, and the hematocrit. Hereditary spherocytosis is not a hyperchromic anemia. \| \| MCV elevated MCH reduced MCHC reduced \| Regenerative anemia (e.g., observed during the first few days of adequate supplementation in anemias due to an underlying deficiency of iron, copper or vitamin B6). \| \| MCV elevated MCH normal MCHC normal/ low \| Folic acid or vitamin B12 deficiency anemia, liver cirrhosis, alcoholism Cancer patients under cytostatic therapy Myelodysplastic syndrome, hereditary stomatocytosis. \| \| MCV elevated MCH elevated MCHC elevated \| Presence of high titer cold agglutinins, that cause erythrocyte agglutination. The erythrocyte number is underestimated, and the MCV is overestimated. In consequence, the erythrocyte number and the hematocrit are erroneously low, and the MCH and MCHC calculations are erroneously high. \| Table 15.3-1 Formation of hemiglobincyanide from hemoglobin /2/ \| Hb (Fe2+) + [FeIII(CN)6]3– → Hi (Fe3+) + [FeII (CN)6]4– \| \| Hi (Fe3+) + CN– → Hi [FeCN]2+ \| Table 15.3-2 Hemoglobin reference intervals \| Fetus /3/ \| \| \| Week 15 \| 109 ± 7 \| \| Week 16 \| 125 ± 8 \| \| Week 17 \| 124 ± 9 \| \| Weeks 18–21 \| 117 ± 13 \| \| Weeks 22–25 \| 122 ± 16 \| \| Weeks 26–29 \| 129 ± 14 \| \| Weeks > 30 \| 136 ± 22 \| \| Children /5, 6/ \| \| \| Umbilical cord \| 135–207 \| \| 1. day \| 152–235 \| \| 2–6 days \| 150–240 \| \| 14–23 days \| 127–187 \| \| 24–37 days \| 103–179 \| \| 40–50 days \| 90–166 \| \| 2.0–2.5 months \| 92–150 \| \| 3.0–3.5 months \| 96–128 \| \| 5–7 months \| 101–129 \| \| 8–10 months \| 105–129 \| \| 11–13.5 months \| 107–113 \| \| 1.5–3 years \| 108–128 \| \| 5 years \| 111–143 \| \| 10 years \| 119–147 \| \| 12 years \| 118–150 \| \| 15 years \| 128–168 \| \| Adults /4/ \| \| \| ♀ \| 115–160 \| \| ♂ \| 135–178 \| Data expressed in g/L. Values are x ± 1 s Values are 2.5th and 97.5th percentiles Values are x ± 2 s. Conversion: – mmol/L = g/L × 0.0621; g/L = mmol/L × 16.1 – mmol/L = g/dL × 0.621; g/dL = mmol/L × 1.61 Table 15.3-3 Proposed low hemoglobin thresholds for white and black adults /9/ \| Group \| Age (years) \| Hb (g/L) \| \| White men \| 20–59 \| 137 \| \| ≥ 60 \| 132 \| \| White women \| 20–49 \| 122 \| \| ≥ 50 \| 122 \| \| Black men \| 20–59 \| 129 \| \| ≥ 60 \| 127 \| \| Black women \| 20–49 \| 115 \| \| ≥ 50 \| 115 \| Based on the Scripps-Kaiser data for the 5th percentile. Ref. 7 Tab. 2. NHANES data are considered to be confirmatory. Table 15.3-4 Proposed low hemoglobin thresholds for children in the USA /10/ \| Age (years) \| \| Hb (g/L) \| HCT (fraction) \| \| 1–1.9 \| \| 110 \| 0.330 \| \| 2–4.9 \| \| 112 \| 0.340 \| \| 5–7.9 \| \| 114 \| 0.345 \| \| 8–11.9 \| \| 116 \| 0.350 \| \| 12–14.9 \| ♀ \| 118 \| 0.355 \| \| ♂ \| 123 \| 0.370 \| \| 15–17.9 \| ♀ \| 120 \| 0.360 \| \| ♂ \| 126 \| 0.380 \| \| ≥ 18 \| ♀ \| 120 \| 0.360 \| \| ♂ \| 136 \| 0.410 \| Table 15.3-5 Correction for anemia during an extended stay at high altitude /10/ \| Altitude (feet) \| Hb (g/L) \| HCT (fraction) \| \| < 3,000 \| 0.0 \| 0.0 \| \| 3,000–3,999 \| + 2.0 \| + 0.005 \| \| 4,000–4,999 \| + 3.0 \| + 0.010 \| \| 5,000–5,999 \| + 5.0 \| + 0.015 \| \| 6,000–6,999 \| + 7.0 \| + 0.020 \| \| 7,000–7,999 \| + 10 \| + 0.030 \| \| 8,000–8,999 \| + 13 \| + 0.040 \| \| 9,000–9,999 \| + 16 \| + 0.050 \| \| ≥ 10,000 \| + 20 \| + 0.060 \| The Hb or HCT value must be added to the reference interval value. Table 15.3-6 Correction for anemia in smokers /10/ \| Smoker status \| Hb (g/L) \| HCT (fraction) \| \| Non-smokers \| 0.0 \| 0.0 \| \| Smokers (all) \| + 3.0 \| + 0.0010 \| \| 0.5–1 packs/day \| + 3.0 \| + 0.0010 \| \| 1–2 packs/day \| + 5.0 \| + 0.0015 \| \| > 2 packs/day \| + 7.0 \| + 0.0020 \| The Hb or HCT value must be added to the reference interval value. Table 15.3-7 Classification of anemia according to erythropoietic activity of the bone marrow \| Clinical and laboratory findings in hyporegenerative anemia \| \| Ineffective erythropoiesis may be present due to inadequately low erythropoietin synthesis or to intrinsic hypo proliferation of erythropoiesis; however, a combination of both causes usually occurs. Causes are: Deficiency states; iron, vitamin B12, folate, erythropoietin (chronic kidney disease), hormones (thyroid, hypophysis or adrenal gland) Disorder of proliferation and/or stem cell differentiation; chronic disease (inflammation, autoimmune disease, tumor, liver disease), toxic (alcohol, cytostatics), radiation, aplastic anemia Displacement of erythropoiesis; acute leukemia, myelodysplastic syndrome, myeloproliferative syndrome, malignant lymphoma, metastases of solid tumors in the bone marrow, storage diseases. Sarcopenia is a disease characterized by decreased skeletal muscle mass and function in elderly persons. There is an association between anemia and sarcopenia /80/. \| \| Clinical and laboratory findings in hyperregenerative anemia \| \| In these anemias the erythropoietic response corresponds to the extent of the anemia. The increase in erythropoietin secretion is adequate with respect to the fall in Hb and the erythropoietic tissue is healthy. The following may be the underlying cause: Therapy-associated regeneration of erythropoiesis (therapy with iron, folate or Vitamin B12) Subacute bleeding (maglinant tumor) Enhanced erythrocyte degradation; hypersplenism, hemolytic anemia ,direct toxic (e.g., malaria, Wilson’s disease), mechanical (e.g., artificial heart valves, vascular prostheses, disseminated intravascular coagulation) Erythrocyte membrane defects (e.g., hereditary spherocytosis, elliptocytosis, paroxysmal nocturnal hemoglobinuria) Hemoglobinopathies, enzyme defects \| Table 15.3-8 Blood cell status in certain groups of individuals and patients \| Clinical and laboratory findings \| \| Anemia in children In the first weeks of life, newborns undergo a decline in erythrocyte mass, which leads to a decrease in Hb value and in the hematocrit. In healthy infants aged 10–12 weeks, the Hb nadir seldom falls below 90 g/L. The fall is more marked in premature infants, in whom the nadir may be as low as 80 g/L or 70 g/L with birth weights of 1.0–1.5 kg and of below 1.0 kg, respectively. While full-term newborns can tolerate the fall in Hb and, therefore, one speaks of physiological anemia, premature infants may manifest symptoms that require blood transfusion or therapy with erythropoiesis-stimulating agents (ESA) /20/. The cause of the marked decline in Hb in premature newborns is iron deficiency, due to insufficient iron storage caused by the pre term birth, earlier postnatal activation of erythropoiesis, faster postnatal growth, and frequent blood sampling for diagnostic purposes. The reference intervals for the blood count and iron metabolism parameters in full term neonates are also valid for premature infants during the first year of life /21/. Hb concentration and hematocrit increase continuously with age. In the USA, the Centers for Disease Control defined the lower limits for Hb shown in Tab. 15.3-4 – Proposed low hemoglobin thresholds for children in the USA. \| \| Sports anemia /16/ Endurance athletes often show reduced levels of Hb and HCT. Sports anemia is caused mainly by an increase in plasma volume relative to the red blood cell mass causing hemodilution although erythropoiesis is stimulated. Acute heavy exercise decreases the arterial O2 saturation and changes renal hemodynamics; both are triggers for the erythropoietin production. \| \| Anemia in the elderly /22/ The mean prevalence of anemia in individuals over the age of 65 is 17%. Approximately 60% of individuals over the age of 65 have diseases that lead to diminished Hb levels. In the Cardiovascular Health Study cohort, individuals older than 65 years had a poorer clinical outcome with Hb values of below 137 g/L (men) or 126 g/L (women). Only 1% of elderly individuals who are not hospitalized have Hb concentrations of less than 100 g/L. The anemia may possibly be masked in some cases due to diuretics-associated hemoconcentration. According to the Chianti study, the prevalence of anemia is considerably higher in individuals with a creatinine clearance of below 30 [mL × min–1 × (1.73 m2)–1] /23/. Elderly individuals with low Hb values are at elevated risk of hospitalization and mortality, in comparison with those whose values are within the reference interval. This also holds for elderly individuals with mild anemia (women Hb 100–119 g/L, men 100–129 g/L). The three-year risks for hospitalization and mortality were associated with hazard ratios of 1.32 and 1.86, respectively /24/. One fourth of anemia cases in the elderly are not explained by nutrient deficiency (iron, vitamin B12, folic acid), renal insufficiency, chronic inflammation or myelodysplastic syndrome /25/. \| \| Gestational anemia Erythropoiesis undergoes substantial changes during pregnancy /26/. Starting from the 10th week of gestation, increases in blood volume by 30–40%, in plasma volume by 40–50%, and in erythrocyte mass by 20–30% occur. As of the 10th week of gestation this increase can be observed, with a peak between the 32nd and 34th week of gestation, thereafter it remains rather constant. WHO recommends that the Hb concentration should not fall below 110 g/L at any time during pregnancy and not below 100 g/L in the puerperium /8/. In the puerperium, Hb decreases by 8 g/L and the HCT by 0.10 relative to the levels before delivery, while the reticulocytes already increase during delivery. The Hb concentration and the hematocrit increase continuously from the 4th day postpartum onwards /27/. \| \| Vegetarian adults A variety of different definitions exist /76/: Vegetarians are individuals who consume red meat no more than once a month, consume fish and chicken no more than once a week. Someone who never consumes red meat and consumes chicken or fish no more than once a week. Lacto-ovo vegetarians and lacto-vegetarians eat no meat products, fish or poultry but consume both diary products and eggs or dairy products, respectively. Vegans exclude meat, fish, poultry, dairy products and eggs. Adult vegetarians, especially females in the reproductive age, have a high prevalence of depleted iron stores (ferritin < 15 ug/L) and iron deficiency anemia (Hb < 120 g/L) compared to nonvegetarians. In children on a vegetarian diet serum ferritin concentration was significantly lower in nonvegetarians, but the concentration of the soluble transferrin receptor (sTfR) was significantly higher and the serum hepcidin concentration significantly lower than in nonvegetarians /77/. \| \| Anemia in heart failure /16/ A high proportion of patients with chronic cardiac insufficiency have anemia and Hb values of 100–120 g/L. Anemia in this order of magnitude causes an increase in exercise cardiac output. This results in an increased preload wall stress and left ventricular work, which in turn increases oxygen consumption and accelerates myocyte loss. The hypoxia at the tissue level and decreased blood viscosity cause arterial vasodilatation, which decreases after load. A chronic volume overload state induced by anemia causes the addition and lengthening of myofibrils causing ventricular dilatation and an increase in wall tension (see Section 2.7 – Chronic heart failure). Anemia predicts mortality in severe heart failure . In the Prospective Randomized Amlodipine Survival Evaluation (PRAISE) the relationships between baseline hemoglobin/hematocrit and mortality in patients with left ventricular ejection fraction of less than 30% and a NYHA classification functional class IIIb and IV were evaluated over the course of 15 months. Patients in the lowest quintile (Hb 116 ± 9 g/L) had a 52% higher mortality risk (hazard ratio 1.52) than the patients in the highest quintile (Hb 162 ± 9 g/L) /28/. \| \| Anemia in the critically ill /29/ Critically ill patients in intensive care units often have anemia and Hb value ≤ 100 g/L. This is based upon the fact that these patients may have substantial changes in plasma volume (hyper hydration, dehydration), which render the assessment of the Hb concentration more difficult. About 62% of patients admitted to intensive care units have an Hb value of less than 120 g/L, and 29% below 100 g/L. Generally, normocytic, normochromic anemia is found, associated with hemorrhage (elevated reticulocytes), reduced erythrocyte formation due to inflammation (decreased reticulocytes), or hemolysis or sequestration of erythrocytes (increased reticulocytes). Reactive hemophagocytic syndrome may also be present as a result of an infectious trigger or secondary to malignant disease. The blood cells are phagocytized by histiocytes, resulting in pancytopenia. \| \| Postoperative anemia /31/ For the postoperative management and diagnosis of anemia in surgical patients the following diagnostic management is recommended: whenever assess the iron status within 24 h postoperatively, if it has not been already performed in the pre-operative assessment. Monitor hemoglobin (Hb) 3-4 days postoperatively. Iron treatment should be considered if postoperative ferritin is below 100 μg/L or below 300 μg/L and transferrin saturation below 20%, or reticulocyte hemoglobin content is below 28 pg. The calculation of total iron deficiency is as follows: total iron deficiency (mg) = (target Hb – actual Hb) × weight (kg) × 0.24. Hb value is measured in g/L. Add another 10 mg/kg for replenishing iron stores. \| \| Anemia of chronic kidney disease (ACD) Renal anemia is normocytic, normochromic, and hypo regenerative. See also Tab. 15.10-2 – Diseases and conditions associated with an adequate rise in EPO concentration. The reasons are diminished erythropoietin (EPO) synthesis in relation to the Hb level and the cumulation of toxic substances which are excreted normally in the urine. Both factors cause intrinsic hypo proliferation of erythropoiesis and are possibly also responsible for the slight reduction in the erythrocyte life span. The National Kidney Foundation Disease Outcomes Quality Initiative (KDOQI) recommends that all patients with a GFR of below 60 [mL × min–1 × (1.73 m2)–1] be evaluated for anemia. According to the results of the Third National Health and Nutrition Examination Survey (NHANES III) /30/, a Hb value of below 110 g/L was diagnosed in 42.2% of patients with an eGFR of less than 30 [mL × min–1 × (1.73 m2)–1], and in 3.5% of those with a GFR of 30–59 [mL × min–1 × (1.73 m2)–1]. A relevant cause is an inadequate response of EPO to the Hb decrease. This is not the case in polycystic kidney disease, in which the synthesis of EPO is maintained. In these patients it is important to evaluate iron metabolism. See Section 7.3 – Ferritin, Section 7.4 – Soluble transferrin receptor (sTfR), and Section 7.5 – Transferrin saturation (TfS). Treatment for patients undergoing hemodialysis consists of administration of ESA . Functioning iron metabolism is important in order to assure effective erythropoiesis. Therefore, prior to the start of treatment, ferritin level should be at least 100 μg/L (better 200–500 μg/L), TfS should be ≥ 20% (better 30–40%), %HYPO should be below 10% (better below 2.5%), and for monitoring of iron demand of erythropoiesis under ESA therapy, CHr (Ret-HE) is determined. An increase to ≥ 29 pg indicates no iron demand of erythropoiesis /32/. In a study a high-dose intravenous iron regimen administered pro actively (400 mg monthly unless the ferritin concentration was > 700 μg/L or the transferrin saturation was 40% or higher), resulted in lower dose of ESA being administered /33/. \| \| Anemia in diabetes Approximately one fifth of non-hospitalized diabetics has anemia, defined by Hb levels of below 120 g/L (women) and 130 g/L (men). According to one study /34/: Patients with a GFR above 60 [mL × min–1 × (1.73 m2)–1] and without albuminuria have higher Hb values than those with albuminuria Patients with macro albuminuria have lower Hb values than those with micro albuminuria Patients with a GFR below 60 [mL × min–1 × (1.73 m2)–1] and normo albuminuria have similar Hb values to those with a GFR of greater than 60 [mL × min–1 × (1.73 m2)–1] and macro albuminuria. In diabetics, therefore, the evaluation of anemia should be performed not only in impaired GFR, but also in the presence of albuminuria in patients with a normal GFR. Clinically, the anemia can emerge earlier than the renal insufficiency. The pathophysiology of anemia in diabetic nephropathy is multifactorial. Relevant factors are damage to the interstitial cells and the renal vascular architecture, with consequent interstitial fibrosis. The more marked this is, the less erythropoietin is produced. There is limited evidence to suggest that correction of anemia, specifically in patients with complications of diabetes, might be beneficial. However, while correction of anemia in patients with advanced kidney disease may stabilize left ventricular hypertrophy, earlier intervention may lead to regression. \| \| Alcoholism Alcoholism induced anemia is multifactorial. Important pathophysiological factors are: Toxic suppression of erythropoiesis due to inhibition of ribosomal and mitochondrial protein synthesis Megaloblastic changes due to deficient folic acid metabolism caused by alcohol Deficient vitamin B6 metabolism and, consequently, mitochondrial accumulation of iron in erythroblasts and formation of ring sideroblasts Changes in the erythrocyte membrane due to alcohol, resulting in hemolysis and reduced erythrocyte life span. Laboratory findings: the Hb level is 80–120 g/L, the mean MCV is 5–10% higher than in healthy controls, and the RDW manifests dimorphism. In the blood smear, acanthocytosis and stomatocytosis and ring sideroblasts can be present. \| \| Cancer Patients receiving ongoing chemotherapy who present with anemia (Hb below 110 g/L or Hb decrease more than 20 g/L from a baseline level below 120 g/L) and absolute iron deficiency (serum ferritin below 100 ug/L) should receive iron treatment with an i.v. iron preparation to correct iron deficiency. If ESA (erythropoietin stimulating agent) treatment is considered, iron treatment should be given before the initiation of ESA therapy in the case of functional iron deficiency (transferrin saturation < 20% and serum ferritin > 100 ug/l) /61/. Red blood cell (RBC) transfusion should be considered in patients with Hb < 70–80 g/L or severe anemia related symptoms (even at higher Hb levels) and the need for immediate and Hb symptom improvement, the administration of RBC transfusion without delay is justified /61/. Studies reporting the outcome of patients receiving transfusions during radical surgery for non-metastatic cancer report that RBC transfusions were associated with an increased risk of death (hazard ratio 1.36; 95% CI 1.26–1.46). The survival was reduced even in cancer at survival and increased the risk of relapse /74/. \| Table 15.3-9 Classification and differentiation of microcytic anemia \| Clinical and laboratory findings \| \| Microcytic anemia Depending on the hematology analyzer, an MCV below 80–83 fL is characteristic. It is always necessary to assess the MCV, in association with the erythrocyte distribution width, in order to rule out erythrocyte dimorphism or to assess erythrocyte morphology in the blood smear. If a low MCV has been confirmed it is differentiated, based upon the MCHC, whether hypochromia (MCHC below 320 g/L) or normochromia (MCHC ≥ 320 g/L) are present. Microcytic, hypochromic anemia can result from a defect in the globin genes (hemoglobinopathy or thalassemia), a defect in heme synthesis, or erythroid precursor cell demand for iron. Usually iron restricted erythropoiesis, which is due to nutritive iron deficiency or chronic or acute bleeding is the main cause /35/. \| \| Iron deficiency anemia In women up to the age of 50, menstrual bleeding, in men of the same age group nutritional iron deficiency, and in both genders above the age of 50 gastrointestinal bleeding due to tumors are most common causes of iron deficiency /36/. In children, reduced iron intake is the main cause of iron deficiency. Depending upon the extent and duration of iron deficiency, Hb, MCV, MCH and MCHC are reduced. The erythrocyte count is usually low-normal or only moderate decreased. The hypochromia usually predominates relative to the microcytosis /7/. RDW is increased to ≥ 15%. It takes some 8 weeks from the time of empty iron stores, with transition to iron-restricted erythropoiesis, until the occurrence of an elevated percentage of hypochromic erythrocytes. If the proportion of hypochromic erythrocytes (%HYPO) is determined, this time is reduced to 2–3 weeks, and with the determination of reticulocyte Hb content (CHr, RetHE), it is reduced still further, to some 4 days. Erythrocyte poikilocytosis (irregular shape) and anisocytosis (unequal size) are seen in the blood smear. In iron-restricted erythropoiesis, the %HYPO increases to ≥ 5%, and the CHr decreases to below 28 pg. Under effective oral iron therapy, the Hb value increases within 4 weeks to over 10 g/L, as do the CHr after one week. For further diagnostic investigation, see Chapter 7 – Iron metabolism and Section 15.2 – Erythrocytes (cell count and indices). In iron deficiency anemia, erythropoiesis is hypo regenerative to normo regenerative. \| \| β-thalassemia In β-thalassemia, the biochemical defect is the lack of synthesis of β-globin chains of the erythroid precursor cells (see Section 15.7 – Hemoglobinopathies). In heterozygous trait carriers (thalassemia minor), this leads to deficient erythrocyte hemoglobinization. In homozygous patients (thalassemia major), not only is there a lack of β-globin chains but also, erythropoiesis is impaired due to a relative excess of α-globin chains. These excessive chains, which cannot bind to HbA, precipitate in the erythroid precursor cell and leads to cell death and dyserythropoiesis /37/. Thalassemia major is associated with severe hypochromic anemia in childhood. In uncomplicated cases of thalassemia minor, Hb concentrations are 110–120 g/L. This is maintained by means of a high erythrocyte count of (5–7) × 1012/L. MCV is below 80 fL, usually even below 75 fL, and it can even decrease to 55 fL. MCH is significantly decreased, MCHC is greater than 310 g/L, RDW is normal since the microcytosis is uniform. Microcytosis predominates in relation to hypochromia and a %MICRO/%HYPO ratio above 0.9 with a hypochromic erythrocyte fraction ≥ 19% is indicative of heterozygous β-thalassemia /38/. This can be associated with mild reticulocytosis and elevated concentrations of soluble transferrin receptor. Hematological values in heterozygous and homozygous β-thalassemia are shown in Tab. 15.3-10 – Hematological data in patients with β-thalassemia. \| \| HbE syndrome After HbS, HbE is the second most common Hb variant. The gene occurs with a prevalence of some 10% in Southeast Asia. In the region of Cambodia, Laos and Thailand, the prevalence of HbE is 20–40%. HbE is a mutation that is due to the substitution of glutamine with lysine at position 26 of the β-globin chain. The cause is the exchange of guanine with adenine at codon 26 of the β-globin gene. Since, in addition, this mutation leads to a change in the splicing site, a decrease in functional β-globin mRNA occurs with, in consequence, reduced formation of β-chains, resulting in a milder thalassemia phenotype. There exist three HbE syndromes; common to all of them is microcytosis and a varying HbE proportion /39/: HbEE is the homozygous phenotype. This is a benign condition with no, or only minimal, anemia. The MCV is 20–25 fL lower than normal, and the life span of the erythrocytes is normal. The erythrocyte count is elevated, while the reticulocyte count is usually normal. In the blood smear, up to 75% target cells can be detected, on Hb electrophoresis no HbA is found, and HbE is detected only in combination with a low percentage of HbF. The heterozygous phenotype is HbAE. Individuals of this type are, for the most part, asymptomatic; anemia is not present. The erythrocyte count is slightly elevated, the MCV is 10–15 fL below normal. A few target cells are visible in the blood smear. Hb electrophoresis shows an HbE fraction of some 30%; the rest is HbA. HbE/β-thalassemia phenotype. Heterozygous HbE and β-thalassemia is present, with a clinical picture that is similar to that of β-thalassemia. The clinical symptomatology is variable; it can be associated with mild microcytic anemia and can, also, require transfusion. \| \| Hereditary spherocytosis (HS) /40/ HS is a hemolytic anemia in Caucasians that is often genetically determined, with an incidence of 1 in 2000–5000 births. Some 75% of the cases are inherited as an autosomal dominant disease, the rest are inherited in a recessive manner or a novel mutation may be involved. The molecular defects are very heterogeneous, and include genes that code for the erythrocyte proteins spectrin, ankyrin, and band 3. All of these proteins are involved in the structure of the red blood cell cytoskeleton, and a deficiency or dysfunction leads to the formation of spheroid, fragile and osmotically labile erythrocytes which are captured and destroyed by the spleen. The defective protein can be detected with SDS polyacrylamide gel electrophoresis. The spectrin deficiency phenotype is usually associated with severe anemia and more spherocytes in the smear than the band 3 deficiency phenotype. In one study /41/, HS manifested as mild, moderate and severe forms in 38%, 11%, and 9% of the cases, respectively. The membrane defect (spectrin, ankyrin, band 3) was not clearly associated with the severity of the HS. The diagnosis is usually already made in children, based upon anemia, icterus and splenomegaly; in adults, the disease is diagnosed on the basis of splenomegaly and bile stones. The severity of the anemia correlates with the extent of hemolysis and the splenomegaly. Splenectomy leads to reversal of the anemia. Laboratory findings: Hb low or normal, MCV low (particularly good to determine with hematology analyzers that use flow cytometric techniques), MCHC elevated, RDW increased, percentage of hyper dense red blood cells increased, reticulocytosis, elevated erythropoietin. Following splenectomy, normalization of Hb, reticulocytes and erythropoietin. HbS typically comprises 35% to 45% of the total hemoglobin, unless concurrent iron deficiency or alpha thalassemia is present Iron deficiency and alpha thalassemia reduce the percentage of hemoglobin comprised by HbS in sickle trait. \| \| Sideroblastic anemia /42/ Sideroblastic anemia is characterized by anemia with the emergence of ring sideroblasts in the bone marrow. Ring sideroblasts are erythroblasts characterized by iron accumulation in perinuclear mitochondria due to impaired iron utilization. There are two forms of sideroblastic anemia: The inherited sideroblastic anemia is a rare and heterogenous disease caused by mutations of genes involved in heme biosynthesis, iron-sulfur (Fe-S) cluster biogenesis, Fe-S cluster transport, and mitochondrial metabolism. The acquired sideroblastic anemia is relatively common and caused by drugs or alcohol; however, the best known acquired sideroblastic anemia is refractory anemia with ring sideroblasts (RARS), a subtype of myelodysplastic syndrome. \| \| X-linked sideroblastic anemia (XLSA) /42/ XLSA is the most common inherited sideroblastic anemia. A mutation in the gene ALAS2 of the hematopoiesis-specific aminolevulinate synthase (ALAS) occurs. The enzyme ALAS2 catalyzes the first step in heme synthesis, the formation of aminolevulinic acid from glycine and succinyl-CoA (see also Fig. 7.1-6 – First step of heme synthesis). Many mutations occur, most of which are missense mutations. Hemizygous men manifest microcytic anemia and iron overload. The disease can already be diagnosed in newborns, but it often only becomes clinically apparent in middle age. Sideroblastic anemia due to SLC25A38 gene mutation is the next most common inherited sideroblastic anemia. The iron-transferrin containing endosomes of the cell deliver iron directly to mitochondria. The SLC25A38 gene regulates the iron importer protein mitoferrin 1 (Mfrn1; SLC25A38), which is a member of the solute carrier family localized in the inner mitochondrial membrane, and is used for heme synthesis and iron-sulfur cluster biogenesis. Laboratory findings: the most important characteristics are microcytic, hypochromic anemia, often with Pappenheim bodies (iron-positive inclusions), erythrocyte dimorphism (one microcytic and one normocytic population), ring sideroblasts in bone marrow, in particular iron accumulation in orthochromatic erythroblasts, secondary iron overload due to ineffective erythropoiesis. The disease is manifested in all age groups; it is more common in men than in women, but the anemia is more severe in women. \| \| Iron refractory iron deficiency anemia (IRIDA) A new cause of hereditary anemia called IRIDA (OMIM 206200) has recently been described. IRIDA is due to a mutation in the TMPRSS6 gene encoding the membrane-bound serine protease matriptase-2. Matriptase-2 cleaves hemojuvelin, a major regulator of hepcidin expression and that this function is altered in this genetic form of anemia /71/. In contrast to low hepcidin levels, observed in acquired iron deficiency, in patients with matriptase-2 deficiency, serum hepcidin is inappropriately high for the low iron status and accounts for the absent/delayed response to oral iron treatment. Refer to Fig. 7.6-4 – Signals and pathways for the regulation of hepcidin expression by stimulating factors. The prevalence of this condition is not known, but it has certainly been under-diagnosed up to now. Clinically, IRIDA patients are chacterized by a hypochromic microcytic anemia. The clinical phenotype develops only after the neonatal period. Laboratory findings: hypochromic microcytic anemia, very low serum and transferrin saturation levels, ferritin concentrations are mostly within the reference range. Normal to high hepcidin levels characterize IRIDA. In inflammatory conditions a laboratory diagnostic pattern is analyzed similar to what is observed in IRIDA. However, the anemia is normo- or moderately microcytic as opposed to the marked microcytosis in IRIDA. \| \| Post-transplantation anemia (PTA) /43/ PTA (Hb in women below 120 g/L, in men below 130 g/L) is a common problem that may hinder patient’s quality of life. It occurs in 12–94% of patients, and is most common in the immediate post transplant period. Thus, in one study of 240 kidney transplant recipients, the prevalence of PTA was 76%; one year later it was 21%, and after 4 years it was 36% /44/. During the first 6 months, predominantly microcytic anemia was present. One cause of microcytosis is therapy with sirolimus. The extent of the anemia is dependent upon the chronic kidney disease (CKD) stage following transplantation. The anemia prevalence in CKD stages 1, 2, 3, 4 and 5 was, in each case, 0%, 3%, 7%, 27% and 33%. Patients who are disease-free after two or five years after hematopoietic cell transplantation (HCT) have a greater than 80% subsequent 10-year survival rate. Many studies show that HCT survivors suffer from significant late effects that adversely affect morbidity, mortality, working status, and quality of live. Laboratory tests for detection of late effects are TSH (hypothyroidism), glucose and HbA1c (diabetes), lipids (dyslipidemia), Cortisol-stimulation test (adrenal insufficiency), testosterone (gonadal dysfunction), FSH, LH (gonadal dysfunction), ferritin (iron overload), ALT (hepatitis) /69/. Post-transplantation lymphoproliferative disorders (PTLD) account for 21% of all cancers in recipients of solid organ transplants. In the adult population the incidences of recipients are: kidney transplants 0.8–2,5%, pancreatic transplants 0.5–5%, Liver transplants 1–5.5%, heart transplants 2–8%, lung transplants 3–10%, and multi organ and intestinal transplants ≤ 20%. PLTD is a consequence of therapeutic immunosuppression /70/ \| \| Hemolytic anemia Hemolysis occurring in hematologic diseases is often associated with an iron loading anemia. The iron overload is the result of entrance of free hemoglobin in the circulation. Massive hemolysis is confirmed by a complete decrease of haptoglobin and hemopexin, increased lactate dehydrogenase, an increase in red cell distribution width, a reduced half-life of red blood cells, an increase in ferroportin, and a decrease in hepcidin. Tissue iron overload derived from heme or hemoglobin is primarily localized in the liver and spleen macrophages rather than hepatocytes. Because of depressed hepcidin formation of the hepatocytes enteral iron absorption is increased. Serum ferritin is increased but transferrin saturation remains in the normal range. The absence of hepatocyte iron overload is a consequence of both the huge increase in erythroblast production and urinary iron losses /72/. \| Table 15.3-10 Hematological data in patients with β-thalassemia /37/ \| Parameter \| Normal \| Hetero- zygous \| Homo- zygous \| \| Hb (g/L) \| 110–150 \| 80–110 \| 40–70 \| \| MCH (pg) \| 27–33 \| 16–24 \| 16–24 \| \| MCV (fL) \| 75–90 \| 60–75 \| 60–75 \| \| Reticulocytes (%) \| 1–2 \| 1–2 \| 2–10 \| \| HbA (%) \| 96–98 \| 90–95 \| 0–20 \| \| HbF (%) \| 0–2 \| 1–5 \| 60–90 \| \| HbA2 (%) \| 1–3 \| 4–6 \| 0–10 \| \| Erythroblasts \| 0 \| 0 \| Many \| Table 15.3-11 Classification and differentiation of normocytic anemia \| Clinical and laboratory findings \| \| Normocytic normochromic anemia – Generally In normocytic normochromic anemia, MCV, MCH, and MCHC values are normal. These anemias are subdivided into hypo regenerative and hyper regenerative forms, corresponding to their regenerative erythropoietic activity. Major hyper regenerative anemias are posthemorrhagic anemia and hemolytic anemia. Corresponding to the regenerative erythropoietic activity, an increase in reticulocytes and in sTfR is to be expected. The pathophysiology of hypo regenerative normocytic normochromic anemias is different and is associated mainly with inflammation, malignancies, or diminished erythropoietin synthesis /7/. \| \| Hyper regenerative normocytic anemia Hyper regenerative anemias are due to: Erythrocyte destruction (hemolysis; erythrocyte life span is decreased) External or internal bleeding (e.g., hemorrhage or gastrointestinal bleeding) Treatment of deficiency anemia (e.g., during the regeneration phase following iron substitution or substitution with vitamin B12 or folate). Laboratory findings: characteristic findings are reticulocytosis and, in the blood smear, polychromasia of the erythrocytes. The regenerative responses (i.e., reticulocytosis and polychromasia) are more marked in hemolytic anemia than in anemia due to bleeding. In hemolytic anemia, this is due to recirculation of iron stored in macrophages and of factors exerting stimulatory effects on erythropoiesis /45/. \| \| Hemolytic anemia (HA) – Generally HA comprises some 5% of anemias and can have the causes listed in Tab. 15.3-12 – Causes of hemolytic anemia. Hemolysis can be associated with normocytic or microcytic anemia. The hemolysis can take place in the vascular system (intravascular hemolysis) or in the spleen (extravascular hemolysis) in the presence of splenomegaly. This pathological process is termed hypersplenism. Both types of hemolysis can cause hyperbilirubinemia. While reticulocytosis is an indicator of both extravascular and intravascular hemolysis, decreased haptoglobin (Hp) mainly occurs with intravascular hemolysis. The half-life time of Hp is 4 days, but the Hp-Hb complex survives for only a few minutes. Since, in hemolysis, the Hp catabolic rate is much greater than that of its synthesis, the Hp concentration is reduced. Intravascular and extravascular hemolysis often occur together, but one form predominates. Plasma total protein concentration does not change in HA /46/. HA is associated with hypercoagulability /47/. HA is hyperregenerative; on days 2–3 reticulocytosis of 15% is reached, and this can be as high as 50% after one week, particularly in immune mediated hemolysis. For further information on hemolytic anemia, see Sections LD, reticulocytes and haptoglobin as well. Hereditary HA is essentially caused by an erythrocyte membrane disorder (hereditary spherocytosis, hereditary ellipocytosis), by erythrocyte enzyme defects, by diminished formation of normal Hb (thalassemia syndrome) or by the synthesis of qualitatively abnormal Hb (HbE syndrome, sickle cell anemia). With the Hb disorders, those with α-globin chain abnormality are already manifested clinically at birth, those with β-globin chain disorders first become obvious at the age of 4–6 months /39/. Refer also to Tab. 7.6-2 – Disorders of hepcidin and ferroportin regulation. \| \| – Hereditary elliptocytosis This disease is occasionally discovered during the assessment of a blood smear. These patients have an elliptocyte proportion of 15–70% relative to the total erythrocyte population. They are, normally, not anemic, and their automated blood count is usually normal. Approximately 5–20% of the patients have compensated or mild anemia and reticulocytosis of about 20%. \| \| – Enzyme defects Enzyme defect anemia makes up some 18% of non-spherocytic hemolytic anemia. Pyruvate kinase and glucose-6-phosphate dehydrogenase deficiencies affect, respectively, 54% and 30% of these anemias (see Section 15.8 – Erythrocyte enzymes). \| \| – Sickling disease (SCD) /48/ SCD comprise a group of autosomal co-dominant hemoglobinopathies characterized by mutations in the gene encoding the hemoglobin ß-subunit on chromosome 11. The gene codes the assembly of two β-globin chains to HbA. The mutant β-allel (βS) codes for the production of the variant hemoglobin S(HbS). In the mutated sickle cell gene (βS), adenine is substituted with thymine and thus, in amino acid position 6 of the β-globin chain, valine is incorporated instead of glutamic acid. The heterozygous carrier state, known as sickle cell trait, results in the production of both hemoglobin A and S and has predominately a benign clinical picture. Five haplotypes of the βS-allele specific to different regions of the world exist. The haplotypes differ from one another in the severity of the clinical picture to which they give rise, and in the frequency and the pattern of the clinical symptoms. SCD patients are virtually resistant to malaria, since the intraerythrocytic parasites are degraded during hemolysis. SCD is the most common hemoglobinopathy in the world, with a high incidence in Mediterranean countries, the near east, central Africa, and certain parts of India. In peoples of African origin, there are three major beta globin haplotypes (Benin, Bantu, and Senegal) which dominates in North America and the Caribbean. The Asian haplotype accounts for the disease in India and the Eastern Province of Saudi Arabia /79/. Polymerization of deoxygenated HbS (PO2 < 40 mmHg) is the primary event in the pathogenesis of SCD. HbS is transferred from solid single molecules into a gel by binding of valine in position 6 to phenylalanine in position 85 in the adjacent HbS resulting in a typical fibre structure. The sickle cell gets its typical shape by polymerization of the fibre structures inside the erythrocyte /49/. If the HbS fraction is greater than 50%, the erythrocytes loose their plasticity and clog the capillaries of almost all organs, whereupon trophic disturbances and necrosis occur. Pain crises, which can last for minutes or days, result from the shift in the micro flow pathways. Four sickle cell syndromes are distinguished; their hematological findings are summarized in Tab. 15.3-13 – Laboratory findings in sickle cell anemia /39/: HbSS is the homozygous phenotype. In Africa, some 120,000 children of this type are born annually; in the USA 3000–4000. Patients with this phenotype have normocytic normochromic anemia with Hb values of 60–100 g/L and reticulocytosis of 10–25%. Erythrocyte survival time is shortened to 5–20 days. While normally the sickle shape reverts to the discoid shape with oxygenation of the cells, with the homozygous phenotype up to 30% of the sickle cells are irreversibly altered and hemolyzed. The membrane rigidity of sickle cells is associated with the increased activity of the Ca2+ sensitive K+-channel and potassium-chloride co transporters cause dehydration and generate the dense cells (MCHC ≥ 460 g/L) /49/. The cells are visible in the blood smear and contain Howell-Jolly bodies. Patients with this phenotype manifest the clinical symptomatology outlined in the foregoing. HbAS is a heterozygous phenotype consisting of 35–45% HbS, and the remainder HbA. In certain regions of West Africa, 20–25% of the population are carriers of the sickle cell gene; in the USA, this is the case for 10% of the black population. HbAS phenotypes do not manifest anemia; they have normal blood cell morphology. HbSC phenotype: heterozygous inheritance of HbS and HbC, in the proportion of 45–55%, gives rise to symptoms milder than the HbSS disease. The anemia is mild, the MCV is reduced, and reticulocytosis of 3–6% is measured. In Africa, its prevalence is one fifth that of HbSS disease. The HbSC phenotype, like the HbSS disease, also manifests pain crises, and necrosis of the renal papillae, necrosis of the femoral head, and retinopathy are frequently observed. In contrast to HbSS disease, in which splenic atrophy occurs as of the age of 5 years, the HbSC phenotype manifests lifelong splenomegaly. HbS/β-thalassemia phenotype: heterozygosity for HbS and β-thalassemia is present. This phenotype occurs preferentially in Mediterranean countries. The HbS fraction is 60–90%, and the remainder is HbF or, in a few cases, HbA. Anemia is mild (Hb approximately 100 g/L) and MCV and MCH are decreased. The clinical course is milder than in phenotypes HbSS and HbSC, since the increased erythrocyte HbF content prevents the formation of sickle cells. Clinical findings: SCD is characterized by the painful recurrence of vaso-occlusive events, multi-organ morbidity and an increased risk of early death. There is strong evidence that elevated HbF level is associated with relatively mild clinical manifestations of SCD /66/. Young SCD patients have a diminished urinary concentrating ability, electrolyte regulation, and enhanced proximal tubular function. Because of increased renal plasma flow, the glomeruli are hypertrophic and the GFR is elevated. Proteinuria develops in the second decade of life; 10–20% of young patients often already manifest macroalbuminuria and, later, focal glomerular sclerosis. In the Baby Hug trial, it was shown that an incipient disorder of kidney function can be diagnosed early with DPTA-GFR, and that organ damage can be prevented with early hydroxyurea treatment /50/. Chronic administration of blood transfusion is frequent in SCD patients, and this leads to iron overload. Ferritin values of < 1,500 μg/L signal acceptable iron overload; values > 3,000 μg/L are associated with liver. The spleen is one of the first organs to be damaged in SCD. The damage is often silent and progressive, resulting in functional asplenia with a high risk of infections. Some children with SCD experience acute splenic sequestration and chronic hypersplenism. Laboratory findings: RBC, hemoglobin, and hematocrit are low in hemolytic anemia patients with heterozygous S/0 and homozygous SS mutations White blood cell and platelets are increased and fluctuating. However reticulocyte counts vary and are affected by a variety of variables. In patients taking hydroxyurea, mean corpuscular volume (MCV) is increased. Because of the various RBC subpopulations, higher red cell distribution width is observed /51/. Gene therapy with LentiGlobin consists of autologous transplantation of hematopoietic stem and progenitor cells transduced with the BB305 lentiviral vector a modified β-globin gene, which produces the antisickling HbAT87. One-time treatment with LentiGlobin resulted in sustained production of HbAT87Q in most red cells, leading to reduced hemolysis and severe vaso-occlusive events /80/. \| \| – Paroxysmal nocturnal hemoglobinuria (PNH) PNH is a rare, acquired non-malignant clonal stem cell disease, caused by a mutation of the Phosphatidyl inositol glycan-class (PIG)-A gene. PNH results from hematpoietic stem cells and its progeny that are unable to synthesize the glycosylphosphatidylinositol (GPI) anchors, which serves to tether the cell surface complement inhibitory proteins CD55 and CD59. Both of the proteins are are key regulators of the complement pathway and protect host cells from complement-mediated removal. The defect, or deficient GPI anchor protein, does not support the transfer of N-acetylglucosamine to glycosylphosphatidylinositol for the synthesis of the intact GPI anchor. The proteins that are to be anchored by GPI are synthesized, but they are not integrated into the cell membrane and are detectable in high concentrations in the plasma. The GPI anchor also permits the influx of GPI-specific phospholipase C for the separation of the anchored proteins from the cell membrane. The prevalence of PNH is 1 : 100,000 to 1 : 500,000. The complement dysregulation leads to chronic hemolysis and thrombosis characteristic for PNH /78/. Clinical symptoms: PNH is characterized by a unique clinical triad: intravascular hemolysis, a high risk of thrombosis, and bone marrow impairment. By frequency: thrombosis 40%, anemia 35%, hemoglobinuria 26%, hemorrhage 18%, aplastic anemia 13%, gastrointestinal symptoms 10%, icterus 9%, iron deficiency anemia 6%. Hemolysis: the somatic mutation in the X-linked gene PIGA makes the mutant hematopoietic stem cell and its progeny unable to synthesize the glycosylphosphatidyl inositol anchor. Therefore CD55 and CD59 do not protect hematopoetic cells from activated complement, the consequences are hemolytic anemia and hemoglobinuria due to chronic intravascular hemolysis. Other complaints are stomach pain, esophageal spasm, difficulties in swallowing. Thromboembolism (TE) /52/: thromboembolism in PNH is associated with a 7-fold higher mortality risk. In 46% of patients, TE develops in spite of anticoagulation. The TE risk is 7-fold higher in patients with lactate dehydrogenase (LD) ≥ 1.5-fold the upper reference interval value in comparison to patients with lower levels. The TE risk is 3-fold increased in the presence of abdominal or thoracic pain, or in dyspnea. With LD ≥ 1.5-fold and thoracic pain, abdominal pain, dyspnea and hemoglobinuria, TE risks are increased 19-fold, 18-fold, 10-fold and 10-fold, respectively. Bone marrow impairment: aplastic anemia and myelodysplatic syndrome may result. Laboratory findings: increased LD, increased bilirubin, decreased hemoglobin, decreased haptoglobin, hemoglobinuria. Confirmation of the diagnosis with flow cytometry. The GPI anchored surface proteins are missing in all of the hematopoietic cell lines (CD55/CD59 deficiency). Cells with normal expression or with partial or complete loss of expression are termed, respectively, type I, type II and type III cells. Whereas in untreated patients the direct Coombs’ test is negative (because red cells with bound C3 are rapidly lysed) patients treated suboptimal by the C5 inhibitor eculizumab can have a positive Coombs’ test /81/. \| \| – Hemolytic uremic syndrome (HUS) /53/ HUS is a thrombotic micro angiopathy, characterized by hemolytic anemia, thrombocytopenia and acute kidney failure. A distinction is made between the following types: The D(+)HUS type that is associated with diarrhea, triggered by intestinal infection with Shiga toxin-producing E. coli O157:H7 but also other subtypes like O103; O111, O26. These bacteria/toxins are responsible for approximately 95% of all HUS cases. The source of these enterohemorrhagic E. coli (EHEC) is infected cattle. The prevalence in Germany and Austria is 0.7/100,000 children below the age of 15 years, and 1.5–1.9/100,000 children below the age of 5. The infective dose is some 100 germs. Orally ingested germs colonize the intestine, permeate the intestinal barrier, and reach the blood. An extremely rare type of HUS (SPA-HUS), associated with Streptococcus pneumonia, which can be accompanied by sepsis, meningitis, and pneumonia with pleural empyema. Atypical HUS D(–). It is based upon excessive complement activation in renal glomeruli and arteriolar cells. The cause is a mutation in the factor B gene. The risk of HUS following infection with Shiga toxin-producing E. coli is 10–15%; it is often self-limiting and improves within 1–3 weeks of disease onset. However, a permanent reduction in GFR occurs in 10–20% of the cases. Laboratory findingss:’ hemolytic anemia with Hb ≤ 100 g/L, fragmented erythrocytes, reticulocytosis, thrombocytopenia < 150 × 109/L, creatinine above the 97th percentile for the corresponding age, increased bilirubin and LD, which is markedly elevated in atypical HUS. \| \| – Malaria /54/ The disease is triggered by an anopheles mosquito bite; sporozoites are released into the blood of the affected individual, from where they take hold in the liver. Schizonts mature in the liver, and merozoites from disintegrating schizonts pass into the blood and colonize in the erythrocytes. In the erythrocytic phase, they can take two reproductive pathways: The sexual pathway which, following a repeated mosquito bite, makes transmission to other individuals possible The asexual pathway. Accordingly, the merozoite develops to a ring form, which matures to the trophozoite and then to the schizont. This disintegrates and releases numerous merozoites, which again affect additional erythrocytes. In P. falciparum infection, this is a synchronous process that lasts 48 hours in each case. Laboratory findings: anemia, haptoglobin decrease, LD elevation. Decrease in serum iron and inappropriately low reticulocytosis relative to the anemia, rise in ferritin. In chronic forms, the erythropoietic response is impaired, possibly due to a diminished response to erythropoietin and enhanced erythrophagocytosis. See Chapter 44 – Parasitic infections for further malaria findings. \| \| – Warm autoimmune hemolytic anemia (wAIHA) The wAIHA is a rare autoantibody mediated immune disorder. The pathology of the disease is caused by IgG-, IgM- or IgA-type autoantibodies associated, or not, with molecules of the complement system, and directed against self-erythrocytes. Red blood cell (RBC) destruction is induced by activation of FcγR-bearing effector cells, mainly in the spleen after sequestration and trapping by splenic macrophages. The most common RBC antigen targets are membrane proteins. Data of a study /68/ demonstrate that the ability of anti-RBC autoantibodies to induce phagocytosis, trogocytosis and antigen dependent cytotoxicity is not related to their antigen specificity, but the clinical severity may be dependent on the functional activity of the anti-RBC antibodies. \| \| Acute hemorrhage In acute hemorrhage erythrocytes and plasma proteins are lost, but not to the same proportion /45/. Immediately following acute hemorrhage, the erythrocytes are redistributed from the small vessels, and the spleen, particularly from the vessels in the region of the nervus splanchnicus. Therefore, blood levels of Hb and erythrocytes in the large vessels are normal or show a moderate decrease. Only after 12–24 hours a stronger decline occurs, and the nadir is reached 48–72 hours after the acute hemorrhage. Following 2–4 weeks, the values recover to pre-bleeding levels. Total protein concentration in serum drops after 4–6 hours, particularly if the patients drink following the volume loss. The nadir of total protein is reached within 12–24 hours, and the concentrations recover to pre-bleeding levels 1–3 weeks later. Based upon total protein, hemolytic anemia can be differentiated from acute hemorrhagic anemia within the first 24 hours. Total protein is normal and/or remains constant in the former case, while in the latter, a decrease occurs. Laboratory findings: in acute hemorrhage, Hb level, erythrocyte count and the Hct fall proportionately. Reticulocytosis can be detected on the second or the third day and reaches 15% following 1–2 weeks. With comparable bleeding, regeneration of anemia following internal blood loss occurs more rapidly than with external bleeding, since iron and proteins are conserved in internal blood loss. \| \| – Postpartum hemorrhage Wordwide postpartum hemorrhage accounts for 8% of maternal deaths in developed regions of the world and 20% of maternal deaths in developing regions. The traditional definition of postpartum hemorrhage is blood loss of more than 500 mL after a vaginal delivery or more than 1.000 mL after cesarean delivery. Typical clinical signs and symptoms of hypovolemia e.g., tachycardia and hypovolemia due to postpartum hemorrhage may not appear until blood loss exceeds 25% of total blood volume (> 1,500 mL during pregnancy). Postpartum hemorrhage is considered to be primary when it occurs within the first 24 hours after delivery and secondary when it occurs between 24 hours and up to 12 weeks after delivery. The cause of postpartum hemorrhage is summarized by the four “Ts” tone (utrine atony), trauma (lacerations or uterine rupture), tissue (retained placenta or clots), and thrombin (clotting factor deficiency) /75/. The most common cause is uterine atony (70% of cases).Transfusion is typically begun when the blood loss exceeds 1,500 mL. The goals are to maintain a hemoglobin level more than 80 g/L (tranfusion protocol 450 mL red cells, fresh frozen plasma 1 Unit, cryoprecipitate 1 Unit (factor VIII, fibrinogen), perhaps 1 U apheresis platelet concentrate. \| \| Hypo regenerative normocytic anemia – Generally Hypo regenerative normocytic anemia is usually normochromic, with minimal variability in erythrocyte morphology. The reticulocyte count is usually below 40 × 109/L. \| \| – Anemia of chronic disease (ACD) ACD occurs in infection, chronic inflammation, chronic liver disease and malignant tumors /55/. The trigger of the anemia is IL-6 stimulated hepcidin expression within the framework of an acute phase response. The effects are: inhibition of proliferation of erythropoiesis, reduction in iron availability due to inhibition of the cellular iron exporter ferroportin in macrophages and enterocytes. Since the erythroblast takes up less iron via the transferrin receptor (TfR) and releases none due to the ferroportin blockade, normocytic normochromic anemia is common during the first years of ACD. The Hb level is 100–120 g/L, transferrin saturation is > 16%, ferritin is > 100 μg/L, sTfR is normal, CRP is often > 5 mg/L. Among internal medicine patients, ACD comprises 40–70% of all cases of anemia. Approximately 10–20% of ACD patients develop a mild, hypochromic anemia, due to iron-restricted erythropoiesis (ACD/IRE). In ACD/IRE either bleeding or hepcidin induced long term reduced intestinal iron absorption are the causes of decreased iron supply for erythropoiesis (see also Section 7.6 – Hepcidin). Sensitive indicators for the identification of IRE in ACD are a decrease of the reticulocyte Hb content (CHr, Ret-He) below 28 pg or a %HYPO of > 5% /56/. Hb can be normalized by means of treatment of the underlying disease or therapy with erythroid-stimulating agents. ACD includes forms of anemia that are due to renal insufficiency, disorders of liver function, and endocrine disorders. \| \| – Anemia of cancer /57/ The prevalence of anemia in patients with different cancer types (39% at enrollment and 68% becoming anemic at least once during a 6-month period) has been shown in the European Cancer Anemia Survey /58/. Mean Hb values are 100–120 g/L, and they seldom decline to below 80 g/L. In cancer patients, ACD and chemotherapy are the main causes of anemia, which becomes more severe due to chronic loss of blood and nutritional iron deficiency. The mortality risk in cancer patients with anemia is increased by 65% /59/, and average annual health care cost per patient is 4-fold increased. Of the patients with solid tumors, those with colorectal, lung and ovarian carcinoma have the highest prevalence of anemia and the greatest requirement for transfusion. In patients undergoing chemotherapy, there is a direct correlation between the Hb value and quality of life. The following factors can cause anemia of cancer /60/: Directly tumor-related (e.g., loss of blood, hemolysis, hypersplenism, bone marrow infiltration) Indirectly, due to chemotherapy, radiation therapy, or due to the ACD, which is particularly the case with solid metastatic tumors. Approximately one third of tumor patients have elevated CRP levels, defined as values greater than 8 mg/L. Management of anemia and iron deficiency /61/: Red blood cell transfusion for patients with Hb-levels below 70–80 g/l Intravenous iron therapy if ferritin below 100 μg/L and transferrin saturation below 20% for patients with Hb levels 80–100 g/L Intravenous iron therapy if ferritin below 100 μg/L or transferrin saturation below 20% for patients with Hb levels 100–110 g/L \| \| Aplastic anemia /62/ Aplastic anemia is featured by bone marrow hypo cellularity and peripheral pancytopenia. Aplastic anemia can be due to congenital (20%) or acquired causes (80%). All hematopoietic cells are diminished. Symptoms are fatigue, tachycardia, dyspnea, bruising, mucocutaneous bleeding, and nose bleeds. The incidence is 2–6 cases per million per year. The age distribution is biphasic, with peaks at 15–25 as well as at ≥ 60 years of age. The bone marrow is hypo cellular, and hematopoietic cells are replaced by fat cells. Causes are: Idiopathic (70%), SLE, Sjögren syndrome, rheumatoid arthritis, myasthenia gravis, medications (phenytoin, azathioprine, isoniazid, thyreostatics), Parvovirus B19, B cell lymphoproliferative diseases. Laboratory findings: marrow cellularity of less than 25% of normal or less than 50% with hematopoietic cells representing less than 30% of the residual cells and at least two of the following blood counts: leukocyte count ≤ 3.5 × 109/L (neutrophil count ≤ 0.5 × 109/L), platelets < 20 × 109/L, reticulocytes < 1%. \| \| AIDS Cytopenia is infrequent in the early stages of HIV infection; it occurs only in clinically manifest AIDS disease. Laboratory findings: more than 70% of patients have predominantly normochromic normocytic anemia. Leukopenia and thrombocytopenia are present in greater than 70% and 40% of the cases. Hb concentrations are, as a rule, 90–100 g/L. Disorders of iron metabolism are present, as in ACD, with reduced iron, elevated serum ferritin and diminished transferrin saturation /63/. \| \| Dock 11 deficiency /83/ Dedicator of cytokinesis (DOCK 11) participates in actin cytoskeleton dynamics through its guanine nucleotide exchange factor (GEF) activity, resulting in activation of the small Rho guanosine triphosphatases (GTPases) RAC and cell division cycle 42 (CDC42). Clinically DOCK 11 deficiency includes phenotypes associated with CD42 mutations, such as recurrent infections, immune dysregulation, a normocytic anemia, thrombocyte anomalies and neurodevelopmental abnormalities. Unlike patients carrying CD42 mutations patients with DOCK 11 deficiency have no detectable facial abnormalities, neutropenia, monocytopenia, or hemophagocytic lymphohistiocytosis. The erythroid hypoplasia of the patients is consistent with a bone marrow defect and the observed erythroid-differentiation defect in DOCK11 knockdown CD34+ cord blood cells. DOCK 11-deficient T-cells show reduced transendothelial migration in vitro, yet also increased migration speed under confinement in vivo and in vitro, which suggests that DOCK 11 is involved in leukocyte diapedesis and interstitial migation.Autoinflammation has been observed in several actin-related deficiencies and has been linked to increased inflammasome-mediated secretion of interleukin 1β and interleukin 18. \| Table 15.3-12 Causes of hemolytic anemia \| Antibody-mediated: idiopathic, medications, infections, isoimmunization, transfusion reaction, lymphoproliferative disease, rheumatic disease Mechanical: prosthetic heart valves, implants, vasculitis (micro angiopathic hemolysis), arteriovenous malformations Disorders of hemoglobin synthesis or structure Erythrocyte enzyme defects, erythrocyte membrane defects Miscellaneous: march hemoglobinuria, parasitic infection, snake poison, thermic damage. \| Table 15.3-13 Laboratory findings in sickle cell anemia, courtesy of Ref. /39/ \| Hb variants \| Hb (g/L) \| HCT \| MCV (fL) \| Retic. (%) \| ISC \| Fractions of Hb variants \| \| HbSS \| 60–100 \| 0.20–0.30 \| 80–90 \| 10– 15 \| 4+ \| 80–95% HbS 2–20% HbF 2–4% HbA2 \| \| HbS/β0 thal. \| 60–100 \| 0.20–0.30 \| 60–70 \| 10– 15 \| 3+ \| 75–95% HbS 2–20% HbF 3–6% HbA2 \| \| HbS/β+ thal. \| 80–120 \| 0.30–0.36 \| 65–75 \| 3– 6 \| 1+ \| 50–85% HbS 10–30% HbA 2–20% HbF 3% HbA2 \| \| HbSC \| 100–120 \| 0.30– 0.36 \| 70–80 \| 5– 10 \| 1+ \| 50% HbS 50% HbC \| ISC, irreversibly sickled cells, microscopically detectable Table 15.3-14 Classification and differentiation of macrocytic anemia \| Clinical and laboratory findings \| \| Folic acid deficiency, vitamin B12 deficiency, antimetabolites (folic acid, purine and pyrimidine analogs) in cancer therapy /7/ The vitamin deficiency and medication with antimetabolites results in retarded purine and pyrimidine (notably thymidilate) synthesis. Therefore DNA does not replicate, but RNA and hemoglobin synthesis continues leading to typical changes, especially in rapidly dividing tissues such as the bone marrow. With constant cytoplasm formation, the diminished cell division leads to the development of larger cells. Typical changes in the peripheral blood are pancytopenia, elevated MCV, MCH and red cell diameter indexes, MCHC is reduced or low-normal. The granulocytes have polylobulated nuclei. Since the Hb content has not increased in proportion with the cell volume, the MCH is high-normal but the MCHC is low-normal. Therefore, macrocytic erythrocytes are hyperchromic relative to the MCH, and hypochromic relative to the MCHC. LD is elevated due to the ineffective erythropoiesis. The reticulocyte count is reduced, but it can be elevated in bleeding. The combination of iron, vitamin B12 or folate deficiency simulates a normal MCV value, if the erythrocyte distribution width is not evaluated. \| \| Chronic alcoholism, liver disease In chronic alcoholism and chronic active liver disease, the impairment of DNA synthesis is also due to insufficient availability of folic acid and vitamin B12. \| \| Congenital dyserythropoietic anemia (CDA) /64, 65/ CDA belongs to a group of inherited conditions characterized by maturation arrest during erythropoiesis with a reduced reticulocyte production in contrast with erythroid hyperplasia in bone marrow. Three classical types have defined on the basis of bone marrow morphology. CDA I and II are autosomal recessive diseases. Whereas CDA I displays abnormalities in chromatin structure, CDA II patients have a marked increase in bi- and multi-nucleated erythroblasts in their bone marrow. CDA type III is an autosomal dominant disease with giant multi-nucleated erythroblasts in bone marrow. CDA I: the Hb value is 90–110 g/L, and the MCV is 100–120 fL. Hemolytic anemia with an inadequately low reticulocyte count, mild icterus, and low haptoglobin concentration are present. The bone marrow is hyper cellular, containing 30–60% early and late polychromatic erythroblasts with nuclei of abnormal size and shape. The proerythroblasts and the early basophil erythroblasts are normal. Typical are incompletely divided cells with chromatin bridges between the erythroblasts. CDA II: these patients have anemia, icterus (90%), splenomegaly (70%), and hepatomegaly (45%). The anemia is normocytic with anisocytosis and poikilocytosis, in children there is a slight elevation in the MCV, the reticulocyte count is normal to slightly elevated. The bone marrow is hyper cellular but normoblastic, and contains 10–15% binuclear erythroblasts. Diagnosis of a CDA relies on examinations of the erythroblasts in a bone marrow biopsy which demonstrates the presence of chromatin bridges between erythroblast nuclei and a spongy Swiss cheese appearance of the chromatin in CDA I, binucleated cells and endoplasmic reticulum remnants in CDA II, and medullary hyperplasia with the presence of giant erythroblasts containing up to 10 nuclei in CDA III. Laboratory findings: hypo regenerative anemia, macrocytosis, intermittent hyperbilirubinemia, decrease in haptoglobin, increase in LD, atypic electrophoresis of erythrocyte membrane protein. \| Table 15.4-1 Equations for HCT calculation \| Calculation of relative centrifugal force (RCF): RCF (gn) = 0.00001118 × r × N2 gn, gravity; r, rotor diameter in cm; N, number of revolutions in rpm \| \| Calculation of the HCT: Some hematology analyzer determine the sum of electrical impulses and divide them by the number of impulses. The calculation is performed according to the equation: \| Table 15.4-2 Hematocrit reference intervals \| Adults \| Fraction \| Relative % \| \| Caucasians /2/ \| 0.42 (0.36–0.48) \| 42 (36–48) \| \| 0.46 (0.40–0.53) \| 46 (40–53) \| \| Blacks /1/ \| 0.38 (0.34–0.43) \| 38 (34–43) \| \| 0.42 (0.34–0.48) \| 41 (34–48) \| \| Athletes /3/ \| 0.41 (0.37–0.45) \| 41 (37–45) \| \| 0.45 (0.40–0.50) \| 45 (40–50) \| \| Data expressed as fraction and in relative percent. Values are x ± 2s or the 2.5th and 97.5th percentiles. \| \| \| \| Fetus (week of pregnancy) /4/ \| \| \| \| 15 \| 0.28–0.42 \| (28–42) \| \| 16 \| 0.34–0.42 \| (34–42) \| \| 17 \| 0.31–0.43 \| (31–43) \| \| 18–21 \| 0.31–0.45 \| (31–45) \| \| 22–25 \| 0.31–0.47 \| (31–47) \| \| 26–29 \| 0.32–0.50 \| (32–50) \| \| ≥ 30 \| 0.30–0.58 \| (30–58) \| \| Neonates /5, 6/ \| \| \| \| Umbilical cord blood \| 0.48–0.56 \| (48–56) \| \| Venous blood (2 h after delivery) \| 0.49–0.71 \| (49–71) \| \| Venous blood (6 h after delivery) \| 0.44–0.68 \| (44–68) \| \| Children /7, 8/ \| \| \| \| 2–6 days \| 0.40–0.70 \| (40–70) \| \| 1–2 weeks \| 0.38–0.70 \| (38–70) \| \| 2–3 weeks \| 0.38–0.60 \| (38–60) \| \| 3–7 weeks \| 0.36–0.46 \| (36–46) \| \| 7–12 weeks \| 0.30–0.38 \| (30–38) \| \| 10–12 mo \| 0.35–0.43 \| (35–43) \| \| 4–5 yrs \| 0.32–0.40 \| (32–40) \| \| 6–8 yrs \| 0.32–0.41 \| (32–41) \| \| 10–13 yrs \| 0.34–0.44 \| (34–44) \| \| 14–16 yrs \| ♀ 0.35–0.43 \| (35–43) \| \| ♂ 0.38–0.49 \| (38–49) \| Data expressed as fraction and in brackets in relative percent. Values are x ± 2s or the 2.5th and 97.5th percentiles. Table 15.4-3 Diseases and conditions with decreased HCT \| Clinical and laboratory findings \| \| Pregnancy During the first two trimesters of pregnancy, hydremia and hemodilution develop with a disproportionate increase in plasma volume. The red cell mass and the plasma volume increase by 30% and 50%, respectively. The result is a relative fall in the HCT of 25–30%, and a fall in Hb of 20–40 g/L. In one study /26/, the HCT was 0.34–0.43 in non-pregnant women, 0.31–0.41 during the first trimester of pregnancy, 0.30–0.38 during the second trimester, and 0.28–0.39 during the third. \| \| Hemorrhage Immediately after acute blood loss, the HCT is normal with a decreased blood volume. Following 12–36 hours the blood volume is, once again, normal, the lost volume having been replaced by fluids from the interstitial and intracellular spaces. In consequence, the HCT decreases continuously and in parallel with the Hb value. A decrease in the HCT of 0.03 corresponds to an fall in Hb of 10 g/L. A single HCT determination provides only a limited information of the volume loss; the monitoring of the course is more valid /27/. In hemorrhage with a decline in the HCT to below 0.25 to 0.30, a decision must be made with regard to the transfusion of erythrocytes particularly in the elderly and in patients with cardiac disease. Young patients with normal cardiac function often tolerate, without hypoxic organ damage, an isovolemic decrease in the HCT to 0.20 to 0.25, corresponding to a Hb value of 60–75 g/L. With the transfusion of 300 mL of packed red cells containing an erythrocyte fraction of 200 mL to a 70 kg adult, a rise in the HCT of 0.03 may be expected. The number units of packed cells that must be transfused, and the HCT that can be reached after transfusion, can be calculated by means of the equations that are shown in Tab. 15.4-4 – Calculation of desired packed red cells in dependence of the HCT /28/. Following transfusions or infusions, varying extents of fluid displacement depending upon the type of blood product or infusion occur between the intra- and extracellular space. A corresponding equilibrium can be observed following a minimum of 24 hours, at which time the HCT permits the assessment of the composition of the blood. Acute gastrointestinal bleeding is the second-most common indication for transfusion of blood products in the UK. Clinical evidence now favors restrictive transfusion of erythrocytes in the hemodynamically stable bleeding patient. It is specifically important to consider the specific case of the cirrhotic patient with gastrointestinal bleeding. Deranged laboratory measures of coagulation may not be associated with increased bleeding risks. \| \| Increase in plasma volume Conditions associated with an increase in plasma volume, for e.g., intraoperative hyperhydration with plasma expanders, or internal diseases that cause an expansion of the plasma volume, such as primary hyperaldosteronism, lead to a decrease in the HCT and anemia. In approximately half the cases anemia can be explained as a result of hemodilution, rather than by a reduction in red cell mas, especially in chronic heart and liver diseases /39/. \| \| Pre-operative anemia (see also Section 19.2 – Oxidative stress) The 30-day post-operative mortality in patients undergoing non-cardiac surgery was investigated as a function of the pre-operative HCT in comparison with operated patients with a normal HCT. In mild anemia (HCT in women < 0.36 to > 0.29; in men < 0.39 to > 0.29) as well as in moderate to severe anemia (HCT ≤ 0.29 in both men and women), the odds ratio for post-operative mortality was 1.41. If additional risk factors were present, the odds ratios for morbidity and mortality were higher than in cases of anemia or a risk factor alone /29/. Each perioperative transfusion, even of only one unit packed blood, increased morbidity and mortality /30/. \| Table 15.4-4 Calculation of desired packed red cells (PC) in dependence of the HCT /28/ \| Number of PC (nPC) that has to be transfused at the actual HCT (HCTa), in order to reach the desired HCT (HCTd): \| \| \| Expected value of the HCT (%) [HCTd] following administration of a certain number of PC (nPC): \| \| \| Explanation: the volume of one PC is 300 mL, and it contains 200 mL of red cells. HCTa, actual HCT; HCTd, desired HCT; HCT is expressed in %. BV is the estimated normal blood volume of an individual, and is calculated as follows: \| \| \| ♀ \| BV (mL) = Body weight in kg × 67 \| \| ♂ \| BV (mL) = Body weight in kg × 77 \| Table 15.4-5 Diseases and conditions with increased hematocrit (HCT) \| Clinical and laboratory findings \| \| Polycythemia vera (PV) /31/ The myeloproliferative neoplasms (PV, essential thrombocytosis and primary myelofibrosis) are unique hematopoietic stem-cell disorders that share mutations that constitutively activate physiologic signal-transduction pathways responsible for hematopoiesis. In addition to phenotypic mimicry, each type of myeloproliferative neoplasm is capable of evolving into another type, making diagnosis, risk assessment, and therapeutic choices difficult /32/. PV occurs in patients of all ethnic backgrounds, however it is much more common in those of European than Asian descent, having an annual incidence ranging from 2 to 10 cases per million population per year. PV is generally a disease of older individuals, occurring most frequently between 50 and 70 years of age; childhood and adolescent disease is exceptionally rare. The most common complaints are thrombosis (deep vein thrombosis, stroke, myocardial ischemia, Budd-Chiari syndrome, mesenteric ischemia) and bleeding complications (epistaxis, oral mucosal hemorrhage, gastrointestinal hemorrhage, or nonspecific ecchymosis). Hyper viscosity syndrome due to sluggish blood flow and micro thrombi may cause hypertension, headache, stupor, disturbances of vision, dizziness, tinnitus, claudication and erythromelalgia. Pruritus occurs following prolonged exposure to warm water, due to histamine release from activated basophil granulocytes. More than half of the patients have splenomegaly. Gouty arthritis is not uncommon. World Health Organization diagnostic criteria for PV are shown in Tab. 15.4-6 – World Health Organization diagnostic criteria for polycythemia vera. The pathogenesis of PV is the result of different mutations: Some 95% of patients with PV have the JAK2 V617F mutation in exon 14 of the JAK2 gene. The enzyme (Janus kinase) encoded by this gene is a cytoplasmic tyrosine kinase that is involved in signal transmission of different cytokine receptors, including the EPO receptor. The mutation enhances the Janus kinase activity and leads to an EPO-independent stimulation of erythropoiesis. The mutation is acquired, and can also be detectable in primary thrombocythemia and primary myelofibrosis. An abnormal clone in the bone marrow leads to the disease. Other patients have mutations in exon 12 of the JAK2 gene (JAK2 F537-K539delinsL, JAK2 H538QK539L, JAK2N542-E543del, JAK2 K539L). These patients have a predominantly erythroid phenotype. Laboratory findings /19/: suspicion of PV exists with HCT > 0.52 (Hb > 185 g/L) or 0.48 (Hb > 165 g/L) in men and women, respectively, along with normocytic normochromic erythrocytes. More than 40% of patients have leukocytosis > 12 × 109/L or neutrophil granulocytosis > 10 × 109/L. If oxygen saturation is normal, EPO is determined. Leukocytes are investigated for the JAK2 V617F mutation because erythrocytes lacks the nucleus. PV is present if a homozygous mutation is found. More than 60% of the patients have thrombocytosis > 400× 109/L and more than 70% have an elevation in vitamin B12 concentration of 25% above the upper reference interval value. EPO concentrations are low or subnormal. The erythrocyte sedimentation rate is markedly reduced. Bilirubin and uric acid can be increased. Bleeding time is normal, but the global coagulation tests can be pathological due to the high HCT, since the anticoagulant to plasma ratio is no longer correct. \| \| Primary erythrocytosis /33, 34/ Primary erythrocytosis (also known as primary familial and congenital polycythemia, PFCP) is a pathology of erythroid progenitors, which display hypersensitivity to EPO. However, the erythropoiesis is polyclonal and PFCP patients do not show any risk of their disease transforming into myelofibrosis or leukemia /34/. The erythropoietin receptor (EPOR) is expressed by erythroid progenitors. Transcription of the EPO gene is regulated by mechanisms that are dependent upon the cellular response to changes in oxygen supply. The binding of EPO to its receptor leads to the transient activation of preformed EPOR-JAK2 complexes and down streaming signaling pathways, including signal transducer and activators of transcription (STAT), phosphatidylinositol 3 kinase/akt and mitogen-activated protein kinase pathways. Activated STAT causes homodimerization of the EPOR in the presence of EPO and auto phosphorylation of JAK2 occurs. Once JAK2 is activated, specific EPOR tyro sines are phosphorylated and form docking sites for adaptor molecules (See Fig. 15.4-2 – Homodimer erythropoietin receptor showing phosphorylated tyro sines) Around 20 germ line heterozygous nonsense and frame shift mutations located in exon 8 of the gene EPOR have been described so far in PFCP, all leading to the truncation of the C-terminal part of EPOR. The C-terminal part of the EPOR includes several conserved tyrosine residues that are docking sites for positive and negative regulators of EPOR signaling. The EPO hypersensitivity of PFCP progenitors is, therefore, usually explained by the disappearance of negative regulatory domains located in the truncated part of the receptor. EPO hypersensitivity induced by the EPOR mutant p.Gln434Profs11 is explained by the appearance of a new C-terminal tail /34/. The latter, by increasing EPOR dimerization, stability and cell-surface localization, causes pre activation of EPOR and JAK2, constitutive signaling and hypersensitivity to EPO. Laboratory findings: Erythrocytosis with subnormal EPO concentration. Gene panel sequencing improves the diagnostic work-up of patients with erythrocytosis /38/. \| \| High oxygen-affinity hemoglobins /19/ High oxygen affinity hemoglobins may be stable or unstable, and as they do not release oxygen readily to the tissues the resulting hypoxia drives EPO production and thus erythrocytosis. Mutations in the genes that code for the α- and β-chains can lead to hemoglobins with high-affinity binding of O2. The hemoglobin-oxygen dissociation curve is shifted to the left indicating abnormal Hb which has increased affinity for O2. As a rule, with oxygen partial pressure of 20 mm Hg, O2 saturation in the capillaries is 35%. With Hb variants with high oxygen affinity this can, however, reach 60%. Laboratory findings: erythrocytosis, high or, relative to the Hb value, inappropriately high EPO concentrations. The oxygen partial pressure at which Hb saturation of 50% is reached is low. Example: control 27 mm Hg, patient 16.5 mm Hg. \| \| VHL gene mutation All affected patients are homozygous for a single mutation C598T. The amino acid 200 arginine changed to tryptophan in the von Hippel-Lindau (VHL) gene. The VHL gene plays an important role in hypoxia sensing where cells sense a decrease in O2 and allow the organism to adapt. In the renal peri tubular fibroblasts, the VHL protein regulates the degradation of hypoxia-inducible factor 1 (HIF-1), which is made up of one α- and one β subunit and exists in three isoforms (HIF-1α, HIF-1β and HIF-1γ). O2 activates prolyl hydrolase which then hydroxylates HIF-α. If the O2 saturation is normal, this leads to the binding of VHL protein and ubiquitination of HIF-α and the destruction of the proteins. EPO synthesis is reduced. In hypoxia HIF-α associates with its β-subunit, the complex binds to hypoxia responsive elements leading to increased production of EPO. With a homozygous mutation of the VHL gene, the normal mechanism is disrupted, HIF-α is driven down the hypoxia pathway resulting in increased EPO production and erythrocytosis (Fig. 15.4-3 – Increased synthesis of erythropoietin in von Hippel-Lindau gene mutation). The disease is also called Chuvash polycythemia. Laboratory findings: the oxygen partial pressure at which Hb saturation of 50% is reached is low. Inappropriate elevated EPO concentrations, identification of the VHL gene mutation. \| \| Bisphosphoglycerate mutase deficiency /19/ During glycolysis the 2,3-bisphosphoglycerate (2,3-BPG) shunt involves conversion of 1,3-BPG to 2,3-BPG catalyzed by the enzyme bisphosphoglycerate mutase. In the erythrocyte 2,3-BPG binds to the hemoglobin tetramer and converts the Hb molecule to a low oxygen affinity state (Fig. 15.3-2 – The O2 dissociation curves in dependence of PO2 and the red cell content of 2,3-DPG). The binding of 2,3-BPG therefore shifts the oxygen affinity curve to the right. Lack of 2,3-BPG shifts the oxygen affinity curve to the left as the Hb tetramer is kept in a high O2 affinity state. Due to a mutation in the gene that encodes for bisphosphoglycerate mutase, the activity of the enzyme is low or absent. Little or no 2,3-BPG is formed and results in a left shifted oxygen affinity curve (Fig. 15.4-4 – Effect of 2,3-diphosphoglycerate (2,3-DPG) on O2 saturation curve) a high affinity Hb and a compensatory erythrocytosis. Laboratory findings: erythrocytosis, oxygen partial pressure at which Hb saturation of 50% is reached is low. Inappropriately elevated EPO concentrations, demonstration of bisphosphoglycerate mutase gene mutation. \| \| Neonates Immediately after delivery, neonates have a HCT ≤ 0.56 in umbilical cord blood and ≤ 0.70 in venous blood 2 hours later /5, 6/. Higher values are indicative of abnormal flow kinetics (hyper viscosity). Other authors select a threshold HCT of 0.65. According to one study /35/, only 47% of neonates with an umbilical cord blood HCT above 0.65 have hyper viscosity, and only 23% of those with hyper viscosity have erythrocytosis. 2,3-BPG concentrations in fetal erythrocytes are lower than in adults, and this leads to a left shift of the oxygen affinity curve. In consequence, HbF can bind O2 with greater affinity, and can accept the O2 from the maternal blood in the placenta. \| \| Smoker Smokers’ polycythemia is caused by CO intoxication, which leads to proliferation of erythropoiesis due to hypoxia and in which, additionally and for unknown reasons, the plasma volume is diminished /21/. Smokers often have leukocytosis of above 10 × 109/L. \| \| erythrocytosis in congenital heart defects Acquired heart defects are characterized by an increased red blood cell production because of an congenital erythropoetin (Epo) receptor mutation or an acquired bone marrow defect. Serum concentrations of Epo are usually lower than normal because of negative feed back that high blood oxygen levels exert on Epo producing cells. \| \| Acquired erythrocytosis – Generally Acquired erythrocytoses are differentiated into central and local hypoxia driven processes. They are associated with elevated EPO concentrations, and do not lead like polycythemia vera to an increase in platelets and leukocytes. \| \| – Hypoxia at altitude /21/ The HCT and the Hb values increase linearly to an altitude of approximately 4000 meters. Above that there is a disproportional increase up to about 6000 meters, from where on a decrease in erythropoiesis is found. The erythropoiesis is diminished dramatically when the arterial O2 saturation decreases to below 60%. An arterial O2 saturation of less than 92% is considered low. In blood with HCT values higher than 0.55 the blood viscosity is increased. This enhances the energy demand for the heart to pump blood and may impair the microcirculation. \| \| – Local hypoxic causes Inadequate O2 supply to the renal O2 sensors, due to impaired renal function, results in augmented EPO synthesis and erythrocytosis. This is the case in renal artery stenosis, end-stage renal disease, polycystic kidney disease, hydronephrosis, and following kidney transplantation. \| \| – Post-transplantation erythrocytosis (PTE) PTE is a complication of renal transplantation; the incidence is 5–17% /36/. The HCT rises to as high as above 0.60 many months following transplantation, making phlebotomy necessary. In such cases, thromboembolic complications can occur. The erythrocytosis is often transient in nature. The cause of the inappropriately high EPO secretion relative to the HCT is unknown. \| \| – Autonomous EPO synthesis Paraneoplastic polycythemia is found in cerebellar hemangioblastoma, meningioma, parathyroid gland hyperplasia, hepatocellular and renal cell carcinoma, pheochromocytoma, and leiomyosarcoma of the uterus /19/. \| \| – ESA therapy The administration of erythropoiesis-stimulating agents (ESA) for the purpose of EPO doping, leads to erythrocytosis. Androgen doping also leads to erythrocytosis. \| \| Thrombotic risk Certain disorders with elevated hematocrit, such as polycythemia, Chuvash erythrocytosis, primary familial and congenital polycythemia or erythrocytosis (EPOR mutation) and EPAS1 gain-of-function mutations, are associated with thrombotic complications. These conditions are characterized by diverse cellular and metabolic changes that could be directly associated with thrombotic risk, irrespective of hematocrit level /33/. \| Table 15.4-6 World Health Organization diagnostic criteria for polycythemia vera /20/ \| \| Criteria \| \| A1 \| Elevated RBC mass > 25% above mean normal predicted value, or Hb > 185 g/L in men, > 165 g/L in women \| \| A2 \| No case of secondary erythrocytosis, including absence of familial erythrocytosis; no elevation of erythropoietin due to hypoxia (arterial PO2 ≤ 92%), high affinity hemoglobin, truncated erythropoietin receptor, inappropriate erythropoietin production by tumor \| \| A3 \| Splenomegaly \| \| A4 \| Clonal genetic abnormality other than Ph chromosome or BCR/ABL fusion gene in marrow cells \| \| A5 \| Endogenous erythroid colony formation in vitro \| \| B1 \| Thrombocytosis > 400 × 109/L \| \| B2 \| WBC > 12 × 109/L \| \| B3 \| Bone marrow biopsy showing pan myelosis with prominent erythroid and megakaryocytic proliferation \| \| B4 \| Low serum erythropoietin level \| Polycythemia vera is diagnosed when A1+A2 + any other A, or A1+A2 + any 2B. RBC, red blood cell; Hb, hemoglobin; Ph, Philadelphia; WBC, white blood cell Table 15.5-1 Agents capable of inducing methemoglobinemia /5/ \| Local anesthetics \| Nitrites/ nitrates \| Analgesics \| \| Benzocaine Lidocaine Procaine \| Sodium nitrite Nitroglycerin Amyl nitrite Butyl nitrite Isobutyl nitrite Primaquine \| Phenazo- pyridine Phenacetin \| \| Anti- microbials \| Miscellaneous \| \| \| Dapsone Sulfanilamide Sulfathiazole \| Aniline dyes Chlorates Nitrobenzene Aminophenol Celecoxib (Cox-2 inhibitor) \| \| Table 15.5-2 Correlation of clinical symptoms with increasing proportion of metHb /7/ \| MetHb fraction (%) \| Symptomatology \| \| < 15 \| Usually asymptomatic \| \| 15–20 \| Cyanosis, headache, drowsiness \| \| 20–45 \| Marked cyanosis, nausea \| \| 45–70 \| Severe cyanosis, vomiting, confusion, seizures \| \| > 70 \| Usually lethal \| Table 15.5-3 Clinical symptoms in dependence of the COHb concentration /8/ \| COHb (%) \| Clinical and laboratory symptoms \| \| 0–10 \| No relevant symptoms (smokers) \| \| > 10–15 \| No relevant symptoms, possibly shortness of breath on physical exertion (heavy smokers) \| \| > 15–25 \| Usually no effect at rest, shortness of breath on physical exertion, possibly dizziness and headache; dilatation of cutaneous capillaries \| \| > 25–35 \| Headache, dizziness, vomiting, pulse acceleration, irritability, disturbances of judgment, mild fatigability, vision disorders \| \| > 35–45 \| Similar to 25–35%, but more severe, additionally state of confusion, signs of paralysis, fainting even on mild exertion \| \| > 45–55 \| Marked impairment of consciousness or unconsciousness, increased respiratory and pulse rate, collapse, danger of death with prolonged exposure \| \| > 55–65 \| Additionally seizures, respiratory paralysis \| \| > 65 \| Immediate danger of death \| Table 15.5-4 Hemolysis and plasma free hemoglobin \| Clinical and laboratory findings \| \| Venovenous and venoarterial extracorporal membrane oxigenation (ECMO) /20/ ECMO is a lifesaving support that is used in critically ill patients. Three main pathophysiologic aspects in the short-term ECMO setting are in patients with free hemoglobin: Regulation vascular tone. Free Hb interacts with NO and catalyzes transformation of NO into nitrate if oxygen-Hb is involved or into a ferric-NO complex if deoxy-Hb is involved. Subsequent NO depletion is then responsible for a loss in a major negative regulator of the vascular smooth muscle tone, inducing a potent vasoconstriction. The low NO-dependent vasoconstriction is responsible for both an increase in systemic and vascular pulmonary vascular resistance /25/. Abnormal coagulation and platelet activation. NO depletion is responsible for platelet function in endothelial behavior towards platelet aggregation and coagulation. Free Hb increases the procoagulant state in diseases such as paroxysmal nocturnal hemoglobinuria and sickle cell disease. Renal tubular toxicity: Heme-containing proteins are known to be toxic to kidneys. All the pathologic changes caused by free Hb in the bloodstream are associated with an increased morbidity and perhaps mortality. Severe hemolysis may be present during venovenous ECMO. A peak in free Hb and creatinine can be a predictor of acute renal failure. Severe increase in free Hb (> 500 mg/L) in the first 24 hours of ECMO onset is an independent predictor of mortality. In a study /26/ patients with free Hb > 300 mg/L had a 10-fold increase in hospital mortality and seemed associated with renal impairment. Two factors are recognized to promote and generate hemolysis in ECMO: thrombosis (head pump and oxygenator) and strong mechanical stress of erythrocytes (head pump) /20/. Hemolysis is a common phenomenon during ECMO support in most cases. Usually the intensity of hemolysis is low, generating tolerable free Hb concentration. \| \| Severe sepsis Results of a study are /27/: In non-survivors of severe sepsis free Hb concentration was twice the value compared to survivors (57 mg/L vs. 147 mg/L, Harboe method). In postoperative patients the free Hb concentration was 45–78 mg/L (ELISA method). Thirty-day survival of patients was markedly lower in patients with high free Hb than in patients with low free Hb. \| \| Massive Transfusion and/or transfusion of stored blood Aged stored blood hemolyzes after massive transfusion. Free Hb led to vascular and kidney injury indicating that intravascular hemolysis is a pathological mechanism in several human diseases /28/. \| Table 15.6-1 Hematocrit (HCT) correction and shift correction /9/ Table 15.6-2 Expected lower RPI thresholds and reticulocyte count as function of HCT /16/ \| HCT \| Reticuloc. × 109 \| RPI \| \| 0.35 \| 150 \| 2–3 \| \| 0.30 \| 250 \| > 3 \| \| < 0.25 \| > 250 \| > 4 \| RPI, reticulocyte production index Table 15.6-3 Reticulocyte reference intervals \| \| Relative (%) \| Absolute (109/L) \| \| Children \| \| \| \| Pre-term delivery /11/ \| 2.6 ± 1.0 \| 72 ± 29 \| \| Full-term neonates /12/ \| 2.4 ± 1.2 \| 74 ± 26 \| \| 0–14 days /12/ \| 5.3 ± 2.0 \| 318 ± 50 \| \| 14 days–1 yr /12/ \| 2.0 ± 0.5 \| 89 ± 23 \| \| 1–3 yrs /12/ \| 2.0 ± 0.7 \| 88 ± 20 \| \| 3–8 yrs /12/ \| 2.2 ± 0.7 \| 99 ± 28 \| \| 8–12 yrs /12/ \| 2.1 ± 0.8 \| 90 ± 27 \| \| 12 yrs to adult /12/ \| 2.3 ± 1.2 \| 92 ± 25 \| \| Adults /13/ manual \| 0.4–2.3 \| (19–111) \| \| \| \| Count (109/L) \| \| Adults /14/ automated \| \| \| \| Abbott CD 4000 \| ♀ \| 30–117 \| \| ♂ \| 31–115 \| \| Advia 120 \| ♀ \| 25.9–97.5 \| \| ♂ \| 36–101.1 \| \| Coulter LH 750 \| ♀ \| 13.9–98.3 \| \| ♂ \| 21.7–114.5 \| \| Horiba ABX, Pentra 12 \| ♀ \| 27–91 \| \| ♂ \| 39–113 \| \| Sysmex XE 2100 \| ♀ \| 19.8–80.7 \| \| ♂ \| 24.8–96.2 \| Values expressed as x ± s; Values are 2.5th and 97.5th percentiles. The values provided in Ref. /12/ are valid for flow cytometry with the Becton Dickinson FACSCalibur. Since reference intervals depend on the hematology analyzer used, intervals for adults from 5 well-known manufacturers are shown. Table 15.6-4 Diseases and conditions with reticulocytosis /17/ \| Clinical and laboratory findings \| \| Acute hemorrhage After an acute hemorrhage, the timing of the onset of reticulocytosis and its extent depend on the amount of blood loss. The rise usually occurs on the 2nd–3rd day and the reticulocytosis ranges from 5–15%. \| \| Hemolytic anemia Acute hemolytic anemias are associated with high reticulocyte counts but clinical signs may be absent. Reticulocytoses of > 50% are observed in immune hemolytic anemias. In efficient treatment with corticosteroids, the reticulocyte count declines within a few days. Acquired, mild hemolytic syndromes may be associated with a normal Hb concentration and a reticulocytosis of 2–5%. In congenital hemolytic anemias, with the exception of G-6PD deficiency, the RBC volume is changed and slightly elevated reticulocyte counts are present. In pyruvate kinase deficiency, a hereditary chronic hemolytic anemia with an autosomal mode of inheritance, the severity of the clinical manifestations is subject to great variability and reticulocytosis of up to 50%. After splenectomy, persistent reticulocytosis occurs which may range from 50–100% /22/. \| \| Hypersplenism A normocytic, slightly hemolytic anemia is present. The reticulocyte count is slightly elevated. In the peripheral blood smear, polychromasia and erythroblasts are found in some of the cases. \| \| Treatment of deficiency anemia In the case of deficiency of iron, folate, vitamin B6 or vitamin B12, the regeneration of the hypo regenerative erythropoiesis is indicated within one 1–2 weeks by a rise in the reticulocyte count, if there is efficient supplementation. After 2–4 weeks, a maximal reticulocytosis of 10–20% is measured. \| \| Athletes, injection of erythropoiesis-stimulating agents (ESA) /23/ Top-level athletes have a normal reticulocyte count but often an elevated immature reticulocyte fraction (IRF), dependent upon activation of erythropoiesis due to the mild hemolysis caused by participation in sports. The IRF increases 36 hours following bolus injection of ESA, reaching a maximal value after 3 to 4 days and normalizing after 7 days. The reticulocyte life span in circulation is prolonged from 1.7 to 3.4 days. The overall result is a doubling of the reticulocyte count. Following repeated ESA administration, reticulocyte values are increased from days 7–24 and remain elevated for a further 7 days after the last ESA dose. \| \| Aplasiogenic cytostatic therapy An increase in the reticulocyte count is an early surrogate marker of a recovery of hematopoiesis following chemotherapy /24/. \| \| Allogenic bone marrow transplantation The reticulocyte maturity index (RMI) is a good indicator of transplant function and is as valuable a transplantation marker as the rise in neutrophile granulocytes. A rise in both of these parameters is seen during the early post transplantation phase, if the transplant functions. In clinical or subclinical infections, the rise in neutrophil granulocytes may be absent. The RMI, however, is not influenced by infectious processes and increases in the presence of a functioning transplant /25/. \| Table 15.6-5 RMI and IRF reference intervals \| \| LFR% \| MFR% \| HFR% \| \| RMI adults /29/ \| \| \| \| \| Sysmex 9500 \| 85–97 \| 3–14 \| < 2,5 \| \| Advia 120 \| 88–98 \| 2–11 \| ≤ 2,0 \| \| IRF adults /29/ \| \| \| \| \| Cell DYN 4000: proportion of reticulocytes with LFR, MFR and HFR in each case 0.14–0.35. \| \| \| \| RMI, reticulocyte maturity index; IMR, immature reticiulocyte fraction; LFR, low size fraction reticulocytes; MFR, medium size fraction reticulocytes; HFR, high size fraction reticulocytes Table 15.6-6 Diagnostic significance of reticulocyte count and reticulocyte maturity index (RMI) /32/ \| Anemia \| Reticulocyte count \| RMI \| \| Aplastic anemia \| ↓ \| ↓ \| \| Aplastic crisis \| ↓ \| ↓/N \| \| Intrinsic marrow hypoplasia \| ↓ \| N/↑ \| \| Bone marrow regeneration \| ↓ \| N/↑ \| \| Anemia of chronic disease \| ↓/N \| N \| \| Iron deficiency \| ↓/N \| ↑ \| \| Thalassemia \| N/↑ \| N/↑ \| \| Myelodysplastic syndrome \| ↓/N/↑ \| N/↑ \| \| Folate/vitamin B12 deficiency \| ↓/N \| ↑ \| \| Hemolytic anemia \| ↑ \| ↑ \| \| Hemorrhage \| N/↑ \| ↑ \| Immature reticulocyte fraction: N, normal; ↓ low; ↑ high Table 15.6-7 Reference interval for reticulocyte hemoglobin (CHr, Ret-He) \| Adults \| CHr \| 28–35 pg /37/ \| \| Ret-He \| 28–35 pg /36/ \| \| CHCMr \| 270–330 g/l /34/ \| \| RFHb \| 1.76 ± 0.59 g/L /38/ \| \| Neonates \| CHr \| 33–38 pg /39/ \| \| Children \| CHr \| 27.5–33 pg /40/ \| RFHb, Hb content of the reticulocyte fraction; CHCMr, Reticulocyte Hb content Table 15.7-1 Normal human hemoglobins \| Hb type \| Chain formula \| Occurrence \| Quantitative proportion \| \| HbA \| α2β2 \| Children > ½ year Adults \| 97.0% \| \| HbA \| α2δ2 \| Children > 1 year Adults \| 2.5–3.0% \| \| HbF \| α2γ2 \| Fetal/neonatal peroid Children < 1 year \| Children age-dependent, adults up to 0.5% \| \| Hb (embryo) \| Normally no longer detectable after birth \| \| \| Table 15.7-2 Prevalence of hemoglobinopathy gene carriers in the world’s population /4/ \| Country \| Gene carrier \| \| African nations \| 5–30% \| \| Arab nations \| 5–40% \| \| Up to 60% regionally \| \| Central Asia and India \| 10–20% \| \| Southeast Asia \| 5–40% \| \| Up to 70% regionally \| \| USA and Central America \| 5–20% \| \| Italy \| 7–9% \| \| Greece \| 6–7% \| \| Turkey \| 7–10% \| \| Germany, Great Britain, Spain, France, The Netherlands, Belgium, Scandinavian countries \| Among total population 0.5 to 1% \| \| Among immigrants 5% \| \| Balkan countries \| 2–5% \| \| Russia \| Rare \| \| Caucasus countries \| Up to 5% \| Table 15.7-3 The most important hemoglobinopathies in Germany \| β-thalassemia syndromes \| 65.8% \| \| HbS \| 21.2% \| \| α-thalassemia syndromes \| 7.8% \| \| HbE \| 2.1% \| \| HbC \| 1.8% \| \| Unstable Hb abnormalities \| 0.2% \| \| Reference value: 35,831 laboratory diagnoses \| 100% \| Table 15.7-4 Program for the laboratory diagnostic investigation of hemoglobinopathies \| Hematology \| Complete blood count and, possibly, reticulocyte count \| \| Clinical chemistry \| Iron status: ferritin, transferrin saturation Hemolysis markers: haptoglobin, LD, bilirubin \| \| Hemoglobin analysis \| Electrophoresis and/or HPLC HbF determination, HbF cell staining Hb solubility test, sickle cell test for HbS \| \| Molecular genetics \| DNA purification PCR amplification DNA sequencing MLPA deletion diagnostics \| Table 15.7-5 Basic information on molecular genetic methodology \| DNA purification \| Original material \| \| Polymerase chain amplification \| Technique for rapid amplification (duplication) of specific DNA sequences of different size. \| \| DNA sequencing \| Analysis of the nucleotide sequence of DNA molecules or fragments using enzymatic or chemical methods. The analysis is performed using automated DNA sequencers. \| \| MLPA deletions \| Multiplex ligation-dependent probe amplification, molecular genetic method with which, following large genomic deletions or duplications, individual gene regions or an entire gene can be investigated. \| Table 15.7-7 Basic forms of hemoglobinopathies \| Hemoglobinopathy \| Basic defect \| \| 1. Thalassemia syndromes α-, β-, δβ-, γδβ-, γ and δ-thalassemia. Thalassemic hemoglobinopathies 2. Abnormalities of hemoglobin structure: Structural defects of α- or β globin chains (e.g., HbS, HbC, HbE) 3. Types of combination and interaction (e.g., HbSC disease HbS-β thalassemia HbE-α thalassemia) \| ad 1) Reduced or lack of synthesis of α-, β-, γ- or δ-globin chains ad 2) Abnormal structure of α-chains or β-chains: Combination of different forms of thalassemia with one another or with abnormal hemoglobins ad 3) Combinations of different abnormal hemoglobins with one another \| Table 15.7-6 Reference intervals for hemoglobins /2, 3/ \| \| Full-term newborn infants \| Children 1–2 yrs \| Children > 2 yrs and adults \| \| HbA \| 17.7 (13.1–22.3) 15–40 \| 96.0 (95.0–98.2) \| 97.5 (97–98.5) \| \| HbA2 \| 0.25 (0.05–0.45) \| 1.8 (1.5–3.0) \| 2.5 (1.5–3.2) \| \| HbF \| 81.7 (77.5–86.9) 60–85 \| 1.5 (≤ 2.0) 0.8 (0.4–1.0) \| 0.5 (≤ 0.5) \| Reference intervals for children < 1 year are age-dependent. Data expressed in percent. Values expressed as x ± s. Methods: if not otherwise specified: electrophoresis; chromatography; alkali denaturation. Table 15.7-8 Compilation of the most important criteria of α-thalassemias \| Hereditary status/ diagnosis \| Arrangement of α-globulin gene cluster \| \| Erythrogram, hemoglobin pattern (hp) \| Cardinal symptoms \| \| Normal findings \| ▲▲/▲▲ \| αα/αα \| Hb norm. MCH norm. hp norm. \| No symptoms \| \| Heterozygous α+ thalassemia (α thalassemia minima) \| ▼▲/▼▼ \| –α/αα \| Hb norm. MCH < 27 pg hp norm. \| No symptoms, slight blood count changes \| \| Homozygous α+ thalassemia (α thalassemia minor) \| ▼▲/▲▼ \| –α/–α \| Hb norm. or ↓ MCH < 26 pg hp norm. \| Mild anemia, substantial blood count changes \| \| Heterozygous α0 thalassemia (α thalassemia minor) \| ▼▼/▲▲ \| ––/αα \| Hb norm. or ↓ MCH < 24 pg hp norm. \| Mild anemia, substantial blood count changes \| \| Mixed heterozygosity α+/α0 thalassemia (HbH disease) \| ▼▼/▼▲ \| ––/–α \| Hb 80–100 g/L MCH < 22 pg hp: HbH ≈ 10–20% \| Variable chronic hemolytic anemia \| \| Homozygous α0 thalassemia (Hydrops fetalis syndrome) \| ▼▼/▼▼ \| ––/–– \| Hb < 60 g/L MCH < 20 pg hp: Hb Bart’s 80–90% Hb Portland ≈ 10–20% \| Life-threatening fetal anemia, generalized hydrops \| Table 15.7-9 Compilation of the most important criteria of β-thalassemias \| Hereditary status Diagnosis \| \| Erythrogram \| Hb pattern \| Cardinal symptoms \| \| Heterozygous β-thalassemia (β thalassemia minor) \| β++ β+ β0 \| Hb (g/L) ♂ 90–150 ♀ 90–130 MCV 55–75 fL MCH 19–25 pg \| HbA2 > 3.2% HbF 0.5–6% \| Mild anemia without disease \| \| Homozygous β-thalassemia Compound heterozygous β-thalassemia \| β+/β+ β0/β0 β+/β0 \| Hb < 70 MCV 50–60 fL MCH 14–20 pg \| HbA2 variable HbF 70–90% \| Severe disease with permanently transfusion-dependent anemia \| \| Mild homozygous or compound heterozygous β-thalassemia (β-thalassemia intermedia) \| β+/β+ I β+/β++ β+/β0 β0/β0 \| Hb 60–100 MCV 55–70 fL MCH 15–23 pg \| HbA2 variable HbF to 100% \| Moderately severe disease, variably transfusion-dependent \| β-thalassemia major; , Influencing variables Table 15.7-10 Clinical classification of the most important pathological hemoglobin variants \| Variants associated with vascular occlusion and hemolysis \| e.g., HbS, HbC \| \| Variants associated with the thalassemia phenotype \| e.g., HbE, Hb Lepore \| \| Unstable hemoglobins associated with hemolysis \| e.g., Hb Köln, Hb Leiden \| \| Variants associated with disturbed oxygen transport function (familial erythrocytosis) \| e.g., Hb Ohio \| \| Variants associated with pathological methemoglobin (familial cyanosis) \| e.g., HbM Iwate \| Table 15.7-11 Diagnostic characteristics of sickle cell β-thalassemia \| HbS-β0-thalassemia \| HbS-β+-thalassemia \| \| Appearance Similar to severe sickle cell disease \| Appearance Depending upon the residual β-globulin activity = mild to severe sickle cell disease \| \| Laboratory diagnostic investigation and findings Hematology: Hb 60–100 g/L MCH < 22 pg \| Laboratory diagnostic investigation and findings Hematology: Hb 90–120 g/L MCH < 24 pg \| \| Clinical chemistry Hemolysis parameters positive \| Clinical chemistry Hemolysis parameters positive \| \| Hemoglobin analysis HbA2 > 3.5% HbS > 80% HbF < 20% HbA 0% \| Hemoglobin analysis HbA2 3.5% HbA 3–30% HbS 55–75% HbF ≈ 20% \| \| Molecular genetics Demonstration of the following mutations: 1. HbS = β codon 6 GAG → GTG heterozygous and 2. β0-thalassemia mutation heterozygous = compound heterozygosity \| Molecular genetics Demonstration of the following mutations: 1. HbS = β codon 6 GAG → GTG heterozygous and 2. β+-thalassemia mutation heterozygous = compound heterozygosity \| Table 15.7-12 Characteristic features of the most important hemoglobin abnormalities \| Sickness, Genotype \| Erythrogram \| Hemoglobin pattern \| Cardinal symptoms \| \| Sickle cell disease HbSS \| Hb 60–90 g/L Normochromic sickle cells Positive hemolysis markers \| HbS 55–90% HbA2 > 3.5% HbF < 10 to > 20% \| Sickle cell crises/pain crises, acute organ syndromes, chronic hemolytic anemia \| \| HbS heterozygosity HbAS \| Normal \| HbS 35–40% HbA2 ≥ 3.5% \| No manifestation of disease \| \| Sickle cell β+-thalassemia HbSβ+-thalassemia \| Hb 90–120 g/L Hypochromia Microcytosis \| HbS > 55% HbF > 20% HbA2 > 3.5% \| Variable, mild sickle cell disease \| \| Sickle cell β0-thalassemia HbSβ0-thalas- semia \| Hb 60–100 g/L Hypochromia Microcytosis \| HbS > 80% HbF < 20% HbA2 > 3.5% \| Severe sickle cell disease \| \| HbSC disease HbSC \| Hb 100–130 g/L Target cells MCV < 75 fL \| HbS 50% HbC 50% HbF < 5% \| Attenuated symptoms of sickle cell disease, chronic hemolytic anemia \| \| HbC disease HbCC \| Hb 100–120 g/L Target cells MCV < 75 fL MCHC > 350 g/L \| HbC > 95% HbA2 2,5% HbF 0.5% \| Pain crises, organ dysfunction, chronic hemolytic anemia \| \| HbC heterozygosity HbAC \| Normal \| HbC 50% HbA 47% HbA2 3% \| No manifestation of disease \| \| HbE heterozygosity HbAE \| Hb normal or low Hypochromia \| HbE 25–35% \| Mild hypochromic anemia \| \| HbE disease HbEE \| Hb 100–140 g/L High red cell count MCH 20 pg MCV 65 fL Target cells \| HbE > 95% HbA2 2.5% HbF < 3% \| Mild anemia, hemolysis due to infection/medication \| \| HbE-β+-thalassemia HbE-β+-thalas-semia \| Hb variable ↓ Hypochromia Microcytosis \| HbE + HbA2 25–80% HbF 6–50% HbA 5–60% \| Variable intermediate hypochromic anemia \| \| HbE-β0-thalassemia HbE-β0-thalas-semia \| Hb < 80 g/L MCV < 60 fL MCH < 22 pg \| HbE to 85% HbA2 < 5% HbF 15–25% \| Same as β-thalassemia major \| \| Hemoglobino- pathy with unstable Hb HbX; some 150 different variants HbX/HbA \| Hb variable – significantly anemic; Heinz bodies, hemolysis due to viral infection/ medication \| HbX 20% HbA2 3–4% HbF < 5% \| Variable, partially transfusion-dependent chronic hemolytic anemia \| \| Abnormal hemoglobin with disturbed O2 transport/ function Different variants \| Polyglobulia Increase in metHb \| Different, according to the type of abnormality \| Congenital cyanosis associated with HbM abnormalities, congenital polyglobulia associated with Hb abnormalities with increased O2 affinity \| Table 15.8-1 Reference intervals for erythrocyte enzymes \| Enzymes /1, 6/ \| Substrate turnover (g Hb/min.) \| Substrate turnover (1011 red cells/min.) \| \| Glycolysis \| \| \| \| Pyruvate kinase \| 20.2 ± 2.2 \| 41 ± 10 \| \| Hexokinase \| 1.0 ± 0.1 \| 2.3 ± 0.5 \| \| Glucose phosphate isomerase \| 44.7 ± 4.8 \| 124 ± 13 \| \| Triosephosphate isomerase \| 2,180 ± 254 \| 6,055 ± 705 \| \| Pentose phosphate cycle and glutathione metabolism \| \| \| \| Glucose-6-phosphate dehydrogenase \| 11.0 ± 1.6 \| 30.6 ± 4.5 \| \| 6-phosphogluconate dehydrogenase \| 9.5 ± 1.5 \| 26.2 ± 4.1 \| \| Glutathione reductase \| 4.6 ± 0.8 \| 25.7 ± 3.0 \| Values expressed as x ± s; Substrate turnover data in μmol Table 15.8-2 Findings and enzymatic characteristics associated with the most important erythrocyte enzymopathies \| Disease \| Findings \| Enzyme characteristics \| \| Glucose-6- phosphate dehydrogenase (G6PD) deficiency \| Types of presentation: acute, potentially life-threatening hemolytic crisis following oxidative stress; most severe anemia, formation of Heinz bodies, basket cells, punched out cells; hemoglobinemia, hemoglobinuria; in neonates hyperbilirubinemia \| Heterozygous, hemizygous or homozygous G6PD deficiency of the Mediterranean, African or Asian type. Very different enzyme activities. \| \| Chronic hemolytic anemia \| Homozygosity or heterozygosity of various G6PD deficiency types. Occurring mostly in the Caucasian population of North America and Europe. \| \| Pyruvate kinase (PK) deficiency \| Variable symptomatology from mild, compensated hemolytic anemia to moderate severity, and to severe forms of hemolysis with very pronounced anemia. Hepatosplenomegaly, macrocytosis, pyknocytosis, poikilocytosis, relatively few reticulocytes. \| Homozygous enzyme deficiency. Very variable residual enzyme activity in the different pathological enzyme types. Special types: double heterozygosity for two enzyme variants. \| \| Heterozygous PK deficiency \| Not clinically and hematologically apparent, except possibly for neonatal hyperbilirubinemia. \| Enzyme activity of heterozygous defect carriers approximately 50% of normal. \| Table 15.8-3 Medication and chemicals that cause hemolysis in G6PD deficiency \| Acetanilide Methylene blue Naphthalene Niridazole Nitrofurantoin Phenylhydrazine Pamaquine Pentaquine \| Primaquine Sulfanilamide Sulfacetamide Sulfapyridine Sulfamethoxazole Thiazolsulfone Toluidine blue Trinitrotoluol \| Table 15.10-1 Reference intervals for erythropoietin in dependence of the hemoglobin concentration \| \| EPO (U/L) \| Hemoglobin (g/L) \| \| Adults /1/ \| 6–32 \| 125–180 \| \| Children /4/ \| \| \| \| 0–6 days \| 33.0 ± 31.4 \| 156 ± 22 \| \| 7–50 days \| 11.7 ± 3.6 \| 128 ± 11 \| \| 51–100 days \| 21.1 ± 5.5 \| 114 ± 10 \| \| 101–150 days \| 15.1 ± 3.9 \| 112 ± 12 \| \| 151–200 days \| 17.8 ± 6.3 \| 125 (n = 2) \| \| > 201 days \| 23.1 ± 9.7 \| 118 ± 8 \| Values are the 2.5th and 97.5th percentiles. Values expressed as x ± s. The children’s cohorts (normal weight at birth) include only 2–13 individuals. Table 15.10-2 Diseases and conditions associated with an adequate rise in EPO concentration \| Clinical and laboratory findings \| \| Hypoxia In healthy individuals, a rise in serum EPO concentration occurs following 2 hours in an environment containing 11% O2, and after 8 hours the basal values are approximately doubled. A similar rise in EPO occurs 6 h following the second of two ascents to an altitude of 4,300 meters with an overnight stay at 1,200 meters after the first ascent. Nonetheless, with repeated climbs the EPO rise is more moderate /25/. Hypoxic conditions, caused by severe cyanotic heart disease or hemoglobinopathy with a mutation resulting in high affinity to O2, such as Hb Köln or Hb York, also lead to a rise in EPO. Hypoxia causes a reduction in hepcidin expression, resulting in increased enteral iron absorption and increased iron release from macrophages. The resulting elevated plasma iron concentrations are the prerequisite for an increase in erythropoiesis in chronic hypoxia. Refer also to Tab. 15.4-5 – Diseases and conditions with increased HCT. \| \| Chronic pulmonary disease Serum EPO is elevated in 69% of patients with chronic hypoxic pulmonary disease and secondary polycythemia /26/. \| \| Pregnancy The red blood cell mass falls during the first trimester of pregnancy, increases during the second trimester, and reaches its maximal volume close to the due date. At that time the red blood cell mass is some 125% of that seen in non-pregnant women. The plasma volume increases by 150%. EPO levels first increase significantly during the second trimester. In comparison with non-pregnant women with EPO values of 16.4 ± 4.1 U/L these are, during the 1st, 2nd and 3rd trimesters and postpartum, respectively 19.1 ± 6.2, 28.4 ± 15.5, 37.7 ± 24.2 and 35.1 ± 34.7 /27/. \| \| Acute fetal distress EPO concentrations in fetal umbilical cord blood are low, on average 1.6 U/L, increase, however, continuously from the 37th week of gestation onwards. In fetal anemia, the fetus responds from around the 24th week of pregnancy with a marked rise in EPO. This is also the case in all other conditions of deficient O2 supply to the fetus /8/. With the determination of EPO in the amniotic fluid, it is possible to distinguish acute and chronic causes of intrauterine fetal death. In acute intrauterine fetal death, amniotic fluid EPO concentrations are below 20 U/L, while with chronic causes they are 49.9–391 U/L /28/. \| \| Autologous blood donation With a collection of 550 mL of blood, a mass of 75 g of Hb is lost; it is replaced within, on the average, 36 days. In one study of autologous blood donation for hip replacement, consisting of five blood collections on days 1, 3, 7, 14 and 21, the Hb value dropped from 141 ± 10 g/L to 110 ± 9 g/L /29/. Serum EPO rose following each phlebotomy procedure, with a peak value of 35.8 ± 15 U/L on day 16. The rise that occurred with each donation was followed by a drop to a somewhat lower plateau. Overall, the EPO levels increased in a step-wise manner until day 16. \| \| Autologous re transfusion of blood /30/ The collection of 20% of one’s own blood and its replacement with hydroxy ethyl starch raised EPO concentration by 4-fold within one day, before it then decreased exponentially. The reticulocyte count increased 2.4-fold within one week, and remained high for an additional week. The Hb value declined by 15% for a period of 2 weeks. Soluble transferrin receptor (sTfR) rose within 14 days by 60%. The re transfusion of 800 mL of erythrocytes 4 weeks after blood collection acutely increased the Hb by 8%; 7 days later, the initial value prior to blood collection was reached. EPO concentration remained unchanged, the reticulocyte count decreased by 30%, sTfR remained elevated for 3 days until re transfusion and then decreased. \| \| Anemia In anemia with adequate EPO response, there is an inverse logarithmic relationship between the Hb level and the HCT, on the one hand, and EPO concentrations, on the other hand (Fig. 15.10-2 – Expected range of EPO concentration as a function of the hematocrit value) /8/. Aplastic anemia: in patients with a Hb value of 90 ± 14 g/L (x ± s) the mean EPO concentration was 408 U/L (range: 180–2700 U/L) /25/. Iron deficiency anemia, hemolytic anemia: in uncomplicated iron deficiency anemia and in auto-immune hemolytic anemia, with a Hb value of some 90 g/L, EPO concentrations were 50–100 U/L. In the study, mean EPO concentrations in healthy volunteers with a Hb value of 130–150 g/L were 11–12 U/L /32/. Sickle cell anemia: in chronic non-compensated hemolytic anemia in adults, including sickle cell anemia, EPO values are lower than in acute hemolytic syndromes. The reason is that in chronic hemolytic anemia, the tissue oxygen supply is better accounted for by means of a reduction in O2 affinity for heme or by compensatory mechanisms such as heart rate increase than in the acute hemolytic syndromes /31/. Refer also to Section 15.3.5.8 – Oxygen binding by EPO. β-thalassemia syndrome: high EPO values are measured in β-thalassemia/HbE disease which, in comparison with isolated β-thalassemia or isolated HbE disease, causes severe anemia. Thus, in a study /32/ in children aged 4–10 years, the following mean EPO values were determined: β-thalassemia 24 U/L, HbE disease 16 U/L, β-thalassemia/HbE disease 372 U/L, healthy controls 19 U/L. Myelodysplastic syndrome: EPO values are always greater than 100 U/L. Cancer-related anemia: in children with malignant tumors, anemia is not based upon inadequately low EPO production. This is true of solid tumors as well as of hemoblastosis. In adults, however, decreased EPO synthesis is believed to be present /32/. \| Table 15.10-3 Diseases and conditions with inadequately low increase in EPO concentration \| Clinical and laboratory findings \| \| Chronic kidney disease The kidneys are the key site of EPO synthesis. Chronic kidney disease (CKD) is associated with normocytic, normochromic anemia if the glomerular filtration rate is below approximately 30 [mL × min–1 × (1.73 m2)–1]. This is usually the case with serum creatinine values of greater than 1,5–2,0 mg/dL (133–177 μmol/L). Due to damage to the renal cells that are responsible for EPO synthesis, no adequate rise in serum EPO occurs in patients with anemia caused by CKD /25/. Patients with renal insufficiency and anemia have normal EPO concentrations. However, this is not sufficient to stimulate the erythroid precursor cells to an extent allowing 2 × 1011 senescent erythrocytes to be replaced daily in the circulation, as required. In the absence iron deficiency, the hemoglobinization of the red cells proceeds normally. The feedback mechanism between the O2 sensor and EPO synthesis is, however, still functional in patients with CKD (Fig. 15.10-7 – Erythropoietin feedback control loop). Thus, acute bleeding and acute hypoxia result in an increase in EPO concentration. Similarly, viral, drug-associated or toxic hepatic disease lead to a temporary rise in serum EPO concentrations, due to the cell regeneration /30/. Patients with renal insufficiency associated with polycystic kidney have significantly higher Hb and EPO values than other hemodialysis patients. In acute kidney failure, EPO concentrations are low during the anuric phase, but they normalize once again following restoration of renal function. With a functional graft, following renal transplantation, EPO is normalized if the HCT exceeds 0.32. The values are lower in cases of chronic rejection reaction. In one study /34/, EPO values in situations of stable grafts, acute rejection and chronic rejection were, respectively, 24.0 ± 19.7 U/L, 35.6 ± 33.9 U/L and 6.2 ± 3.4 U/L. \| \| Anemia of chronic disease /35/ In anemia of chronic disease (ACD), which includes anemia of infection, anemia of chronic inflammation, renal anemia and cancer-related anemia (CRA), systemic elevation of pro-inflammatory cytokines such as TNF-α, IL-1 and IL-6 are believed to inhibit messenger-RNA synthesis for EPO /36/. Impairment of EPO binding to its receptor on the erythroid precursor cells is also suspected. Altogether, the result is EPO secretion that is inadequately low in relation to the extent of the anemia. The feedback mechanism between the O2 sensor and EPO synthesis is, however, functional. \| \| Multiple trauma The pathogenesis of anemia in patients with multiple trauma is based on a combination of gastrointestinal blood loss, renal and hepatic insufficiency, and reduced erythrocyte life span. Relative to the anemia, serum EPO concentrations in these patients are inadequately low in the presence of elevated pro-inflammatory cytokines /37/. \| \| Neonatal anemia During the first week of life, all newborns experience a physiological decrease in the red blood cell mass, with the development of anemia. In healthy newborns, the nadir of the Hb value is seldom below 90 g/L at the age of 10–12 weeks. This fall occurs earlier and is more marked in premature infants, particularly in those who are ill. The Hb value usually falls to 80 g/L in those with a birth weight of 1–1.5 kg, and to 70 g/L in those with a birth weight of less than 1.0 kg. While the Hb decline is well tolerated in full-term babies, a blood transfusion is frequently required in pre term neonates. The cause of anemia is diminished EPO synthesis. At term, 70–75% of the EPO is formed in the liver. However, the liver is believed to respond in a less sensitive manner than the kidneys to anemia and tissue hypoxia, and this in particular in premature infants, resulting in inadequate stimulation of erythropoiesis /38/. \| \| Primary familial and congenital polycythemia (PFCP) Primary familial and congenital polycythemia (PFCP), also known as benign erythrocytosis or familial erythrocytosis, is relatively rare; its inheritance is autosomal dominant. It is characterized by increased erythroid mass, elevated Hb value, increased erythroid precursor cell sensitivity for EPO, and decreased serum EPO concentrations /39/. \| \| Polycythemia vera (PV) Low EPO concentrations are measured in PV, before and after treatment. At presentation, patients with Hb levels of ≥ 180 g/L usually have an EPO value of below 2.9 U/L. This can, however, also be the case in relative polycythemia. In this apparent erythrocytosis, however, normalization of the EPO value, or even a slight increase thereof, occurs following phlebotomy accompanied by a decrease in the HCT. In PV, on the other hand, the EPO value remains low /5/. \| \| Multiple myeloma (MM) Anemia is a frequent complication in MM, especially in advanced disease. The cause of the anemia is less a reduction in erythropoiesis due to plasma cell infiltration of the marrow than intrinsic hypo proliferation of erythropoiesis and inadequately low EPO synthesis. This occurs in 60% of MM patients with renal insufficiency, but it is also seen with normal renal function /40/. \| Table 15.10-4 Diseases and conditions with a disproportionate rise in EPO concentration \| Clinical and laboratory findings \| \| Malignant tumor EPO that is formed ectopically can, as a para neoplastic syndrome, trigger polycythemia /41/. This is the case in 63% of patients with renal cell carcinoma and 23% of patients with hepatocellular carcinoma /42/. Likewise, Wilms tumor, cerebellar hemangiomas and uterine fibromyomas can be associated with elevated EPO levels. Polycythemia does not necessarily have to emerge clinically, due to chronic hemorrhage or tumor-associated anemia. Following successful surgery, the excessive EPO formation usually normalizes; if however, there is a recurrence of the tumor, EPO formation will again be elevated. Nonetheless, the EPO value cannot be taken as an indicator of complete surgical resection, since clearly defined threshold values do not exist. EPO is, however, suitable for monitoring purposes; a continuous increase allows for the identification of a relapse /41/. \| \| Secondary erythrocytosis Secondary erythrocytosis must be differentiated from polycythemia vera. Elevated EPO values which go against a diagnosis of polycythemia vera (primary erythrocytosis) have to be excluded /5/: Presence of a high-affinity Hb mutation (Hb Köln, Hb York): more than 50 mutants of the αHb gene and βHb gene have been described. Due to their enhanced O2 binding the high-affinity mutants on the α1/β2 region cause diminished O2 release in the tissues, resulting in a compensatory elevation of EPO and the erythroid mass. See Fig. 15.3-3 – Structure and function of hemoglobin. Hereditary methemoglobinemia: O2 binds reversibly to deoxyhemoglobin Fe2+, but not, however, to methemoglobin (Fe3+). Furthermore, in some subunits of the Hb tetramer, Fe3+-heme increases the O2 affinity of Fe 2+-heme. In consequence, the O2 dissociation curve is displaced to the left (Fig. 15.3-2 – O2 dissociation curves), and less O2 is released to the tissues. This leads to a compensatory increase in EPO synthesis and in the erythroid mass. Hereditary 2,3-diphosphoglycerate (2,3-DPG) deficiency due to a defect in 2,3-diphosphoglycerate mutase (DPGM): the O2 affinity for Hb is increased, with the result that less O2 is released to the peripheral tissues. A compensatory elevation in EPO occurs, and the erythroid mass in increased. \| Table 15.10-5 Tests for the assessment of erythropoiesis under ESA therapy \| Clinical and laboratory findings \| \| Blood count Basic investigation: the complete blood count serves for the assessment of the extent of the anemia and its classification. Findings which may lead to consideration of ESA therapy include normocytic, normochromic anemia with an Hb concentration of 80–110 g/L, a platelet count of > 100 × 109/L, an EPO concentration < 100 U/L or an O/P ratio of less than 1.2, and absence of leukocytosis. Monitoring: a rise in the Hb concentration is a criterion of erythroid proliferation and Hb formation and, thereby, an indicator of therapeutic success. Chronic kidney disease /20/: the Hb value should be determined every 1–2 weeks, at the start of treatment as well as with changes in the ESA dose, until a stable value has been reached. The goal is an increase in the Hb concentration of 10–20 g/L/month. If, following the initiation of therapy or an increase of the ESA dose, no increase by 7 g/L is achieved within 2–4 weeks, the ESA dose should be raised by 50%. If, following the initiation of therapy or an increase of the ESA dose, the Hb value increases by more than 25 g/L/month, or if the Hb level exceeds the target value, the weekly ESA dose should be reduced by 25–50%. The lack of an increase of at least 10 g/L/month is a sign of an inadequate response. The most important reasons for this is functional or absolute iron deficiency or inflammation. Cancer patients /12/: with adequate ESA therapy, an Hb increase of 10–20 g within 6–8 weeks is expected. Patients who do not meet this target are considered to be non-responders. The causes are, primarily, functional or absolute iron deficiency, or tumor progression. \| \| Reticulocyte count, reticulocyte Hb (CHr, Ret-He) Basic investigation: reticulocytes are released by the bone marrow 18–36 hours before they mature into erythrocytes. They provide insight into the functional status of the bone marrow (hypo-, normo- or hyper regenerative erythropoiesis). The CHr or Ret-He provide information as to whether normochromic or hypochromic red blood cells will be formed. Monitoring: in the early phase of ESA stimulated erythropoiesis (days 1–5) reticulocytosis, reflecting the release of immature reticulocytes from the marrow rather than an increase in erythropoiesis, occurs. An increase in the reticulocyte count of > 40 × 109/L following 4 weeks of therapy is an indication of a positive erythroid response /10/. If functional iron deficiency develops under ESA, the CHr or Ret-He decrease within 10 days after the start of treatment. \| \| Soluble transferrin receptor (sTfR) Under ESA therapy, the serum concentration of sTfR is primarily an indicator of erythroid proliferation rather than of functional iron deficiency. Proliferation of erythropoiesis leads to a rise in sTfR concentration /42/. Basic investigation: the most important therapeutic effect of ESA is the enhancement of erythropoiesis. In normal iron supply, there is a positive correlation between erythropoiesis and the sTfR concentration. The basal sTfR value is, therefore, an indicator for the assessment of erythropoiesis prior to the start of therapy. Elevated values are indicative of hyper proliferative erythropoiesis, that is to say, of an increase in the erythron. The absence of an elevation relative to the basal value indicates no response to ESA. Monitoring: a rise of the sTfR value of > 20% within 4 weeks is an indicator of responsiveness; the increase often takes place as early as within 10 days following the start of therapy. \| \| Erythropoietin (EPO) The determination of endogenous EPO formation should be performed, for the assessment of erythroid proliferation, as the basic evaluation prior to a planned ESA therapy in all patients, with the exception of those with renal anemia. The EPO values in anemic patients cannot simply be compared with the reference interval in healthy individuals. The relation to the Hb concentration or the HCT, and the determination of the expected (predicted = P) value by means of a reference curve (Fig. 15.10-2 – Relationship between HCT and EPO concentration) is important. In individual patients, with the generation of the O/P ratio, the measured (observed = O) value is then set in relation to the expected value. The 95% confidence interval for the O/P ratio is 0.8–1.2 /10/. An O/P ratio of below 0.8 is indicative of inadequately low endogenous EPO formation and apparently good responsiveness to ESA. In myelodysplastic syndrome the O/P ratio is high, and the response to ESA is low. \| \| Ferritin Ferritin is a marker of iron stores. With ESA therapy, ferritin levels fall in comparison with the basal value. Under ESA therapy, the ferritin value reflects the provisioning of the functional iron compartment (which can increase 3-fold) with iron. Many patients have inappropriately high ferritin values, relative to their available iron stores because the synthesis of ferritin is increased as a result of an acute phase response, or because ferritin is released due to chronic damage of the hepatocytes. Conditions such as these are seen, for (e.g., infection, chronic inflammation, hepatocellular injury, with malignant tumors, in hyperthyreosis, alcoholism, and with the ingestion of oral contraceptives). In order that EPO resistance due to iron deficiency does not develop with EPO therapy, ferritin concentrations must always be sufficiently high. Basic investigation: in patients with renal insufficiency, ferritin levels before the start of ESA therapy must be > 100 μg/L (optimally 200–500 μg/L) /20/. Monitoring: in the USA, 50% of hemodialysis patients who are receiving ESA therapy have ferritin values of > 800 μg/L; in Europe, the corresponding value is 500 μg/L. In routine therapy, ferritin values should be > 500 μg/L. In order to prevent iron overload, values of > 800 μg/L should be avoided /43/. At the initiation of ESA therapy, ferritin should be monitored every 3 months in patients receiving iron intravenously, and every 4–6 weeks in those not being treated with iron. \| \| Transferrin saturation (TfS) TfS is an indicator of functional iron, but it is applicable only in the absence of an acute phase response (APR), because transferrin is a negative acute phase protein and TfS is low in patients with APR. This is the case with CRP levels of > 5 mg/L. Basic investigation: in order to assure a sufficient iron supply for erythropoiesis, TfS should be > 20% prior to treatment with ESA; 30–40% is better, and under ESA therapy TfS should be some 30%. Monitoring: TfS of < 20% is an indication of a need for iron. In order to continuously maintain values > 20%, the therapeutic goal of iron substitution in patients with CKD is a TfS value of 30–40% /20/. Monitoring intervals as described for ferritin. \| \| %HYPO The proportion of hypochromic erythrocytes (%HYPO) is a valuable marker for iron demand of erythropoiesis and the erythropoietic response to ESA after intravenous administration of iron /20/. %HYPO is one of the APR independent markers of the iron demand. Basic investigation: %HYPO should be below 10 and, optimally, below 2.5 as an indicator of an adequate iron supply prior to ESA therapy. Monitoring: %HYPO > 10 points to iron demand for erythropoiesis. In infection or chronic inflammation, ferritin may be normal and TfS may not be useful. In this situation, a %HYPO of > 10 is indicative for functional iron deficiency. However, significant increases are not seen earlier than 3 weeks after the onset of the functional iron deficiency. Monitoring intervals as described for ferritin. \| \| Reticulocyte Hb Reticulocyte Hb content (CHr, Ret-He) is a real-time APR independent indicator of the iron demand for erythropoiesis. Basic investigation: a CHr or Ret-He ≥ 28 pg rule out iron-restricted erythropoiesis. At the start of ESA therapy, intravenous iron should also be substituted /44/. Monitoring: in patients receiving ESA treatment, a decrease in CHr or Ret-He to below 28 pg, or a relative decrease of 2 pg with values ≥ 28 pg, after some 5 days, is indicative of functional iron deficiency. This may be due to over stimulation of erythropoiesis relative to the iron supply. An increase in the CHr or Ret-He , or an unchanging normal CHr or Ret-He under the therapy of ESA and intravenous iron, are suggestive of adequate iron supply. In dialysis patients, CHr or Ret-He are better markers of the response to iron therapy than ferritin, TfS and %HYPO /44/. Monitoring intervals as described for ferritin. \| \| C-reactive protein (CRP) CRP is an indicator of APR, which leads to a hepcidin induced disorder of iron distribution, and inhibits adequate erythroid proliferation via an inflammatory blockade /45/. Basic investigation: apart from iron deficiency, APR is the most important cause of ESA resistance. If the CRP value is > 20 mg/L in patients with CKD, a 30–70% increase in the ESA dose must be taken into account /45/. Monitoring: the determination of CRP is necessary if the expected response to ESA therapy does not materialize. In such a case, an ESA dose increase is imperative. In patients with chronic renal failure and ESA treatment, this means a dose increase of 1.5–3 fold /20/. \| \| Diagnostic diagram (Iron plot) /46/ The iron plot (see Fig. 15.6-5 – Diagnostic diagram for the assessment, monitoring and therapy of the iron status) has been evaluated for the identification of patients who require ESA therapy, for the early detection of responders, for monitoring under therapy and for the identification of the balance between ESA dose and iron requirement. Out of 286 patients /47/: Two hundred and four with anemia of chronic disease (ACD), adequate storage iron and a CHr ≥ 28 pg were treated, according to recommendations of the plot, with ESA monotherapy; 56% of these patients manifested a rise in Hb ≥ 10 g/L Twenty two patients with ACD and iron-restricted erythropoiesis (CHr < 28 pg) were treated with ESA and intravenous iron. All of these manifested a rise in Hb of > 10 g/L. Sixty patients with empty iron stores (estimated as an elevated ferritin index) and a CHr < 28 pg were treated with iron only, orally in most cases. An Hb increase was seen in 73% of these patients. The study /47/ showed that the diagnostic diagram is an effective concept for the differential diagnosis and treatment of anemia in cancer patients receiving cytostatic therapy. \| Table 15.10-6 Erythropoieses stimulating agents /49/ \| Recombinant human Erythropoietin: rHuEPO is the artificial equivalent of the endogenous hormone erythropoietin (EPO). RHuEPO has the capacity to initiate the production of red blood cells and is commercially available for the treatment of anemia. RHuEPO stimulates the erythropoietic bone marrow compartment and is used for therapy of anemia of chronic kidney disease and myelodysplastic syndrome. RHuEPO products differ in their biological composition which results in differences in their clinical efficiencies and adverse events. They not only have benefits (reducing red cell transfusion requirements, improved quality of life) but also, they have the risks of thromboembolic complications of arterio-venous fistulas, apoplexia, and increased mortality in malignant disease. \| \| Epoetin alpha: Epoetin alpha is produced in cell culture using rDNA technology. Epoetin alpha is synthesized in Chinese hamster ovary cells, and the amino acid sequence is similar to endogenes erythropoietin. The half-life time of these erythropoietins is relatively short (8 hours intravenous and 19–24 hours subcutaneous administration).The administration is 2–3 times weekly via intravenous or subcutaneous route. Epoetin alpha is a first generation erythropoietin. \| \| Epoetin Beta: This erythropoietin is effective to stimulate the red cell compartment and reduces transfusion requirements. Epoetin beta does not increase mortality in malignant disese. Epoetin beta is a first generation erythropoietin. \| \| Darbopoetin alpha: This erythropoiesis stimulating agent (ESA) is synthesized in Chinese hamster ovary cells by means of rDNA technology and its amino acid sequence is rHuEPO. Darbopoetin alpha is a long-acting ESA and has a higher molecular weight than endogenous erythropoietin because it has a greater carbohydrate content. Its half-life time is 3 to 4 times longer than rHuEPO because its affinity to the erythropoietin receptor is 4,3-times higher than of rHuEPO. The half-life time ist 3–4 fold as long as of Epoetin alpha or Epoetin beta. Darbopoetin alpha is administerd once every 2 weeks. \| \| Epoetin DELta: This erythropoietin is sythesized in a human cell lines using rDNA technology. Epoetin delta has a similar amino acid sequence to endogenous human erythropoietin. In comparison to other erythropoietins epoetin delta has lower concentrations of certain nonhuman carbohydrate residues (N-glycolylneuraminic acid). Epoetin delta is a third generation erythropoietin. \| \| Continuous ErythropoIetin Receptor Agonist (CERA): CERA is an ESA product produced by adding a large water-soluble polyethylene glycol moiety to epoetin beta. CERA is a continuous erythropoietin receptor agonist. It has 50 to 100 times lower affinity and much slower binding rate to the erythropoietin receptor compared to epoetin beta. CERA promises both lower dosing and less frequent injections for patients (a 5-fold longer half-life time than recombinant human erythropoietin). Hemoglobin levels can be maintained with the target range (100–135 g/L) using beweekly or even monthly CERA administration. In patients undergoing maintenance hemodialysis the hemoglobin concentrations were 109 ± 8 g/L in the weekly group. Ferritin and hepcidin concentrations were maintained /50/. Some authors reported that CERAs suppress hepcidin levels, inducing the release of stored iron and effective erythropoiesis. \| \| ESA Mechanisms Of Action: The mechanims of action of ESAs are comparable to Hypoxia-inducible factor (HIF) prolyl hydroxylase inhibitors (PHIs), which effectively increase the level of hemoglobin in patients undergoing maintenance hemodialysis /51/ and in patients with chronic kidney disease for various reasons /52/. \| \| Hypoxia-inducible factor (HIF) prolyl hydroxylase inhibitor (PHI): These factors (e.g.,daprodustat) are a new class of compounds that stabilize HIF, which in turn stimulates endogenous erythropoeitin (EPO) production and thus increases hemoglobin (Hb) concentrations in patients with chronic kidney disease (CKD). HIFPHIs are as effective as ESAs in increasing Hb levels. Among patients with chronic kidney disease undergoing dialysis daprodustat was noninferior to ESAs regarding the change in the Hb level from baseline and cardiovascular outcomes /53/. \| Table 15.11-1 Thrombocyte reference intervals \| \| ♀ \| ♂ \| \| Platelet count adults (109/L), instrument dependent /8/ \| \| \| \| Abbott CD 4000 \| 168–405 \| 150–346 \| \| Advia 120 \| 203–445 \| 166–389 \| \| Coulter LH 750 \| 166–387 \| 137–327 \| \| Horiba ABX, Pentra 12 \| 179–443 \| 168–355 \| \| Sysmex XE -2100 \| 176–391 \| 146–328 \| \| Values are 2.5th and 97.5th percentiles Premature infants, neonates, small children: values as in adults /9/ School children: values as in adults /10/ \| \| \| \| Mean Platelet Volume (MPV), data in fL /11/ \| \| \| \| Abbott CD 4000 \| 6.9–10.6 \| \| \| Advia 120 \| 6.4–9.7 \| \| \| Coulter Gen S \| 7.6–10.7 \| \| \| Horiba ABX, Pentra 12 \| 6.8–10.0 \| \| \| Sysmex SE-9500 \| 9.4–12.9 \| \| \| Values are 2.5th and 97.5th percentiles \| \| \| Table 15.11-2 Differences between primary and secondary thrombocytosis /12/ \| Criterion \| Thrombocytosis \| \| \| Primary \| Secondary \| \| Age (years) \| Frequently > 20, often > 40 \| All ages \| \| Complications \| Thrombosis, hemorrhage \| Very rare \| \| Splenomegaly \| Often \| Rare \| \| Thrombocyte count \| Usually ≥ 1000 × 109/L \| Frequently < 1,000 × 109/L \| \| Duration \| > 2 years \| Days to weeks \| \| Thrombocytes \| \| \| \| Function \| Disturbed \| Normal \| \| Morphology \| Large, dysmorphic \| Large, normal \| \| Cause \| Genetic defect \| Reactive \| Table 15.11-3 Differentiation of the mechanism of thrombocytopenia /15/ \| Examination \| Increased consumption \| Reduced formation \| \| MPV \| Elevated \| Normal \| \| PDW \| Increased \| Normal \| \| Thrombocyte life span \| Decreased \| Normal \| \| Thrombocyte- bound IgG \| Detectable \| Undetectable or hardly detectable \| \| Bleeding time \| Usually normal \| Prolonged \| \| Other blood cells \| Usually normal \| Often normal \| \| Megakaryocytes \| Normal/increased \| Decreased \| \| Other cell lines in the bone marrow \| Normal \| Frequently decreased or abnormal \| Table 15.11-4 Medication-associated autoimmune thrombocytopenia, according to Ref. /16/ \| Abciximab Acetaminophen Acetazolamide Allopurinol Aminoglutethimide Amiodarone Amrinone Aspirin Carbamazepine Carbenicillin Cefotiam Cephalexin Cefalotin Cefamandole Quinine Quinidine Chlordiazepoxide Chlorpheniramine Chlorthalidone Chlorthiazide Cimetidine Cocaine Codeine Danazol Desipramine Diazepam \| Diclofenac Difunisal Digitoxin Digoxin Diltiazem Eptifitabid Erythromycin Fenoprofen Furosemide Gentamicin Gold salts Heparin Hydrochlorothiazide Hydroxychloroquine Imipramine Interferon Isoniazide Ketoprofen Levamisole Lidocaine Meperidine Meprobamate Methicillin Methyldopa Minoxidil Morphine \| Naproxen Nitrofurantoin Nomifensine p-Aminosalicylic acid Penicillamine Penicillin Pentamidine Phenylbutazone Phenytoin Piperacillin Piroxicam Procainamide Prochlorperazine Prothetazine Ranitidine Rifampicin Spironolactone Sulfasalazine Sulfonyureas Tetracycline Thioguanine Tirofiban Ticlopidine Trimethoprim/sulfamethoxazole Valproic acid Vancomycin \| Table 15.11-5 Recommendation for thrombocyte transfusion /17/ \| Surgical interventions \| Thres- hold \| Hematology/ Oncology \| Limit value \| \| Low risk of bleeding \| 10 \| No hemorrhagic complications \| 5 \| \| Major surgery \| 50 \| Solid tumors \| 10 \| \| Cardiac surgery \| 20 \| Leukemia \| 10 \| \| Neurosurgery \| 70–100 \| Chemotherapy \| 10 \| \| Epidural anesthesia \| 80 \| Tumors and thrombopenia \| 20 \| \| Spinal anesthesia \| 50 \| Necrotic tumor \| 50 \| \| Diagnostic interventions \| \| Miscellaneous \| \| \| Lumbar puncture \| 10 \| Hepatic insufficiency \| 10 \| \| Needle biopsy of liver \| 10 \| Liver failure \| 20 \| \| Arthrocentesis \| 20 \| Central vein catheter \| 20 \| \| Dental treatment \| 20 \| \| \| \| Major dental surgery \| 50 \| Hemorrhage requiring transfusion \| 100 \| \| Gastrointestinal endoscopy \| 20 \| Coagulopathy due to loss of blood (prophylaxis) \| 100 \| \| Bronchoscopy \| 20 \| Preterm infants \| 50 \| \| Transbroncial biopsy \| 50 \| Neonates \| 30 \| \| Neonates ill \| 50 \| \| Angiography \| 20 \| \| \| Threshold in 109/L below which a thrombocyte transfusion should be provided; Limit value, bleeding is likely Table 15.11-6 Diseases and conditions with primary thrombocytosis \| Clinical and laboratory findings \| \| Myeloproliferative disease – Generally /23/ The myeloproliferative diseases polycythemia vera (PV), essential thrombocythemia (ET), and primary myelofibrosis (MF) are clonal diseases of pluripotent hematopoietic stem cells. This leads to an uncontrolled proliferation of erythropoiesis, leukopoiesis and thrombopoiesis, alone or in combination. Clinically, the tendency for arterial or venous thrombosis, bone marrow fibrosis, splenomegaly, or transformation to acute leukemia, is observed with varying frequency. Hemorrhagic and thrombotic complications are common in the myeloproliferative diseases of ET, PV, MF and chronic myeloid leukemia (CML). One reason for the commonality of PV, ET and MF is an activating mutation (V617F) in the JAK2 (Janus kinase 2) gene, which codes for tyrosine kinase. JAK2 is a member of the Janus kinase family. In the hematopoietic system, these enzymes phosphorylate tyrosine residues on receptors (erythropoietin, thrombopoietin and GM-CSF) following binding of the ligand, thereby ensuring signal transmission to the cell nucleus (Fig. 15.4-2 – Homodimer erythropoietin receptor showing phosphorylated tyrosines). As a rule, JAK2 is activated when the ligands (for e.g., erythropoietin) bind to their corresponding receptors. In myoproliferative syndrome a point mutation leads to the exchange of phenyl alanine with valine (V617F) at the regulatory domain, thereby inducing a long-term activation of signal transmission in the absence of ligand-receptor binding. The activated kinases stimulate cell proliferation and are resistant to apoptosis. In this way, cells expressing a mutant JAK2 have a growth and survival advantage over normal cells. The result of the V617F mutation is that thrombocyte formation in ET and PV is increased by 2 to 15-fold and, in MF, by 2 to 8-fold. In spite of the fact that an elevated blood thrombocyte count results from the increased thrombocyte formation, there is no direct relationship between megakaryopoiesis and the peripheral thrombocyte count. This is because of the enlarged spleen and the abbreviated life span of the thrombocytes in myeloproliferative disease. It is particularly striking that in MF, the spleen is very large and the thrombocyte life span is markedly shortened. The thrombocyte counts (109/L) are the average, 660 (in PV), 910 (in ET), 610 (in CML) and 290 (in MF). Laboratory findings: the demonstration of the JAK2 V617F mutation confirms the tentative clinical diagnosis of myeloproliferative syndrome. However, since the mutation occurs in PV, ET and MF, differentiation by means of molecular biological determination is not possible. Since, nonetheless, the mutation is present is almost all PV patients, PV should be considered in the first place upon demonstration of the mutation, and the presence of erythrocytosis should be chosen as an important diagnostic criterion. Since it has become possible to determine the JAK2 V617F mutation, the incidence of diagnosed ET has increased significantly /24/. \| \| – Essential thrombocythemia (ET) The incidence of ET is 2.5 per 1 million in the general population. ET is the most indolent myeloproliferative neoplasm and is characterized by thrombocytosis with an otherwise normal complete blood count. The age distribution is bimodal, with a peak at 60 years of age in both women and men, and a smaller peak at early adulthood that concerns mainly women. ET is caused by JAK2 V617F, CALR or MPL mutations and infrequently by germline single-nucleotide variants. Diagnosis is made according to the positive criteria and the exclusion criteria of the WHO /26/. ES can present as artery thrombosis and elevated concentration of lactate dehydrogenase (LD). Laboratory findings in ET: Thrombocyte count ≥ 600 × 109/L Bone marrow with predominant proliferation of megakaryopoiesis and a significantly increased number of mature megakaryocytes. ET exclusion criteria: No evidence of polycythemia vera (normal red blood cell mass or Hb below 185 g/L in men and 165 g/L in women, detectable iron in the bone marrow, normal serum ferritin or normal MCV) No evidence of chronic myeloid leukemia (absence of the Philadelphia chromosome or the BCR/ABL fusion gene) No evidence of chronic idiopathic myelofibrosis No evidence of myelodysplastic syndrome; no del(5q)-, t(3;3)(q21;q26.1)-, inv(3;3)(q21;q26.1) abnormality and no significant granulocytic dysplasia; few, if any, scattered micromegakaryocytes No evidence of secondary (reactive) thrombocytosis Megakaryocyte morphology 80% Splenomegaly (ultrasound or CT) 26% Abnormal thrombocyte function 83%. The clinical course of ET is, like that of PV, characterized by an elevated incidence of thrombosis, and progression to myelofibrosis and to acute myeloid leukemia. The basal leukocyte count at the time of disease diagnosis is a risk factor for severe thrombosis, particularly acute coronary syndrome. Thus, ET patients with the JAK2 V617F mutation anda leukocyte count of below 8.0 × 109/L had a hazard ratio (HR) for severe thrombosis of 1.5, while the HR in those patients with a leukocyte count greater than 11.0 × 109/L was 2.0. In ET patients, thrombosis occurs at an increased rate with thrombocyte counts of ≥ 1000 × 109/L. However, the prognosis depends exclusively upon the leukocyte count, and is better if the basal leukocyte count is below 11 × 109/L. Thus, the rates of vascular events (data expressed as 109/L) were /27/: Leukocytes below 11, thrombocytes above 1,000; rate 1.59 Leukocytes below 11, thrombocytes below 1,000; rate 2.26 Leukocytes above 11, thrombocytes above 1,000; rate 2.88 Leukocytes above 11, thrombocytes below 1,000; rate 2.95. Response criteria following therapy of ET /28/: Complete response: thrombocyte count ≤ 400 × 109/L, leukocytes ≤ 10 × 109/L, no disease-related symptoms, no enlarged spleen. Reduction of the molecular abnormalities to an undetectable level. Partial response: thrombocyte count ≤ 600 × 109/L or a decrease of over 50% compared to the starting value, leukocytes ≤ 10 × 109/L, no disease-related symptoms. Reduction of the molecular abnormalities of more than 50%. Laboratory findings /29/: 1. Thrombocyte count ≥ 600 × 109/L; 2. Hematocrit below 0.40; 3. Bone marrow storage iron present or normal MCV value; 4. No Philadelphia chromosome or bcr/abl gene rearrangement; 5. Bone marrow fibrosis (a) absent or (b) less than one third and without significant splenomegaly or leukoerythroblastosis in the blood smear; 6. No morphological or cytogenetic indication of myelodysplastic syndrome; 7, No cause for reactive thrombocytosis. The prevalence of the findings is /30/: Thrombocyte count below 600 × 109/L = 4%; (600–999) × 109/L = 59%; (1000–1499) × 109/L = 29%; over 1500 × 109/L = 8%. With thrombocyte count (> 1000–1500) × 109/L, there is an elevated risk of bleeding, since von Willebrand syndrome is acquired due to a decrease in high molecular weight multimers of VWF. The daily ingestion of over 300 mg of aspirin represents a further danger of bleeding /31/. Leukocytosis of greater than 10 × 109/L = 32%. Bone marrow puncture; increased cellularity 17%, elevated megakaryocyte count 79%, abnormal megakaryocyte morphology 80% Splenomegaly (ultrasound or CT) 26% Abnormal platelet function 83%. \| \| – Familial essential thrombocythemia (FET) /32/ FET is a rare myeloproliferative disease, characterized by autonomously activated megakaryopoiesis and associated with excessive thrombocyte production. The gene c-MPL that codes for the thrombopoietin receptor and its ligand, thrombopoietin (TPO), regulate the proliferation and differentiation of megakaryocytes. FET is caused by an activating point mutation in the TPO receptor gene. This leads to the exchange of asparagine with serine at position 505 of the transmembrane domain (Ser505Asn). \| \| – Hereditary thrombocythemia /33/ The hereditary thrombocythemia is an autosomal dominant disease with the clinical features of sporadic ET. It is characterized by proliferation of megakaryopoiesis and overproduction of thrombocytes. Mutations in the TPHO gene, which codes for TPO, lead to enhanced translation of TPHO mRNA, thereby resulting in enhanced TPO synthesis and increased formation of thrombocytes. \| \| – Polycythemia vera (PV) /23/ PV presents with erythrocytosis, often microcytic, leukocytosis, and thrombocytosis. It is also present if only the combination of erythrocytosis and leukocytosis, or of erythrocytosis and thrombocytosis, occurs along with splenomegaly. However, 7–20% of PV patients do not have thrombocytosis, and up to 17% have only erythrocytosis. The determination of the JAK2 V617F mutation, which is demonstrable in some 92% of the patients, is, therefore, a diagnostically important criterion. For further information, see Tab. 15.4-5 – Diseases and conditions with increased HCT. \| Table 15.11-7 Diseases and conditions with secondary thrombocytosis \| Clinical and laboratory findings \| \| Secondary thrombocytoses – Generally In adults, the most frequent causes of thrombocytoses above 500 × 109/L /34/ are infections 21.9%, reactive processes (e.g., after acute bleeding) 19.4%, tissue damage 17.9%, chronic inflammatory diseases 13.1%, malignant tumors 5.9%, and multiple factors 6.1%. The following etiologies are relevant in children up to the age of 16 years: infections 30.6%, hemolytic anemia 19.3%, tissue damage 15.2%, reactive thrombocytosis 14.8%, chronic inflammation 4.1%, kidney disease 4.1%, malignant diseases 2% /35/. In inflammatory disease, the elevation in thrombocytes is the result of enhanced synthesis of TPO in the liver, stimulated by IL-6. \| \| – Physical exertion Upon vigorous physical exertion, rise of 50% compared with the basal value following approximately 15 min., return to basal value following some 30 min. This is, for example, also the case with delivery. \| \| – Surgical intervention With major surgical interventions, decrease in thrombocyte count during intervention, normalization within 2–6 days, then, in the following days, increase by 35–150% with peak values between postoperative weeks 1 and 2 /36/. Thrombocytosis can also occur following cardiopulmonary bypass surgery and corrective surgery of fractures. \| \| – Acute infection This is the most frequent cause of thrombocytosis. Most prominent are purulent infections of the respiratory, gastrointestinal and urogenital tracts, hepatobiliary infections, meningitis, sepsis and pyogenic abscesses /34, 35/. In children, thrombocytosis of over 700 × 109/L favors the likelihood of an infection /37/. \| \| – Chronic infection In adults, most prominent are rheumatoid arthritis, inflammatory intestinal diseases, tuberculosis, and autoimmune diseases; with reference to the latter, Kawasaki syndrome in particular. In children, rheumatoid arthritis, rheumatic fever, Henoch-Schönlein purpura and, again, Kawasaki syndrome, are to be mentioned /34, 35/. \| \| – Reactive thrombocytosis Acute bleeding, iron deficiency anemia, and hypoxemia (anemic, pulmonary, cardiac) are the most frequent causes in adults and children /34, 35/. \| \| – Hemolytic anemia In children, hemolytic anemia is a frequently diagnosed cause of thrombocytosis. The majority of these children have sickle cell anemia or thalassemia syndrome /12/. In adults, hemolytic anemia is less relevant than in children for the etiogenesis of thrombocytosis. \| \| – Renal disease Approximately 5% of adults with thrombocytosis have acute or chronic kidney disease, or nephrotic syndrome /12/. In children, the frequency of the etiogenesis is similar /13/. \| \| – Malignant disease Within the context of malignant disease, thrombocytosis occurs more frequently in Hodgkin’s- and non-Hodgkin lymphoma than in solid tumors /12/. Approximately 30–60% of patients with advanced solid tumors with metastases manifest thrombocytosis, thrombopathy or thromboembolic events /38/. \| \| – Splenectomy Post splenectomy thrombocytosis is not a risk factor for thrombosis. \| Table 15.11-8 Thrombocytopenia of various origins \| Clinical and laboratory findings \| \| Pseudo thrombocytopenia In pseudo thrombocytopenia, erroneously low thrombocyte count is determined with hematology analyzers. The explanation is found in the blood smear assessment. Causes of pseudo thrombocytopenia are: Formation of thrombocyte aggregates or agglutinates (e.g. EDTA induced thrombocytopenia) Presence of giant thrombocytes, either inherited or acquired Satellite phenomenon. In EDTA blood, thrombocytes are coating the surface of segmented granulocytes, and are not counted Cold agglutinins, which can induce thrombocyte agglutination in an EDTA independent manner. The frequency of occurrence of EDTA induced thrombocytopenia is about 0.1%; it is likely based upon the baring of thrombocyte membrane glycoproteins by EDTA. The glycoproteins react with heterophile antibodies and form agglutinates. If the thrombocyte determination is performed immediately following the blood sampling, the observed thrombocytopenia is usually only slight, or even non-existent. It only occurs when the blood is left standing for an extending period of time. In order to minimize the misdiagnosis of thrombocytopenia, the measurement of EDTA, citrate or heparin blood is always recommended, as is microscopic testing for agglutinates, for the verification of EDTA induced thrombocytopenia \| \| Pregnancy /39/ Mean thrombocyte count decreases during pregnancy in all the women, beginning in the first trimester. In women who have a thrombocyte count less than 100 × 109/L a cause other than pregnancy or its complications should be considered. Refer to Tab. 15.11-9 – Thrombocytopenia in pregnancy. According to a study /39/ at the time of delivery 9.9% of the women with uncomplicated pregnancies had a thrombocyte count below 150 × 109/L. During the course of the uncomplicated pregnancies and deliveries only 1% had a thombocyte count below 100 × 109/L. Thrombocyte counts of less than 150 × 109/L at the time of delivery were more common among women who had pregnancy related complications (11.9% versus 9.9%). \| \| Neonatal thrombocytopenia Neonatal thrombocytopenia is a relatively rare condition with a prevalence of 0.9% in unselected populations. Newborns with severe thrombocytopenia may suffer from hemorrhage that leads to death or to severe lifelong defects. In one study /40/, in which umbilical cord blood from over 15,000 newborn infants and their mothers was investigated, 19 neonates (0.12%) had a thrombocyte count of below 50 × 109/L. Of 756 mothers with incidental thrombocytopenia, 1414 with hypertension and 46 with idiopathic thrombocytopenic purpura (ITP), 1,5 and 4 newborns in each group manifested thrombocytopenia of (20–50) × 109/L. Six newborns had values below 20 × 109/L; all were children who were at risk of neonatal alloimmune thrombocytopenia. The study showed that severe thrombocytopenia in newborns of mothers with incidental thrombocytopenia, hypertension and ITP are rare; if thrombocytopenia occurs, it is often associated with fetal and neonatal alloimmune thrombocytopenia. \| \| Fetal and neonatal alloimmune thrombocytopenia (FNAIT) /41/ FNAIT occurs when a women becomes alloimmunized against fetal platelet antigens inherited from the fetus’s father (which are absent on maternal platelets), leading to fetal thrombocytopenia (below 150 × 109/L). Mothers who do not possess this antigens develop specific IgG antibodies. As of the 14th week of pregnancy the antibodies cross the placenta and bind to fetal thrombocytes, which are then removed from the blood prematurely. The incidence is 1 in 800–1000 births. About 24 human thrombocyte-specific allo antigens are recognized. In white populations (approximately 80%) the most common antibody is anti-HPA-1a. The antibody binds to the polymorphic leu/pro residue of glycoprotein IIIa on the thrombocyte membrane. With a proportion of 15%, anti-HPA-5b is the second most common antibody. Some 10% of HPA-1a-negative women who have been pregnant with an HPA-1a-positive child have antibodies. The medical history provides an indication of possible FNAIT, if sisters or close relatives have given birth to children with thrombocytopenia and an anti-HPA-1a was determined. The mothers have a normal thrombocyte count. Alloimmunization is associated with the HLA-DRB30101 allele, which is expressed by 90% of immunized women. Clinical findings: Die FNAIT is the most common cause of severe thrombocytopenia in the newborn, accounting for 3% of all fetal and neonatal thrombocytopenia and 27% of severe cases defined as < 50 × 109/L or intracranial hemorrhage (ICH). The clinical symptoms defined as from isolated petechiae to intracranial hemorrhage (ICH), which occurs in 7–26% of the cases. Approximately 80% of ICH occurs intrauterine, and up to 42% of the cases occur before the 32nd week of pregnancy. In newborns with signs of hemorrhage and thrombocyte count below 30 × 109/L, the infusion of banked thrombocytes is indicated. Most of the other causes of neonatal thrombocytopenia are infection, and autoimmune thrombocytopenia Laboratory findings: the maternal HPA antibody titer is determined, for (e.g., in the 22nd and/or the 34th week of pregnancy) using the MAIPA method (see Section 17.4 – Alloimmune thrombocytopenia); in the neonate, the thrombocyte count is determined. There is an association between the severity of the thrombocytopenia and the antibody concentration. In one study /42/, with an anti-HPA-1a cutoff value of > 3.0 IU/mL for severe FNAIT (thrombocyte count below 50 × 109/L) the diagnostic sensitivity was 93% with a specificity of 63%. The positive and negative predictive values were 54% and 95%, respectively. Of the 338 newborns with NAIT, 105 had thrombocyte counts of below 50 × 109/L, and 233 had higher values. The normalization of the thrombocyte count occurred within 4 weeks and, even after only 7–10 days. The determination of the maternal thrombocyte characteristics elicited the HPA-1A1/1A2 genotype; the baby and the father were positive for the HPA-1A1. \| \| Surgical intervention Thrombocyte counts of higher than 50 × 109/l do not necessitate pre-operative thrombocyte substitution, because the bleeding complications that only seldom occur can be treated intraoperatively. With counts of (50–80) × 109/L, surgical interventions, in which hemorrhage can easily be identified or stopped, can be performed. With operations within body cavities (bladder, thorax, abdomen), in which bleeding is only recognized later, it is necessary to increase the pre-, peri- and post-operative thrombocyte level to above 80 × 109/L /43/. \| \| Postoperative thrombocytopenia Postoperative thrombocytopenia is seen particularly following cardiovascular surgery, surgical interventions for aneurysms or peripheral vascular grafting, as well as orthopedic interventions such as hip replacement surgery /44/. Thus, in intensive surgical care units, thrombocytopenia of below 100 × 109/L is seen at least once during a stay in 20–40% of patients, and thrombocytopenia of less than 50 × 109/L is observed in 10–20% of patients /45/. Mild to moderate thrombocytopenia occurs following the transfusion of large quantities of blood. Thus complete replacement of erythrocytes with 11 red blood cell units in a patients weighing 75 kg leads to a decrease in thrombocytes from 250 × 109/L to 80 × 109/L. Generally, however, postoperative hemorrhage only occurs if the thrombocyte count falls to below 60 × 109/L and, additionally, there is a clotting factor deficiency due to vitamin K deficiency, diminished liver function or plasma dilution /44/. Thrombocytopenia can also be of relevance in postoperative thrombosis prophylaxis with heparin. If the patient has a deficiency in platelet factor 4, heparin is neutralized too slowly and accumulates in the blood, with consequential bleeding. A postoperative thrombocyte count below 60 × 109/L raises the suspicion of sepsis, even in the absence of fever, of DIC, and of heparin-associated thrombocytopenia /44/. With the administration of thrombocyte units and in the presence of a mixed population of native and foreign thrombocytes in the circulating blood, the therapeutic effect lasts for a maximum of 4 days. The duration of the substitution effect of transfused thrombocytes is only 1 day. \| \| Sepsis /45/ In sepsis patients, the thrombocyte count decreases during the first 4 days following admission to the intensive care unit. The incidence of sepsis is 35–44% for a thrombocyte count of below 150 × 109/L, 20–25% for below 100 × 109/L, and 12–15% for below 50 × 109/L. \| \| Post-transfusional purpura (PTP) PTP involves a delayed transfusion reaction. Approximately 5–10 days following the transfusion of thrombocytes or of thrombocyte-containing red cell units severe hemorrhage, due to massive thrombocytopenia, occurs. The cause is immunization, often dating to many years earlier, against a thrombocytic allo antigen, usually HPA-1A. The antibodies were formed during pregnancy or following a thrombocyte transfusion. Re-exposure leads to an immune reaction and the elimination of thrombocytes. In this manner, not only are the incompatible thrombocytes damaged, but the autologous platelets are also involved in the reaction as innocent bystanders /46/. Clinical findings: with a delay of 1 week following transfusion, massive hemorrhagic diathesis occurs. Women in particular are affected. Laboratory findings: the thrombocytopenia can be below 10 × 109/L and return to above 100 × 109/L after 3–75 days. \| \| Immune thrombocytopenia (ITP) – Generally ITP is triggered by IgG antibodies that cause destruction of the thrombocytes and inhibition of their formation. The major clinical symptom is bleeding, which correlates with the extent of the thrombocytopenia. Primary or secondary forms of ITP can occur. An international working group /47/ has standardized the terminology and definitions. Primary ITP is an autoimmune disorder characterized by isolated thrombocytopenia (below 100 × 109/L) in the absence of other causes or disorders that may be associated with thrombocytopenia. The diagnosis of primary ITP remains one of exclusion; no robust clinical laboratory parameters are currently available to establish its diagnosis with accuracy. The main clinical problem of primary ITP is an increased risk of bleeding, although bleeding symptoms may not always be present. Secondary ITP: all forms of immune-mediated thrombocytopenia except primary ITP Newly diagnosed ITP: within 3 months of diagnosis Persistent ITP: within 3–12 months of diagnosis. It includes patients without spontaneous or complete therapeutic remission. Chronic ITP: present for longer than 1 year Severe ITP: patients present with newly occurring hemorrhage, or bleeding again following therapy Primary ITP comprises some 80% of all ITP cases, and secondary ITP some 20%, if medication-dependent ITP is not taken into consideration /48/. \| \| – Primary ITP Primary ITP is caused by an immune-mediated mechanism. In some patients the antibodies are directed against a single glycoprotein; in others, against a pattern of glycoproteins. The antigens in question are glycoproteins that are located on the thrombocyte membrane (e.g., GPIIb/IIIa, GP Ia/IIa, GPI/IX). With the demonstration of antibodies against these proteins, it must be ensured that one is dealing with autoantibodies, and not with allo antibodies such as the common HLA class I antibodies. For further information, see Section 17.2 – Autoimmune thrombocytopenia. \| \| – Secondary ITP /48/ The various forms of secondary ITP are caused by different immunological preconditions and infections. \| \| – Acute immune thrombocytopenia of childhood Some two thirds of children with ITP have a prior febrile disease. It is believed that thrombocytes express viral antigens or bind immune complexes against which viral antibodies are formed. As a result, cross reaction with thrombocyte antigens occurs. The thrombocytopenia is also believed to be due to a macrophage infection with resulting hemo phagocytosis and enhanced megakaryocyte clearance. About 80% of the cases of thrombocytopenia normalize within 6–12 months. \| \| – Measles, mumps, and rubella (MMR) vaccine Acute ITP may occur following vaccination against various viruses such as MMR, as well as Streptococcus pneumoniae, Hemophilus influenzae, Hepatitis B virus and Varicella-zoster virus. Following vaccination with the MMR vaccine, the prevalence of ITP is increased by a factor of 6. The thrombocytopenia is often severe; it develops within 72 days, and normalizes after 2 months. \| \| – Helicobacter pylori (HP) There is a relationship between HP infection and ITP. The pathogenesis is based upon molecular mimicry, whereby HP-induced antibodies cross-react with thrombocyte antigens. It is believed that in patients with HP antibodies, altered mucosal permeability and/or bacterial eradication following treatment with proton pump inhibitors and antibiotics initiate the immune reaction against thrombocytes. The thrombocytes are activated by the binding of HP antibodies to FcγIIa, or by an interaction of HP antibody-bound von Willebrand factor and the thrombocyte glycoprotein GPIb. Under bacterial eradication, thrombocytopenia has been reported at a rate of 0–7% in the USA, and 100% in Japan. Patients with ITP and a thrombocyte count of over 30 × 109/L should be observed, while those with a count lower than that should be treated with corticosteroids /49/. \| \| – Cytomegaly (CMV) CMV infection leads to severe congenital IPT, and to delayed initiation of thrombopoiesis in bone marrow transplantation. Immune-suppressed and immune-incompetent individuals with CMV infection can also suffer with ITP-like symptoms. \| \| – Varicella-zoster virus (VZV) VZV-ITP develops in 5% of children with acute Varicella-zoster virus infection within 5 days of the outbreak of exanthema. In non fulminant cases, spontaneous remission occurs within 2 weeks. In rare cases, the ITP does not develop until a number of weeks after the outbreak of exanthema. \| \| – Hepatitis C Some 20% of patients with ITP have HCV antibodies. By binding to CD81, the HCV structural protein E2 is believed to activate polyclonal B cells and to lead to activation of B cells. The result is the development of antibodies against thrombocyte glycoproteins. The thrombocyte count is below 50 × 109/L. \| \| – HIV infection Thrombocyte count below 150 × 109/L is reported in 5–30% of HIV patients. In a study of 36,515 patients, the 1-year incidence of HIV-associated ITP with a thrombocyte count below 50 × 109/L was 3.7%. The causes of HIV-associated ITP are multifactorial. \| \| Chronic lymphocytic leukemia (CLL) CLL-associated ITP develops in 1–5% of CLL patients; its incidence is approximately one tenth that of autoimmune hemolytic anemia. The ITP develops, on the average, 13 months following diagnosis of CLL; it can, however, also precede the CLL, and can develop at any time. \| \| Hodgkin’s lymphoma The incidence of ITP in Hodgkin’s lymphoma is approximately 0.2–1%. \| \| Large granular lymphocyte leukemia (LGL) This involves the clonal proliferation of CD8+T cells. Anemia, neutropenia, and mild thrombocytopenia occur frequently in these patients. Severe ITP is seen in 1% of the patients. \| \| Common variable immunodeficiency (CVID) More than 10% of CVID patients develop ITP, with a thrombocyte count of about 20 × 109/L. The pathogenesis of the antibody development is unknown. \| \| Autoimmune lymphoproliferative syndrome (ALPS) ALPS is a hereditary disease, characterized by defective apoptosis of T- and B cells. Benign hepatosplenomegaly and lymphadenopathy are found. Approximately 20% of the patients develop ITP. \| \| Evans syndrome In Evans syndrome, ITP and autoimmune hemolytic anemia (AHA), caused by warm antibodies, and usually directed against the Rh system, occur simultaneously. Some 50% of the patients have neutropenia, and 10% have pancytopenia. The ITP can precede the AHA, but it often also occurs subsequent to the AHA. \| \| Antiphospholipid syndrome (APS) Many APS patients have thrombocytopenia. The question arises as to whether the thrombocytopenia is caused by antibodies or by activation of coagulation. In ITP, antibodies to thrombocyte glycoproteins are detected, and their titer correlates better with the extent of the thrombocytopenia than the phospholipid antibodies. The thrombocytopenia is mild to moderate. Severe thrombocytopenia correlates with the risk of thrombosis. The thrombocyte values should be maintained at (40–50) × 109/L. \| \| Systemic lupus erythematosus (SLE) In SLE, thrombocytopenia is generally not severe and is diagnosed in one third of patients. Conversely, 2–5% of ITP patients have SLE. The thrombocytopenia develops during the course of the disease. The pathogenesis of SLE-associated ITP is multifactorial; autoantibodies against thrombocyte glycoproteins and against DNA, phospholipids and phospholipid-binding proteins, CD40 ligand, as well as thrombopoietin and its receptor, can be present. \| \| Post-transplantation ITP ITP can develop following bone marrow and organ transplantation. This may occur as early as after 1 day, or only 1 year later. The course can be fatal, but remission is also possible. \| \| Medication-induced thrombocytopenia (MIT) Medications can induce an immune response that leads to the degradation of thrombocytes. Substances associated with this undesirable side effect are listed in Tab. 15.11-4 – Medication-associated autoimmune thrombocytopenia. It is difficult to diagnose MIT in patients who are taking a number of medications. Approximately 1 out of 100,000 patients taking medicine develop immune thrombocytopenia /50/. A difficulty lies in the differential diagnosis, since immune thrombocytopenia also develops within the framework of disease, and thrombocytopenia often occurs in severely ill patients, even in the absence of exposure to medication. In medication-associated immune thrombocytopenia, antibodies against one or both of the glycoproteins GPIb and GPIIb/IIIa are often present. The thrombocyte count may be below 10 × 109/L, and petechiae or purpura are common. The majority of patients have cutaneous bleeding, gastrointestinal or urogenital hemorrhage occurs in 10%, and 2% manifest intracranial bleeding. The thrombocyte decline usually occurs during the first week following the medication, and a rise is usually seen within 2 weeks of dis continuation of treatment /16/. \| \| Glycoprotein IIb/IIIa-receptor antagonists Thrombocytopenia of below 100 × 109/L or below 50 × 109/L occur under medication /51/: With abcicimab at the rate of 2.5–6% or 0.4–1.6%; following dis continuation and re-exposure thrombocytopenia of below 20 × 109/L may occur With eptifibatid, at the rate of 1.2–6.8% or 0.2%; following dis continuation as early as 2 h following re-exposure, thrombocytopenia of below 20 × 109/L may occur With tirofiban, 1.1–1.9% or 0.2–0.5%. Approximately 1/3 of the cases of severe thrombocytopenia following abcicimab are pseudo thrombocytopenia. \| \| Heparin-induced thrombocytopenia (HIT) For further information, see Section 17.5 – Heparin-induced thrombocytopenia and Lit. /52/. \| \| Cyclic thrombocytopenia Cyclic thrombocytopenia (CTP) is a rare disease. The hallmark of CTP is a periodic fluctuation of the platelet count. The severity of the thrombocytopenia and the duration of the cycles may vary. Mean cycle duration were 14 to 34 days. Bleeding complications can occur and patients suffer from mucocutaneous bleeding. Comorbidities may be hypothyroidism. CTP is often misdiagnosed as immune thrombocytopenia. In some patients thrombocytopenia is followed by rebound thrombocytosis with peak thrombocyte counts well about the upper reference range value /53/ \| \| Malignant disease Thrombocytopenia occurs frequently in patients with solid tumors and in malignant hematological disease /54/. It can also be caused by: Displacement of megakaryocytes in the bone marrow, which is the case in metastatic solid tumors, leukemia, multiple myeloma, and advanced lymphoma Tumor-associated hypersplenism with thrombocyte sequestration or splenic metastases (rare) Tumor-associated disseminated intravascular coagulation in promyelocytic leukemia, mucinous adenocarcinoma of the prostate, or pancreatic carcinoma with the release of enzymes that lead to activation of coagulation Chemotherapy or radiotherapy. This is the most common cause of tumor-associated thrombocytopenia, since, daily, some 300,000 patients worldwide receive a course of chemotherapy that leads to thrombocytopenia. Approximately 1–3 weeks following suppression of the thrombopoiesis, a recovery in the count is observed. The administration of mitomycin C or nitroso ureas may lead to long term thrombocytopenia /55/. \| Table 15.11-9 Thrombocytopenia in pregnancy \| Clinical and laboratory findings \| \| Incidental thrombocytopenia The incidental thrombocytopenia is first diagnosed at the time of delivery; it concerns some 5% of pregnant women, causes 75% of pregnancy-associated thrombocytopenia, and is associated with a platelet count of over 100 × 109/L. Following delivery, the platelet count is normalized once again. It can reoccur during subsequent pregnancies. Bone marrow findings are normal. Hemostasis is not affected. Incidental thrombocytopenia is a diagnosis of exclusion /56/. \| \| Preeclampsia In otherwise normal pregnancy, thrombocytopenia develops in association with pregnancy-associated hypertension (preeclampsia). Preeclampsia and thrombocytopenia are seen in 1–2% of pregnancies; thrombocytopenia occurs in 50% of the cases of preeclampsia. Approximately 13–15% of the cases of thrombocytopenia occurring in pregnancy are due to preeclampsia. The platelet count is above 75 × 109/L. Thrombocytes increase within 72 hours of delivery /56/. Further pathological laboratory findings are an elevation of LD and the demonstration of schistocytes in the blood smear. Schistocytes are also detected in micro angiopathy-associated thrombocytopenia. \| \| Immune thrombocytopenia The incidence of immune thrombocytopenia is 1–2 per 10,000 pregnancies; it comprises 1–2% of the cases of pregnancy-associated thrombocytopenia. The onset of ITP at the beginning of pregnancy is not uncommon. The thrombocyte count is usually below 50 × 109/L, while a count of over 30 × 109/L is seldom associated with bleeding. Caesarean sections can be performed without bleeding complications with thrombocyte counts of greater than 70 × 109/L /57/. \| \| HELLP syndrome This syndrome comprises micro angiopathic hemolytic anemia with edema, hypertension and proteinuria. Laboratory findings: schistocytes in the blood smear, thrombocytopenia of less than 100 × 109/L, in severe cases less than 50 × 109/L, elevated bilirubin, LD and AST. Since many patients with HELLP syndrome have hypertension and proteinuria, there is clinical overlapping with preeclampsia. Approximately 4–12% of preeclampsia patients also manifest the diagnostic criteria of HELLP syndrome /58/. \| \| HIV infection Thrombocytopenia occurs frequently with HIV infection, even early on. In one study /59/, out of 26 HIV positive pregnant women, 18 had thrombocytopenia and the thrombocyte count in 12 of these patients was below 50 × 109/L. Out of 13 caesarean sections, postoperative bleeding complications occurred in 4 cases. \| Table 15.12-1 Leukocyte reference intervals (data expressed in 109/L or %) \| Age \| Total \| Neutro- phils \| Eosino- phils \| Baso- phils \| Lympho- cytes \| Mono- cytes \| \| Fetus /3/ \| 3.4 ± 0.9 \| 0.2 ± 0.1 \| 0.08 ± 0.1 \| 0.02 ± 0.02 \| 2.6 ± 0.7 \| 0.2 ± 0.1 \| \| Umbilical cord blood /4/ \| 6.2–17.7 \| Cesarean section (Cs) \| \| \| \| \| \| 7.3–48.0 \| Spontaneous delivery \| \| \| \| \| \| 4.2–40.3 \| Emergency Cs \| \| \| \| \| \| Children /5, 6/ \| \| \| \| \| \| \| \| 1–14 d \| 6.5–15.0 \| 2.0–10.0 \| 0–0.8 \| \| 3.0–7.5 \| 0–3.0 \| \| 20–60% \| 0–8% \| 0–2.5 \| 18–55% \| 2–20 \| \| 15–180 d \| 6.5–15.0 \| 1.5–6.5 \| 0–0.5 \| \| 2.0–8.0 \| 0–3.0 \| \| 15–60% \| 0–5% \| 0–1.2% \| 18–65% \| 5–20% \| \| 0.5–< 2 yrs \| 6.5–15.0 \| 2.0–9.0 \| 0–0.3 \| \| 1.6–7.0 \| 0.4–2.0 \| \| 20–70% \| 0–4% \| 0–1.1% \| 18–60% \| 5–15% \| \| 2–< 6 yrs \| 5.0–12.0 \| 2.0–8.0 \| 0–0.3 \| \| 1.5–4.5 \| 0.3–1.2 \| \| 30–75% \| 0–4% \| 0–1.0% \| 13–55% \| 4–10% \| \| 6–< 12 yrs \| 4.5–11.0 \| 2.0–7.5 \| 0–0.4 \| \| 1.2–3.6 \| 0.3–0.9 \| \| 40–75% \| 0–5% \| 0–1.0% \| 13–50% \| 4–10% \| \| 12–< 18 yrs \| 4.5–10.5 \| 2.0–7.5 \| 0–0.3 \| \| 1.0–3.2 \| 0.4–1.3 \| \| 40–75% \| 0–5% \| 0–1.0% \| 13–45% \| 4–8% \| \| Adults /7/ \| \| \| ♀ + ♂ \| ♀ + ♂ \| \| \| \| Abbott CD 4000 \| ♀ 4.0–11.2 \| 2.0–7.7 \| 0.03–0.45 \| 0.007–0.09 \| 1.2–3.6 \| 0.23–0.82 \| \| ♂ 3.9–10.1 \| 1.7–7.0 \| \| \| 1.1–3.2 \| 0.25–0.94 \| \| Advia 120 \| ♀ 4.0–11.2 \| 2.1–7.7 \| 0.03–0.47 \| 0.02–0.11 \| 1.2–3.5 \| 0.20–0.65 \| \| ♂ 3.8–10.3 \| 1.8–7.0 \| \| \| 1.1–3.1 \| 0.24–0.73 \| \| Coulter LH 750 \| ♀ 4.0–11.2 \| 2.2–7.5 \| 0.04–0.40 \| 0.00–0.10 \| 1.1–3.5 \| 0.26–0.81 \| \| ♂ 3.9–10.9 \| 2.0–6.7 \| \| \| 1.2–3.0 \| 0.29–0.86 \| \| Horiba Pentra12 \| ♀ 4.0–11.5 \| 2.2–7.9 \| 0.06–0.78 \| 0.02–0.27 \| 1.1–3.5 \| 0.22–0.93 \| \| ♂ 3.9–10.3 \| 1.8–6.7 \| \| \| 1.1–3.1 \| 0.25–0.90 \| \| Sysmex XE-2100 \| ♀ 3.9–10.4 \| 1.9–7.3 \| 0.03–0.44 \| 0.01–0.08 \| 1.2–3.6 \| 0.25–0.85 \| \| ♂ 3.7–9.9 \| 1.8–6.2 \| \| \| 1.1–3.2 \| 0.25–0.85 \| \| Relative % /8/ \| \| 40–75 \| 0.5–7 \| 0.2–1.5 \| 17–47 \| 4–12 \| The values for fetuses aged 20–27 weeks (of pregnancy) are expressed as x ± s. The values for children were averaged from three studies since the variation was considerable. Values are 2.5th and 97.5th percentiles. The childrens’ values were obtained with the following hematology analyzers: Bayer H3, Siemens Advia 120, Coulter STKS. Values for adults are 2.5th and 97.5th percentiles. Table 15.12-2 Kinetics of the neutrophil granulocytes /10/ \| Progenitor cells \| \| \| Mean maturation time in the mitotic pool (myeloblast to myelocyte) \| 7–9 days \| \| Mean time in the post-mitotic and storage pool (metamyelocyte to PMN) \| 3–7 days \| \| Polymorphonuclear neutrophil (PMN) \| \| \| Mean half life time in the circulation \| 6 h \| \| Neutrophil count in the total body pool \| 4.6 × 1011 \| \| Circulating pool in the vascular system \| 2.2 × 1010 \| \| Marginal pool of the vascular system \| 2.3 × 1010 \| \| Daily turnover \| 1.0 × 1011 \| Based upon body weight of 70 kg. Table 15.12-3 Criteria of neonatal sepsis \| Manroe criteria /9/ Two or more criteria must be fulfilled. I/T neutrophil ratio greater than 0.16 Neutrophil count below 7.5 × 109/L or above 14.5 × 109/L Immature neutrophil count above 1.4 × 109/L \| \| Rodwell criteria /14/ Three or more criteria must be fulfilled. Manroe criteria, plus: I/M ratio ≥ 0.30 Leukocyte count ≤ 5.0 × 109/L or above 25 × 109/L Degenerative neutrophil changes Thrombocyte count ≤ 150 × 109/L \| I, immature granulocytes; T, total granulocytes; M, mature granulocytes Table 15.12-4 Diseases and conditions with neutrophilia /10/ \| Clinical and laboratory findings \| \| Physical work, situations of mental and physical stress, normal delivery Neutrophilia is caused by the release of catecholamines, because the proportion of marginated PMNs is reduced. The pulmonary capillary bed is the main site of the marginated PMNs. The lymphocyte and monocyte counts also increase. Leukocyte counts of 20 × 109/L are achieved. In leukocytosis associated with acute infection and glucocorticoid administration, no increase in lymphocytes or monocytes is seen. Following a marathon run, the leukocyte, neutrophil and immature granulocyte counts increase respectively by 230%, 330%, and 150% /34/. \| \| Smoker On the average, the leukocyte count is 10% higher in smokers than in non-smokers. The increase is seen with as few as 4–5 cigarettes per day. Heavy smokers may have values of up to 15 × 109/L. Almost all leukocyte sub populations are increased to the same extent /35/. \| \| Pregnant women /36/ The leukocyte count is elevated in pregnancy, mainly due to the rise in PMNs. Accordingly, the following neutrophil values in 109/L were observed in women: non-pregnant 3.70 ± 1.43; pregnant 1st trimester 6.46 ± 1.64; pregnant 2nd trimester 7.70 ± 1.67; pregnant 3rd trimester 7.37 ± 1.76. \| \| Neonate The leukocyte counts are (5–30) × 109/L. Higher or lower values are indicative of bacterial infection. The I/T ratio is also indicative. The ratio of immature (I) to total (T) leukocytes of ≥ 0.5 on the first day of life, and of > 0.2 from the second day of life onwards can be indicative for sepsis. \| \| Bacterial infection Acute bacterial infections result in neutrophilia primarily the PMNs, but also bands are increased. Metamyelocytes are found only following marked stimulation. Bacterial infections lead to neutrophilia of (15–25) × 109/L, and less often of up to > 50 × 109/L. A left shift and toxic granulation are common. In sepsis, especially sepsis caused by gram negative bacteria, neutropenia may develop. Acute infections that are associated with splenomegaly, such as enteric fever, only lead initially, if at all, to mild neutrophilia. Leukemic reaction: this reaction occurs in severe systemic infection (e.g., sepsis, miliary tuberculosis) as well as in severe hemolysis. The leukocytosis is > 25 × 109/L. Cells such as myelocytes, promyelocytes and myeloblasts pass from the mitotic compartment into the blood. In contrast to chronic myeloid leukemia, all precursor cells of granulopoiesis are present, and there is no leukemic hiatus. \| \| Fungal, parasitic and viral infection These infections lead to leukocytosis of only up to 20 × 109/L; neutrophilia is frequently present only in the early phase. Viral infections are often associated with neutropenia. \| \| Chronic inflammatory disease An increase in PMNs of up to 3-fold the upper reference interval value may occur, for example, in rheumatoid arthritis, rheumatoid fever, bronchitis, colitis, dermatitis, or pyelonephritis /10/. \| \| Metabolic disease Coma diabeticum, coma uremicum, coma hepaticum, acute gout attack, eclampsia, and acute thyrotoxicosis lead to endogenous toxic neutrophilia. \| \| Cushing’s disease /37/ Glucocorticoid receptors are expressed in leukocytes and are known to play a role in cell adhesion and leukocyte recruitment from bone marrow. Approximately 40% of the patients present with leukocytosis of (10.5 ± 2.6) × 109/L. and dropped to (8.4 ± 1.9) × 109/L. Mean neutrophil count was (7.6 ± 2.6) × 109/L and dropped to (5.3 ± 1.7) × 109/L. \| \| Coronary heart disease (CHD) /38/ Leukocytosis is an independent risk factor for CHD. Re infarction studies also show that the basal leukocyte count correlates with the frequency of re infarction. In patients with angina pectoris or myocardial infarction, mortality is increased if the leukocyte count is greater than 10 × 109/L. \| \| Intoxication Corticosteroids, lead, mercury, benzene, and carbon monoxide lead to exogenously or endogenously induced neutrophilia. The neutrophils increase shortly or a few hours after exposure. \| \| Acute loss of blood Post-hemorrhagic neutrophilia. Variable increases in neutrophils up to 25 × 109/L on day 3–5. Frequently concurrent thrombocytosis. Major surgery leads to similar findings. \| \| Malignant tumor Neutrophilia is encountered with many gastrointestinal and pulmonary tumors, particularly those with liver and lung metastases. Neutrophilia is either caused by the inflammatory reaction to destruction of the tumor tissue, or is associated with the production, by some of the malignant tumors, of granulopoiesis-activating colony-stimulating factors (G-CSF). \| \| Chronic myeloid leukemia Proliferative leukocytosis. Leukocyte count increases from 20 × 109/L to higher than 50 × 109/L with left shift (blasts, fraction < 20%), proliferation of eosinophils and basophils. Optional thrombocytosis, demonstration of the Philadelphia chromosome in 90% of the patients, splenomegaly. \| \| Myelofibrosis Proliferative leukocytosis with leukocytes up to 50 × 10 9/L with pathological left shift (appearance of blasts) and extramedullary hematopoiesis. Reduction, to varying degrees, of the erythrocyte and thrombocyte life spans. ANP index low-normal to increased /10/. \| \| Polycythemia vera Mean proliferative neutrophilia of 20 × 109/L due to enhanced granulopoiesis. The ANP index may be elevated. Erythrocytosis and thrombocytosis are characteristic /10/. \| \| Splenectomized patient According to one study /39/, 42.5% of patients have leukocyte counts above 10 × 109/L, and 9% have counts of over 13.5 × 109/L. An disproportionate neutrophil increase occurs. Approximately 46% of the patients have a thrombocyte count of greater than 350 × 109/L, with a normal thrombocyte distribution. \| Table 15.12-5 Drugs that can cause neutropenia \| Analgetics and anti- inflammatory drugs Indomethacin Gold salts Pentazocine Acetaminophen Phenacetin Aminopyrine Butasone Valproic acid Antimalarial drugs Chloroquine Pyrimethamine Dapsone \| Antibiotics Cephalosporins Clindamycin Gentamicin Isoniazide Metronidazole Nitrofurantoin PAS Penicillin Rifampicin Streptomycin Sulfonamides Tetracycline Vancomycin \| Anticonvulsants Carbamazepine Ethosuximide Phenytoin Valproic acid Antidepressants Amitriptyline Desipramine Doxepine Imipramine H2-blockers Cimetidine Ranitidine \| \| Cardiovascular medications Ajmalin Alprenolol Captopril Quinidine Disopyramide Flecainide Hydralazine α-Methyldopa Oxprenolol Procainamide Propafenone Propranolol Tocainide \| Antidiabetics Chlorpropamide Tolbutamide Diuretics Acetazolamide Chlorthalidone Chlorthiazide Ethacrynic acid Hydrochlorothiazide Spironolactone Others Allopurinol Levamisol Penicillamine \| Antithyroid drugs Carbimazole Methimazole Hypnotics, sedatives Benzodiazepines Chlorpromazine Clozapine Chlordiazepoxide Meprobamate Others Pindolol Ticlopidine Interferon-α Ribaverin \| Table 15.12-6 Congenital neutropenia \| Clinical and laboratory findings \| \| Severe congenital neutropenia (SCN) /40/ SCN was first discovered by Kostmann in the 1950s and is therefore also known as Kostmann syndrome. SCN is a heterogeneous group of diseases, with severe, life-threatening bacterial infection beginning at an early age. In addition to a reduced granulocyte count, some patients manifest a spectrum of qualitative changes such as defective production of bactericidal proteins or defective granulocyte activation. Furthermore, a key feature of many forms of SCN is the termination of maturation at the level of myoblast to promyelocyte. The cells manifest atypical nucleoli and cytoplasmic vacuolization. More than 95% of patients react to rHuGCSF with a rise in granulocytes to over 1 × 109/L. The following forms of SCN are distinguished: Kostmann form (autosomal recessive SCN). Functionally inactive mutations in the HAX1 gene are present. The gene codes for the mitochondrial protein of the same name; its role is the prevention of cellular apoptosis. Autosomal dominant gain-of-function mutations in ELANE, which encodes neutrophil elastase, lead to the mislocalization and intracellular accumulation of unfolded proteins, creating endoplasmic reticulum stress The stress in turn affects the survival and differentiation of granulocytic progenitors with few or no mature neutrophils in the peripheral blood. The patients have life-threatening bacterial infections shortly after birth. Approximately 50% of patients with congenital neutropenia harbour autosomal dominant gain-of-function mutations in ELANE. There are more than a hundred mutations in ELANE that span its five exons. Daily subcutaneous administration of recombinant human granulocyte-stimulating factor, bone marrow transplantation or correction of a single mutation in ELANE with the use of CRISPR-Cas9 gene editing of primary hematopoietic stem cells (HSCs) are methods of therapy /71/. Autosomal dominant inherited mutation in the GFI1 gene, a transcription factor. GFI1 is a zinc finger molecule with transcriptional repressor and splicing functions. It regulates the differentiation of hematopoietic and non-hematopoietic cells. Monocytosis, lymphopenia and disturbed lymphocyte function occur in addition to granulocytopenia. X chromosomal recessive inherited mutations in the WAS genes that encode the Wiskott-Aldrich protein. In addition to neutropenia, monocytopenia, T-lymphocyte activation and severe hypoplasia of bone marrow myelopoiesis (which is resistant to GCSF treatment) occur. A mutation in the G6PC3 gene, which encodes the catalytic unit 3 of glucose-6-phosphatase. The result is premature apoptosis of the neutrophils. Apart from that, the patients manifest constitutional problems such as cardiac defects and abnormal urogenital development /41/. Autosomal dominant inherited mutation in the CSF3R gene, which codes for the G-CSF receptor. Shwachman-Diamond syndrome: defect in the SBDS gene. Autosomal recessive disease with enzymatic deficiencies of the gastrointestinal tract and skeletal abnormalities. Barth syndrome: defect in the Taz1 gene. X-linked disease with disturbances of lipid metabolism, cardiomyopathy and muscle weakness. WHIM syndrome: defect in the CXCR4 gene. Autosomal dominant disease with wart formation, hypogammaglobulinemia and recurrent bacterial infection. Chediak-Higashi syndrome: defect in the LYST (CHS1) gene. Autosomal recessive systemic disease with hypopigmentation, prolonged bleeding time and peripheral neuropathy. Clinical findings: overall, SCN is a are rare disease, often associated with severe bacterial infection and fever starting in the first months of life. Some 20% of the children have splenomegaly in the first year of life, around 40% by the age of 10. With rHuGCSF treatment, the neutrophil count can be maintained at above 1 × 10 9/L, and the clinical symptomatology can be reduced significantly. The most important differential diagnosis in children aged 1-3 years are autoantibodies, which can cause destruction of the neutrophils. Nonetheless, these children do not suffer from severe infections. Laboratory findings: often severe neutropenia, usually below 0.2 × 109/L. Monocytes are elevated 3–5-fold, mild anemia, thrombocytes and IgG slightly elevated. In the bone marrow presence of myeloblasts and promyelocytes, but marked reduction in myelocytes and metamyelocytes, bands and neutrophils. Marrow eosinophils and monocytes are increased, other cell lines are normal. \| \| Cyclic neutropenia Genetic molecular and cellular studies have demonstrated that autosomal dominant cyclic neutropenia and the sporadically occurring form involve mutations in the gene ELA2 that codes for neutrophil elastase. This enzyme is formed in the granulation of the precursor cells. In cyclic neutropenia an atypical enzyme that causes apoptosis of the granulocytic precursor cells is formed, leading to effective oscillatory cell production /42/. Clinical findings: suspicion sets in during the first year of life in children with recurrent fever, pharyngitis, mouth ulcers, lymphadenopathy or recurrent cellulitis. Rare adult cases manifest similar symptomatology, often with painful cervical lymphadenopathy. Not infrequently, these patients present with acute peritonitis, ileus or septic shock. Laboratory findings: typically, oscillations of the PMN and monocytes occur every 21 days (i.e., every 21 days the neutrophil count falls to below 0.2 × 109/L for 3–5 days) and 10–15 days later reaches a peak with subnormal numbers of some 2.0 × 109/L. Some patients manifest oscillations of the thrombocytes and the reticulocytes, other, in addition, of the lymphocytes and eosinophils. \| \| Glycogen storage disease Glycogen storage disease (GSD) is a rare autosomal recessive disease that leads to elevated glycogen concentrations or to an abnormal glycogen structure. Based upon the causative defect, GSD is subdivided into 10 types. GSD1b is caused by a glucose-6-phosphatase translocase deficiency. This enzyme transports glucose-6-phosphate to the endoplasmic reticulum, where the conversion of glucose-6-phosphate to glucose and phosphate takes place. A deficiency in this enzyme causes deficient glucose formation from gluconeogenesis and glycogenolysis. The exact mechanism underlying the neutropenia in these patients is unknown /43/. Clinical and laboratory findings: hepatomegaly, seizures, coma, hypoglycemia, hyper lactatemia, acidosis, hyperuricemia, hyperlipidemia. See also Tab. 5.6-6 – Hereditary metabolic disorders causing lactic acidosis and Tab. 3.2-2 – Hypoglycemia syndromes in childhood and infancy). \| \| Primary immunodeficiency syndromes The primary immunodeficiency syndromes are a heterogeneous group of more than 75 diseases, some of which are associated with neutropenia /44/ (see also Tab. 21.2-9 – Frequency and inheritance of severe combined immunodeficiency (SCID) forms). X-linked agammaglobulinemia (XLA): some of these patients have significant neutropenia, and are at risk of suffering from a fungal or a Pneumocystis carinii infection. Hyper-IgM syndrome: according to a European Society for Immunodeficiencies report, 68% of patients with hyper-IgM syndrome have neutropenia. Approximately 45% of these cases are chronic, and a few of these are cyclic. Common Variable Immunodeficiency (CVID): the neutropenia may be autoimmune in nature. IgA deficiency: the basis for the neutropenia is likely autoimmune. Cartilage-hair hypoplasia: moderate to severe neutropenia with cell neutrophil counts of (0.1–2.0) × 109/L. Reticular dysgenesis: this is the most severe form of SCID, with early interruption of differentiation of the granulocyte cell lineage and severe lymphopenia. Moderate to severe neutropenia. \| Table 15.12-7 Secondary neutropenia \| Clinical and laboratory findings \| \| Pre-term delivery Approximately 50% of newborns weighing less than 2 kg have neutropenia below 1.5 × 109/L. In some newborns of mothers with preeclampsia, the neutropenia may be severe, and may be associated with gram negative infection. The presence of a proliferative disorder of regulation of the myeloid lineage, with a reduction of the post mitotic pool, is assumed /45/. \| \| Chronic idiopathic neutropenia This is an acquired selective neutropenia of unknown etiology that occurs in all age groups and lasts for more than 3 months. Some of the patients are symptomatic, while in others the disease course is associated with relatively few infections. A neutrophil count of (1.5–0.5) × 109/L is typical, while in severe cases it may fall to below 0.5 × 109/L. With regard to other cells, the blood count is normal. The bone marrow is also often normal; otherwise, it may show a decrease in post mitotic granulocytes. Long-term studies demonstrate that chronic idiopathic neutropenia does not convert to leukemia, myelodysplastic syndrome or aplastic anemia /46/. \| \| Alloimmune neonatal neutropenia (AINN) AINN is caused by IgG allo antibodies that are directed against neutrophil alloantigens. They are formed by the mother against fetal neutrophil granulocytes, they cross the placenta and are detected in maternal blood and the umbilical cord /47/. The neutropenic newborn is prone to infections during the neonatal period. AINN occurs only infrequently; anti neutrophil antibodies were detected in 4 out of 1016 sera collected from newborn infants, but none of these infants had AINN /48/. Laboratory findings: selective neutropenia in umbilical cord blood, often with compensatory monocytosis and eosinophilia. Evidence for presence of allo antibodies (e.g., anti-HNA-1a, anti-HNA-1b, anti-HNA-2a, anti-HNA-3a, anti-pan-FcRγIIIb; ND1). \| \| Autoimmune neutropenia (AIN) AIN is a chronic neutropenia in which autoantibodies to neutrophils are detected. AIN is a relatively benign, non-familial disease. It is found primarily in children < 2 years of age. AIN begins 1–15 months following birth with recurrent oropharyngeal and cutaneous infections, and ends spontaneously 7–73 months later /49/. The leukocyte count can be normal in spite of neutropenia < 0.25 × 109/L. AIN in adults is rare. Laboratory findings: the diagnostic investigation involves the demonstration of autoantibodies in the neutrophil agglutination test, the neutrophil immunofluorescence test, or the Monoclonal Antibody Immobilization of Neutrophil Antigen (MAINA) assay /47/. \| \| Autoimmune disease Autoimmune diseases such as rheumatoid arthritis (RA), Felty syndrome, large granular lymphocyte (LGL) leukemia and systemic lupus erythematosus (SLE) may be associated with chronic neutropenia /50/. Felty syndrome: this syndrome is defined by arthritis, splenomegaly and neutropenia. Patients who develop Felty syndrome often have long standing RA with nodes, ulcers, splenomegaly and hyper pigmentation of the lower extremities. With disease progression, vasculitis, neuropathy, pulmonary fibrosis and hepatomegaly can develop. Laboratory findings: neutropenia < 1,5 × 109/L, mild anemia, thrombocytopenia and lymphopenia, high level of rheumatoid factor, hypergammaglobulinemia, positive antinuclear antibodies /28/. LGL leukemia: LGL are microscopically large lymphocytes with rough eosinophilic inclusions. LGL can be CD8+T cells or natural killer (NK) cells. Some of the RA patients have clonal proliferation of CD8+LGL and neutropenia. Virtually all patients with Felty syndrome, and the RA patients with LGL leukemia, are HLA-DR4-positive which renders a uniform disease process likely /50/. SLE: neutropenia is common in SLE, although infections due to severe neutropenia are infrequent. In one study /51/, 47% of the patients had neutropenia, 13% had hemolytic anemia, 20% lymphopenia and 27% thrombocytopenia. Causes of neutropenia were neutrophil granulocyte reactive IgG, enhanced apoptosis, and intrinsic hyper proliferation of myelopoiesis /50/. \| \| Medication-dependent neutropenia – Generally The granulopoiesis disorder can affect the mitotic compartment or the pluripotent stem cells; immune-mediated neutropenia may also occur /52/. Some of the medications that can cause neutropenia are shown in Tab. 15.12-5 – Drugs that can cause neutropenia. \| \| – Isolated neutropenia Suppression of the mitotic compartment: the granulocyte storage pool contains a 5-day supply of neutrophil granulocytes. In mitotic compartment dysfunction and with a granulocyte half-life time of 6 hours, a continuous decline in the granulocyte count occurs within 5 days. If the toxic substance is discontinued, recovery takes place within 36–72 hours. \| \| – Aplastic anemia Suppression of the pluripotent stem cells: the result is a decline of the maturation of all hematopoietic cell lines a condition that is called aplasia. In acute aplastic anemia, neutropenia and thrombocytopenia (petechiae and/or ecchymosis) occur during the first 2–3 weeks following the start of medication. Since the life span of the erythrocytes is longer, anemia is not seen at this point in time. Hematopoiesis is diminished in chronic aplastic anemia, but thrombocytes and neutrophils are at levels which prevent hemorrhage and sepsis. \| \| – Immune neutropenia Immune-mediated neutropenia: the medication or a metabolite thereof alters the antigen pattern of the neutrophil membrane by attaching as a hapten. The immunization can occur either following a single administration or only after long term ingestion. The pathogenesis is believed to be based upon an antibody or a cell-mediated immune response. Immune regulation may, possibly, also be influenced through the formation of autoantibodies /52/. Neutropenia may develop rapidly, within hours, but it may also develop slowly. The neutrophil count can stabilize at a low level, but it may also normalize again. \| \| Chemotherapy of malignant tumors Chemotherapy-induced suppression of granulopoiesis is dependent upon the cytotoxic substance. A decrease in neutrophil granulocytes first occurs when the granulocyte storage pool is empty and the maturation of CFU-GM to myeloblasts stops. Following chemotherapy, neither a shortened neutrophil half-life time nor ineffective granulopoiesis are usually observed /53/. Cytotoxic substances which act during the M phase such as vinblastin, cytarabin or methotrexate lead to more rapid neutropenia and thrombocytopenia, followed by quick normalization. Only longer term infusion of these phase-specific cytostatic agents leads to permanent neutropenia. Non-phase-specific cytostatics (anthracyclines and dactomycin) with effects on the mitotic pool are associated with an onset of neutropenia similar to that induced by the phase-specific agents, but the duration of the neutropenia lasts longer. Cytostatics with no effect on the mitotic pool (busulfan) cause a more delayed nadir and recovery of granulocytes. After chemotherapy, granulocyte-colony stimulating factor (G-CSF) such as filgastrim, lenogastrim, and pegylated filgastrim is used to accelerate neutrophil recovery and thus prevent complications due to neutropenia.The European Organisation for Research and Treatment of Cancer, the American Society of Clinical Oncology, and others have established evidence based guidelines for the prophylactic use of G-CSF. However, there is either a lack of use or incorrect application of the guidelines in up to 90% of patients with lung cancer /58/. \| \| Methotrexate (MTX) therapy of chronic polyarthritis In chronic rheumatoid arthritis, MTX is administered as basic therapy at weekly doses of 7.5–25 mg. Aplastic anemia occurs in some cases, with a leukocyte count of (0.2–2.6) × 109/L, a thrombocyte count of (3–63) × 109/L, and an Hb concentration of 49–121 g/L /54/. \| \| Metamizole Metamizole is employed in the initial treatment of hyperthyreosis. Agranulocytosis with fatal consequences may occur. In a treatment study with metamizole / neutropenia or agranulocytosis occurred in 1.6% of the cases treated with a daily dose of 30 mg; at a daily dose of 15 mg, the frequency of these side effects was only 0.47%, with no attenuation of the therapeutic efficacy /55/. \| \| Acute febrile infection Acute febrile infection in adults may be associated with leukopenia and thrombocytopenia (Tab. 15.11-8 – Thrombocytopenia of various origins). Thus, in diarrhea caused by Campylobacter jejuni, 30% of the patients had a leukocyte count below 4 × 109/L, in 25% the thrombocyte count was below 130 × 109/L, and 1 patient (5%) manifested both of these /56/. \| \| Myelodysplastic syndrome (MDS) Obligatory are leukocytes below 4 × 109/L, thrombocytes below 100 × 109/L, and Hb levels below 120 g/L. Reticulocytes are often decreased, LD and ferritin increased. Possibly bone marrow mutation analyses for diagnosis confirmation: tet2, runx1, asxl1, sf3b1, srsf2, tp53, u2af1, dnmt3a, zrsr2, ezh2, nras, kras. See also Section 15.16 – Myelodysplastic syndrome. \| \| Acute leukemia (AL) ALs are malignant diseases, the origins of which are to be found in a myelomonocytic precursor cell (AML), a lymphoid precursor cell (ALL), or in the stem cells. Characteristic laboratory findings in these patients are neutropenia, anemia, and thrombocytopenia. \| \| Cirrhosis of the liver Anemia, leukopenia, and thrombocytopenia occur frequently in chronic liver disease, especially cirrhosis of the liver. Presumed mechanisms are hypersplenism, hemolysis, and humoral inhibition of hematopoietic precursor cells. \| \| Megaloblastic anemia Due to a DNA synthesis disorder, anemias of severe folate and vitamin B12 deficiency are associated with neutropenia and pancytopenia. The marrow is hyperc ellular, and a maturation dissociation of the hematopoietic cell lineages occurs. Hematopoiesis is ineffective, since a considerable fraction of the cells does not mature and become apoptotic. \| Table 15.12-8 Acute febrile infections with leukopenia and/or thrombocytopenia /56, 57/ \| Bacterial infections \| Viral infections \| \| Severe sepsis, particularly gram-negative pathogens \| Herpes viruses: EBV, CMV, HHV6, VZV, HSV \| \| Toxic shock syndrome (staphylococci, streptococci) \| Measles, rubella \| \| Typhoid fever (S. typhi, S. paratyphi) \| Hepatitis A and B HIV and AIDS \| \| Shigella enteritis, campylobacter infections \| Parvovirus B19 Dengue virus infection \| \| Rickettsia infections (spotted fever, typhus) \| Severe acute respiratory tract infection due to Coronavirus \| \| Ehrlichiosis, anaplasmosis \| Influenzavirus \| \| Coxiella burneti infection \| Nonspecific viral infection \| \| Bartonella infection (cat scratch disease) \| Following protective vaccination against viral infections \| \| Brucellosis, leptospirosis, tularemia \| \| \| Borrelia burgdorferi infection \| Parasitosis \| \| Recurrent fever (Borrelia spp.) \| Toxoplasmosis \| \| Tuberculosis (M. tuberculosis) \| Malaria \| \| Mycoplasma pneumoniae infection \| Visceral leishmaniasis \| Table 15.12-9 Diseases and conditions with lymphopenia \| Clinical and laboratory findings \| \| Congenital immunodeficiency Different forms of severe combined immunodeficiency (SCID), ataxia teleangiectasia, malnutrition, and zinc deficiency may be associated with varying degrees of lymphopenia. The formation of lymphocytes is either not in place due to a lack of stem cells, or is disturbed for other reasons /20/. \| \| HIV infection The selective destruction of CD4+T cells and the reversal of the CD4+/CD8+ ratio occurs /58/. \| \| Chemotherapy Chemotherapy can lead to severe lymphopenia. Alkylating substances have very different actions /20/. Thus, cyclophosphamide causes lymphopenia even at low to moderate doses. The lymphopenia may persist for years following dis continuation of therapy. CD4+T cells are more severely affected than other T cells. A decrease in CD4+T cells to below 0.2 × 109/L should not be permitted to occur, due to the severe deficiency in the defense against infections that will occur. \| \| Daily therapy with low radiation doses is more destructive to the lymphocytes than higher twice weekly doses /20/. \| \| Systemic lupus erythematosus (SLE) Not only in SLE, but also in mixed connective tissue disease (MCTD) and dermatomyositis, does lymphopenia occur. Anti-lymphocytic antibodies, which lead to lymphopenia via complement-mediated lysis, are detected in SLE. The antibody titer correlates with the extent of the lymphopenia /20/. \| \| Lymphocytopenia with a particularly marked reduction of CD4+T-cells occurs. Normalization returns under successful therapy. \| \| Influenza virus infection Typically, lymphopenia develops following recovery from infection /57/. \| \| In intensive care patients, lymphopenia and the neutrophil/lymphocyte cell ratio (NLCR) and CRP can be predictors of bacteremia. The lymphocyte count is often below 1.0 × 109/L and the NLCR is above 10. Thus, in a study /59/ in patients with positive blood cultures: CRP with a concentration higher than 50 mg/L demonstrated positive and negative predictive values of 54.3% and 59.6%, respectively Lymphocyte counts below 1.0 × 109/L showed positive and negative predictive values of 63.6% and 59.6%, respectively NLCR of greater than 10 had positive and negative predictive values of 67.6% and 73.4%, respectively. \| \| Miscellaneous Lymphopenia can occur in sarcoidosis, uremia, Cushing’s disease, with corticosteroid treatment, in inflammatory intestinal disease, following snake bites, burns, anesthesia, surgery including cardiopulmonary bypass surgery. The causes are, in some cases, lymphocyte recirculation disorders and, to a lesser extent, a reduction in the organism’s total lymphocyte count /20/. \| Table 15.12-10 Diseases and conditions with monocytosis /61/ \| Clinical and laboratory findings \| \| Infection /60/ Monocytosis is found in: Chronic infections such as tonsillitis, dental infection, liver abscess, endocarditis, tuberculosis, candidiasis Acute bacterial infections in the recovery phase (e.g., in the 2nd week of pneumococcal pneumonia). \| \| Autoimmune disease Mild monocytosis occurs with low prevalence in autoimmune vasculitis, rheumatoid arthritis, myositis, and arteritis temporalis. \| \| Hematological systemic disease Myelodysplastic syndrome, approximately one third of the patients have monocytosis Chronic myeloid leukemia; a high monocyte fraction may be present; if it is considerable, then myelomonocytic leukemia or monocyte leukemia are present Some of the diseases with chronic neutropenia are associated with monocytosis Medication-induced neutropenia; transient monocyte elevations occur in the acute phase, but the elevations during the recovery phase (e.g., following a chemotherapy cycle) are characteristic Hodgkin’s disease; one fourth of the cases is associated with monocytosis, but there is no correlation for purposes of prognosis Histiocytosis, multiple myeloma Non-hematological solid tumors are believed to be associated with monocytosis, independent of metastases, in up to 70% of the cases. \| \| Various causes Glucocorticoid therapy, splenectomy, poisoning, Hand-Schüller-Christian disease, Niemann-Pick disease, M. Gaucher. \| Table 15.12-11 Differential diagnosis of Eosinophilia \| Clinical and laboratory findings \| \| Parasitosis /62/ Following a stay in the tropics, helminthiasis should always be taken into consideration if eosinophilia is present. In infection with intestinal worms, mild eosinophilia is usually found (below 1.5 × 109/L), in infection with tissue-invasive worms, severe eosinophilia occurs (above 5 × 109/L). The positive predictive value of eosinophilia with regard to the diagnostic investigation of helminthiasis is believed to lie in the range of 10%. Tissue-invasive parasitosis includes infection with Strongyloides, Trichinella, Toxocara, Echinococcus, Cysticercus and Schistosoma. Suspicion of a parasitic infection exists if the patient originates from a region in which certain parasites are endemic, or if he returns from a trip to such a region. Parasitosis normally leads to moderate eosinophilia, with a cell count of (1–5) × 109/L. In infection caused by Toxocara species, which humans usually acquire from dogs and cats, eosinophilia of up to 50 × 109/L can occur if visceral Larva migrans is present /63/. Tropical eosinophilia: this disease is a variant of human filarial infection, which results from a hypersensitivity reaction to substances that are released from the micro filariae Wucheria bancrofti and Brugia malayi. Endemic regions are the coastal regions of India, Pakistan, Sri Lanka, Burma, Thailand, Malaysia, the Philippines, Southern China, Korea, and Central Africa. The eosinophil count is greater than 3 × 109/L and in hyperacute cases it can reach (30–50) × 109/L. Total serum IgE is elevated, and filaria-specific IgG and IgE antibodies are detectable /64/. \| \| Allergy Allergic diseases which must be taken into consideration are bronchial asthma, allergic rhinitis, atopic dermatitis, urticaria, and angioneurotic edema. Allergic reactions to aspirin, sulfonamide, allopurinol, and L-tryptophan can also lead to eosinophilia. The eosinophil count is about (0.5–1) × 109/L. \| \| Asthma /65/ Up to 40% of severe asthma start later in life, the eosinophilic form of asthma probably represents less than 5% of the total number of cases of adult-onset asthma. Patients with asthma who have eosinophilia have greater airway remodeling and more exacerbations, whereas those without eosinophilia have more airway obstruction. Eosinophils play a role in exacerbations and contribute to the pathophysiology of asthma. The eosinophils and the eosinophil cationic protein are elevated in the bronchial lavage of eosinophilic asthma /66/. \| \| Eosinophilic pneumonia (EP) /67/ EP includes a broad spectrum of lung diseases characterized by peripheral blood eosinophilia of over 1 × 109/L and/or alveolar eosinophilia over 25%. Blood eosinophilia may be lacking, as in the early phase of idiopathic acute EP or in patients taking oral corticosteroids. EP may present with varying severity, ranging from almost asymptomatic infiltrates to the acute respiratory symptom. Possible causes of EP are drugs and a variety of parasitic infections, however chronic EP remains idiopathic in many cases. \| \| Hyper eosinophilic syndrome (HES) /68/ Hyper eosinophilic syndromes are a group of disorders characterized by persistent and marked hyper eosinophilia of over 1.5 × 109/L not due to an underlying disease to cause eosinophil expansion (allergic drug reaction, parasitic infection), and which is directly implicated in damage or dysfunction of at least one target organ. Pathogenic mechanisms leading to hyper eosinophilia in patients fulfilling classical HES diagnostic criteria involve: Stem cell mutations leading to expression of PDGFRA-containing fusion genes with constitutive tyrosine kinase activity (mainly the FIP1L1-PDGFRA fusion gene) Sustained overproduction of IL-5 by activated T cell subsets with unusual phenotypes, and/or clonal T cell receptor rearrangement patterns Unknown molecular mechanisms leading to acquisition of abnormal phenotype by T cells such as CD3–CD4+, CD3+CD4–CD8– or CD3+CD4+CD7–. \| \| Eosinophilia-associated muscular disorders /69/ Various skeletal muscular pathologies are associated with blood and/or tissue eosinophilia. In addition to L-tryptophan related eosinophilia-myalgia syndrome and toxic oil syndrome, muscle involvement may occur in parasitic infections, Churg-Strauss vasculitis, hematological and non-hematological malignancies, and Shulman’s eosinophilic fasciitis. Idiopathic eosinophilic myopathies have also been described, encompassing three main subtypes: focal eosinophilic myositis, eosinophilic perimyositis and eosinophilic polymyositis. Skin involvement, in the form of an erythematosus rash, subcutaneous indurations and angioedema, is common. In a study /69/, 7 eosinophilic myopathies (without parasitosis) were published; the blood eosinophil count was (132–1464) × 109/L. \| \| Adrenal insufficiency Primary and secondary adrenal insufficiency are associated with eosinophilia. The cell count is usually (0.5–1) × 109/L. This condition is suspected in patients with hyper pigmentation, fatigue, hypotension, anorexia, nausea, and stomach disorder. Laboratory diagnostic evaluation includes cortisol and/or ACTH. \| \| Malignant disease Some 0.5% of malignant tumors are associated with eosinophilia which has been described in the following carcinomas: lung, breast, cervical, ovarian, liver, pancreatic, thyroid, Hodgkin’s, T cell lymphoma, multiple myeloma /70/. \| \| Idiopathic hyper eosinophilia (IHES) IHES is a largely heterogenous group of disorders defined as persistent marked hyper eosinophilia of unknown origin generally complicated by end-organ damage. With the concept of IHES, empirical diagnostic criteria have been defined, including /26/: Eosinophilia of at least 1.5 × 109/L for longer than 6 months No identifiable etiology of the hyper eosinophilia in spite of intensive diagnostic investigation Presence of functional disorders or injury to organs due to the idiopathic hyper eosinophilia. The organs that are most commonly affected are the skin, the heart, and the central nervous system. Observations indicate that distinct primitive hematological disorders involving either myeloid or lymphoid cells account for hyper eosinophilia in patients fulfilling the initial diagnostic criteria of IHES. Many patients can be classified in one of two diverse variants: The myeloproliferative variant with clinical and biological features reminiscent of chronic myelogenous leukemia. Prognosis has been considered poor. The lymphocytic variant. T-cell clones display eosinophilopoietic activity. This variant can be defined as a primitive lymphoid disorder characterized by nonmalignant expansion of a T-cell population able to produce eosinophilopoietic cytokines, especially IL-5. \| \| chronic subdural hematoma (csdh) The mechanism of eosinophil infiltration in CSDH is unknown. Laboratory tests showed a white blood cell count of 7.8 × 109/L and eosinophil percentage of 18,3% /77/. \| \| DRUG RASH WITH EOSINOPHILIA AND SYSTEMIC SYMPTOMS (DRESS) The Drug Rash with Eosinophilia and Systemic Symptoms (DRESS) syndrome is a rare but severe, potentially fatal idiosyncratic, adverse drug reaction with cutaneous and systemic manifestations /74/. Treatment with trimethoprim-sulfamethoxazole of a patient caused eosinophil count of 3028/μL, lymphocytic infiltration, alanine aminotransferase of 989 U/L, and aspartate aminotransferase of 162 U/L /74/. \| Table 15.13-1 Artifacts and morphological changes of leukocytes in the blood smear \| Clinical and laboratory diagnostics \| \| Granulocytes – Generally Morphological changes in the granulocytes are often of diagnostic value. \| \| – Drumstick A drumstick is characteristic of the female gender (Fig. 15.13-3 – Morphological granulocyte alterations). \| \| – Hyper segmentation of segmented granulocytes Polymorphonuclear neutrophils (PMNs) with more than five segments indicate hyper segmentation. Hyper segmented granulocytes in combination with giant bands is typical of a megaloblastic anemia. Hyper segmented granulocytes are often found in body cavity fluids (e.g., ascites) in which the granulocytes are separated from the circulation. \| \| – Toxic granulation, Döhle bodies, cytoplasmic vacuolization Toxic granulation, Döhle bodies and cytoplasmic vacuolization of granulocytes are nonspecific toxic manifestations of infection /11/. Toxic granulation: appearance of increased dark granules in the cytoplasm of neutrophils. They occur following infection. Toxic granulation is found in two thirds of sepsis, but it’s diagnostic specificity is low. Döhle bodies: these pale blue inclusion bodies, located in the peripheral cytoplasm, are aggregates of the endoplasmic reticulum. They are seen in infection and in association with toxic granulation. Approximately one third of sepsis patients have Döhle bodies. They also occur in pregnancy and postpartum. Toxic vacuolization: toxic vacuolization of the cytoplasm of neutrophils occurs in 90% of patients with bacterial sepsis. The vacuoles are large, opaque, and numerous, the entire cell is often deformed. Artificial vacuolization, which occurs when old EDTA blood samples are smeared, must be kept in mind. These vacuoles are usually smaller, more uniform in shape, do not deform the cell structure. \| \| – Pelger-Huet anomaly /12/ Pelger-Huet anomaly is a hereditary peculiarity of the neutrophil granulocytes whose nuclear segmentation does not exceed two segments. Approximately 30% of the cells have a band-shaped or peanut-shaped nucleus (bilobed or dumbbell-shaped nucleus). The prevalence is approximately 1: 6000. The function of the cells is normal, and they have a normal life span. Pelger anomaly is not gender-related. The distinction from an infection-dependent left shift is recognizable by means of the homologous cell morphology in Pelger, in contrast to the polymorphous appearance of bands. The acquired form of the Pelger-Huet anomaly (Pseudo-Pelger) is found in polycythemia vera, myeloid leukemia, osteomyelofibrosis, Hodgkin’s disease, and multiple myeloma, and can occur for the first time following bone marrow transplantation. It is also found in the loss of the short arm of chromosome 17 (17p-syndrome). Under therapy with taxane /8/ and ibuprofen /13/ a transient occurrence of Pseudo-Pelger occurs. The Pseudo-Pelger is distinguished from congenital Pelger-Huet anomaly based on the fact that in Pseudo-Pelger: Only a certain percentage of granulocytes manifests the characteristic changes Neutropenia, and a decrease in neutrophil granules, may be present There are no Pelger-type monocytes as there are in the hereditary form of the disease. \| \| – Chediak-Higashi syndrome In this abnormality, giant lysosomes in the form of eosinophilic inclusions are present in neutrophil granulocytes and other blood cells. Pseudo-Chediak-Higashi syndrome occurs in acute myeloid leukemia and myelodysplastic syndrome /8/. \| \| – Howell-Jolly-body-like inclusions Howell-Jolly-body-like inclusions in the cytoplasm of neutrophils are nuclear fragments that are identical to those found in erythrocytes. They occur in HIV infection, and must be distinguished from other unusual inclusions like those seen in infection or those which occur in Chediak-Higashi syndrome /14/. \| \| – Leukocyte agglutination Easily recognizable granulocyte agglutinates, which cause pseudo leukopenia, can be caused /8/: EDTA induced and affects not only neutrophils, but also eosinophils and basophils, as well as lymphocytes By cold agglutinins. The agglutination can be eliminated by incubation of the samples at 37 °C. \| \| Lymphocytes – Generally For the purpose of lymphocyte differentiation, they are primarily classified as “typical” or “atypical”. Typical means that morphologically, the lymphocytes are normal (cell diameter approximately 10 μm, nuclear diameter approximately 7 μm, heterogeneous lumpy nuclear chromatin, light-basophilic cytoplasm with smooth margins). Large granular lymphocytes (LGL) contain azurophilic granulation, and are considered to be typical if their proportion does not exceed 10%. All lymphocytes, apart from typical lymphocytes, as well as LGL, are designated as atypical if their proportion exceed 10%. The classification of atypical lymphocytes is made according to the categories of: “atypical, presumably reactive” or “atypical, presumably neoplastic”. Lymphocytes that cannot be classified are designated as “diverse”. For these, a commenting description such as, for (e.g., hair cells is recommended). \| \| – Diagnostic investigation of leukemia and lymphoma The role of the blood smear in the diagnosis of leukemia and lymphoma lies in the making of a tentative diagnosis as a basis for immunophenotyping. Examples: In acute leukemia, Auer rods (fused granules) are indicative Bilobular leukemia cells with fine granules and with faggots or bundles of fine Auer rods are indicative of acute promyelocytic leukemia. The same holds true for abnormal promyelocytes that are studded with rough azurophilic granules (hypergranular form) Hair-like cytoplasmic lymphocyte filaments. In the presence of low normal values of leukocytes, thrombocytes, granulocytes and hemoglobin, they are indicative of hairy cell leukemia. Large basophilic vacuolized lymphoma cells are indicative of Burkitt lymphoma. Philadelphia (Ph) chromosome: Ph-chromosome is associated with chronic myeloid leukemia (CML) in about 90% of cases and in acute lymphoblastic leukemia in 20–25% of cases /24/. Ph-chromosome is not seen microscopically in the stained and colored specimen of the bone marrow. Ph-chromosome is a rearrangement between ABL1 of chromosome 9 and BCR of chromosome 22. ABL1 (v-abl; Abelson murine leukemia viral oncogene homolog 1), originally located on chromosome 9 is translocated within the BCR (breakpoint cluster region) gene on chromosome 22 under formation of the fusion gene BCR-ABL1. In the normal location on chromosome 9 the ABL1 proto-oncogene encodes a cytoplasmic and nuclear tyrosine kinase that is implicated in cell differentiation, cell division and cell adhesion. The t (9, 22) translocation results in the fusion of the ABL1 and BCR gene on chromosome 22. The fusion gene encodes an unregulated cytoplasmic tyrosine kinase that allows cells to proliferate without being regulated by cytokines. Patients with CML, treated with tyrosine kinase inhibitors, have a life expectancy similar to that of the general population /25/. \| Table 15.13-2 Artifacts and morphological red blood cell changes in the blood smear \| Clinical and laboratory diagnostics \| \| There are differences in red blood cell staining intensity. These are dependent upon the red blood cell hemoglobin concentration. Hypochromic and normochromic erythrocytes are distinguished from one another. Hypochromic anisocytosis is typical of iron deficiency. Hyperchromic red blood cells do not exist, since erythrocyte hemoglobin concentrations cannot be greater than 360 g/L. Cells more intensively stained as normal appear to be hyperchromic. However, these cells have an atypical form like spherocytes or oval macrocytes in disorders of DNA synthesis /1/. \| \| Polychromatic red blood cells are stained pale blue. These are cells in the transformation of reticulocytes to erythrocytes. An increase of polychromatic erythrocytes indicates an enhanced consumption of erythrocytes (acute hemorrhage) /1/. \| \| Variability in the size of red blood cells. With the determination of red cell distribution width (RDW), hematology analyzers provide more precise information than the blood smear. Anisocytosis is found in iron deficiency anemia, DNA synthesis disorder (vitamin B12 or folic acid deficiency), and in anemia that is associated with reticulocytosis (e.g., acute hemorrhage, acute hemolysis). \| \| Microcytosis Microcytes are cells with a reduced volume. Microcytosis is associated with a decrease in erythrocyte hemoglobin content (MCH). Microcytes are pale in color and may have reduced MCHC. All microcytic red cells are hypochromic and all hypochromic cells are microcytic /1/. Hypochromic microcytic anemias are caused by iron deficiency and thalassemia syndromes. The blood smear is generally less important in the differential diagnosis of the microcytic anemias. However, it is important to note that the presence of Pappenheim bodies and red cell dimorphism in the sideroblastic anemias and of basophilic stippling in cases of lead poisoning and in some types of thalassemia is diagnostically significant /2/. \| \| Macrocytes are erythrocytes with a diameter of greater than 8 μm and normal thickness, while megalocytes have a diameter of over 9 μm. If, along with an increased diameter, the MCV is also elevated, it is necessary to rule out reticulocytosis as well as a sample from a newborn. Pseudo macrocytosis on the blood film is caused by the presence of thin red cells (leptocytes). If macrocytes are seen on the blood film and confirmed by an elevation in MCV, the presence of an increased reticulocyte count should be excluded. If in an adult patient with macrocytosis and without reticulocytosis the macrocytes are round, then the macrocytosis is likely due to chronic liver disease; if they are oval, they have megaloblastic maturation (vitamin B12 deficiency, folate deficiency) /1/. In such cases, anisocytosis, poikilocytosis and hypogranular or hypo lobulated neutrophils are often present. In the elderly, myelodysplastic syndrome is a frequent cause of macrocytic anemia. \| \| Nucleated Red Cells (NRC) The presence of NRC on the peripheral blood smear suggests an abnormality in the bone marrow or in the red cell production. NRC are especially common in hemolytic anemia, massive blood loss, severe oxygen deficiency, acute and chronic leukemias, malignant tumor, osteomyelofibrosis. \| \| Appearance of deformed red blood cells with many possible variations of shape. \| \| Hypochromic microcytic red blood cells, in which the hemoglobin is concentrated in a thin rim around a pale center /1/. Typical for chronic severe iron deficiency. \| \| In healthy individuals, some 1% of the erythrocytes have an elliptical shape. The shape varies from a slightly discernible ellipse to cells with a rod-like shape. Elliptocytes are present in vitamin B12 deficiency or folate deficiency, they can result from chemotherapy agents that disturb DNA and RNA synthesis or occur in patients with dyserythropoietic anemia /1/. Hereditary elliptocytosis: this is a rare autosomal dominant disease, in which up to 90% of the erythrocytes have an elliptical form. Hemolytic anemia is present. In Southeast Asian ovalocytosis, the cells are about twice as large as normal, and in addition, stomatocytes with two Y-shaped or V-shaped stomata are present /2/. \| \| The spherocyte has a smaller diameter than the erythrocyte, but the volume is identical. The cause is the loss of cell membrane. The spherocyte appears dense or darker due to the greater thickness of the cells in comparison to normocytes /1/. Young cells are not spherocytic in appearance. Spherocytosis is indicative of hemolytic anemia. The presence of spherocytes is not diagnostically specific, since this may result from hereditary spherocytosis, autoimmune hemolytic anemia, or alloimmune hemolytic anemia. In hereditary spherocytosis, a cell membrane synthesis disorder is present. Microspherocytes (cells that are both hyperchromic and significantly reduced in size and therefore in diameter) may be present in low numbers in patients with hemolytic anemia, but are also characteristic of burns and of microangiopathic hemolytic anemia /2/. \| \| Burr cells Burr cells frequently have bizarre shapes reminiscent of animal shapes. The finding of Burr cells lacks clinical specificity /1/. \| \| Heinz bodies Heinz bodies are precipitated hemoglobin and may be visible with Giemsa or Wright’s stain but not with ordinary Romanowsky stained films. Heinz bodies may be seen in /1/: Unstable hemoglobin disorders After the use oxidative drugs such as dapsone In red cell metabolic deficiencies (methemoglobin reductase deficiency). \| \| In the diagnosis of acute hemolysis the detection of keratocytes (have a horn-like shape) or bite cells is important /1/. The erythrocytes have changed their shape due to oxidative damage. Due to a drug, hemoglobin of the erythrocytes precipitates as a dense mass, causes Heinz bodies and leads to changes in the shape of the cells. The erythrocytes with irregular outline either have horn-like protuberances or they present as bit cells or the shape is still more or less normal, but they manifest large hemoglobin-free areas /2/. Oxidant-induced hemolysis is often seen in glucose-6-phosphate dehydrogenase (G6PD) deficiency, or disorders of the pentose phosphate shunt. Worldwide, millions of people are affected by G6PD deficiency. See also Section 15.8 – Erythrocyte enzymes). \| \| Thorn apple-shaped erythrocytes. Echinocytes are artifacts of blood collection and blood film preparation. Exposure to an alkaline pH, usually the result of glass effect, results in the formation of echinocytes. The slow drying of the blood smear on the glass slide is the most common cause of echinocyte formation /1/. \| \| Stomatocytes are usually an artifact on blood films, since hereditary stomatocytosis is very rare. Exposure of erythrocytes to an acid pH results in the formation of stomatocytes. This is also the case in the presence of drugs such as phenothiazines or of cationic and amphiphilic substances /1/. Stomatocytes are also present in liver disease, particularly those in which the pathogenesis is related to alcohol. \| \| Acanthocytes are erythrocytes with a lot of spines. They develop from normal red cells in the circulation and the nucleated precursor cells are normal. Acanthocytes occur in hereditary acanthocytosis, which is characterized by abetalipoproteinemia. Acanthocytes also develop in severe liver failure with hypo cholesterolemia /1/. Acanthocytosis can also develop with parenteral nutrition with fish oil; it is reversed following dis continuation /16/. Drugs such as benzodiazepines and 5-fluorouracil have also been described as causes acanthocytosis. \| \| The erythrocytes have the shape of a teardrop. These cells are observed in chronic idiopathic myelofibrosis, and in myelophthisic blood pictures associated with bone marrow replacement by malignant tumor cells /1/. \| \| Sickle cells These cells represent the ultimate spiculated red cells /1/. They are observed in the blood of homozygous hemoglobin S disease. Sickle cells are rarely seen in heterozygous trait carriers /17/. \| \| Schizocytes are erythrocyte fragments or remnants after fragments have been removed. The features are small size, dense appearance, and sharp pointed edges /1/. Many schizocytes are triangular shaped fragments. Schizocytes are formed by mechanical destruction (e.g., following artificial heart valve replacement, in hemolytic uremic syndrome, or in disseminated intravascular coagulation). \| \| Target cells Target cells have the form of a shooting target, because hemoglobin is concentrated in a rim and in the center of the cell. They occur in hypochromic anemia, hemolytic anemia, thalassemia, HbC disease, and following splenectomy. \| \| Erythrocyte agglutinates Irreversible erythrocyte agglutination due to heterophile IgM antibodies. They are formed in infectious mononucleosis (EBV), infection with Mycoplasma pneumoniae, and in the presence cold agglutinins of unknown etiology. \| \| Rouleaux formation Rouleaux formation is also known as the rouleaux phenomenon. In thick parts of the smear, the term rouleaux phenomenon refers to two and more cells that form a chain; in thin areas none are seen. The rouleaux phenomenon is reversible, it resolves when the blood is diluted with saline. For this reason, rouleaux formation does not affect the MCV measurement with hematology analyzers. Hyper fibrinogenemia and hypergammaglobulinemia are causes of rouleaux formation /1/. \| \| Basophilic stippling Basophilic stippling is seen in stained erythrocytes as fine dark inclusions and appear with Wright and Pappenheim staining. The inclusions are precipitated ribosomal RNA that developed during drying and staining of the smear /1/. Basophilic stippling is a sign of impaired regeneration of erythropoiesis; it may be seen in megaloblastic anemia, alcoholism, lead poisoning (impaired heme synthesis), arsenic poisoning, thalassemia (impaired globin synthesis), porphyria, and pyrimidine 5’-nucleotidase deficiency. \| \| Howell-Jolly bodies Small, round, violet-colored chromatin bodies in the erythrocytes. These are nuclear DNA residues which are normally removed by the spleen. They occur in hyper regenerative erythropoiesis with compromised reticulocyte nuclear expulsion. Howell-Jolly bodies are found in splenomegaly and chronic graft versus host disease. \| \| Cabot’s rings Reddish rings in the erythrocytes, in some cases as single rings or complex loops, such as a figure eight. Cabot rings are observed in DNA synthesis disorders (megaloblastic blood films) /1/. If, during erythropoiesis, iron that is taken up by the erythroid precursor cells is not immediately utilized for the synthesis of hemoglobin but rather, is stored as ferritin and hemosiderin, this can be shown using iron staining. Erythocytes containing siderosomes are called siderocytes, and erythroblasts that contain siderosomes are known as sideroblasts. Erythroblast mitochondria are organized as rings around the nucleus. Defects in erythroblast protoporphyrin synthesis lead to the accumulation of iron in the mitochrondria. Ring sideroblasts are seen in the hereditary form of sideroblastic anemia, in its acquired variant, namely, myelodysplastic syndrome (see Section 15.16 – Myelodysplastic syndrome), and in idiopathic forms of sideroblastic anemia /1/. \| Table 15.13-3 Artifacts and morphological platelet changes in the blood smear in thrombocytopenia and thrombocytosis \| Clinical and laboratory diagnostics \| \| Platelet aggregation Platelet aggregation is often EDTA induced, and causes pseudo thrombocytopenia. Depending upon the literature, a frequency of 0.1–1.9% is estimated. Most patients have no characteristic underlying disease. However, EDTA induced platelet aggregation is noted to occur more frequently in autoimmune rheumatoid arthritis, Sjögren’s syndrome, Guillain-Barre syndrome, neoplasms, atherosclerosis and in critically ill patients /8/. Treatment with the glycoprotein IIb/IIIa antibody abciximab can lead to platelet aggregation and pseudo thrombocytopenia. Some 5% of treated patients develop this symptomatology, while one third has EDTA induced thrombocytopenia /18/. \| \| Platelet satellitosis Platelets have coated neutrophil granulocytes, basophils, eosinophils, lymphocytes, and lymphoma cells. Satellitosis is mediated by IgG antibodies that are directed against the glycoprotein IIb/IIIa thrombocyte receptor and the FcRIII receptor on the leukocyte membrane /19/. Mild thrombocytopenia and leukopenia can be measured with the hematology analyzer. \| \| Giant platelets Giant platelets are seen in myeloproliferative disease, hypersplenism, and hereditary conditions like Bernard-Soulier syndrome and May-Hegglin anomaly /8/. When present in largely numbers giant platelets can falsely elevate the automated white blood cell and red blood cell counts while falsely lowering the automated platelet count /4/. \| \| Gray platelet syndrome The platelets look gray or pale blue, due to a deficiency in α-granules. The phenomenom is seen in hereditary syndromes such as Bernard-Soulier syndrome or May-Hegglin anomaly, but also may be present in myeloproliferative disorders, hypersplenism, following cardiopulmonary bypass surgery, and in hairy cell leukemia /8/. \| \| Other particles Erroneously high platelet counts are measured with hematology analyzers in the presence of red blood cell fragments of leukemia cells, and in mycosis-related sepsis /2/. \| Table 15.13-4 Morphological blood smear findings in infectious disease \| Clinical and laboratory diagnostics \| \| Clostridium perfringens The presence of bacteria in routinely prepared blood smears is an uncommon phenomenon in the hematology laboratory. Sepsis caused by bacteria and fungi can lead to intravascular hemolysis, but is a rare phenomenon. Despite its rarity, the most well-known cause of bacterial hemolysis is associated with C. perfringens /11/. Sepsis caused by this pathogen occurs preferentially in patients with gastrointestinal tumors. The hemolysis is dependent upon the formation of bacterial lecithinase with direct destruction of erythrocyte membranes. The C. perfringens sepsis in tumor patients is almost always fatal /20/. In the smear there is prominent spherocytosis and red cell fragmentation /11/. \| \| Malaria See Section 44.5.2.3 – Malaria. In malaria infection, the size of the affected erythrocytes is altered as follows: not enlarged in infection with P. falciparum and P. malariae, significantly enlarged (1.5 to 2-fold) in infection with P. vivax, and enlarged by 1.25 to 1.5-fold in P. ovale infection. \| \| N. meningitidis, S. aureus Generally, bacteria are seldom found in blood smears. If they are found, this speaks for severe sepsis with poor prognosis. The bacteria are located in neutrophil granulocytes, tend to clusters and usually lie within a distinct vacuole. In the routine diagnostic investigation, Staphylococcus aureus and Neisseria meningitidis are most commonly found /11/. If the bacteria are found only outside of the leukocytes, contamination of the dye solution must be taken into consideration. \| \| Ehrlichia /11/ Ehrlichia are obligatory intracellular growing coccoid, pleomorphic, gram negative bacteria belonging to the Rickettsiaceae family. The vectors are ticks. Relevant are human monocytic Ehrlichosis (HME) caused by E. chafeenis, and granulocytic Ehrlichosis (HGE) caused by an unknown species. Ehrlichia are phagocytized by granulocytes and monocytes and multiply to form 1–3-μm, membrane bound clusters of pale-blue staining coccobacillary organisms. These inclusions are known as morulae and their detection establishes the diagnosis of Ehrlichiosis /21/. In 25–80% of patients with HGE but only 1–2% of those with HME detectable morulae in leukocytes are present. The morulae may be easily overlooked. Ehrlichosis is a mild and self-limiting disease. Older persons and patients with severe underlying disease are affected by a symptomatic course, which can be mild, severe, or fatal. This also concerns immune-suppressed patients (HIV, post-organ transplantation, high-dose corticosteroid therapy) /22/. \| \| Babesia /11/ Babesia are protozoa, the vectors are ticks. There are two forms of babesia a mild American form, the pathogen is B. microti, and a severe European form that is caused by B. divergens. Following a tick bite, Babesia merozoites enter the erythrocytes, mature to trophozoites, multiply asexually, and can cause lysis of the erythrocytes. Trophozoites that are found in the erythrocytes are ring-shaped; they can have very small rings, of diameter as small as 2 μm. One cell may contain 1 to 12 rings. In most patients, 1–20% of the erythrocytes are affected; in splenectomized patients, this figure can be as high as 85%. The duration of parasitemia ranges from 3–12 weeks. The main differential diagnosis of babesiosis is malaria. Clinical symptoms present 1–4 weeks following a tick bite, and can lead to a broad spectrum of flu-like symptoms as well as to moderately severe malaria-like symptoms. Fever, weakness, sweating attacks, myalgia, arthralgia and nausea in early summer (tick season) should also bring to mind the possibility of Babeosis /22/. \| \| Spirochetes /11/ During acute febrile episodes of recurrent fever (not in afebrile periods) spirochetes are detected in blood smears in 70% of the cases. Buffy coat investigations with acridine-orange staining has been advocated for detection. The organisms are extracellular; they have a length of 8–40 μm, a width of 0.2–0.6 μm, and 3–10 loose, irregular coils. The recurrent fever can be triggered either by Borrelia recurrentis (vector is the louse) or by any one of 16 Borrelia species (vector is the tick). \| \| Trypanosoma /11/ There are two geographic variants: African trypanosomiasis (sleeping sickness), caused by T. brucei gambiense or T. brucei rhodesiense, and South American trypanosomiasis (Chagas disease), caused by T. cruzi. The vector for the African variant is the tsetse fly, while a bed-bug is the vector for the South American disease. Both disorders begin with a localized inflammatory reaction at the site of the infection bite (chancre). In acute phase of Chagas disease trypomastigotes are detectable in the blood smear. This is not the case in chronic Chagas disease. The circulating parasite in tryposomiasis is the trypomastigote, an extracellular, non invading form. The organisms measure 14–33 μm in length and 1.5–3.5 μm in width. \| \| Five species of filariae produce a circulating microfilarial form in humans: Wucheria bancrofti, Brugia malayi, Loa Loa, Mansonella perstans, Mansonella ozzardi. Microfilaria have a length of 230–300 μm, a thickness of some 7 μm, and a ribbon-like structure. They contain a chain of nuclei extending most of the length of the body. Because their demonstration in a normal blood smear is insensitive, concentration methods are recommended /23/. \| Table 15.14-1 Methods for investigation of bone marrow in primary diagnosis or suspected diagnosis of leukemia and lymphoma /8/ \| \| \| MPN/ CML \| \| Lym- phoma \| Re- lapse \| \| Cyto- morphology \| \| + \| + \| + \| + \| \| Cyto- chemistry \| + \| \| + \| – \| + \| \| \| \| + \| (+) \| + \| (+) \| \| Immuno- phenotyping \| + \| \| (+) \| + \| \| \| \| + \| + \| + \| (+) \| + \| \| \| \| § \| + \| § \| § \| \| 24 color FISH \| § \| § \| § \| § \| § \| \| \| + \| + \| (+) \| # \| # \| +, obligatory; (+), optional, for special issues; #, only within framework of studies with clearly-defined issues; §, if necessary to supplement other methods (e.g., cytogenetics). AL, acute leukemia; CML, chronic myeloid leukemia; MDS, myelodysplastic syndrome; MPN, myeloproliferative neoplasia; blast crisis Table 15.14-2 Percentage of bone marrow cells in healthy individuals according to Ref. /5/ \| Bone marrow cell \| 95% range \| \| Blast cells \| 0–3.0 \| \| \| \| \| \| \| \| \| \| \| Bands and segments \| \| \| Men \| \| \| Women \| \| \| Eosinophils \| \| \| Basophils \| 0–0.4 \| \| Erythroblasts \| \| \| Men \| \| \| Women \| \| \| Lymphocytes \| \| \| Plasma cells \| \| \| Monocytes \| \| \| Macrophages \| \| \| Myelopoiesis/erythropoiesis ratio \| \| \| Men \| \| \| Women \| \| Table 15.15-1 Methods for the diagnostic investigation of acute leukemia \| \| \| \| \| Cytomorphology (blood, bone marrow) \| + \| + \| \| \| + \| + \| \| Histology \| \| 0/+ \| \| \| + \| + \| \| Cytogenetics \| + \| + \| \| FISH \| \| bB \| \| PCR \| bB (+): (+) \| bB \| +, should be part of the routine; 0, must not be part of the routine; bB, as necessary in accordance with other results; FISH, fluorescence in situ hybridization; PCR, polymerase chain reaction Table 15.15-2 Previous FAB classification of AML (can be performed as initial diagnostic step before WHO classification) \| \| Mostly undifferentiated acute myeloid leukemia, immunophenotype necessary for confirmation \| \| \| Granulocytic leukemia without tendency for maturation \| \| \| Granulocytic leukemia with maturation of at least 10% of the leukemia cells to promyelocytes, myelocytes or further mature cells \| \| \| Hypergranular promyelocytic leukemia \| \| \| Similar to M3, with bilobular nuclei and hypogranulated cytoplasm, corresponding to microgranular promyelocyte leukemia \| \| \| Leukemia with evidence for a granulocytic and a monocytic component \| \| \| Similar to M4, but with pathological eosinophil granulocytes \| \| \| Acute monoblastic leukemia \| \| \| Acute monocytic leukemia \| \| \| At least 50% erythroblasts, among the other cells ≥ 30% blasts \| \| \| Megakaryoblastic leukemia, immunophenotype necessary for confirmation \| Table 15.15-3 WHO classification of AML (2008) \| AML with specific genetic aberrations AML with t(8;21)(q22;q22); RUNX1-RUNKX1T1 AML with inv(16)(p13.1q22) or t(16;16)(p13.1;q22) \| \| Acute promyelocytic leukemia with t(15;17)(q22;q12) \| \| AML with t(9;11)(p22;q23); MLLT3-MLL AML with t(6;9)(p23;q34); DEK-NUP214 AML with inv(3)(q21q26.2) or t(3;3)(q21;q26.2);RPN1-EVII Megakaryoblastc AML with t(1;22)(p13;q13) \| \| AML with mutated NPMI AML with mutated CEBPA \| \| AML with myelodysplasia-associated changes AML with prior MDS or MDS/MPN AML with MDS-specific cytogenetic aberrations AML with multilineage dysplasia \| \| Therapy-induced myeloid neoplasia AML following alkylating agents AML following epipodophyllotoxin Post-radiation AML \| \| AML with no other classification criteria Minimally differentiated AML AML without maturation AML with maturation Acute myelomonocytic leukemia Acute monoblastic and monocytic leukemia \| \| Acute erythrocytic leukemia Acute megakaryoblastic leukemia Acute basophilic leukemia Acute panmyelosis with myelofibrosis \| \| \| \| Down syndrome-associated myeloid proliferation \| \| Blastic plasmacytoid dendritic cell neoplasm \| Table 15.15-4 Differentiation of blasts (type I and II) from abnormal promyelocytes (in AML M3) and from normal promyelocytes \| Criterion \| Blasts in AML type I/II \| Abnormal promyelocytes in AML M3 \| Normal promyelocytes \| \| Nuclear position \| \| Central \| \| \| Nuclear chromatin \| \| \| \| \| \| 0–2 \| 0–2 \| \| \| Golgi zone \| None or suggested \| Rare \| \| \| Cytoplasm basophilia \| \| + up to reddish \| + \| \| Azurophilic granules \| \| Very much, emphasized \| Usually > 20 \| \| Auer rods \| \| In bundles (faggot cells) \| \| Table 15.15-5 Laboratory results in acute myeloid leukemia /14/ \| \| In 30–40% of patients, the white blood cell (WBC) count > 100 × 109/L is present diagnosed with acute leukemia and presence of blasts \| \| \| Normocytic to hypochromic anemia \| \| Thrombocytes \| (50–150) × 109 platelets \| \| \| > 38 °C \| \| \| 2–3 mmol/L; leukemic leukocytes have augmented sodium-potassium pump activity resulting in movement of potassium from plasma into leukocytes, decreasing the plasma concentration of potassium \| \| \| 135–145 mmol/L \| \| \| eGFR < 70 [mL × min–1 × (1.73 m2)–1]. In leukemia of monocytic differentiation lysozymuria, is a rare paraneoplastic complication. Lysozyme is released in the blood circulation, filtered by the glomerula and reabsorbed by renal tubules leading to toxic tubular injury and loss of potassium. The use of Cisplatin, aminoglycosides or loop diuretics is another cause of renal potassium loss. \| \| \| May be elevated \| \| Uric acid \| Hyperuricemia; leukemic infiltrates or uric acid nephropathy can reflect kidney injury \| \| \| \| \| \| \| \| \| Normal \| \| \| Greatly increased \| \| \| Increased \| \| \| Increased \| \| \| Normal \| \| \| Normal \| \| \| Slightly increased \| \| pO2 \| PO2 is a major criterion for evaluation of respiratory distress. Because oxygen consumption is proportional to the increase in white blood cells, arterial oxygen tension is decreased. \| \| pCO2 \| Normal \| Table 15.16-1 WHO classification of myelodysplastic syndrome /1/ \| \| Dysplasia (usually only) \| % Blood blasts \| % BM blasts \| % BM Ring-Sb \| Cyto- genetics \| \| \| \| < 5% \| < 5% \| < 15% \| Only 5q \| \| RCUD (RN, RA, RT) \| dysG, dysE or dysM \| < 1% \| < 5% \| < 15% \| \| \| \| dysE \| \| < 5% \| ≥ 15% \| V \| \| \| 2–3 lineages \| Rare \| < 5% \| < 15% \| V \| \| \| 1–3 lineages \| < 5% \| \| < 15% \| V \| \| \| 1–3 lineages \| \| \| < 15% \| V \| \| \| 1 lineage \| None \| < 5% \| < 15% \| V \| BM, bone marrow; V, various; RCUD, refractory cytopenia with single-lineage dysplasia; RN, refractory neutropenia; RA, refractory anemia; RT, refractory thrombocytopenia; RCMD, refractory cytopenia with multi-lineage dysplasia; RAEB, refractory anemia with increased blasts; Ring-Sb, ring sideroblasts Table 15.16-2 International Prognostic Scoring System: grouping according to points /4/ \| \| Score points \| \| \| \| \| \| 0 \| \| \| \| \| \| % Blasts in bone marrow \| < 5 \| \| – \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Favorable karyotype: normal, -Y, del(5q), del(20q); unfavorable karyotype: complex, aberrant (≥ 3 aberrations), aberrations on chromosome 7; all other chromosome findings are intermediat; F, favorable; UF, unfavorable; I, intermediate Table 15.16-3 International Prognostic Scoring System: prognostic classification /4/ \| \| 0 \| 0,5 to 1,0 \| 1,5 to 2,0 \| ≥ 2,5 \| \| Groups at risk \| \| \| \| \| Table 15.16-4 Characteristic cytogenetic alterations in MDS, in comparison with AML \| Chromosomal abnormality \| Frequency in MDS (%) \| Frequency in AML (%) \| \| only del(5q) \| 5–20% \| \| \| -7, only del(7q-) allein \| \| 2% \| \| \| \| \| \| -17, del(17p), iso (17q) \| 5% \| \| \| \| \| \| \| -X, -Y \| 2% \| 0,5% \| \| t(3;3)(q21;q26), inv(3)(q21q26) \| < 1% \| \| \| Complex karyotype \| \| 10–15% \| Table 15.17-1 Clinical characteristics and orienting markers in the differential diagnosis of MPN \| Findings \| \| \| \| \| \| Spleno- megaly \| (+) \| No \| + \| No \| \| Blood count changes in granulopoi- esis and/or erythropoi- esis \| Leukocytosis with left shift, eosinophilia, basophilia \| Hematocrit ↑ \| Red cell teardrops, left shift of granulopoi- esis, normo- blasts \| No \| \| Thrombo- cytes \| \| (↑) Giant forms \| ↓ up to (↑) \| > 450 × 109/L, giant forms \| \| Bone marrow cytology \| Very hypercellular, basophilia, megakaryo- cytes and eosinophils ↑ \| Markedly hypercellular erythropoiesis ↑ \| Usually punctio sicca \| Megakaryo-cytes markedly increased \| \| Bone marrow histology \| Granulopoi- esis ↑ Megakaryo- cytes ↑ \| Cell content ↑ \| Marked fibrosis \| Megakaryo-cytes increased, lying in nests \| \| Philadelphia- chromosom t(9;22); \| \| No \| No \| No \| \| Other chromos. changes \| In acceleration and blast crisis \| del (20q), +8. +9, +1q \| –7, +8, +9, +1q \| Very rare \| \| Vit. B12 > 900 (ng/L) \| ↑ \| ↑ \| Normal \| Normal \| CML, chronic myeloid leukemia; PV, polycythemia vera; PMF, primary myelofibrosis; ET, essential thrombocythemia Table 15.17-2 Methods at the time of diagnosis in chronic myeloproliferative neoplasia \| Methods \| PV \| ET \| PMF \| CML \| \| Cytomorphology (blood and BM) \| + \| + \| + \| + \| \| Histology \| + \| + \| + \| + \| \| Cytogenetics \| (+) \| – \| + \| + \| \| FISH \| – \| – \| – \| + (BCR-ABL1) \| \| PCR \| + \| + \| + \| + (BCR-ABL1) \| +, should be part of the routine; –, must not be part of the routine; FISH, fluorescent in situ hybridization; PCR, polymerase chain reaction; BM, bone marrow Table 15.17-3 WHO classification of CML based upon the morphological findings /1/ \| Chronic phase CML Classic cytomorphological and/or histological picture BCR-ABL1 positive or demonstration of t(9;22). Blasts in blood < 2%, in bone marrow < 5–10%. \| \| CML in accelerated phase (one or more criteria met) Blasts 10–19% in blood and/or bone marrow Basophils in blood > 20% Platelets not therapy-related < 100 × 109/L or > 1,000 × 109/L Leukocytes > 10 × 109/L or spleen increasing and not susceptible to treatment Clonal evolution in cytogenetics. \| \| CML in blast crisis (one or more criteria met) ≥ 20% blasts in blood and/or bone marrow Extramedullary blast cell proliferation Large foci of blast nests seen on bone marrow biopsy. \| Table 15.17-4 Diagnostic criteria of polycythemia vera (PV) \| Main criteria \| \| \| \| Hemoglobin values > 185 g/L in men, > 165 g/L in women, or any further evidence of erythrocytosis \| \| \| Demonstration of JAK2 V617F mutation or other equivalent functional mutation (e.g., JAK2 exon12) \| \| Secondary criteria \| \| \| 1. \| Age nonspecific, hypercellular bone marrow with three lineage growth (panmyelosis) with marked erythrocytic, granulocytic, and megakaryocytic proliferation \| \| 2. \| Serum erythropoietin values below the reference interval \| \| \| Endogenous erythroid colony formation in vitro \| Hemoglobin (Hb) or hematocrit values above the 99th percentile of the method-specific upper reference interval value for age, gender or hemoglobin values > 170 g/L in men or > 150 g/L in women (as long as this is associated with a documented and continuous rise of at least 20 g/L from the individual basal value that is not due to therapy of iron deficiency); or erythrocytosis of greater than 25% of the hypothetical normal value Table 15.17-5 Diagnosis criteria for primary myelofibrosis (PMF) \| Main criteria \| \| \| 1. \| Demonstration of megakaryocytic proliferation and atypiaa, usually associated with reticulin fibrosis and/or collagen fibrosis if relevant reticulin fibrosis cannot be demonstrated the megakaryocyte change must be accompanied by increased bone marrow hypercellularity, characterized by granulocytic proliferation and, often, decreased erythropoiesis (e.g., prefibrotic disease phase). \| \| 2. \| The WHO criteria for polycythemia vera b, BCR-ABL1+ chronic myeloid leukemia c, myelodysplastic syndromed, or other myeloid neoplasia, are not fulfilled. \| \| 3. \| Presence of JAK2 V617F or other clonal markers (e.g., MPL W515K/L) or if a clonal marker is not present, the bone marrow fibrosis or other changes may not assumed to be indicative of an infection, autoimmune disease, other chronic disease such as hairy cell leukemia, or other lymphoid neoplasia, metastatic malignancy or toxic (chronic) myelopathye. \| \| Secondary criteria \| \| \| 1. \| Leukoerythroblastosisf \| \| 2. \| Increase in serum lactate dehydrogenasef \| \| 3. \| Anemiaf \| \| \| Splenomegalyf \| a Small to large megakaryocytes with an abnormal nucleus to cytoplasm ratio and hyperchromatic, round or irregularly-shaped nuclei and dense clustering. b Presumes the failure of treatment with iron to increase hemoglobin levels, in order to bring the value with low serum ferritin up to that seen in polycythemia vera. Polycythemia vera can be ruled out based upon the hemoglobin and hematocrit values, and measurement of the red cell mass is not necessary. c BCR-ABL1 must be negative. d Neither dyserythropoiesis nor dysgranulopoiesis may be demonstrated. e Patients with symptoms resembling those of reactive myelofibrosis are not immune to PMF; the diagnosis should be taken into consideration if other conditions are met. f The degree of variance can be borderline or marked. Table 15.17-6 Diagnostic criteria of essential thrombocythemia (ET) \| 1. \| Persistent a thrombocyte count ≥ 450 × 109/L \| \| 2. \| The bone marrow biopsy manifests mainly proliferation of the megakaryocyte line with an increasing number of enlarged, mature megakaryocytes. No rise worth mentioning or left shift of neutrophilic granulopoiesis or of erythropoiesis. \| \| 3. \| The WHO criteria for polycythemia verab, primary polycythemiac, BCR-ABL1 positive chronic myeloid leukemiad, or myelodysplastic syndromee, or other myeloid neoplasia are not met in each case. \| \| 4. \| Demonstration of JAK2 V617F or other clonal markers, or if JAK2 V617F is not mutated no evidence for reactive thrombocytosisf. \| a During the clarification process. b If the failure of an iron treatment for raising the hemoglobin value is postulated, in order to bring the value in the presence of low serum ferritin to the level in polycythemia vera. Polycythemia vera can be ruled out on the basis of the hemoglobin and hematocrit values; measurement of the red blood cell mass is not necessary. c If an absence of relevant reticulin fibrosis or collagen fibrosis is postulated leukoerythroblastosis in the peripheral blood, or marked hypercellular bone marrow accompanied by megakaryocytic morphology which is typical of primary myelofibrosis, including small to large hyperchromic megakaryocytes with an aberrant nucleus-to-cytoplasm ratio, round or irregularly-shaped nuclei and dense clustering. d BCR-ABL1 must be negative. e Dyserythropoiesis or dysgranulopoiesis must be undetectable. f The following must be considered as putative causes of reactive thrombocytosis: iron deficiency, splenectomy, surgery, infection, inflammation, disease of connective tissue, metastatic carcinoma, and lymphoproliferative disease. If, however, the first three criteria are met, conditions that resemble reactive thrombocytosis cannot rule out the possibility of ET. Table 15.18-1 Basic panel for the initial diagnosis in acute leukemia, recommended by the German Kompetenznetz Akute und chronische Leukämien \| \| \| \| \| \| \| cy 1 \| \| cy-NK \| cy-NK \| cy-NK \| \| cy 2 \| \| \| \| \| \| cy 3 \| \| \| CD45 \| \| \| 1 \| \| NK \| NK \| NK \| \| 2 \| \| \| CD45 \| \| \| \| \| \| CD45 \| \| \| \| \| \| CD45 \| CD19 \| \| 5 \| \| \| CD45 \| \| \| 6 \| CD34 \| \| CD45 \| \| \| \| \| \| CD45 \| \| \| \| \| \| CD45 \| – \| F1–F4, fluorescence 1–4; cy, cytoplasmic staining; NK, negative control Table 15.18-2 Terminology, frequency and phenotype of immunological ALL subtypes \| \| B precursor cell ALL \| \| \| B cell ALL, Burkitt lymphoma \| T precursor cell ALL \| \| \| \| \| \| pro-Ba \| \| \| \| \| \| \| \| \| Incidence (%) children/adultsb \| \| \| \| \| \| \| \| \| \| Progenitor cell-associated antigens \| \| \| \| \| \| \| \| \| \| TdT \| +c \| + \| + \| – \| – \| + \| + \| \| \| HLA-DR \| + \| + \| + \| + \| + \| – \| – \| – \| \| CD10 \| – \| + \| ± \| ± \| ± \| ± \| ± \| – \| \| B cell antigens \| \| \| \| \| \| \| \| \| \| CD19 \| + \| + \| + \| + \| – \| – \| – \| – \| \| CD22d \| + \| + \| + \| + \| – \| – \| – \| – \| \| CD79a \| + \| + \| + \| + \| – \| – \| – \| – \| \| cyIgM \| – \| – \| + \| – \| – \| – \| – \| – \| \| mIgM \| – \| – \| – \| + \| – \| – \| – \| – \| \| T cell antigens \| \| \| \| \| \| \| \| \| \| cyCD3 \| – \| – \| – \| – \| + \| + \| ± \| – \| \| CD7 \| – \| – \| – \| – \| + \| + \| + \| + \| \| CD2 \| – \| – \| – \| – \| – \| + \| + \| + \| \| CD5 \| – \| – \| – \| – \| – \| + \| + \| + \| \| CD1a \| – \| – \| – \| – \| – \| – \| + \| – \| \| mCD3 \| – \| – \| – \| – \| – \| – \| ± \| + \| a Synonym pre-pre-B ALL; b Corresponding to the results of the German Multicenter ALL Therapy Study /4, 21/; c +, positive; ±, positive or negative; –, negative; demonstration of the respective antigen on ≥ 20% of the leukemia cells (membrane-bound antigens), or in ≥ 10% of the leukemia cells (MPO, cyCD3, cyCD79a, TdT); d Intracytoplasmic expression in immature subtypes. Abbreviations: CD, Cluster of Differentiation; Cy, intracytoplasmic; m, membrane-bound; Ig, Immunoglobulin Table 15.18-3 Correlation of the FAB classification, cytogenetics and immunophenotype \| \| FAB classification \| \| \| \| \| \| \| \| \| \| \| \| \| M5 \| \| \| \| \| \| \| \| MPO \| \| + \| + \| + \| \| – \| – \| \| CD2 \| – \| \| \| +/– \| \| \| \| \| CD13 \| +/– \| + \| + \| + \| +/– \| – \| +/– \| \| CD14 \| – \| – \| – \| +/– \| +/– \| – \| – \| \| \| – \| +/– \| –/+ \| +/– \| \| + \| \| \| CD19 \| – \| +/– \| \| \| \| \| \| \| CD33 \| +/– \| +/– \| + \| + \| + \| + \| + \| \| CD34 \| +/– \| +/– \| \| \| \| \| \| \| CD56 \| \| +/– \| –/+ \| –/+ \| \| \| \| \| CD61 \| – \| – \| – \| – \| – \| – \| + \| \| \| – \| – \| +/– \| + \| + \| + \| \| \| \| –/+ \| +/– \| –/+ \| + \| +/– \| + \| +/– \| \| CD117 \| +/– \| +/– \| –/+ \| +/– \| –/+ \| \| \| \| HLA-DR \| +/– \| + \| – \| + \| + \| + \| +/– \| Immunological and cytogenetic characteristics of acute myeloid leukemia in correlation with the FAB subtype. There is no definite correlation, immunophenotype is only indicative, never proof of definite subtyping of AML. –, antigen not expressed; –/+, antigen expressed in < 50% of the cases; +/– , antigen expressed in the majority of the cases; +, antigen expressed. Blank fields show only partial expression, which is nonspecific for the diagnosis, or for which no reliable data is available. Table 15.18-4 Score for the definition of mixed phenotype acute leukemia (MPAL) \| Myeloid cell series MPO or two markers of monocytic differentiation, like CD11c, CD14, CD64, lysozyme, NSE \| \| T cell series cyCD3 or mCD3 (only a few cases known) \| \| B cell series CD19 (strong expression) accompanied by an additional marker such as CD79a, cyCD22, CD10 or CD 19 (weak expression) accompanied by two additional markers such as CD79a, cyCD22, CD10 \| In order to diagnose MPAL, the expression criteria for at least two cell series (myeloid, B or T) must be fulfilled. Abbreviations: CD, cluster of differentiation; cy/m, intracytoplasmic or membrane-bound expression; MPO, myeloperoxidase; NSE, neuron-specific enolase Table 15.18-5 Antigen combinations for the diagnosis of minimal residual disease in patients with acute lymphocytic leukemia \| Subtype \| \| Incidence (%) \| \| B precursor cell ALL \| \| \| \| \| \| \| \| \| \| T precursor cell ALL \| \| 90–95 \| \| \| 30–50 \| \| AML \| \| \| \| \| 20 \| Table 15.18-6 Immunological marker profile of normal B cells and chronic lymphoproliferative diseases of the B cell type \| \| Norm B \| \| \| \| \| \| \| \| \| \| \| \| + \| (+) \| + \| + \| + \| + \| + \| + \| + \| + \| \| \| – \| + \| –/+ \| – \| + \| – \| – \| – \| – \| – \| \| CD10 \| – \| – \| – \| – \| – \| +/– \| – \| – \| – \| –/+ \| \| \| – \| – \| – \| \| – \| – \| + \| + \| –/+ \| – \| \| CD19 \| + \| + \| + \| + \| + \| + \| + \| + \| + \| + \| \| \| +/– \| – \| + \| + \| + \| + \| + \| + \| + \| + \| \| \| –/+ \| + \| –/+ \| –/+ \| – \| – \| – \| – \| –/+ \| \| \| \| + \| + \| + \| +/– \| + \| + \| –/+ \| –/+ \| + \| + \| \| \| – \| – \| – \| –/(+) \| – \| – \| + \| – \| –/+ \| –/+ \| \| \| – \| + \| +/– \| –/+ \| + \| – \| – \| – \| – \| – \| \| \| + \| +/– \| + \| + \| + \| + \| + \| + \| + \| + \| \| \| – \| – \| – \| – \| – \| – \| + \| + \| –/+ \| – \| \| \| – \| + \| N.A. \| N.A. \| – \| N.A. \| N.A. \| N.A. \| N.A. \| N.A. \| \| \| – \| – \| + \| + \| + \| + \| + \| + \| –/+ \| N.A. \| \| Cyclin D1 \| – \| – \| – \| – \| + \| – \| –/(+) \| –/(+) \| – \| – \| Abbreviations: N.A. no data available; Norm B, normal circulating B cells; B-CLL, B cell chronic lymphocytic leukemia; B-PLL, B cell prolymphocyte leukemia; LPL, lymphoplasmacytic lymphoma; MZL, mantel cell lymphoma; FL, follicular lymphoma; HCL, hairy cell leukemia; HCL-V, variant form of hairy cell leukemia; SLVL, splenic lymphoma with villous lymphocytes; sIg, surface immunoglobulin; DLBCL, diffuse large cell B cell lymphoma. Table 15.18-7 Immunological marker profile of normal T cells and T cell type lymphoproliferative diseases \| \| Norm T \| Act T \| LGLL/ T \| LGLL NK \| \| \| \| \| \| 2 \| + \| + \| + \| + \| + \| + \| + \| +/– \| \| 3 \| + \| + \| + \| – \| + \| + \| + \| +/– \| \| 4 \| +/– \| –/+ \| – \| – \| +/– \| + \| + \| +/– \| \| 5 \| + \| + \| – \| – \| – \| + \| +/– \| +/– \| \| 7 \| + \| + \| – \| – \| + \| – \| – \| +/– \| \| 8 \| –/+ \| +/– \| + \| +/– \| –/+ \| – \| – \| +/– \| \| 10 \| – \| –/+ \| – \| – \| – \| – \| – \| +/– \| \| \| – \| – \| + \| + \| – \| – \| – \| – \| \| \| – \| +/– \| – \| – \| – \| – \| + \| + \| \| 30 \| – \| +/– \| – \| – \| – \| – \| +/– \| +/– \| \| \| – \| – \| – \| +/– \| – \| – \| – \| +/– \| \| \| – \| – \| +/– \| +/– \| – \| – \| – \| – \| \| \| – \| + \| – \| – \| –/+ \| – \| +/– \| +/– \| Abbreviations: Norm T, normal circulating T cells; Act T, activated T cells, LGLL/T, T cell large granular lymphocyte leukemia; LGLL/NK, NK cell large granular lymphocyte leukemia; T-PLL, T prolymphocyte leukemia; ATLL, adult T cell leukemia/lymphoma; ALCL, anaplastic large cell lymphoma Figure 15.1-1 Hierarchy of the hematopoietic system with myelopoiesis and lymphopoiesis; courtesy of Ref. /9/. CFU, colony forming unit; BFU, burst forming unit; E, erythrocytic; G, granulocytic; M, macrophagoytic; Meg, megaloblastic. Figure 15.1-2 Erythropoiesis develops a proliferation pool (upper row) and a maturation pool (lower row). In the proliferation pool, the committed progenitor cells proliferate and differentiate stimulated by growth factors, particularly erythropoietin (EPO). EPO deficiency leads to apoptosis of CFU-E cells. The cells form hemoglobin (Hb) in the maturation pool; Hb synthesis is diminished in iron deficiency. Courtesy of Ref. /29/. Figure 15.2-1 Erythrogram. The hemoglobin content (HC) of the erythrocytes is plotted on the abscissa against their volume (V) on the ordinate. A normal erythrogram is depicted. In iron deficiency anemia, the point cloud is shifted to the bottom left, in pernicious anemia it is displaced toward the upper field, and in hereditary spherocytosis it is moved towards the center right. Figure 15.2-2 Histogram of red cell distribution width (RDW). The erythrocyte volume is plotted on the abscissa in fL against the relative frequency (REL%) on the ordinate. I. Normocytic anemia with normal RDW. In thalassemia the RDW is normal or only slightly elevated. II. In iron deficiency the distribution peak is shifted further to the right. III. Macrocytic anemia. The RDW is broadened in the upper MCV range. IV. Distribution in the presence of cold agglutinins at a high titer, the arrow shows the characteristic change. Figure 15.3-1 Frequent causes of anemia. Modified from Ref. /18/. The hypoproliferative anemias are highlighted in dark gray. Figure 15.3-2 The O2 dissociation curves in dependence of PO2 and the O2 saturation of hemoglobin. With increasing PO2, the O2 saturation (SO2) reaches a plateau at high PO2. In this portion the O2 dissociation curve is flat indicating that PO2 values in arterial blood lead to full O2 saturation of hemoglobin. The steep part of the O2 dissociation curve at low PO2 is important for O2 release from hemoglobin to the tissues, since with an only small decline in tissue PO2 a great amount of O2 is unloaded from hemoglobin /16/. The curve A shows the normal O2 dissociation curve The curve B depicts the situation in the presence of decreased erythrocyte 2,3-DPG content The curve C presents the O2 dissociation curve if 2,3-DPG in the erythrocytes is completely absent The curve D shows the situation if erythrocyte 2,3-DPG content is increased. Figure 15.3-3 Structure and function of hemoglobin (Hb), modified according to Ref. /19/. The α and β components first combine to αβ-dimers and the αβ-dimers aggregate spontaneously to form the hemoglobin tetramer α2β2. The region where the two dimers come into contact is known as the α1β2-region. Both dimers can modify their gap in the α1β2-region, depending upon the 2,3-diphosphoglycerate (DPG) content. Each monomer contains a heme molecule (+) for the binding of O2. The binding site for DPG is located between the two β-units. The erythrocyte DPG content is maintained by diphosphoglycerate mutase. The Hb molecule oscillates between two conformations. In the deoxygenated conformation (T, taut form), the affinity for O2 is low; in the oxygenated conformation (R, relaxed form) it is high. DPG is the regulator of O2 binding. The affinity of Hb for O2 decreases, and O2 is released, if DPG binds to Hb in the peripheral tissues. The dimers move closer together, and the tetramer takes on the T configuration. If the T conformation reaches the lungs, the reverse process takes place and O2 is bound. A mutation in the Diphosphoglycerate mutase gene leads to a decrease in intra erythrocyte DPG, whereby O2 binding of the Hb molecule remains high and insufficient O2 is released to the tissues; the result is compensatory erythropoietin mediated erythrocytosis. Figure 15.4-1 Classification of erythrocytosis. EPO, erythropoietin; 2,3-DPG-Mutase, 2,3-diphosphoglycerate mutase; HVL gene, von Hippel-Lindau gene; PHD, prolylhydroxylase; HIF2α, transcription factor Figure 15.4-2 Homodimer erythropoietin (EPO) receptor showing phosphorylated tyrosines. The tyrosines form binding sites for the components of the EPO signaling pathway. The tyrosines are phosphorylated by Janus kinase 2 (JAK2) and bind positive and negative regulators of the receptor to nucleus activated by EPO /33/. Positive and negative regulators respectively stimulate and suppress signal transmission. Figure 15.4-3 Increased synthesis of erythropoietin (EPO) in von Hippel-Lindau gene mutation. The relationship between the proteins prolylhydroxylase (PHD), von Hippel Lindau protein (VHL), transcription factor HIF2α and normoxia and hypoxia are shown. Modified according to Ref. /19/. Normally O2 activates PHD, which then hydroxylates HIF2α. This leads to the binding of VHL protein and to ubiquitination and degradation of HIF2α. In hypoxia, HIF2α associates with its β subunit (not shown in figure), the complex binds to hypoxia-responsible elements (HREs) and leads to gene activation for enhanced synthesis of EPO. Figure 15.4-4 Effect of 2,3-diphosphoglycerate (2,3-DPG) on O2 saturation curve. Courtesy of Ref. /37/ modified. A, normal erythrocyte 2,3-DPG content; B, low erythrocyte 2,3-DPG content; C, no 2,3-DPG present in the erythrocytes; D, elevated erythrocyte 2,3-DPG content. Figure 15.6-1 Differentiation of normocytic anemia by reticulocyte count /17/. Figure 15.6-2 Differentiation of macrocytic anemia by reticulocyte count /17/. Figure 15.6-3 Differentiation of microcytic anemia by reticulocyte count /17/. Figure 15.6-4 Differentiation of reticulocytes in maturation stages of low fluorescence reticulocytes (LFR), medium fluorescence reticulocytes (MFR), and high fluorescence reticulocytes (HFR) of a healthy individual. Figure 15.6-5 Diagnostic diagram for the assessment, monitoring and therapy of the iron status /37/. The ferritin index (sTfR/log10 ferritin) is a marker of iron supply, CHr is an indicator of the iron demand of erythropoiesis. Ferritin indices of greater than 1.5 and 0.8 in patients without and with inflammation, respectively, reflect deficient iron supply, a CHr below 28 pg indicates the iron demand of erythropoiesis. Over 95% of ACD patients are found in quadrants 1 and 4, those with iron-restricted erythropoiesis or functional iron deficiency in quadrant 4 and those without in quadrant 1. The data points of patients with latent iron deficiency (iron deficiency without anemia) are found in quadrant 2, those with classic iron deficiency in quadrant 3. If sTfR is determined with the Siemens assay, the threshold values of the ferritin indices are 1.5 and 0.8., for the Roche assay 3.2 and 2.0. Treatment recommendations are shown in the 4 quadrants. Figure 15.7-1 Clinical laboratory diagnosis of thalassemia syndromes and hemoglobinopathies. Figure 15.7-2 Use of RDW values in diagnostic investigation of thalassemia minor. Figure 15.7-3 Separation schema for normal and abnormal hemoglobin on micro zone electrophoresis. Figure 15.10-1 Compensation of O2 deficiency in patients with acute hemorrhage induced anemia. O2 deficiency in the kidney activates hypoxia inducible factor HIF2α to stimulate the production of erythropoietin (EPO). Erythropoietic activity is stimulated by EPO. The erythroid regulator erythroferrone is produced by erythroid precursors and facilitates compensatory acquisition of iron. Erythroferrone acts directly on the liver to decrease hepcidin production, and thereby increases iron availability for new erythrocyte synthesis. Figure 15.10-2 Expected range of serum EPO level as a function of the hematocrit (HCT) value. With a decrease in the HCT, the EPO level increases exponentially. The 95% confidence interval is shown. Modified from Ref. /5/. Figure 15.10-3 Relationship between HCT and serum EPO level /7/. 1, β-thalassemia; 2, hemolytic anemia; 3, renal anemia; 4, anemia of chronic disease; 5, pregnancy; 6, polycythemia vera; 7, normal individuals; 8, renal cell carcinoma. The line represents the standard EPO increase in patients with acute bleeding. Figure 15.10-4 Activation of the immune system in acute phase response (APR) as the result of the formation of inflammatory cytokines (IFN-γ, IL-6) by activated immune cells (TH-lymphocytes, monocytes/macrophages). Erythroid precursor cells of the bone marrow proliferation pool need the binding of some 10 EPO molecules per cell in order to enter the maturation pool and not to become apoptotic. EPO exerts anti-apoptotic effects, while inflammatory cytokines such as IFN-γ are pro-apoptotic. The survival of the erythroid precursors is dependent upon the balance between EPO and inflammatory cytokines. In severe inflammation (high CRP values), apoptosis predominates and erythropoiesis becomes hypoproliferative, mediated by hepcidin, and the result is normochromic anemia /17/. Figure 15.10-5 Site of action of erythropoietin (EPO) and other hematopoietic growth factors in erythropoiesis. Modified from Ref. /21/. BFU-E , Burst forming unit erythroid; CFU-E, Colony forming unit erythroid; IL-3, interleukin-3; GM-CSF, granulocyte-macrophage colony-stimulating factor; SCF, stem cell factor; RBC, red blood cells. Figure 15.10-6 Erythropoietin feedback control loop modified according to Ref. /24/. Figure 15.10-7 Homodimer erythropoietin (EPO) receptor showing phosphorylated tyrosines. The tyrosines form binding sites for the components of the EPO signaling pathway. The tyrosines are phosphorylated by Janus kinase 2 (JAK2) and bind positive and negative regulators of the receptor to nucleus activated by EPO /34/. Positive and negative regulators respectively stimulate and suppress signal transmission. Figure 15.12-1 Hematogram of leukocyte differentiation using volume, conductivity and scatter technology. N, neutrophils; E, eosinophils; B, basophils; L, lymphocytes; M, monocytes. Figure 15.12-2 Hematogram of leukocyte differentiation with combined flow cytometry and cytochemistry. In the leukogram (left diagram), the following leukocytes are counted: 1, neutrophils; 2, eosinophils; 3, monocytes; 4, lymphocytes and blasts; 5, large unstained cells (LUC). In the nucleogram (right diagram), the following leukocytes are differentiated after stripping of cell cytoplasm: 6, mononuclear cells; 7, polynuclear cells; 8, basophils. Figure 15.12-3 Neutrophile count in the newborn, according to Ref. /9/. The upper and lower reference interval numbers are shown. Figure 15.12-4 Age-dependency of the lymphocyte count in children according to Ref. /21/. Values are 2.5th and 97.5th percentiles. Figure 15.13-1 Blood smear method. Figure 15.13-2 Relevant areas for the evaluation of the blood smear. Figure 15.13-3 Morphological granulocyte alterations. Figure 15.13-4 Morphological lymphocyte alterations. Figure 15.13-5 Morphological thrombocyte alterations. Figure 15.13-6 Morphological erythrocyte alterations. Figure 15.14-1 Transverse section at the level of the biopsy site. Figure 15.14-2 Jamshidi needle (above) in comparison with a usual puncture needle. Figure 15.14-3 Preparation of a smear of bone marrow fragments. Figure 15.15-1 Previous FAB classification: bone marrow blast threshold between AML and MDS is now 20%. Figure 15.15-2 Standard diagnostic investigation of acute lymphocytic leukemia. Figure 15.17-1 Algorithm for molecular genetic testing in myeloproliferative neoplasia. PV, polycythemia vera; ET, essential thrombocythemia; PMF, primary myelofibrosis; MPN, myeloproliferative neoplasia; CMML, chronic myelomonocytic leukemia; TET2, tet oncogene family member 2; MPL, myeloproliferative leukemia virus oncogene/thrombopoietin receptor; CBL, Casitas B-lineage lymphoma proto-oncogene; VHL, von Hippel-Lindau tumor suppressor gene; EPO, erythropoietin; HIF2A, HIF-1-alpha-like factor; PHD2, HIF prolyl hydroxylase 2; THPO, thrombopoietin; PDGFRB, Platelet-derived growth factor receptor beta; PDGFRA, platelet-derived growth factor receptor alpha; EZH2, Enhancer of zeste homolog 2 (Drosophila); RUNX1, runt-related transcription factor 1. Figure 15.18-1 Flow diagram for immunophenotyping in acute leukemia. Abbreviations: CD, cluster of differentiation; cy/m, intracytoplasmic or cell surface expression; MPO, myeloperoxidase; Ig, Immunoglobulin; TCR, T-cell receptor; TdT, terminal deoxynucleotidyl transferase \| Acute leukemia – Diagnosis according to FAB criteria ↓ Assignment to B or T lymph cell or myeloid cell series /20/ B cell lineage: CD19, (cy) CD22, (cy) CD79a T cell lineage: cyCD3, CD7 Myeloid lineage: CD13, CD33, CD65, CD117, MPO NK celllineage CD56, CD57 ↓ \| Diagnosis: B precursor cell ALL, B-ALL/Burkitt’s lymphoma, T-lineage ALL, AML \| ↓ Definition of stage of maturity or immunological subtype (see Tab. 15.17-1 – Clinical characteristics and orienting markers in the differential diagnosis of MPN) B-lineage ALL: CD10, CD20, CD24, cyIgM, mIg; CD34, HLA-DR, TdT T-lineage ALL: CD1a, CD2, mCD3, CD4, CD5, CD8, CD10, TCRα/β or γ/δ; CD34, HLA-DR, TdT AML: CD14, CD15, CD41, CD61, CD64, Glycophorin A; CD34, HLA-DR, TdT \|
14538	https://thirdspacelearning.com/us/blog/fraction-bar-model/	NEW LOWER-COST TUTORING Introducing Skye, your students’ AI voice tutor Adaptive, dialogue-driven one-on-one tutoring built by math teachers Unlimited sessions for as many grade 3-8 students as need it One fixed low yearly cost no matter how many sessions you schedule Meet Skye A personal math tutor for every student that needs it "This innovative one-on-one math tutoring solution offers a cost-effective alternative to traditional one-on-one tutoring." Meet Skye Free ready-to-use math resources Hundreds of free math resources created by experienced math teachers to save time, build engagement and accelerate growth Explore all resources Contents Instruction Strategies The Fraction Bar Model: How To Teach Fractions Using Bar Models March 31, 2025 \| 2 min read Vanessa Sipple-Asher The fraction bar model is a great way to introduce fractions to your young math learners. Fractions can be a tricky concept for students to grasp and the visual representations of the bar model method can really help. What is the fraction bar model? When do children learn to use the fraction bar model? Why teach bar model fractions? How to draw a fraction bar model Finding fractions using the fraction bar model Improper fractions and the fraction bar model Equivalent fractions and the fraction bar model More fractions and bar models resources What is the fraction bar model? What is a bar model and how can it be used to teach fractions? Bar models, also known as strip diagrams, are part of the concrete pictorial abstract approach of teaching mathematical concepts and are an essential step in understanding the meaning of fractions and fraction concepts. The fraction bar model uses a whole bar broken up into equal parts, a part whole model, to illustrate fractions of a whole. For example, a whole bar broken up into eight equal parts would be used to demonstrate eighths, and that eight eighths make up one whole. A fraction wall uses a series of bar models to illustrate fractions and how different fractions relate to a whole. When do children learn to use the fraction bar model? According to the Common Core and Texas standards for math, children will begin learning simple fractions, such as halves and quarters in 2nd and 3rd grade. As children progress in elementary school into 4th grade and 5th grade, they will be introduced to increasingly complex math problems using fractions where a bar model could help. Fraction Operations Check for Understanding Quiz Printable quiz to check your grade 4th, 5th and 6th graders’ understanding of fraction operations. Download Free Now! Why teach bar model fractions? Using the bar model is a great way to support childrens’ understanding of fractions and problem solving. Bar models can be used to teach both unit fractions and non-unit fractions. Bar models help children to visualize the math in front of them, organize their thoughts and solve problems. Using the bar model can also be used to teach number bonds, place value,decimals, addition, subtraction, percentage, solve word problems and for bar model multiplication and division. Read more: 5th grade word problems Meet Skye, the voice-based AI tutor making math success possible for every student. Built by teachers and math experts, Skye uses the same pedagogy, curriculum and lesson structure as our traditional tutoring. But, with more flexibility and a low cost, schools can scale online math tutoring to support every student who needs it. Watch Skye in action How to draw a fraction bar model Drawing a fraction bar model is simple: Draw a rectangle, it does not matter what size. Look at the denominator of the fraction you wish to represent. Divide the rectangle into the same number of equal parts as the denominator of the fraction. Shade in the same number of pieces as the numerator of the fraction. Finding fractions using the fraction bar model Using the bar model can be used to help children find fractions of an amount, for example ⅕ of 25. Instead of starting with the fraction, students start with the whole number which they then must divide into fifths. The bar model helps to illustrate that five equal parts are added together to make the whole. Improper fractions and the fraction bar model Improper fractions are fractions that have a larger numerator than the denominator. Children will often be asked to convert improper fractions into mixed numbers. In this case, using the bar model can be very useful. A mixed number is a whole number combined with a fraction, for example, 3 ⅔. This mixed number and its equivalent improper fraction can be represented using the fraction bar model. Equivalent fractions and the fraction bar model Some math problems may require children to identify equivalent fractions. Bar models can make this a lot easier as it removes the abstract and instead uses a concrete visualization, enabling children to compare visuals and identify equivalent fractions. More fractions and bar models resources Third Space Learning’s math hub provides a wide range of elementary school resources such as printable worksheets, for teaching fractions and bar models. Our online one-to-one tuition also uses bar models to help children to gain a deep understanding of mathematical concepts. Do you have students who need extra support in math? Skye—our AI math tutor built by experienced teachers—provides students with personalized one-on-one, spoken instruction that helps them master concepts, close skill gaps, and gain confidence. Since 2013, we’ve delivered over 2 million hours of math lessons to more than 170,000 students, guiding them toward higher math achievement. Discover how our AI math tutoring can boost student success, or see how our math programs can support your school’s goals: – 3rd grade tutoring – 4th grade tutoring – 5th grade tutoring – 6th grade tutoring – 7th grade tutoring – 8th grade tutoring The content in this article was originally written by content team Vanessa Sipple-Asher and has since been revised and adapted for US schools by elementary math teacher Jaclyn Wassell. Share: Related articles How To Teach Math: 10 Effective Strategies For Teaching Math In The Classroom 9 min read Building Thinking Classrooms: Effective Strategies For Your Math Classroom 10 min read Differentiation Strategies In The Classroom: 8 Methods For Every Teacher 10 min read Retrieval Practice: A Foolproof Method To Improve Student Retention and Recall 7 min read x [FREE] Ultimate Math Vocabulary Lists (K-5) An essential guide for your Kindergarten to Grade 5 students to develop their knowledge of important terminology in math. Use as a prompt to get students started with new concepts, or hand it out in full and encourage use throughout the year. Download free
14539	https://www.matthewdevaney.com/powerapps-collections-cookbook/calculate-the-sum-count-average-max-min/	Skip to content Matthew Devaney No Ads, No Fluff, Just Power Platform Stuff Home Power Platform Guides Power Apps Standards Power Automate Standards Power Platform Pipelines & ALM Setup Guide Power Platform Functions Power Apps Functions Power Automate Functions Collections Cookbook Free Power Apps Icons About Me Calculate the SUM, COUNT, AVERAGE, MAX, MIN Input collection: myTravelExpenses11 \| \| \| \| --- \| Date \| Item \| Value \| \| 1/1/2020 \| Hotel \| 1050 \| \| 1/1/2020 \| Food \| 30 \| \| 1/2/2020 \| Food \| 75 \| \| 1/3/2020 \| Hotel \| 1300 \| \| 1/3/2020 \| Food \| 50 \| \| 1/4/2020 \| Flight \| 800 \| Output value: SUM of Value \| \| \| 3305 \| Solution code: //Create a collection ClearCollect(myTravelExpenses11, {Date: Date(2020,1,1), Item: "Hotel", Value: 1050}, {Date: Date(2020,1,1), Item: "Food", Value: 30 }, {Date: Date(2020,1,2), Item: "Food", Value: 75 }, {Date: Date(2020,1,3), Item: "Hotel",Value: 1300}, {Date: Date(2020,1,3), Item: "Food", Value: 50}, {Date: Date(2020,1,4), Item: "Flight", Value: 800} ); //Get the sum code Sum(myTravelExpenses11,Value) Output value: COUNT of Value \| \| \| 6 \| Solution code: //Get the count code CountRows(myTravelExpenses11) Output value: AVERAGE of Value \| \| \| 550.8333 \| Solution code: //Get the average code Average(myTravelExpenses11,Value) Output value: MAX of Value \| \| \| 1300 \| Solution code: //Get the maximum code Max(myTravelExpenses11,Value) Output value: MIN of Value \| \| \| 30 \| Solution code: //Get the minimum code Min(myTravelExpenses11,Value)
14540	https://brighterly.com/math/inverse-operations/	What are Inverse Operations ⭐ Definition Facts & Examples Book free lesson Math program Math tutors by grade 1st grade math tutor 2nd grade math tutor 3rd grade math tutor 4th grade math tutor 5th grade math tutor 6th grade math tutor 7th grade math tutor 8th grade math tutor 9th grade math tutor See all Math programs Online math classes Homeschool math online Afterschool math program Scholarship program Reading Program Reading tutor by grade 1st grade reading tutor 2nd grade reading tutor 3rd Grade Reading Tutors 4th Grade Reading Tutors 5th Grade Reading Tutors 6th Grade Reading Tutors 7th Grade Reading Tutors 8th Grade Reading Tutors 9th Grade Reading Tutors See all Our Library Math worksheets 1st grade worksheet 2nd grade worksheet 3rd grade worksheet 4th grade worksheet 5th grade worksheet 6th grade worksheet 7th grade worksheet 8th grade worksheet 9th grade worksheet See all Reading worksheets 1st grade worksheet 2nd grade worksheet 3rd grade worksheet 4th grade worksheet 5th grade worksheet 6th grade worksheet 7th grade worksheet 8th grade worksheet 9th grade worksheet See all Resources Blog Knowledge base Math questions Math tests Reading tests Parent’s reviews Pricing Book free lessonLog in Book free lessonLog in Math tutors / Knowledge Base / Inverse Operations – Definition with Examples Inverse Operations – Definition with Examples Jo-ann Caballes Updated on January 15, 2024 Table of Contents Welcome to yet another thrilling adventure into the world of mathematics with Brighterly – where we make learning math engaging and enjoyable for children of all ages! In today’s action-packed journey, we’ll be delving deep into the captivating realm of inverse operations. So, sharpen those pencils, gather your enthusiasm, and join us as we embark on this remarkable expedition together! What Are Inverse Operations? Inverse operations are mathematical operations that “undo” each other. They are like superheroes with opposite powers, working together to maintain balance in the world of numbers! For example, addition and subtraction are inverse operations, as are multiplication and division. If you add a number and then subtract the same number, you’ll end up with the original number. The same goes for multiplication and division. What Are the Properties of Inverses? Inverses have some fascinating properties that help us better understand and solve math problems. Let’s explore these properties one by one. Property 1: Undoing Operations The primary purpose of inverse operations is to “undo” the effect of another operation. If you perform an operation and its inverse, you’ll get back to the original value. This property can be helpful in solving equations and other mathematical problems. Property 2: Commutativity Addition and multiplication are both commutative operations. This means that the order in which you perform these operations doesn’t matter. The inverses of commutative operations (subtraction and division) are also commutative. Property 3: Associativity Another essential property of inverse operations is associativity. In associative operations, the grouping of numbers doesn’t affect the result. Both addition and multiplication are associative, so their inverses (subtraction and division) are associative as well. Property 4: Identity Element An identity element is a special number that, when combined with another number using a particular operation, leaves the other number unchanged. For addition, the identity element is 0, and for multiplication, it’s 1. The inverses (subtraction and division) also have their identity elements. Property 5: Inverse Element For every number, there is an inverse element that, when combined using a specific operation, results in the identity element. For addition, the inverse element is the negative of the number, and for multiplication, it’s the reciprocal of the number. Addition and Subtraction Addition and subtraction are inverse operations that help us solve various mathematical problems. When we add a number and its negative, we get 0, the identity element for addition. Similarly, when we subtract a number from itself, we get 0. This inverse relationship helps us solve equations that involve addition and subtraction. Multiplication and Division Just like addition and subtraction, multiplication and division are inverse operations. When we multiply a number by its reciprocal, we get 1, the identity element for multiplication. Similarly, when we divide a number by itself, we get 1. This inverse relationship is useful for solving equations involving multiplication and division. Solved Examples on Inverse Operations Let’s look at some examples to understand inverse operations better. If you have the equation x + 5 = 10, you can use the inverse operation of subtraction to solve for x. Subtracting 5 from both sides gives you x = 5. For the equation 3x = 9, you can use the inverse operation of division to solve for x. Dividing both sides by 3 gives you x = 3. Practice Problems on Inverse Operations Ready to test your understanding of inverse operations? Try solving these practice problems! Solve for y: y – 7 = 15 Solve for z: 4z = 20 Conclusion Inverse operations are essential tools in mathematics, helping us solve equations and understand the relationships between different operations. By recognizing and using inverse operations, we can simplify and tackle various mathematical problems with ease. Remember, addition and subtraction are inverses, as are multiplication and division. Their unique properties, such as undoing operations, commutativity, associativity, identity elements, and inverse elements, all contribute to their effectiveness in problem-solving. Now that you’ve learned about inverse operations and their properties, we hope you’ll find solving math problems even more enjoyable! Keep practicing, and remember, the world of mathematics is always full of exciting discoveries waiting for you! Frequently Asked Questions on Inverse Operations What are inverse operations in math? Inverse operations are mathematical operations that “undo” each other. For example, addition and subtraction are inverse operations, as are multiplication and division. If you perform an operation followed by its inverse, you’ll get back to the original value. What are the properties of inverses? Inverses have several properties, including undoing operations, commutativity, associativity, identity elements, and inverse elements. Why are inverse operations important? Inverse operations are important because they help us solve mathematical problems and understand the relationships between different operations. By using inverse operations, we can simplify equations and tackle various mathematical problems more easily. Information Sources Wikipedia: Inverse function National Council of Educational Research and Training – Inverses Wolfram MathWorld: Inverse Operation Jo-ann Caballes 14 articles As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable. Table of Contents What Are Inverse Operations? What Are the Properties of Inverses? Property 1: Undoing Operations Property 2: Commutativity Property 3: Associativity Property 4: Identity Element Property 5: Inverse Element Addition and Subtraction Multiplication and Division Solved Examples on Inverse Operations Practice Problems on Inverse Operations Conclusion Frequently Asked Questions on Inverse Operations What are inverse operations in math? What are the properties of inverses? Why are inverse operations important? Math & reading from 1st to 9th grade Looking for homework support for your child? Choose kid's grade Grade 1 Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Math & reading from 1st to 9th grade Looking for homework support for your child? Book free lesson Related math 58000 in Words In words, 58000 is spelled as “fifty-eight thousand”. This number is fifty-eight sets of one thousand each. If you have fifty-eight thousand books, it means you have fifty-eight thousand books in total. Thousands Hundreds Tens Ones 58 0 0 0 How to Write 58000 in Words? The number 58000 is written as ‘Fifty-Eight Thousand’ in […] Read more 93 in Words The number 93 is expressed as “ninety-three” in words. It comes after ninety-two. For instance, if you have ninety-three stars, you have ninety-two stars and one additional star. Tens Ones 9 3 How to Write 93 in Words? Writing the number 93 in words involves recognizing its place value. The number 93 has a ‘9’ […] Read more 15 Times Table – Multiplication Table Multiplication Table of 15 The multiplication table of 15 is a fundamental building block in the field of mathematics, especially for young learners. It’s a sequence that follows a simple pattern, multiplying the number 15 by the integers from 1 to 10. Understanding this pattern forms the basis for more complex arithmetic calculations. Here’s what […] Read more Want your kid to excel in math and reading? Kid’s grade Select grade Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Parent’s email Successfully sent Close a child’s math gaps with a tutor! Book a free demo lesson with our math tutor and see your kid fill math gaps with interactive lessons Book demo lesson Get full test results See Your Child’s Test Results Enter your name and email to view your child’s test results now! 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14541	https://www.health.state.mn.us/diseases/cardiovascular/documents/mnacstoolkitaccessible.pdf	Minnesota Chest Pain/Acute Coronary Syndrome Toolkit Patient with Chest Pain or Potential Acute Coronary Syndrome (Page 1) 1. STEMI, Non-STEMI Chest Pain? a. If STEMI, follow MN STEMI Guideline (Pages 3-4) b. If Non-STEMI, follow MN Non-STEMI Guideline (Page 8) c. If Chest Pain, follow MN ED Chest Pain Guideline (Page 10) 2. This ACS/Chest Pain “Tool-Kit” was created with coordination from the Minnesota Department of Health, in conjunction with the American Heart Association Minnesota Mission: Lifeline™ Workgroup. This information is intended only as a guideline. Please use your judgement or newly published literature in the treatment of patients. The Minnesota Department of Health is not responsible for inaccuracies contained herein. No responsibility is assumed for damages or liabilities arising from accuracy, content error, or omission. Minnesota Chest Pain / Acute Coronary Syndrome Toolkit Table of Contents (Page 2) 1. Cover Minnesota Chest Pain / Acute Coronary Syndrome Tool-Kit 2. Table of Contents 3. Minnesota STEMI Inter-Facility Transfer Guideline page 1 of 2 4. Minnesota STEMI Inter-Facility Transfer Guideline page 2 of 2 5. Minnesota EMS STEMI Transfer Guideline page 1 of 2 6. Minnesota EMS STEMI Transfer Guideline page 2 of 2 7. Minnesota EMS STEMI Transport Flowchart 8. Minnesota Non-STEMI Guideline 9. Minnesota Non-STEMI Flowchart 10. Minnesota ED Chest Pain Protocol 11. Minnesota ED Cheat Pain Flowchart 12. Minnesota Low Risk Chest Pain Shared Decision-Making Tool 13. Minnesota Moderate Risk Chest Pain Shared Decision-Making Tool 14. Minnesota High Risk Chest Pain Shared Decision-Making Tool 15. Who Needs a 12-Lead ECG? (Symptom and Age Algorithm Minnesota STEMI Guideline (Pages 3-4) Minnesota Mission: Lifeline Statewide STEMI Transfer Guideline 1. Identify/Confirm STEMI a. Signs & Symptoms suspect for AMI (Acute Myocardial Infarction) Duration > 15 min < 12 hours b. ST Elevation defined by criteria on Page 2 (Number 14 below) c. Pre-Hospital STEMI criteria on Page 2 (Number 15 below) 2. Activate Transport a. Establish availability and ETA of Air or Ground ALS EMS for Interfacility Transfer to Primary PCI Hospital 3. Active your Internal STEMI Alert a. Alert appropriate provider(s) and team members 4. Establish Key Times: a. Symptom onset b. First Medical Contact (FMC) c. ETA to arrival at PCI Hospital 5. Estimate FMC (First Medical Contact) to Potential PCI a. Allow approximately 20 minutes after arrival to PCI Capable hospital 6. If Estimated FMC to PCI ≤ 120 minutes OR FMC > 120 minutes and one of the following then Do NOT give Lytic/TNK! a. Fibrinolytic Ineligible b. Resuscitated out-of-hospital cardiac arrest patients whose initial ECG shows STEMI c. Evidence of either Cardiogenic Shock or Acute Severe CHF 7. For all patients in Number 6 above a. Aspirin 325 mg PO chewed b. Heparin IV Bolus 60 Units/kg, max 4,000 Units (No IV Heparin Drip) c. Ticagrelor 180 mg PO (If Ticagrelor not available, then give Clopidogrel 600 mg PO) 8. If Estimated FMC to PCI 120-180 minutes then a. Establish if Fibrinolytic appropriate (See Page 2, Numbers 16 and 17 below, for contraindications) b. Goal: Door to Needle < 30 min c. For all patients transferring not utilizing pharmaco-invasive strategy proceed to Full Dose Fibrinolytic Strategy (Number 9 below) d. For patient transferring to Abbott Northwestern/MHI utilizing pharmaco-invasive strategy, administer HALF-Dose TNK IV and transfer for PCI (Dosing table on Page 2, Number 18 below) e. Aspirin 81 mg x 4 (324 mg) chewed f. Heparin IV Bolus 60 Units/kg, max 4,000 Units (No IV Heparin Drip) M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 3 g. Clopidogrel 600 mg PO h. TNK “Half-Dose” IV 9. If Estimated FMC to PCI > 120 minutes and for all ages transferred with an estimated FMC to PCI > 180 minutes) a. Establish if Fibrinolytic appropriate (See page 2 for contraindications, Numbers 16 and 17 below) b. Goal: Door to Needle < 30 minutes c. Consider consultation with PCI receiving center Cardiology prior to administration of fibrinolytic d. Aspirin 325 mg PO chewed e. Heparin IV Bolus 60 Units/kg, max 4,000 Units f. Heparin IV Drip 12 Units/kg/hr, max 1,000 Units/hr g. For Age ≤ 75 years then: i. Clopidogrel 300 mg PO ii. TNK “Full-Dose” IV (See Dosing table page 2, Number 18 below) h. For Age > 75 years then: i. Clopidogrel 75 mg PO ii. TNK “Half-Dose IV (See Dosing table page 2, Number 18 below) 10. For all patients, Activate Code STEMI/STEMI Alert at PCI Hospital (Follow your regional STEMI protocol) 11. Transport Patient as Soon as Possible! a. Fax or Transmit ECG and other pertinent records (EMS reports, allergies, past medical history, etc.) 12. Top Patient Care Priorities: a. Establish DNR/Resuscitation Status b. Obtain vital signs and access pain level on scale of 1-10 c. Cardiac Monitor and attach hands-free defibrillator pads d. Establish Saline Lock – large bore needle (left arm preferred) e. Oxygen PRN at 2 L/min and titrate to SpO2 > 90% f. Assess Allergies (Note if reaction to IV Contrast?) g. Section for user to add notes 13. Patient Care When Time Allows: a. Establish 2nd large bore IV with Normal Saline @TKO (Left Arm preferred) b. Obtain Appropriate Labs: Troponin, CBC, Potassium, Creatinine, PT/INR, aPTT c. Nitroglycerin 0.4 mg SL every 5 min or Nitropaste PRN for chest pain (hold for SBP < 90) d. Evaluate if erectile dysfunction or pulmonary hypertension medications taken in the past 48 hours including: Sildenafil (Viagra, Revatio), Vardenafil (Levitra, Staxyn), Avanafil (Stendra), or Tadalafil (Cialis, Adcirca), and if so, hold nitrates for 48 hours. M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 4 14. STEMI (ST Elevation Myocardial Infarction) Diagnostic Criteria a. ST elevation a the J point in at least 2 contiguous leads of ≥ 2 mm (0.2 mV) in men or ≥ 1.5 mm (0.15 mV) in women in leads V2-V3 and/or of ≥ 1 mm (0.1 mV) in other contiguous chest leads or the limb leads b. Signs & Symptoms of discomfort suspect for AMI (Acute Myocardial Infarction) or STEMI with a duration of > 15 minutes < 12 hours c. Although new, or presumably new, LBBB at presentation occurs infrequently and may interfere with ST-elevation analysis, care should be exercised in not considering this an acute myocardial infarction (MI) in isolation. If in doubt, immediate consultation with PCI receiving center is recommended. d. ECG demonstrates evidence of ST depression suspect of a Posterior MI. Consult with PCI receiving center. e. If initial ECG is not diagnostic but suspicion is high for STEMI, obtain serial 12 Lead ECG’s at 5-10 minute intervals. 15. Pre-Hospital STEMI confirmed by 12 Lead ECG trained ALS EMS, recognize ST segment elevation of ≥ 1 mm in 2 contiguous leads, confirmed interpretation of STEMI transmitted, or ECG monitor interpretive statement infers: “Acute Myocardial Infarction” with patient signs and symptoms suspect of AMI. 16. Absolute Contraindications for Fibrinolysis a. Chest Pain/Symptom Onset > 12 hours b. Suspected aortic dissection c. Any prior intracranial hemorrhage d. Structural cerebral vascular lesion or malignant intracranial neoplasm e. Any active bleeding (excluding menses) f. Ischemic stroke within 3 months g. Significant closed-head or facial trauma within 3 months h. Pregnancy 17. Relative Contraindications for Fibrinolysis a. Chest Pain/Symptom Onset > 6 hours b. Current use of oral anticoagulants (Warfarin, Dabigatran, Rivaroxaban, Apixaban, etc.) c. Uncontrolled hypertension on presentation (SBP > 180 or DBP > 90 mm Hg) d. History of ischemic stroke more than 3 months, dementia, or known intracranial pathology not covered in contraindications e. Traumatic or prolonged CPR (over 10 minutes) f. Major surgery within last 3 weeks g. Recent internal bleeding (within last 2-4 weeks) M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 5 18. Tenecteplase (TNKase) Dosing Chart Patient Weight FULL-DOSE HALF-DOSE 59 kg or less 30 mg = 6 mL 15 mg = 3 mL 60-69 kg 35 mg = 7 mL 18 mg = 3.5 mL 70-79 kg 40 mg = 8 mL 20 mg = 4 mL 80-89 kg 45 mg = 9 mL 23 mg = 4.5 mL 90 kg or more 50 mg = 10 mL 25 mg = 5 mL 19. AHA Mission:Lifeline STEMI Recommendations a. FMS (First Medical Contact)-to-First ECG time ≤ 10 minutes unless pre-hospital ECG obtained b. All eligible STEMI patients receiving a Reperfusion Therapy (Primary PCI or Fibrinolysis) c. Fibrinolytic eligible STEMI patients with a Door-to-Needle time ≤ 30 minutes d. Primary PCI eligible patients transferred to a PCI receiving center with referring center Door in-Door out (Length of Stay) ≤ 45 minutes e. Referring Center ED or Pre-Hospital First Medical Contact-to-PCI time ≤ 120 minutes (including transport time) f. All STEMI patients without a contraindication receiving Aspirin prior to referring center ED discharge 20. Section for user to add notes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 6 Minnesota Mission: Lifeline EMS STEMI Transport Guideline (Pages 5-6) 1. Obtain 12 L ECG with Initial Assessment & Vital Sign a. Goal: First Medical contact to ECG ≤ 10 min, Scene time: ≤ 15 min – to provide early identification and pre-hospital arrival notification for suspected myocardial infarction or STEMI b. Chest pain, pressure, tightness or persistent discomfort above the waist in patients ≥ 35 years of age c. “Heartburn” or epigastric pain d. Complaints of “heart racing” (HR > 150 or irregular and > 120) or “heart too slow” (HR < 50 and symptomatic) e. A syncopal episode, severe weakness, or unexplained fatigue f. New onset stroke symptoms (<24 hours old) g. Difficulty breathing or shortness of breath (with no obvious non-cardiac cause) h. ROSC (return of spontaneous circulation) post cardiac arrest i. Recent Cocaine, stimulant and/or other illicit drug use (patients of any age) j. If initial ECG is not diagnostic but suspicion remains high for ACS (acute coronary syndrome) and symptoms persist, obtain serial ECG’s at 5-10 minute intervals 2. Pre-Hospital STEMI ALERT Activation Criteria: a. Goal: Identify potential ACS patients, Recognize STEMI, Alert Receiving Facility b. Activate STEMI Alert when any one or more of the following criteria are met and patient demonstrates signs and symptoms suspect of (AMI) acute myocardial infarction as described above with a duration of ≥ 15 minutes ≤ 24 hours i. EMS personnel trained in 12 L ECG interpretation recognize ST segment elevation of ≥ 1 mm in 2 contiguous leads ii. Interpretation of ECG transmitted and reviewed by a provider (Physician, NP, PA) confirmed to be diagnostic of STEMI iii. 12 Lead ECG Monitor Algorithm Interpretative statement reads: “Acute Myocardial Infarction” 3. Determine Transport Destination a. If Transport time estimated to be ≤ 60 min (Goal FMC to PCI ≤ 120 min) then i. Notify medical control of STEMI and consider transport via the most expedient method available to the nearest PCI Capable Receiving Hospital for Primary PCI ii. Active STEMI Alert at Receiving Facility and transmit 12 L ECG as able iii. Consider Air transport b. If Transport time estimated to be ≥ 60 min (Goal Door to Thrombolysis administration ≤ 30 min) then: i. Notify medical control and consider transport to the closest appropriate non-PCI capable referring hospital for possible thrombolytic therapy and subsequent urgent transfer to a PCI Capable Receiving Facility for reperfusion. ii. Initiate thrombolytic contraindication checklist per protocol M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 7 iii. Activate STEMI Alert at Receiving Facility and transmit 12 L ECG as able for provider confirmation iv. Consider Air Transport c. Diversion Criteria: If patient demonstrates instability and/or has any of the following criteria that may require ED evaluation and treatment by a practitioner proceed to the nearest appropriate hospital. i. Symptoms suggestive of acute stroke or neurological evaluation ii. Respiratory or Circulatory Instability iii. Chest trauma or MVC victims iv. DNR status v. Consider Left Bundle Branch Block 4. BLS & ALS: a. Administer Oxygen to maintain SpO2 90% - 94% titrate as needed started at 2 LPM per nasal cannula b. Obtain Systolic/Diastolic blood pressure (BP) in both arms c. Administer Chewable Aspirin 81 mg x 4 by mouth d. Evaluate if Erectile Dysfunction or Pulmonary hypertension medications taken in past 24 hours including: Sildenafil (Viagra, Revatio), Vardenafil (Levitra, Staxyn), Avanafil (Stendra), or Tadalafil (Cialis, Adcirca). Hold nitrates for 48 hours following the last dose. e. Administer Nitroglycerin Sublingual 0.4 mg every 5 minutes up to 3 doses if chest discomfort present and SBP > 100. Check BP prior to each administering dose. Hold if SBP ≤ 90. f. BLS Only: Request ALS intercept per local protocol (if transport time exceeds 15 min) g. Establish large bore IV Access (L) upper extremity preferred. Establish a 2nd IV line as time allows. 5. ALS: a. If available consider one of the following: i. Ticagrelor (Brilinta) 180 mg by mouth if transferring for PPCI with confirmation by PCI Receiving Facility and local medical control per protocol. Do Not Administer both Clopidogrel and Ticagrelor. ii. If Ticagrelor not available, then give Clopidogrel 600 mg by mouth if transferring for PPCI with confirmation by PCI Receiving Facility and local medical control per protocol. b. Heparin IV Bolus 60 Units/kg, max 4,000 Units (No IV Heparin Drip) if transferring for PPCI after confirmation by PCI Receiving Facility and local medical control per protocol. c. Establish a Nitroglycerine IV Drip if chest discomfort is unrelieved. Initiate @ 56 mcg/min & titrate in increments of 5 mcg/min every 5 minutes for chest discomfort per protocol. Maintain a systolic BP of ≥ 90 mm Hg or greater. Hold nitrates as indicated for criteria above. d. Administer Analgesia as needed per protocol. 6. Documentation Reminders: a. Provide a printed copy of EMS Run Sheet, and 12 L ECG with Report to the receiving hospital ED staff b. Document Date and Time of: i. EMS dispatch ii. First Medical Patient Contact M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 8 iii. Scene Departure iv. STEMI alert requested 7. AHA Mission: Lifeline EMS Best Practice Goals a. All patients with non-traumatic chest discomfort, ≥ 35 years of age, treated and transported by EMS receive a pre-hospital 12-lead electrocardiogram b. All STEMI patients transported directly to a STEMI receiving center, receive a first (pre-hospital) medical contact to PCI time ≤ 90 minutes directly or ≤ 120 minutes for Interfacility hospital transfers c. All thrombolytic eligible STEMI patients treated and transported to a referring hospital for fibrinolytic therapy receive a door to needle time ≤ 30 minutes 8. AHA Mission: Lifeline EMS Recognition Achievement Measures a. Percentage of patients with non-traumatic chest pain, ≥ 35 years of age, treated and transported by EMS who receive a pre-hospital 12-lead electrocardiogram b. Percentage of STEMI patients treated and transported directly to a STEMI receiving center, with pre-hospital first medical contact to device time ≤ 90 minutes c. Percentage of lytic eligible STEMI patients treated and transported to a STEMI referring hospital for thrombolytic therapy with a door to administration time ≤ 30 minutes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 9 EMS STEMI Mission: Lifeline Minnesota Transport Flowchart (Page 7) 1. Suspected Cardiac Event 2. Obtain 12 Lead ECG with Initial Assessment & Vital Signs a. Goal: First Medical Contact to ECG ≤ 10 min 3. Provide All Additional Treatments per Chest Pain or Appropriate Protocols 4. Pre-Hospital STEMI ALERT Activation Criteria a. Signs & symptoms suspect of AMI with duration ≥ 15 min but ≤ 24 hrs, and one or more of the following are met: i. ECG transmitted & reviewed by a provider (Physician, NP, PA) confirmed to be diagnostic of STEMI ii. EMS personnel trained in 12 L ECG interpretation recognize ST elevation of ≥ 1 mm in 2 contiguous Leads iii. 12-Lead ECG Monitor Algorithm Interpretative statement (i.e. Acute Myocardial Infarction ) 5. Question: Is STEMI Alert Activation Criteria Met? a. If no, then if suspicion high for ACS & symptoms persist, continue protocol treatments and obtain serial ECG’s at 5-10 minute intervals and return to Number 2 b. If yes, then go to Number 6 6. Determine Transport Destination a. Consider Air Transport b. Consider ALS Intercept if BLS transport > 15 min 7. Question: Is Transport Time to PCI Capable Hospital ≤ 60 min? a. If yes, then Notify Medical Control at nearest appropriate PCI Hospital of STEMI ALERT i. Request Cath Lab Activation ii. Consider transport via the most expedient ALS method available for Primary PCI iii. Question: Instability or Diversion Criteria? 1. If yes, patient demonstrates instability and/or has any one of the following criteria that may require ED evaluation and treatment by a practitioner; Symptoms suggestive of acute stroke or neurological evaluation, Respiratory or Circulatory Instability, Chest trauma or MVC victims, DNR status, Consider Left Bundle Branch Block, then Proceed to Nearest Appropriate Hospital 2. If no, then Call PCI Hospital with Patient Identifiers and Proceed to PCI Hospital b. If no, then Notify medical control of STEMI ALERT and consider transport to the closest appropriate non-PCI capable referring hospital for possible Fibrinolytic therapy and/or subsequent urgent transfer to a PCI Capable Receiving Facility for Reperfusion i. Proceed to Nearest Appropriate Hospital 8. For all transported patients, Documentation Reminders upon Arrival at Hospital M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 10 a. Provide a printed copy of EMS Run Sheet & 12 Lead ECG with Report to the receiving hospital staff b. Document Date and Time of: i. EMS dispatch ii. First Medical Patient Contact iii. 12-Lead ECG iv. Scene departure v. STEMI alert requested vi. Destination Arrival vii. EMS agency number, and run number M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 11 Minnesota Non-STEMI Guideline (Page 8) 1. Patient meets any of the following criteria: a. HEART Score of 7-10 b. ST depression or dynamic T-wave inversion strongly suspicious for ischemia c. Otherwise identified Non-ST elevation acute coronary syndrome (Non-STEMI) 2. Next Step: a. Admit to CCU or appropriate unit with cardiac telemetry (may require transfer) b. Consider Cardiology consult 3. Medications: a. Start adjunctive treatments (as indicated/if no contraindications): i. Aspirin 324 mg PO (give suppository if unable to take PO) ii. Ticagrelor 180 mg PO or Clopidogrel 600 mg PO (loading doses) 1. Prasugrel 60 mg could also be considered but note warnings in Number 9 below) iii. Heparin 60 Units/kg (max 4,000 Units) IV bolus iv. Heparin 12 Units/kg/hr (max 1,000 Units) IV infusion b. Other medications as indicated per institutional AMI order set 4. Assess Criteria for Early Invasive Strategy (Cath Lab) a. High-risk features & patient a candidate for invasive angiography (PCI)? b. Persistent or recurrent symptoms? c. New ST-segment depression and positive serum Troponin(s)? d. Depressed LV functional study that suggests multi-vessel CAD? e. Hemodynamic instability or VT? 5. Choose Treatment Strategy a. If Early Invasive Strategy (Cath Lab) then go to Number 6 below b. If Ischemia-Guided Strategy (Medical Therapy) then go to Number 10 below 6. If Early Invasive Strategy (Cath Lab) then a. Prepare for Cath Lab i. Transfer if necessary by ground ambulance (Air transfers should be reserved for STEMI) b. Insert 2 large bore peripheral saline lock IV’s in left arm c. Continue adjunctive treatments as above d. Consult Cardiology for additional treatments (i.e. Beta-Blocker, Nitro, Morphine, O2, etc.) 7. If CABG surgery is required then a. Continue Aspirin b. Consult CT surgeon about stopping other therapies and timing (i.e. when to hold antiplatelet) 8. P2Y12 Inhibitor Maintenance Dosing & Considerations a. Ticagrelor 90 mg PO twice daily or b. Clopidogrel 75 mg once daily or c. Prasugrel 10 mg once daily (5 mg if ≤ 60 kg) (Note warnings in Number 9 below) M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 12 d. Continue up to 12 months if medically treated e. Continue at least 12 months if treated with drug eluting stent, or per Cardiologist Discretion f. If switching to a different P2Y12 inhibitor, consider a full loading dose at the time the next dose would be due 9. Prasugrel Warnings: a. Do NOT use if history of stroke or TIA b. Avoid in patients ≥ 75 years old or < 60 kg c. Do NOT start if patient likely to undergo urgent CABG 10. If Ischemia-Guided Strategy (Medial Therapy) then a. Continue adjunctive treatments as indicated b. Continue serum Troponins q 3 hours x 3 c. Continue serial ECG’s i. Repeat PRN for recurring/worsening symptoms d. Obtain cardiac imaging study i. Consult Cardiology for appropriate test (i.e. Echocardiography, CTA, Radionuclide, etc.) e. If therapy not effective, or pending results of imaging study, reconsider if Invasive Strategy (Cath Lab) would be appropriate (Number 6 above) 11. Late/Hospital/Posthospital Care a. Aspirin 81 mg PO once daily b. ACE inhibitor or ARB c. Beta-Blocker d. High Intensity Statin e. P2Y12 Inhibitor per Cardiology f. Cardiac Rehab Referral 12. This Guideline is part of the ACS/Chest Pain “Tool-Kit” created with coordination from the Minnesota Department of Health, in conjunction with the American Heart Association Minnesota Mission: Lifeline™ Workgroup. This information is intended only as a guideline. Please use your best judgement or newly published literature in the treatment of patients. The Minnesota Department of Health is not responsible for inaccuracies contained herein. No responsibility is assumed for damages or liabilities arising from accuracy, content error, or omission. Minnesota Non-STEMI Guideline – Flowchart (Page 9) This page is an image of a flowchart depicting the Minnesota Non-STEMI Guideline described on Page 8. M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 13 Minnesota ED Chest Pain Protocol (Page 10) 1. For patients presenting to an emergency department with chest pain or equivalent symptoms of a potential Acute Coronary Syndrome (ACS) 2. Obtain STAT 12-lead ECG and IV blood draw for serum Troponin level a. If ECG or Troponin is positive for ACS, patient is no longer low risk, follow appropriate ACS protocols b. Repeat 12-Lead ECG immediately if symptoms change c. Once the first Troponin is resulted, calculate the HEART Score (See Calculate HEART Score below) 3. If the Heart Score is zero to three, patient is considered Low Risk a. Use the Low Risk Shared Decision-Making Tool (Page 12) b. Inform patient at this point, there is a 1.7% risk of adverse cardiac event in the next 4-6 weeks c. Advise patient to stay for another Troponin and ECG at hour of ED admission d. If second Troponin and ECG are negative: i. Inform the patient that now there is a 0.6% risk of an adverse cardiac event in the next 4 weeks ii. Advise that patient can be ruled out for ACS without a stress test iii. Advise patient to follow up with a provider within 1 week, or per local standard of care 4. If the Heart Score is four to six, patient is considered Moderate Risk (Page 13) a. Use the Moderate Risk Shared Decision-Making Tool b. Inform patient at this point there is a 13% risk of an adverse cardiac event in the next 4-6 weeks c. Advise patient to be admitted for observation d. Obtain serial ECG’s and Troponins at hours 3 and 6 e. Evaluate need for admission or a provocative cardiac Stress Test within the next 72 hours f. Follow appropriate ACS protocols, depending on findings 5. If the Heart Score is seven to ten, patient is considered High Risk (Page 14) a. Use the High Risk Shared Decision-Making Tool b. Inform patient at this point there is at least a 50% risk of an adverse cardiac event in the next 4-6 weeks c. Advise patient to be admitted to PCI capable hospital and follow appropriate ACS protocols d. Obtain serial ECG’s and Troponins at hours 3 and 6 e. Post Cardiology for consult 6. Section for user to add notes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 14 Calculate HEART Score 1. Calculate History a. 2 points for High Suspicious (mostly high-risk features) i. High-risk features include: 1. Middle or left-sided 2. Heavy chest pain 3. Diaphoresis 4. Radiation 5. Nausea/Vomiting 6. Exertional 7. Relief of symptoms by sublingual nitrates b. 1 point for Moderately Suspicious (mixture of high-risk and low-risk features) c. 0 points for Slightly Suspicious (mostly low-risk features) i. Low-risk features include: 1. Well localized 2. Sharp pain 3. Non-exertional 4. No diaphoresis 5. No nausea/vomiting 2. Calculate ECG a. 2 points for New Ischemic changes (Ischemic ST segment depression (≥ 1 mm) b. 1 point for Non-specific changes i. Repolarization abnormalities ii. Non-specific T-wave changes iii. Non-specific ST-segment depression or elevation iv. Bundle branch blocks v. Pacemaker rhythms vi. LVH vii. Early repolarization viii. Digoxin effect c. 0 points for completely normal 3. Calculate Age a. 2 points for ≥ 65 years b. 1 point for 45 to 64 years c. 0 points for < 45 years 4. Calculate Risk Factors a. 2 points for any of one of the following: i. Known CAD ii. Prior stroke iii. Peripheral arterial disease b. Or 2 points for 3 or more of the following risk factors: i. Obesity (BMI ≥ 30) M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 15 ii. Current or recent smoke (< 90 days) iii. Currently treated diabetes mellitus iv. Family history of CAD (siblings and/or parents < 55 years old) v. Hypertension vi. Hypercholesterolemia c. 1 point for 1 or 2 of the above risk factors d. 0 points for none of the above risk factors 5. Calculate Troponin a. 2 points for ≥ 3 times Normal Limit b. 1 point >1 to < 3 times Normal Limit c. 0 points for ≤ Normal Limit 6. Sum points from HEART Score Categories -- History, ECG, Age, Risk Factors, and Troponin -- to get Patient Score a. Total of 0 to 3 points is Low Risk b. Total of 4 to 6 points is Moderate Risk c. Total of 7 to 10 points is High Risk Minnesota ED Chest Pain Protocol Flowchart (Page 11) This page is an image of a flowchart depicting the Minnesota ED Chest Pain Protocol described on Page 10. M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 16 Low Risk Shared Decision Making Tool (Page 12) 1. For Chest Pain patients whom: Initial ECG and Troponin are negative, and HEART Score is Low Risk a. Of every 100 people with factors like yours who came to the Emergency Department with chest pain, and had 2 negative ECG and Troponin tests only 1 had a heart attack or a heart complication, while 99 did not, within 30 days of their Emergency Department visit. What to Expect Next? 1. Your Chest Pain Diagnosis a. Initial testing has NOT shown any evidence of a heart attack. This is based on a blood test, an electrocardiogram (ECG), your exam, and your risk factors. b. It is recommended that a repeat blood test (Troponin), and electrocardiogram (ECG) both be performed approximately 2 to 3 hours after initial tests to further rule out a heart attack. c. However, even if everything is normal, your chest pain may still be an early warning sign of a possible FUTURE heart attack or heart complication. 2. Further Evaluation a. Further evaluation and testing will help check if your heart is working correctly. b. Understanding your risk of having a heart attack or heart complication can help decide how to best proceed with your care in the Emergency Department. 3. Your Personal Risk Evaluation a. If a second Troponin blood test and ECG are both negative, your risk of having a heart attack or heart complication within the next 30 days can be determined by comparing you to people with similar factors who also came to an Emergency Department with chest pain. 4. The Next Step a. Another ECG and Troponin blood test should be repeated 2 to 3 hours after your initial blood test, and if they are also negative, your Emergency Department Provider and you may both decide that you could be discharged to home, and recommend you follow up with a primary care provider or cardiologist. b. If you refuse, and go home before a second set of tests, your risk for a heart attack may be doubled, up to 2 out of every 100 patients. 5. This shared decision tool was intended to help you understand your Personal Risk Evaluation. Even though you might be going home, you need to understand the importance of following up with your primary provider, or a cardiologist within 1 week. If your chest pain or heart related symptoms return or worsen, you should call 911 or return to the Emergency Department immediately. 6. Factors used to determine your risk using the HEART Score a. H means History M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 17 b. E means ECG c. A means Age d. R means Risk Factors for Heart Disease e. T means Troponin 7. Section for user to add notes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 18 Moderate Risk Shared Decision Making Tool (Page 13) 1. For Chest Pain patients whom: Initial ECG and Troponin are negative, and HEART Score is Moderate Risk a. Of every 100 people with factors like yours who came to the Emergency Department with chest pain, and had 2 negative ECG and Troponin tests 13 had a heart attack or a heart complication, while 87 did not, within 30 days of their Emergency Department visit. What to Expect Next? 1. Your Chest Pain Diagnosis b. Initial testing has NOT shown any evidence of a heart attack. This is based on a blood test, an electrocardiogram (ECG), your exam, and your risk factors. c. It is recommended that a repeat blood test (Troponin), and electrocardiogram (ECG) both be performed approximately 2 to 3 hours after initial tests to further rule out a heart attack, and possibly again 3 hours later. d. However, even if everything is normal, your chest pain may still be an early warning sign of a possible FUTURE heart attack or heart complication. 2. Further Evaluation a. Further evaluation and testing will help check if your heart is working correctly. b. Understanding your risk of having a heart attack or heart complication can help decide how to best proceed with your care in the Emergency Department. 3. Your Personal Risk Evaluation a. If your second Troponin blood test and ECG are both negative, your risk of having a heart attack or heart complication within the next 30 days can be determined by comparing you to people with similar factors who also came to an Emergency Department with chest pain. 4. The Next Step a. You have a moderate (intermediate) risk of a heart attack or complication in the near future. Your Emergency Department Provider may want you to agree to stay for observation and further testing. b. If you decline repeated tests and go home now, your current risk for a heart attack may be even greater than 13 out of 100 patients. 5. This shared decision tool was intended to help you understand your Personal Risk Evaluation. Further observation and testing may be necessary during this visit to the Emergency Department. If you do end up going home, you may still need further testing as an out-patient. You need to understand the importance of following up with your primary provider, or a cardiologist within 1 week, or whatever is recommended by your Emergency Department provider. If you do end up going home, and your chest pain or heart related symptoms return or worsen, you should call 911 or return to the Emergency Department immediately. M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 19 6. Factors used to determine your risk using the HEART Score a. H means History b. E means ECG c. A means Age d. R means Risk Factors for Heart Disease e. T means Troponin 7. Section for user to add notes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 20 High Risk Shared Decision Making Tool (Page 14) 1. For Chest Pain patients whom: Initial ECG and Troponin are negative, and HEART Score is High Risk a. Of every 100 people with factors like yours who came to the Emergency Department with chest pain, and had 2 negative ECG and Troponin tests 50 had a heart attack or a heart complication, while 50 did not, within 30 days of their Emergency Department visit. What to Expect Next? 1. Your Chest Pain Diagnosis b. Our testing so far has NOT shown any evidence of a heart attack. This is based on a blood test, an electrocardiogram (ECG), your exam, and your risk factors. c. It is recommended that a repeat blood test (Troponin), and electrocardiogram (ECG) both be performed approximately 2 to 3 hours after initial tests to further rule out a heart attack, and likely again 3 hours later. d. However, even if everything is normal, your chest pain may still be an early warning sign of a possible FUTURE heart attack or heart complication. 2. Further Evaluation a. Further evaluation and testing will help check if your heart is working correctly. b. Understanding your risk of having a heart attack or heart complication can help decide how to best proceed with your care in the Emergency Department. 3. Your Personal Risk Evaluation a. If your second Troponin blood test and ECG are both negative, your risk of having a heart attack or heart complication within the next 30 days can be determined by comparing you to people with similar factors who also came to an Emergency Department with chest pain. 4. The Next Step a. You have a high risk of a heart attack or complication in the near future. Your Emergency Department Provider will likely recommend you stay for observation and further testing. b. If you decline repeated tests and go home now, your current risk for a heart attack may be even greater than 50 out of 100 patients. 5. This shared decision tool was intended to help you understand your Personal Risk Evaluation. Further observation and testing may be necessary during this visit to the Emergency Department. If you do end up going home, you may still need further testing as an out-patient. You need to understand the importance of following up with your primary provider, or a cardiologist within 1 week, or whatever is recommended by your Emergency Department provider. If you do end up going home, and your chest pain or heart related symptoms return or worsen, you should call 911 or return to the Emergency Department immediately. 6. Factors used to determine your risk using the HEART Score M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 21 a. H means History b. E means ECG c. A means Age d. R means Risk Factors for Heart Disease e. T means Troponin 7. Section for user to add notes M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 22 Who Needs a Stat 12-Lead ECG? (Page 15) 1. If Age ≥ 18 years and either of the following: a. Cardiac Chest Pain b. Or Clinical Judgment suggests need for ECG i. Clinical judgment requires assessment beyond the chief complaint. This list is simply a guide. Clinical history, and evaluation of multiple symptoms beyond chest pain, may be present that should trigger concern for potential Acute Coronary Syndrome. Some of these include things like: Pressure, Discomfort, Tightness, Radiating Pain, Pounding, Racing, Beating Fast, Sweating, etc. Be suspicious of patients with cardiac risk factors, like high blood pressure, high cholesterol, diabetes, smoking history, and patients with a known cardiac history or with recent cardiac surgery or intervention. If in doubt, always err on the side of caution, and obtain a STAT 12-lead ECG. c. Then 12 Lead ECG Needed STAT! 2. If No and Age ≥ 30 and Any Chest Pain a. Then 12 Lead ECG Needed STAT! 3. If No and Age ≥ 50 and any of the following: a. Shortness of Breath b. Weakness c. Altered Mental Status d. Syncope e. Upper Extremity Pain – Nose to Navel (Arm, Back, Jaw, Neck, etc.) f. Then 12 Lead ECG Needed STAT! 4. If No and Age ≥ 80 and either of the following: a. Abdominal Pain b. Nausea/Vomiting c. Then 12 Lead ECG Needed STAT! 5. If No, 12 Lead ECG Not Needed STAT, then continue to monitor and assess patient for any changes that would trigger the need for an ECG!!! 6. This flowchart adapted from: Development and Validation of a Prioritization Rule for Obtaining an Immediate 12-lead Electrocardiogram in the Emergency Department to Identify ST-elevation Myocardial Infarction. Seth W. Glickman, MD, MBA, Frances S. Shofer, PhD, Michael C. Wu, PhD, Matthew J. Scholer, MD, PhD, Adanma Ndubuizu, MD, MPH, Eric D. Petersen, MD, MPH, Christopher B. Granger, MD, Charles B. Cairns, MD, Lawrence T. Glickman, VMD, DrPH, Chapel Hill, Durham, NC Am Heart J. 2012; 163(3):372-382. M I N N E S O TA C H ES T P A I N / AC UT E C O RO NA RY S Y N D RO ME T O O L K IT 23 Cardiovascular Health Unit PO Box 640882 St. Paul, MN 55164-0882 651-201-5405 health.heart@state.mn.us www.health.state.mn.us 06/30/2019 To obtain this information in a different format, call: 651-201-5405. Printed on recycled paper.
14542	https://artofproblemsolving.com/wiki/index.php/Characteristic_polynomial?srsltid=AfmBOor2O2ZAP0Zk_scLSHuKFGUni4cB-bOTfjDtrwTf7g9dn5_-2lSN	Art of Problem Solving Characteristic polynomial - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Characteristic polynomial Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Characteristic polynomial The characteristic polynomial of a linear operator refers to the polynomial whose roots are the eigenvalues of the operator. It carries much information about the operator. In the context of problem-solving, the characteristic polynomial is often used to find closed forms for the solutions of linear recurrences. Contents [hide] 1 Definition 2 Properties 3 Linear recurrences 3.1 Proof 1 (Linear Algebra) 3.2 Proof 2 (Induction) 3.3 Proof 3 (Partial fractions) 4 Differential equations 4.1 Proof 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Hints/Solutions 6.1 Introductory 6.2 Intermediate 6.3 Olympiad Definition Suppose is a matrix (over a field ). Then the characteristic polynomial of is defined as , which is a th degree polynomial in . Here, refers to the identity matrix. Written out, the characteristic polynomial is the determinant Properties An eigenvector is a non-zero vector that satisfies the relation , for some scalar . In other words, applying a linear operator to an eigenvector causes the eigenvector to dilate. The associated number is called the eigenvalue. There are at most distinct eigenvalues, whose values are exactly the roots of the characteristic polynomial of the square matrix. To prove this, we use the fact that the determinant of a matrix is iff the column vectors of the matrix are linearly dependent. Observe that if satisfies , then the column vectors of the matrix are linearly dependent. Indeed, if we define and let denote the column vectors of , this is equivalent to saying that there exists not all zero scalars such that Hence, there exists a non-zero vector such that . Distributing and re-arranging, , as desired. In the other direction, if , then . But then, the column vectors of are linearly dependent, so it follows that . Note that if , then . Hence, the characteristic polynomial encodes the determinant of the matrix. Also, the coefficient of the term of gives the negative of the trace of the matrix (which follows from Vieta's formulas). By the Hamilton-Cayley Theorem, the characteristic polynomial of a square matrix applied to the square matrix itself is zero, that is . The minimal polynomial of thus divides the characteristic polynomial . Linear recurrences Let be a sequence of real numbers. Consider a monic homogenouslinear recurrence of the form where are real constants. The characteristic polynomial of this recurrence is defined as the polynomial For example, let be the th Fibonacci number defined by , and Then, its characteristic polynomial is . The roots of the polynomial can be used to write a closed form for the recurrence. If the roots of this polynomial are distinct, then suppose the roots are . Then, there exists real constants such that If we evaluate different values of (typically ), we can find a linear system in the s that can be solved for each constant. Refer to the introductory problems below to see an example of how to do this. In particular, for the Fibonacci numbers, this yields Binet's formula. If there are roots with multiplicity greater than , suppose . Then we would replace the term with the expression That is, there is now a polynomial in multiplied with the exponent. Note that there are still the same number of constants that we must solve for. For example, consider the recurrence relation . It’s characteristic polynomial, , has a double root. Then, its closed form solution is of the type . Given a linear recurrence of the form , we often try to find a new sequence such that is a homogenous linear recurrence. Then, we can find a closed form for , and then the answer is given by . Of the following proofs, the second and third are more approachable. Proof 1 (Linear Algebra) Note: The ideas expressed in this section can be transferred to the next section about differential equations. This requires some knowledge of linear algebra (upto the Spectral Theorem). Let . Then, we can express our linear recurrence as the matrix so that (try to verify this). The characteristic polynomial of is precisely . This is not difficult to show via Laplace's expansion and induction, and is left as an exercise to the reader. If the roots of are distinct, then there exists a basis (of ) consisting of eigenvectors of (since eigenvectors of different eigenvalues are linearly independent). That is, applying a change of bases, we can write for a matrix and a diagonal matrix . Then, , and in general, . Thus, . Here, and are fixed (note that to find the values of , we may need to trace the recurrence backwards. We only take the th index for simplicity). It follows that is a linear combination of the diagonal elements of , namely . Suppose now that that the roots of are not distinct. Then, we can write the matrix in the Jordan normal form. For simplicity, first consider just a single root repeated times. The corresponding Jordan form of the matrix is given by (that is, the matrix is similar to the following): . Exponentiating this matrix to the th power will yield binomial coefficients as follows We can treat the binomial coefficient as a polynomial in . Furthermore, we can scale away the powers of the eigenvalue (which is a constant); after taking the appropriate linear combinations of the binomial coefficients and a little bit of work, the result follows. See also a viewtopic.php?t=290351 graph-theoretic approach. Proof 2 (Induction) There are a couple of lower-level ways to prove this. One is by induction, though the proof is not very revealing; we can explicitly check that a sequence , for real numbers , satisfies the linear recurrence relation . If the two sequences are the same for the first values of the sequence, it follows by induction that the two sequences must be exactly the same. In particular, for , we can check this by using the identity It is also possible to reduce the recurrence to a telescoping sum. However, the details are slightly messy. Proof 3 (Partial fractions) Another method uses partial fractions and generating functions, though not much knowledge of each is required for this proof. Let be a linear recurrence. Consider the generating function given by Then, writing the following expressing out and carefully comparing coefficients (try it), where is a remainder polynomial with degree . Re-arranging, This polynomial in the denominator is the reverse of the characteristic polynomial of the recurrence. Hence, its roots are , assuming that the roots are distinct. Using partial fraction decomposition (essentially the reverse of the process of adding fractions by combining denominators, except we now pull the denominators apart), we can write for some constants . Using the geometric series formula, we have . Thus, Comparing coefficients with our original definition of gives , as desired. The generalization to characteristic polynomials with multiple roots is not difficult from here, and is left to the reader. The partial fraction decomposition will have terms of the form for . A note about this argument: all of the power series used here are defined formally, and so we do not actually need to worry whether or not they converge. See these blogposts for further ideas. Differential equations Given a monic linear homogenousdifferential equation of the form , then the characteristic polynomial of the equation is the polynomial Here, is short-hand for the differential operator. If the roots of the polynomial are distinct, say , then the solutions of this differential equation are precisely the linear combinations . Similarly, if there is a set of roots with multiplicity greater than , say , then we would replace the term with the expression . In general, given a linear differential equation of the form , where is a linear differential operator, then the set of solutions is given by the sum of any solution to the homogenous equation and a specific solution to . Proof We can apply an induction technique similar to the section on linear recurrences above. From linear algebra, we can use the following vector space decomposition theorem. Let be a linear operator for a vector spaces over a field . Suppose that there exists a polynomial such that , where and are non-zero polynomials such that , and such that . Then . This allows us to reduce the differential equation into finding the solutions to the equation , which has a basis of functions . Problems Introductory Prove Binet's formula. Find a similar closed form equation for the Lucas sequence, defined with the starting terms , and satisfying the recursion . Let denote the sequence defined by the recursion , and . Find a closed form expression for . Given , , and the general relation for . Find a linear recurrence for . (AHSME 1958, Problem 40) Intermediate Let denote the number of ternary sequences (consisting of ,, and s) of length , such that they do not contain a substring of "10", "01", or "11". Find a closed form expression for . Let and be sequences defined as follows: Let be the largest integer that satisfies all of the following conditions:(i) , for some positive integer ;(ii) , for some positive integer ;(iii) .Find the remainder when is divided by . (2007 iTest, #47) Let , , and be geometric sequences with different common ratios and let for all integers . If , , , , , and , find . (Mock AIME 1 2006-2007, Problem 13) Find all possible values of and such that the sequence defined by: contains infinitely many natural numbers. Show that can be written as a linear combination of the elementary symmetric polynomials. In general, prove Newton's sums. Olympiad Let be a real number, and let be a sequence such that , and for . For which values of does for all positive integers ? (WOOT) Let be the polynomial , and the polynomial . Prove that we can find a polynomial which is identically equal to . For example, . (1976 Putnam, Problem A2). are distinct positive integers such that holds for all positive integer . Prove that there exists such that for . (2010 Chinese MO, Problem 6) Let and be opposite vertices of an octagon. A frog starts at vertex From any vertex except it jumps to one of the two adjacent vertices. When it reaches it stops. Let be the number of distinct paths of exactly jumps ending at . Prove that: (1979 IMO Problems/Problem 6) Hints/Solutions Introductory A proof of Binet's formula may be found in that link. The characteristic polynomial of the Lucas sequence is exactly the same. Hence, the only thing we have to change are the coefficients. We will work out this problem in full detail. The recurrence relation is , and its characteristic polynomial is given by . The roots of this polynomial are and . Thus, a closed form solution is given by . For , we get , and for , we get . Solving gives . Thus, our answer is . viewtopic.php?t=282744 Discussion (1958 AHSME, 40) Intermediate viewtopic.php?t=335138 Discussion Answer (2007 iTest 47): The answer is (before taking ). viewtopic.php?t=287441 Discussion Hint (Newton’s Sum): Let us work backwards. Suppose are the roots of the characteristic polynomial of a linear recurrence. Then apply Vieta's formulas. Olympiad Hint (WOOT): substitute the closed form solution. It is true for all . Hint (1976 Putnam A2): do the even and odd cases separately. viewtopic.php?search_id=804457492&t=327474 Discussion (2010 Chinese MO, 6) Retrieved from " Category: Linear algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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14544	https://math.stackexchange.com/questions/4635886/how-would-you-simplify-this-expression-involving-complex-numbers-and-a-logarithm	How would you simplify this expression involving complex numbers and a logarithm? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How would you simplify this expression involving complex numbers and a logarithm? Ask Question Asked 2 years, 7 months ago Modified2 years, 7 months ago Viewed 86 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. The expression is log(∣∣∣z+i a z−i a∣∣∣)log⁡(\|z+i a z−i a\|) Here, a>0 a>0 is real, i i is the imaginary unit, and z=x+i y z=x+i y. The expression comes from extracting the imaginary part of an expression so it should end up being purely real, however I can't manage to remove the imaginary unit from any of my expressions. I know that it should end up being something like: 1 2 log(x 2+(y+a)2 x 2+(y−a)2)1 2 log⁡(x 2+(y+a)2 x 2+(y−a)2) complex-analysis complex-numbers absolute-value potential-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Feb 9, 2023 at 16:34 ChillpaddeChillpadde asked Feb 9, 2023 at 16:21 ChillpaddeChillpadde 45 4 4 bronze badges 5 What sort of simplification are you looking for. It clearly is real (unless y=−a y=−a) since the expression is ≥0≥0. But without more structure, this might be all you get QC_QAOA –QC_QAOA 2023-02-09 16:29:05 +00:00 Commented Feb 9, 2023 at 16:29 I want to reach the expression at the end of my question, in terms of x x and y y and no imaginary unit. I want to know how.Chillpadde –Chillpadde 2023-02-09 16:31:20 +00:00 Commented Feb 9, 2023 at 16:31 x,y∈R x,y∈R since they describe the real and imaginary parts of z z respectively QC_QAOA –QC_QAOA 2023-02-09 16:35:32 +00:00 Commented Feb 9, 2023 at 16:35 Of course. I want to know how to get from my first expression to my second expression.Chillpadde –Chillpadde 2023-02-09 16:38:09 +00:00 Commented Feb 9, 2023 at 16:38 In the answer below you express confusion since "Your first line holds only if x x is real, no? In my case what you are treating like x x is in fact z z, a complex number". I am saying that x x is in fact a real number. So what is your confusion with the answer then?QC_QAOA –QC_QAOA 2023-02-09 16:43:25 +00:00 Commented Feb 9, 2023 at 16:43 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Note that \|z\|=\|x+i y\|=x 2+y 2−−−−−−√\|z\|=\|x+i y\|=x 2+y 2 when both x x and y y are assumed to be real numbers. Therefore, we have ∣∣∣z+i a z−i a∣∣∣=x 2+(y+a)2 x 2+(y−a)2−−−−−−−−−−−√\|z+i a z−i a\|=x 2+(y+a)2 x 2+(y−a)2 Can you finish now? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Feb 9, 2023 at 21:28 amWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges answered Feb 9, 2023 at 16:27 Mark ViolaMark Viola 185k 12 12 gold badges 154 154 silver badges 264 264 bronze badges 5 Your first line holds only if x x is real, no? In my case what you are treating like x x is in fact z z, a complex number.Chillpadde –Chillpadde 2023-02-09 16:33:06 +00:00 Commented Feb 9, 2023 at 16:33 @Chillpadde You're mistaken. Complex numbers z z can always be expressed in the form x+i y,x,y∈R x+i y,x,y∈R. The answer is exactly correct.Robert Shore –Robert Shore 2023-02-09 16:49:52 +00:00 Commented Feb 9, 2023 at 16:49 @RobertShore So does the following statement hold? \|z+i a\|=x 2+(y+a)2−−−−−−−−−−−√\|z+i a\|=x 2+(y+a)2.Chillpadde –Chillpadde 2023-02-09 16:54:51 +00:00 Commented Feb 9, 2023 at 16:54 That was understood from the beginning. I had some other stupid tangle in my head that is cleared up now. I was having trouble seeing why \|x+i y+i a\|=x 2+(y+a)2−−−−−−−−−−−√\|x+i y+i a\|=x 2+(y+a)2, but on second look it is glaringly obvious. You've simply taken i i common and are left with \|x+i(y+a)\|\|x+i(y+a)\|, from which point the answer is obvious.Chillpadde –Chillpadde 2023-02-09 17:04:37 +00:00 Commented Feb 9, 2023 at 17:04 @Chillpadde Yes, you see it now.Robert Shore –Robert Shore 2023-02-09 20:01:42 +00:00 Commented Feb 9, 2023 at 20:01 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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14545	https://fr.scribd.com/document/326624524/Density-Worksheet	Density Worksheet: Mass V M \| PDF \| Density \| Volume Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0%(1)0% found this document useful (1 vote) 3K views 2 pages Density Worksheet: Mass V M The document discusses density and provides examples of density problems and their solutions. It defines density as the relationship between the mass and volume of an object. Several density… Full description Uploaded by Tahir Saeed AI-enhanced title and description Go to previous items Go to next items Download Save Save Density Worksheet For Later Share 0%0% found this document useful, undefined 100%, undefined Print Embed Ask AI Report Download Save Density Worksheet For Later You are on page 1/ 2 Search Fullscreen Name Per: Date: Density Worksheet Chemistry in Context 2008-9 Perhaps someone has tried to trick you with this question: “Which is heavier, a pound of lead or a pound of feathers?” Many people would instinctively answer “lead.” When they give this incorrect answer, these people are really thinking o f density. If a piece of lead and a feather of the same volume are weighed, the lead would have a greater mas s than the feather. It would take a much larger volume of feathers to equal the mass of a given volume of lead. Density is the relationship of the mass of an object to its volume. Density is usually reported in units of grams per cubic centimeter (g/cm 3 ). For example, water has a density of 1.00 g/cm 3 . Since a cubic centimeter contains the same volume as a milliliter, in some cases you may see d ensity expressed as g/mL. Density = volume mass or D = V M To solve density problems, list the known and unknown values, then use one of the following. ! When a problem requires you to calculate density, use the density equation, D = V M ! You can solve for mass by multiplying both sides of the density equation by volume. D V = V MV or M = D V ! You can solve for volume by dividing both sides of the equation above by density. D M = D V D or V = D M Example: What is the mas s of an object that has a density of 8 g/cm 3 and a volume of 64 cm 3 ? Known: D = 8 g/cm 3 V = 64 cm 3 Unknown: M = ? Equation to use: M = D V “Plug and chug”: M = (8 g/cm 3 ) (64 cm 3 ) = 512 g PROBLEMS List the known and unknown values; try to derive the equation without looking above. 1. A piece of tin has a mass of 16.52 g and a volume of 2.26 cm 3 . What is the density of tin? Known: Unknown: 2. A man has a 50.0 cm 3 bottle completely filled wit h 163 g of a sli my green liquid. What is the density of the liquid? Known: Unknown: adDownload to read ad-free 3. A sealed 2500 cm 3 flask is full to capacity with 0.36 g of a substance. Determine the density o f the substance. Guess if the subs tance is a gas, a liquid, or a s olid. Known: Unknown: 4. Different kinds of wood have dif ferent densities. The density of oak wood is generally 0.7 g/cm 3 . If a 35 cm 3 piece of wood has a mass of 25 g, is the wood likely to be oak? Known: Unknown: 5. The density of pine is generally about 0.5 g/cm 3 . What is the mass of a 800 cm 3 piece of pine? Known: Unknown: 6. What is the volume of 325 g of metal with a density of 9.0 g/cm 3 ? Known: Unknown: 7. Diamonds have a density of 3.5 g/cm 3 . How big is a diamond that has a mass of 0.10 g? Known: Unknown: 8. What mass of water in grams will fill a tank 10 0 cm long, 50 cm wide, and 30 cm high? Known: Unknown: 9. A graduated cylinder is filled with water to a level of 40.0 mL. When a piece of copper is lowe red into the cylinder, the water level ris es to 63.4 m L. Find the volume of the copper sample. If the density of the copper is 8.9 g/cm 3 , what is its mass? 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14546	http://lya.fciencias.unam.mx/jele/EDOs2k15.2/Libros/A%20First%20Course%20in%20Differential%20Equations%20with%20Modeling%20Applications%2010e%202012%20Zill.pdf	REVIEW OF DIFFERENTIATION a a a a a Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. BRIEF TABLE OF INTEGRALS Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Tenth Edition A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Tenth Edition A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications DENNIS G. ZILL Loyola Marymount University Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. A First Course in Differential Equations with Modeling Applications, Tenth Edition Dennis G. Zill Publisher: Richard Stratton Senior Sponsoring Editor: Molly Taylor Development Editor: Leslie Lahr Assistant Editor: Shaylin Walsh Hogan Editorial Assistant: Alex Gontar Media Editor: Andrew Coppola Marketing Manager: Jennifer Jones Marketing Coordinator: Michael Ledesma Marketing Communications Manager: Mary Anne Payumo Content Project Manager: Alison Eigel Zade Senior Art Director: Linda May Manufacturing Planner: Doug Bertke Rights Acquisition Specialist: Shalice Shah-Caldwell Production Service: MPS Limited, a Macmillan Company Text Designer: Diane Beasley Projects Piece Designer: Rokusek Design Cover Designer: One Good Dog Design Cover Image: ©Stocktrek Images Compositor: MPS Limited, a Macmillan Company Section 4.8 of this text appears in Advanced Engineering Mathematics, Fourth Edition, Copyright 2011, Jones & Bartlett Learning, Burlington, MA 01803 and is used with the permission of the publisher. © 2013, 2009, 2005 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. 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Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. The publisher reserves the right to remove content from this title at any time if subsequent rights restrictions require it. For valuable information on pricing, previous editions, changes to current editions, and alternate formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3 v Contents 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 1 Preface ix Projects P-1 1.1 Deﬁnitions and Terminology 2 1.2 Initial-Value Problems 13 1.3 Differential Equations as Mathematical Models 20 Chapter 1 in Review 33 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 35 2.1 Solution Curves Without a Solution 36 2.1.1 Direction Fields 36 2.1.2 Autonomous First-Order DEs 38 2.2 Separable Equations 46 2.3 Linear Equations 54 2.4 Exact Equations 63 2.5 Solutions by Substitutions 71 2.6 A Numerical Method 75 Chapter 2 in Review 80 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS 83 3.1 Linear Models 84 3.2 Nonlinear Models 95 3.3 Modeling with Systems of First-Order DEs 106 Chapter 3 in Review 113 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5 4 vi ● CONTENTS HIGHER-ORDER DIFFERENTIAL EQUATIONS 116 4.1 Preliminary Theory—Linear Equations 117 4.1.1 Initial-Value and Boundary-Value Problems 117 4.1.2 Homogeneous Equations 119 4.1.3 Nonhomogeneous Equations 124 4.2 Reduction of Order 129 4.3 Homogeneous Linear Equations with Constant Coefﬁcients 132 4.4 Undetermined Coefﬁcients—Superposition Approach 139 4.5 Undetermined Coefﬁcients—Annihilator Approach 149 4.6 Variation of Parameters 156 4.7 Cauchy-Euler Equation 162 4.8 Green’s Functions 169 4.8.1 Initial-Value Problems 169 4.8.2 Boundary-Value Problems 176 4.9 Solving Systems of Linear DEs by Elimination 180 4.10 Nonlinear Differential Equations 185 Chapter 4 in Review 190 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 192 5.1 Linear Models: Initial-Value Problems 193 5.1.1 Spring/Mass Systems: Free Undamped Motion 193 5.1.2 Spring/Mass Systems: Free Damped Motion 197 5.1.3 Spring/Mass Systems: Driven Motion 200 5.1.4 Series Circuit Analogue 203 5.2 Linear Models: Boundary-Value Problems 210 5.3 Nonlinear Models 218 Chapter 5 in Review 228 SERIES SOLUTIONS OF LINEAR EQUATIONS 231 6.1 Review of Power Series 232 6.2 Solutions About Ordinary Points 238 6.3 Solutions About Singular Points 247 6.4 Special Functions 257 Chapter 6 in Review 271 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CONTENTS ● vii 7 THE LAPLACE TRANSFORM 273 7.1 Deﬁnition of the Laplace Transform 274 7.2 Inverse Transforms and Transforms of Derivatives 281 7.2.1 Inverse Transforms 281 7.2.2 Transforms of Derivatives 284 7.3 Operational Properties I 289 7.3.1 Translation on the s-Axis 290 7.3.2 Translation on the t-Axis 293 7.4 Operational Properties II 301 7.4.1 Derivatives of a Transform 301 7.4.2 Transforms of Integrals 302 7.4.3 Transform of a Periodic Function 307 7.5 The Dirac Delta Function 312 7.6 Systems of Linear Differential Equations 315 Chapter 7 in Review 320 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 325 8.1 Preliminary Theory—Linear Systems 326 8.2 Homogeneous Linear Systems 333 8.2.1 Distinct Real Eigenvalues 334 8.2.2 Repeated Eigenvalues 337 8.2.3 Complex Eigenvalues 342 8.3 Nonhomogeneous Linear Systems 348 8.3.1 Undetermined Coefﬁcients 348 8.3.2 Variation of Parameters 351 8.4 Matrix Exponential 356 Chapter 8 in Review 360 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 362 9.1 Euler Methods and Error Analysis 363 9.2 Runge-Kutta Methods 368 9.3 Multistep Methods 373 9.4 Higher-Order Equations and Systems 375 9.5 Second-Order Boundary-Value Problems 380 Chapter 9 in Review 384 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. viii ● CONTENTS APPENDIXES I Gamma Function APP-1 II Matrices APP-3 III Laplace Transforms APP-21 Answers for Selected Odd-Numbered Problems ANS-1 Index I-1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ix TO THE STUDENT Authors of books live with the hope that someone actually reads them. Contrary to what you might believe, almost everything in a typical college-level mathematics text is written for you, and not the instructor. True, the topics covered in the text are chosen to appeal to instructors because they make the decision on whether to use it in their classes, but everything written in it is aimed directly at you, the student. So I want to encourage you—no, actually I want to tell you—to read this textbook! But do not read this text like you would a novel; you should not read it fast and you should not skip anything. Think of it as a workbook. By this I mean that mathemat-ics should always be read with pencil and paper at the ready because, most likely, you will have to work your way through the examples and the discussion. Before attempt-ing any of the exercises, work all the examples in a section; the examples are con-structed to illustrate what I consider the most important aspects of the section, and therefore, reﬂect the procedures necessary to work most of the problems in the exer-cise sets. I tell my students when reading an example, copy it down on a piece of paper, and do not look at the solution in the book. Try working it, then compare your results against the solution given, and, if necessary, resolve any differences. I have tried to include most of the important steps in each example, but if something is not clear you should always try—and here is where the pencil and paper come in again—to ﬁll in the details or missing steps. This may not be easy, but that is part of the learning process. The accumulation of facts followed by the slow assimilation of understanding simply cannot be achieved without a struggle. Specifically for you, a Student Resource Manual (SRM) is available as an optional supplement. In addition to containing worked-out solutions of selected problems from the exercises sets, the SRM contains hints for solving problems, extra examples, and a review of those areas of algebra and calculus that I feel are particu-larly important to the successful study of differential equations. Bear in mind you do not have to purchase the SRM; by following my pointers given at the beginning of most sections, you can review the appropriate mathematics from your old precalculus or calculus texts. In conclusion, I wish you good luck and success. I hope you enjoy the text and the course you are about to embark on—as an undergraduate math major it was one of my favorites because I liked mathematics with a connection to the physical world. If you have any comments, or if you ﬁnd any errors as you read/work your way through the text, or if you come up with a good idea for improving either it or the SRM, please feel free to contact me through my editor at Cengage Learning: molly.taylor@cengage.com TO THE INSTRUCTOR In case you are examining this textbook for the ﬁrst time, A First Course in Differential Equations with Modeling Applications, Tenth Edition, is intended for either a one-semester or a one-quarter course in ordinary differential equations. The longer version of the textbook, Differential Equations with Boundary-Value Problems, Eighth Edition, can be used for either a one-semester course, or a two-semester course covering ordinary and partial differential equations. This longer book includes six additional chapters that cover plane autonomous systems of differential equations, Preface Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. stability, Fourier series, Fourier transforms, linear partial differential equations and boundary-value problems, and numerical methods for partial differential equations. For a one semester course, I assume that the students have successfully completed at least two semesters of calculus. Since you are reading this, undoubtedly you have already examined the table of contents for the topics that are covered. You will not ﬁnd a “suggested syllabus” in this preface; I will not pretend to be so wise as to tell other teachers what to teach. I feel that there is plenty of material here to pick from and to form a course to your liking. The textbook strikes a reasonable balance be-tween the analytical, qualitative, and quantitative approaches to the study of differ-ential equations. As far as my “underlying philosophy” it is this: An undergraduate textbook should be written with the student’s understanding kept ﬁrmly in mind, which means to me that the material should be presented in a straightforward, read-able, and helpful manner, while keeping the level of theory consistent with the notion of a “ﬁrst course.” For those who are familiar with the previous editions, I would like to mention a few of the improvements made in this edition. • Eight new projects appear at the beginning of the book. Each project includes a related problem set, and a correlation of the project material with a section in the text. • Many exercise sets have been updated by the addition of new problems— especially discussion problems—to better test and challenge the students. In like manner, some exercise sets have been improved by sending some prob-lems into retirement. • Additional examples have been added to many sections. • Several instructors took the time to e-mail me expressing their concerns about my approach to linear ﬁrst-order differential equations. In response, Section 2.3, Linear Equations, has been rewritten with the intent to simplify the discussion. • This edition contains a new section on Green’s functions in Chapter 4 for those who have extra time in their course to consider this elegant application of variation of parameters in the solution of initial-value and boundary-value problems. Section 4.8 is optional and its content does not impact any other section. • Section 5.1 now includes a discussion on how to use both trigonometric forms in describing simple harmonic motion. • At the request of users of the previous editions, a new section on the review of power series has been added to Chapter 6. Moreover, much of this chapter has been rewritten to improve clarity. In particular, the discussion of the modiﬁed Bessel functions and the spherical Bessel functions in Section 6.4 has been greatly expanded. STUDENT RESOURCES • Student Resource Manual (SRM), prepared by Warren S. Wright and Carol D. Wright (ISBN 9781133491927 accompanies A First Course in Differential Equations with Modeling Applications, Tenth Edition and ISBN 9781133491958 accompanies Differential Equations with Boundary-Value Problems, Eighth Edition), provides important review material from algebra and calculus, the solution of every third problem in each exercise set (with the exception of the Discussion Problems and Computer Lab Assignments), relevant command syntax for the computer algebra systems Mathematica and Maple, lists of important concepts, as well as helpful hints on how to start certain problems. y Asin(vt f) and y Acos(vt f) x ● PREFACE Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. PREFACE ● xi INSTRUCTOR RESOURCES • Instructor’s Solutions Manual (ISM) prepared by Warren S. Wright and Carol D. Wright (ISBN 9781133602293) provides complete, worked-out solutions for all problems in the text. • Solution Builder is an online instructor database that offers complete, worked-out solutions for all exercises in the text, allowing you to create customized, secure solutions printouts (in PDF format) matched exactly to the problems you assign in class. Access is available via www.cengage.com/solutionbuilder • ExamView testing software allows instructors to quickly create, deliver, and customize tests for class in print and online formats, and features automatic grading. Included is a test bank with hundreds of questions customized di-rectly to the text, with all questions also provided in PDF and Microsoft Word formats for instructors who opt not to use the software component. • Enhanced WebAssign is the most widely used homework system in higher education. Available for this title, Enhanced WebAssign allows you to assign, collect, grade, and record assignments via the Web. This proven homework system includes links to textbook sections, video examples, and problem spe-ciﬁc tutorials. Enhanced WebAssign is more than a homework system—it is a complete learning system for students. ACKNOWLEDGMENTS I would like to single out a few people for special recognition. Many thanks to Molly Taylor (senior sponsoring editor), Shaylin Walsh Hogan (assistant editor), and Alex Gontar (editorial assistant) for orchestrating the development of this edition and its component materials. Alison Eigel Zade (content project manager) offered the resourcefulness, knowledge, and patience necessary to a seamless production process. Ed Dionne (project manager, MPS) worked tirelessly to provide top-notch publishing services. And ﬁnally, I thank Scott Brown for his superior skills as accuracy reviewer. Once again an especially heartfelt thank you to Leslie Lahr, developmental editor, for her support, sympathetic ear, willingness to communicate, suggestions, and for obtaining and organizing the excellent projects that appear at the front of the text. I also extend my sincerest appreciation to those individuals who took the time out of their busy schedules to submit a project: Ivan Kramer, University of Maryland—Baltimore County Tom LaFaro, Gustavus Adolphus College Jo Gascoigne, Fisheries Consultant C. J. Knickerbocker, Sensis Corporation Kevin Cooper, Washington State University Gilbert N. Lewis, Michigan Technological University Michael Olinick, Middlebury College Finally, over the years these textbooks have been improved in a countless num-ber of ways through the suggestions and criticisms of the reviewers. Thus it is ﬁtting to conclude with an acknowledgement of my debt to the following wonderful people for sharing their expertise and experience. REVIEWERS OF PAST EDITIONS William Atherton, Cleveland State University Philip Bacon, University of Florida Bruce Bayly, University of Arizona William H. Beyer, University of Akron R. G. Bradshaw, Clarkson College Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Dean R. Brown, Youngstown State University David Buchthal, University of Akron Nguyen P. Cac, University of Iowa T. Chow, California State University—Sacramento Dominic P. Clemence, North Carolina Agricultural and Technical State University Pasquale Condo, University of Massachusetts—Lowell Vincent Connolly, Worcester Polytechnic Institute Philip S. Crooke, Vanderbilt University Bruce E. Davis, St. Louis Community College at Florissant Valley Paul W. Davis, Worcester Polytechnic Institute Richard A. DiDio, La Salle University James Draper, University of Florida James M. Edmondson, Santa Barbara City College John H. Ellison, Grove City College Raymond Fabec, Louisiana State University Donna Farrior, University of Tulsa Robert E. Fennell, Clemson University W. E. Fitzgibbon, University of Houston Harvey J. Fletcher, Brigham Young University Paul J. Gormley, Villanova Layachi Hadji, University of Alabama Ruben Hayrapetyan, Kettering University Terry Herdman, Virginia Polytechnic Institute and State University Zdzislaw Jackiewicz, Arizona State University S. K. Jain, Ohio University Anthony J. John, Southeastern Massachusetts University David C. Johnson, University of Kentucky—Lexington Harry L. Johnson, V.P.I & S.U. Kenneth R. Johnson, North Dakota State University Joseph Kazimir, East Los Angeles College J. Keener, University of Arizona Steve B. Khlief, Tennessee Technological University (retired) C. J. Knickerbocker, Sensis Corporation Carlon A. Krantz, Kean College of New Jersey Thomas G. Kudzma, University of Lowell Alexandra Kurepa, North Carolina A&T State University G. E. Latta, University of Virginia Cecelia Laurie, University of Alabama James R. McKinney, California Polytechnic State University James L. Meek, University of Arkansas Gary H. Meisters, University of Nebraska—Lincoln Stephen J. Merrill, Marquette University Vivien Miller, Mississippi State University Gerald Mueller, Columbus State Community College Philip S. Mulry, Colgate University C. J. Neugebauer, Purdue University Tyre A. Newton, Washington State University Brian M. O’Connor, Tennessee Technological University J. K. Oddson, University of California—Riverside Carol S. O’Dell, Ohio Northern University A. Peressini, University of Illinois, Urbana—Champaign J. Perryman, University of Texas at Arlington Joseph H. Phillips, Sacramento City College Jacek Polewczak, California State University Northridge Nancy J. Poxon, California State University—Sacramento Robert Pruitt, San Jose State University xii ● PREFACE Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. K. Rager, Metropolitan State College F. B. Reis, Northeastern University Brian Rodrigues, California State Polytechnic University Tom Roe, South Dakota State University Kimmo I. Rosenthal, Union College Barbara Shabell, California Polytechnic State University Seenith Sivasundaram, Embry-Riddle Aeronautical University Don E. Soash, Hillsborough Community College F. W. Stallard, Georgia Institute of Technology Gregory Stein, The Cooper Union M. B. Tamburro, Georgia Institute of Technology Patrick Ward, Illinois Central College Jianping Zhu, University of Akron Jan Zijlstra, Middle Tennessee State University Jay Zimmerman, Towson University REVIEWERS OF THE CURRENT EDITIONS Bernard Brooks, Rochester Institute of Technology Allen Brown, Wabash Valley College Helmut Knaust, The University of Texas at El Paso Mulatu Lemma, Savannah State University George Moss, Union University Martin Nakashima, California State Polytechnic University—Pomona Bruce O’Neill, Milwaukee School of Engineering Dennis G. Zill Los Angeles PREFACE ● xiii Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Tenth Edition A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Is AIDS an Invariably Fatal Disease? by Ivan Kramer This essay will address and answer the question: Is the acquired immunodeﬁciency syndrome (AIDS), which is the end stage of the human immunodeﬁciency virus (HIV) infection, an invariably fatal disease? Like other viruses, HIV has no metabolism and cannot reproduce itself outside of a living cell. The genetic information of the virus is contained in two identical strands of RNA. To reproduce, HIV must use the reproductive apparatus of the cell it invades and infects to produce exact copies of the viral RNA. Once it penetrates a cell, HIV transcribes its RNA into DNA using an enzyme (reverse transcriptase) contained in the virus. The double-stranded viral DNAmigrates into the nucleus of the invaded cell and is inserted into the cell’s genome with the aid of another viral enzyme (integrase). The viral DNA and the invaded cell’s DNA are then integrated, and the cell is infected. When the infected cell is stimulated to reproduce, the proviral DNA is transcribed into viral DNA, and new viral particles are synthesized. Since anti-retroviral drugs like zi-dovudine inhibit the HIV enzyme reverse transcriptase and stop proviral DNA chain synthesis in the laboratory, these drugs, usually administered in combination, slow down the progression to AIDS in those that are infected with HIV (hosts). What makes HIV infection so dangerous is the fact that it fatally weakens a host’s immune system by binding to the CD4 molecule on the surface of cells vital for defense against disease, including T-helper cells and a subpopulation of natural killer cells. T-helper cells (CD4 T-cells, or T4 cells) are arguably the most important cells of the immune system since they organize the body’s defense against antigens. Modeling suggests that HIV infection of natural killer cells makes it impossible for even modern antiretroviral therapy to clear the virus . In addition to the CD4 molecule, a virion needs at least one of a handful of co-receptor molecules (e.g., CCR5 and CXCR4) on the surface of the target cell in order to be able to bind to it, pene-trate its membrane, and infect it. Indeed, about 1% of Caucasians lack coreceptor molecules, and, therefore, are completely immune to becoming HIV infected. Once infection is established, the disease enters the acute infection stage, lasting a matter of weeks, followed by an incubation period, which can last two decades or more! Although the T-helper cell density of a host changes quasi-statically during the incubation period, literally billions of infected T4 cells and HIV particles are destroyed—and replaced—daily. This is clearly a war of attrition, one in which the immune system invariably loses. A model analysis of the essential dynamics that occur during the incubation period to invariably cause AIDS is as follows . Because HIV rapidly mutates, its ability to infect T4 cells on contact (its infectivity) eventually increases and the rate T4 cells become infected increases. Thus, the immune system must increase the destruction rate of infected T4 cells as well as the production rate of new, uninfected ones to replace them. There comes a point, however, when the production rate of T4 cells reaches its maximum possible limit and any further increase in HIV’s infectiv-ity must necessarily cause a drop in the T4 density leading to AIDS. Remarkably, about 5% of hosts show no sign of immune system deterioration for the ﬁrst ten years of the infection; these hosts, called long-term nonprogressors, were originally Project for Section 3.1 P-1 Cell infected with HIV Thomas Deerinck, NCMIR/Photo Researchers, Inc. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. thought to be possibly immune to developing AIDS, but modeling evidence suggests that these hosts will also develop AIDS eventually . In over 95% of hosts, the immune system gradually loses its long battle with the virus. The T4 cell density in the peripheral blood of hosts begins to drop from normal levels (between 250 over 2500 cells/mm3) towards zero, signaling the end of the incubation period. The host reaches the AIDS stage of the infection either when one of the more than twenty opportunistic infections characteristic of AIDS develops (clinical AIDS) or when the T4 cell density falls below 250 cells/mm3 (an additional deﬁnition of AIDS promulgated by the CDC in 1987). The HIV infection has now reached its potentially fatal stage. In order to model survivability with AIDS, the time t at which a host develops AIDS will be denoted by t 0. One possible survival model for a cohort of AIDS patients postulates that AIDS is not a fatal condition for a fraction of the cohort, denoted by Si, to be called the immortal fraction here. For the remaining part of the cohort, the probability of dying per unit time at time t will be assumed to be a con-stant k, where, of course, k must be positive. Thus, the survival fraction S(t) for this model is a solution of the linear ﬁrst-order differential equation (1) Using the integrating-factor method discussed in Section 2.3, we see that the solution of equation (1) for the survival fraction is given by (2) Instead of the parameter k appearing in (2), two new parameters can be deﬁned for a host for whom AIDS is fatal: the average survival time Taver given by Taver k1 and the survival half-life T12 given by T12 ln(2)k. The survival half-life, deﬁned as the time required for half of the cohort to die, is completely analogous to the half-life in radioactive nuclear decay. See Problem 8 in Exercise 3.1. In terms of these parameters the entire time-dependence in (2) can be written as (3) Using a least-squares program to ﬁt the survival fraction function in (2) to the actual survival data for the 159 Marylanders who developed AIDS in 1985 produces an immortal fraction value of Si 0.0665 and a survival half life value of T12 0.666 year, with the average survival time being Taver 0.960 years . See Figure 1. Thus only about 10% of Marylanders who developed AIDS in 1985 survived three years with this condition. The 1985 Maryland AIDS survival curve is virtually iden-tical to those of 1983 and 1984. The ﬁrst antiretroviral drug found to be effective against HIV was zidovudine (formerly known as AZT). Since zidovudine was not known to have an impact on the HIV infection before 1985 and was not common ekt et>Taver 2t>T1>2 S(t) Si [1 Si]ekt. dS(t) dt k[S(t) Si]. P-2 ● PROJECTS IS AIDS AN INVARIABLY FATAL DISEASE? 1.0 0.8 0.6 0.4 0.2 0 16 16 48 80 112 144 176 208 240 272 Survival time t(w) S(t) Survival fraction data Two-parameter model fit FIGURE 1 Survival fraction curve S(t). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. therapy before 1987, it is reasonable to conclude that the survival of the 1985 Maryland AIDS patients was not signiﬁcantly inﬂuenced by zidovudine therapy. The small but nonzero value of the immortal fraction Si obtained from the Maryland data is probably an artifact of the method that Maryland and other states use to determine the survivability of their citizens. Residents with AIDS who changed their name and then died or who died abroad would still be counted as alive by the Maryland Department of Health and Mental Hygiene. Thus, the immortal fraction value of Si 0.0665 (6.65%) obtained from the Maryland data is clearly an upper limit to its true value, which is probably zero. Detailed data on the survivability of 1,415 zidovudine-treated HIV-infected hosts whose T4 cell densities dropped below normal values were published by Easterbrook et al. in 1993 . As their T4 cell densities drop towards zero, these peo-ple develop clinical AIDS and begin to die. The longest survivors of this disease live to see their T4 densities fall below 10 cells/mm3. If the time t 0 is redeﬁned to mean the moment the T4 cell density of a host falls below 10 cells/mm3, then the survivability of such hosts was determined by Easterbrook to be 0.470, 0.316, and 0.178 at elapsed times of 1 year, 1.5 years, and 2 years, respectively. A least-squares ﬁt of the survival fraction function in (2) to the Easterbrook data for HIV-infected hosts with T4 cell densities in the 0–10 cells/mm3 range yields a value of the immortal fraction of Si 0 and a survival half-life of T12 0.878 year ; equivalently, the average survival time is Taver 1.27 years. These results clearly show that zidovudine is not effective in halting replication in all strains of HIV, since those who receive this drug eventually die at nearly the same rate as those who do not. In fact, the small difference of 2.5 months between the survival half-life for 1993 hosts with T4 cell densities below 10 cells/mm3 on zidovudine therapy (T12 0.878 year) and that of 1985 infected Marylanders not taking zidovudine (T12 0.666 year) may be entirely due to improved hospitalization and improve-ments in the treatment of the opportunistic infections associated with AIDS over the years. Thus, the initial ability of zidovudine to prolong survivability with HIV dis-ease ultimately wears off, and the infection resumes its progression. Zidovudine therapy has been estimated to extend the survivability of an HIV-infected patient by perhaps 5 or 6 months on the average . Finally, putting the above modeling results for both sets of data together, we ﬁnd that the value of the immortal fraction falls somewhere within the range 0 Si 0.0665 and the average survival time falls within the range 0.960 years Taver 1.27 years. Thus, the percentage of people for whom AIDS is not a fatal disease is less than 6.65% and may be zero. These results agree with a 1989 study of hemophilia-associated AIDS cases in the USA which found that the median length of survival after AIDS diagno-sis was 11.7 months . A more recent and comprehensive study of hemophiliacs with clinical AIDS using the model in (2) found that the immortal fraction was Si 0, and the mean survival times for those between 16 to 69 years of age varied be-tween 3 to 30 months, depending on the AIDS-deﬁning condition . Although bone marrow transplants using donor stem cells homozygous for CCR5 delta32 deletion may lead to cures, to date clinical results consistently show that AIDS is an invariably fatal disease. Related Problems 1. Suppose the fraction of a cohort of AIDS patients that survives a time t after AIDS diagnosis is given by S(t) exp(kt). Show that the average survival time Taver after AIDS diagnosis for a member of this cohort is given by Taver 1k. 2. The fraction of a cohort of AIDS patients that survives a time t after AIDS diagnosis is given by S(t) exp(kt). Suppose the mean survival for a cohort of hemophiliacs diagnosed with AIDS before 1986 was found to be Taver 6.4 months. What fraction of the cohort survived 5 years after AIDS diagnosis? PROJECTS IS AIDS AN INVARIABLY FATAL DISEASE? ● P-3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3. The fraction of a cohort of AIDS patients that survives a time t after AIDS diag-nosis is given by S(t) exp(kt). The time it takes for S(t) to reach the value of 0.5 is deﬁned as the survival half-life and denoted by T12. (a) Show that S(t) can be written in the form . (b) Show that T12 Taver ln(2), where Taver is the average survival time deﬁned in problem (1). Thus, it is always true that T12 Taver. 4. About 10% of lung cancer patients are cured of the disease, i.e., they survive 5 years after diagnosis with no evidence that the cancer has returned. Only 14% of lung cancer patients survive 5 years after diagnosis. Assume that the fraction of incurable lung cancer patients that survives a time t after diagnosis is given by exp(kt). Find an expression for the fraction S(t) of lung cancer patients that survive a time t after being diagnosed with the disease. Be sure to determine the values of all of the constants in your answer. What fraction of lung cancer patients survives two years with the disease? References 1. Kramer, Ivan. What triggers transient AIDS in the acute phase of HIV infection and chronic AIDS at the end of the incubation period? Computational and Mathematical Methods in Medicine, Vol. 8, No. 2, June 2007: 125–151. 2. Kramer, Ivan. Is AIDS an invariable fatal disease?: A model analysis of AIDS survival curves. Mathematical and Computer Modelling 15, no. 9, 1991: 1–19. 3. Easterbrook, Philippa J., Emani Javad, Moyle, Graham, Gazzard, Brian G. Progressive CD4 cell depletion and death in zidovudine-treated patients. JAIDS, Aug. 6, 1993, No. 8: 927–929. 4. Kramer, Ivan. The impact of zidovudine (AZT) therapy on the survivability of those with progressive HIV infection. Mathematical and Computer Modelling, Vol. 23, No. 3, Feb. 1996: 1–14. 5. Stehr-Green, J. K., Holman, R. C., Mahoney, M. A. Survival analysis of hemophilia-associated AIDS cases in the US. Am J Public Health, Jul. 1989, 79 (7): 832–835. 6. Gail, Mitchel H., Tan, Wai-Yuan, Pee, David, Goedert, James J. Survival after AIDS diagnosis in a cohort of hemophilia patients. JAIDS, Aug. 15, 1997, Vol. 15, No. 5: 363–369. ABOUT THE AUTHOR Ivan Kramer earned a BS in Physics and Mathematics from The City College of New York in 1961 and a PhD from the University of California at Berkeley in theo-retical particle physics in 1967. He is currently associate professor of physics at the University of Maryland, Baltimore County. Dr. Kramer was Project Director for AIDS/HIV Case Projections for Maryland, for which he received a grant from the AIDS Administration of the Maryland Department of Health and Hygiene in 1990. In addition to his many published articles on HIV infection and AIDS, his current research interests include mutation models of cancers, Alzheimers disease, and schizophrenia. S(t) 2t>T1>2 P-4 ● PROJECTS IS AIDS AN INVARIABLY FATAL DISEASE? Courtesy of Ivan Kramer Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The Allee Effect by Jo Gascoigne The top ﬁve most famous Belgians apparently include a cyclist, a punk singer, the in-ventor of the saxophone, the creator of Tintin, and Audrey Hepburn. Pierre François Verhulst is not on the list, although he should be. He had a fairly short life, dying at the age of 45, but did manage to include some excitement—he was deported from Rome for trying to persuade the Pope that the Papal States needed a written constitu-tion. Perhaps the Pope knew better even then than to take lectures in good gover-nance from a Belgian. . . . Aside from this episode, Pierre Verhulst (1804–1849) was a mathematician who concerned himself, among other things, with the dynamics of natural populations— ﬁsh, rabbits, buttercups, bacteria, or whatever. (I am prejudiced in favour of ﬁsh, so we will be thinking ﬁsh from now on.) Theorizing on the growth of natural popula-tions had up to this point been relatively limited, although scientists had reached the obvious conclusion that the growth rate of a population (dNdt, where N(t) is the population size at time t) depended on (i) the birth rate b and (ii) the mortality rate m, both of which would vary in direct proportion to the size of the population N: (1) After combining b and m into one parameter r, called the intrinsic rate of natural increase—or more usually by biologists without the time to get their tongues around that, just r—equation (1) becomes (2) This model of population growth has a problem, which should be clear to you—if not, plot dNdt for increasing values of N. It is a straightforward exponential growth curve, suggesting that we will all eventually be drowning in ﬁsh. Clearly, something eventually has to step in and slow down dNdt. Pierre Verhulst’s insight was that this something was the capacity of the environment, in other words, How many fish can an ecosystem actually suppor ? He formulated a differential equation for the population N(t) that included both r and the carrying capacity K: (3) Equation (3) is called the logistic equation, and it forms to this day the basis of much of the modern science of population dynamics. Hopefully, it is clear that the term (1 NK), which is Verhulst’s contribution to equation (2), is (1 NK) 1 when N 0, leading to exponential growth, and (1 NK) : 0 as N : K, hence it causes the growth curve of N(t) to approach the horizontal asymptote N(t) K. Thus the size of the population cannot exceed the carrying capacity of the environment. dN dt rN1 N K, r 0. dN dt rN. dN dt bN mN. P-5 Project for Section 3.2 Dr Jo with Queenie; Queenie is on the left Courtesy of Jo Gasoigne Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. P-6 ● PROJECTS THE ALLEE EFFECT The logistic equation (3) gives the overall growth rate of the population, but the ecology is easier to conceptualize if we consider per capita growth rate—that is, the growth rate of the population per the number of individuals in the population—some measure of how “well” each individual in the population is doing. To get per capita growth rate, we just divide each side of equation (3) by N: This second version of (3) immediately shows (or plot it) that this relationship is a straight line with a maximum value of (assuming that negative popu-lation sizes are not relevant) and dNdt 0 at N K. Er, hang on a minute . . . “a maximum value of ” Each shark in the population does best when there are . . . zero sharks? Here is clearly a ﬂaw in the logistic model. (Note that it is now a model—when it just presents a relationship be-tween two variables dNdt and N, it is just an equation. When we use this equation to try and analyze how populations might work, it becomes a model.) The assumption behind the logistic model is that as population size decreases, indi-viduals do better (as measured by the per capita population growth rate). This assump-tion to some extent underlies all our ideas about sustainable management of natural resources—a ﬁsh population cannot be ﬁshed indeﬁnitely unless we assume that when a population is reduced in size, it has the ability to grow back to where it was before. This assumption is more or less reasonable for populations, like many ﬁsh pop-ulations subject to commercial ﬁsheries, which are maintained at 50% or even 20% of K. But for very depleted or endangered populations, the idea that individuals keep doing better as the population gets smaller is a risky one. The Grand Banks popula-tion of cod, which was ﬁshed down to 1% or perhaps even 0.1% of K, has been pro-tected since the early 1990s, and has yet to show convincing signs of recovery. Warder Clyde Allee (1885–1955) was an American ecologist at the University of Chicago in the early 20th century, who experimented on goldﬁsh, brittlestars, ﬂour beetles, and, in fact, almost anything unlucky enough to cross his path. Allee showed that, in fact, individuals in a population can do worse when the population becomes very small or very sparse. There are numerous ecological reasons why this might be—for example, they may not ﬁnd a suitable mate or may need large groups to ﬁnd food or express social behavior, or in the case of goldﬁsh they may alter the water chemistry in their favour. As a result of Allee’s work, a population where the per capita growth rate declines at low population size is said to show an Allee effect. The jury is still out on whether Grand Banks cod are suffering from an Allee effect, but there are some possible mechanisms—females may not be able to ﬁnd a mate, or a mate of the right size, or maybe the adult cod used to eat the ﬁsh that eat the juvenile cod. On the other hand, there is nothing that an adult cod likes more than a snack of baby cod—they are not ﬁsh with very picky eating habits—so these arguments may not stack up. For the moment we know very little except that there are still no cod. Allee effects can be modelled in many ways. One of the simplest mathematical models, a variation of the logistic equation, is: (4) where A is called the Allee threshold. The value N (t) A is the population size below which the population growth rate becomes negative due to an Allee effect—situated at dN dt rN1 N K N A 1. 1 N dN dt at N 0?! 1 N dN dt at N 0 1 N dN dt r1 N K r r K N. Population size and population density are mathematically interchangeable, assuming a ﬁxed area in which the population lives (although they may not necessarily be interchangeable for the individuals in question). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. PROJECTS THE ALLEE EFFECT ● P-7 a value of N somewhere between N 0 and N K, that is, 0 A K, depending on the species (but for most species a good bit closer to 0 than K, luckily). Equation (4) is not as straightforward to solve for N(t) as (3), but we don’t need to solve it to gain some insights into its dynamics. If you work through Problems 2 and 3, you will see that the consequences of equation (4) can be disastrous for endan-gered populations. Related Problems 1. (a) The logistic equation (3) can be solved explicitly for N(t) using the technique of partial fractions. Do this, and plot N(t) as a function of t for 0 t 10. Appropriate values for r, K, and N(0) are r 1, K 1, N(0) 0.01 (ﬁsh per cubic metre of seawater, say). The graph of N(t) is called a sigmoid growth curve. (b) The value of r can tell us a lot about the ecology of a species—sardines, where females mature in less than one year and have millions of eggs, have a high r, while sharks, where females bear a few live young each year, have a low r. Play with r and see how it affects the shape of the curve. Question: If a marine protected area is put in place to stop overﬁshing, which species will recover quickest—sardines or sharks? 2. Find the population equilibria for the model in (4). [Hint: The population is at equilibrium when dNdt 0, that is, the population is neither growing nor shrinking. You should ﬁnd three values of N for which the population is at equi-librium.] 3. Population equilibria can be stable or unstable. If, when a population deviates a bit from the equilibrium value (as populations inevitably do), it tends to return to it, this is a stable equilibrium; if, however, when the population deviates from the equilibrium it tends to diverge from it ever further, this is an unstable equi-librium. Think of a ball in the pocket of a snooker table versus a ball balanced on a snooker cue. Unstable equilibria are a feature of Allee effect models such as (4). Use a phase portrait of the autonomous equation (4) to determine whether the nonzero equilibria that you found in Problem 2 are stable or unstable. [Hint: See Section 2.1 of the text.] 4. Discuss the consequences of the result above for a population N(t) ﬂuctuating close to the Allee threshold A. References 1. Courchamp, F., Berec L., and Gascoigne, J. 2008. Allee Effects in Ecology and Conservation. Oxford University Press. 2. Hastings, A. 1997. Population Biology—Concepts and Models. Springer-Verlag, New York. ABOUT THE AUTHOR After a degree in Zoology, Jo Gascoigne thought her ﬁrst job, on conservation in East Africa, would be about lions and elephants—but it turned out to be about ﬁsh. Despite the initial crushing disappointment, she ended up loving them—so much, in fact, that she went on to complete a PhD in marine conservation biology at the College of William and Mary, in Williamsburg, Virginia, where she studied lobster and Caribbean conch, and also spent 10 days living underwater in the Aquarius habitat in Florida. After graduating, she returned to her native Britain and studied the mathemat-ics of mussel beds at Bangor University in Wales, before becoming an independent consultant on ﬁsheries management. She now works to promote environmentally sus-tainable ﬁsheries. When you buy seafood, make good choices and help the sea! Copper sharks and bronze whaler sharks feeding on a bait ball of sardines off the east coast of South Africa Courtesy of Jo Gascoigne Doug Perrine/Getty Images Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. P-8 ● PROJECTS THE ALLEE EFFECT Project for Section 3.3 Wolf Population Dynamics by C.J. Knickerbocker Early in 1995, after much controversy, public debate, and a 70-year absence, gray wolves were re introduced into Yellowstone National Park and Central Idaho. During this 70-year absence, signiﬁcant changes were recorded in the populations of other predator and prey animals residing in the park. For instance, the elk and coyote pop-ulations had risen in the absence of inﬂuence from the larger gray wolf. With the reintroduction of the wolf in 1995, we anticipated changes in both the predator and prey animal populations in the Yellowstone Park ecosystem as the success of the wolf population is dependent upon how it inﬂuences and is inﬂuenced by the other species in the ecosystem. For this study, we will examine how the elk (prey) population has been inﬂu-enced by the wolves (predator). Recent studies have shown that the elk population has been negatively impacted by the reintroduction of the wolves. The elk population fell from approximately 18,000 in 1995 to approximately 7,000 in 2009. This article asks the question of whether the wolves could have such an effect and, if so, could the elk population disappear? Let’s begin with a more detailed look at the changes in the elk population inde-pendent of the wolves. In the 10 years prior to the introduction of wolves, from 1985 to 1995, one study suggested that the elk population increased by 40% from 13,000 in 1985 to 18,000 in 1995. Using the simplest differential equation model for popu-lation dynamics, we can determine the growth rate for elks (represented by the vari-able r) prior to the reintroduction of the wolves. (1) In this equation, E(t) represents the elk population (in thousands) where t is measured in years since 1985. The solution, which is left as an exercise for the reader, ﬁnds the combined birth/death growth rate r to be approximately 0.0325 yielding: In 1995, 21 wolves were initially released, and their numbers have risen. In 2007, biologists estimated the number of wolves to be approximately 171. To study the interaction between the elk and wolf populations, let’s consider the following predator-prey model for the interaction between the elk and wolf within the Yellowstone ecosystem: (2) where E(t) is the elk population and W(t) is the wolf population. All populations are measured in thousands of animals. The variable t represents time measured in years from 1995. So, from the initial conditions, we have 18,000 elk and 21 wolves in the year 1995. The reader will notice that we estimated the growth rate for the elk to be the same as that estimated above r 0.0325. E(0) 18.0, W(0) 0.021 dW dt 0.6W 0.05EW dE dt 0.0325E 0.8EW E(t) 13.0 e0.0325t E(10) 18.0 E(0) 13.0, dE dt rE, A gray wolf in the wild P-8 Damien Richard/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Before we attempt to solve the model (2), a qualitative analysis of the system can yield a number of interesting properties of the solutions. The first equation shows that the growth rate of the elk is positively impacted by the size of the herd (0.0325E). This can be interpreted as the probability of breeding in-creases with the number of elk. On the other hand the nonlinear term (0.8EW) has a negative impact on the growth rate of the elk since it measures the interaction between predator and prey. The second equation shows that the wolf population has a negative effect on its own growth which can be interpreted as more wolves create more competition for food. But, the interac-tion between the elk and wolves (0.05EW) has a positive impact since the wolves are finding more food. Since an analytical solution cannot be found to the initial-value problem (2), we need to rely on technology to ﬁnd approximate solutions. For example, below is a set of instructions for ﬁnding a numerical solution of the initial-value problem using the computer algebra system MAPLE. e1 := diff(e(t),t)- 0.0325 e(t) + 0.8 e(t)w(t) : e2 := diff(w(t),t)+ 0.6 w(t) - 0.05 e(t)w(t) : sys := {e1,e2} : ic := {e(0)=18.0,w(0)=0.021} : ivp := sys union ic : H:= dsolve(ivp,{e(t),w(t)},numeric) : The graphs in Figures 1 and 2 show the populations for both species between 1995 and 2009. As predicted by numerous studies, the reintroduction of wolves into Yellowstone had led to a decline in the elk population. In this model, we see the popula-tion decline from 18,000 in 1995 to approximately 7,000 in 2009. In contrast, the wolf population rose from an initial count of 21 in 1995 to a high of approximately 180 in 2004. dW>dt 0.6W 0.05EW (dE>dt) The alert reader will note that the model also shows a decline in the wolf popu-lation after 2004. How might we interpret this? With the decline in the elk population over the ﬁrst 10 years, there was less food for the wolves and therefore their popula-tion begins to decline. Figure 3 below shows the long-term behavior of both populations. The interpre-tation of this graph is left as an exercise for the reader. Information on the reintroduction of wolves into Yellowstone Park and central Idaho can be found on the Internet. For example, read the U.S. Fish and Wildlife Service news release of November 23, 1994, on the release of wolves into Yellowstone National Park. 20000 18000 16000 14000 12000 10000 8000 6000 4000 2000 1995 1997 1999 2001 Year Elk population 2003 2005 2007 2009 0 200 180 160 140 120 100 80 60 40 20 1995 1997 1999 2001 Year Wolf population 2003 2005 2007 2009 0 PROJECTS WOLF POPULATION DYNAMICS ● P-9 FIGURE 1 Elk population FIGURE 2 Wolf population Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. P-10 ● PROJECTS WOLF POPULATION DYNAMICS Related Problems 1. Solve the pre-wolf initial-value problem (1) by ﬁrst solving the differential equation and applying the initial condition. Then apply the terminal condition to ﬁnd the growth rate. 2. Biologists have debated whether the decrease in the elk from 18,000 in 1995 to 7,000 in 2009 is due to the reintroduction of wolves. What other factors might account for the decrease in the elk population? 3. Consider the long-term changes in the elk and wolf populations. Are these cyclic changes reasonable? Why is there a lag between the time when the elk begins to decline and the wolf population begins to decline? Are the minimum values for the wolf population realistic? Plot the elk population versus the wolf population and interpret the results. 4. What does the initial-value problem (1) tell us about the growth of the elk pop-ulation without the inﬂuence of the wolves? Find a similar model for the intro-duction of rabbits into Australia in 1859 and the impact of introducing a prey population into an environment without a natural predator population. 20000 18000 16000 14000 12000 10000 8000 6000 4000 2000 0 200 180 160 140 120 100 80 60 40 20 0 1990 2000 2010 2020 2030 2040 2050 2060 2070 2080 Year Elk and Wolf Populations Elk Wolf ABOUT THE AUTHOR C. J. Knickerbocker Professor of Mathematics and Computer Science (retired) St. Lawrence University Principal Research Engineer Sensis Corporation C. J. Knickerbocker received his PhD in mathematics from Clarkson University in 1984. Until 2008 he was a professor of mathematics and computer science at St. Lawrence University, where he authored numerous articles in a variety of topics, including nonlinear partial differential equations, graph theory, applied physics, and psychology. He has also served as a consultant for publishers, software companies, and government agencies. Currently, Dr. Knickerbocker is a principal research engi-neer for the Sensis Corporation, where he studies airport safety and efﬁciency. FIGURE 3 Long-term behavior of the populations Courtesy of C. J. Knickerbocker Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Bungee Jumping by Kevin Cooper Suppose that you have no sense. Suppose that you are standing on a bridge above the Malad River canyon. Suppose that you plan to jump off that bridge. You have no sui-cide wish. Instead, you plan to attach a bungee cord to your feet, to dive gracefully into the void, and to be pulled back gently by the cord before you hit the river that is 174 feet below. You have brought several different cords with which to afﬁx your feet, including several standard bungee cords, a climbing rope, and a steel cable. You need to choose the stiffness and length of the cord so as to avoid the unpleasantness associated with an unexpected water landing. You are undaunted by this task, because you know math! Each of the cords you have brought will be tied off so as to be 100 feet long when hanging from the bridge. Call the position at the bottom of the cord 0, and measure the position of your feet below that “natural length” as x(t), where x increases as you go down and is a function of time t. See Figure 1. Then, at the time you jump, x(0) = -100, while if your six-foot frame hits the water head ﬁrst, at that time x(t) = 174 - 100 - 6 = 68. Notice that distance increases as you fall, and so your velocity is positive as you fall and negative when you bounce back up. Note also that you plan to dive so your head will be six feet below the end of the chord when it stops you. You know that the acceleration due to gravity is a constant, called g, so that the force pulling downwards on your body is mg. You know that when you leap from the bridge, air resistance will increase proportionally to your speed, providing a force in the opposite direction to your motion of about bv, where b is a constant and v is your velocity. Finally, you know that Hooke’s law describing the action of springs says that the bungee cord will eventually exert a force on you proportional to its distance past its natural length. Thus, you know that the force of the cord pulling you back from destruction may be expressed as The number k is called the spring constant, and it is where the stiffness of the cord you use inﬂuences the equation. For example, if you used the steel cable, then k would be very large, giving a tremendous stopping force very suddenly as you passed the natural length of the cable. This could lead to discomfort, injury, or even a Darwin award. You want to choose the cord with a k value large enough to stop you above or just touching the water, but not too suddenly. Consequently, you are inter-ested in ﬁnding the distance you fall below the natural length of the cord as a func-tion of the spring constant. To do that, you must solve the differential equation that we have derived in words above: The force mx on your body is given by mx mg + b(x) - bx . Here mg is your weight, 160 lb., and x is the rate of change of your position below the equilibrium with respect to time; i.e., your velocity. The constant b for air resis-tance depends on a number of things, including whether you wear your skin-tight pink spandex or your skater shorts and XXL T-shirt, but you know that the value today is about 1.0. b(x) 0 kx x 0 x 0 P-11 Project for Section 5.1 100 ft Bridge Bungee 174 ft x = −100 x = 0 x = 74 Water FIGURE 1 The bungee setup Bungee jumping from a bridge Sean Nel/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. This is a nonlinear differential equation, but inside it are two linear differential equations, struggling to get out. We will work with such equations more extensively in later chapters, but we already know how to solve such equations from our past experience. When x 0, the equation is mx = mg - bx , while after you pass the natural length of the cord it is mx = mg - kx - bx . We will solve these separately, and then piece the solutions together when x(t) = 0. In Problem 1 you ﬁnd an expression for your position t seconds after you step off the bridge, before the bungee cord starts to pull you back. Notice that it does not depend on the value for k, because the bungee cord is just falling with you when you are above x(t) = 0. When you pass the natural length of the bungee cord, it does start to pull back, so the differential equation changes. Let t1 denote the ﬁrst time for which x(t1) = 0, and let v1 denote your speed at that time. We can thus describe the motion for x(t) 0 using the problem x = g - kx - bx , x(t1) = 0, x (t1) = v1. An illustration of a solution to this problem in phase space can be seen in Figure 2. This will yield an expression for your position as the cord is pulling on you. All we have to do is to ﬁnd out the time t2 when you stop going down. When you stop going down, your velocity is zero, i.e., x (t2) = 0. As you can see, knowing a little bit of math is a dangerous thing. We remind you that the assumption that the drag due to air resistance is linear applies only for low speeds. By the time you swoop past the natural length of the cord, that approx-imation is only wishful thinking, so your actual mileage may vary. Moreover, springs behave nonlinearly in large oscillations, so Hooke’s law is only an approx-imation. Do not trust your life to an approximation made by a man who has been dead for 200 years. Leave bungee jumping to the professionals. Related Problems 1. Solve the equation mx + bx = mg for x(t), given that you step off the bridge—no jumping, no diving! Stepping off means x(0) = -100, x (0) = 0. You may use mg = 160, b = 1, and g = 32. 2. Use the solution from Problem 1 to compute the length of time t1 that you freefall (the time it takes to go the natural length of the cord: 100 feet). 3. Compute the derivative of the solution you found in Problem 1 and evaluate it at the time you found in Problem 2. Call the result v1. You have found your down-ward speed when you pass the point where the cord starts to pull. 4. Solve the initial-value problem For now, you may use the value k = 14, but eventually you will need to replace that with the actual values for the cords you brought. The solution x(t) repre-sents the position of your feet below the natural length of the cord after it starts to pull back. 5. Compute the derivative of the expression you found in Problem 4 and solve for the value of t where it is zero. This time is t2. Be careful that the time you compute is greater than t1—there are several times when your motion stops at the top and bottom of your bounces! After you ﬁnd t2, substitute it back into the solution you found in Problem 4 to ﬁnd your lowest position. 6. You have brought a soft bungee cord with k = 8.5, a stiffer cord with k = 10.7, and a climbing rope for which k = 16.4. Which, if any, of these may you use safely under the conditions given? 7. You have a bungee cord for which you have not determined the spring constant. To do so, you suspend a weight of 10 lb. from the end of the 100-foot cord, caus-ing the cord to stretch 1.2 feet. What is the k value for this cord? You may neglect the mass of the cord itself. x (t1) v1. x(t1) 0, mx bx kx mg, FIGURE 2 An example plot of x(t) against x (t) for a bungee jump 60 40 20 20 40 _20 _20 0 _40 _60 _80 _100 _40 x(t) x'(t) P-12 ● PROJECTS BUNGEE JUMPING Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. PROJECTS BUNGEE JUMPING ● P-13 ABOUT THE AUTHOR Kevin Cooper, PhD, Colorado State University, is the Computing Coordinator for Mathematics at Washington State University, Pullman, Washington. His main inter-est is numerical analysis, and he has written papers and one textbook in that area. Dr. Cooper also devotes considerable time to creating mathematical software compo-nents, such as DynaSys, a program to analyze dynamical systems numerically. Courtesy of Kevin Cooper Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The Collapse of the Tacoma Narrows Suspension Bridge by Gilbert N. Lewis In the summer of 1940, the Tacoma Narrows Suspension Bridge in the State of Washington was completed and opened to trafﬁc. Almost immediately, observers no-ticed that the wind blowing across the roadway would sometimes set up large verti-cal vibrations in the roadbed. The bridge became a tourist attraction as people came to watch, and perhaps ride, the undulating bridge. Finally, on November 7, 1940, dur-ing a powerful storm, the oscillations increased beyond any previously observed, and the bridge was evacuated. Soon, the vertical oscillations became rotational, as ob-served by looking down the roadway. The entire span was eventually shaken apart by the large vibrations, and the bridge collapsed. Figure 1 shows a picture of the bridge during the collapse. See and for interesting and sometimes humorous anec-dotes associated with the bridge. Or, do an Internet search with the key words “Tacoma Bridge Disaster” in order to ﬁnd and view some interesting videos of the collapse of the bridge. The noted engineer von Karman was asked to determine the cause of the col-lapse. He and his coauthors claimed that the wind blowing perpendicularly across the roadway separated into vortices (wind swirls) alternately above and below the roadbed, thereby setting up a periodic, vertical force acting on the bridge. It was this force that caused the oscillations. Others further hypothesized that the frequency of this forcing function exactly matched the natural frequency of the bridge, thus lead-ing to resonance, large oscillations, and destruction. For almost ﬁfty years, resonance was blamed as the cause of the collapse of the bridge, although the von Karman group denied this, stating that “it is very improbable that resonance with alternating vortices plays an important role in the oscillations of suspension bridges” . As we can see from equation (31) in Section 5.1.3, resonance is a linear phe-nomenon. In addition, for resonance to occur, there must be an exact match between the frequency of the forcing function and the natural frequency of the bridge. Furthermore, there must be absolutely no damping in the system. It should not be surprising, then, that resonance was not the culprit in the collapse. If resonance did not cause the collapse of the bridge, what did? Recent research provides an alternative explanation for the collapse of the Tacoma Narrows Bridge. Lazer and McKenna contend that nonlinear effects, and not linear resonance, were the main factors leading to the large oscillations of the bridge (see for a good review article). The theory involves partial differential equations. However, a simpli-ﬁed model leading to a nonlinear ordinary differential equation can be constructed. The development of the model below is not exactly the same as that of Lazer and McKenna, but it results in a similar differential equation. This example shows an-other way that amplitudes of oscillation can increase. Consider a single vertical cable of the suspension bridge. We assume that it acts like a spring, but with different characteristics in tension and compression, and with no damping. When stretched, the cable acts like a spring with Hooke’s constant, b, while, when compressed, it acts like a spring with a different Hooke’s constant, a. We assume that the cable in compression exerts a smaller force on the roadway than when stretched the same distance, so that 0 a b. Let the vertical deﬂection (positive direction downward) of the slice of the roadbed attached to this cable be The rebuilt Tacoma Narrows bridge (1950) and new parallel bridge (2009) Collapse of the Tacoma Narrows Bridge Project for Section 5.3 AP Photo P-14 Sir Armstrong/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. denoted by y(t), where t represents time, and y 0 represents the equilibrium posi-tion of the road. As the roadbed oscillates under the inﬂuence of an applied vertical force (due to the von Karman vortices), the cable provides an upward restoring force equal to by when y 0 and a downward restoring force equal to ay when y 0. This change in the Hooke’s Law constant at y 0 provides the nonlinearity to the differ-ential equation. We are thus led to consider the differential equation derived from Newton’s second law of motion my f(y) g(t), where f(y) is the nonlinear function given by g(t) is the applied force, and m is the mass of the section of the roadway. Note that the differential equation is linear on any interval on which y does not change sign. Now, let us see what a typical solution of this problem would look like. We will assume that m 1 kg, b 4 N/m, a 1N/m, and g(t) sin(4t) N. Note that the fre-quency of the forcing function is larger than the natural frequencies of the cable in both tension and compression, so that we do not expect resonance to occur. We also assign the following initial values to y: y(0) 0, y (0) 0.01, so that the roadbed starts in the equilibrium position with a small downward velocity. Because of the downward initial velocity and the positive applied force, y(t) will initially increase and become positive. Therefore, we ﬁrst solve this initial-value problem y 4y sin(4t), y(0) 0, y (0) 0.01. (1) The solution of the equation in (1), according to Theorem 4.1.6, is the sum of the complementary solution, yc(t), and the particular solution, yp(t). It is easy to see that yc(t) c1cos(2t) c2sin(2t) (equation (9), Section 4.3), and yp(t) (Table 4.4.1, Section 4.4). Thus, y(t) c1cos(2t) c2 sin(2t) . (2) The initial conditions give y(0) 0 c1, y (0) 0.01 2c2 , so that c2 (0.01 )2. Therefore, (2) becomes (3) We note that the ﬁrst positive value of t for which y(t) is again equal to zero is . At that point, Therefore, equation (3) holds on [0, ]. After becomes negative, so we must now solve the new problem (4) Proceeding as above, the solution of (4) is (5) cost0.01 2 5 4 15 sint cos(2t). y(t) 0.01 2 5cost 1 15sin(4t) y y sin(4t), y p 2 0, y p 2 0.01 2 3. t p 2, y p>2 y (p 2) (0.01 2 3). t p 2 sin(2t) 1 20.01 1 3 1 6 cos(2t). y(t) 1 20.01 1 3sin(2t) 1 12sin(4t) 1 3 1 3 1 12sin(4t) 1 12sin(4t) f(y) by if y 0 ay if y 0 , PROJECTS THE COLLAPSE OF THE TACOMA NARROWS SUSPENSION BRIDGE ● P-15 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. P-16 ● PROJECTS THE COLLAPSE OF THE TACOMA NARROWS SUSPENSION BRIDGE The next positive value of t after at which y(t) 0 is at which point so that equation (5) holds on . At this point, the solution has gone through one cycle in the time interval During this cycle, the section of the roadway started at the equilibrium with positive velocity, became positive, came back to the equilibrium position with negative ve-locity, became negative, and ﬁnally returned to the equilibrium position with positive velocity. This pattern continues indeﬁnitely, with each cycle covering time units. The solution for the next cycle is (6) It is instructive to note that the velocity at the beginning of the second cycle is (0.01 ), while at the beginning of the third cycle it is (0.01 ). In fact, the velocity at the beginning of each cycle is greater than at the beginning of the pre-vious cycle. It is not surprising then that the amplitude of oscillations will increase over time, since the amplitude of (one term in) the solution during any one cycle is directly related to the velocity at the beginning of the cycle. See Figure 2 for a graph of the deflection function on the interval [0, 3p]. Note that the maximum deﬂection on [3p2, 2p] is larger than the maximum deﬂection on [0, p2], while the maximum deﬂection on [2p, 3p] is larger than the maximum deﬂection on [p2, 3p2]. It must be remembered that the model presented here is a very simpliﬁed one-dimensional model that cannot take into account all of the intricate interactions of real bridges. The reader is referred to the account by Lazer and McKenna for a more complete model. More recently, McKenna has reﬁned that model to provide a different viewpoint of the torsional oscillations observed in the Tacoma Bridge. Research on the behavior of bridges under forces continues. It is likely that the models will be refined over time, and new insights will be gained from the research. However, it should be clear at this point that the large oscillations caus-ing the destruction of the Tacoma Narrows Suspension Bridge were not the result of resonance. 2 15 4 15 2 15 y(t) sint 0.01 8 15 4 15 cost cos(2t) on [2p, 3p]. y(t) sin(2t) 1 20.01 7 15 1 6 cos(2t) on [3p>2, 2p], 3p 2 [0, 3p 2 ]. [p>2, 3p>2] y (3p 2 ) 0.01 2 15, t 3p 2 , t p 2 0.2 y t 0.0 2 4 6 8 −0.2 −0.4 −0.6 FIGURE 2 Graph of deﬂection function y(t) Related Problems 1. Solve the following problems and plot the solutions for 0 t 6p. Note that reso-nance occurs in the ﬁrst problem but not in the second. (a) (b) y y cos(2t), y(0) 0, y (0) 0. y y cost, y(0) 0, y (0) 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. PROJECTS THE COLLAPSE OF THE TACOMA NARROWS SUSPENSION BRIDGE ● P-17 2. Solve the initial-value problem where and (a) b 1, a 4, (Compare your answer with the example in this project.) (b) b 64, a 4, (c) b 36, a 25. Note that, in part (a), the condition b a of the text is not satisﬁed. Plot the solu-tions. What happens in each case as t increases? What would happen in each case if the second initial condition were replaced with y (0) 0.01? Can you make any conclusions similar to those of the text regarding the long-term solution? 3. What would be the effect of adding damping (cy , where c 0) to the system? How could a bridge design engineer incorporate more damping into the bridge? Solve the problem where and (a) c 0.01 (b) c 0.1 (c) c 0.5 References 1. Lewis, G. N., “Tacoma Narrows Suspension Bridge Collapse” in A First Course in Differential Equations, Dennis G. Zill, 253–256. Boston: PWS-Kent, 1993. 2. Braun, M., Differential Equations and Their Applications, 167–169. New York: Springer-Verlag, 1978. 3. Amman, O. H., T. von Karman, and G. B. Woodruff, The Failure of the Tacoma Narrows Bridge. Washington D.C.: Federal Works Agency, 1941. 4. Lazer, A. C., and P. J. McKenna. Large amplitude periodic oscillations in sus-pension bridges: Some new connections with nonlinear analysis. SIAM Review 32 (December 1990): 537–578. 5. Peterson, I., Rock and roll bridge. Science News 137 (1991): 344–346. 6. McKenna, P. J., Large torsional oscillations in suspension bridges revisited: Fixing an old approximation. American Mathematical Monthly 106 (1999):1–18. f(y) 4y if y 0 y if y 0 , y cy f(y) sin(4t), y(0) 0, y (0) 1, f(y) by if y 0 ay if y 0 , y f(y) sin(4t), y(0) 0, y (0) 1, ABOUT THE AUTHOR Dr. Gilbert N. Lewis is professor emeritus at Michigan Technological University, where he has taught and done research in Applied Math and Differential Equations for 34 years. He received his BS degree from Brown University and his MS and PhD degrees from the University of Wisconsin-Milwaukee. His hobbies include travel, food and wine, ﬁshing, and birding, activities that he intends to continue in retirement. Courtesy of Gilbert N. Lewis Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Murder at the Mayfair Diner by Tom LoFaro Dawn at the Mayfair Diner. The amber glow of streetlights mixed with the violent red ﬂash of police cruisers begins to fade with the rising of a furnace orange sun. Detective Daphne Marlow exits the diner holding a steaming cup of hot joe in one hand and a summary of the crime scene evidence in the other. Taking a seat on the bumper of her tan LTD, Detective Marlow begins to review the evidence. At 5:30 a.m. the body of one Joe D. Wood was found in the walk in refrigerator in the diner’s basement. At 6:00 a.m. the coroner arrived and determined that the core body temperature of the corpse was 85 degrees Fahrenheit. Thirty minutes later the coroner again measured the core body temperature. This time the reading was 84 degrees Fahrenheit. The thermostat inside the refrigerator reads 50 degrees Fahrenheit. Daphne takes out a fading yellow legal pad and ketchup-stained calculator from the front seat of her cruiser and begins to compute. She knows that Newton’s Law of Cooling says that the rate at which an object cools is proportional to the difference between the temperature T of the body at time t and the temperature Tm of the envi-ronment surrounding the body. She jots down the equation (1) where k is a constant of proportionality, T and Tm are measured in degrees Fahrenheit, and t is time measured in hours. Because Daphne wants to investigate the past using positive values of time, she decides to correspond t 0 with 6:00 a.m., and so, for example, t 4 is 2:00 a.m. After a few scratches on her yellow pad, Daphne realizes that with this time convention the constant k in (1) will turn out to be positive. She jots a reminder to herself that 6:30 a.m. is now t 12. As the cool and quiet dawn gives way to the steamy midsummer morning, Daphne begins to sweat and wonders aloud, “But what if the corpse was moved into the fridge in a feeble attempt to hide the body? How does this change my estimate?” She re-enters the restaurant and ﬁnds the grease-streaked thermostat above the empty cash register. It reads 70 degrees Fahrenheit. “But when was the body moved?” Daphne asks. She decides to leave this ques-tion unanswered for now, simply letting h denote the number of hours the body has been in the refrigerator prior to 6:00 a.m. For example, if h 6, then the body was moved at midnight. Daphne ﬂips a page on her legal pad and begins calculating. As the rapidly cooling coffee begins to do its work, she realizes that the way to model the environmental tem-perature change caused by the move is with the unit step function (t). She writes Tm(t) 50 20(t h) (2) and below it the differential equation (3) Daphne’s mustard-stained polyester blouse begins to drip sweat under the blaze of a midmorning sun. Drained from the heat and the mental exercise, she ﬁres up her cruiser and motors to Boodle’s Café for another cup of java and a heaping plate dT dt k(T – Tm(t)). dT dt k(T Tm), t 0, Project for Section 7.3 P-18 The Mayfair diner in Philadelphia, PA © Ronald C. Saari Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. of scrapple and fried eggs. She settles into the faux leather booth. The intense air-conditioning conspires with her sweat-soaked blouse to raise goose ﬂesh on her rapidly cooling skin. The intense chill serves as a gruesome reminder of the tragedy that occurred earlier at the Mayfair. While Daphne waits for her breakfast, she retrieves her legal pad and quickly reviews her calculations. She then carefully constructs a table that relates refrigeration time h to time of death while eating her scrapple and eggs. Shoving away the empty platter, Daphne picks up her cell phone to check in with her partner Marie. “Any suspects?” Daphne asks. “Yeah,” she replies, “we got three of ’em. The ﬁrst is the late Mr. Wood’s ex-wife, a dancer by the name of Twinkles. She was seen in the Mayfair between 5 and 6 p.m. in a shouting match with Wood.” “When did she leave?” “A witness says she left in a hurry a little after six. The second suspect is a South Philly bookie who goes by the name of Slim. Slim was in around 10 last night having a whispered conversation with Joe. Nobody overheard the conversation, but witnesses say there was a lot of hand gesturing, like Slim was upset or something.” “Did anyone see him leave?” “Yeah. He left quietly around 11. The third suspect is the cook.” “The cook?” “Yep, the cook. Goes by the name of Shorty. The cashier says he heard Joe and Shorty arguing over the proper way to present a plate of veal scaloppine. She said that Shorty took an unusually long break at 10:30 p.m. He took off in a huff when the restaurant closed at 2:00 a.m. Guess that explains why the place was such a mess.” “Great work, partner. I think I know who to bring in for questioning.” Related Problems 1. Solve equation (1), which models the scenario in which Joe Wood is killed in the refrigerator. Use this solution to estimate the time of death (recall that normal liv-ing body temperature is 98.6 degrees Fahrenheit). 2. Solve the differential equation (3) using Laplace transforms. Your solution T(t) will depend on both t and h. (Use the value of k found in Problem 1.) 3. (CAS) Complete Daphne’s table. In particular, explain why large values of h give the same time of death. h time body moved time of death 12 6:00 p.m. 11 10 9 8 7 6 5 4 3 2 4. Who does Daphne want to question and why? 5. Still Curious? The process of temperature change in a dead body is known as algor mortis (rigor mortis is the process of body stiffening), and although it is not PROJECTS MURDER AT THE MAYFAIR DINER ● P-19 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. perfectly described by Newton’s Law of Cooling, this topic is covered in most forensic medicine texts. In reality, the cooling of a dead body is determined by more than just Newton’s Law. In particular, chemical processes in the body con-tinue for several hours after death. These chemical processes generate heat, and thus a near constant body temperature may be maintained during this time before the exponential decay due to Newton’s Law of Cooling begins. A linear equation, known as the Glaister equation, is sometimes used to give a preliminary estimate of the time t since death. The Glaister equation is (4) where T0 is measured body temperature (98.4 F is used here for normal living body temperature instead of 98.6 F). Although we do not have all of the tools to derive this equation exactly (the 1.5 degrees per hour was determined experimen-tally), we can derive a similar equation via linear approximation. Use equation (1) with an initial condition of T(0) T0 to compute the equa-tion of the tangent line to the solution through the point (0, T0). Do not use the values of Tm or k found in Problem 1. Simply leave these as parameters. Next, let T 98.4 and solve for t to get (5) ABOUT THE AUTHOR Tom LoFaro is a professor and chair of the Mathematics and Computer Science Department at Gustavus Adolphus College in St. Peter, Minnesota. He has been involved in developing differential modeling projects for over 10 years, including being a princi-pal investigator of the NSF-funded IDEA project ( and a contributor to CODEE's ODE Architect (Wiley and Sons). Dr. LoFaro’s nonacade-mic interests include ﬂy ﬁshing and coaching little league soccer. His oldest daughter (age 12) aspires to be a forensic anthropologist much like Detective Daphne Marlow. t 98.4 T0 k(T0 Tm). t 98.4 T0 1.5 P-20 ● PROJECTS MURDER AT THE MAYFAIR DINER Courtesy of Tom LoFaro Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Earthquake Shaking of Multistory Buildings by Gilbert N. Lewis Large earthquakes typically have a devastating effect on buildings. For example, the famous 1906 San Francisco earthquake destroyed much of that city. More re-cently, that area was hit by the Loma Prieta earthquake that many people in the United States and elsewhere experienced second-hand while watching on televi-sion the Major League Baseball World Series game that was taking place in San Francisco in 1989. In this project, we attempt to model the effect of an earthquake on a multi-story building and then solve and interpret the mathematics. Let xi represent the horizontal displacement of the ith ﬂoor from equlibrium. Here, the equilibrium position will be a ﬁxed point on the ground, so that x0 0. During an earthquake, the ground moves horizontally so that each ﬂoor is considered to be displaced relative to the ground. We assume that the ith ﬂoor of the building has a mass mi, and that successive ﬂoors are connected by an elastic connector whose effect resembles that of a spring. Typically, the structural elements in large buildings are made of steel, a highly elastic material. Each such connector supplies a restoring force when the ﬂoors are displaced relative to each other. We assume that Hooke’s Law holds, with propor-tionality constant ki between the ith and the (i 1)st ﬂoors. That is, the restoring force between those two ﬂoors is F ki(xi1 xi), where xi1 xi is the displacement (shift) of the (i 1)st ﬂoor relative to the ith ﬂoor. We also assume a similar reaction between the ﬁrst ﬂoor and the ground, with pro-portionality constant k0 . Figure 1 shows a model of the building, while Figure 2 shows the forces acting on the ith ﬂoor. mn mn1 m2 m1 ground kn1 kn2 k1 k0 FIGURE 1 Floors of building mi1 mi mi1 ki(xi1 xi) FIGURE 2 Forces on ith ﬂoor ki1(xi xil) Project for Section 8.2 Collapsed apartment building in San Francisco, October 18, 1989, the day after the massive Loma Prieta earthquake P-21 s76/ZUMA Press/Newscom Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. We can apply Newton’s second law of motion (Section 5.1), F ma, to each ﬂoor of the building to arrive at the following system of linear differential equations. As a simple example, consider a two-story building with each ﬂoor having mass m 5000 kg and each restoring force constant having a value of k 10000 kg/s2. Then the differential equations are The solution by the methods of Section 8.2 is where and Now suppose that the following initial conditions are applied: (0) 0. These correspond to a building in the equilibrium position with the ﬁrst ﬂoor being given a horizontal speed of 0.2 m/s. The solution of the initial value problem is where See Figures 3 and 4 for graphs of x1(t) and x2(t). Note that initially x1 moves to the right but is slowed by the drag of x2, while x2 is initially at rest, but accelerates, due to the pull of x1, to over-take x1 within one second. It continues to the right, eventually pulling x1 along until the two-second mark. At that point, the drag of x1 has slowed x2 to a stop, after which x2 moves left, passing the equilibrium point at 3.2 seconds and continues moving left, draging x1 along with it. This back-and-forth motion continues. There is no damping in the system, so that the oscillatory behavior continues forever. c2 4 v2 20.1>[v2 1 v2 2v1] 0.0317 c4. x2(t) 4 v2 1c2 sin v1t 4 v2 2c4 sin v2t, x1(t) 2c2 sin v1t 2c4 sin v2t, x2 x1(0) 0, x1 (0) 0.2, x2(0) 0, v2 23 15 0.874. v1 23 15 2.288, 4 v2 2c4 sin v2t, x2(t) 4 v2 1c1 cos v1t 4 v2 1c2 sin v1t 4 v2 2c3 cos v2t x1(t) 2c1 cos v1t 2c2 sin v1t 2c3 cos v2t 2c4 sin v2t, d2x2 dt2 2x1 2x2. d2x1 dt2 4x1 2x2 m1 d 2x1 dt 2 k0x1 k1(x2 x1) m2 d 2x2 dt 2 k1(x2 x1) k2(x3 x2) o o mn d 2xn dt2 kn1(xn xn1). P-22 ● PROJECTS EARTHQUAKE SHAKING OF MULTISTORY BUILDINGS 0.1 0.2 t 1 2 3 4 5 −0.1 x2(t) 0.05 0.10 x1(t) t 1 2 3 4 5 −0.05 −0.10 FIGURE 3 Graph of x1(t) FIGURE 4 Graph of x2(t) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. If a horizontal oscillatory force of frequency or is applied, we have a sit-uation analogous to resonance discussed in Section 5.1.3. In that case, large oscil-lations of the building would be expected to occur, possibly causing great damage if the earthquake lasted an appreciable length of time. Let’s deﬁne the following matrices and vector: Then the system of differential equations can be written in matrix form Note that the matrix M is a diagonal matrix with the mass of the ith ﬂoor being the ith diagonal element. Matrix M has an inverse given by . We can therefore represent the matrix differential equation by Where the matrix M is called the mass matrix, and the matrix K is the stiffness matrix. The eigenvalues of the matrix A reveal the stability of the building during an earth-quake. The eigenvalues of A are negative and distinct. In the ﬁrst example, the eigen-values are and The natural frequencies of the building are the square roots of the negatives of the eigenvalues. If is the ith eigen-value, then is the ith frequency, for i 1, 2, . . . , n. During an earth-quake, a large horizontal force is applied to the ﬁrst ﬂoor. If this is oscillatory in nature, say of the form F(t) G cosgt, then large displacements may develop in the building, especially if the frequency g of the forcing term is close to one of the natural frequencies of the building. This is reminiscent of the resonance phenomenon studied in Section 5.1.3. vi 1li li 3 15 5.236. 3 15 0.764 A M1K, X (M1K)X or X AX. M1 m1 1 0 o 0 0 m1 2 0 0 0 0 . . . . . . . . . 0 0 o m1 n M d2X dt2 KX or MX KX. X(t) x1(t) x2(t) o xn(t) K (k0 k1) k1 0 o 0 0 k1 (k1 k2) k2 0 0 0 k2 (k2 k3) 0 0 0 0 k3 0 0 . . . . . . . . . . . . . . . 0 0 0 kn2 0 0 0 0 (kn2 kn1) kn1 0 0 0 o kn1 kn1 M m1 0 o 1 0 m2 0 0 0 0 . . . . . . . . . 0 0 o mn , v2 v1 PROJECTS EARTHQUAKE SHAKING OF MULTISTORY BUILDINGS ● P-23 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. P-24 ● PROJECTS EARTHQUAKE SHAKING OF MULTISTORY BUILDINGS As another example, suppose we have a 10-story building, where each ﬂoor has a mass 10000 kg, and each ki value is 5000 kg/s2. Then The eigenvalues of A are found easily using Mathematica or another similar computer package. These values are 1.956, 1.826, 1.623, 1.365, 1.075, 0.777, 0.5, 0.267, 0.099, and 0.011, with corresponding frequencies 1.399, 1.351, 1.274, 1.168, 1.037, 0.881, 0.707, 0.517, 0.315, and 0.105 and periods of oscillation (2p/v) 4.491, 4.651, 4.932, 5.379, 6.059, 7.132, 8.887, 12.153, 19.947, and 59.840. During a typical earthquake whose period might be in the range of 2 to 3 seconds, this building does not seem to be in any danger of developing resonance. However, if the k values were 10 times as large (multiply A by 10), then, for example, the sixth period would be 2.253 seconds, while the ﬁfth through seventh are all on the order of 2–3 seconds. Such a building is more likely to suffer damage in a typical earthquake of period 2–3 seconds. Related Problems 1. Consider a three-story building with the same m and k values as in the ﬁrst exam-ple. Write down the corresponding system of differential equations. What are the matrices M, K, and A? Find the eigenvalues for A. What range of frequencies of an earthquake would place the building in danger of destruction? 2. Consider a three-story building with the same m and k values as in the second example. Write down the corresponding system of differential equations. What are the matrices M, K, and A? Find the eigenvalues for A. What range of fre-quencies of an earthquake would place the building in danger of destruction? 3. Consider the tallest building on your campus. Assume reasonable values for the mass of each ﬂoor and for the proportionality constants between ﬂoors. If you have trouble coming up with such values, use the ones in the example problems. Find the matrices M, K, and A, and ﬁnd the eigenvalues of A and the frequen-cies and periods of oscillation. Is your building safe from a modest-sized period-2 earthquake? What if you multiplied the matrix K by 10 (that is, made the building stiffer)? What would you have to multiply the matrix K by in order to put your building in the danger zone? 4. Solve the earthquake problem for the three-story building of Problem 1: , where F(t) = G cosgt, G = EB, B = [1 0 0]T, E = 10,000 lbs is the amplitude of the earthquake force acting at ground level, and g = 3 is the frequency of the earthquake (a typical earthquake frequency). See Section 8.3 for the method of solving nonhomogeneous matrix differential equations. Use initial conditions for a building at rest. MX KX F(t) A M1K 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 1 0.5 0 0 0 0 0 0 0 0 0.5 0.5 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Modeling Arms Races by Michael Olinick The last hundred years have seen numerous dangerous, destabilizing, and expensive arms races. The outbreak of World War I climaxed a rapid buildup of armaments among rival European powers. There was a similar mutual accumulation of conven-tional arms just prior to World War II. The United States and the Soviet Union en-gaged in a costly nuclear arms race during the forty years of the Cold War. Stockpiling of ever-more deadly weapons is common today in many parts of the world, including the Middle East, the Indian subcontinent, and the Korean peninsula. British meteorologist and educator Lewis F. Richardson (1881–1953) developed several mathematical models to analyze the dynamics of arms races, the evolution over time of the process of interaction between countries in their acquisition of weapons. Arms race models generally assume that each nation adjusts its accumula-tion of weapons in some manner dependent on the size of its own stockpile and the armament levels of the other nations. Richardson’s primary model of a two country arms race is based on mutual fear: A nation is spurred to increase its arms stockpile at a rate proportional to the level of armament expenditures of its rival. Richardson’s model takes into account internal constraints within a nation that slow down arms buildups: The more a nation is spending on arms, the harder it is to make greater increases, because it becomes increasingly difﬁcult to divert society’s resources from basic needs such as food and housing to weapons. Richardson also built into his model other factors driving or slowing down an arms race that are independent of levels of arms expen-ditures. The mathematical structure of this model is a linked system of two ﬁrst-order linear differential equations. If x and y represent the amount of wealth being spent on arms by two nations at time t, then the model has the form where a, b, m, and n are positive constants while r and s are constants which can be positive or negative. The constants a and b measure mutual fear; the constants m and n represent proportionality factors for the “internal brakes” to further arms increases. Positive values for r and s correspond to underlying factors of ill will or distrust that would persist even if arms expenditures dropped to zero. Negative values for r and s indicate a contribution based on goodwill. The dynamic behavior of this system of differential equations depends on the relative sizes of ab and mn together with the signs of r and s. Although the model is a relatively simple one, it allows us to consider several different long-term out-comes. It’s possible that two nations might move simultaneously toward mutual disarmament, with x and y each approaching zero. A vicious cycle of unbounded increases in x and y is another possible scenario. A third eventuality is that the arms expenditures asymptotically approach a stable point (x, y) regardless of the initial level of arms expenditures. In other cases, the eventual outcome depends on the starting point. Figure 1 shows one possible situation with four different initial dy dt bx ny s dx dt ay mx r P-25 Project for Section 8.3 Weapons and ammunition recovered during military operations against Taliban militants in South Waziristan in October 2009 NICOLAS ASFOURI/AFP/Getty Images/Newscom Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. levels, each of which leads to a “stable outcome,” the intersection of the nullclines dxdt 0 and dydt 0. Although “real world” arms races seldom match exactly with Richardson’s model, his pioneering work has led to many fruitful applications of differential equation models to problems in international relations and political science. As two leading researchers in the ﬁeld note in , “The Richardson arms race model constitutes one of the most impor-tant models of arms race phenomena and, at the same time, one of the most inﬂuential formal models in all of the international relations literature.” Arms races are not limited to the interaction of nation states. They can take place between a government and a paramilitary terrorist group within its borders as, for ex-ample, the Tamil Tigers in Sri Lanka, the Shining Path in Peru, or the Taliban in Afghanistan. Arms phenomena have also been observed between rival urban gangs and between law enforcement agencies and organized crime. The “arms” need not even be weapons. Colleges have engaged in “amenities arms races,” often spending millions of dollars on more luxurious dormitories, state- of-the-art athletic facilities, epicurean dining options, and the like, to be more competitive in attracting student applications. Biologists have identiﬁed the possi-bility of evolutionary arms races between and within species as an adaptation in one lineage may change the selection pressure on another lineage, giving rise to a counter-adaptation. Most generally, the assumptions represented in a Richardson-type model also characterize many competitions in which each side perceives a need to stay ahead of the other in some mutually important measure. Related Problems 1. (a) By substituting the proposed solutions into the differential equations, show that the solution of the particular Richardson arms model dy dt 2x 4y 8 dx dt y 3x 3 P-26 ● PROJECTS MODELING ARMS RACES dx/dt = 0 y x dy/dt = 0 1 1 0 2 3 4 5 6 2 3 4 5 6 FIGURE 1 Expenditures approaching a stable point Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. with initial condition x(0) 12, y(0) 15 is What is the long-term behavior of this arms race? (b) For the Richardson arms race model (a) with arbitrary initial conditions x(0) A, y(0) B, show that the solution is given by x(t) Ce5t De2t 2 C (A B 1)3 where y(t) 2Ce5t De2t 3 D (2A B 7)3 Show that this result implies that the qualitative long-term behavior of such an arms race is the same (x(t) : 2, y(t) : 3), no matter what the initial values of x and y are. 2. The qualitative long-term behavior of a Richardson arms race model can, in some cases, depend on the initial conditions. Consider, for example, the system For each of the given initial conditions below, verify that the proposed solu-tion works and discuss the long-term behavior: (a) x(0) 1, y(0) 1 : x(t) 10 9et, y(t) 10 9et (b) x(0) 1, y(0) 22 : x(t) 10 9e6t, y(t) 10 12e6t (c) x(0) 1, y(0) 29 : x(t) 12e6t 3et 10, y(t) 16e6t 3et 10 (d) x(0) 10, y(0) 10 : x(t) 10, y(t) 10 for all t 3. (a) As a possible alternative to the Richardson model, consider a stock adjustment model for an arms race. The assumption here is that each country sets a desired level of arms expenditures for itself and then changes its weapons stock pro-portionally to the gap between its current level and the desired one. Show that this assumption can be represented by the system of differential equations where x and y are desired constant levels and a, b are positive constants. How will x and y evolve over time under such a model? (b) Generalize the stock adjustment model of (a) to a more realistic one where the desired level for each country depends on the levels of both countries. In particular, suppose x has the form x c dy where c and d are positive constants and that y has a similar format. Show that, under these assump-tions, the stock adjustment model is equivalent to a Richardson model. 4. Extend the Richardson model to three nations, deriving a system of linear differen-tial equations if the three are mutually fearful: each one is spurred to arm by the ex-penditures of the other two. How might the equations change if two of the nations are close allies not threatened by the arms buildup of each other, but fearful of the armaments of the third. Investigate the long-term behavior of such arms races. 5. In the real world, an unbounded runaway arms race is impossible since there is an absolute limit to the amount any country can spend on weapons; e.g. gross na-tional product minus some amount for survival. Modify the Richardson model to incorporate this idea and analyze the dynamics of an arms race governed by these new differential equations. dx dt b(y y) dx dt a(x x) dy dt 4x 3y 10 dx dt 3y 2x 10 y(t) 32 3 e2t 4 3e5t 3 x(t) 32 3 e2t 2 3e5t 2 PROJECTS MODELING ARMS RACES ● P-27 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. References 1. Richardson, Lewis F., Arms and Insecurity: A Mathematical Study of the Causes and Origins of War. Pittsburgh: Boxwood Press, 1960. 2. Olinick, Michael, An Introduction to Mathematical Models in the Social and Life Sciences. Reading, MA: Addison-Wesley, 1978. 3. Intriligator, Michael D., and Dagobert L. Brito, “Richardsonian Arms Race Models” in Manus I. Midlarsky, ed., Handbook of War Studies. Boston: Unwin Hyman, 1989. P-28 ● PROJECTS MODELING ARMS RACES ABOUT THE AUTHOR After earning a BA in mathematics and philosophy at the University of Michigan and an MA and PhD from the University of Wisconsin (Madison), Michael Olinick moved from the Midwest to New England where he joined the Middlebury College faculty in 1970 and now serves as Professor of Mathematics. Dr. Olinick has held visiting positions at University College Nairobi, University of California at Berkeley, Wesleyan University, and Lancaster University in Great Britain. He is the author or co-author of a number of books on single and multivariable calculus, mathematical modeling, probability, topology, and principles and practice of mathematics. He is currently developing a new textbook on mathematical models in the humanities, social, and life sciences. Courtesy of Michael Olinick Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1 1.1 Deﬁnitions and Terminology 1.2 Initial-Value Problems 1.3 Differential Equations as Mathematical Models Chapter 1 in Review The words differential and equations certainly suggest solving some kind of equation that contains derivatives y, y, . . . . Analogous to a course in algebra and trigonometry, in which a good amount of time is spent solving equations such as x2 5x 4 0 for the unknown number x, in this course one of our tasks will be to solve differential equations such as y 2y y 0 for an unknown function y (x). The preceding paragraph tells something, but not the complete story, about the course you are about to begin. As the course unfolds, you will see that there is more to the study of differential equations than just mastering methods that mathematicians over past centuries devised to solve them. But ﬁrst things ﬁrst. In order to read, study, and be conversant in a specialized subject, you have to master some of the terminology of that discipline. This is the thrust of the ﬁrst two sections of this chapter. In the last section we brieﬂy examine the link between differential equations and the real world. Practical questions such as How fast does a disease spread ? How fast does a population change? involve rates of change or derivatives. And so the mathematical description—or mathematical model—of phenomena, experiments, observations, or theories may be a differential equation. 1 Introduction to Differential Equations Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS A Definitio The equation that we made up in (1) is called a differential equation. Before proceeding any further, let us consider a more precise deﬁnition of this concept. DEFINITION 1.1.1 Differential Equation An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE). To talk about them, we shall classify differential equations according to type, order, and linearity. Classification by Type If a differential equation contains only ordinary de-rivatives of one or more unknown functions with respect to a single independent variable, it is said to be an ordinary differential equation (ODE). An equation in-volving partial derivatives of one or more unknown functions of two or more inde-pendent variables is called a partial differential equation (PDE). Our ﬁrst example illustrates several of each type of differential equation. EXAMPLE 1 Types of Differential Equations (a) The equations an ODE can contain more than one unknown function (2) are examples of ordinary differential equations. dy dx 5y ex, d2y dx2 dy dx 6y 0, and dx dt dy dt 2x y b b DEFINITIONS AND TERMINOLOGY REVIEW MATERIAL ●The deﬁnition of the derivative ●Rules of differentiation ●Derivative as a rate of change ●Connection between the ﬁrst derivative and increasing/decreasing ●Connection between the second derivative and concavity INTRODUCTION The derivative dydx of a function y (x) is itself another function (x) found by an appropriate rule. The exponential function is differentiable on the interval (, ), and, by the Chain Rule, its ﬁrst derivative is . If we replace on the right-hand side of the last equation by the symbol y, the derivative becomes . (1) Now imagine that a friend of yours simply hands you equation (1)—you have no idea how it was constructed—and asks, What is the function represented by the symbol y? You are now face to face with one of the basic problems in this course: How do you solve such an equation for the function y (x)? dy dx 0.2xy e0.1x2 dy>dx 0.2xe0.1x2 y e0.1x2 1.1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (b) The following equations are partial differential equations: (3) Notice in the third equation that there are two unknown functions and two indepen-dent variables in the PDE. This means u and v must be functions of two or more independent variables. Notation Throughout this text ordinary derivatives will be written by using 2u x2 2u y2 0, 2u x2 2u t2 2u t, u y v x. 1.1 DEFINITIONS AND TERMINOLOGY ● 3 Except for this introductory section, only ordinary differential equations are considered in A First Course in Differential Equations with Modeling Applications, Tenth Edition. In that text the word equation and the abbreviation DE refer only to ODEs. Partial differential equations or PDEs are considered in the expanded volume Differential Equations with Boundary-Value Problems, Eighth Edition. either the Leibniz notation dydx, d2ydx2, d3ydx3, . . . or the prime notation y, y, y , . . . . By using the latter notation, the ﬁrst two differential equations in (2) can be written a little more compactly as y 5y ex and y y 6y 0. Actually, the prime notation is used to denote only the ﬁrst three derivatives; the fourth derivative is written y(4) instead of y. In general, the nth derivative of y is written dnydxn or y(n). Although less convenient to write and to typeset, the Leibniz notation has an ad-vantage over the prime notation in that it clearly displays both the dependent and independent variables. For example, in the equation it is immediately seen that the symbol x now represents a dependent variable, whereas the independent variable is t. You should also be aware that in physical sciences and engineering, Newton’s dot notation (derogatorily referred to by some as the “ﬂyspeck” notation) is sometimes used to denote derivatives with respect to time t. Thus the differential equation d2sdt2 32 becomes ¨ s 32. Partial derivatives are often denoted by a subscript notation indicating the indepen-dent variables. For example, with the subscript notation the second equation in (3) becomes uxx utt 2ut. Classification by Order The order of a differential equation (either ODE or PDE) is the order of the highest derivative in the equation. For example, is a second-order ordinary differential equation. In Example 1, the ﬁrst and third equations in (2) are ﬁrst-order ODEs, whereas in (3) the ﬁrst two equations are second-order PDEs. First-order ordinary differential equations are occasionally writ-ten in differential form M(x, y) dx N(x, y) dy 0. For example, if we assume that y denotes the dependent variable in (y x) dx 4x dy 0, then y dydx, so by dividing by the differential dx, we get the alternative form 4xy y x. In symbols we can express an nth-order ordinary differential equation in one dependent variable by the general form , (4) where F is a real-valued function of n 2 variables: x, y, y, . . . , y(n). For both prac-tical and theoretical reasons we shall also make the assumption hereafter that it is possible to solve an ordinary differential equation in the form (4) uniquely for the F(x, y, y, . . . , y(n)) 0 first order second order 5( ) 3 4y e x dy ––– dx d 2y –––– dx2 d 2x ––– dt2 16x 0 unknown function or dependent variable independent variable Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. highest derivative y(n) in terms of the remaining n 1 variables. The differential equation , (5) where f is a real-valued continuous function, is referred to as the normal form of (4). Thus when it suits our purposes, we shall use the normal forms to represent general ﬁrst- and second-order ordinary differential equations. For example, the normal form of the ﬁrst-order equation 4xy y x is y (x y)4x; the normal form of the second-order equation y y 6y 0 is y y 6y. See (iv) in the Remarks. Classification by Linearity An nth-order ordinary differential equation (4) dy dx f (x, y) and d 2y dx2 f (x, y, y) d ny dxn f (x, y, y, . . . , y(n1)) 4 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS EXAMPLE 2 Linear and Nonlinear ODEs (a) The equations are, in turn, linear ﬁrst-, second-, and third-order ordinary differential equations. We have just demonstrated that the ﬁrst equation is linear in the variable y by writing it in the alternative form 4xy y x. (b) The equations are examples of nonlinear ﬁrst-, second-, and fourth-order ordinary differential equa-tions, respectively. nonlinear term: coefﬁcient depends on y nonlinear term: nonlinear function of y nonlinear term: power not 1 (1 y)y 2y ex, sin y 0, and d 2y –––– dx2 y 2 0 d 4y –––– dx 4 (y x)dx 4xy dy 0, y 2y y 0, x3 d 3y dx3 x dy dx 5y ex is said to be linear if F is linear in y, y, . . . , y(n). This means that an nth-order ODE is linear when (4) is an(x)y(n) an1(x)y(n1) a1(x)y a0(x)y g(x) 0 or . (6) Two important special cases of (6) are linear ﬁrst-order (n 1) and linear second-order (n 2) DEs: . (7) In the additive combination on the left-hand side of equation (6) we see that the char-acteristic two properties of a linear ODE are as follows: • The dependent variable y and all its derivatives y, y, . . . , y(n) are of the ﬁrst degree, that is, the power of each term involving y is 1. • The coefﬁcients a0, a1, . . . , an of y, y, . . . , y(n) depend at most on the independent variable x. A nonlinear ordinary differential equation is simply one that is not linear. Nonlinear functions of the dependent variable or its derivatives, such as sin y or , cannot appear in a linear equation. ey a1(x) dy dx a0(x)y g(x) and a2(x) d 2y dx2 a1(x) dy dx a0(x)y g(x) an(x) d ny dxn an1(x) d n1y dxn1 a1(x) dy dx a0(x)y g(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Solutions As was stated before, one of the goals in this course is to solve, or ﬁnd solutions of, differential equations. In the next deﬁnition we consider the con-cept of a solution of an ordinary differential equation. 1.1 DEFINITIONS AND TERMINOLOGY ● 5 DEFINITION 1.1.2 Solution of an ODE Any function , deﬁned on an interval I and possessing at least n derivatives that are continuous on I, which when substituted into an nth-order ordinary dif-ferential equation reduces the equation to an identity, is said to be a solution of the equation on the interval. In other words, a solution of an nth-order ordinary differential equation (4) is a func-tion that possesses at least n derivatives and for which We say that satisfie the differential equation on I. For our purposes we shall also assume that a solution is a real-valued function. In our introductory discussion we saw that is a solution of dydx 0.2xy on the interval (, ). Occasionally, it will be convenient to denote a solution by the alternative symbol y(x). Interval of Definitio You cannot think solution of an ordinary differential equation without simultaneously thinking interval. The interval I in Deﬁnition 1.1.2 is variously called the interval of definition the interval of existence, the interval of validity, or the domain of the solution and can be an open interval (a, b), a closed interval [a, b], an inﬁnite interval (a, ), and so on. y e0.1x2 F(x, (x), (x), . . . , (n)(x)) 0 for all x in I. EXAMPLE 3 Verification of a Solutio Verify that the indicated function is a solution of the given differential equation on the interval (, ). (a) (b) SOLUTION One way of verifying that the given function is a solution is to see, after substituting, whether each side of the equation is the same for every x in the interval. (a) From we see that each side of the equation is the same for every real number x. Note that is, by deﬁnition, the nonnegative square root of . (b) From the derivatives y xex ex and y xex 2ex we have, for every real number x, right-hand side: 0. left-hand side: y 2y y (xex 2ex) 2(xex ex) xex 0, 1 16 x4 y1/2 1 4 x2 right-hand side: xy1/2 x 1 16 x4 1/2 x 1 4 x2 1 4 x3, left-hand side: dy dx 1 16 (4 x3) 1 4 x3, y 2y y 0; y xex dy>dx xy1/2; y 1 16 x4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Note, too, that in Example 3 each differential equation possesses the constant so-lution y 0, x . A solution of a differential equation that is identically zero on an interval I is said to be a trivial solution. Solution Curve The graph of a solution of an ODE is called a solution curve. Since is a differentiable function, it is continuous on its interval I of deﬁni-tion. Thus there may be a difference between the graph of the function and the graph of the solution . Put another way, the domain of the function need not be the same as the interval I of deﬁnition (or domain) of the solution . Example 4 illustrates the difference. 6 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 1 x y 1 (a) function y 1/x, x 0 (b) solution y 1/x, (0, ) 1 x y 1 FIGURE 1.1.1 In Example 4 the function y 1x is not the same as the solution y 1x EXAMPLE 4 Function versus Solution The domain of y 1x, considered simply as a function, is the set of all real numbers x except 0. When we graph y 1x, we plot points in the xy-plane cor-responding to a judicious sampling of numbers taken from its domain. The ratio-nal function y 1x is discontinuous at 0, and its graph, in a neighborhood of the origin, is given in Figure 1.1.1(a). The function y 1x is not differentiable at x 0, since the y-axis (whose equation is x 0) is a vertical asymptote of the graph. Now y 1x is also a solution of the linear ﬁrst-order differential equation xy y 0. (Verify.) But when we say that y 1x is a solution of this DE, we mean that it is a function deﬁned on an interval I on which it is differentiable and satisﬁes the equation. In other words, y 1x is a solution of the DE on any inter-val that does not contain 0, such as (3, 1), , (, 0), or (0, ). Because the solution curves deﬁned by y 1x for 3 x 1 and are sim-ply segments, or pieces, of the solution curves deﬁned by y 1x for x 0 and 0 x , respectively, it makes sense to take the interval I to be as large as possible. Thus we take I to be either (, 0) or (0, ). The solution curve on (0, ) is shown in Figure 1.1.1(b). Explicit and Implicit Solutions You should be familiar with the terms explicit functions and implicit functions from your study of calculus. A solution in which the dependent variable is expressed solely in terms of the independent variable and constants is said to be an explicit solution. For our purposes, let us think of an explicit solution as an explicit formula y (x) that we can manipulate, evaluate, and differentiate using the standard rules. We have just seen in the last two examples that , y xex, and y 1x are, in turn, explicit solutions of dydx xy1/2, y 2y y 0, and xy y 0. Moreover, the trivial solu-tion y 0 is an explicit solution of all three equations. When we get down to the business of actually solving some ordinary differential equations, you will see that methods of solution do not always lead directly to an explicit solution y (x). This is particularly true when we attempt to solve nonlinear ﬁrst-order differential equations. Often we have to be content with a relation or expression G(x, y) 0 that deﬁnes a solution implicitly. y 1 16 x4 1 2 x 10 (1 2, 10) DEFINITION 1.1.3 Implicit Solution of an ODE A relation G(x, y) 0 is said to be an implicit solution of an ordinary differential equation (4) on an interval I, provided that there exists at least one function that satisﬁes the relation as well as the differential equation on I. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. It is beyond the scope of this course to investigate the conditions under which a relation G(x, y) 0 deﬁnes a differentiable function . So we shall assume that if the formal implementation of a method of solution leads to a relation G(x, y) 0, then there exists at least one function that satisﬁes both the relation (that is, G(x, (x)) 0) and the differential equation on an interval I. If the implicit solution G(x, y) 0 is fairly simple, we may be able to solve for y in terms of x and obtain one or more explicit solutions. See (i) in the Remarks. 1.1 DEFINITIONS AND TERMINOLOGY ● 7 y x 5 5 y x 5 5 y x 5 5 −5 (a) implicit solution x2 y2 25 (b) explicit solution y1 25 x2, 5 x 5 (c) explicit solution y2 25 x2, 5 x 5 (a) FIGURE 1.1.2 An implicit solution and two explicit solutions of (8) in Example 5 FIGURE 1.1.3 Some solutions of DE in part (a) of Example 6 y x c>0 c<0 c=0 EXAMPLE 5 Verification of an Implicit Solutio The relation x2 y2 25 is an implicit solution of the differential equation (8) on the open interval (5, 5). By implicit differentiation we obtain . Solving the last equation for the symbol dydx gives (8). Moreover, solving x2 y2 25 for y in terms of x yields . The two functions and satisfy the relation (that is, x2 1 2 25 and x2 2 2 25) and are explicit solutions deﬁned on the interval (5, 5). The solution curves given in Figures 1.1.2(b) and 1.1.2(c) are segments of the graph of the implicit solution in Figure 1.1.2(a). Any relation of the form x2 y2 c 0 formally satisﬁes (8) for any constant c. However, it is understood that the relation should always make sense in the real number system; thus, for example, if c 25, we cannot say that x2 y2 25 0 is an implicit solution of the equation. (Why not?) Because the distinction between an explicit solution and an implicit solution should be intuitively clear, we will not belabor the issue by always saying, “Here is an explicit (implicit) solution.” Families of Solutions The study of differential equations is similar to that of integral calculus. In some texts a solution is sometimes referred to as an integral of the equation, and its graph is called an integral curve. When evaluating an anti-derivative or indeﬁnite integral in calculus, we use a single constant c of integration. Analogously, when solving a ﬁrst-order differential equation F(x, y, y) 0, we usually obtain a solution containing a single arbitrary constant or parameter c. A solution containing an arbitrary constant represents a set G(x, y, c) 0 of solutions called a one-parameter family of solutions. When solving an nth-order differential equation F(x, y, y, . . . , y(n)) 0, we seek an n-parameter family of solutions G(x, y, c1, c2, . . . , cn) 0. This means that a single differential equation can possess an infinite number of solution corresponding to the unlimited number of choices for the parameter(s). A solution of a differential equation that is free of arbitrary parameters is called a particular solution. y 2(x) 125 x2 y 1(x) 125 x2 y 225 x2 d dx x2 d dx y2 d dx 25 or 2x 2y dy dx 0 dy dx x y EXAMPLE 6 Particular Solutions (a) The one-parameter family is an explicit solution of the linear ﬁrst-order equation on the interval (, ). (Verify.) Figure 1.1.3 shows the graphs of some particular solutions in this family for various choices of c. The solution y x cos x, the blue graph in the ﬁgure, is a particular solution corresponding to c 0. xy y x2 sin x y cx x cos x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (b) The two-parameter family y c1ex c2xex is an explicit solution of the linear second-order equation y 2y y 0 in part (b) of Example 3. (Verify.) In Figure 1.1.4 we have shown seven of the “dou-ble inﬁnity” of solutions in the family. The solution curves in red, green, and blue are the graphs of the particular solutions y 5xex (cl 0, c2 5), y 3ex (cl 3, c2 0), and y 5ex 2xex (c1 5, c2 2), respectively. Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation—that is, a solution that cannot be obtained by spe-cializing any of the parameters in the family of solutions. Such an extra solution is called a singular solution. For example, we have seen that and y 0 are solutions of the differential equation dydx xy1/2 on (, ). In Section 2.2 we shall demonstrate, by actually solving it, that the differential equation dydx xy1/2 possesses the one-parameter family of solutions . When c 0, the resulting particular solution is . But notice that the trivial solution y 0 is a singular solution, since it is not a member of the family ; there is no way of assigning a value to the constant c to obtain y 0. In all the preceding examples we used x and y to denote the independent and dependent variables, respectively. But you should become accustomed to seeing and working with other symbols to denote these variables. For example, we could denote the independent variable by t and the dependent variable by x. y (1 4 x2 c) 2 y 1 16 x4 y (1 4 x2 c) 2 y 1 16 x4 8 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS FIGURE 1.1.5 Some solutions of DE in Example 8 (a) two explicit solutions (b) piecewise-defined solution c = 1 c = −1 x y c = 1, x 0 ≤ c = −1, x < 0 x y FIGURE 1.1.4 Some solutions of DE in part (b) of Example 6 y x EXAMPLE 7 Using Different Symbols The functions x c1 cos 4t and x c2 sin 4t, where c1 and c2 are arbitrary constants or parameters, are both solutions of the linear differential equation For x c1 cos 4t the ﬁrst two derivatives with respect to t are x 4c1 sin 4t and x 16c1 cos 4t. Substituting x and x then gives In like manner, for x c2 sin 4t we have x 16c2 sin 4t, and so Finally, it is straightforward to verify that the linear combination of solutions, or the two-parameter family x c1 cos 4t c2 sin 4t, is also a solution of the differential equation. The next example shows that a solution of a differential equation can be a piecewise-deﬁned function. x 16x 16c2 sin 4t 16(c2 sin 4t) 0. x 16x 16c1 cos 4t 16(c1 cos 4t) 0. x 16x 0. EXAMPLE 8 Piecewise-Defined Solutio The one-parameter family of quartic monomial functions y cx4 is an explicit solu-tion of the linear ﬁrst-order equation xy 4y 0 on the interval (, ). (Verify.) The blue and red solution curves shown in Figure 1.1.5(a) are the graphs of y x4 and y x4 and correspond to the choices c 1 and c 1, respectively. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The piecewise-deﬁned differentiable function is also a solution of the differential equation but cannot be obtained from the family y cx4 by a single choice of c. As seen in Figure 1.1.5(b) the solution is constructed from the family by choosing c 1 for x 0 and c 1 for x 0. Systems of Differential Equations Up to this point we have been discussing single differential equations containing one unknown function. But often in theory, as well as in many applications, we must deal with systems of differential equations. A system of ordinary differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable. For example, if x and y denote dependent variables and t denotes the independent variable, then a system of two ﬁrst-order differential equations is given by (9) A solution of a system such as (9) is a pair of differentiable functions x 1(t), y 2(t), deﬁned on a common interval I, that satisfy each equation of the system on this interval. dy dt g(t, x, y). dx dt f(t, x, y) y x4, x 0 x4, x 0 1.1 DEFINITIONS AND TERMINOLOGY ● 9 REMARKS (i) A few last words about implicit solutions of differential equations are in order. In Example 5 we were able to solve the relation x2 y2 25 for y in terms of x to get two explicit solutions, and , of the differential equation (8). But don’t read too much into this one example. Unless it is easy or important or you are instructed to, there is usually no need to try to solve an implicit solution G(x, y) 0 for y explicitly in terms of x. Also do not misinterpret the second sentence following Deﬁnition 1.1.3. An implicit solution G(x, y) 0 can deﬁne a perfectly good differentiable function that is a solution of a DE, yet we might not be able to solve G(x, y) 0 using analytical methods such as algebra. The solution curve of may be a segment or piece of the graph of G(x, y) 0. See Problems 45 and 46 in Exercises 1.1. Also, read the discussion following Example 4 in Section 2.2. (ii) Although the concept of a solution has been emphasized in this section, you should also be aware that a DE does not necessarily have to possess a solution. See Problem 39 in Exercises 1.1. The question of whether a solution exists will be touched on in the next section. (iii) It might not be apparent whether a ﬁrst-order ODE written in differential form M(x, y)dx N(x, y)dy 0 is linear or nonlinear because there is nothing in this form that tells us which symbol denotes the dependent variable. See Problems 9 and 10 in Exercises 1.1. (iv) It might not seem like a big deal to assume that F(x, y, y, . . . , y(n)) 0 can be solved for y(n), but one should be a little bit careful here. There are exceptions, and there certainly are some problems connected with this assumption. See Problems 52 and 53 in Exercises 1.1. (v) You may run across the term closed form solutions in DE texts or in lectures in courses in differential equations. Translated, this phrase usually 2(x) 125 x2 1(x) 125 x2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 10 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS refers to explicit solutions that are expressible in terms of elementary (or familiar) functions: ﬁnite combinations of integer powers of x, roots, exponen-tial and logarithmic functions, and trigonometric and inverse trigonometric functions. (vi) If every solution of an nth-order ODE F(x, y, y, . . . , y(n)) 0 on an inter-val I can be obtained from an n-parameter family G(x, y, c1, c2, . . . , cn) 0 by appropriate choices of the parameters ci, i 1, 2, . . . , n, we then say that the family is the general solution of the DE. In solving linear ODEs, we shall im-pose relatively simple restrictions on the coefﬁcients of the equation; with these restrictions one can be assured that not only does a solution exist on an interval but also that a family of solutions yields all possible solutions. Nonlinear ODEs, with the exception of some ﬁrst-order equations, are usually difﬁcult or impos-sible to solve in terms of elementary functions. Furthermore, if we happen to obtain a family of solutions for a nonlinear equation, it is not obvious whether this family contains all solutions. On a practical level, then, the designation “general solution” is applied only to linear ODEs. Don’t be concerned about this concept at this point, but store the words “general solution” in the back of your mind—we will come back to this notion in Section 2.3 and again in Chapter 4. EXERCISES 1.1 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1–8 state the order of the given ordinary differ-ential equation. Determine whether the equation is linear or nonlinear by matching it with (6). 1. (1 x)y 4xy 5y cos x 2. 3. t5y(4) t3y 6y 0 4. 5. 6. 7. (sin )y (cos )y 2 8. In Problems 9 and 10 determine whether the given ﬁrst-order differential equation is linear in the indicated dependent variable by matching it with the ﬁrst differential equation given in (7). 9. (y2 1) dx x dy 0; in y; in x 10. u dv (v uv ueu) du 0; in v; in u ¨ x 1 x . 2 3x . x 0 d 2R dt 2 k R2 d 2y dx 2 B1 dy dx 2 d 2u dr 2 du dr u cos(r u) x d3y dx3 dy dx 4 y 0 In Problems 11–14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of deﬁnition for each solution. 11. 2y y 0; y ex/2 12. 13. y 6y 13y 0; y e3x cos 2x 14. y y tan x; y (cos x)ln(sec x tan x) In Problems 15–18 verify that the indicated function y (x) is an explicit solution of the given ﬁrst-order differential equation. Proceed as in Example 2, by consider-ing simply as a function, give its domain. Then by consid-ering as a solution of the differential equation, give at least one interval I of deﬁnition. 15. 16. y 25 y2; y 5 tan 5x 17. y 2xy2; y 1(4 x2) 18. 2y y3 cos x; y (1 sin x)1/2 In Problems 19 and 20 verify that the indicated expression is an implicit solution of the given ﬁrst-order differential equa-tion. Find at least one explicit solution y (x) in each case. (y x)y y x 8; y x 42x 2 dy dt 20y 24; y 6 5 6 5 e20t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Use a graphing utility to obtain the graph of an explicit solu-tion. Give an interval I of deﬁnition of each solution . 19. 20. 2xy dx (x2 y) dy 0; 2x2y y2 1 In Problems 21–24 verify that the indicated family of func-tions is a solution of the given differential equation. Assume an appropriate interval I of deﬁnition for each solution. 21. 22. 23. 24. 25. Verify that the piecewise-deﬁned function is a solution of the differential equation xy 2y 0 on (, ). 26. In Example 5 we saw that y 1(x) and are solutions of dydx xy on the interval (5, 5). Explain why the piecewise-deﬁned function is not a solution of the differential equation on the interval (5, 5). In Problems 27–30 ﬁnd values of m so that the function y emx is a solution of the given differential equation. 27. y 2y 0 28. 5y 2y 29. y 5y 6y 0 30. 2y 7y 4y 0 In Problems 31 and 32 ﬁnd values of m so that the function y xm is a solution of the given differential equation. 31. xy 2y 0 32. x2y 7xy 15y 0 y 125 x2, 125 x2, 5 x 0 0 x 5 y 2(x) 125 x2 125 x2 y x2, x 0 x2, x 0 y c1x1 c2x c3x ln x 4x2 x3 d 3y dx3 2x2 d 2y dx2 x dy dx y 12x2; d 2y dx2 4 dy dx 4y 0; y c1e2x c2xe2x dy dx 2xy 1; y ex2 x 0 et2 dt c1ex2 dP dt P(1 P); P c1et 1 c1et dX dt (X 1)(1 2X); ln 2X 1 X 1 t 1.1 DEFINITIONS AND TERMINOLOGY ● 11 In Problems 33–36 use the concept that y c, x , is a constant function if and only if y 0 to determine whether the given differential equation possesses constant solutions. 33. 3xy 5y 10 34. y y2 2y 3 35. (y 1)y 1 36. y 4y 6y 10 In Problems 37 and 38 verify that the indicated pair of functions is a solution of the given system of differential equations on the interval (, ). 37. 38. , Discussion Problems 39. Make up a differential equation that does not possess any real solutions. 40. Make up a differential equation that you feel conﬁdent possesses only the trivial solution y 0. Explain your reasoning. 41. What function do you know from calculus is such that its ﬁrst derivative is itself? Its ﬁrst derivative is a constant multiple k of itself? Write each answer in the form of a ﬁrst-order differential equation with a solution. 42. What function (or functions) do you know from calcu-lus is such that its second derivative is itself? Its second derivative is the negative of itself? Write each answer in the form of a second-order differential equation with a solution. 43. Given that y sin x is an explicit solution of the first-order differential equation . Find an in-terval I of deﬁnition. [Hint: I is not the interval (, ).] 44. Discuss why it makes intuitive sense to presume that the linear differential equation y 2y 4y 5 sin t has a solution of the form y A sin t B cos t, where A and B are constants. Then ﬁnd speciﬁc constants A and B so that y A sin t B cos t is a particular solu-tion of the DE. dy dx 11 y2 y cos 2t sin 2t 1 5 et y e2t 5e6t x cos 2t sin 2t 1 5 et x e2t 3e6t, d 2y dt 2 4x et; dy dt 5x 3y; d 2x dt 2 4y et dx dt x 3y Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In Problems 45 and 46 the given ﬁgure represents the graph of an implicit solution G(x, y) 0 of a differential equation dydx f (x, y). In each case the relation G(x, y) 0 implicitly deﬁnes several solutions of the DE. Carefully reproduce each ﬁgure on a piece of paper. Use different colored pencils to mark off segments, or pieces, on each graph that correspond to graphs of solutions. Keep in mind that a solution must be a function and differentiable. Use the solution curve to estimate an interval I of deﬁnition of each solution . 45. 12 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 51. Discuss, and illustrate with examples, how to solve differential equations of the forms dydx f (x) and d2ydx2 f (x). 52. The differential equation x(y)2 4y 12x3 0 has the form given in (4). Determine whether the equation can be put into the normal form dydx f (x, y). 53. The normal form (5) of an nth-order differential equa-tion is equivalent to (4) whenever both forms have exactly the same solutions. Make up a ﬁrst-order differ-ential equation for which F(x, y, y) 0 is not equiva-lent to the normal form dydx f (x, y). 54. Find a linear second-order differential equation F(x, y, y, y) 0 for which y c1x c2x2 is a two-parameter family of solutions. Make sure that your equa-tion is free of the arbitrary parameters c1 and c2. Qualitative information about a solution y (x) of a differential equation can often be obtained from the equation itself. Before working Problems 55–58, recall the geometric signiﬁcance of the derivatives dydx and d2ydx2. 55. Consider the differential equation . (a) Explain why a solution of the DE must be an increasing function on any interval of the x-axis. (b) What are What does this suggest about a solution curve as (c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up. (d) Sketch the graph of a solution y (x) of the dif-ferential equation whose shape is suggested by parts (a)–(c). 56. Consider the differential equation dydx 5 y. (a) Either by inspection or by the method suggested in Problems 33–36, ﬁnd a constant solution of the DE. (b) Using only the differential equation, ﬁnd intervals on the y-axis on which a nonconstant solution y (x) is increasing. Find intervals on the y-axis on which y (x) is decreasing. 57. Consider the differential equation dydx y(a by), where a and b are positive constants. (a) Either by inspection or by the method suggested in Problems 33–36, ﬁnd two constant solutions of the DE. (b) Using only the differential equation, ﬁnd intervals on the y-axis on which a nonconstant solution y (x) is increasing. Find intervals on which y (x) is decreasing. (c) Using only the differential equation, explain why y a2b is the y-coordinate of a point of inﬂection of the graph of a nonconstant solution y (x). x : ? lim x : dy>dx and lim x : dy>dx? dy>dx ex2 FIGURE 1.1.6 Graph for Problem 45 FIGURE 1.1.7 Graph for Problem 46 y x 1 1 1 x 1 y 46. 47. The graphs of members of the one-parameter family x3 y3 3cxy are called folia of Descartes. Verify that this family is an implicit solution of the ﬁrst-order differential equation 48. The graph in Figure 1.1.7 is the member of the family of folia in Problem 47 corresponding to c 1. Discuss: How can the DE in Problem 47 help in ﬁnding points on the graph of x3 y3 3xy where the tangent line is vertical? How does knowing where a tangent line is vertical help in determining an interval I of deﬁnition of a solution of the DE? Carry out your ideas, and compare with your estimates of the intervals in Problem 46. 49. In Example 5 the largest interval I over which the explicit solutions y 1(x) and y 2(x) are deﬁned is the open interval (5, 5). Why can’t the interval I of deﬁnition be the closed interval [5, 5]? 50. In Problem 21 a one-parameter family of solutions of the DE P P(1 P) is given. Does any solution curve pass through the point (0, 3)? Through the point (0, 1)? dy dx y(y3 2x3) x(2y3 x3). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1.2 INITIAL-VALUE PROBLEMS ● 13 (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the xy-plane into three regions. In each region, sketch the graph of a non-constant solution y (x) whose shape is sug-gested by the results in parts (b) and (c). 58. Consider the differential equation y y2 4. (a) Explain why there exist no constant solutions of the DE. (b) Describe the graph of a solution y (x). For example, can a solution curve have any relative extrema? (c) Explain why y 0 is the y-coordinate of a point of inﬂection of a solution curve. (d) Sketch the graph of a solution y (x) of the differential equation whose shape is suggested by parts (a)–(c). Computer Lab Assignments In Problems 59 and 60 use a CAS to compute all derivatives and to carry out the simpliﬁcations needed to verify that the indicated function is a particular solution of the given differ-ential equation. 59. y(4) 20y 158y 580y 841y 0; y xe5x cos 2x 60. y 20 cos(5 ln x) x 3 sin(5 ln x) x x3y 2x2y 20xy 78y 0; Geometric Interpretation of IVPs The cases n 1 and n 2 in (1), (2) Subject to: y(x0) y0 Solve: dy dx f (x, y) INITIAL-VALUE PROBLEMS REVIEW MATERIAL ●Normal form of a DE ●Solution of a DE ●Family of solutions INTRODUCTION We are often interested in problems in which we seek a solution y(x) of a differential equation so that y(x) also satisﬁes certain prescribed side conditions—that is, conditions that are imposed on the unknown function y(x) and its derivatives at a point x0. On some interval I containing x0 the problem of solving an nth-order differential equation subject to n side conditions speciﬁed at x0: (1) where y0, y1, . . . , yn1 are arbitrary real constants, is called an nth-order initial-value problem (IVP). The values of y(x) and its ﬁrst n 1 derivatives at x0, y(x0) y0, y(x0) y1, . . . , y(n1)(x0) yn1 are called initial conditions (IC). Solving an nth-order initial-value problem such as (1) frequently entails ﬁrst ﬁnding an n-parameter family of solutions of the given differential equation and then using the initial-conditions at x0 to determine the n constants in this family. The resulting particular solution is deﬁned on some interval I containing the initial point x0. Subject to: y(x0) y0, y(x0) y1, . . . , y(n1)(x0) yn1, Solve: d ny dxn f x, y, y, . . . , y(n1) 1.2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. and (3) are examples of first and second-order initial-value problems, respectively. These two problems are easy to interpret in geometric terms. For (2) we are seeking a solution y(x) of the differential equation y f(x, y) on an interval I containing x0 so that its graph passes through the speciﬁed point (x0, y0). A solution curve is shown in blue in Figure 1.2.1. For (3) we want to ﬁnd a solution y(x) of the differential equation y f(x, y, y) on an interval I containing x0 so that its graph not only passes through (x0, y0) but the slope of the curve at this point is the number y1. A solution curve is shown in blue in Figure 1.2.2. The words initial conditions derive from physical sys-tems where the independent variable is time t and where y(t0) y0 and y(t0) y1 rep-resent the position and velocity, respectively, of an object at some beginning, or initial, time t0. Subject to: y(x0) y0, y(x0) y1 Solve: d2y dx2 f (x, y, y) 14 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS FIGURE 1.2.3 Solution curves of two IVPs in Example 1 y x (0, 3) (1, −2) FIGURE 1.2.1 Solution curve of ﬁrst-order IVP FIGURE 1.2.2 Solution curve of second-order IVP x I solutions of the DE ( x 0 , y 0 ) y m = y 1 x I solutions of the DE ( x 0 , y 0 ) y EXAMPLE 1 Two First-Order IVPs (a) In Problem 41 in Exercises 1.1 you were asked to deduce that y cex is a one-parameter family of solutions of the simple ﬁrst-order equation y y. All the solutions in this family are deﬁned on the interval (, ). If we impose an initial condition, say, y(0) 3, then substituting x 0, y 3 in the family determines the constant 3 ce0 c. Thus y 3ex is a solution of the IVP (b) Now if we demand that a solution curve pass through the point (1, 2) rather than (0, 3), then y(1) 2 will yield 2 ce or c 2e1. In this case y 2ex1 is a solution of the IVP The two solution curves are shown in dark blue and dark red in Figure 1.2.3. The next example illustrates another ﬁrst-order initial-value problem. In this example notice how the interval I of deﬁnition of the solution y(x) depends on the initial condition y(x0) y0. y y, y(1) 2. y y, y(0) 3. EXAMPLE 2 Interval I of Definition of a Solutio In Problem 6 of Exercises 2.2 you will be asked to show that a one-parameter family of solutions of the ﬁrst-order differential equation y 2xy2 0 is y 1(x2 c). If we impose the initial condition y(0) 1, then substituting x 0 and y 1 into the family of solutions gives 1 1c or c 1. Thus y 1(x2 1). We now emphasize the following three distinctions: • Considered as a function, the domain of y 1(x2 1) is the set of real numbers x for which y(x) is deﬁned; this is the set of all real numbers except x 1 and x 1. See Figure 1.2.4(a). • Considered as a solution of the differential equation y 2xy2 0, the interval I of deﬁnition of y 1(x2 1) could be taken to be any interval over which y(x) is deﬁned and differentiable. As can be seen in Figure 1.2.4(a), the largest intervals on which y 1(x2 1) is a solution are (,1), (1, 1), and (1, ). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. • Considered as a solution of the initial-value problem y 2xy2 0, y(0) 1, the interval I of deﬁnition of y 1(x2 1) could be taken to be any interval over which y(x) is deﬁned, differentiable, and contains the initial point x 0; the largest interval for which this is true is (1, 1). See the red curve in Figure 1.2.4(b). See Problems 3–6 in Exercises 1.2 for a continuation of Example 2. 1.2 INITIAL-VALUE PROBLEMS ● 15 FIGURE 1.2.4 Graphs of function and solution of IVP in Example 2 (0, −1) x y 1 −1 x y 1 −1 (a) function defined for all x except x = ±1 (b) solution defined on interval containing x = 0 y y = 0 y = x4/16 (0, 0) 1 x FIGURE 1.2.5 Two solutions curves of the same IVP in Example 4 EXAMPLE 3 Second-Order IVP In Example 7 of Section 1.1 we saw that x c1 cos 4t c2 sin 4t is a two-parameter family of solutions of x 16x 0. Find a solution of the initial-value problem (4) SOLUTION We ﬁrst apply x(2) 2 to the given family of solutions: c1 cos 2 c2 sin 2 2. Since cos 2 1 and sin 2 0, we ﬁnd that c1 2. We next apply x(2) 1 to the one-parameter family x(t) 2 cos 4t c2 sin 4t. Differentiating and then setting t 2 and x 1 gives 8 sin 2 4c2 cos 2 1, from which we see that . Hence is a solution of (4). Existence and Uniqueness Two fundamental questions arise in considering an initial-value problem: Does a solution of the problem exist? If a solution exists, is it unique? For the ﬁrst-order initial-value problem (2) we ask: Existence { Does the differential equation dydx f(x, y) possess solutions? Do any of the solution curves pass through the point (x0, y0)? Uniqueness { When can we be certain that ther e is precisely one solution curve passing through the point (x0, y0)? Note that in Examples 1 and 3 the phrase “a solution” is used rather than “the solu-tion” of the problem. The indeﬁnite article “a” is used deliberately to suggest the possibility that other solutions may exist. At this point it has not been demonstrated that there is a single solution of each problem. The next example illustrates an initial-value problem with two solutions. x 2 cos 4t 1 4 sin 4t c2 1 4 x 16x 0, x 2 2, x 2 1. EXAMPLE 4 An IVP Can Have Several Solutions Each of the functions y 0 and satisﬁes the differential equation dydx xy1/2 and the initial condition y(0) 0, so the initial-value problem has at least two solutions. As illustrated in Figure 1.2.5, the graphs of both functions, shown in red and blue pass through the same point (0, 0). Within the safe conﬁnes of a formal course in differential equations one can be fairly conﬁdent that most differential equations will have solutions and that solutions of initial-value problems will probably be unique. Real life, however, is not so idyllic. Therefore it is desirable to know in advance of trying to solve an initial-value problem dy dx xy1/2, y(0) 0 y 1 16 x4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. whether a solution exists and, when it does, whether it is the only solution of the prob-lem. Since we are going to consider ﬁrst-order differential equations in the next two chapters, we state here without proof a straightforward theorem that gives conditions that are sufﬁcient to guarantee the existence and uniqueness of a solution of a ﬁrst-order initial-value problem of the form given in (2). We shall wait until Chapter 4 to address the question of existence and uniqueness of a second-order initial-value problem. 16 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS x I0 R a b c d (x0, y0) y FIGURE 1.2.6 Rectangular region R THEOREM 1.2.1 Existence of a Unique Solution Let R be a rectangular region in the xy-plane deﬁned by a x b, c y d that contains the point (x0, y0) in its interior. If f(x, y) and fy are continuous on R, then there exists some interval I0: (x0 h, x0 h), h 0, contained in [a, b], and a unique function y(x), deﬁned on I0, that is a solution of the initial-value problem (2). The foregoing result is one of the most popular existence and uniqueness theo-rems for ﬁrst-order differential equations because the criteria of continuity of f(x, y) and fy are relatively easy to check. The geometry of Theorem 1.2.1 is illustrated in Figure 1.2.6. EXAMPLE 5 Example 4 Revisited We saw in Example 4 that the differential equation dydx xy1/2 possesses at least two solutions whose graphs pass through (0, 0). Inspection of the functions shows that they are continuous in the upper half-plane deﬁned by y 0. Hence Theorem 1.2.1 enables us to conclude that through any point (x0, y0), y0 0 in the upper half-plane there is some interval centered at x0 on which the given differential equation has a unique solution. Thus, for example, even without solving it, we know that there exists some interval centered at 2 on which the initial-value problem dydx xy1/2, y(2) 1 has a unique solution. In Example 1, Theorem 1.2.1 guarantees that there are no other solutions of the initial-value problems y y, y(0) 3 and y y, y(1) 2 other than y 3ex and y 2ex1, respectively. This follows from the fact that f(x, y) y and fy 1 are continuous throughout the entire xy-plane. It can be further shown that the interval I on which each solution is deﬁned is (, ). Interval of Existence/Uniqueness Suppose y(x) represents a solution of the initial-value problem (2). The following three sets on the real x-axis may not be the same: the domain of the function y(x), the interval I over which the solution y(x) is deﬁned or exists, and the interval I0 of existence and uniqueness. Example 2 of Section 1.1 illustrated the difference between the domain of a function and the interval I of deﬁnition. Now suppose (x0, y0) is a point in the interior of the rectan-gular region R in Theorem 1.2.1. It turns out that the continuity of the function f (x, y) on R by itself is sufﬁcient to guarantee the existence of at least one solution of dydx f (x, y), y(x0) y0, deﬁned on some interval I. The interval I of deﬁni-tion for this initial-value problem is usually taken to be the largest interval contain-ing x0 over which the solution y(x) is deﬁned and differentiable. The interval I depends on both f (x, y) and the initial condition y(x0) y0. See Problems 31–34 in Exercises 1.2. The extra condition of continuity of the ﬁrst partial derivative fy f (x, y) xy1/2 and f y x 2y1/2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1.2 INITIAL-VALUE PROBLEMS ● 17 on R enables us to say that not only does a solution exist on some interval I0 con-taining x0, but it is the only solution satisfying y(x0) y0. However, Theorem 1.2.1 does not give any indication of the sizes of intervals I and I0; the interval I of definition need not be as wide as the r egion R, and the interval I 0 of existence and uniqueness may not be as lar ge as I. The number h 0 that deﬁnes the interval I0: (x0 h, x0 h) could be very small, so it is best to think that the solution y(x) is unique in a local sense —that is, a solution deﬁned near the point (x0, y0). See Problem 50 in Exercises 1.2. REMARKS (i) The conditions in Theorem 1.2.1 are sufﬁcient but not necessary. This means that when f(x, y) and fy are continuous on a rectangular region R, it must always follow that a solution of (2) exists and is unique whenever (x0, y0) is a point interior to R. However, if the conditions stated in the hypothesis of Theorem 1.2.1 do not hold, then anything could happen: Problem (2) may still have a solution and this solution may be unique, or (2) may have several solu-tions, or it may have no solution at all. Arereading of Example 5 reveals that the hypotheses of Theorem 1.2.1 do not hold on the line y 0 for the differential equation dydx xy1/2, so it is not surprising, as we saw in Example 4 of this section, that there are two solutions deﬁned on a common interval h x h satisfying y(0) 0. On the other hand, the hypotheses of Theorem 1.2.1 do not hold on the line y 1 for the differential equation dydx y 1. Nevertheless it can be proved that the solution of the initial-value problem dydx y 1, y(0) 1, is unique. Can you guess this solution? (ii) You are encouraged to read, think about, work, and then keep in mind Problem 49 in Exercises 1.2. (iii) Initial conditions are prescribed at a single point x0. But we are also inter-ested in solving differential equations that are subject to conditions speciﬁed on y(x) or its derivative at two different points x0 and x1. Conditions such as and called boundary conditions. A differential equation together with bound-ary conditions is called a boundary-value problem (BVP). For example, is a boundary-value problem. See Problems 39–44 in Exercises 1.2. When we start to solve differential equations in Chapter 2 we will solve only ﬁrst-order equations and ﬁrst-order initial-value problems. The mathe-matical description of many problems in science and engineering involve second-order IVPs or two-point BVPs. We will examine some of these prob-lems in Chapters 4 and 5. y ly 0, y(0) 0, y(p) 0 y(1) 0, y(5) 0 or y(p>2) 0, y(p) 1 EXERCISES 1.2 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1 and 2, y 1(1 c1ex) is a one-parameter family of solutions of the ﬁrst-order DE y y y2. Find a solution of the ﬁrst-order IVP consisting of this differential equation and the given initial condition. 1. 2. y(1) 2 In Problems 3–6, y 1(x2 c) is a one-parameter family of solutions of the ﬁrst-order DE y 2xy2 0. Find a y(0) 1 3 solution of the ﬁrst-order IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution is deﬁned. 3. 4. 5. y(0) 1 6. In Problems 7–10, x c1 cos t c2 sin t is a two-parameter family of solutions of the second-order DE x x 0. Find y(1 2) 4 y(2) 1 2 y(2) 1 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. a solution of the second-order IVP consisting of this differ-ential equation and the given initial conditions. 7. x(0) 1, x(0) 8 8. x(2) 0, x(2) 1 9. 10. In Problems 11–14, y c1ex c2ex is a two-parameter family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differ-ential equation and the given initial conditions. 11. 12. y(1) 0, y(1) e 13. y(1) 5, y(1) 5 14. y(0) 0, y(0) 0 In Problems 15 and 16 determine by inspection at least two solutions of the given ﬁrst-order IVP. 15. y 3y2/3, y(0) 0 16. xy 2y, y(0) 0 In Problems 17–24 determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. 17. 18. 19. 20. 21. (4 y2)y x2 22. (1 y3)y x2 23. (x2 y2)y y2 24. (y x)y y x In Problems 25–28 determine whether Theorem 1.2.1 guar-antees that the differential equation pos-sesses a unique solution through the given point. 25. (1, 4) 26. (5, 3) 27. (2, 3) 28. (1, 1) 29. (a) By inspection ﬁnd a one-parameter family of solu-tions of the differential equation xy y. Verify that each member of the family is a solution of the initial-value problem xy y, y(0) 0. (b) Explain part (a) by determining a region R in the xy-plane for which the differential equation xy y would have a unique solution through a point (x0, y0) in R. y 1y2 9 dy dx y x x dy dx y dy dx 1xy dy dx y2/3 y(0) 2 y(0) 1, x(>4) 12, x(>4) 212 x(>6) 1 2, x(>6) 0 18 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS (c) Verify that the piecewise-deﬁned function satisﬁes the condition y(0) 0. Determine whether this function is also a solution of the initial-value problem in part (a). 30. (a) Verify that y tan (x c) is a one-parameter family of solutions of the differential equation y 1 y2. (b) Since f (x, y) 1 y2 and fy 2y are continu-ous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to ﬁnd an explicit solution of the ﬁrst-order initial-value problem y 1 y2, y(0) 0. Even though x0 0 is in the interval (2, 2), explain why the solution is not deﬁned on this interval. (c) Determine the largest interval I of deﬁnition for the solution of the initial-value problem in part (b). 31. (a) Verify that y 1(x c) is a one-parameter family of solutions of the differential equation y y2. (b) Since f(x, y) y2 and fy 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Find a solution from the family in part (a) that satisﬁes y(0) 1. Then ﬁnd a solution from the family in part (a) that satisﬁes y(0) 1. Determine the largest interval I of deﬁnition for the solution of each initial-value problem. (c) Determine the largest interval I of deﬁnition for the solution of the ﬁrst-order initial-value problem y y2, y(0) 0. [Hint: The solution is not a mem-ber of the family of solutions in part (a).] 32. (a) Show that a solution from the family in part (a) of Problem 31 that satisﬁes y y2, y(1) 1, is y 1(2 x). (b) Then show that a solution from the family in part (a) of Problem 31 that satisﬁes y y2, y(3) 1, is y 1(2 x). (c) Are the solutions in parts (a) and (b) the same? 33. (a) Verify that 3x2 y2 c is a one-parameter fam-ily of solutions of the differential equation y dydx 3x. (b) By hand, sketch the graph of the implicit solution 3x2 y2 3. Find all explicit solutions y (x) of the DE in part (a) deﬁned by this relation. Give the interval I of deﬁnition of each explicit solution. (c) The point (2, 3) is on the graph of 3x2 y2 3, but which of the explicit solutions in part (b) satis-ﬁes y(2) 3? 34. (a) Use the family of solutions in part (a) of Problem 33 to ﬁnd an implicit solution of the initial-value y 0, x 0 x, x 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1.2 INITIAL-VALUE PROBLEMS ● 19 FIGURE 1.2.7 Graph for Problem 35 y x 5 −5 5 In Problems 39–44, is a two-parameter family of solutions of the second-order DE If possible, ﬁnd a solution of the differential equation that satisﬁes the given side conditions. The condi-tions speciﬁed at two different points are called boundary conditions. 39. 40. 41. 42. 43. 44. Discussion Problems In Problems 45 and 46 use Problem 51 in Exercises 1.1 and (2) and (3) of this section. 45. Find a function y f (x) whose graph at each point (x, y) has the slope given by 8e2x 6x and has the y-intercept (0, 9). 46. Find a function y f (x) whose second derivative is y 12x 2 at each point (x, y) on its graph and y x 5 is tangent to the graph at the point corre-sponding to x 1. 47. Consider the initial-value problem y x 2y, . Determine which of the two curves shown in Figure 1.2.11 is the only plausible solution curve. Explain your reasoning. y(0) 1 2 y(p>2) 1, y(p) 0 y(0) 0, y(p) 2 y(0) 1, y(p) 5 y(0) 0, y(p>6) 0 y(0) 0, y(p) 0 y(0) 0, y(p>4) 3 y 4y 0. y c1 cos 2x c2 sin 2x FIGURE 1.2.10 Graph for Problem 38 y x 5 −5 5 36. 37. 38. FIGURE 1.2.8 Graph for Problem 36 FIGURE 1.2.9 Graph for Problem 37 y x 5 −5 5 y x 5 −5 5 FIGURE 1.2.11 Graphs for Problem 47 (0, ) 1 2 1 1 x y problem y dydx 3x, y(2) 4. Then, by hand, sketch the graph of the explicit solution of this problem and give its interval I of deﬁnition. (b) Are there any explicit solutions of y dydx 3x that pass through the origin? In Problems 35–38 the graph of a member of a family of solutions of a second-order differential equation d2ydx2 f (x, y, y) is given. Match the solution curve with at least one pair of the following initial conditions. (a) y(1) 1, y(1) 2 (b) y(1) 0, y(1) 4 (c) y(1) 1, y(1) 2 (d) y(0) 1, y(0) 2 (e) y(0) 1, y(0) 0 (f) y(0) 4, y(0) 2 35. 48. Determine a plausible value of x0 for which the graph of the solution of the initial-value problem y 2y 3x 6, y(x0) 0 is tangent to the x-axis at (x0, 0). Explain your reasoning. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 20 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 49. Suppose that the ﬁrst-order differential equation dydx f (x, y) possesses a one-parameter family of solutions and that f (x, y) satisﬁes the hypotheses of Theorem 1.2.1 in some rectangular region R of the xy-plane. Explain why two different solution curves cannot intersect or be tangent to each other at a point (x0, y0) in R. 50. The functions and have the same domain but are clearly different. See Figures 1.2.12(a) and 1.2.12(b), respectively. Show that both functions are solutions of the initial-value problem dydx xy1/2, y(2) 1 on the interval (, ). Resolve the apparent contradiction between this fact and the last sentence in Example 5. y(x) 0, 1 16 x4, x 0 x 0 y(x) 1 16 x4, x Mathematical Model 51. Population Growth Beginning in the next section we will see that differential equations can be used to describe or model many different physical systems. In this problem suppose that a model of the growing popu-lation of a small community is given by the initial-value problem where P is the number of individuals in the community and time t is measured in years. How fast—that is, at what rate—is the population increasing at t 0? How fast is the population increasing when the population is 500? dP dt 0.15P(t) 20, P(0) 100, FIGURE 1.2.12 Two solutions of the IVP in Problem 50 (a) (2, 1) y x (b) (2, 1) y x Mathematical Models It is often desirable to describe the behavior of some real-life system or phenomenon, whether physical, sociological, or even economic, in mathematical terms. The mathematical description of a system of phenomenon is called a mathematical model and is constructed with certain goals in mind. For ex-ample, we may wish to understand the mechanisms of a certain ecosystem by study-ing the growth of animal populations in that system, or we may wish to date fossils by analyzing the decay of a radioactive substance, either in the fossil or in the stra-tum in which it was discovered. DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS REVIEW MATERIAL ●Units of measurement for weight, mass, and density ●Newton’s second law of motion ●Hooke’s law ●Kirchhoff’s laws ●Archimedes’ principle INTRODUCTION In this section we introduce the notion of a differential equation as a mathematical model and discuss some speciﬁc models in biology, chemistry, and physics. Once we have studied some methods for solving DEs in Chapters 2 and 4, we return to, and solve, some of these models in Chapters 3 and 5. 1.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Construction of a mathematical model of a system starts with (i) identification of the variables that are responsible for changing the system. We may choose not to incorporate all these variables into the model at first. In this step we are specifying the level of resolution of the model. Next (ii) we make a set of reasonable assumptions, or hypotheses, about the system we are trying to describe. These assumptions will also include any empirical laws that may be applicable to the system. For some purposes it may be perfectly within reason to be content with low-resolution models. For example, you may already be aware that in beginning physics courses, the retarding force of air friction is sometimes ignored in modeling the motion of a body falling near the surface of the Earth, but if you are a scientist whose job it is to accurately predict the ﬂight path of a long-range projectile, you have to take into account air resistance and other factors such as the curvature of the Earth. Since the assumptions made about a system frequently involve a rate of change of one or more of the variables, the mathematical depiction of all these assumptions may be one or more equations involving derivatives. In other words, the mathemat-ical model may be a differential equation or a system of differential equations. Once we have formulated a mathematical model that is either a differential equation or a system of differential equations, we are faced with the not insigniﬁcant problem of trying to solve it. If we can solve it, then we deem the model to be reason-able if its solution is consistent with either experimental data or known facts about the behavior of the system. But if the predictions produced by the solution are poor, we can either increase the level of resolution of the model or make alternative as-sumptions about the mechanisms for change in the system. The steps of the model-ing process are then repeated, as shown in the diagram in Figure 1.3.1. 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 21 Express assumptions in terms of DEs Display predictions of model (e.g., graphically) Solve the DEs If necessary, alter assumptions or increase resolution of model Assumptions and hypotheses Mathematical formulation Obtain solutions Check model predictions with known facts FIGURE 1.3.1 Steps in the modeling process with differential equations Of course, by increasing the resolution, we add to the complexity of the mathemati-cal model and increase the likelihood that we cannot obtain an explicit solution. A mathematical model of a physical system will often involve the variable time t. A solution of the model then gives the state of the system; in other words, the values of the dependent variable (or variables) for appropriate values of t describe the system in the past, present, and future. Population Dynamics One of the earliest attempts to model human pop-ulation growth by means of mathematics was by the English clergyman and econo-mist Thomas Malthus in 1798. Basically, the idea behind the Malthusian model is the assumption that the rate at which the population of a country grows at a certain time is Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. proportional to the total population of the country at that time. In other words, the more people there are at time t, the more there are going to be in the future. In mathematical terms, if P(t) denotes the total population at time t, then this assumption can be expressed as , (1) where k is a constant of proportionality. This simple model, which fails to take into account many factors that can inﬂuence human populations to either grow or decline (immigration and emigration, for example), nevertheless turned out to be fairly accu-rate in predicting the population of the United States during the years 1790–1860. Populations that grow at a rate described by (1) are rare; nevertheless, (1) is still used to model growth of small populations over short intervals of time (bacteria growing in a petri dish, for example). Radioactive Decay The nucleus of an atom consists of combinations of protons and neutrons. Many of these combinations of protons and neutrons are unstable—that is, the atoms decay or transmute into atoms of another substance. Such nuclei are said to be radioactive. For example, over time the highly radioactive radium, Ra-226, transmutes into the radioactive gas radon, Rn-222. To model the phenomenon of radioactive decay, it is assumed that the rate dAdt at which the nuclei of a sub-stance decay is proportional to the amount (more precisely, the number of nuclei) A(t) of the substance remaining at time t: . (2) Of course, equations (1) and (2) are exactly the same; the difference is only in the in-terpretation of the symbols and the constants of proportionality. For growth, as we expect in (1), k 0, and for decay, as in (2), k 0. The model (1) for growth can also be seen as the equation dSdt rS, which describes the growth of capital S when an annual rate of interest r is compounded continuously. The model (2) for decay also occurs in biological applications such as determining the half-life of a drug—the time that it takes for 50% of a drug to be eliminated from a body by excretion or metabolism. In chemistry the decay model (2) appears in the mathematical description of a ﬁrst-order chemical reaction. The point is this: A single differential equation can serve as a mathematical model for many different phenomena. Mathematical models are often accompanied by certain side conditions. For ex-ample, in (1) and (2) we would expect to know, in turn, the initial population P0 and the initial amount of radioactive substance A0 on hand. If the initial point in time is taken to be t 0, then we know that P(0) P0 and A(0) A0. In other words, a mathematical model can consist of either an initial-value problem or, as we shall see later on in Section 5.2, a boundary-value problem. Newton’s Law of Cooling/W arming According to Newton’s empirical law of cooling/warming, the rate at which the temperature of a body changes is proportional to the difference between the temperature of the body and the temper-ature of the surrounding medium, the so-called ambient temperature. If T(t) repre-sents the temperature of a body at time t, Tm the temperature of the surrounding dA dt A or dA dt kA dP dt P or dP dt kP 22 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS If two quantities u and v are proportional, we write u v. This means that one quantity is a constant multiple of the other: u kv. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. medium, and dTdt the rate at which the temperature of the body changes, then Newton’s law of cooling/warming translates into the mathematical statement , (3) where k is a constant of proportionality. In either case, cooling or warming, if Tm is a constant, it stands to reason that k 0. Spread of a Disease A contagious disease—for example, a ﬂu virus—is spread throughout a community by people coming into contact with other people. Let x(t) denote the number of people who have contracted the disease and y(t) denote the number of people who have not yet been exposed. It seems reasonable to assume that the rate dxdt at which the disease spreads is proportional to the number of encoun-ters, or interactions, between these two groups of people. If we assume that the num-ber of interactions is jointly proportional to x(t) and y(t)—that is, proportional to the product xy—then , (4) where k is the usual constant of proportionality. Suppose a small community has a ﬁxed population of n people. If one infected person is introduced into this commu-nity, then it could be argued that x(t) and y(t) are related by x y n 1. Using this last equation to eliminate y in (4) gives us the model . (5) An obvious initial condition accompanying equation (5) is x(0) 1. Chemical Reactions The disintegration of a radioactive substance, governed by the differential equation (1), is said to be a first-orde reaction. In chemistry a few reactions follow this same empirical law: If the molecules of substance A decompose into smaller molecules, it is a natural assumption that the rate at which this decomposition takes place is proportional to the amount of the ﬁrst substance that has not undergone conversion; that is, if X(t) is the amount of substance A remaining at any time, then dXdt kX, where k is a negative constant since X is decreasing. An example of a ﬁrst-order chemical reaction is the conversion of t-butyl chloride, (CH3)3CCl, into t-butyl alcohol, (CH3)3COH: Only the concentration of the t-butyl chloride controls the rate of reaction. But in the reaction one molecule of sodium hydroxide, NaOH, is consumed for every molecule of methyl chloride, CH3Cl, thus forming one molecule of methyl alcohol, CH3OH, and one molecule of sodium chloride, NaCl. In this case the rate at which the reaction proceeds is proportional to the product of the remaining concentrations of CH3Cl and NaOH. To describe this second reaction in general, let us suppose one molecule of a substance A combines with one molecule of a substance B to form one molecule of a substance C. If X denotes the amount of chemical C formed at time t and if and are, in turn, the amounts of the two chemicals A and B at t 0 (the initial amounts), then the instantaneous amounts of A and B not converted to chemical C are X and X, respectively. Hence the rate of formation of C is given by , (6) where k is a constant of proportionality. A reaction whose model is equation (6) is said to be a second-order reaction. dX dt k( X)( X) CH3Cl NaOH : CH3OH NaCl (CH3)3CCl NaOH : (CH3)3COH NaCl. dx dt kx(n 1 x) dx dt kxy dT dt T Tm or dT dt k(T Tm) 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 23 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Mixtures The mixing of two salt solutions of differing concentrations gives rise to a ﬁrst-order differential equation for the amount of salt contained in the mix-ture. Let us suppose that a large mixing tank initially holds 300 gallons of brine (that is, water in which a certain number of pounds of salt has been dissolved). Another brine solution is pumped into the large tank at a rate of 3 gallons per minute; the concentration of the salt in this inﬂow is 2 pounds per gallon. When the solution in the tank is well stirred, it is pumped out at the same rate as the entering solution. See Figure 1.3.2. If A(t) denotes the amount of salt (measured in pounds) in the tank at time t, then the rate at which A(t) changes is a net rate: . (7) The input rate Rin at which salt enters the tank is the product of the inﬂow concentra-tion of salt and the inﬂow rate of ﬂuid. Note that Rin is measured in pounds per minute: Now, since the solution is being pumped out of the tank at the same rate that it is pumped in, the number of gallons of brine in the tank at time t is a constant 300 gal-lons. Hence the concentration of the salt in the tank as well as in the outﬂow is c(t) A(t)300 lb/gal, so the output rate Rout of salt is The net rate (7) then becomes (8) If rin and rout denote general input and output rates of the brine solutions, then there are three possibilities: rin rout, rin rout, and rin rout. In the analysis lead-ing to (8) we have assumed that rin rout. In the latter two cases the number of gal-lons of brine in the tank is either increasing (rin rout) or decreasing (rin rout) at the net rate rin rout. See Problems 10–12 in Exercises 1.3. Draining a Tank In hydrodynamics, Torricelli’s law states that the speed v of efﬂux of water though a sharp-edged hole at the bottom of a tank ﬁlled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h—that is, , where g is the acceleration due to gravity. This last expression comes from equating the kinetic energy with the potential energy mgh and solving for v. Suppose a tank ﬁlled with water is allowed to drain through a hole under the inﬂuence of gravity. We would like to ﬁnd the depth h of water remaining in the tank at time t. Consider the tank shown in Figure 1.3.3. If the area of the hole is Ah (in ft2) and the speed of the water leaving the tank is (in ft/s), then the volume of water leaving the tank per second is (in ft3/s). Thus if V(t) denotes the volume of water in the tank at time t, then , (9) dV dt Ah12gh Ah12gh v 12gh 1 2mv2 v 12gh dA dt 6 A 100 or dA dt 1 100 A 6. Rout ( lb/gal) (3 gal/min) lb/min. A(t) –––– 300 A(t) –––– 100 concentration of salt in outﬂow output rate of brine output rate of salt concentration of salt in inﬂow input rate of brine input rate of salt Rin (2 lb/gal) (3 gal/min) (6 lb/min). dA dt input rate of salt output rate of salt Rin Rout 24 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS input rate of brine 3 gal/min output rate of brine 3 gal/min constant 300 gal FIGURE 1.3.2 Mixing tank h Aw Ah FIGURE 1.3.3 Draining tank Don’t confuse these symbols with Rin and Rout, which are input and output rates of salt. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. where the minus sign indicates that V is decreasing. Note here that we are ignoring the possibility of friction at the hole that might cause a reduction of the rate of ﬂow there. Now if the tank is such that the volume of water in it at time t can be written V(t) Awh, where Aw (in ft2) is the constant area of the upper surface of the water (see Figure 1.3.3), then dVdt Aw dhdt. Substituting this last expression into (9) gives us the desired differential equation for the height of the water at time t: . (10) It is interesting to note that (10) remains valid even when Aw is not constant. In this case we must express the upper surface area of the water as a function of h—that is, Aw A(h). See Problem 14 in Exercises 1.3. Series Circuits Consider the single-loop LRC-series circuit shown in Fig-ure 1.3.4(a), containing an inductor, resistor, and capacitor. The current in a circuit after a switch is closed is denoted by i(t); the charge on a capacitor at time t is de-noted by q(t). The letters L, R, and C are known as inductance, resistance, and capac-itance, respectively, and are generally constants. Now according to Kirchhoff’s second law, the impressed voltage E(t) on a closed loop must equal the sum of the voltage drops in the loop. Figure 1.3.4(b) shows the symbols and the formulas for the respective voltage drops across an inductor, a capacitor, and a resistor. Since current i(t) is related to charge q(t) on the capacitor by i dqdt, adding the three voltages inductor resistor capacitor and equating the sum to the impressed voltage yields a second-order differential equation (11) We will examine a differential equation analogous to (11) in great detail in Section 5.1. Falling Bodies To construct a mathematical model of the motion of a body moving in a force ﬁeld, one often starts with the laws of motion formulated by the English mathematician Isaac Newton (1643–1727). Recall from elementary physics that Newton’s first law of motio states that a body either will remain at rest or will continue to move with a constant velocity unless acted on by an external force. In each case this is equivalent to saying that when the sum of the forces — that is, the net or resultant force—acting on the body is zero, then the acceleration a of the body is zero. Newton’s second law of motion indicates that when the net force acting on a body is not zero, then the net force is proportional to its accelera-tion a or, more precisely, F ma, where m is the mass of the body. Now suppose a rock is tossed upward from the roof of a building as illustrated in Figure 1.3.5. What is the position s(t) of the rock relative to the ground at time t? The acceleration of the rock is the second derivative d2sdt2. If we assume that the upward direction is positive and that no force acts on the rock other than the force of gravity, then Newton’s second law gives . (12) In other words, the net force is simply the weight F F1 W of the rock near the surface of the Earth. Recall that the magnitude of the weight is W mg, where m is m d 2s dt2 mg or d 2s dt2 g F Fk L d 2q dt2 R dq dt 1 C q E(t). L di dt L d 2q dt2, iR R dq dt, and 1 C q dh dt Ah Aw 12gh 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 25 (a) (b) E(t) L C R (a) LRC-series circuit (b) L R Inductor inductance L: henries (h) voltage drop across: L di dt i Capacitor capacitance C: farads (f) voltage drop across: 1 C i Resistor resistance R: ohms (Ω) voltage drop across: iR i q C FIGURE 1.3.4 Symbols, units, and voltages. Current i(t) and charge q(t) are measured in amperes (A) and coulombs (C), respectively ground building rock s(t) s0 v0 FIGURE 1.3.5 Position of rock measured from ground level Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. the mass of the body and g is the acceleration due to gravity. The minus sign in (12) is used because the weight of the rock is a force directed downward, which is opposite to the positive direction. If the height of the building is s0 and the initial velocity of the rock is v0, then s is determined from the second-order initial-value problem . (13) Although we have not been stressing solutions of the equations we have con-structed, note that (13) can be solved by integrating the constant g twice with respect to t. The initial conditions determine the two constants of integration. From elementary physics you might recognize the solution of (13) as the formula Falling Bodies and Air Resistance Before the famous experiment by the Italian mathematician and physicist Galileo Galilei (1564–1642) from the leaning tower of Pisa, it was generally believed that heavier objects in free fall, such as a can-nonball, fell with a greater acceleration than lighter objects, such as a feather. Obviously, a cannonball and a feather when dropped simultaneously from the same height do fall at different rates, but it is not because a cannonball is heavier. The dif-ference in rates is due to air resistance. The resistive force of air was ignored in the model given in (13). Under some circumstances a falling body of mass m, such as a feather with low density and irregular shape, encounters air resistance proportional to its instantaneous velocity v. If we take, in this circumstance, the positive direction to be oriented downward, then the net force acting on the mass is given by F F1 F2 mg kv, where the weight F1 mg of the body is force acting in the positive direction and air resistance F2 kv is a force, called viscous damping, acting in the opposite or upward direction. See Figure 1.3.6. Now since v is related to acceleration a by a dvdt, Newton’s second law becomes F ma m dvdt. By equating the net force to this form of Newton’s second law, we obtain a ﬁrst-order differential equation for the velocity v(t) of the body at time t, . (14) Here k is a positive constant of proportionality. If s(t) is the distance the body falls in time t from its initial point of release, then v dsdt and a dvdt d2sdt2. In terms of s, (14) is a second-order differential equation (15) Suspended Cables Suppose a ﬂexible cable, wire, or heavy rope is sus-pended between two vertical supports. Physical examples of this could be one of the two cables supporting the roadbed of a suspension bridge as shown in Figure 1.3.7(a) or a long telephone wire strung between two posts as shown in Figure 1.3.7(b). Our goal is to construct a mathematical model that describes the shape that such a cable assumes. To begin, let’s agree to examine only a portion or element of the cable between its lowest point P1 and any arbitrary point P2. As drawn in blue in Figure 1.3.8, this element of the cable is the curve in a rectangular coordinate system with y-axis cho-sen to pass through the lowest point P1 on the curve and the x-axis chosen a units below P1. Three forces are acting on the cable: the tensions T1 and T2 in the cable that are tangent to the cable at P1 and P2, respectively, and the portion W of the total vertical load between the points P1 and P2. Let T1 T1, T2 T2, and W W denote the magnitudes of these vectors. Now the tension T2 resolves into horizontal and vertical components (scalar quantities) T2 cos and T2 sin . m d 2s dt 2 mg k ds dt or m d 2s dt 2 k ds dt mg. m dv dt mg kv s(t) 1 2gt2 v0t s0. d 2s dt 2 g, s(0) s0, s(0) v0 26 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS positive direction air resistance gravity kv mg FIGURE 1.3.6 Falling body of mass m (a) suspension bridge cable (b) telephone wires FIGURE 1.3.7 Cables suspended between vertical supports FIGURE 1.3.8 Element of cable cos wire T2 θ θ sin T2 T2 P2 T1 W P1 θ y x (x, 0) (0, a) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Because of static equilibrium we can write By dividing the last equation by the ﬁrst, we eliminate T2 and get tan WT1. But because dydx tan , we arrive at (16) This simple ﬁrst-order differential equation serves as a model for both the shape of a ﬂexible wire such as a telephone wire hanging under its own weight and the shape of the cables that support the roadbed of a suspension bridge. We will come back to equation (16) in Exercises 2.2 and Section 5.3. What Lies Ahead Throughout this text you will see three different types of approaches to, or analyses of, differential equations. Over the centuries differential equations would often spring from the efforts of a scientist or engineer to describe some physical phenomenon or to translate an empirical or experimental law into mathematical terms. As a consequence, a scientist, engineer, or mathematician would often spend many years of his or her life trying to ﬁnd the solutions of a DE. With a solution in hand, the study of its properties then followed. This quest for so-lutions is called by some the analytical approach to differential equations. Once they realized that explicit solutions are at best difﬁcult to obtain and at worst impossible to obtain, mathematicians learned that a differential equation itself could be a font of valuable information. It is possible, in some instances, to glean directly from the dif-ferential equation answers to questions such as Does the DE actually have solutions? If a solution of the DE exists and satisfie an initial condition, is it the only such so-lution? What are some of the properties of the unknown solutions? What can we say about the geometry of the solution curves? Such an approach is qualitative analysis. Finally, if a differential equation cannot be solved by analytical methods, yet we can prove that a solution exists, the next logical query is Can we somehow approxi-mate the values of an unknown solution? Here we enter the realm of numerical analysis. An afﬁrmative answer to the last question stems from the fact that a differ-ential equation can be used as a cornerstone for constructing very accurate approxi-mation algorithms. In Chapter 2 we start with qualitative considerations of ﬁrst-order ODEs, then examine analytical stratagems for solving some special ﬁrst-order equations, and conclude with an introduction to an elementary numerical method. See Figure 1.3.9. dy dx W T1 . T1 T2 cos and W T2 sin . 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 27 (a) analytical (b) qualitative (c) numerical y'=f(y) FIGURE 1.3.9 Different approaches to the study of differential equations Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 28 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS REMARKS Each example in this section has described a dynamical system—a system that changes or evolves with the ﬂow of time t. Since the study of dynamical systems is a branch of mathematics currently in vogue, we shall occasionally relate the terminology of that ﬁeld to the discussion at hand. In more precise terms, a dynamical system consists of a set of time-dependent variables, called state variables, together with a rule that enables us to determine (without ambiguity) the state of the system (this may be a past, present, or future state) in terms of a state prescribed at some time t0. Dynamical systems are classiﬁed as either discrete-time systems or continuous-time systems. In this course we shall be concerned only with continuous-time systems— systems in which all variables are deﬁned over a continuous range of time. The rule, or mathematical model, in a continuous-time dynamical system is a differ-ential equation or a system of differential equations. The state of the system at a time t is the value of the state variables at that time; the speciﬁed state of the system at a time t0 is simply the initial conditions that accompany the math-ematical model. The solution of the initial-value problem is referred to as the response of the system. For example, in the case of radioactive decay, the rule is dAdt kA. Now if the quantity of a radioactive substance at some time t0 is known, say A(t0) A0, then by solving the rule we ﬁnd that the response of the system for t t0 is (see Section 3.1). The response A(t) is the single state variable for this system. In the case of the rock tossed from the roof of a building, the response of the system—the solution of the differential equation d2sdt2 g, subject to the initial state s(0) s0, s(0) v0, is the function , where T represents the time when the rock hits the ground. The state variables are s(t) and s(t), which are the vertical position of the rock above ground and its velocity at time t, respectively. The acceleration s(t) is not a state variable, since we have to know only any initial position and initial velocity at a time t0 to uniquely determine the rock’s position s(t) and velocity s(t) v(t) for any time in the interval t0 t T. The acceleration s(t) a(t) is, of course, given by the differential equation s(t) g, 0 t T. One last point: Not every system studied in this text is a dynamical system. We shall also examine some static systems in which the model is a differential equation. s(t) 1 2gt2 v0t s0, 0 t T A(t) A0e(tt0) EXERCISES 1.3 Answers to selected odd-numbered problems begin on page ANS-1. Population Dynamics 1. Under the same assumptions that underlie the model in (1), determine a differential equation for the population P(t) of a country when individuals are allowed to immigrate into the country at a constant rate r 0. What is the differential equation for the population P(t) of the country when individuals are allowed to emigrate from the country at a constant rate r 0? 2. The population model given in (1) fails to take death into consideration; the growth rate equals the birth rate. In another model of a changing population of a commu-nity it is assumed that the rate at which the population changes is a net rate—that is, the difference between the rate of births and the rate of deaths in the commu-nity. Determine a model for the population P(t) if both the birth rate and the death rate are proportional to the population present at time t 0. 3. Using the concept of net rate introduced in Problem 2, determine a model for a population P(t) if the birth rate is proportional to the population present at time t but the death rate is proportional to the square of the population present at time t. 4. Modify the model in Problem 3 for net rate at which the population P(t) of a certain kind of ﬁsh changes by also assuming that the ﬁsh are harvested at a constant rate h 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 29 Newton’s Law of Cooling/Warming 5. A cup of coffee cools according to Newton’s law of cooling (3). Use data from the graph of the temperature T(t) in Figure 1.3.10 to estimate the constants Tm, T0, and k in a model of the form of a ﬁrst-order initial-value problem: dTdt k(T Tm), T(0) T0. number of people x(t) who have adopted the innovation at time t if it is assumed that the rate at which the innova-tions spread through the community is jointly propor-tional to the number of people who have adopted it and the number of people who have not adopted it. Mixtures 9. Suppose that a large mixing tank initially holds 300 gal-lons of water in which 50 pounds of salt have been dis-solved. Pure water is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differen-tial equation for the amount of salt A(t) in the tank at time t 0. What is A(0)? 10. Suppose that a large mixing tank initially holds 300 gal-lons of water is which 50 pounds of salt have been dissolved.Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation for the amount of salt A(t) in the tank at time t 0. 11. What is the differential equation in Problem 10, if the well-stirred solution is pumped out at a faster rate of 3.5 gal/min? 12. Generalize the model given in equation (8) on page 24 by assuming that the large tank initially contains N0 number of gallons of brine, rin and rout are the input and output rates of the brine, respectively (measured in gal-lons per minute), cin is the concentration of the salt in the inﬂow, c(t) the concentration of the salt in the tank as well as in the outﬂow at time t (measured in pounds of salt per gallon), and A(t) is the amount of salt in the tank at time t 0. Draining a Tank 13. Suppose water is leaking from a tank through a circular hole of area Ah at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to where c (0 c 1) is an empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown in Figure 1.3.12. The radius of the hole is 2 in., and g 32 ft/s2. cAh12gh, FIGURE 1.3.10 Cooling curve in Problem 5 FIGURE 1.3.11 Ambient temperature in Problem 6 t T 25 50 minutes 75 100 100 50 200 0 150 t Tm(t) 12 noon 24 midnight 36 noon 48 midnight 0 midnight 20 40 60 80 100 120 6. The ambient temperature Tm in (3) could be a function of time t. Suppose that in an artiﬁcially controlled environment, Tm(t) is periodic with a 24-hour period, as illustrated in Figure 1.3.11. Devise a mathematical model for the temperature T(t) of a body within this environment. Spread of a Disease/Technology 7. Suppose a student carrying a ﬂu virus returns to an iso-lated college campus of 1000 students. Determine a dif-ferential equation for the number of people x(t) who have contracted the ﬂu if the rate at which the disease spreads is proportional to the number of interactions between the number of students who have the ﬂu and the number of students who have not yet been exposed to it. 8. At a time denoted as t 0 a technological innovation is introduced into a community that has a ﬁxed population of n people. Determine a differential equation for the h circular hole 10 ft Aw FIGURE 1.3.12 Cubical tank in Problem 13 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 14. The right-circular conical tank shown in Figure 1.3.13 loses water out of a circular hole at its bottom. Determine a differential equation for the height of the water h at time t 0. The radius of the hole is 2 in., g 32 ft/s2, and the friction/contraction factor introduced in Problem 13 is c 0.6. 30 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Newton’s Second Law and Archimedes’ Principle 18. A cylindrical barrel s feet in diameter of weight w lb is ﬂoating in water as shown in Figure 1.3.17(a). After an initial depression the barrel exhibits an up-and-down bobbing motion along a vertical line. Using Figure 1.3.17(b), determine a differential equation for the vertical displacement y(t) if the origin is taken to be on the vertical axis at the surface of the water when the barrel is at rest. Use Archimedes’principle: Buoyancy, or upward force of the water on the barrel, is equal to the weight of the water displaced. Assume that the downward direction is positive, that the weight density of water is 62.4 lb/ft3, and that there is no resistance between the barrel and the water. FIGURE 1.3.13 Conical tank in Problem 14 L R E FIGURE 1.3.14 LR-series circuit in Problem 15 FIGURE 1.3.15 RC-series circuit in Problem 16 FIGURE 1.3.16 Air resistance proportional to square of velocity in Problem 17 R C E FIGURE 1.3.17 Bobbing motion of ﬂoating barrel in Problem 18 8 ft circular hole h 20 ft Aw mg kv2 SKYDIVING MADE EASY 0 y(t) 0 surface s/2 (b) (a) s/2 Series Circuits 15. A series circuit contains a resistor and an inductor as shown in Figure 1.3.14. Determine a differential equa-tion for the current i(t) if the resistance is R, the induc-tance is L, and the impressed voltage is E(t). 16. A series circuit contains a resistor and a capacitor as shown in Figure 1.3.15. Determine a differential equa-tion for the charge q(t) on the capacitor if the resis-tance is R, the capacitance is C, and the impressed voltage is E(t). Falling Bodies and Air Resistance 17. For high-speed motion through the air—such as the skydiver shown in Figure 1.3.16, falling before the parachute is opened—air resistance is closer to a power of the instantaneous velocity v(t). Determine a differen-tial equation for the velocity v(t) of a falling body of mass m if air resistance is proportional to the square of the instantaneous velocity. Newton’s Second Law and Hooke’s Law 19. After a mass m is attached to a spring, it stretches it s units and then hangs at rest in the equilibrium position as shown in Figure 1.3.18(b). After the spring/mass FIGURE 1.3.18 Spring/mass system in Problem 19 unstretched spring equilibrium position m x = 0 x(t) > 0 x(t) < 0 m s (a) (b) (c) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 31 system has been set in motion, let x(t) denote the di-rected distance of the mass beyond the equilibrium po-sition. As indicated in Figure 1.3.17(c), assume that the downward direction is positive, that the motion takes place in a vertical straight line through the center of gravity of the mass, and that the only forces acting on the system are the weight of the mass and the restoring force of the stretched spring. Use Hooke’s law: The restoring force of a spring is proportional to its total elongation. Determine a differential equation for the displacement x(t) at time t 0. 20. In Problem 19, what is a differential equation for the displacement x(t) if the motion takes place in a medium that imparts a damping force on the spring/ mass system that is proportional to the instantaneous velocity of the mass and acts in a direction opposite to that of motion? Newton’s Second Law and Rocket Motion When the mass m of a body is changing with time, Newton’s second law of motion becomes (17) where F is the net force acting on the body and mv is its momentum. Use (17) in Problems 21 and 22. 21. A small single-stage rocket is launched vertically as shown in Figure 1.3.19. Once launched, the rocket con-sumes its fuel, and so its total mass m(t) varies with time t 0. If it is assumed that the positive direction is up-ward, air resistance is proportional to the instantaneous velocity v of the rocket, and R is the upward thrust or force generated by the propulsion system, then con-struct a mathematical model for the velocity v(t) of the rocket. [Hint: See (14) in Section 1.3.] F d dt (mv), FIGURE 1.3.20 Satellite in Problem 23 FIGURE 1.3.21 Hole through Earth in Problem 24 Earth of mass M R satellite of mass m r s u rf a c e satellite of mass m surface m R r 24. Suppose a hole is drilled through the center of the Earth and a bowling ball of mass m is dropped into the hole, as shown in Figure 1.3.21. Construct a mathematical model that describes the motion of the ball. At time t let r de-note the distance from the center of the Earth to the mass m, M denote the mass of the Earth, Mr denote the mass of that portion of the Earth within a sphere of radius r, and denote the constant density of the Earth. FIGURE 1.3.19 Single-stage rocket in Problem 21 22. In Problem 21, the mass m(t) is the sum of three differ-ent masses: where mp is the constant mass of the payload, mv is the constant mass of the vehicle, and mf(t) is the variable amount of fuel. (a) Show that the rate at which the total mass m(t) of the rocket changes is the same as the rate at which the mass mf(t) of the fuel changes. (b) If the rocket consumes its fuel at a constant rate , ﬁnd m(t). Then rewrite the differential equation in Problem 21 in terms of and the initial total mass m(0) m0. (c) Under the assumption in part (b), show that the burnout time tb 0 of the rocket, or the time at which all the fuel is consumed, is where mf(0) is the initial mass of the fuel. Newton’s Second Law and the Law of Universal Gravitation 23. By Newton’s universal law of gravitation the free-fall acceleration a of a body, such as the satellite shown in Figure 1.3.20, falling a great distance to the surface is not the constant g. Rather, the acceleration a is inversely pro-portional to the square of the distance from the center of the Earth, a kr2, where k is the constant of proportion-ality. Use the fact that at the surface of the Earth r R and a g to determine k. If the positive direction is upward, use Newton’s second law and his universal law of gravita-tion to ﬁnd a differential equation for the distance r. tb mf(0)>l, l l m(t) mp mv mf(t), James L. Davidson/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. equation that describes the shape of the curve C. Such a curve C is important in applications ranging from con-struction of telescopes to satellite antennas, automobile headlights, and solar collectors. [Hint: Inspection of the ﬁgure shows that we can write 2. Why? Now use an appropriate trigonometric identity.] Discussion Problems 30. Reread Problem 41 in Exercises 1.1 and then give an explicit solution P(t) for equation (1). Find a one-parameter family of solutions of (1). 31. Reread the sentence following equation (3) and assume that Tm is a positive constant. Discuss why we would ex-pect k 0 in (3) in both cases of cooling and warming. You might start by interpreting, say, T(t) Tm in a graphical manner. 32. Reread the discussion leading up to equation (8). If we assume that initially the tank holds, say, 50 lb of salt, it stands to reason that because salt is being added to the tank continuously for t 0, A(t) should be an increas-ing function. Discuss how you might determine from the DE, without actually solving it, the number of pounds of salt in the tank after a long period of time. 33. Population Model The differential equation where k is a positive constant, is a model of human population P(t) of a certain commu-nity. Discuss an interpretation for the solution of this equation. In other words, what kind of population do you think the differential equation describes? 34. Rotating Fluid As shown in Figure 1.3.24(a), a right-circular cylinder partially ﬁlled with ﬂuid is rotated with a constant angular velocity about a vertical y-axis through its center. The rotating ﬂuid forms a surface of revolution S. To identify S, we ﬁrst establish a coordinate system consisting of a vertical plane determined by the y-axis and an x-axis drawn perpendicular to the y-axis such that the point of intersection of the axes (the origin) is located at the lowest point on the surface S. We then seek a function y f(x) that represents the curve C of in-tersection of the surface S and the vertical coordinate plane. Let the point P(x, y) denote the position of a parti-cle of the rotating ﬂuid of mass m in the coordinate plane. See Figure 1.3.23(b). (a) At P there is a reaction force of magnitude F due to the other particles of the ﬂuid which is normal to the surface S. By Newton’s second law the magnitude of the net force acting on the particle is m2x. What is this force? Use Figure 1.3.24(b) to discuss the na-ture and origin of the equations (b) Use part (a) to ﬁnd a ﬁrst-order differential equation that deﬁnes the function y f (x). F cos mg, F sin m2x. dP dt (k cos t)P, 32 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Additional Mathematical Models 25. Learning Theory In the theory of learning, the rate at which a subject is memorized is assumed to be pro-portional to the amount that is left to be memorized. Suppose M denotes the total amount of a subject to be memorized and A(t) is the amount memorized in time t 0. Determine a differential equation for the amount A(t). 26. Forgetfulness In Problem 25 assume that the rate at which material is forgotten is proportional to the amount memorized in time t 0. Determine a differential equa-tion for the amount A(t) when forgetfulness is taken into account. 27. Infusion of a Drug A drug is infused into a patient’s bloodstream at a constant rate of r grams per second. Simultaneously, the drug is removed at a rate proportional to the amount x(t) of the drug present at time t. Determine a differential equation for the amount x(t). 28. Tractrix Aperson P, starting at the origin, moves in the direction of the positive x-axis, pulling a weight along the curve C, called a tractrix, as shown in Figure 1.3.22. The weight, initially located on the y-axis at (0, s), is pulled by a rope of constant length s, which is kept taut throughout the motion. Determine a differential equation for the path C of motion. Assume that the rope is always tangent to C. FIGURE 1.3.22 Tractrix curve in Problem 28 FIGURE 1.3.23 Reﬂecting surface in Problem 29 y x (0, s) (x, y) y s P θ C C θ θ φ x y O P(x, y) L tangent 29. Reflecting Surface Assume that when the plane curve C shown in Figure 1.3.23 is revolved about the x-axis, it generates a surface of revolution with the property that all light rays L parallel to the x-axis strik-ing the surface are reﬂected to a single point O (the origin). Use the fact that the angle of incidence is equal to the angle of reﬂection to determine a differential Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 1 IN REVIEW ● 33 35. Falling Body In Problem 23, suppose r R s, where s is the distance from the surface of the Earth to the falling body. What does the differential equation obtained in Problem 23 become when s is very small in comparison to R? [Hint: Think binomial series for (R s)2 R2 (1 sR)2.] ω P y x θ θ mg 2x mω P(x, y) F tangent line to curve C at P curve C of intersection of xy-plane and surface of revolution (a) (b) y FIGURE 1.3.24 Rotating ﬂuid in Problem 34 36. Raindrops Keep Falling In meteorology the term virga refers to falling raindrops or ice particles that evaporate before they reach the ground. Assume that a typical raindrop is spherical. Starting at some time, which we can designate as t 0, the raindrop of radius r0 falls from rest from a cloud and begins to evaporate. (a) If it is assumed that a raindrop evaporates in such a manner that its shape remains spherical, then it also makes sense to assume that the rate at which the rain-drop evaporates—that is, the rate at which it loses mass—is proportional to its surface area. Show that this latter assumption implies that the rate at which the radius r of the raindrop decreases is a constant. Find r(t). [Hint: See Problem 51 in Exercises 1.1.] (b) If the positive direction is downward, construct a mathematical model for the velocity v of the falling raindrop at time t 0. Ignore air resistance. [Hint: Use the form of Newton’s second law given in (17).] 37. Let It Snow The “snowplow problem” is a classic and appears in many differential equations texts, but it was probably made famous by Ralph Palmer Agnew: One day it started snowing at a heavy and steady rate. A snowplow started out at noon, going 2 miles the first hour and 1 mile the second hou . What time did it start snowing? Find the textbook Differential Equations, Ralph Palmer Agnew, McGraw-Hill Book Co., and then discuss the construction and solution of the mathematical model. 38. Reread this section and classify each mathematical model as linear or nonlinear. CHAPTER 1 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1 and 2 ﬁll in the blank and then write this result as a linear ﬁrst-order differential equation that is free of the symbol c1 and has the form dydx f (x, y). The symbol c1 represents a constant. 1. 2. In Problems 3 and 4 ﬁll in the blank and then write this result as a linear second-order differential equation that is free of the symbols c1 and c2 and has the form F(y, y) 0. The symbols c1, c2, and k represent constants. 3. 4. d 2 dx2 (c1 cosh kx c2 sinh kx) d 2 dx2 (c1 cos kx c2 sin kx) d dx (5 c1e2x) d dx c1e10x In Problems 5 and 6 compute y and y and then combine these derivatives with y as a linear second-order differential equation that is free of the symbols c1 and c2 and has the form F(y, y y) 0. The symbols c1 and c2 represent constants. 5. y c1ex c2xex 6. y c1ex cos x c2ex sin x In Problems 7–12 match each of the given differential equa-tions with one or more of these solutions: (a) y 0, (b) y 2, (c) y 2x, (d) y 2x2. 7. xy 2y 8. y 2 9. y 2y 4 10. xy y 11. y 9y 18 12. xy y 0 In Problems 13 and 14 determine by inspection at least one solution of the given differential equation. 13. y y 14. y y(y 3) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In Problems 27–30 verify that the indicated expression is an implicit solution of the given differential equation. 27. 28. 29. 30. In Problems 31–34, y c1e3x c2ex 2x is a two-parameter family of the second-order DE y 2y 3y 6x 4. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 31. y (0) 0, y(0) 0 32. y (0) 1, y(0) 3 33. y (1) 4, y(1) 2 34. y (1) 0, y(1) 1 35. The graph of a solution of a second-order initial-value problem d2ydx2 f (x, y, y), y(2) y0, y(2) y1, is given in Figure 1.R.1. Use the graph to estimate the val-ues of y0 and y1. (1 xy)y y2; y exy y 2y(y)3; y3 3y 1 3x dy dx 2 1 1 y2 ; (x 5)2 y2 1 x dy dx y 1 y2; x3y3 x3 1 34 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS In Problems 15 and 16 interpret each statement as a differen-tial equation. 15. On the graph of y (x) the slope of the tangent line at a point P(x, y) is the square of the distance from P(x, y) to the origin. 16. On the graph of y (x) the rate at which the slope changes with respect to x at a point P(x, y) is the nega-tive of the slope of the tangent line at P(x, y). 17. (a) Give the domain of the function y x2/3. (b) Give the largest interval I of deﬁnition over which y x2/3 is solution of the differential equation 3xy 2y 0. 18. (a) Verify that the one-parameter family y2 2y x2 x c is an implicit solution of the differential equation (2y 2)y 2x 1. (b) Find a member of the one-parameter family in part (a) that satisﬁes the initial condition y(0) 1. (c) Use your result in part (b) to ﬁnd an explicit function y (x) that satisﬁes y(0) 1. Give the domain of the function . Is y (x) a solution of the initial-value problem? If so, give its interval I of deﬁnition; if not, explain. 19. Given that y x 2x is a solution of the DE xy y 2x. Find x0 and the largest interval I for which y(x) is a solution of the ﬁrst-order IVP xy y 2x, y(x0) 1. 20. Suppose that y(x) denotes a solution of the first-order IVP y x2 y2, y(1) 1 and that y(x) possesses at least a second derivative at x 1. In some neigh-borhood of x 1 use the DE to determine whether y(x) is increasing or decreasing and whether the graph y(x) is concave up or concave down. 21. A differential equation may possess more than one fam-ily of solutions. (a) Plot different members of the families y 1(x) x2 c1 and y 2(x) x2 c2. (b) Verify that y 1(x) and y 2(x) are two solutions of the nonlinear ﬁrst-order differential equation (y)2 4x2. (c) Construct a piecewise-deﬁned function that is a solution of the nonlinear DE in part (b) but is not a member of either family of solutions in part (a). 22. What is the slope of the tangent line to the graph of a solution of that passes through (1, 4)? In Problems 23–26 verify that the indicated function is an explicit solution of the given differential equation. Give an interval of deﬁnition I for each solution. 23. y y 2 cos x 2 sin x; y x sin x x cos x 24. y y sec x; y x sin x (cos x)ln(cos x) 25. x2y xy y 0; y sin(ln x) 26. x2y xy y sec(ln x); y cos(ln x) ln(cos(ln x)) (ln x) sin(ln x) y 61y 5x3 y x 5 −5 5 FIGURE 1.R.1 Graph for Problem 35 36. A tank in the form of a right-circular cylinder of radius 2 feet and height 10 feet is standing on end. If the tank is initially full of water and water leaks from a circular hole of radius inch at its bottom, determine a differen-tial equation for the height h of the water at time t 0. Ignore friction and contraction of water at the hole. 37. The number of ﬁeld mice in a certain pasture is given by the function 200 10t, where time t is measured in years. Determine a differential equation governing a population of owls that feed on the mice if the rate at which the owl population grows is proportional to the difference between the number of owls at time t and number of ﬁeld mice at time t 0. 38. Suppose that dAdt 0.0004332 A(t) represents a mathematical model for the radioactive decay of radium-226, where A(t) is the amount of radium (measured in grams) remaining at time t (measured in years). How much of the radium sample remains at the time t when the sample is decaying at a rate of 0.002 gram per year? 1 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 35 2 2.1 Solution Curves Without a Solution 2.1.1 Direction Fields 2.1.2 Autonomous First-Order DEs 2.2 Separable Equations 2.3 Linear Equations 2.4 Exact Equations 2.5 Solutions by Substitutions 2.6 A Numerical Method Chapter 2 in Review The history of mathematics is rife with stories of people who devoted much of their lives to solving equations—algebraic equations at ﬁrst and then eventually differential equations. In Sections 2.2–2.5 we will study some of the more important analytical methods for solving ﬁrst-order DEs. However, before we start solving anything, you should be aware of two facts: It is possible for a differential equation to have no solutions, and a differential equation can possess solutions, yet there might not exist any analytical method for solving it. In Sections 2.1 and 2.6 we do not solve any DEs but show how to glean information about solutions directly from the equation itself. In Section 2.1 we see how the DE yields qualitative information about graphs that enables us to sketch renditions of solution curves. In Section 2.6 we use the differential equation to construct a procedure, called a numerical method, for approximating solutions. First-Order Differential Equations Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 36 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS solution curv e (a) lineal element at a point (b) lineal element is tangent to solution curve that passes through the point slope = 1.2 (2, 3) x y tangent (2, 3) x y FIGURE 2.1.1 A solution curve is tangent to lineal element at (2, 3) SOLUTION CURVES WITHOUT A SOLUTION REVIEW MATERIAL ●The ﬁrst derivative as slope of a tangent line ●The algebraic sign of the ﬁrst derivative indicates increasing or decreasing INTRODUCTION Let us imagine for the moment that we have in front of us a ﬁrst-order differ-ential equation dydx f (x, y), and let us further imagine that we can neither ﬁnd nor invent a method for solving it analytically. This is not as bad a predicament as one might think, since the dif-ferential equation itself can sometimes “tell” us speciﬁcs about how its solutions “behave.” We begin our study of ﬁrst-order differential equations with two ways of analyzing a DE qual-itatively. Both these ways enable us to determine, in an approximate sense, what a solution curve must look like without actually solving the equation. 2.1 2.1.1 DIRECTION FIELDS Some Fundamental Questions We saw in Section 1.2 that whenever f (x, y) and fy satisfy certain continuity conditions, qualitative questions about existence and uniqueness of solutions can be answered. In this section we shall see that other qualitative questions about properties of solutions—How does a solution behave near a certain point? How does a solution behave as ?—can often be an-swered when the function f depends solely on the variable y. We begin, however, with a simple concept from calculus: A derivative dydx of a differentiable function y y(x) gives slopes of tangent lines at points on its graph. Slope Because a solution y y(x) of a ﬁrst-order differential equation (1) is necessarily a differentiable function on its interval I of deﬁnition, it must also be con-tinuous on I. Thus the corresponding solution curve on I must have no breaks and must possess a tangent line at each point (x, y(x)). The function f in the normal form (1) is called the slope function or rate function. The slope of the tangent line at (x, y(x)) on a solution curve is the value of the ﬁrst derivative dydx at this point, and we know from (1) that this is the value of the slope function f(x, y(x)). Now suppose that (x, y) represents any point in a region of the xy-plane over which the function f is deﬁned. The value f(x, y) that the function f assigns to the point represents the slope of a line or, as we shall envision it, a line segment called a lineal element. For example, consider the equation dydx 0.2xy, where f(x, y) 0.2xy. At, say, the point (2, 3) the slope of a lineal element is f(2, 3) 0.2(2)(3) 1.2. Figure 2.1.1(a) shows a line segment with slope 1.2 passing though (2, 3). As shown in Figure 2.1.1(b), if a solution curve also passes through the point (2, 3), it does so tangent to this line segment; in other words, the lineal element is a miniature tangent line at that point. Direction Field If we systematically evaluate f over a rectangular grid of points in the xy-plane and draw a line element at each point (x, y) of the grid with slope f (x, y), then the collection of all these line elements is called a direction fiel or a slope fiel of the differential equation dydx f (x, y). Visually, the direction ﬁeld suggests the appearance or shape of a family of solution curves of the differential equation, and consequently, it may be possible to see at a glance certain qualitative aspects of the solutions—regions in the plane, for example, in which a dy dx f (x, y) x : Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.1 SOLUTION CURVES WITHOUT A SOLUTION ● 37 solution exhibits an unusual behavior. A single solution curve that passes through a direction ﬁeld must follow the ﬂow pattern of the ﬁeld; it is tangent to a lineal element when it intersects a point in the grid. Figure 2.1.2 shows a computer-generated direc-tion ﬁeld of the differential equation dydx sin(x y) over a region of the xy-plane. Note how the three solution curves shown in color follow the ﬂow of the ﬁeld. EXAMPLE 1 Direction Field The direction ﬁeld for the differential equation dydx 0.2xy shown in Figure 2.1.3(a) was obtained by using computer software in which a 5 5 grid of points (mh, nh), m and n integers, was deﬁned by letting 5 m 5, 5 n 5, and h 1. Notice in Figure 2.1.3(a) that at any point along the x-axis (y 0) and the y-axis (x 0), the slopes are f (x, 0) 0 and f (0, y) 0, respectively, so the lineal elements are horizontal. Moreover, observe in the ﬁrst quadrant that for a ﬁxed value of x the values of f (x, y) 0.2xy increase as y increases; similarly, for a ﬁxed y the values of f (x, y) 0.2xy increase as x increases. This means that as both x and y increase, the lineal elements almost become vertical and have positive slope ( f(x, y) 0.2xy 0 for x 0, y 0). In the second quadrant, f (x, y) increases as x and y increase, so the lineal elements again become almost vertical but this time have negative slope ( f (x, y) 0.2xy 0 for x 0, y 0). Reading from left to right, imagine a solution curve that starts at a point in the second quadrant, moves steeply downward, becomes ﬂat as it passes through the y-axis, and then, as it enters the ﬁrst quadrant, moves steeply upward—in other words, its shape would be concave upward and similar to a horseshoe. From this it could be surmised that y : as x : . Now in the third and fourth quadrants, since f (x, y) 0.2xy 0 and f (x, y) 0.2xy 0, respectively, the situation is reversed: Asolution curve increases and then decreases as we move from left to right. We saw in (1) of Section 1.1 that is an explicit solution of the differential equation dydx 0.2xy; you should verify that a one-parameter family of solutions of the same equation is given by . For purposes of comparison with Figure 2.1.3(a) some representative graphs of members of this family are shown in Figure 2.1.3(b). y ce0.1x2 y e0.1x2 EXAMPLE 2 Direction Field Use a direction ﬁeld to sketch an approximate solution curve for the initial-value problem dydx sin y, . SOLUTION Before proceeding, recall that from the continuity of f (x, y) sin y and fy cos y, Theorem 1.2.1 guarantees the existence of a unique solution curve passingthroughanyspeciﬁedpoint(x0, y0)intheplane.Nowwesetourcomputersoft-ware again for a 5 5 rectangular region and specify (because of the initial condition) points in that region with vertical and horizontal separation of unit—that is, at points (mh, nh), , m and n integers such that 10 m 10, 10 n 10. The result is shown in Figure 2.1.4. Because the right-hand side of dydx sin y is 0 at y 0, and at y , the lineal elements are horizontal at all points whose second coordinates are y 0 or y . It makes sense then that a solution curve passing through the initial point (0, has the shape shown in the ﬁgure. Increasing/Decreasing Interpretation of the derivative dydx as a function that gives slope plays the key role in the construction of a direction ﬁeld. Another telling property of the ﬁrst derivative will be used next, namely, if dydx 0 (or dydx 0) for all x in an interval I, then a differentiable function y y(x) is increasing (or decreasing) on I. 3 2) h 1 2 1 2 y(0) 3 2 c>0 c<0 x y 4 _4 _4 _2 2 4 _4 _2 2 4 _2 2 4 _4 _2 2 x y c=0 (b) some solution curves in the family y ce0.1x2 (a) direction fiel for dy/dx 0.2xy FIGURE 2.1.3 Direction ﬁeld and solution curves in Example 1 FIGURE 2.1.2 Solution curves following ﬂow of a direction ﬁeld Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. x y _4 _2 2 4 4 _4 _2 2 FIGURE 2.1.4 Direction ﬁeld in Example 2 on page 37 38 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS REMARKS Sketching a direction ﬁeld by hand is straightforward but time consuming; it is probably one of those tasks about which an argument can be made for doing it once or twice in a lifetime, but it is overall most efﬁciently carried out by means of computer software. Before calculators, PCs, and software the method of isoclines was used to facilitate sketching a direction ﬁeld by hand. For the DE dydx f (x, y), any member of the family of curves f (x, y) c, c a constant, is called an isocline. Lineal elements drawn through points on a speciﬁc iso-cline, say, f(x, y) c1 all have the same slope c1. In Problem 15 in Exercises 2.1 you have your two opportunities to sketch a direction ﬁeld by hand. 2.1.2 AUTONOMOUS FIRST-ORDER DEs Autonomous First-Order DEs In Section 1.1 we divided the class of ordi-nary differential equations into two types: linear and nonlinear. We now consider brieﬂy another kind of classiﬁcation of ordinary differential equations, a classiﬁca-tion that is of particular importance in the qualitative investigation of differential equations. An ordinary differential equation in which the independent variable does not appear explicitly is said to be autonomous. If the symbol x denotes the indepen-dent variable, then an autonomous ﬁrst-order differential equation can be written as f (y, y) 0 or in normal form as . (2) We shall assume throughout that the function f in (2) and its derivative f are contin-uous functions of y on some interval I. The ﬁrst-order equations f (y) f (x, y) p p are autonomous and nonautonomous, respectively. Many differential equations encountered in applications or equations that are models of physical laws that do not change over time are autonomous. As we have already seen in Section 1.3, in an applied context, symbols other than y and x are rou-tinely used to represent the dependent and independent variables. For example, if t represents time then inspection of , where k, n, and Tm are constants, shows that each equation is time independent. Indeed, all of the ﬁrst-order differential equations introduced in Section 1.3 are time independent and so are autonomous. Critical Points The zeros of the function f in (2) are of special importance. We say that a real number c is a critical point of the autonomous differential equation (2) if it is a zero of f—that is, f(c) 0. A critical point is also called an equilibrium point or stationary point. Now observe that if we substitute the constant function y(x) c into (2), then both sides of the equation are zero. This means: If c is a critical point of (2), then y(x) c is a constant solution of the autonomous differential equation. A constant solution y(x) c of (2) is called an equilibrium solution; equilibria are the only constant solutions of (2). dA dt kA, dx dt kx(n 1 x), dT dt k(T Tm), dA dt 6 1 100 A dy dx 1 y2 and dy dx 0.2xy dy dx f (y) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. As was already mentioned, we can tell when a nonconstant solution y y(x) of (2) is increasing or decreasing by determining the algebraic sign of the derivative dydx; in the case of (2) we do this by identifying intervals on the y-axis over which the function f (y) is positive or negative. 2.1 SOLUTION CURVES WITHOUT A SOLUTION ● 39 EXAMPLE 3 An Autonomous DE The differential equation where a and b are positive constants, has the normal form dPdt f(P), which is (2) with t and P playing the parts of x and y, respectively, and hence is autonomous. From f(P) P(a bP) 0 we see that 0 and ab are critical points of the equation, so the equilibrium solutions are P(t) 0 and P(t) ab. By putting the critical points on a vertical line, we divide the line into three intervals deﬁned by P 0, 0 P ab, ab P . The arrows on the line shown in Figure 2.1.5 indicate the algebraic sign of f(P) P(a bP) on these intervals and whether a nonconstant solution P(t) is increasing or decreasing on an interval. The following table explains the ﬁgure. Interval Sign of f (P) P(t) Arrow (, 0) minus decreasing points down (0, ab) plus increasing points up (ab, ) minus decreasing points down Figure 2.1.5 is called a one-dimensional phase portrait, or simply phase portrait, of the differential equation dPdt P(a bP). The vertical line is called a phase line. Solution Curves Without solving an autonomous differential equation, we can usually say a great deal about its solution curves. Since the function f in (2) is independent of the variable x, we may consider f deﬁned for x or for 0 x . Also, since f and its derivative f are continuous functions of y on some interval I of the y-axis, the fundamental results of Theorem 1.2.1 hold in some hori-zontal strip or region R in the xy-plane corresponding to I, and so through any point (x0, y0) in R there passes only one solution curve of (2). See Figure 2.1.6(a). For the sake of discussion, let us suppose that (2) possesses exactly two critical points c1 and c2 and that c1 c2. The graphs of the equilibrium solutions y(x) c1 and y(x) c2 are horizontal lines, and these lines partition the region R into three subregions R1, R2, and R3, as illustrated in Figure 2.1.6(b). Without proof here are some conclusions that we can draw about a nonconstant solution y(x) of (2): • If (x0, y0) is in a subregion Ri, i 1, 2, 3, and y(x) is a solution whose graph passes through this point, then y(x) remains in the subregion Ri for all x. As illustrated in Figure 2.1.6(b), the solution y(x) in R2 is bounded below by c1 and above by c2, that is, c1 y(x) c2 for all x. The solution curve stays within R2 for all x because the graph of a nonconstant solution of (2) cannot cross the graph of either equilibrium solution y(x) c1 or y(x) c2. See Problem 33 in Exercises 2.1. • By continuity of f we must then have either f (y) 0 or f (y) 0 for all x in a subregion Ri, i 1, 2, 3. In other words, f (y) cannot change signs in a subregion. See Problem 33 in Exercises 2.1. dP dt P(a bP), P-axis a 0 b FIGURE 2.1.5 Phase portrait of DE in Example 3 R I R1 R2 (x0, y0) (x0, y0) y(x) = c2 y(x) = c1 R3 y y x x (a) region R (b) subregions R1, R2, and R3 of R FIGURE 2.1.6 Lines y(x) c1 and y(x) c2 partition R into three horizontal subregions Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. R1 R2 P0 P0 P0 P P a b 0 t R3 phase line decreasing decreasing increasing tP-plane FIGURE 2.1.7 Phase portrait and solution curves in Example 4 40 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS • Since dydx f (y(x)) is either positive or negative in a subregion Ri, i 1, 2, 3, a solution y(x) is strictly monotonic—that is, y(x) is either increasing or decreasing in the subregion Ri. Therefore y(x) cannot be oscillatory, nor can it have a relative extremum (maximum or minimum). See Problem 33 in Exercises 2.1. • If y(x) is bounded above by a critical point c1 (as in subregion R1 where y(x) c1 for all x), then the graph of y(x) must approach the graph of the equilibrium solution y(x) c1 either as x : or as x : . If y(x) is bounded—that is, bounded above and below by two consecutive critical points (as in subregion R2 where c1 y(x) c2 for all x)—then the graph of y(x) must approach the graphs of the equilibrium solutions y(x) c1 and y(x) c2, one as x : and the other as x : . If y(x) is bounded below by a critical point (as in subregion R3 where c2 y(x) for all x), then the graph of y(x) must approach the graph of the equilibrium solution y(x) c2 either as x : or as x : . See Problem 34 in Exercises 2.1. With the foregoing facts in mind, let us reexamine the differential equation in Example 3. EXAMPLE 4 Example 3 Revisited The three intervals determined on the P-axis or phase line by the critical points 0 and ab now correspond in the tP-plane to three subregions deﬁned by: R1: P 0, R2: 0 P ab, and R3: ab P , where t . The phase portrait in Figure 2.1.7 tells us that P(t) is decreasing in R1, increasing in R2, and decreasing in R3. If P(0) P0 is an initial value, then in R1, R2, and R3 we have, respectively, the following: (i) For P0 0, P(t) is bounded above. Since P(t) is decreasing, P(t) decreases without bound for increasing t, and so P(t) : 0 as t : . This means that the negative t-axis, the graph of the equilibrium solution P(t) 0, is a horizontal asymptote for a solution curve. (ii) For 0 P0 ab, P(t) is bounded. Since P(t) is increasing, P(t) : ab as t : and P(t) : 0 as t : . The graphs of the two equilibrium solutions, P(t) 0 and P(t) ab, are horizontal lines that are horizontal asymptotes for any solution curve starting in this subregion. (iii) For P0 ab, P(t) is bounded below. Since P(t) is decreasing, P(t) : ab as t : . The graph of the equilibrium solution P(t) ab is a horizontal asymptote for a solution curve. In Figure 2.1.7 the phase line is the P-axis in the tP-plane. For clarity the origi-nal phase line from Figure 2.1.5 is reproduced to the left of the plane in which the subregions R1, R2, and R3 are shaded. The graphs of the equilibrium solutions P(t) ab and P(t) 0 (the t-axis) are shown in the ﬁgure as blue dashed lines; the solid graphs represent typical graphs of P(t) illustrating the three cases just discussed. In a subregion such as R1 in Example 4, where P(t) is decreasing and unbounded below, we must necessarily have P(t) : . Do not interpret this last statement to mean P(t) : as t : ; we could have P(t) : as t : T, where T 0 is a ﬁnite number that depends on the initial condition P(t0) P0. Thinking in dynamic terms, P(t) could “blow up” in ﬁnite time; thinking graphically, P(t) could have a vertical asymptote at t T 0. A similar remark holds for the subregion R3. The differential equation dydx sin y in Example 2 is autonomous and has an inﬁnite number of critical points, since sin y 0 at y n, n an integer. Moreover, we now know that because the solution y(x) that passes through is bounded (0, 3 2) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. above and below by two consecutive critical points ( y(x) 0) and is decreasing (sin y 0 for y 0), the graph of y(x) must approach the graphs of the equilibrium solutions as horizontal asymptotes: y(x) : as x : and y(x) : 0 as x : . 2.1 SOLUTION CURVES WITHOUT A SOLUTION ● 41 EXAMPLE 5 Solution Curves of an Autonomous DE The autonomous equation dydx (y 1)2 possesses the single critical point 1. From the phase portrait in Figure 2.1.8(a) we conclude that a solution y(x) is an increasing function in the subregions deﬁned by y 1 and 1 y , where x . For an initial condition y(0) y0 1, a solution y(x) is increasing and bounded above by 1, and so y(x) : 1 as x : ; for y(0) y0 1 a solution y(x) is increasing and unbounded. Now y(x) 1 1(x c) is a one-parameter family of solutions of the differ-ential equation. (See Problem 4 in Exercises 2.2.) A given initial condition deter-mines a value for c. For the initial conditions, say, y(0) 1 1 and y(0) 2 1, we ﬁnd, in turn, that y(x) 1 1(x , and y(x) 1 1(x 1). As shown in Figures 2.1.8(b) and 2.1.8(c), the graph of each of these rational functions possesses 1 2) c c c c y0 (d) y0 (c) y0 (b) y0 (a) FIGURE 2.1.9 Critical point c is an attractor in (a), a repeller in (b), and semi-stable in (c) and (d). 1 increasing y increasing (a) phase line (0, −1) y 1 1 2 x x y (b) xy-plane y(0) 1 (0, 2) y 1 x 1 x y (c) xy-plane y(0) 1 = = = = − FIGURE 2.1.8 Behavior of solutions near y 1 in Example 5 a vertical asymptote. But bear in mind that the solutions of the IVPs are deﬁned on special intervals. They are, respectively, The solution curves are the portions of the graphs in Figures 2.1.8(b) and 2.1.8(c) shown in blue. As predicted by the phase portrait, for the solution curve in Figure 2.1.8(b), y(x) : 1 as x : ; for the solution curve in Figure 2.1.8(c), y(x) : as x : 1 from the left. Attractors and Repellers Suppose that y(x) is a nonconstant solution of the autonomous differential equation given in (1) and that c is a critical point of the DE. There are basically three types of behavior that y(x) can exhibit near c. In Figure 2.1.9 we have placed c on four vertical phase lines. When both arrowheads on either side of the dot labeled c point toward c, as in Figure 2.1.9(a), all solutions y(x) of (1) that start from an initial point (x0, y0) sufﬁciently near c exhibit the asymp-totic behavior . For this reason the critical point c is said to be limx: y(x) c y(x) 1 1 x 1 2 , 1 2 x and y(x) 1 1 x 1 , x 1. dy dx ( y 1)2, y(0) 1 and dy dx (y 1)2, y(0) 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. slopes of lineal elements on a vertical line vary slopes of lineal elements on a horizontal line are all the same x y y FIGURE 2.1.10 Direction ﬁeld for an autonomous DE x y y = 0 y = 3 FIGURE 2.1.11 Translated solution curves of an autonomous DE 42 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS asymptotically stable. Using a physical analogy, a solution that starts near c is like a charged particle that, over time, is drawn to a particle of opposite charge, and so c is also referred to as an attractor. When both arrowheads on either side of the dot labeled c point away from c, as in Figure 2.1.9(b), all solutions y(x) of (1) that start from an initial point (x0, y0) move away from c as x increases. In this case the critical point c is said to be unstable. An unstable critical point is also called a repeller, for obvious reasons. The critical point c illustrated in Figures 2.1.9(c) and 2.1.9(d) is neither an attractor nor a repeller. But since c exhibits characteristics of both an attractor and a repeller—that is, a solution starting from an initial point (x0, y0) sufﬁ-ciently near c is attracted to c from one side and repelled from the other side—we say that the critical point c is semi-stable. In Example 3 the critical point ab is asymptotically stable (an attractor) and the critical point 0 is unstable (a repeller). The critical point 1 in Example 5 is semi-stable. Autonomous DEs and Direction Fields If a ﬁrst-order differential equa-tion is autonomous, then we see from the right-hand side of its normal form dydx f(y) that slopes of lineal elements through points in the rectangular grid used to con-struct a direction ﬁeld for the DE depend solely on the y-coordinate of the points. Put another way, lineal elements passing through points on any horizontal line must all have the same slope and therefore are parallel; slopes of lineal elements along any vertical line will, of course, vary. These facts are apparent from inspection of the hor-izontal yellow strip and vertical blue strip in Figure 2.1.10. The ﬁgure exhibits a di-rection ﬁeld for the autonomous equation dydx 2(y 1). The red lineal elements in Figure 2.1.10 have zero slope because they lie along the graph of the equilibrium solution y 1. Translation Property You may recall from precalculus mathematics that the graph of a function , where k is a constant, is the graph of rigidly translated or shifted horizontally along the x-axis by an amount the trans-lation is to the right if k 0 and to the left if k 0. It turns out that under the condi-tions stipulated for (2), solution curves of an autonomous ﬁrst-order DE are related by the concept of translation. To see this, let’s consider the differential equation dydx y(3 y), which is a special case of the autonomous equation considered in Examples 3 and 4. Because y 0 and y 3 are equilibrium solutions of the DE, their graphs divide the xy-plane into three subregions In Figure 2.1.11 we have superimposed on a direction ﬁeld of the DE six solutions curves. The ﬁgure illustrates that all solution curves of the same color, that is, solu-tion curves lying within a particular subregion Ri, all look alike. This is no coinci-dence but is a natural consequence of the fact that lineal elements passing through points on any horizontal line are parallel. That said, the following translation prop-erty of an automonous DE should make sense: If y(x) is a solution of an autonomous differ ential equation dydx f(y), then y1(x) y(x k), k a constant, is also a solution. Thus, if y(x) is a solution of the initial-value problem dydx f(y), y(0) y0, then y1(x) y(x x0) is a solution of the IVPdydx f(y), y(x0) y0. For example, it is easy to verify that is a solution of the IVP dydx y, y(0) 1 and so a solution y1(x) of, say, dydx y, y(5) 1 is y(x) ex translated 5 units to the right: y1(x) y(x 5) ex5, x . y(x) ex, x , R1: y 0, R2: 0 y 3, and R3: 3 y . R1, R2, and R3: k; y f(x) y f(x k) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. x 3 2 1 _1 _2 _3 _3 _2 _1 1 2 3 y FIGURE 2.1.12 Direction ﬁeld for Problem 1 x y 8 _8 _8 _4 4 8 _4 4 FIGURE 2.1.13 Direction ﬁeld for Problem 2 2. (a) y(6) 0 (b) y(0) 1 (c) y(0) 4 (d) y(8) 4 dy dx e0.01xy2 In Problems 5–12 use computer software to obtain a direc-tion ﬁeld for the given differential equation. By hand, sketch an approximate solution curve passing through each of the given points. 5. y x 6. y x y (a) y(0) 0 (a) y(2) 2 (b) y(0) 3 (b) y(1) 3 7. 8. (a) y(1) 1 (a) y(0) 1 (b) y(0) 4 (b) y(2) 1 9. 10. (a) (a) y(0) 2 (b) y(2) 1 (b) y(1) 2.5 y(0) 1 2 dy dx xey dy dx 0.2x2 y dy dx 1 y y dy dx x 2.1 SOLUTION CURVES WITHOUT A SOLUTION ● 43 EXERCISES 2.1 Answers to selected odd-numbered problems begin on page ANS-1. 2.1.1 DIRECTION FIELDS In Problems 1–4 reproduce the given computer-generated direction ﬁeld. Then sketch, by hand, an approximate solu-tion curve that passes through each of the indicated points. Use different colored pencils for each solution curve. 1. (a) y(2) 1 (b) y(3) 0 (c) y(0) 2 (d) y(0) 0 dy dx x2 y2 3. (a) y(0) 0 (b) y(1) 0 (c) y(2) 2 (d) y(0) 4 dy dx 1 xy 4. (a) y(0) 1 (b) y(1) 0 (c) y(3) 3 (d) y(0) 5 2 dy dx (sin x) cos y x y 4 _4 _4 _2 2 4 _2 2 FIGURE 2.1.14 Direction ﬁeld for Problem 3 x y 4 _4 _4 _2 2 4 _2 2 FIGURE 2.1.15 Direction ﬁeld for Problem 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 44 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 13. 11. 12. (a) y(2) 2 (a) (b) y(1) 0 (b) In Problems 13 and 14 the given ﬁgure represents the graph of f(y) and f(x), respectively. By hand, sketch a direction ﬁeld over an appropriate grid for dydx f(y) (Problem 13) and then for dydx f(x) (Problem 14). y(3 2) 0 y(1 2) 2 dy dx 1 y x y y cos 2 x f 1 y 1 FIGURE 2.1.16 Graph for Problem 13 f x 1 1 FIGURE 2.1.17 Graph for Problem 14 14. 15. In parts (a) and (b) sketch isoclines f(x, y) c (see the Remarks on page 38) for the given differential equation using the indicated values of c. Construct a direction ﬁeld over a grid by carefully drawing lineal elements with the appropriate slope at chosen points on each isocline. In each case, use this rough direction ﬁeld to sketch an ap-proximate solution curve for the IVPconsisting of the DE and the initial condition y(0) 1. (a) dydx x y; c an integer satisfying 5 c 5 (b) dydx x2 y2; Discussion Problems 16. (a) Consider the direction ﬁeld of the differential equa-tion dydx x(y 4)2 2, but do not use tech-nology to obtain it. Describe the slopes of the lineal elements on the lines x 0, y 3, y 4, and y 5. (b) Consider the IVPdydx x(y 4)2 2, y(0) y0, where y0 4. Can a solution y(x) : as x : ? Based on the information in part (a), discuss. c 1 4, c 1, c 9 4, c 4 17. For a ﬁrst-order DE dydx f (x, y) a curve in the plane deﬁned by f (x, y) 0 is called a nullcline of the equa-tion,sincealinealelementatapointonthecurvehaszero slope. Use computer software to obtain a direction ﬁeld over a rectangular grid of points for dydx x2 2y, and then superimpose the graph of the nullcline over the direction ﬁeld. Discuss the behavior of solution curves in regions of the plane deﬁned by and by . Sketch some approximate solution curves. Try to generalize your observations. 18. (a) Identify the nullclines (see Problem 17) in Problems 1, 3, and 4. With a colored pencil, circle any lineal elements in Figures 2.1.12, 2.1.14, and 2.1.15 that you think may be a lineal element at a point on a nullcline. (b) What are the nullclines of an autonomous ﬁrst-order DE? 2.1.2 AUTONOMOUS FIRST-ORDER DEs 19. Consider the autonomous ﬁrst-order differential equa-tion dydx y y3 and the initial condition y(0) y0. By hand, sketch the graph of a typical solution y(x) when y0 has the given values. (a) y0 1 (b) 0 y0 1 (c) 1 y0 0 (d) y0 1 20. Consider the autonomous ﬁrst-order differential equation dydx y2 y4 and the initial condition y(0) y0. By hand, sketch the graph of a typical solution y(x) when y0 has the given values. (a) y0 1 (b) 0 y0 1 (c) 1 y0 0 (d) y0 1 In Problems 21–28 ﬁnd the critical points and phase portrait of the given autonomous ﬁrst-order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the xy-plane determined by the graphs of the equilibrium solutions. 21. 22. 23. 24. 25. 26. 27. 28. In Problems 29 and 30 consider the autonomous differential equation dydx f(y), where the graph of f is given. Use the graph to locate the critical points of each differential dy dx yey 9y ey dy dx y ln(y 2) dy dx y(2 y)(4 y) dy dx y2(4 y2) dy dx 10 3y y2 dy dx (y 2)4 dy dx y2 y3 dy dx y2 3y y 1 2 x2 y 1 2 x2 y 1 2 x2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 29. f c y FIGURE 2.1.18 Graph for Problem 29 30. f y 1 1 FIGURE 2.1.19 Graph for Problem 30 Discussion Problems 31. Consider the autonomous DE dydx (2)y sin y. Determine the critical points of the equation. Discuss a way of obtaining a phase portrait of the equation. Classify the critical points as asymptotically stable, unstable, or semi-stable. 32. A critical point c of an autonomous ﬁrst-order DE is said to be isolated if there exists some open interval that contains c but no other critical point. Can there exist an autonomous DE of the form given in (2) for which every critical point is nonisolated? Discuss; do not think profound thoughts. 33. Suppose that y(x) is a nonconstant solution of the autonomous equation dydx f (y) and that c is a critical point of the DE. Discuss: Why can’t the graph of y(x) cross the graph of the equilibrium solution y c? Why can’t f (y) change signs in one of the subregions discussed on page 39? Why can’t y(x) be oscillatory or have a relative extremum (maximum or minimum)? 34. Suppose that y(x) is a solution of the autonomous equa-tion dydx f (y) and is bounded above and below by two consecutive critical points c1 c2, as in subregion R2 of Figure 2.1.6(b). If f (y) 0 in the region, then limx: y(x) c2. Discuss why there cannot exist a num-ber L c2 such that limx: y(x) L. As part of your discussion, consider what happens to y(x) as x : . 2.1 SOLUTION CURVES WITHOUT A SOLUTION ● 45 equation. Sketch a phase portrait of each differential equa-tion. By hand, sketch typical solution curves in the subre-gions in the xy-plane determined by the graphs of the equi-librium solutions. 35. Using the autonomous equation (2), discuss how it is possible to obtain information about the location of points of inﬂection of a solution curve. 36. Consider the autonomous DE dydx y2 y 6. Use your ideas from Problem 35 to ﬁnd intervals on the y-axis for which solution curves are concave up and intervals for which solution curves are concave down. Discuss why each solution curve of an initial-value problem of the form dydx y2 y 6, y(0) y0, where 2 y0 3, has a point of inﬂection with the same y-coordinate. What is that y-coordinate? Carefully sketch the solution curve for which y(0) 1. Repeat for y(2) 2. 37. Suppose the autonomous DE in (2) has no critical points. Discuss the behavior of the solutions. Mathematical Models 38. Population Model The differential equation in Exam-ple 3 is a well-known population model. Suppose the DE is changed to , where a and b are positive constants. Discuss what happens to the population P as time t increases. 39. Population Model Another population model is given by , where h and k are positive constants. For what initial values P(0) P0 does this model predict that the popu-lation will go extinct? 40. Terminal Velocity In Section 1.3 we saw that the auto-nomous differential equation , where k is a positive constant and g is the acceleration due to gravity, is a model for the velocity v of a body of mass m that is falling under the inﬂuence of gravity. Because the term kv represents air resistance, the velocity of a body falling from a great height does not in-crease without bound as time t increases. Use a phase portrait of the differential equation to ﬁnd the limiting, or terminal, velocity of the body. Explain your reasoning. 41. Suppose the model in Problem 40 is modiﬁed so that air resistance is proportional to v2, that is, . See Problem 17 in Exercises 1.3. Use a phase portrait to ﬁnd the terminal velocity of the body. Explain your reasoning. m dv dt mg kv2 m dv dt mg kv dP dt kP h dP dt P(aP b) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 46 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 42. Chemical Reactions When certain kinds of chemicals are combined, the rate at which the new compound is formed is modeled by the autonomous differential equation where k 0 is a constant of proportionality and 0. Here X(t) denotes the number of grams of the new compound formed in time t. (a) Use a phase portrait of the differential equation to predict the behavior of X(t) as t : . dX dt k( X)( X), (b) Consider the case when . Use a phase portrait of the differential equation to predict the behavior of X(t) as t : when X(0) . When X(0) . (c) Verify that an explicit solution of the DE in the case when k 1 and is X(t) 1(t c). Find a solution that satisﬁes X(0) 2. Then ﬁnd a solution that satisﬁes X(0) 2. Graph these two solutions. Does the behavior of the solutions as t : agree with your answers to part (b)? Solution by Integration Consider the ﬁrst-order differential equation dydx f (x, y). When f does not depend on the variable y, that is, f(x, y) g(x), the differen-tial equation (1) can be solved by integration. If g(x) is a continuous function, then integrating both sides of (1) gives , where G(x) is an antiderivative (indeﬁ-nite integral) of g(x). For example, if dydx 1 e2x, then its solution is or . A Definitio Equation (1), as well as its method of solution, is just a special case when the function f in the normal form dydx f (x, y) can be factored into a function of x times a function of y. y x 1 2e2x c y (1 e2x) dx y g(x) dx G(x) c dy dx g(x) DEFINITION 2.2.1 Separable Equation A ﬁrst-order differential equation of the form is said to be separable or to have separable variables. dy dx g(x)h(y) For example, the equations dy dx y2xe3x4y and dy dx y sin x SEPARABLE EQUATIONS REVIEW MATERIAL ●Basic integration formulas (See inside front cover) ●Techniques of integration: integration by parts and partial fraction decomposition ● See also the Student Resource Manual. INTRODUCTION We begin our study of how to solve differential equations with the simplest of all differential equations: ﬁrst-order equations with separable variables. Because the method in this section and many techniques for solving differential equations involve integration, you are urged to refresh your memory on important formulas (such as duu) and techniques (such as integration by parts) by consulting a calculus text. 2.2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. are separable and nonseparable, respectively. In the ﬁrst equation we can factor f (x, y) y2xe3x4y as g(x) h(y) p p , but in the second equation there is no way of expressing y sin x as a product of a function of x times a function of y. Observe that by dividing by the function h(y), we can write a separable equation dydx g(x)h(y) as , (2) where, for convenience, we have denoted 1h(y) by p(y). From this last form we can see immediately that (2) reduces to (1) when h(y) 1. Now if y (x) represents a solution of (2), we must have p((x))(x) g(x), and therefore . (3) But dy (x) dx, and so (3) is the same as , (4) where H(y) and G(x) are antiderivatives of p(y) 1h(y) and g(x), respectively. Method of Solution Equation (4) indicates the procedure for solving separable equations. A one-parameter family of solutions, usually given implicitly, is obtained by integrating both sides of p(y) dy g(x) dx. Note There is no need to use two constants in the integration of a separable equation, because if we write H(y) c1 G(x) c2, then the difference c2 c1 can be replaced by a single constant c, as in (4). In many instances throughout the chap-ters that follow, we will relabel constants in a manner convenient to a given equation. For example, multiples of constants or combinations of constants can sometimes be replaced by a single constant. p(y) dy g(x) dx or H(y) G(x) c p( (x))(x) dx g(x) dx p(y) dy dx g(x) f (x, y) y2xe3x4y (xe3x)( y2e4y) 2.2 SEPARABLE EQUATIONS ● 47 EXAMPLE 1 Solving a Separable DE Solve (1 x) dy y dx 0. SOLUTION Dividing by (1 x)y, we can write dyy dx(1 x), from which it follows that ; laws of exponents Relabeling as c then gives y c(1 x). ec1 ec1(1 x). 1 x ec1 y eln1xc1 eln1x ec1 ln y ln 1 x c1 dy y dx 1 x ; 1 x 1 x, 1 x (1 x), x 1 x <1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. FIGURE 2.2.1 Solution curve for the IVP in Example 2 x y (4, −3) 48 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS ALTERNATIVE SOLUTION Because each integral results in a logarithm, a judicious choice for the constant of integration is lnc rather than c. Rewriting the second line of the solution as lny ln1 x lnc enables us to combine the terms on the right-hand side by the properties of logarithms. From lny lnc(1 x) we immediately get y c(1 x). Even if the indeﬁnite integrals are not all logarithms, it may still be advantageous to use lnc. However, no ﬁrm rule can be given. In Section 1.1 we saw that a solution curve may be only a segment or an arc of the graph of an implicit solution G(x, y) 0. EXAMPLE 2 Solution Curve Solve the initial-value problem . SOLUTION Rewriting the equation as y dy x dx, we get . We can write the result of the integration as x2 y2 c2 by replacing the constant 2c1 by c2. This solution of the differential equation represents a family of concentric circles centered at the origin. Now when x 4, y 3, so 16 9 25 c2. Thus the initial-value problem determines the circle x2 y2 25 with radius 5. Because of its simplicity we can solve this implicit solution for an explicit solution that satisﬁes the initial condition. We saw this solution as y 2(x) or in Example 3 of Section 1.1. A solution curve is the graph of a differentiable function. In this case the solution curve is the lower semicircle, shown in dark blue in Figure 2.2.1 containing the point (4, 3). Losing a Solution Some care should be exercised in separating variables, since the variable divisors could be zero at a point. Speciﬁcally, if r is a zero of the function h(y), then substituting y r into dydx g(x)h(y) makes both sides zero; in other words, y r is a constant solution of the differential equation. But after variables are separated, the left-hand side of g(x) dx is undeﬁned at r. As a consequence, y r might not show up in the family of solutions that are obtained after integration and simpliﬁcation. Recall that such a solution is called a singular solution. dy h(y) y 125 x2, 5 x 5 y dy x dx and y2 2 x2 2 c1 dy dx x y, y(4) 3 EXAMPLE 3 Losing a Solution Solve . SOLUTION We put the equation in the form . (5) The second equation in (5) is the result of using partial fractions on the left-hand side of the ﬁrst equation. Integrating and using the laws of logarithms gives . ln y 2 y 2 4x c2 or y 2 y 2 e4xc2 1 4 ln y 2 1 4 ln y 2 x c1 dy y2 4 dx or 1 4 y 2 1 4 y 2 dy dx dy dx y2 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.2 SEPARABLE EQUATIONS ● 49 EXAMPLE 4 An Initial-Value Problem Solve . SOLUTION Dividing the equation by ey cos x gives . Before integrating, we use termwise division on the left-hand side and the trigono-metric identity sin 2x 2 sin x cos x on the right-hand side. Then integration by parts : yields ey yey ey 2 cos x c. (7) The initial condition y 0 when x 0 implies c 4. Thus a solution of the initial-value problem is ey yey ey 4 2 cos x. (8) Use of Computers The Remarks at the end of Section 1.1 mentioned that it may be difﬁcult to use an implicit solution G(x, y) 0 to ﬁnd an explicit solution y (x). Equation (8) shows that the task of solving for y in terms of x may present more problems than just the drudgery of symbol pushing—sometimes it simply cannot be done! Implicit solutions such as (8) are somewhat frustrating; nei-ther the graph of the equation nor an interval over which a solution satisfying y(0) 0 is deﬁned is apparent. The problem of “seeing” what an implicit solution looks like can be overcome in some cases by means of technology. One way of proceeding is to use the contour plot application of a computer algebra system (CAS). Recall from multivariate calculus that for a function of two variables z G(x, y) the two-dimensional curves deﬁned by G(x, y) c, where c is constant, are called the level curves of the function. With the aid of a CAS, some of the level curves of the func-tion G(x, y) ey yey ey 2 cos x have been reproduced in Figure 2.2.2. The family of solutions deﬁned by (7) is the level curves G(x, y) c. Figure 2.2.3 illus-trates the level curve G(x, y) 4, which is the particular solution (8), in blue color. The other curve in Figure 2.2.3 is the level curve G(x, y) 2, which is the member of the family G(x, y) c that satisﬁes y(2) 0. If an initial condition leads to a particular solution by yielding a speciﬁc value of the parameter c in a family of solutions for a ﬁrst-order differential equation, there is (ey yey) dy 2 sin x dx e2y y ey dy sin 2x cos x dx (e2y y) cos x dy dx ey sin 2x, y(0) 0 Here we have replaced 4c1 by c2. Finally, after replacing by c and solving the last equation for y, we get the one-parameter family of solutions . (6) Now if we factor the right-hand side of the differential equation as dydx (y 2)(y 2), we know from the discussion of critical points in Section 2.1 that y 2 and y 2 are two constant (equilibrium) solutions. The solution y 2 is a member of the family of solutions deﬁned by (6) corresponding to the value c 0. However, y 2 is a singular solution; it cannot be obtained from (6) for any choice of the parameter c. This latter solution was lost early on in the solution process. Inspection of (5) clearly indicates that we must preclude y 2 in these steps. y 2 1 ce4x 1 ce4x ec2 In Section 2.6 we will discuss several other ways of proceeding that are based on the concept of a numerical solver. x y 2 _2 _2 _1 1 2 _1 1 FIGURE 2.2.2 Level curves of G(x, y) ey yey ey 2 cos x FIGURE 2.2.3 Level curves c 2 and c 4 (0, 0) /2,0) ( π x y 2 _2 _2 _1 1 2 _1 1 c=4 c=2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. a = > 0 a 0 (0, 0) x y FIGURE 2.2.4 Piecewise-deﬁned solutions of (9) 50 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS a natural inclination for most students (and instructors) to relax and be content. However, a solution of an initial-value problem might not be unique. We saw in Example 4 of Section 1.2 that the initial-value problem (9) has at least two solutions, y 0 and . We are now in a position to solve the equation. Separating variables and integrating y1/2 dy x dx gives When x 0, then y 0, so necessarily, c 0. Therefore . The trivial solution y 0 was lost by dividing by y1/2. In addition, the initial-value problem (9) possesses inﬁnitely many more solutions, since for any choice of the parameter a 0 the piecewise-deﬁned function satisﬁes both the differential equation and the initial condition. See Figure 2.2.4. Solutions Defined by Integral If g is a function continuous on an open in-terval I containing a, then for every x in I, You might recall that the foregoing result is one of the two forms of the fundamental theorem of calculus. In other words, is an antiderivative of the function g. There are times when this form is convenient in solving DEs. For example, if g is continuous on an interval I containing x0 and x, then a solution of the simple initial-value problem , that is deﬁned on I is given by You should verify that y(x) deﬁned in this manner satisﬁes the initial condition. Since an antiderivative of a continuous function g cannot always be expressed in terms of elementary functions, this might be the best we can do in obtaining an explicit solution of an IVP. The next example illustrates this idea. y(x) y0 x x0 g(t) dt dy>dx g(x), y(x0) y0 x a g(t) dt d dx x a g(t) dt g(x). y 0, 1 16 (x2 a2)2, x a x a y 1 16 x4 2y1/2 x2 2 c1 or y x2 4 c 2 , c 0. y 1 16 x4 dy dx xy1/2, y(0) 0 EXAMPLE 5 An Initial-Value Problem Solve SOLUTION The function is continuous on , but its antideriva-tive is not an elementary function. Using t as dummy variable of integration, we can write y(x) y(3) x 3 et2 dt. y(x) y(3) x 3 et2 dt y(t)] x 3 x 3 et2 dt x 3 dy dt dt x 3 et2 dt (, ) g(x) ex2 dy dx ex2, y(3) 5. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Using the initial condition y(3) 5, we obtain the solution The procedure demonstrated in Example 5 works equally well on separable equations where, say, f (y) possesses an elementary antiderivative but g(x) does not possess an elementary antiderivative. See Problems 29 and 30 in Exercises 2.2. dy>dx g(x) f (y) y(x) 5 x 3 et2 dt. 2.2 SEPARABLE EQUATIONS ● 51 REMARKS (i) As we have just seen in Example 5, some simple functions do not possess an antiderivative that is an elementary function. Integrals of these kinds of functions are called nonelementary. For example, and are nonelementary integrals. We will run into this concept again in Section 2.3. (ii) In some of the preceding examples we saw that the constant in the one-parameter family of solutions for a ﬁrst-order differential equation can be rela-beled when convenient. Also, it can easily happen that two individuals solving the same equation correctly arrive at dissimilar expressions for their answers. For example, by separation of variables we can show that one-parameter families of solutions for the DE (1 y2) dx (1 x2) dy 0 are . As you work your way through the next several sections, bear in mind that fami-lies of solutions may be equivalent in the sense that one family may be obtained from another by either relabeling the constant or applying algebra and trigonom-etry. See Problems 27 and 28 in Exercises 2.2. arctan x arctan y c or x y 1 xy c sin x2 dx x 3 et2 dt EXERCISES 2.2 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1–22 solve the given differential equation by separation of variables. 1. 2. 3. dx e3xdy 0 4. dy (y 1)2dx 0 5. 6. 7. 8. 9. 10. 11. csc y dx sec2x dy 0 12. sin 3x dx 2y cos33x dy 0 13. (ey 1)2ey dx (ex 1)3ex dy 0 14. x(1 y2)1/2 dx y(1 x2)1/2 dy dy dx 2y 3 4x 5 2 y ln x dx dy y 1 x 2 exy dy dx ey e2xy dy dx e3x2y dy dx 2xy2 0 x dy dx 4y dy dx (x 1)2 dy dx sin 5x 15. 16. 17. 18. 19. 20. 21. 22. In Problems 23–28 ﬁnd an explicit solution of the given initial-value problem. 23. 24. 25. x2 dy dx y xy, y(1) 1 dy dx y2 1 x2 1 , y(2) 2 dx dt 4(x2 1), x( >4) 1 (ex ex) dy dx y2 dy dx x11 y2 dy dx xy 2y x 2 xy 3y x 3 dy dx xy 3x y 3 xy 2x 4y 8 dN dt N Ntet2 dP dt P P2 dQ dt k(Q 70) dS dr kS Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 52 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 26. 27. 28. (1 x4) dy x(1 4y2) dx 0, y(1) 0 In Problems 29 and 30 proceed as in Example 5 and ﬁnd an explicit solution of the given initial-value problem. 29. 30. In Problems 31–34 ﬁnd an explicit solution of the given initial-value problem. Determine the exact interval I of deﬁ-nition by analytical methods. Use a graphing utility to plot the graph of the solution. 31. 32. 33. 34. 35. (a) Find a solution of the initial-value problem consist-ing of the differential equation in Example 3 and each of the initial-conditions: y(0) 2, y(0) 2, and . (b) Find the solution of the differential equation in Example 4 when ln c1 is used as the constant of integration on the left-hand side in the solution and 4 ln c1 is replaced by ln c. Then solve the same initial-value problems in part (a). 36. Find a solution of that passes through the indicated points. (a) (0, 1) (b) (0, 0) (c) (d) 37. Find a singular solution of Problem 21. Of Problem 22. 38. Show that an implicit solution of is given by ln(x2 10) csc y c. Find the constant solutions, if any, that were lost in the solution of the dif-ferential equation. Often a radical change in the form of the solution of a differen-tial equation corresponds to a very small change in either the initial condition or the equation itself. In Problems 39–42 ﬁnd an explicit solution of the given initial-value problem. Use a graphing utility to plot the graph of each solution. Compare each solution curve in a neighborhood of (0, 1). 39. dy dx (y 1)2, y(0) 1 2x sin2 y dx (x2 10) cos y dy 0 (2, 1 4) (1 2, 1 2) x dy dx y2 y y(1 4) 1 sin x dx ydy 0, y(0) 1 ey dx ex dy 0, y(0) 0 (2y 2) dy dx 3x2 4x 2, y(1) 2 dy dx 2x 1 2y , y(2) 1 dy dx y 2 sin x2, y(2) 1 3 dy dx yex2, y(4) 1 11 y2 dx 11 x2 dy 0, y(0) 13 2 dy dt 2y 1, y(0) 5 2 40. 41. 42. 43. Every autonomous ﬁrst-order equation dydx f(y) is separable. Find explicit solutions y1(x), y2(x), y3(x), and y4(x) of the differential equation dydx y y3 that satisfy, in turn, the initial conditions y1(0) 2, , , and y4(0) 2. Use a graphing utility to plot the graphs of each solution. Compare these graphs with those predicted in Problem 19 of Exercises 2.1. Give the exact interval of deﬁnition for each solution. 44. (a) The autonomous ﬁrst-order differential equation dydx 1(y 3) has no critical points. Nevertheless, place 3 on the phase line and obtain a phase portrait of the equation. Compute d2ydx2 to determine where solution curves are concave up and where they are concave down (see Problems 35 and 36 in Exercises 2.1). Use the phase portrait and concavity to sketch, by hand, some typical solution curves. (b) Find explicit solutions y1(x), y2(x), y3(x), and y4(x) of the differential equation in part (a) that satisfy, in turn, the initial conditions y1(0) 4, y2(0) 2, y3(1) 2, and y4(1) 4. Graph each solution and compare with your sketches in part (a). Give the exact interval of deﬁnition for each solution. In Problems 45–50 use a technique of integration or a substi-tution to ﬁnd an explicit solution of the given differential equation or initial-value problem. 45. 46. 47. 48. 49. 50. Discussion Problems 51. (a) Explain why the interval of deﬁnition of the explicit solution y 2(x) of the initial-value problem in Example 2 is the open interval (5, 5). (b) Can any solution of the differential equation cross the x-axis? Do you think that x2 y2 1 is an implicit solution of the initial-value problem dydx xy, y(1) 0? 52. (a) If a 0, discuss the differences, if any, between the solutions of the initial-value problems consist-ing of the differential equation dydx xy and dy dx x tan1 x y , y(0) 3 dy dx e1x y , y(1) 4 dy dx y2/3 y (1x x)dy dx 1y y dy dx sin1x 1y dy dx 1 1 sin x y3(0) 1 2 y2(0) 1 2 dy dx (y 1)2 0.01, y(0) 1 dy dx (y 1)2 0.01, y(0) 1 dy dx (y 1)2, y(0) 1.01 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. each of the initial conditions y(a) a, y(a) a, y(a) a, and y(a) a. (b) Does the initial-value problem dydx xy, y(0) 0 have a solution? (c) Solve dydx xy, y(1) 2 and give the exact interval I of deﬁnition of its solution. 53. In Problems 43 and 44 we saw that every autonomous ﬁrst-order differential equation dydx f (y) is separable. Does this fact help in the solution of the initial-value problem ? Discuss. Sketch, by hand, a plausible solution curve of the problem. 54. (a) Solve the two initial-value problems: and (b) Show that there are more than 1.65 million digits in the y-coordinate of the point of intersection of the two solution curves in part (a). 55. Find a function whose square plus the square of its derivative is 1. 56. (a) The differential equation in Problem 27 is equiva-lent to the normal form in the square region in the xy-plane deﬁned by x 1, y 1. But the quantity under the radical is nonnegative also in the regions deﬁned by x 1, y 1. Sketch all regions in the xy-plane for which this differential equation possesses real solutions. (b) Solve the DE in part (a) in the regions deﬁned by x 1, y 1. Then ﬁnd an implicit and an explicit solution of the differential equation subject to y(2) 2. Mathematical Model 57. Suspension Bridge In (16) of Section 1.3 we saw that a mathematical model for the shape of a ﬂexible cable strung between two vertical supports is , (10) where W denotes the portion of the total vertical load between the points P1 and P2 shown in Figure 1.3.7. The dy dx W T1 dy dx B 1 y2 1 x2 dy dx y y x ln x, y(e) 1. dy dx y, y(0) 1 dy dx 11 y2 sin2 y, y(0) 1 2 DE (10) is separable under the following conditions that describe a suspension bridge. Let us assume that the x- and y-axes are as shown in Figure 2.2.5—that is, the x-axis runs along the horizon-tal roadbed, and the y-axis passes through (0, a), which is the lowest point on one cable over the span of the bridge, coinciding with the interval [L2, L2]. In the case of a suspension bridge, the usual assumption is that the vertical load in (10) is only a uniform roadbed dis-tributed along the horizontal axis. In other words, it is assumed that the weight of all cables is negligible in comparison to the weight of the roadbed and that the weight per unit length of the roadbed (say, pounds per horizontal foot) is a constant . Use this information to set up and solve an appropriate initial-value problem from which the shape (a curve with equation y (x)) of each of the two cables in a suspension bridge is determined. Express your solution of the IVP in terms of the sag h and span L. See Figure 2.2.5. 2.2 SEPARABLE EQUATIONS ● 53 FIGURE 2.2.5 Shape of a cable in Problem 57 L/2 L (span) h (sag) cable roadbed (load) x (0, a) L/2 y Computer Lab Assignments 58. (a) Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation . Experiment with different numbers of level curves as well as various rectangular regions deﬁned by a x b, c y d. (b) On separate coordinate axes plot the graphs of the particular solutions corresponding to the initial conditions: y(0) 1; y(0) 2; y(1) 4; y(1) 3. 59. (a) Find an implicit solution of the IVP (b) Use part (a) to ﬁnd an explicit solution y (x) of the IVP. (c) Consider your answer to part (b) as a function only. Use a graphing utility or a CAS to graph this func-tion, and then use the graph to estimate its domain. (d) With the aid of a root-ﬁnding application of a CAS, determine the approximate largest interval I of (2y 2) dy (4x3 6x) dx 0, y(0) 3. dy dx 8x 5 3y2 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 54 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS deﬁnition of the solution y (x) in part (b). Use a graphing utility or a CAS to graph the solution curve for the IVP on this interval. 60. (a) Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation . Experiment with different numbers of level curves as well as various rectan-gular regions in the xy-plane until your result resembles Figure 2.2.6. (b) On separate coordinate axes, plot the graph of the implicit solution corresponding to the initial condi-tion . Use a colored pencil to mark off that segment of the graph that corresponds to the solu-tion curve of a solution that satisﬁes the initial y(0) 3 2 dy dx x(1 x) y(2 y) condition. With the aid of a root-ﬁnding application of a CAS, determine the approximate largest inter-val I of deﬁnition of the solution . [Hint: First ﬁnd the points on the curve in part (a) where the tangent is vertical.] (c) Repeat part (b) for the initial condition y(0) 2. x y FIGURE 2.2.6 Level curves in Problem 60 LINEAR EQUATIONS REVIEW MATERIAL ●Review the deﬁnitions of linear DEs in (6) and (7) of Section 1.1. INTRODUCTION We continue our quest for solutions of ﬁrst-order differential equations by next examining linear equations. Linear differential equations are an especially “friendly” family of differential equations, in that, given a linear equation, whether ﬁrst order or a higher-order kin, there is always a good possibility that we can ﬁnd some sort of solution of the equation that we can examine. 2.3 A Definitio The form of a linear ﬁrst-order DE was given in (7) of Sec-tion 1.1. This form, the case when in (6) of that section, is reproduced here for convenience. n 1 DEFINITION 2.3.1 Linear Equation A ﬁrst-order differential equation of the form , (1) is said to be a linear equation in the variable y. a1(x) dy dx a0(x)y g(x) Standard Form By dividing both sides of (1) by the lead coefﬁcient we obtain a more useful form, the standard form, of a linear equation: (2) We seek a solution of (2) on an interval I for which both coefﬁcient functions P and f are continuous. dy dx P(x)y f(x). a1(x), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.3 LINEAR EQUATIONS ● 55 Before we examine a general procedure for solving equations of form (2) we note that in some instances (2) can be solved by separation of variables. For exam-ple, you should verify that the equations and are both linear and separable, but that the linear equation is not separable. Method of Solution The method for solving (2) hinges on a remarkable fact that the left-hand side of the equation can be recast into the form of the exact deriva-tive of a product by multiplying the both sides of (2) by a special function It is relatively easy to ﬁnd the function because we want left hand side of product product rule (2) multipled by m(x) 2 4 4 c c these must be equal The equality is true provided that The last equation can be solved by separation of variables. Integrating gives . Even though there are an inﬁnite choices of (all constant multiples of ), all produce the same desired result. Hence we can simplify life and choose The function (3) is called an integrating factor for equation (2). Here is what we have so far: We multiplied both sides of (2) by (3) and, by construction, the left-hand side is the derivative of a product of the integrating factor and y: Finally, we discover why (3) is called an integrating factor. We can integrate both sides of the last equation, and solve for y. The result is a one-parameter family of solutions of (2): (4) We emphasize that you should not memorize formula (4). The following proce-dure should be worked through each time. y eP(x)dx eP(x)dx f(x)dx ceP(x)dx. eP(x)dx y eP(x)dx f(x) c d dx [eP(x)dx y] eP(x)dx f(x). eP(x)dx dy dx P(x)eP(x)dx y eP(x)dx f(x) (x) eP(x)dx c2 1. eP(x)dx (x) (x) c2eP(x)dx d Pdx and solving ln(x) P(x)dx c1 d dx P. d dx [(x)y] dy dx d dxy dy dx Py. (x) (x). dy dx y x dy dx y 5 dy dx 2xy 0 We match each equation with (2). In the first equation P(x) 2x, f(x) 0 and in the second P(x) 1, f(x) 5. See Problem 50 in Exercises 2.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 56 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS SOLVING A LINEAR FIRST-ORDER EQUATION (i) Remember to put a linear equation into the standard form (2). (ii) From the standard form of the equation identify P(x) and then ﬁnd the integrating factor eP(x)dx. No constant need be used in evaluating the indeﬁnite integral P(x)dx. (iii) Multiply the both sides of the standard form equation by the integrating factor. The left-hand side of the resulting equation is automatically the derivative of the product of the integrating factor eP(x)dx and y: (iv) Integrate both sides of the last equation and solve for y. d dx [eP(x)dxy] eP(x)dxf(x). EXAMPLE 1 Solving a Linear Equation Solve SOLUTION This linear equation can be solved by separation of variables. Alternatively, since the differential equation is already in standard form (2), we iden-tify and so the integrating factor is . We then multiply the given equation by this factor and recognize that Integration of the last equation, then yields or y ce3x, x . e3xy c d dx [e3x y] dx 0 dx e3x dy dx 3e3x y e3x 0 is the same as d dx [e3x y] 0. e(3)dx e3x P(x) 3, dy dx 3y 0. EXAMPLE 2 Solving a Linear Equation Solve SOLUTION This linear equation, like the one in Example 1, is already in standard form with Thus the integrating factor is again . This time multiply-ing the given equation by this factor gives Integrating the last equation, or When a1, a0, and g in (1) are constants, the differential equation is autonomous. In Example 2 you can verify from the normal form dydx 3(y 2) that 2 is a critical point and that it is unstable (a repeller). Thus a solution curve with an y 2 ce3x, x . d dx [e3x y] dx 6e3x dx gives e3x y 6 e3x 3 c, e3x dy dx 3e3x y 6e3x and so d dx [e3x y] 6e3x. e3x P(x) 3. dy dx 3y 6. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. initial point either above or below the graph of the equilibrium solution y 2 pushes away from this horizontal line as x increases. Figure 2.3.1, obtained with the aid of a graphing utility, shows the graph of y 2 along with some addi-tional solution curves. General Solution Suppose again that the functions P and f in (2) are con-tinuous on a common interval I. In the steps leading to (4) we showed that if (2) has a solution on I, then it must be of the form given in (4). Conversely, it is a straight-forward exercise in differentiation to verify that any function of the form given in (4) is a solution of the differential equation (2) on I. In other words, (4) is a one-parameter family of solutions of equation (2) and every solution of (2) defined on I is a member of this family . Therefore we call (4) the general solution of the differential equation on the interval I. (See the Remarks at the end of Section 1.1.) Now by writing (2) in the normal form y F (x, y), we can identify F (x, y) P(x)y f (x) and Fy P(x). From the continuity of P and f on the interval I we see that F and Fy are also continuous on I. With Theorem 1.2.1 as our justiﬁcation, we conclude that there exists one and only one solution of the initial-value problem (5) deﬁned on some interval I0 containing x0. But when x0 is in I, ﬁnding a solution of (5) is just a matter of ﬁnding an appropriate value of c in (4)—that is, to each x0 in I there corresponds a distinct c. In other words, the interval I0 of existence and uniqueness in Theorem 1.2.1 for the initial-value problem (5) is the entire interval I. dy dx P(x)y f(x), y(x0) y0 2.3 LINEAR EQUATIONS ● 57 FIGURE 2.3.1 Solution curves of DE in Example 2 1 _1 2 3 4 _2 _1 1 _3 x y y =_2 EXAMPLE 3 General Solution Solve . SOLUTION Dividing by x, the standard form of the given DE is . (6) From this form we identify P(x) 4x and f (x) x5ex and further observe that P and f are continuous on (0, ). Hence the integrating factor is Here we have used the basic identity . Now we multiply (6) by x4 and rewrite It follows from integration by parts that the general solution deﬁned on the interval (0, ) is x4y xex ex c or y x5ex x4ex cx4. Except in the case in which the lead coefficient is 1, the recasting of equation (1) into the standard form (2) requires division by a1(x). Values of x for which a1(x) 0 are called singular points of the equation. Singular points are poten-tially troublesome. Specifically, in (2), if P(x) (formed by dividing a0(x) by a1(x)) is discontinuous at a point, the discontinuity may carry over to solutions of the differential equation. x4 dy dx 4x5y xex as d dx [x4y] xex. blogbN N, N 0 e4dx/x e4ln x eln x4 x4. we can use ln x instead of ln x since x 0 dy dx 4 x y x5ex x dy dx 4y x6ex In case you are wondering why the interval (0, ) is important in Example 3, read this paragraph and the paragraph following Example 4. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 58 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 4 General Solution Find the general solution of . SOLUTION We write the differential equation in standard form (7) and identify P(x) x(x2 9). Although P is continuous on (, 3), (3, 3), and (3, ), we shall solve the equation on the ﬁrst and third intervals. On these intervals the integrating factor is . After multiplying the standard form (7) by this factor, we get . Integrating both sides of the last equation gives Thus for either x 3 or x 3 the general solution of the equation is . Notice in Example 4 that x 3 and x 3 are singular points of the equation and that every function in the general solution is discontinuous at these points. On the other hand, x 0 is a singular point of the differential equation in Example 3, but the general solution y x5ex x4ex cx4 is noteworthy in that every function in this one-parameter family is continuous at x 0 and is deﬁned on the interval (, ) and not just on (0, ), as stated in the solution. However, the family y x5ex x4ex cx4 deﬁned on (, ) cannot be considered the gen-eral solution of the DE, since the singular point x 0 still causes a problem. See Problems 45 and 46 in Exercises 2.3. y c1x2 9 y c 1x2 9 1x2 9 y c. d dx 1x2 9 y 0 ex dx/(x29) e 1 2 2x dx/(x29) e 1 2 lnx29 1x2 9 dy dx x x2 9 y 0 (x2 9) dy dx xy 0 EXAMPLE 5 An Initial-Value Problem Solve . SOLUTION The equation is in standard form, and P(x) 1 and f (x) x are contin-uous on (, ). The integrating factor is edx ex, so integrating gives exy xex ex c. Solving this last equation for y yields the general solution y x 1 cex. But from the initial condition we know that y 4 when x 0. Substituting these values into the general solution implies that c 5. Hence the solution of the problem is y x 1 5ex, x . (8) Figure 2.3.2, obtained with the aid of a graphing utility, shows the graph of the solution (8) in dark blue along with the graphs of other members of the one-parameter family of solutions y x 1 cex. It is interesting to observe that as x increases, the graphs of all members of this family are close to the graph of the solu-tion The last solution corresponds to in the family and is shown in c 0 y x 1. d dx [exy] xex dy dx y x, y(0) 4 x y 4 _4 _4 _2 2 4 _2 2 c=0 c>0 c=5 c<0 FIGURE 2.3.2 Solution curves of DE in Example 5 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. dark green in Figure 2.3.2. This asymptotic behavior of solutions is due to the fact that the contribution of cex, , becomes negligible for increasing values of x. We say that cex is a transient term, since ex : 0 as x : . While this behavior is not characteristic of all general solutions of linear equations (see Example 2), the notion of a transient is often important in applied problems. Discontinuous Coefficient In applications, the coefﬁcients P(x) and f (x) in (2) may be piecewise continuous. In the next example f (x) is piecewise continuous on [0, ) with a single discontinuity, namely, a (ﬁnite) jump discontinuity at x 1. We solve the problem in two parts corresponding to the two intervals over which f is deﬁned. It is then possible to piece together the two solutions at x 1 so that y(x) is continuous on [0, ). c 0 2.3 LINEAR EQUATIONS ● 59 EXAMPLE 6 An Initial-Value Problem Solve SOLUTION The graph of the discontinuous function f is shown in Figure 2.3.3. We solve the DE for y(x) ﬁrst on the interval [0, 1] and then on the interval (1, ). For 0 x 1 we have . Integrating this last equation and solving for y gives y 1 c1ex. Since y(0) 0, we must have c1 1, and therefore y 1 ex, 0 x 1. Then for x 1 the equation leads to y c2ex. Hence we can write By appealing to the deﬁnition of continuity at a point, it is possible to determine c2 so that the foregoing function is continuous at x 1. The requirement that implies that c2e1 1 e1 or c2 e 1. As seen in Figure 2.3.4, the function (9) is continuous on (0, ). It is worthwhile to think about (9) and Figure 2.3.4 a little bit; you are urged to read and answer Problem 48 in Exercises 2.3. Functions Defined by Integrals At the end of Section 2.2 we discussed the fact that some simple continuous functions do not possess antiderivatives that are elementary functions and that integrals of these kinds of functions are called nonelementary. For example, you may have seen in calculus that and sin x2 dx are nonelementary integrals. In applied mathematics some important func-tions are define in terms of nonelementary integrals. Two such special functions are the error function and complementary error function: . (10) erf(x) 2 1 x 0 et2 dt and erfc(x) 2 1 x et2 dt ex2 dx y 1 ex, (e 1)ex, 0 x 1, x 1 limx:1 y(x) y(1) y 1 ex, c2ex, 0 x 1, x 1. dy dx y 0 dy dx y 1 or, equivalently, d dx [exy] ex dy dx y f (x), y(0) 0 where f (x) 1, 0, 0 x 1, x 1. FIGURE 2.3.3 Discontinuous f(x) in Example 6 x y 1 x y FIGURE 2.3.4 Graph of (9) in Example 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. This result is usually proved in the third semester of calculus. †Certain commands have the same spelling, but in Mathematica commands begin with a capital letter (DSolve), whereas in Maple the same command begins with a lower case letter (dsolve). When discussing such common syntax, we compromise and write, for example, dsolve. See the Student Resource Manual for the complete input commands used to solve a linear ﬁrst-order DE. FIGURE 2.3.5 Solution curves of DE in Example 7 x y 60 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS From the known result we can write Then from it is seen from (10) that the complementary error func-tion erfc(x) is related to erf(x) by erf(x) erfc(x) 1. Because of its importance in probability, statistics, and applied partial differential equations, the error func-tion has been extensively tabulated. Note that erf(0) 0 is one obvious function value. Values of erf(x) can also be found by using a CAS. 0 x 0 x (21 ) 0 et2 dt 1. 0 et2 dt 1 2 EXAMPLE 7 The Error Function Solve the initial-value problem . SOLUTION Since the equation is already in standard form, we see that the integrat-ing factor is , so from . (11) Applying y(0) 1 to the last expression then gives c 1. Hence the solution of the problem is The graph of this solution on the interval (, ), shown in dark blue in Figure 2.3.5 among other members of the family deﬁned in (11), was obtained with the aid of a computer algebra system. Use of Computers The computer algebra systems Mathematica and Maple are capable of producing implicit or explicit solutions for some kinds of differential equations using their dsolve commands.† y 2ex2 x 0 et2 dt ex2 or y ex2 [1 1 erf(x)]. d dx [ex2y] 2ex2 we get y 2ex2 x 0 et2 dt cex2 ex2 dy dx 2xy 2, y(0) 1 REMARKS (i) A linear ﬁrst-order differential equation is said to be homogeneous, whereas the equation with not identically zero is said to be nonhomogeneous. For example, the linear equations and are, in turn, homogeneous and nonhomogeneous. As can be seen in this example the trivial solution is always a solution of a homogeneous linear DE. Store this terminology in the back of your mind because it becomes important when we study linear higher-order ordinary differential equations in Chapter 4. y 0 xy y ex xy y 0 g(x) a1(x) dy dx a0(x)y g(x) a1(x) dy dx a0(x)y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.3 LINEAR EQUATIONS ● 61 (ii) Occasionally, a ﬁrst-order differential equation is not linear in one variable but in linear in the other variable. For example, the differential equation is not linear in the variable y. But its reciprocal is recognized as linear in the variable x. You should verify that the integrating factor and integration by parts yield the explicit solution for the second equation. This expression is then an implicit solution of the ﬁrst equation. (iii) Mathematicians have adopted as their own certain words from engineer-ing, which they found appropriately descriptive. The word transient, used ear-lier, is one of these terms. In future discussions the words input and output will occasionally pop up. The function f in (2) is called the input or driving func-tion; a solution of the differential equation for a given input is called the output or response. (iv) The term special functions mentioned in conjunction with the error func-tion also applies to the sine integral function and the Fresnel sine integral introduced in Problems 55 and 56 in Exercises 2.3. “Special Functions” is actually a well-deﬁned branch of mathematics. More special functions are studied in Section 6.4. y(x) x y2 2y 2 cey e(1)dy ey dx dy x y2 or dx dy x y2 dy dx 1 x y2 EXERCISES 2.3 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1–24 ﬁnd the general solution of the given dif-ferential equation. Give the largest interval I over which the general solution is deﬁned. Determine whether there are any transient terms in the general solution. 1. 2. 3. 4. 5. y 3x2y x2 6. y 2xy x3 7. x2y xy 1 8. y 2y x2 5 9. 10. 11. 12. 13. x2y x(x 2)y ex 14. xy (1 x)y ex sin 2x 15. y dx 4(x y6) dy 0 16. y dx (yey 2x) dy 17. cos x dy dx (sin x)y 1 (1 x) dy dx xy x x2 x dy dx 4y x3 x x dy dx 2y 3 x dy dx y x2 sin x 3 dy dx 12y 4 dy dx y e3x dy dx 2y 0 dy dx 5y 18. 19. 20. 21. 22. 23. 24. In Problems 25–36 solve the given initial-value problem. Give the largest interval I over which the solution is deﬁned. 25. 26. 27. xy y ex, y(1) 2 dy dx 2x 3y, y(0) 1 3 dy dx x 5y, y(0) 3 (x2 1) dy dx 2y (x 1)2 x dy dx (3x 1)y e3x dP dt 2tP P 4t 2 dr d r sec cos (x 2)2 dy dx 5 8y 4xy (x 1) dy dx (x 2)y 2xex cos2x sin x dy dx (cos3x)y 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 62 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 28. 29. 30. 31. 32. 33. 34. 35. 36. In Problems 37–40 proceed as in Example 6 to solve the given initial-value problem. Use a graphing utility to graph the continuous function y(x). 37. where 38. where 39. where 40. where 41. Proceed in a manner analogous to Example 6 to solve the initial-value problem y P(x)y 4x, y(0) 3, where Use a graphing utility to graph the continuous function y(x). 42. Consider the initial-value problem y exy f (x), y(0) 1. Express the solution of the IVP for x 0 as a nonelementary integral when f(x) 1. What is the so-lution when f (x) 0? When f (x) ex? 43. Express the solution of the initial-value problem y 2xy 1, y(1) 1, in terms of erf(x). P(x) 2, 2>x, 0 x 1, x 1. f (x) x, x, 0 x 1 x 1 (1 x2) dy dx 2xy f (x), y(0) 0, f (x) x, 0, 0 x 1 x 1 dy dx 2xy f (x), y(0) 2, f (x) 1, 1, 0 x 1 x 1 dy dx y f (x), y(0) 1, f (x) 1, 0, 0 x 3 x 3 dy dx 2y f (x), y(0) 0, y (tan x)y cos2 x, y(0) 1 y (sin x)y 2 sin x, y( >2) 1 x(x 1)dy dx xy 1, y(e) 1 (x 1) dy dx y ln x, y(1) 10 y 4xy x3ex2, y(0) 1 x dy dx y 4x 1, y(1) 8 dT dt k(T Tm), T(0) T0, k, Tm, T0 constants L di dt Ri E, i(0) i0, L, R, E, i0 constants y dx dy x 2y2, y(1) 5 Discussion Problems 44. Reread the discussion following Example 2. Construct a linear ﬁrst-order differential equation for which all nonconstant solutions approach the horizontal asymp-tote y 4 as x : . 45. Reread Example 3 and then discuss, with reference to Theorem 1.2.1, the existence and uniqueness of a solution of the initial-value problem consisting of xy 4y x6ex and the given initial condition. (a) y(0) 0 (b) y(0) y0, y0 0 (c) y(x0) y0, x0 0, y0 0 46. Reread Example 4 and then ﬁnd the general solution of the differential equation on the interval (3, 3). 47. Reread the discussion following Example 5. Construct a linear ﬁrst-order differential equation for which all solu-tions are asymptotic to the line y 3x 5 as x : . 48. Reread Example 6 and then discuss why it is technically incorrect to say that the function in (9) is a “solution” of the IVP on the interval [0, ). 49. (a) Construct a linear ﬁrst-order differential equation of the form xy a0(x)y g(x) for which yc cx3 and yp x3. Give an interval on which y x3 cx3 is the general solution of the DE. (b) Give an initial condition y(x0) y0 for the DE found in part (a) so that the solution of the IVP is y x3 1x3. Repeat if the solution is y x3 2x3. Give an interval I of deﬁnition of each of these solutions. Graph the solution curves. Is there an initial-value problem whose solution is deﬁned on (, )? (c) Is each IVP found in part (b) unique? That is, can there be more than one IVP for which, say, y x3 1x3, x in some interval I, is the solution? 50. In determining the integrating factor (3), we did not use a constant of integration in the evaluation of P(x) dx. Explain why using P(x) dx c1 has no effect on the solution of (2). 51. Suppose P(x) is continuous on some interval I and a is a number in I. What can be said about the solution of the initial-value problem y P(x)y 0, y(a) 0? Mathematical Models 52. Radioactive Decay Series The following system of differential equations is encountered in the study of the decay of a special type of radioactive series of elements: where 1 and 2 are constants. Discuss how to solve this system subject to x(0) x0, y(0) y0. Carry out your ideas. dy dt 1x 2y, dx dt 1x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 53. Heart Pacemaker A heart pacemaker consists of a switch, a battery of constant voltage E0, a capacitor with constant capacitance C, and the heart as a resistor with constant resistance R. When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage E across the heart satisﬁes the linear differential equation Solve the DE, subject to E(4) E0. Computer Lab Assignments 54. (a) Express the solution of the initial-value problem y 2xy 1, , in terms of erfc(x). (b) Use tables or a CAS to ﬁnd the value of y(2). Use a CAS to graph the solution curve for the IVP on (, ). 55. (a) The sine integral function is deﬁned by , where the integrand is Si(x) x 0 (sin t>t) dt y(0) 1 2 dE dt 1 RC E. deﬁned to be 1 at t 0. Express the solution y(x) of the initial-value problem x3y 2x2y 10 sin x, y(1) 0 in terms of Si(x). (b) Use a CAS to graph the solution curve for the IVP for x 0. (c) Use a CAS to ﬁnd the value of the absolute maxi-mum of the solution y(x) for x 0. 56. (a) The Fresnel sine integral is deﬁned by . Express the solution y(x) of the initial-value problem y (sin x2)y 0, y(0) 5, in terms of S(x). (b) Use a CAS to graph the solution curve for the IVP on (, ). (c) It is known that S(x) : as x : and S(x) : as x : . What does the solution y(x) approach as x : ? As x : ? (d) Use a CAS to ﬁnd the values of the absolute maximum and the absolute minimum of the solution y(x). 1 2 1 2 S(x) x 0 sin(pt2>2) dt 2.4 EXACT EQUATIONS ● 63 EXACT EQUATIONS REVIEW MATERIAL ●Multivariate calculus ●Partial differentiation and partial integration ●Differential of a function of two variables INTRODUCTION Although the simple ﬁrst-order equation y dx x dy 0 is separable, we can solve the equation in an alternative manner by recognizing that the expression on the left-hand side of the equality is the differential of the function f (x, y) xy; that is, d(xy) y dx x dy. In this section we examine ﬁrst-order equations in differential form M(x, y) dx N(x, y) dy 0. By applying a simple test to M and N, we can determine whether M(x, y) dx N(x, y) dy is a differen-tial of a function f(x, y). If the answer is yes, we can construct f by partial integration. 2.4 Differential of a Function of Two Variables If z f (x, y) is a function of two variables with continuous ﬁrst partial derivatives in a region R of the xy-plane, then its differential is . (1) In the special case when f (x, y) c, where c is a constant, then (1) implies . (2) f x dx f y dy 0 dz f x dx f y dy Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 64 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS In other words, given a one-parameter family of functions f (x, y) c, we can generate a ﬁrst-order differential equation by computing the differential of both sides of the equality. For example, if x2 5xy y3 c, then (2) gives the ﬁrst-order DE . (3) A Definitio Of course, not every ﬁrst-order DE written in differential form (2x 5y) dx (5x 3y2) dy 0 M(x, y) dx N(x, y) dy 0 corresponds to a differential of f (x, y) c. So for our purposes it is more important to turn the foregoing example around; namely, if we are given a ﬁrst-order DE such as (3), is there some way we can recognize that the differential expression (2x 5y) dx (5x 3y2) dy is the differential d(x2 5xy y3)? If there is, then an implicit solution of (3) is x2 5xy y3 c. We answer this question after the next deﬁnition. DEFINITION 2.4.1 Exact Equation A differential expression M(x, y) dx N(x, y) dy is an exact differential in a region R of the xy-plane if it corresponds to the differential of some function f (x, y) deﬁned in R. A ﬁrst-order differential equation of the form is said to be an exact equation if the expression on the left-hand side is an exact differential. M(x, y) dx N(x, y) dy 0 For example, x2y3 dx x3y2 dy 0 is an exact equation, because its left-hand side is an exact differential: . Notice that if we make the identiﬁcations M(x, y) x2y3 and N(x, y) x3y2, then My 3x2y2 Nx. Theorem 2.4.1, given next, shows that the equality of the partial derivatives My and Nx is no coincidence. d 1 3 x3 y3 x2y3 dx x3y2 dy THEOREM 2.4.1 Criterion for an Exact Differential Let M(x, y) and N(x, y) be continuous and have continuous ﬁrst partial derivatives in a rectangular region R deﬁned by a x b, c y d. Then a necessary and sufﬁcient condition that M(x, y) dx N(x, y) dy be an exact differential is . (4) M y N x PROOF OF THE NECESSITY For simplicity let us assume that M(x, y) and N(x, y) have continuous ﬁrst partial derivatives for all (x, y). Now if the expression M(x, y) dx N(x, y) dy is exact, there exists some function f such that for all x in R, . Therefore , M(x, y) f x , N(x, y) f y M(x, y) dx N(x, y) dy f x dx f y dy Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. and . The equality of the mixed partials is a consequence of the continuity of the ﬁrst par-tial derivatives of M(x, y) and N(x, y). The sufﬁciency part of Theorem 2.4.1 consists of showing that there exists a function f for which f x M(x, y) and fy N(x, y) whenever (4) holds. The construction of the function f actually reﬂects a basic procedure for solving exact equations. Method of Solution Given an equation in the differential form M y y f x 2 f y x x f y N x 2.4 EXACT EQUATIONS ● 65 EXAMPLE 1 Solving an Exact DE Solve 2xy dx (x2 1) dy 0. SOLUTION With M(x, y) 2xy and N(x, y) x2 1 we have . M y 2x N x M(x, y) dx N(x, y) dy 0, determine whether the equality in (4) holds. If it does, then there exists a function f for which . We can ﬁnd f by integrating M(x, y) with respect to x while holding y constant: , (5) where the arbitrary function g(y) is the “constant” of integration. Now differentiate (5) with respect to y and assume that fy N(x, y): This gives . (6) Finally, integrate (6) with respect to y and substitute the result in (5). The implicit solution of the equation is f (x, y) c. Some observations are in order. First, it is important to realize that the expres-sion N(x, y) (y) M(x, y) dx in (6) is independent of x, because . Second, we could just as well start the foregoing procedure with the assumption that f y N(x, y). After integrating N with respect to y and then differentiating that result, we would ﬁnd the analogues of (5) and (6) to be, respectively, . In either case none of these formulas should be memorized. f (x, y) N(x, y) dy h(x) and h(x) M(x, y) x N(x, y) dy x N(x, y) y M(x, y) dx N x y x M(x, y) dx N x M y 0 g(y) N(x, y) y M(x, y) dx f y y M(x, y) dx g(y) N(x, y). f (x, y) M(x, y) dx g(y) f x M(x, y) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 66 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS Thus the equation is exact, and so by Theorem 2.4.1 there exists a function f (x, y) such that . From the ﬁrst of these equations we obtain, after integrating, . Taking the partial derivative of the last expression with respect to y and setting the result equal to N(x, y) gives . ; N(x, y) It follows that g(y) 1 and g(y) y. Hence f (x, y) x2y y, so the solution of the equation in implicit form is x2y y c. The explicit form of the solution is easily seen to be y c(1 x2) and is deﬁned on any interval not containing either x 1 or x 1. Note The solution of the DE in Example 1 is not f (x, y) x2y y. Rather, it f y x2 g(y) x2 1 f (x, y) x2y g(y) f x 2xy and f y x2 1 EXAMPLE 2 Solving an Exact DE Solve (e2y y cos xy) dx (2xe2y x cos xy 2y) dy 0. SOLUTION The equation is exact because . Hence a function f(x, y) exists for which . Now, for variety, we shall start with the assumption that f y N(x, y); that is, . Remember, the reason x can come out in front of the symbol is that in the integra-tion with respect to y, x is treated as an ordinary constant. It follows that , ;M(x, y) and so h(x) 0 or h(x) c. Hence a family of solutions is xe2y sin xy y2 c 0. f x e2y y cos xy h(x) e2y y cos xy f(x, y) xe2y sin xy y2 h(x) f (x, y) 2x e2y dy x cos xy dy 2 y dy f y 2xe2y x cos xy 2y M(x, y) f x and N(x, y) f y M y 2e2y xy sin xy cos xy N x is f (x, y) c; if a constant is used in the integration of g(y), we can then write the solution as f (x, y) 0. Note, too, that the equation could be solved by separation of variables. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.4 EXACT EQUATIONS ● 67 ear equation y P(x)y f (x) can be transformed into a derivative when we multi-ply the equation by an integrating factor. The same basic idea sometimes works for a nonexact differential equation M(x, y) dx N(x, y) dy 0. That is, it is sometimes possible to ﬁnd an integrating factor (x, y) so that after multiplying, the left-hand side of (x, y)M(x, y) dx (x, y)N(x, y) dy 0 (8) is an exact differential. In an attempt to ﬁnd , we turn to the criterion (4) for exact-ness. Equation (8) is exact if and only if (M)y (N)x, where the subscripts denote partial derivatives. By the Product Rule of differentiation the last equation is the same as My yM Nx xN or xN yM (My Nx). (9) Although M, N, My, and Nx are known functions of x and y, the difﬁculty here in determining the unknown (x, y) from (9) is that we must solve a partial differential EXAMPLE 3 An Initial-Value Problem Solve . SOLUTION By writing the differential equation in the form (cos x sin x xy2) dx y(1 x2) dy 0, we recognize that the equation is exact because . Now The last equation implies that h(x) cos x sin x. Integrating gives . Thus , (7) where 2c1 has been replaced by c. The initial condition y 2 when x 0 demands that 4(1) cos2 (0) c, and so c 3. An implicit solution of the problem is then y2(1 x2) cos2 x 3. The solution curve of the IVP is the curve drawn in blue in Figure 2.4.1; it is part of an interesting family of curves. The graphs of the members of the one-parameter family of solutions given in (7) can be obtained in several ways, two of which are using software to graph level curves (as discussed in Section 2.2) and using a graphing utility to carefully graph the explicit functions obtained for var-ious values of c by solving y2 (c cos2 x)(1 x2) for y. Integrating Factors Recall from Section 2.3 that the left-hand side of the lin-y2 2 (1 x2) 1 2 cos2 x c1 or y2(1 x2) cos2 x c h(x) (cos x)(sin x dx) 1 2 cos2 x f x xy2 h(x) cos x sin x xy2. f(x, y) y2 2 (1 x2) h(x) f y y(1 x2) M y 2xy N x dy dx xy2 cos x sin x y(1 x2) , y(0) 2 x y FIGURE 2.4.1 Solution curves of DE in Example 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 68 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS equation. Since we are not prepared to do that, we make a simplifying assumption. Suppose is a function of one variable; for example, say that depends only on x. In this case, x ddx and y 0, so (9) can be written as . (10) We are still at an impasse if the quotient (My Nx)N depends on both x and y. However, if after all obvious algebraic simpliﬁcations are made, the quotient (My Nx)N turns out to depend solely on the variable x, then (10) is a ﬁrst-order ordinary differential equation. We can ﬁnally determine because (10) is separa-ble as well as linear. It follows from either Section 2.2 or Section 2.3 that (x) e(( )/N)dx. In like manner, it follows from (9) that if depends only on the variable y, then . (11) In this case, if (Nx My)M is a function of y only, then we can solve (11) for . We summarize the results for the differential equation M(x, y) dx N(x, y) dy 0. (12) • If (My Nx)N is a function of x alone, then an integrating factor for (12) is . (13) • If (Nx My)M is a function of y alone, then an integrating factor for (12) is . (14) (y) eNxMy M dy (x) eMyNx N dx d dy Nx My M Nx My d dx My Nx N EXAMPLE 4 A Nonexact DE Made Exact The nonlinear ﬁrst-order differential equation xy dx (2x2 3y2 20) dy 0 is not exact. With the identiﬁcations M xy, N 2x2 3y2 20, we ﬁnd the partial derivatives My x and Nx 4x. The ﬁrst quotient from (13) gets us nowhere, since depends on x and y. However, (14) yields a quotient that depends only on y: . The integrating factor is then e3dy/y e3lny eln y3. After we multiply the given DE by (y) y3, the resulting equation is xy4 dx (2x2y3 3y5 20y3) dy 0. You should verify that the last equation is now exact as well as show, using the method of this section, that a family of solutions is . 1 2 x2y4 1 2 y6 5y4 c y3 Nx My M 4x x xy 3x xy 3 y My Nx N x 4x 2x2 3y2 20 3x 2x2 3y2 20 REMARKS (i) When testing an equation for exactness, make sure it is of the precise form M(x, y) dx N(x, y) dy 0. Sometimes a differential equation is written G(x, y) dx H(x, y) dy. In this case, ﬁrst rewrite it as G(x, y) dx H(x, y) dy 0 and then identify M(x, y) G(x, y) and N(x, y) H(x, y) before using (4). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.4 EXACT EQUATIONS ● 69 (ii) In some texts on differential equations the study of exact equations precedes that of linear DEs. Then the method for ﬁnding integrating factors just discussed can be used to derive an integrating factor for y P(x)y f (x). By rewriting the last equation in the differential form (P(x)y f (x)) dx dy 0, we see that . From (13) we arrive at the already familiar integrating factor eP(x)dx used in Section 2.3. My Nx N P(x) EXERCISES 2.4 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1–20 determine whether the given differential equation is exact. If it is exact, solve it. 1. (2x 1) dx (3y 7) dy 0 2. (2x y) dx (x 6y) dy 0 3. (5x 4y) dx (4x 8y3) dy 0 4. (sin y y sin x) dx (cos x x cos y y) dy 0 5. (2xy2 3) dx (2x2y 4) dy 0 6. 7. (x2 y2) dx (x2 2xy) dy 0 8. 9. (x y3 y2 sin x) dx (3xy2 2y cos x) dy 10. (x3 y3) dx 3xy2 dy 0 11. 12. (3x2y ey) dx (x3 xey 2y) dy 0 13. 14. 15. 16. (5y 2x)y 2y 0 17. (tan x sin x sin y) dx cos x cos y dy 0 18. (x sin2 x 4xyexy2) dy (2y sin x cos x y 2y2exy2) dx x2y3 1 1 9x2 dx dy x3y2 0 1 3 y x dy dx y 3 x 1 x dy dx 2xex y 6x2 (y ln y exy) dx 1 y x ln y dy 0 1 ln x y x dx (1 ln x) dy 2y 1 x cos 3x dy dx y x2 4x3 3y sin 3x 0 19. (4t3y 15t2 y) dt (t4 3y2 t) dy 0 20. In Problems 21–26 solve the given initial-value problem. 21. (x y)2 dx (2xy x2 1) dy 0, y(1) 1 22. (ex y) dx (2 x yey) dy 0, y(0) 1 23. (4y 2t 5) dt (6y 4t 1) dy 0, y(1) 2 24. 25. (y2 cos x 3x2y 2x) dx (2y sin x x3 ln y) dy 0, y(0) e 26. , In Problems 27 and 28 ﬁnd the value of k so that the given differential equation is exact. 27. (y3 kxy4 2x) dx (3xy2 20x2y3) dy 0 28. (6xy3 cos y) dx (2kx2y2 x sin y) dy 0 In Problems 29 and 30 verify that the given differential equation is not exact. Multiply the given differential equa-tion by the indicated integrating factor (x, y) and verify that the new equation is exact. Solve. 29. (xy sin x 2y cos x) dx 2x cos x dy 0; (x, y) xy 30. (x2 2xy y2) dx (y2 2xy x2) dy 0; (x, y) (x y)2 In Problems 31–36 solve the given differential equation by ﬁnding, as in Example 4, an appropriate integrating factor. 31. (2y2 3x) dx 2xy dy 0 32. y(x y 1) dx (x 2y) dy 0 y(0) 1 1 1 y2 cos x 2xy dy dx y(y sin x) 3y2 t2 y5 dy dt t 2y4 0, y(1) 1 1 t 1 t2 y t2 y2 dt yey t t2 y2 dy 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 33. 6xy dx (4y 9x2) dy 0 34. 35. (10 6y e3x) dx 2 dy 0 36. (y2 xy3) dx (5y2 xy y3 sin y) dy 0 In Problems 37 and 38 solve the given initial-value problem by ﬁnding, as in Example 4, an appropriate integrating factor. 37. x dx (x2y 4y) dy 0, y(4) 0 38. (x2 y2 5) dx (y xy) dy, y(0) 1 39. (a) Show that a one-parameter family of solutions of the equation (4xy 3x2) dx (2y 2x2) dy 0 is x3 2x2y y2 c. (b) Show that the initial conditions y(0) 2 and y(1) 1 determine the same implicit solution. (c) Find explicit solutions y1(x) and y2(x) of the dif-ferential equation in part (a) such that y1(0) 2 and y2(1) 1. Use a graphing utility to graph y1(x) and y2(x). Discussion Problems 40. Consider the concept of an integrating factor used in Problems 29–38. Are the two equations M dx N dy 0 and M dx N dy 0 necessarily equivalent in the sense that a solution of one is also a solution of the other? Discuss. 41. Reread Example 3 and then discuss why we can con-clude that the interval of deﬁnition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) is (1, 1). 42. Discuss how the functions M(x, y) and N(x, y) can be found so that each differential equation is exact. Carry out your ideas. (a) (b) 43. Differential equations are sometimes solved by having a clever idea. Here is a little exercise in cleverness: Although the differential equation (x ) dx y dy 0 is not exact, show how the rearrangement (x dx y dy) dx and the observation d(x2 y2) x dx y dy can lead to a solution. 44. True or False: Every separable ﬁrst-order equation dydx g(x)h(y) is exact. 1 2 1x2 y2 1x2 y2 x1/2y1/2 x x2 y dx N(x, y) dy 0 M(x, y) dx xexy 2xy 1 x dy 0 cos x dx 1 2 y sin x dy 0 Mathematical Model 45. Falling Chain A portion of a uniform chain of length 8 ft is loosely coiled around a peg at the edge of a high horizontal platform, and the remaining portion of the chain hangs at rest over the edge of the platform. See Figure 2.4.2. Suppose that the length of the overhang-ing chain is 3 ft, that the chain weighs 2 lb/ft, and that the positive direction is downward. Starting at t 0 seconds, the weight of the overhanging portion causes the chain on the table to uncoil smoothly and to fall to the ﬂoor. If x(t) denotes the length of the chain over-hanging the table at time t 0, then v dxdt is its velocity. When all resistive forces are ignored, it can be shown that a mathematical model relating v to x is given by . (a) Rewrite this model in differential form. Proceed as in Problems 31–36 and solve the DE for v in terms of x by ﬁnding an appropriate integrating factor. Find an explicit solution v(x). (b) Determine the velocity with which the chain leaves the platform. xv dv dx v2 32x 70 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS x(t) platform edge peg FIGURE 2.4.2 Uncoiling chain in Problem 45 Computer Lab Assignments 46. Streamlines (a) The solution of the differential equation is a family of curves that can be interpreted as streamlines of a ﬂuid ﬂow around a circular object whose boundary is described by the equation x2 y2 1. Solve this DE and note the solution f(x, y) c for c 0. (b) Use a CAS to plot the streamlines for c 0, 0.2, 0.4, 0.6, and 0.8 in three different ways. First, use the contourplot of a CAS. Second, solve for x in terms of the variable y. Plot the resulting two functions of y for the given values of c, and then combine the graphs. Third, use the CAS to solve a cubic equation for y in terms of x. 2xy (x2 y2)2 dx 1 y2 x2 (x2 y2)2 dy 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.5 SOLUTIONS BY SUBSTITUTIONS ● 71 SOLUTIONS BY SUBSTITUTIONS REVIEW MATERIAL ●Techniques of integration ●Separation of variables ●Solution of linear DEs INTRODUCTION We usually solve a differential equation by recognizing it as a certain kind of equation (say, separable, linear, or exact) and then carrying out a procedure, consisting of equation-specific mathematical steps, that yields a solution of the equation. But it is not uncommon to be stumped by a differential equation because it does not fall into one of the classes of equations that we know how to solve. The procedures that are discussed in this section may be helpful in this situation. 2.5 Substitutions Often the ﬁrst step in solving a differential equation consists of transforming it into another differential equation by means of a substitution. For example, suppose we wish to transform the ﬁrst-order differential equation dydx f (x, y) by the substitution y g(x, u), where u is regarded as a function of the variable x. If g possesses ﬁrst-partial derivatives, then the Chain Rule . If we replace dydx by the foregoing derivative and replace y in f(x, y) by g(x, u), then the DE dydx f(x, y) becomes gx(x, u) gu(x, u) f(x, g(x, u)), which, solved for dudx, has the form F(x, u). If we can determine a solution u (x) of this last equation, then a solution of the original differential equation is y g(x, (x)). In the discussion that follows we examine three different kinds of ﬁrst-order differential equations that are solvable by means of a substitution. Homogeneous Equations If a function f possesses the property f(tx, ty) du dx du dx dy dx g x dx dx g u du dx gives dy dx gx(x, u) gu(x, u) du dx tf(x, y) for some real number , then f is said to be a homogeneous function of degree . For example, f (x, y) x3 y3 is a homogeneous function of degree 3, since f (tx, ty) (tx)3 (ty)3 t3(x3 y3) t3f(x, y), whereas f (x, y) x3 y3 1 is not homogeneous. A ﬁrst-order DE in differential form M(x, y) dx N(x, y) dy 0 (1) is said to be homogeneous if both coefﬁcient functions M and N are homogeneous functions of the same degree. In other words, (1) is homogeneous if . In addition, if M and N are homogeneous functions of degree , we can also write , (2) M(x, y) xM(1, u) and N(x, y) xN(1, u), where u y>x M(tx, ty) tM(x, y) and N(tx, ty) tN(x, y) Here the word homogeneous does not mean the same as it did in the Remarks at the end of Section 2.3. Recall that a linear ﬁrst-order equation a1(x)y a0(x)y g(x) is homogeneous when g(x) 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 72 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS and . (3) See Problem 31 in Exercises 2.5. Properties (2) and (3) suggest the substitutions that can be used to solve a homogeneous differential equation. Speciﬁcally, either of the substi-tutions y ux or x vy, where u and v are new dependent variables, will reduce a homogeneous equation to a separable ﬁrst-order differential equation.To show this, ob-serve that as a consequence of (2) a homogeneous equation M(x, y) dx N(x, y) dy 0 can be rewritten as , where u yx or y ux. By substituting the differential dy u dx x du into the last equation and gathering terms, we obtain a separable DE in the variables u and x: or . At this point we offer the same advice as in the preceding sections: Do not memorize anything here (especially the last formula); rather, work through the procedure each time. The proof that the substitutions x vy and dx v dy y dv also lead to a separable equation follows in an analogous manner from (3). dx x N(1, u) du M(1, u) uN(1, u) 0 [M(1, u) uN(1, u)] dx xN(1, u) du 0 M(1, u) dx N(1, u)[u dx x du] 0 xM(1, u) dx xN(1, u) dy 0 or M(1, u) dx N(1, u) dy 0 M(x, y) yM(v, 1) and N(x, y) yN(v, 1), where v x>y EXAMPLE 1 Solving a Homogeneous DE Solve (x2 y2) dx (x2 xy) dy 0. SOLUTION Inspection of M(x, y) x2 y2 and N(x, y) x2 xy shows that these coefﬁcients are homogeneous functions of degree 2. If we let y ux, then dy u dx x du, so after substituting, the given equation becomes . After integration the last line gives . Using the properties of logarithms, we can write the preceding solution as . Although either of the indicated substitutions can be used for every homoge-neous differential equation, in practice we try x vy whenever the function M(x, y) is simpler than N(x, y). Also it could happen that after using one substitution, we may encounter integrals that are difﬁcult or impossible to evaluate in closed form; switch-ing substitutions may result in an easier problem. ln (x y)2 cx y x or (x y)2 cxey/x y x 2 ln1 y x ln x lnc u 2 ln 1 u ln x ln c 1 2 1 u du dx x 0 1 u 1 u du dx x 0 x2(1 u) dx x3(1 u) du 0 (x2 u2x2) dx (x2 ux2)[u dx x du] 0 ; long division ; resubstituting u yx Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Bernoulli’s Equation The differential equation , (4) where n is any real number, is called Bernoulli’s equation. Note that for n 0 and n 1, equation (4) is linear. For n 0 and n 1 the substitution u y1n reduces any equation of form (4) to a linear equation. dy dx P(x)y f (x)yn 2.5 SOLUTIONS BY SUBSTITUTIONS ● 73 EXAMPLE 2 Solving a Bernoulli DE Solve SOLUTION We ﬁrst rewrite the equation as by dividing by x. With n 2 we have u y1 or y u1. We then substitute into the given equation and simplify. The result is . The integrating factor for this linear equation on, say, (0, ) is . Integrating gives x1u x c or u x2 cx. Since u y1, we have y 1u, so a solu-tion of the given equation is y 1(x2 cx). Note that we have not obtained the general solution of the original nonlinear dif-ferential equation in Example 2, since y 0 is a singular solution of the equation. Reduction to Separation of Variables A differential equation of the form (5) can always be reduced to an equation with separable variables by means of the sub-stitution u Ax By C, B 0. Example 3 illustrates the technique. dy dx f(Ax By C) d dx [x1u] 1 edx/x eln x eln x1 x1 du dx 1 x u x dy dx dy du du dx u2 du dx dy dx 1 x y xy2 x dy dx y x2y2. ; Chain Rule EXAMPLE 3 An Initial-Value Problem Solve SOLUTION If we let u 2x y, then dudx 2 dydx, so the differential equation is transformed into . du dx 2 u2 7 or du dx u2 9 dy dx (2x y)2 7, y(0) 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ; replace by c e6c1 x y FIGURE 2.5.1 Solutions of DE in Example 3 74 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS The last equation is separable. Using partial fractions and then integrating yields . Solving the last equation for u and then resubstituting gives the solution . (6) Finally, applying the initial condition y(0) 0 to the last equation in (6) gives c 1. Figure 2.5.1, obtained with the aid of a graphing utility, shows the graph of the particular solution in dark blue, along with the graphs of some other members of the family of solutions (6). y 2x 3(1 e6x) 1 e6x u 3(1 ce6x) 1 ce6x or y 2x 3(1 ce6x) 1 ce6x 1 6 ln u 3 u 3 x c1 or u 3 u 3 e6x6c1 ce6x du (u 3)(u 3) dx or 1 6 1 u 3 1 u 3 du dx EXERCISES 2.5 Answers to selected odd-numbered problems begin on page ANS-2. Each DE in Problems 1–14 is homogeneous. In Problems 1–10 solve the given differential equation by using an appropriate substitution. 1. (x y) dx x dy 0 2. (x y) dx x dy 0 3. x dx (y 2x) dy 0 4. y dx 2(x y) dy 5. (y2 yx) dx x2 dy 0 6. (y2 yx) dx x2 dy 0 7. 8. 9. 10. In Problems 11–14 solve the given initial-value problem. 11. 12. 13. (x yey/x) dx xey/x dy 0, y(1) 0 14. y dx x(ln x ln y 1) dy 0, y(1) e (x2 2y2) dx dy xy, y(1) 1 xy2 dy dx y3 x3, y(1) 2 x dy dx y 1x2 y2, x 0 y dx (x 1xy) dy 0 dy dx x 3y 3x y dy dx y x y x Each DE in Problems 15–22 is a Bernoulli equation. In Problems 15–20 solve the given differential equation by using an appropriate substitution. 15. 16. 17. 18. 19. 20. In Problems 21 and 22 solve the given initial-value problem. 21. 22. Each DE in Problems 23–30 is of the form given in (5). In Problems 23–28 solve the given differential equation by using an appropriate substitution. 23. 24. 25. 26. 27. 28. dy dx 1 eyx5 dy dx 2 1y 2x 3 dy dx sin(x y) dy dx tan2(x y) dy dx 1 x y x y dy dx (x y 1)2 y1/2 dy dx y3/2 1, y(0) 4 x2 dy dx 2xy 3y4, y(1) 1 2 3(1 t2) dy dt 2ty(y3 1) t2 dy dt y2 ty x dy dx (1 x)y xy2 dy dx y(xy3 1) dy dx y exy2 x dy dx y 1 y2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.6 A NUMERICAL METHOD ● 75 In Problems 29 and 30 solve the given initial-value problem. 29. 30. Discussion Problems 31. Explain why it is always possible to express any homoge-neous differential equation M(x, y) dx N(x, y) dy 0 in the form . You might start by proving that . 32. Put the homogeneous differential equation (5x2 2y2) dx xy dy 0 into the form given in Problem 31. 33. (a) Determine two singular solutions of the DE in Problem 10. (b) If the initial condition y(5) 0 is as prescribed in Problem 10, then what is the largest interval I over which the solution is deﬁned? Use a graphing util-ity to graph the solution curve for the IVP. 34. In Example 3 the solution y(x) becomes unbounded as x : . Nevertheless, y(x) is asymptotic to a curve as x : and to a different curve as x : . What are the equations of these curves? 35. The differential equation dydx P(x) Q(x)y R(x)y2 is known as Riccati’s equation. (a) A Riccati equation can be solved by a succession of two substitutions provided that we know a M(x, y) xaM(1, y>x) and N(x, y) xaN(1, y>x) dy dx F y x dy dx 3x 2y 3x 2y 2, y(1) 1 dy dx cos(x y), y(0) >4 particular solution y1 of the equation. Show that the substitution y y1 u reduces Riccati’s equation to a Bernoulli equation (4) with n 2. The Bernoulli equation can then be reduced to a linear equation by the substitution w u1. (b) Find a one-parameter family of solutions for the differential equation where y1 2x is a known solution of the equation. 36. Determine an appropriate substitution to solve xy y ln(xy). Mathematical Models 37. Falling Chain In Problem 45 in Exercises 2.4 we saw that a mathematical model for the velocity v of a chain slipping off the edge of a high horizontal platform is . In that problem you were asked to solve the DE by con-verting it into an exact equation using an integrating factor. This time solve the DE using the fact that it is a Bernoulli equation. 38. Population Growth In the study of population dy-namics one of the most famous models for a growing but bounded population is the logistic equation , where a and b are positive constants. Although we will come back to this equation and solve it by an alternative method in Section 3.2, solve the DE this ﬁrst time using the fact that it is a Bernoulli equation. dP dt P(a bP) xv dv dx v2 32x dy dx 4 x2 1 x y y2 A NUMERICAL METHOD INTRODUCTION A ﬁrst-order differential equation dydx f (x, y) is a source of information. We started this chapter by observing that we could garner qualitative information from a ﬁrst-order DE about its solutions even before we attempted to solve the equation. Then in Sections 2.2–2.5 we examined ﬁrst-order DEs analytically—that is, we developed some procedures for obtaining explicit and implicit solutions. But a differential equation can a possess a solution, yet we may not be able to obtain it analytically. So to round out the picture of the different types of analyses of differential equations, we conclude this chapter with a method by which we can “solve” the differential equa-tion numerically—this means that the DE is used as the cornerstone of an algorithm for approximat-ing the unknown solution. In this section we are going to develop only the simplest of numerical methods—a method that utilizes the idea that a tangent line can be used to approximate the values of a function in a small neighborhood of the point of tangency. A more extensive treatment of numerical methods for ordi-nary differential equations is given in Chapter 9. 2.6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 76 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS Using the Tangent Line Let us assume that the ﬁrst-order initial-value problem (1) possesses a solution. One way of approximating this solution is to use tangent lines. For example, let y(x) denote the unknown solution of the ﬁrst-order initial-value problem The nonlinear differential equation in this IVP cannot be solved directly by any of the methods considered in Sections 2.2, 2.4, and 2.5; nevertheless, we can still ﬁnd approximate numerical values of the unknown y(x). Speciﬁcally, suppose we wish to know the value of y(2.5). The IVP has a solution, and as the ﬂow of the direction ﬁeld of the DE in Figure 2.6.1(a) sug-gests, a solution curve must have a shape similar to the curve shown in blue. The direction ﬁeld in Figure 2.6.1(a) was generated with lineal elements passing through points in a grid with integer coordinates. As the solution curve passes through the initial point (2, 4), the lineal element at this point is a tangent line with slope given by As is apparent in Figure 2.6.1(a) and the “zoom in” in Figure 2.6.1(b), when x is close to 2, the points on the solution curve are close to the points on the tangent line (the lineal element). Using the point (2, 4), the slope f (2, 4) 1.8, and the point-slope form of a line, we ﬁnd that an equa-tion of the tangent line is y L(x), where L(x) 1.8x 0.4. This last equation, called a linearization of y(x) at x 2, can be used to approximate values of y(x) within a small neighborhood of x 2. If y1 L(x1) denotes the y-coordinate on the tangent line and y(x1) is the y-coordinate on the solution curve corresponding to an x-coordinate x1 that is close to x 2, then y(x1) y1. If we choose, say, x1 2.1, then y1 L(2.1) 1.8(2.1) 0.4 4.18, so y(2.1) 4.18. f (2, 4) 0.114 0.4(2)2 1.8. y 0.11y 0.4x2, y(2) 4. y f(x, y), y(x0) y0 2 (2, 4) slope m = 1.8 x y 2 4 _2 (a) direction fiel for y 0 (b) lineal element at (2, 4) e solution curv FIGURE 2.6.1 Magniﬁcation of a neighborhood about the point (2, 4) solution curv e x y x1 = + x0 h x0 L(x) (x0, y0) (x1, y1) h (x1, y(x1)) slope = f(x0, y0) error FIGURE 2.6.2 Approximating y(x1) using a tangent line Euler’s Method To generalize the procedure just illustrated, we use the lin-This is not an actual tangent line, since (x1, y1) lies on the ﬁrst tangent and not on the solution curve. earization of the unknown solution y(x) of (1) at x x0: . (2) The graph of this linearization is a straight line tangent to the graph of y y(x) at the point (x0, y0). We now let h be a positive increment of the x-axis, as shown in Figure 2.6.2. Then by replacing x by x1 x0 h in (2), we get , where y1 L(x1). The point (x1, y1) on the tangent line is an approximation to the point (x1, y(x1)) on the solution curve. Of course, the accuracy of the approxima-tion L(x1) y(x1) or y1 y(x1) depends heavily on the size of the increment h. Usually, we must choose this step size to be “reasonably small.” We now repeat the process using a second “tangent line” at (x1, y1). By identifying the new starting L(x1) y0 f (x0, y0)(x0 h x0) or y1 y0 hf(x1, y1) L(x) y0 f (x0, y0)(x x0) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. TABLE 2.6.1 h 0.1 xn yn 2.00 4.0000 2.10 4.1800 2.20 4.3768 2.30 4.5914 2.40 4.8244 2.50 5.0768 TABLE 2.6.2 h 0.05 xn yn 2.00 4.0000 2.05 4.0900 2.10 4.1842 2.15 4.2826 2.20 4.3854 2.25 4.4927 2.30 4.6045 2.35 4.7210 2.40 4.8423 2.45 4.9686 2.50 5.0997 2.6 A NUMERICAL METHOD ● 77 point as (x1, y1) with (x0, y0) in the above discussion, we obtain an approximation y2 y(x2) corresponding to two steps of length h from x0, that is, x2 x1 h x0 2h, and . Continuing in this manner, we see that y1, y2, y3, . . . , can be deﬁned recursively by the general formula , (3) where xn x0 nh, n 0, 1, 2, . . . . This procedure of using successive “tangent lines” is called Euler’s method. yn1 yn hf (xn, yn) y(x2) y(x0 2h) y(x1 h) y2 y1 hf (x1, y1) EXAMPLE 1 Euler’s Method Consider the initial-value problem Use Euler’s method to obtain an approximation of y(2.5) using ﬁrst h 0.1 and then h 0.05. SOLUTION With the identiﬁcation (3) becomes . Then for h 0.1, x0 2, y0 4, and n 0 we ﬁnd , which, as we have already seen, is an estimate to the value of y(2.1). However, if we use the smaller step size h 0.05, it takes two steps to reach x 2.1. From we have y1 y(2.05) and y2 y(2.1). The remainder of the calculations were carried out by using software. The results are summarized in Tables 2.6.1 and 2.6.2, where each entry has been rounded to four decimal places. We see in Tables 2.6.1 and 2.6.2 that it takes ﬁve steps with h 0.1 and 10 steps with h 0.05, respectively, to get to x 2.5. Intuitively, we would expect that y10 5.0997 corresponding to h 0.05 is the better approximation of y(2.5) than the value y5 5.0768 corre-sponding to h 0.1. In Example 2 we apply Euler’s method to a differential equation for which we have already found a solution. We do this to compare the values of the approxima-tions yn at each step with the true or actual values of the solution y(xn) of the initial-value problem. y2 4.09 0.05(0.114.09 0.4(2.05)2) 4.18416187 y1 4 0.05(0.114 0.4(2)2) 4.09 y1 y0 h(0.11y0 0.4x0 2) 4 0.1(0.114 0.4(2)2) 4.18 yn1 yn h(0.11yn 0.4xn 2) f (x, y) 0.11y 0.4x2, y 0.11y 0.4x2, y(2) 4. EXAMPLE 2 Comparison of Approximate and Actual Values Consider the initial-value problem y 0.2xy, y(1) 1. Use Euler’s method to obtain an approximation of y(1.5) using ﬁrst h 0.1 and then h 0.05. SOLUTION With the identiﬁcation f (x, y) 0.2xy, (3) becomes where x0 1 and y0 1. Again with the aid of computer software we obtain the values in Tables 2.6.3 and 2.6.4 on page 78. yn1 yn h(0.2xn yn) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 78 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS In Example 1 the true or actual values were calculated from the known solution (Verify.) The absolute error is deﬁned to be . The relative error and percentage relative error are, in turn, . It is apparent from Tables 2.6.3 and 2.6.4 that the accuracy of the approximations improves as the step size h decreases. Also, we see that even though the percentage relative error is growing with each step, it does not appear to be that bad. But you should not be deceived by one example. If we simply change the coefﬁcient of the right side of the DE in Example 2 from 0.2 to 2, then at xn 1.5 the percentage relative errors increase dramatically. See Problem 4 in Exercises 2.6. A Caveat Euler’s method is just one of many different ways in which a solu-tion of a differential equation can be approximated. Although attractive for its sim-plicity, Euler’s method is seldom used in serious calculations. It was introduced here simply to give you a ﬁrst taste of numerical methods. We will go into greater detail in discussing numerical methods that give signiﬁcantly greater accuracy, no-tably the fourth order Runge-Kutta method, referred to as the RK4 method, in Chapter 9. Numerical Solvers Regardless of whether we can actually ﬁnd an explicit or implicit solution, if a solution of a differential equation exists, it represents a smooth curve in the Cartesian plane. The basic idea behind any numerical method for ﬁrst-order ordinary differential equations is to somehow approximate the y-values of a solution for preselected values of x. We start at a speciﬁed initial point (x0, y0) on a solution curve and proceed to calculate in a step-by-step fashion a sequence of points (x1, y1), (x2, y2), . . . , (xn, yn) whose y-coordinates yi approxi-mate the y-coordinates y(xi) of points (x1, y(x1)), (x2, y(x2)), . . . , (xn, y(xn)) that lie on the graph of the usually unknown solution y(x). By taking the x-coordinates close together (that is, for small values of h) and by joining the points (x1, y1), (x2, y2), . . . , (xn, yn) with short line segments, we obtain a polygonal curve whose qualitative characteristics we hope are close to those of an actual solution curve. Drawing curves is something that is well suited to a computer. A computer program written to either implement a numerical method or render a visual representation of an approximate solution curve ﬁtting the numerical data produced by this method is referred to as a numerical solver. Many different numerical solvers are commer-cially available, either embedded in a larger software package, such as a computer absolute error actual value and absolute error actual value 100 actual value approximation y e0.1(x21). TABLE 2.6.3 h 0.1 xn yn Actual value Abs. error % Rel. error 1.00 1.0000 1.0000 0.0000 0.00 1.10 1.0200 1.0212 0.0012 0.12 1.20 1.0424 1.0450 0.0025 0.24 1.30 1.0675 1.0714 0.0040 0.37 1.40 1.0952 1.1008 0.0055 0.50 1.50 1.1259 1.1331 0.0073 0.64 TABLE 2.6.4 h 0.05 xn yn Actual value Abs. error % Rel. error 1.00 1.0000 1.0000 0.0000 0.00 1.05 1.0100 1.0103 0.0003 0.03 1.10 1.0206 1.0212 0.0006 0.06 1.15 1.0318 1.0328 0.0009 0.09 1.20 1.0437 1.0450 0.0013 0.12 1.25 1.0562 1.0579 0.0016 0.16 1.30 1.0694 1.0714 0.0020 0.19 1.35 1.0833 1.0857 0.0024 0.22 1.40 1.0980 1.1008 0.0028 0.25 1.45 1.1133 1.1166 0.0032 0.29 1.50 1.1295 1.1331 0.0037 0.32 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. algebra system, or provided as a stand-alone package. Some software packages simply plot the generated numerical approximations, whereas others generate hard numerical data as well as the corresponding approximate or numerical solution curves. By way of illustration of the connect-the-dots nature of the graphs pro-duced by a numerical solver, the two colored polygonal graphs in Figure 2.6.3 are the numerical solution curves for the initial-value problem y 0.2xy, y(0) 1 on the interval [0, 4] obtained from Euler’s method and the RK4 method using the step size h 1. The blue smooth curve is the graph of the exact solution of the IVP. Notice in Figure 2.6.3 that, even with the ridiculously large step size of h 1, the RK4 method produces the more believable “solution curve.” The numer-ical solution curve obtained from the RK4 method is indistinguishable from the actual solution curve on the interval [0, 4] when a more typical step size of h 0.1 is used. Using a Numerical Solver Knowledge of the various numerical methods is not necessary in order to use a numerical solver. Asolver usually requires that the dif-ferential equation be expressed in normal form dydx f (x, y). Numerical solvers that generate only curves usually require that you supply f (x, y) and the initial data x0 and y0 and specify the desired numerical method. If the idea is to approximate the nu-merical value of y(a), then a solver may additionally require that you state a value for h or, equivalently, give the number of steps that you want to take to get from x x0 to x a. For example, if we wanted to approximate y(4) for the IVP illustrated in Figure 2.6.3, then, starting at x 0 it would take four steps to reach x 4 with a step size of h 1; 40 steps is equivalent to a step size of h 0.1. Although we will not delve here into the many problems that one can encounter when attempting to ap-proximate mathematical quantities, you should at least be aware of the fact that a nu-merical solver may break down near certain points or give an incomplete or mislead-ing picture when applied to some ﬁrst-order differential equations in the normal form. Figure 2.6.4 illustrates the graph obtained by applying Euler’s method to a cer-tain ﬁrst-order initial-value problem dydx f (x, y), y(0) 1. Equivalent results were obtained using three different commercial numerical solvers, yet the graph is hardly a plausible solution curve. (Why?) There are several avenues of recourse when a numerical solver has difﬁculties; three of the more obvious are decrease the step size, use another numerical method, and try a different numerical solver. y e0.1x2 FIGURE 2.6.4 A not-very helpful numerical solution curve x y 1 2 3 4 5 _1 1 2 3 4 5 6 _2 _1 2.6 A NUMERICAL METHOD ● 79 exact solution (0,1) Euler’s method RK4 method _1 1 2 3 4 5 y x 4 5 3 2 1 _1 FIGURE 2.6.3 Comparison of the Runge-Kutta (RK4) and Euler methods EXERCISES 2.6 Answers to selected odd-numbered problems begin on page ANS-3. In Problems 1 and 2 use Euler’s method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) by hand, ﬁrst using h 0.1 and then using h 0.05. 1. y 2x 3y 1, y(1) 5; y(1.2) 2. y x y2, y(0) 0; y(0.2) In Problems 3 and 4 use Euler’s method to obtain a four-decimal approximation of the indicated value. First use h 0.1 and then use h 0.05. Find an explicit solution for each initial-value problem and then construct tables similar to Tables 2.6.3 and 2.6.4. 3. y y, y(0) 1; y(1.0) 4. y 2xy, y(1) 1; y(1.5) In Problems 5–10 use a numerical solver and Euler’s method to obtain a four-decimal approximation of the indi-cated value. First use h 0.1 and then use h 0.05. 5. y ey, y(0) 0; y(0.5) 6. y x2 y2, y(0) 1; y(0.5) 7. y (x y)2, y(0) 0.5; y(0.5) 8. 9. 10. y y y2, y(0) 0.5; y(0.5) In Problems 11 and 12 use a numerical solver to obtain a nu-merical solution curve for the given initial-value problem. First use Euler’s method and then the RK4 method. Use y xy2 y x, y(1) 1; y(1.5) y xy 1y, y(0) 1; y(0.5) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 80 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS h 0.25 in each case. Superimpose both solution curves on the same coordinate axes. If possible, use a different color for each curve. Repeat, using h 0.1 and h 0.05. 11. y 2(cos x)y, y(0) 1 12. y y(10 2y), y(0) 1 Discussion Problems 13. Use a numerical solver and Euler’s method to approximate y(1.0), where y(x) is the solution to y 2xy2, y(0) 1. First use h 0.1 and then use h 0.05. Repeat, using the RK4 method. Discuss what might cause the approximations to y(1.0) to differ so greatly. Computer Lab Assignments 14. (a) Use a numerical solver and the RK4 method to graph the solution of the initial-value problem y 2xy 1, y(0) 0. (b) Solve the initial-value problem by one of the analytic procedures developed earlier in this chapter. (c) Use the analytic solution y(x) found in part (b) and a CAS to ﬁnd the coordinates of all relative extrema. CHAPTER 2 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-3. Answer Problems 1–12 without referring back to the text. Fill in the blanks or answer true or false. 1. The linear DE, y ky A, where k and A are constants, is autonomous. The critical point of the equa-tion is a(n) (attractor or repeller) for k 0 and a(n) (attractor or repeller) for k 0. 2. The initial-value problem x 4y 0, y(0) k, has an inﬁnite number of solutions for k and no solution for k . 3. The linear DE, y k1y k2, where k1 and k2 are nonzero constants, always possesses a constant solution. 4. The linear DE, a1(x)y a0(x)y 0 is also separable. 5. An example of a nonlinear third-order differential equa-tion in normal form is . 6. The ﬁrst-order DE is not separa-ble. 7. Every autonomous DE is separable. 8. By inspection, two solutions of the differential equation are . 9. If then . 10. If a differentiable function satisﬁes then . y(x) y(1) 2, y x, y(x) y y exy, y y 2 dy>dx f(y) dr du ru r u 1 dy dx 11. is a solution of the linear ﬁrst-order differential equation . 12. An example of an autonomous linear ﬁrst-order DE with a single critical point is , whereas an autonomous nonlinear ﬁrst-order DE with a single criti-cal point is . In Problems 13 and 14 construct an autonomous ﬁrst-order differential equation dydx f (y) whose phase portrait is consistent with the given ﬁgure. 13. 3 3 y ecos x x 0 tecos t dt 1 3 y FIGURE 2.R.1 Graph for Problem 13 14. 0 2 4 y FIGURE 2.R.2 Graph for Problem 14 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 2 IN REVIEW ● 81 15. The number 0 is a critical point of the autonomous dif-ferential equation dxdt xn, where n is a positive in-teger. For what values of n is 0 asymptotically stable? Semi-stable? Unstable? Repeat for the differential equa-tion dxdt xn. 16. Consider the differential equation The function f (P) has one real zero, as shown in Figure 2.R.3. Without attempting to solve the differen-tial equation, estimate the value of limt: P(t). f (P) 0.5P 3 1.7P 3.4. dP>dt f (P), where (g) y dx (y xy2) dy (h) (i) xy y y2 2x (j) 2xy y y2 2x2 (k) y dx x dy 0 (l) (m) (n) In Problems 19–26 solve the given differential equation. 19. (y2 1) dx y sec2 x dy 20. y(ln x ln y) dx (x ln x x ln y y) dy 21. 22. 23. 24. (2x y 1)y 1 25. (x2 4) dy (2x 8xy) dx 26. (2r2 cos sin r cos ) d (4r sin 2r cos2 ) dr 0 In Problems 27 and 28 solve the given initial-value problem and give the largest interval I on which the solution is deﬁned. 27. 28. 29. (a) Without solving, explain why the initial-value problem has no solution for y0 0. (b) Solve the initial-value problem in part (a) for y0 0 and ﬁnd the largest interval I on which the solution is deﬁned. 30. (a) Find an implicit solution of the initial-value problem . (b) Find an explicit solution of the problem in part (a) and give the largest interval I over which the solution is deﬁned. A graphing utility may be helpful here. dy dx y2 x2 xy , y(1) 12 dy dx 1y, y(x0) y0 dy dt 2(t 1)y2 0, y(0) 1 8 sin x dy dx (cos x)y 0, y (7p>6) 2 t dQ dt Q t 4 ln t dx dy 4y2 6xy 3y2 2x (6x 1)y2 dy dx 3x2 2y3 0 y x2 dy dx e2x3y2 0 dy dx x y y x 1 x2 2y x dx (3 ln x2) dy x dy dx ye x/y x FIGURE 2.R.4 Portion of a direction ﬁeld for Problem 17 P 1 1 f FIGURE 2.R.3 Graph for Problem 16 17. Figure 2.R.4 is a portion of a direction ﬁeld of a differ-ential equation dydx f (x, y). By hand, sketch two different solution curves—one that is tangent to the lin-eal element shown in black and one that is tangent to the lineal element shown in red. 18. Classify each differential equation as separable, exact, linear, homogeneous, or Bernoulli. Some equations may be more than one kind. Do not solve. (a) (b) (c) (d) (e) (f) dy dx 5y y2 dy dx y2 y x2 x dy dx 1 x(x y) (x 1) dy dx y 10 dy dx 1 y x dy dx x y x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. integers, 7 m 7, 7 n 7. In each direction ﬁeld, sketch by hand an approximate solution curve that passes through each of the solid points shown in red. Discuss: Does it appear that the DE possesses critical points in the interval 3.5 y 3.5? If so, classify the critical points as asymp-totically stable, unstable, or semi-stable. x 3 2 1 _1 _2 _3 _3 _2 _1 1 2 3 y FIGURE 2.R.6 Portion of a direction ﬁeld for Problem 33 x y FIGURE 2.R.5 Graph for Problem 31 34. FIGURE 2.R.7 Portion of a direction ﬁeld for Problem 34 x 3 2 1 _1 _2 _3 _3 _2 _1 1 2 3 y 82 ● CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS 31. Graphs of some members of a family of solutions for a ﬁrst-order differential equation dydx f(x, y) are shown in Figure 2.R.5. The graphs of two implicit solutions, one that passes through the point (1, 1) and one that passes through (1, 3), are shown in blue. Reproduce the ﬁgure on a piece of paper. With colored pencils trace out the solution curves for the solutions y y1(x) and y y2(x) deﬁned by the implicit solu-tions such that y1(1) 1 and y2(1) 3, respectively. Estimate the intervals on which the solutions y y1(x) and y y2(x) are deﬁned. 32. Use Euler’s method with step size h 0.1 to approxi-mate y(1.2), where y(x) is a solution of the initial-value problem , y(1) 9. In Problems 33 and 34 each ﬁgure represents a portion of a direction ﬁeld of an autonomous ﬁrst-order differential equa-tion dydx f(y). Reproduce the ﬁgure on a separate piece of paper and then complete the direction ﬁeld over the grid. The points of the grid are (mh, nh), where m and n h 1 2, y 1 x1y 33. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 83 3 Modeling with First-Order Differential Equations 3.1 Linear Models 3.2 Nonlinear Models 3.3 Modeling with Systems of First-Order DEs Chapter 3 in Review In Section 1.3 we saw how a ﬁrst-order differential equation could be used as a mathematical model in the study of population growth, radioactive decay, continuous compound interest, cooling of bodies, mixtures, chemical reactions, ﬂuid draining from a tank, velocity of a falling body, and current in a series circuit. Using the methods of Chapter 2, we are now able to solve some of the linear DEs in Section 3.1 and nonlinear DEs in Section 3.2 that commonly appear in applications. The chapter concludes with the natural next step. In Section 3.3 we examine how systems of ﬁrst-order differential equations can arise as mathematical models in coupled physical systems (for example, electrical networks, and a population of predators such as foxes interacting with a population of prey such as rabbits). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 84 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS Growth and Decay The initial-value problem , (1) where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. We saw in Section 1.3 that in biological applica-tions the rate of growth of certain populations (bacteria, small animals) over short periods of time is proportional to the population present at time t. Knowing the pop-ulation at some arbitrary initial time t0, we can then use the solution of (1) to predict the population in the future—that is, at times t t0. The constant of proportional-ity k in (1) can be determined from the solution of the initial-value problem, using a subsequent measurement of x at a time t1 t0. In physics and chemistry (1) is seen in the form of a first-o der reaction—that is, a reaction whose rate, or velocity, dxdt is directly proportional to the amount x of a substance that is unconverted or remaining at time t. The decomposition, or decay, of U-238 (uranium) by radioactivity into Th-234 (thorium) is a ﬁrst-order reaction. EXAMPLE 1 Bacterial Growth Aculture initially has P0 number of bacteria. At t 1 h the number of bacteria is mea-sured to be . If the rate of growth is proportional to the number of bacteria P(t) pre-sent at time t, determine the time necessary for the number of bacteria to triple. SOLUTION We ﬁrst solve the differential equation in (1), with the symbol x replaced by P. With t0 0 the initial condition is P(0) P0. We then use the empirical obser-vation that to determine the constant of proportionality k. Notice that the differential equation dPdt kP is both separable and linear. When it is put in the standard form of a linear ﬁrst-order DE, , we can see by inspection that the integrating factor is ekt. Multiplying both sides of the equation by this term and integrating gives, in turn, . Therefore P(t) cekt. At t 0 it follows that P0 ce0 c, so P(t) P0ekt. At t 1 we have or . From the last equation we get , so P(t) P0e0.4055t. To ﬁnd the time at which the number of bac-teria has tripled, we solve 3P0 P0e0.4055t for t. It follows that 0.4055t ln 3, or . See Figure 3.1.1. t ln 3 0.4055 2.71 h k ln 3 2 0.4055 ek 3 2 3 2P0 P0ek d dt [ektP] 0 and ektP c dP dt kP 0 P(1) 3 2P0 3 2P0 dx dt kx, x(t0) x0 t P 3P0 P0 t = 2.71 P(t) = P0e0.4055t FIGURE 3.1.1 Time in which population triples in Example 1 LINEAR MODELS REVIEW MATERIAL ●A differential equation as a mathematical model in Section 1.3 ●Reread “Solving a Linear First-Order Equation” on page 56 in Section 2.3 INTRODUCTION In this section we solve some of the linear ﬁrst-order models that were introduced in Section 1.3. 3.1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.1 LINEAR MODELS ● 85 Notice in Example 1 that the actual number P0 of bacteria present at time t 0 played no part in determining the time required for the number in the culture to triple. The time necessary for an initial population of, say, 100 or 1,000,000 bacteria to triple is still approximately 2.71 hours. As shown in Figure 3.1.2, the exponential function ekt increases as t increases for k 0 and decreases as t increases for k 0. Thus problems describing growth (whether of populations, bacteria, or even capital) are characterized by a positive value of k, whereas problems involving decay (as in radioactive disintegration) yield a negative k value. Accordingly, we say that k is either a growth constant (k 0) or a decay constant (k 0). Half-Life In physics the half-life is a measure of the stability of a radioactive substance. The half-life is simply the time it takes for one-half of the atoms in an initial amount A0 to disintegrate, or transmute, into the atoms of another element. The longer the half-life of a substance, the more stable it is. For example, the half-life of highly radioactive radium, Ra-226, is about 1700 years. In 1700 years one-half of a given quantity of Ra-226 is transmuted into radon, Rn-222. The most commonly occurring uranium isotope, U-238, has a half-life of approximately 4,500,000,000 years. In about 4.5 billion years, one-half of a quantity of U-238 is transmuted into lead, Pb-206. EXAMPLE 2 Half-Life of Plutonium A breeder reactor converts relatively stable uranium-238 into the isotope plutonium-239. After 15 years it is determined that 0.043% of the initial amount A0 of plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration is pro-portional to the amount remaining. SOLUTION Let A(t) denote the amount of plutonium remaining at time t. As in Example 1 the solution of the initial-value problem is A(t) A0ekt. If 0.043% of the atoms of A0 have disintegrated, then 99.957% of the substance remains. To ﬁnd the decay constant k, we use 0.99957A0 A(15)—that is, 0.99957A0 A0e15k. Solving for k then gives ln 0.99957 0.00002867. Hence A(t) A0e0.00002867t. Now the half-life is the corresponding value of time at which . Solving for t gives , or . The last equation yields . Carbon Dating About 1950, a team of scientists at the University of Chicago led by the chemist Willard Libby devised a method using a radioactive isotope of car-bon as a means of determining the approximate ages of carbonaceous fossilized mat-ter. The theory of carbon dating is based on the fact that the radioisotope carbon-14 is produced in the atmosphere by the action of cosmic radiation on nitrogen-14. The ratio of the amount of C-14 to the stable C-12 in the atmosphere appears to be a con-stant, and as a consequence the proportionate amount of the isotope present in all liv-ing organisms is the same as that in the atmosphere. When a living organism dies, the absorption of C-14, by breathing, eating, or photosynthesis, ceases. By comparing the proportionate amount of C-14, say, in a fossil with the constant amount ratio found in the atmosphere, it is possible to obtain a reasonable estimation of its age. The method is based on the knowledge of the half-life of C-14. Libby’s calculated t ln 2 0.00002867 24,180 yr 1 2 e0.00002867t 1 2 A0 A0e0.00002867t A(t) 1 2 A0 k 1 15 dA dt kA, A(0) A0 t ekt, k > 0 growth ekt, k < 0 decay y FIGURE 3.1.2 Growth (k 0) and decay (k 0) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. value of the half-life of C-14 was approximately 5600 years, but today the commonly accepted value of the half-life is approximately 5730 years. For his work, Libby was awarded the Nobel Prize for chemistry in 1960. Libby’s method has been used to date wooden furniture found in Egyptian tombs, the woven ﬂax wrappings of the Dead Sea Scrolls, a recently discovered copy of the Gnostic Gospel of Judas written on papyrus, and the cloth of the enigmatic Shroud of Turin. See Figure 3.1.3 and Problem 12 in Exercises 3.1. 86 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS The number of disintegrations per minute per gram of carbon is recorded by using a Geiger counter. The lower level of detectability is about 0.1 disintegrations per minute per gram. EXAMPLE 3 Age of a Fossil A fossilized bone is found to contain 0.1% of its original amount of C-14. Determine the age of the fossil. SOLUTION The starting point is again A(t) A0ekt. To determine the value of the decay constant k we use the fact that A0 A(5730) or A0 A0e5730k. The last equation implies 5730k ln ln2 and so we get k (ln2)5730 0.00012097. Therefore A(t) A0e0.00012097t. With A(t) 0.001A0 we have 0.001A0 A0e0.00012097t and 0.00012097t ln(0.001) ln 1000. Thus . The date found in Example 3 is really at the border of accuracy for this method. The usual carbon-14 technique is limited to about 10 half-lives of the isotope, or roughly years. One reason for this limitation is that the chemical analysis needed to obtain an accurate measurement of the remaining C-14 becomes somewhat formidable around the point Also, this analysis demands the destruction of a rather large sample of the specimen. If this measurement is accomplished indi-rectly, based on the actual radioactivity of the specimen, then it is very difﬁcult to distinguish between the radiation from the specimen and the normal background radiation. But recently the use of a particle accelerator has enabled scientists to separate the C-14 from the stable C-12 directly. When the precise value of the ratio of C-14 to C-12 is computed, the accuracy can be extended to 70,000 to 100,000 years. Other isotopic techniques, such as using potassium-40 and argon-40, can give dates of several million years. Nonisotopic methods based on the use of amino acids are also sometimes possible. Newton’s Law of Cooling/Warming In equation (3) of Section 1.3 we saw that the mathematical formulation of Newton’s empirical law of cooling/warming of an object is given by the linear ﬁrst-order differential equation , (2) where k is a constant of proportionality, T(t) is the temperature of the object for t 0, and Tm is the ambient temperature—that is, the temperature of the medium around the object. In Example 4 we assume that Tm is constant. dT dt k(T Tm) 0.001A0. 60,000 t ln 1000 0.00012097 57,100 years 1 2 1 2 1 2 EXAMPLE 4 Cooling of a Cake When a cake is removed from an oven, its temperature is measured at 300° F. Three minutes later its temperature is 200° F. How long will it take for the cake to cool off to a room temperature of 70° F? FIGURE 3.1.3 A page of the Gnostic Gospel of Judas NGS Image Collection Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.1 LINEAR MODELS ● 87 SOLUTION In (2) we make the identiﬁcation Tm 70. We must then solve the initial-value problem (3) and determine the value of k so that T(3) 200. Equation (3) is both linear and separable. If we separate variables, , yields ln T 70 kt c1, and so T 70 c2ekt. When t 0, T 300, so 300 70 c2 gives c2 230; therefore T 70 230ekt. Finally, the measurement T(3) 200 leads to , or ln . Thus . (4) We note that (4) furnishes no ﬁnite solution to T(t) 70, since . Yet we intuitively expect the cake to reach room temperature after a reasonably long period of time. How long is “long”? Of course, we should not be disturbed by the fact that the model (3) does not quite live up to our physical intuition. Parts (a) and (b) of Figure 3.1.4 clearly show that the cake will be approximately at room temperature in about one-half hour. The ambient temperature in (2) need not be a constant but could be a function Tm(t) of time t. See Problem 18 in Exercises 3.1. Mixtures The mixing of two ﬂuids sometimes gives rise to a linear ﬁrst-order differential equation. When we discussed the mixing of two brine solutions in Section 1.3, we assumed that the rate A(t) at which the amount of salt in the mixing tank changes was a net rate: . (5) In Example 5 we solve equation (8) of Section 1.3. dA dt (input rate of salt) (output rate of salt) Rin Rout lim t : T(t) 70 T(t) 70 230e0.19018t 13 23 0.19018 k 1 3 e3k 13 23 dT T 70 k dt dT dt k(T 70), T(0) 300 t T 15 30 300 150 T = 70 (a) T(t) t (min) 75 20.1 74 21.3 73 22.8 72 24.9 71 28.6 70.5 32.3 (b) FIGURE 3.1.4 Temperature of cooling cake in Example 4 EXAMPLE 5 Mixture of Two Salt Solutions Recall that the large tank considered in Section 1.3 held 300 gallons of a brine solution. Salt was entering and leaving the tank; a brine solution was being pumped into the tank at the rate of 3 gal/min; it mixed with the solution there, and then the mixture was pumped out at the rate of 3 gal/min. The concentration of the salt in the inﬂow, or solution entering, was 2 lb/gal, so salt was entering the tank at the rate Rin (2 lb/gal)(3 gal/min) 6 lb/min and leaving the tank at the rate Rout (A300 lb/gal)(3 gal/min) A100 lb/min. From this data and (5) we get equa-tion (8) of Section 1.3. Let us pose the question: If 50 pounds of salt were dissolved initially in the 300 gallons, how much salt is in the tank after a long time? SOLUTION To ﬁnd the amount of salt A(t) in the tank at time t, we solve the initial-value problem . Note here that the side condition is the initial amount of salt A(0) 50 in the tank and not the initial amount of liquid in the tank. Now since the integrating factor of the linear differential equation is et/100, we can write the equation as . d dt [et/100A] 6et/100 dA dt 1 100 A 6, A(0) 50 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Integrating the last equation and solving for A gives the general solution A(t) 600 cet/100. When t 0, A 50, so we ﬁnd that c 550. Thus the amount of salt in the tank at time t is given by . (6) The solution (6) was used to construct the table in Figure 3.1.5(b). Also, it can be seen from (6) and Figure 3.1.5(a) that A(t) : 600 as t : . Of course, this is what we would intuitively expect; over a long time the number of pounds of salt in the solution must be (300 gal)(2 lb/gal) 600 lb. In Example 5 we assumed that the rate at which the solution was pumped in was the same as the rate at which the solution was pumped out. However, this need not be the case; the mixed brine solution could be pumped out at a rate rout that is faster or slower than the rate rin at which the other brine solution is pumped in. The next example illustrates the case when the mixture is pumped out at rate that is slower than the rate at which the brine solution is being pumped into the tank. A(t) 600 550et/100 88 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS t A A = 600 500 (a) t (min) A (lb) 50 266.41 100 397.67 150 477.27 200 525.57 300 572.62 400 589.93 (b) FIGURE 3.1.5 Pounds of salt in the tank in Example 5 FIGURE 3.1.7 LR-series circuit E L R FIGURE 3.1.6 Graph of A(t) in Example 6 t A 50 250 500 100 EXAMPLE 6 Example 5 Revisited If the well-stirred solution in Example 5 is pumped out at a slower rate of, say, rout 2 gal/min, then liquid will accumulate in the tank at the rate of rin rout (3 2) gal/min 1 gal/min. After t minutes, (1 gal/min) . (t min) t gal will accumulate, so the tank will contain 300 t gallons of brine. The concentration of the outﬂow is then c(t) A(300 t) lb/gal, and the output rate of salt is Rout c(t) . rout, or . Hence equation (5) becomes . The integrating factor for the last equation is and so after multiplying by the factor the equation is cast into the form Integrating the last equation gives By applying the initial condition and solving for A yields the solution A(t) 600 2t (4.95 107)(300 t)2. As Figure 3.1.6 shows, not unexpectedly, salt builds up in the tank over time, that is, Series Circuits For a series circuit containing only a resistor and an inductor, Kirchhoff’s second law states that the sum of the voltage drop across the inductor (L(didt)) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. See Figure 3.1.7. Thus we obtain the linear differential equation for the current i(t), , (7) where L and R are constants known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system. L di dt Ri E(t) A : as t : . A(0) 50 (300 t)2A 2(300 t)3 c. d dt[(300 t)2 A] 6(300 t)2. e2dt>(300t) e2 ln(300t) eln(300t)2 (300 t)2 dA dt 6 2A 300 t or dA dt 2 300 t A 6 Rout A 300 t lb/gal (2 gal/min) 2A 300 t lb/min Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.1 LINEAR MODELS ● 89 R C E FIGURE 3.1.8 RC-series circuit The voltage drop across a capacitor with capacitance C is given by q(t)C, where q is the charge on the capacitor. Hence, for the series circuit shown in Figure 3.1.8, Kirchhoff’s second law gives . (8) But current i and charge q are related by i dqdt, so (8) becomes the linear differ-ential equation . (9) R dq dt 1 C q E(t) Ri 1 C q E(t) FIGURE 3.1.9 Population growth is a discrete process t 1 t1 t2 P P0 t 1 P P0 (a) (b) (c) t 1 P P0 EXAMPLE 7 Series Circuit A 12-volt battery is connected to a series circuit in which the inductance is henry and the resistance is 10 ohms. Determine the current i if the initial current is zero. SOLUTION From (7) we see that we must solve , subject to i(0) 0. First, we multiply the differential equation by 2 and read off the integrating factor e20t. We then obtain . Integrating each side of the last equation and solving for i gives i(t) ce20t. Now i(0) 0 implies that 0 c or c . Therefore the response is i(t) e20t. From (4) of Section 2.3 we can write a general solution of (7): . (10) In particular, when E(t) E0 is a constant, (10) becomes . (11) Note that as t : , the second term in equation (11) approaches zero. Such a term is usu-ally called a transient term; any remaining terms are called the steady-state part of the solution. In this case E0R is also called the steady-state current; for large values of time it appears that the current in the circuit is simply governed by Ohm’s law (E iR). i(t) E0 R ce(R/L)t i(t) e(R/L)t L e(R/L)tE(t) dt ce(R/L)t 6 5 6 5 6 5 6 5 6 5 d dt [e20ti] 24e20t 1 2 di dt 10i 12 1 2 REMARKS The solution P(t) P0e0.4055t of the initial-value problem in Example 1 described the population of a colony of bacteria at any time t 0. Of course, P(t) is a continuous function that takes on all real numbers in the interval P0 P . But since we are talking about a population, common sense dictates that P can take on only positive integer values. Moreover, we would not expect the population to grow continuously—that is, every second, every microsecond, and so on—as predicted by our solution; there may be intervals of time [t1, t2] over which there is no growth at all. Perhaps, then, the graph shown in Figure 3.1.9(a) is a more realistic description of P than is the graph of an exponential function. Using a continuous function to describe a discrete phenomenon is often more a matter of convenience than of accuracy. However, for some purposes we may be satisﬁed if our model describes the system fairly closely when viewed macroscopically in time, as in Figures 3.1.9(b) and 3.1.9(c), rather than microscopically, as in Figure 3.1.9(a). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 90 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS EXERCISES 3.1 Answers to selected odd-numbered problems begin on page ANS-3. Growth and Decay 1. The population of a community is known to increase at a rate proportional to the number of people present at time t. If an initial population P0 has doubled in 5 years, how long will it take to triple? To quadruple? 2. Suppose it is known that the population of the commu-nity in Problem 1 is 10,000 after 3 years. What was the initial population P0? What will be the population in 10 years? How fast is the population growing at t 10? 3. The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 15% in 10 years. What will be the pop-ulation in 30 years? How fast is the population growing at t 30? 4. The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria? 5. The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present ini-tially, how long will it take for 90% of the lead to decay? 6. Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of decay is proportional to the amount of the substance present at time t, ﬁnd the amount remaining after 24 hours. 7. Determine the half-life of the radioactive substance described in Problem 6. 8. (a) Consider the initial-value problem dAdt kA, A(0) A0 as the model for the decay of a radioac-tive substance. Show that, in general, the half-life T of the substance is T (ln 2)k. (b) Show that the solution of the initial-value problem in part (a) can be written A(t) A02t/T. (c) If a radioactive substance has the half-life T given in part (a), how long will it take an initial amount A0 of the substance to decay to ? 9. When a vertical beam of light passes through a trans-parent medium, the rate at which its intensity I decreases is proportional to I(t), where t represents the thickness of the medium (in feet). In clear seawater, the intensity 3 feet below the surface is 25% of the initial intensity I0 of the incident beam. What is the intensity of the beam 15 feet below the surface? 10. When interest is compounded continuously, the amount of money increases at a rate proportional to the amount 1 8 A0 S present at time t, that is, dSdt rS, where r is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when $5000 is deposited in a savings account drawing 5 % annual interest compounded continuously. (b) In how many years will the initial sum deposited have doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount S 5000(1 (0.0575))5(4) that is accrued when interest is compounded quarterly. Carbon Dating 11. Archaeologists used pieces of burned wood, or char-coal, found at the site to date prehistoric paintings and drawings on walls and ceilings of a cave in Lascaux, France. See Figure 3.1.10. Use the information on page 86 to determine the approximate age of a piece of burned wood, if it was found that 85.5% of the C-14 found in living trees of the same type had decayed. 1 4 3 4 FIGURE 3.1.10 Cave wall painting in Problem 11 Prehistoric/Getty Images 12. The Shroud of Turin, which shows the negative image of the body of a man who appears to have been cruciﬁed, is believed by many to be the burial shroud of Jesus of Nazareth. See Figure 3.1.11. In 1988 the Vatican granted permission to have the shroud carbon-dated. Three inde-pendent scientiﬁc laboratories analyzed the cloth and concluded that the shroud was approximately 660 years old, an age consistent with its historical appearance. FIGURE 3.1.11 Shroud image in Problem 12 © Bettmann/CORBIS Some scholars have disagreed with this ﬁnding. For more information on this fascinating mystery see the Shroud of Turin home page at Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.1 LINEAR MODELS ● 91 Using this age, determine what percentage of the origi-nal amount of C-14 remained in the cloth as of 1988. Newton’s Law of Cooling/Warming 13. A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. After one-half minute the ther-mometer reads 50° F. What is the reading of the ther-mometer at t 1 min? How long will it take for the thermometer to reach 15° F? 14. A thermometer is taken from an inside room to the out-side, where the air temperature is 5° F. After 1 minute the thermometer reads 55° F, and after 5 minutes it reads 30° F. What is the initial temperature of the inside room? 15. A small metal bar, whose initial temperature was 20° C, is dropped into a large container of boiling water. How long will it take the bar to reach 90° C if it is known that its temperature increases 2° in 1 second? How long will it take the bar to reach 98° C? 16. Two large containers A and B of the same size are ﬁlled with different ﬂuids. The ﬂuids in containers A and B are maintained at 0° C and 100° C, respectively. A small metal bar, whose initial temperature is 100° C, is low-ered into container A. After 1 minute the temperature of the bar is 90° C. After 2 minutes the bar is removed and instantly transferred to the other container. After 1 minute in container B the temperature of the bar rises 10°. How long, measured from the start of the entire process, will it take the bar to reach 99.9° C? 17. A thermometer reading 70° F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110° F after minute and 145° F after 1 minute. How hot is the oven? 18. At t 0 a sealed test tube containing a chemical is immersed in a liquid bath. The initial temperature of the chemical in the test tube is 80° F. The liquid bath has a controlled temperature (measured in degrees Fahrenheit) given by Tm(t) 100 40e0.1t, t 0, where t is measured in minutes. (a) Assume that k 0.1 in (2). Before solving the IVP, describe in words what you expect the temper-ature T(t) of the chemical to be like in the short term. In the long term. (b) Solve the initial-value problem. Use a graphing util-ity to plot the graph of T(t) on time intervals of var-ious lengths. Do the graphs agree with your predictions in part (a)? 19. A dead body was found within a closed room of a house where the temperature was a constant 70° F. At the time of discovery the core temperature of the body was determined to be 85° F. One hour later a second mea-1 2 surement showed that the core temperature of the body was 80° F. Assume that the time of death corresponds to t 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found. [Hint: Let t1 0 denote the time that the body was discovered.] 20. The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modiﬁ-cation of (2) is where k 0 and Tm is a constant. Suppose that two cups A and B are ﬁlled with coffee at the same time. Initially, the temperature of the coffee is 150° F. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of the coffee in cup A is 100° F. If Tm 70° F, then what is the temperature of the coffee in cup B after 30 min? Mixtures 21. A tank contains 200 liters of ﬂuid in which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t. 22. Solve Problem 21 assuming that pure water is pumped into the tank. 23. A large tank is ﬁlled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t. 24. In Problem 23, what is the concentration c(t) of the salt in the tank at time t? At t 5 min? What is the concen-tration of the salt in the tank after a long time, that is, as t : ? At what time is the concentration of the salt in the tank equal to one-half this limiting value? 25. Solve Problem 23 under the assumption that the solu-tion is pumped out at a faster rate of 10 gal/min. When is the tank empty? 26. Determine the amount of salt in the tank at time t in Example 5 if the concentration of salt in the inﬂow is variable and given by cin(t) 2 sin(t4) lb/gal. Without actually graphing, conjecture what the solution curve of the IVP should look like. Then use a graphing utility to plot the graph of the solution on the interval [0, 300]. Repeat for the interval [0, 600] and compare your graph with that in Figure 3.1.5(a). 27. A large tank is partially ﬁlled with 100 gallons of ﬂuid in which 10 pounds of salt is dissolved. Brine containing dT dt kS(T Tm), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. pound of salt per gallon is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out at a slower rate of 4 gal/min. Find the num-ber of pounds of salt in the tank after 30 minutes. 28. In Example 5 the size of the tank containing the salt mixture was not given. Suppose, as in the discussion following Example 5, that the rate at which brine is pumped into the tank is 3 gal/min but that the well-stirred solution is pumped out at a rate of 2 gal/min. It stands to reason that since brine is accumulating in the tank at the rate of 1 gal/min, any ﬁnite tank must even-tually overﬂow. Now suppose that the tank has an open top and has a total capacity of 400 gallons. (a) When will the tank overﬂow? (b) What will be the number of pounds of salt in the tank at the instant it overﬂows? (c) Assume that although the tank is overﬂowing, brine solution continues to be pumped in at a rate of 3 gal/min and the well-stirred solution continues to be pumped out at a rate of 2 gal/min. Devise a method for determining the number of pounds of salt in the tank at t 150 minutes. (d) Determine the number of pounds of salt in the tank as t : . Does your answer agree with your intuition? (e) Use a graphing utility to plot the graph of A(t) on the interval [0, 500). Series Circuits 29. A 30-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t) if i(0) 0. Determine the current as t : . 30. Solve equation (7) under the assumption that E(t) E0 sin vt and i(0) i0. 31. A 100-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the capacitance is 104 farad. Find the charge q(t) on the capacitor if q(0) 0. Find the current i(t). 32. A 200-volt electromotive force is applied to an RC-series circuit in which the resistance is 1000 ohms and the capacitance is 5 106 farad. Find the charge q(t) on the capacitor if i(0) 0.4. Determine the charge and current at t 0.005 s. Determine the charge as t : . 33. An electromotive force is applied to an LR-series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current i(t) if i(0) 0. E(t) 120, 0, 0 t 20 t 20 1 2 92 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS 34. Suppose an RC-series circuit has a variable resistor. If the resistance at time t is given by R k1 k2t, where k1 and k2 are known positive constants, then (9) becomes . If E(t) E0 and q(0) q0, where E0 and q0 are constants, show that . Additional Linear Models 35. Air Resistance In (14) of Section 1.3 we saw that a differential equation describing the velocity v of a falling mass subject to air resistance proportional to the instantaneous velocity is , where k 0 is a constant of proportionality. The positive direction is downward. (a) Solve the equation subject to the initial condition v(0) v0. (b) Use the solution in part (a) to determine the limit-ing, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 40 in Exercises 2.1. (c) If the distance s, measured from the point where the mass was released above ground, is related to ve-locity v by dsdt v(t), ﬁnd an explicit expression for s(t) if s(0) 0. 36. How High?—No Air Resistance Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure 3.1.12, with an initial veloc-ity v0 300 ft/s. The answer to the question “How high does the cannonball go?” depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by d2sdt2 g (equation (12) of Section 1.3). Since dsdt v(t) the last m dv dt mg kv q(t) E0C (q0 E0C) k1 k1 k2t 1/Ck2 (k1 k2t) dq dt 1 C q E(t) FIGURE 3.1.12 Find the maximum height of the cannonball in Problem 36 ground level −mg Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.1 LINEAR MODELS ● 93 differential equation is the same as dvdt g, where we take g 32 ft/s2. Find the velocity v(t) of the cannonball at time t. (b) Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level. Find the maximum height attained by the cannonball. 37. How High?—Linear Air Resistance Repeat Prob-lem 36, but this time assume that air resistance is proportional to instantaneous velocity. It stands to reason that the maximum height attained by the cannon-ball must be less than that in part (b) of Problem 36. Show this by supposing that the constant of proportion-ality is k 0.0025. [Hint: Slightly modify the DE in Problem 35.] 38. Skydiving A skydiver weighs 125 pounds, and her parachute and equipment combined weigh another 35 pounds. After exiting from a plane at an altitude of 15,000 feet, she waits 15 seconds and opens her para-chute. Assume that the constant of proportionality in the model in Problem 35 has the value k 0.5 during free fall and k 10 after the parachute is opened. Assume that her initial velocity on leaving the plane is zero. What is her velocity and how far has she traveled 20 seconds after leaving the plane? See Figure 3.1.13. How does her velocity at 20 seconds compare with her terminal velocity? How long does it take her to reach the ground? [Hint: Think in terms of two distinct IVPs.] and the downward direction is taken to be the positive direction. (a) Solve for v(t) if the raindrop falls from rest. (b) Reread Problem 36 of Exercises 1.3 and then show that the radius of the raindrop at time t is r(t) (kr)t r0. (c) If r0 0.01 ft and r 0.007 ft 10 seconds after the raindrop falls from a cloud, determine the time at which the raindrop has evaporated completely. 40. Fluctuating Population The differential equation dPdt (k cos t)P, where k is a positive constant, is a mathematical model for a population P(t) that under-goes yearly seasonal ﬂuctuations. Solve the equation subject to P(0) P0. Use a graphing utility to graph the solution for different choices of P0. 41. Population Model In one model of the changing population P(t) of a community, it is assumed that , where dBdt and dDdt are the birth and death rates, respectively. (a) Solve for P(t) if dBdt k1P and dDdt k2P. (b) Analyze the cases k1 k2, k1 k2, and k1 k2. 42. Constant-Harvest Model A model that describes the population of a ﬁshery in which harvesting takes place at a constant rate is given by where k and h are positive constants. (a) Solve the DE subject to P(0) P0. (b) Describe the behavior of the population P(t) for in-creasing time in the three cases P0 hk, P0 hk, and 0 P0 hk. (c) Use the results from part (b) to determine whether the ﬁsh population will ever go extinct in ﬁnite time, that is, whether there exists a time T 0 such that P(T) 0. If the population goes extinct, then ﬁnd T. 43. Drug Dissemination A mathematical model for the rate at which a drug disseminates into the bloodstream is given by where r and k are positive constants. The function x(t) describes the concentration of the drug in the blood-stream at time t. (a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to ﬁnd the limiting value of x(t) as t : . dx dt r kx, dP dt kP h, dP dt dB dt dD dt FIGURE 3.1.13 Find the time to reach the ground in Problem 38 free fall parachute opens air resistance is 0.5v air resistance is 10v t = 20 s 39. Evaporating Raindrop As a raindrop falls, it evapo-rates while retaining its spherical shape. If we make the further assumptions that the rate at which the raindrop evaporates is proportional to its surface area and that air resistance is negligible, then a model for the velocity v(t) of the raindrop is . Here r is the density of water, r0 is the radius of the rain-drop at t 0, k 0 is the constant of proportionality, dv dt 3(k/) (k/)t r0 v g Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (b) Solve the DE subject to x(0) 0. Sketch the graph of x(t) and verify your prediction in part (a). At what time is the concentration one-half this limiting value? 44. Memorization When forgetfulness is taken into account, the rate of memorization of a subject is given by , where k1 0, k2 0, A(t) is the amount memorized in time t, M is the total amount to be memorized, and M A is the amount remaining to be memorized. (a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to ﬁnd the limiting value of A(t) as t : . Interpret the result. (b) Solve the DE subject to A(0) 0. Sketch the graph of A(t) and verify your prediction in part (a). 45. Heart Pacemaker A heart pacemaker, shown in Figure 3.1.14, consists of a switch, a battery, a capacitor, and the heart as a resistor. When the switch S is at P, the capacitor charges; when S is at Q, the capacitor dis-charges, sending an electrical stimulus to the heart. In Problem 53 in Exercises 2.3 we saw that during this time the electrical stimulus is being applied to the heart, the voltage E across the heart satisﬁes the linear DE . (a) Let us assume that over the time interval of length t1, 0 t t1, the switch S is at position P shown in Figure 3.1.14 and the capacitor is being charged. When the switch is moved to position Q at time t1 the capacitor discharges, sending an impulse to the heart over the time interval of length t2: t1 t t1 t2. Thus over the initial charging/discharging interval 0 t t1 t2 the voltage to the heart is actually modeled by the piecewise-defined differential equation . dE dt 0, 1 RC E, 0 t t1 t1 t t1 t2 dE dt 1 RC E dA dt k1(M A) k2A 94 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS By moving S between P and Q, the charging and discharging over time intervals of lengths t1 and t2 is repeated indeﬁnitely. Suppose t1 4 s, t2 2 s, E0 12 V, and E(0) 0, E(4) 12, E(6) 0, E(10) 12, E(12) 0, and so on. Solve for E(t) for 0 t 24. (b) Suppose for the sake of illustration that R C 1. Use a graphing utility to graph the solution for the IVP in part (a) for 0 t 24. 46. Sliding Box (a) A box of mass m slides down an inclined plane that makes an angle u with the hori-zontal as shown in Figure 3.1.15. Find a differential equation for the velocity v(t) of the box at time t in each of the following three cases: (i) No sliding friction and no air resistance (ii) With sliding friction and no air resistance (iii) With sliding friction and air resistance In cases (ii) and (iii), use the fact that the force of friction opposing the motion of the box is mN, where m is the coefﬁcient of sliding friction and N is the normal component of the weight of the box. In case (iii) assume that air resistance is propor-tional to the instantaneous velocity. (b) In part (a), suppose that the box weighs 96 pounds, that the angle of inclination of the plane is u 30°, that the coefﬁcient of sliding friction is , and that the additional retarding force due to air resistance is numerically equal to v. Solve the dif-ferential equation in each of the three cases, assum-ing that the box starts from rest from the highest point 50 ft above ground. 1 4 134 heart C Q P S switch E0 R FIGURE 3.1.14 Model of a pacemaker in Problem 45 FIGURE 3.1.15 Box sliding down inclined plane in Problem 46 θ 50 ft motion friction W = mg 47. Sliding Box—Continued (a) In Problem 46 let s(t) be the distance measured down the inclined plane from the highest point. Use dsdt v(t) and the solution for each of the three cases in part (b) of Problem 46 to ﬁnd the time that it takes the box to slide completely down the inclined plane. A root-ﬁnding application of a CAS may be useful here. (b) In the case in which there is friction (m 0) but no air resistance, explain why the box will not slide down the plane starting from rest from the highest Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. point above ground when the inclination angle u satisﬁes tan u m. (c) The box will slide downward on the plane when tan u m if it is given an initial velocity v(0) v0 0. Suppose that and u 23°. Verify that tan u m. How far will the box slide down the plane if v0 1 ft/s? (d) Using the values and u 23°, approxi-mate the smallest initial velocity v0 that can be given to the box so that, starting at the highest point 50 ft above ground, it will slide completely down the in-clined plane. Then ﬁnd the corresponding time it takes to slide down the plane. 134 134 3.2 NONLINEAR MODELS ● 95 48. What Goes U p . . . (a) It is well known that the model in which air resistance is ignored, part (a) of Problem 36, predicts that the time ta it takes the cannonball to attain its maximum height is the same as the time td it takes the cannonball to fall from the maximum height to the ground. Moreover, the magnitude of the impact velocity vi will be the same as the initial velocity v0 of the cannonball. Verify both of these results. (b) Then, using the model in Problem 37 that takes air resistance into account, compare the value of ta with td and the value of the magnitude of vi with v0. Aroot-ﬁnding application of a CAS (or graphic cal-culator) may be useful here. NONLINEAR MODELS REVIEW MATERIAL ●Equations (5), (6), and (10) of Section 1.3 and Problems 7, 8, 13, 14, and 17 of Exercises 1.3 ●Separation of variables in Section 2.2 INTRODUCTION We ﬁnish our study of single ﬁrst-order differential equations with an exam-ination of some nonlinear models. 3.2 Population Dynamics If P(t) denotes the size of a population at time t, the P f(P) r K FIGURE 3.2.1 Simplest assumption for f (P) is a straight line (blue color) model for exponential growth begins with the assumption that dPdt kP for some k 0. In this model, the relative, or specific, g owth rate deﬁned by (1) is a constant k. True cases of exponential growth over long periods of time are hard to ﬁnd because the limited resources of the environment will at some time exert restrictions on the growth of a population. Thus for other models, (1) can be expected to decrease as the population P increases in size. The assumption that the rate at which a population grows (or decreases) is dependent only on the number P present and not on any time-dependent mechanisms such as seasonal phenomena (see Problem 33 in Exercises 1.3) can be stated as . (2) The differential equation in (2), which is widely assumed in models of animal populations, is called the density-dependent hypothesis. Logistic Equation Suppose an environment is capable of sustaining no more than a ﬁxed number K of individuals in its population. The quantity K is called the carrying capacity of the environment. Hence for the function f in (2) we have f (K) 0, and we simply let f (0) r. Figure 3.2.1 shows three functions f that sat-isfy these two conditions. The simplest assumption that we can make is that f (P) is linear—that is, f (P) c1P c2. If we use the conditions f (0) r and f (K) 0, dP>dt P f (P) or dP dt Pf (P) dP>dt P Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 96 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS we ﬁnd, in turn, c2 r and c1 rK, and so f takes on the form f(P) r (rK)P. Equation (2) becomes . (3) With constants relabeled, the nonlinear equation (3) is the same as . (4) Around 1840 the Belgian mathematician-biologist P. F. Verhulst (1804–1849) was concerned with mathematical models for predicting the human populations of various countries. One of the equations he studied was (4), where a 0 and b 0. Equation (4) came to be known as the logistic equation, and its solution is called the logistic function. The graph of a logistic function is called a logistic curve. The linear differential equation dPdt kP does not provide a very accurate model for population when the population itself is very large. Overcrowded condi-tions, with the resulting detrimental effects on the environment such as pollution and excessive and competitive demands for food and fuel, can have an inhibiting effect on population growth. As we shall now see, the solution of (4) is bounded as t : . If we rewrite (4) as dPdt aP bP2, the nonlinear term bP2, b 0, can be in-terpreted as an “inhibition” or “competition” term. Also, in most applications the positive constant a is much larger than the constant b. Logistic curves have proved to be quite accurate in predicting the growth patterns, in a limited space, of certain types of bacteria, protozoa, water ﬂeas (Daphnia), and fruit ﬂies (Drosophila). Solution of the Logistic Equation One method of solving (4) is separation dP dt P(a bP) dP dt Pr r KP of variables. Decomposing the left side of dPP(a bP) dt into partial fractions and integrating gives It follows from the last equation that . If P(0) P0, P0 ab, we ﬁnd c1 P0(a bP0), and so after substituting and simplifying, the solution becomes . (5) Graphs of P(t) The basic shape of the graph of the logistic function P(t) can be obtained without too much effort. Although the variable t usually represents time and we are seldom concerned with applications in which t 0, it is nonetheless of some in-terest to include this interval in displaying the various graphs of P. From (5) we see that . P(t) : aP0 bP0 a b as t : and P(t) : 0 as t : P(t) aP0 bP0 (a bP0)eat P(t) ac1eat 1 bc1eat ac1 bc1 eat P a bP c1eat. ln P a bP at ac 1 a ln P 1 a ln a bP t c 1>a P b>a a bPdP dt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The dashed line P a2b shown in Figure 3.2.2 corresponds to the ordinate of a point of inﬂection of the logistic curve. To show this, we differentiate (4) by the Product Rule: . From calculus recall that the points where d2Pdt2 0 are possible points of inﬂec-tion, but P 0 and P ab can obviously be ruled out. Hence P a2b is the only possible ordinate value at which the concavity of the graph can change. For 0 P a2b it follows that P 0, and a2b P ab implies that P 0. Thus, as we read from left to right, the graph changes from concave up to concave down at the point corresponding to P a2b. When the initial value satisﬁes 0 P0 a2b, the graph of P(t) assumes the shape of an S, as we see in Figure 3.2.2(a). For a2b P0 ab the graph is still S-shaped, but the point of inﬂection occurs at a negative value of t, as shown in Figure 3.2.2(b). We have already seen equation (4) in (5) of Section 1.3 in the form dxdt kx(n 1 x), k 0. This differential equation provides a reasonable model for describing the spread of an epidemic brought about initially by introduc-ing an infected individual into a static population. The solution x(t) represents the number of individuals infected with the disease at time t. 2b2P P a bP a 2b P(a bP)(a 2bP) d 2P dt2 P b dP dt (a bP) dP dt dP dt (a 2bP) 3.2 NONLINEAR MODELS ● 97 P P0 a/2b a/b t P P0 a/2b a/b t (a) (a) (b) FIGURE 3.2.2 Logistic curves for different initial conditions (a) t x x = 1000 10 500 5 (a) t (days) x (number infected) 4 50 (observed) 5 124 6 276 7 507 8 735 9 882 10 953 (b) FIGURE 3.2.3 Number of infected students in Example 1 EXAMPLE 1 Logistic Growth Suppose a student carrying a ﬂu virus returns to an isolated college campus of 1000 students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the number of students not infected, determine the number of infected students after 6 days if it is further observed that after 4 days x(4) 50. SOLUTION Assuming that no one leaves the campus throughout the duration of the disease, we must solve the initial-value problem . By making the identiﬁcation a 1000k and b k, we have immediately from (5) that . Now, using the information x(4) 50, we determine k from We ﬁnd . Thus . Finally, . Additional calculated values of x(t) are given in the table in Figure 3.2.3(b). Note that the number of infected students x(t) approaches 1000 as t increases. x(6) 1000 1 999e5.9436 276 students x(t) 1000 1 999e0.9906t 1000k 1 4 ln 19 999 0.9906 50 1000 1 999e4000k. x(t) 1000k k 999ke1000kt 1000 1 999e1000kt dx dt kx(1000 x), x(0) 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Modifications of the Logistic Equation There are many variations of the logistic equation. For example, the differential equations and (6) could serve, in turn, as models for the population in a ﬁshery where ﬁsh are harvested or are restocked at rate h. When h 0 is a constant, the DEs in (6) can be readily an-alyzed qualitatively or solved analytically by separation of variables. The equations in (6) could also serve as models of the human population decreased by emigration or increased by immigration, respectively. The rate h in (6) could be a function of time t or could be population dependent; for example, harvesting might be done peri-odically over time or might be done at a rate proportional to the population P at time t. In the latter instance, the model would look like P P(a bP) cP, c 0. The human population of a community might change because of immigration in such a manner that the contribution due to immigration was large when the population P of the community was itself small but small when P was large; a reasonable model for thepopulationof thecommunitywouldthenbe P P(a bP) cekP, c 0, k 0. See Problem 24 in Exercises 3.2. Another equation of the form given in (2), , (7) is a modiﬁcation of the logistic equation known as the Gompertz differential equa-tion named after the English mathematician Benjamin Gompertz (1779–1865). This DE is sometimes used as a model in the study of the growth or decline of pop-ulations, the growth of solid tumors, and certain kinds of actuarial predictions. See Problem 8 in Exercises 3.2. Chemical Reactions Suppose that a grams of chemical A are combined with b grams of chemical B. If there are M parts of A and N parts of B formed in the com-pound and X(t) is the number of grams of chemical C formed, then the number of grams of chemical A and the number of grams of chemical B remaining at time t are, respectively, . The law of mass action states that when no temperature change is involved, the rate at which the two substances react is proportional to the product of the amounts of A and B that are untransformed (remaining) at time t: . (8) If we factor out M(M N) from the ﬁrst factor and N(M N) from the second and introduce a constant of proportionality k 0, (8) has the form , (9) where a a(M N)M and b b(M N)N. Recall from (6) of Section 1.3 that a chemical reaction governed by the nonlinear differential equation (9) is said to be a second-order reaction. dX dt k( X)( X) dX dt a M M N Xb N M N X a M M N X and b N M N X dP dt P(a b ln P) dP dt P(a bP) h dP dt P(a bP) h 98 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 2 Second-Order Chemical Reaction A compound C is formed when two chemicals A and B are combined. The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B is used. It is observed that 30 grams of the compound C is formed in 10 minutes. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Determine the amount of C at time t if the rate of the reaction is proportional to the amounts of A and B remaining and if initially there are 50 grams of A and 32 grams of B. How much of the compound C is present at 15 minutes? Interpret the solution as t : . SOLUTION Let X(t) denote the number of grams of the compound C present at time t. Clearly, X(0) 0 g and X(10) 30 g. If, for example, 2 grams of compound C is present, we must have used, say, a grams of A and b grams of B, so a b 2 and b 4a. Thus we must use a 2 g of chemical A and b 2 g of B. In general, for X grams of C we must use . The amounts of A and B remaining at time t are then , respectively. Now we know that the rate at which compound C is formed satisﬁes . To simplify the subsequent algebra, we factor from the ﬁrst term and from the second and then introduce the constant of proportionality: . By separation of variables and partial fractions we can write . Integrating gives (10) When t 0, X 0, so it follows at this point that c2 . Using X 30 g at t 10, we ﬁnd 210k ln 0.1258. With this information we solve the last equation in (10) for X: . (11) From (11) we ﬁnd X(15) 34.78 grams. The behavior of X as a function of time is displayed in Figure 3.2.4. It is clear from the accompanying table and (11) that X : 40 as t : . This means that 40 grams of compound C is formed, leaving . 50 1 5 (40) 42 g of A and 32 4 5 (40) 0 g of B X(t) 1000 1 e0.1258t 25 4e0.1258t 88 25 1 10 25 4 ln 250 X 40 X 210kt c1 or 250 X 40 X c2e210kt. 1 210 250 X dX 1 210 40 X dX k dt dX dt k(250 X)(40 X) 4 5 1 5 dX dt 50 1 5 X32 4 5 X 50 1 5 X and 32 4 5 X 1 5 X grams of A and 4 5 X grams of B (4 5) 8 5 (1 5) 2 5 3.2 NONLINEAR MODELS ● 99 10 20 30 40 t X X = 40 (a) t (min) X (g) 10 30 (measured) 15 34.78 20 37.25 25 38.54 30 39.22 35 39.59 (b) FIGURE 3.2.4 Number of grams of compound C in Example 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 100 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS REMARKS The indeﬁnite integral du(a2 u2) can be evaluated in terms of logarithms, the inverse hyperbolic tangent, or the inverse hyperbolic cotangent. For example, of the two results (12) (13) (12) may be convenient in Problems 15 and 26 in Exercises 3.2, whereas (13) may be preferable in Problem 27. du a 2 u 2 1 2a ln a u a u c, u a, du a 2 u 2 1 a tanh1 u a c, u a EXERCISES 3.2 Answers to selected odd-numbered problems begin on page ANS-3. Logistic Equation 1. The number N(t) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem . (a) Use the phase portrait concept of Section 2.1 to pre-dict how many supermarkets are expected to adopt the new procedure over a long period of time. By hand, sketch a solution curve of the given initial-value problem. (b) Solve the initial-value problem and then use a graph-ing utility to verify the solution curve in part (a). How many companies are expected to adopt the new technology when t 10? 2. The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially, N(0) 500, and it is observed that N(1) 1000. Solve for N(t) if it is pre-dicted that the limiting number of people in the commu-nity who will see the advertisement is 50,000. 3. A model for the population P(t) in a suburb of a large city is given by the initial-value problem , where t is measured in months. What is the limiting value of the population? At what time will the popula-tion be equal to one-half of this limiting value? 4. (a) Census data for the United States between 1790 and 1950 are given in Table 3.2.1. Construct a logistic population model using the data from 1790, 1850, and 1910. dP dt P(101 107 P), P(0) 5000 dN dt N(1 0.0005N), N(0) 1 (b) Construct a table comparing actual census popula-tion with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair. Modifications of the Logistic Mode 5. (a) If a constant number h of ﬁsh are harvested from a ﬁshery per unit time, then a model for the popula-tion P(t) of the ﬁshery at time t is given by , where a, b, h, and P0 are positive constants. Suppose a 5, b 1, and h 4. Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves dP dt P(a bP) h, P(0) P0 TABLE 3.2.1 Year Population (in millions) 1790 3.929 1800 5.308 1810 7.240 1820 9.638 1830 12.866 1840 17.069 1850 23.192 1860 31.433 1870 38.558 1880 50.156 1890 62.948 1900 75.996 1910 91.972 1920 105.711 1930 122.775 1940 131.669 1950 150.697 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.2 NONLINEAR MODELS ● 101 corresponding to the cases P0 4, 1 P0 4, and 0 P0 1. Determine the long-term behavior of the population in each case. (b) Solve the IVP in part (a). Verify the results of your phase portrait in part (a) by using a graphing utility to plot the graph of P(t) with an initial condition taken from each of the three intervals given. (c) Use the information in parts (a) and (b) to determine whether the ﬁshery population becomes extinct in ﬁnite time. If so, ﬁnd that time. 6. Investigate the harvesting model in Problem 5 both qualitatively and analytically in the case a 5, b 1, h . Determine whether the population becomes extinct in ﬁnite time. If so, ﬁnd that time. 7. Repeat Problem 6 in the case a 5, b 1, h 7. 8. (a) Suppose a b 1 in the Gompertz differential equation (7). Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch rep-resentative solution curves corresponding to the cases P0 e and 0 P0 e. (b) Suppose a 1, b 1 in (7). Use a new phase por-trait to sketch representative solution curves corre-sponding to the cases P0 e1 and 0 P0 e1. (c) Find an explicit solution of (7) subject to P(0) P0. Chemical Reactions 9. Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 10 grams of C is formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of C after a long time? How much of chemicals A and B remains after a long time? 10. Solve Problem 9 if 100 grams of chemical A is present initially. At what time is chemical C half-formed? Additional Nonlinear Models 11. Leaking Cylindrical Tank A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height h of water in the tank is described by , where Aw and Ah are the cross-sectional areas of the water and the hole, respectively. (a) Solve the DE if the initial height of the water is H. By hand, sketch the graph of h(t) and give its interval dh dt Ah Aw 12gh 25 4 I of deﬁnition in terms of the symbols Aw, Ah, and H. Use g 32 ft/s2. (b) Suppose the tank is 10 feet high and has radius 2 feet and the circular hole has radius inch. If the tank is initially full, how long will it take to empty? 12. Leaking Cylindrical Tank—Continued When fric-tion and contraction of the water at the hole are taken into account, the model in Problem 11 becomes , where 0 c 1. How long will it take the tank in Problem 11(b) to empty if c 0.6? See Problem 13 in Exercises 1.3. 13. Leaking Conical Tank A tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom. (a) Suppose the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches. In Problem 14 in Exercises 1.3 you were asked to show that the differential equation governing the height h of water leaking from a tank is . In this model, friction and contraction of the water at the hole were taken into account with c 0.6, and g was taken to be 32 ft/s2. See Figure 1.3.12. If the tank is initially full, how long will it take the tank to empty? (b) Suppose the tank has a vertex angle of 60° and the circular hole has radius 2 inches. Determine the dif-ferential equation governing the height h of water. Use c 0.6 and g 32 ft/s2. If the height of the water is initially 9 feet, how long will it take the tank to empty? 14. Inverted Conical Tank Suppose that the conical tank in Problem 13(a) is inverted, as shown in Figure 3.2.5, and that water leaks out a circular hole of radius 2 inches in the center of its circular base. Is the time it takes to empty a full tank the same as for the tank with vertex down in Problem 13? Take the friction/contraction coef-ﬁcient to be c 0.6 and g 32 ft/s2. dh dt 5 6h3/2 dh dt c Ah Aw 12gh 1 2 8 ft Aw h 20 ft FIGURE 3.2.5 Inverted conical tank in Problem 14 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 15. Air Resistance A differential equation for the veloc-ity v of a falling mass m subjected to air resistance pro-portional to the square of the instantaneous velocity is , where k 0 is a constant of proportionality. The posi-tive direction is downward. (a) Solve the equation subject to the initial condition v(0) v0. (b) Use the solution in part (a) to determine the limit-ing, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 41 in Exercises 2.1. (c) If the distance s, measured from the point where the mass was released above ground, is related to velocity v by dsdt v(t), ﬁnd an explicit expres-sion for s(t) if s(0) 0. 16. How High?—Nonlinear Air Resistance Consider the 16-pound cannonball shot vertically upward in Prob-lems 36 and 37 in Exercises 3.1 with an initial velocity v0 300 ft/s. Determine the maximum height attained by the cannonball if air resistance is assumed to be propor-tional to the square of the instantaneous velocity. Assume that the positive direction is upward and take k 0.0003. [Hint: Slightly modify the DE in Problem 15.] 17. That Sinking Feeling (a) Determine a differential equation for the velocity v(t) of a mass m sinking in water that imparts a resistance proportional to the square of the instantaneous velocity and also exerts an upward buoyant force whose magnitude is given by Archimedes’ principle. See Problem 18 in Exercises 1.3. Assume that the positive direction is downward. (b) Solve the differential equation in part (a). (c) Determine the limiting, or terminal, velocity of the sinking mass. 18. Solar Collector The differential equation describes the shape of a plane curve C that will reﬂect all incoming light beams to the same point and could be a model for the mirror of a reﬂecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises 1.3. There are several ways of solving this DE. (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution y ux yields . u du 11 u2 (1 11 u2) dx x dy dx x 1x2 y2 y m dv dt mg kv2 102 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve C must be a parabola with focus at the ori-gin and is symmetric with respect to the x-axis. (b) Show that the ﬁrst differential equation can also be solved by means of the substitution u x2 y2. 19. Tsunami (a) A simple model for the shape of a tsunami is given by , where W(x) 0 is the height of the wave expressed as a function of its position relative to a point off-shore. By inspection, ﬁnd all constant solutions of the DE. (b) Solve the differential equation in part (a). A CAS may be useful for integration. (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition W(0) 2. 20. Evaporation An outdoor decorative pond in the shape of a hemispherical tank is to be ﬁlled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R 10 ft, that water is pumped in at a rate of p ft3/min, and that the tank is initially empty. See Figure 3.2.6. As the tank ﬁlls, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k 0.01. (a) The rate of change dVdt of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water shown in the ﬁgure is V pRh2 ph3, where R 10. Express the area of the surface of the water A pr 2 in terms of h. (b) Solve the differential equation in part (a). Graph the solution. (c) If there were no evaporation, how long would it take the tank to ﬁll? (d) With evaporation, what is the depth of the water at the time found in part (c)? Will the tank ever be ﬁlled? Prove your assertion. 1 3 dW dx W 14 2W FIGURE 3.2.6 Decorative pond in Problem 20 Output: water evaporates at rate proportional to area A of surface Input: water pumped in at rate ft3 A V Output: water evaporates at rate proportional to area A of surface Input: water pumped in at rate 3/min π (a) hemispherical tank (b) cross-section of tank R r h Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 21. Doomsday Equation Consider the differential equation where and In Section 3.1 we saw that in the case the linear differential equation is a mathematical model of a population P(t) that exhibits unbounded growth over the inﬁnite time interval , that is, See Example 1 on page 84. (a) Suppose for that the nonlinear differential equation is a mathematical model for a population of small animals, where time t is measured in months. Solve the differential equation subject to the initial condi-tion and the fact that the animal popula-tion has doubled in 5 months. (b) The differential equation in part (a) is called a doomsday equation because the population exhibits unbounded growth over a finite time interval that is, there is some time T such Find T. (c) From part (a), what is 22. Doomsday or Extinction Suppose the population model (4) is modiﬁed to be (a) If show by means of a phase portrait (see page 39) that, depending on the initial condi-tion the mathematical model could in-clude a doomsday scenario or an extinc-tion scenario (b) Solve the initial-value problem Show that this model predicts a doomsday for the population in a ﬁnite time T. (c) Solve the differential equation in part (b) subject to the initial condition Show that this model predicts extinction for the population as Project Problems 23. Regression Line Read the documentation for your CAS on scatter plots (or scatter diagrams) and least-squares linear fit The straight line that best ﬁts a set of t : . P(0) 100. dP dt P(0.0005P 0.1), P(0) 300. (P(t) : 0). (P(t) : ) P(0) P0, a 0, b 0 dP dt P(bP a). P(50)? P(100)? P(t) : as t : T. (0, T), P(t) P(0) 10 dP dt kP1.01, k 0, c 0.01 P(t) : as t : . [0, ) dP>dt kP c 0 c 0. k 0 dP dt kP1c, 3.2 NONLINEAR MODELS ● 103 data points is called a regression line or a least squares line. Your task is to construct a logistic model for the population of the United States, deﬁning f (P) in (2) as an equation of a regression line based on the population data in the table in Problem 4. One way of doing this is to approximate the left-hand side of the ﬁrst equation in (2), using the forward difference quotient in place of dPdt: . (a) Make a table of the values t, P(t), and Q(t) using t 0, 10, 20, . . . , 160 and h 10. For example, the ﬁrst line of the table should contain t 0, P(0), and Q(0). With P(0) 3.929 and P(10) 5.308, . Note that Q(160) depends on the 1960 census popu-lation P(170). Look up this value. (b) Use a CAS to obtain a scatter plot of the data (P(t), Q(t)) computed in part (a). Also use a CAS to ﬁnd an equation of the regression line and to superimpose its graph on the scatter plot. (c) Construct a logistic model dPdt Pf(P), where f(P) is the equation of the regression line found in part (b). (d) Solve the model in part (c) using the initial condi-tion P(0) 3.929. (e) Use a CAS to obtain another scatter plot, this time of the ordered pairs (t, P(t)) from your table in part (a). Use your CAS to superimpose the graph of the solution in part (d) on the scatter plot. (f) Look up the U.S. census data for 1970, 1980, and 1990. What population does the logistic model in part (c) predict for these years? What does the model pre-dict for the U.S. population P(t) as t : ? 24. Immigration Model (a) In Examples 3 and 4 of Section 2.1 we saw that any solution P(t) of (4) possesses the asymptotic behavior P(t) : ab as t : for P0 ab and for 0 P0 ab; as a consequence the equilibrium solution P ab is called an attractor. Use a root-ﬁnding application of a CAS (or a graphic calculator) to approximate the equilibrium solution of the immigration model . (b) Use a graphing utility to graph the function F(P) P(1 P) 0.3eP. Explain how this graph can be used to determine whether the number found in part (a) is an attractor. dP dt P(1 P) 0.3eP Q(0) 1 P(0) P(10) P(0) 10 0.035 Q(t) 1 P(t) P(t h) P(t) h 1 P dP dt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (c) Use a numerical solver to compare the solution curves for the IVPs for P0 0.2 and P0 1.2 with the solution curves for the IVPs for P0 0.2 and P0 1.2. Superimpose all curves on the same coordinate axes but, if possible, use a differ-ent color for the curves of the second initial-value problem. Over a long period of time, what percentage increase does the immigration model predict in the population compared to the logistic model? 25. What Goes Up . . . In Problem 16 let ta be the time it takes the cannonball to attain its maximum height and let td be the time it takes the cannonball to fall from the maximum height to the ground. Compare the value of ta with the value of td and compare the magnitude of the impact velocity vi with the initial velocity v0. See Problem 48 in Exercises 3.1. A root-ﬁnding application of a CAS might be useful here. [Hint: Use the model in Problem 15 when the cannonball is falling.] 26. Skydiving A skydiver is equipped with a stopwatch and an altimeter. As shown in Figure 3.2.7, he opens his parachute 25 seconds after exiting a plane ﬂying at an altitude of 20,000 feet and observes that his altitude is 14,800 feet. Assume that air resistance is proportional to the square of the instantaneous velocity, his initial ve-locity on leaving the plane is zero, and g 32 ft/s2. (a) Find the distance s(t), measured from the plane, the skydiver has traveled during freefall in time t. [Hint: The constant of proportionality k in the model given in Problem 15 is not speciﬁed. Use the expression for terminal velocity vt obtained in part (b) of Problem 15 to eliminate k from the IVP. Then eventually solve for vt.] (b) How far does the skydiver fall and what is his velocity at t 15 s? dP dt P(1 P) 0.3eP, P(0) P0 dP dt P(1 P), P(0) P0 104 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS 27. Hitting Bottom A helicopter hovers 500 feet above a large open tank full of liquid (not water). A dense com-pact object weighing 160 pounds is dropped (released from rest) from the helicopter into the liquid. Assume that air resistance is proportional to instantaneous ve-locity v while the object is in the air and that viscous damping is proportional to v2 after the object has en-tered the liquid. For air take k , and for the liquid take k 0.1. Assume that the positive direction is downward. If the tank is 75 feet high, determine the time and the impact velocity when the object hits the bottom of the tank. [Hint: Think in terms of two distinct IVPs. If you use (13), be careful in removing the ab-solute value sign. You might compare the velocity when the object hits the liquid—the initial velocity for the second problem—with the terminal velocity vt of the object falling through the liquid.] 28. Old Man River . . . In Figure 3.2.8(a) suppose that the y-axis and the dashed vertical line x 1 represent, re-spectively, the straight west and east beaches of a river that is 1 mile wide. The river ﬂows northward with a velocity vr, where mi/h is a constant. A man enters the current at the point (1, 0) on the east shore and swims in a direction and rate relative to the river given by the vector vs, where the speed mi/h is a constant. The man wants to reach the west beach exactly at (0, 0) and so swims in such a manner that keeps his velocity vector vs always directed toward the point (0, 0). Use Figure 3.2.8(b) as an aid in showing that a mathematical model for the path of the swimmer in the river is [Hint: The velocity v of the swimmer along the path or curve shown in Figure 3.2.8 is the resultant v vs vr. Resolve vs and vr into components in the x- and dy dx vsy vr1x2 y2 vsx . \|vs\| vs \|vr\| vr 1 4 s(t) 25 s 14,800 ft FIGURE 3.2.7 Skydiver in Problem 26 y (0, 0) (1, 0) y(t) x(t) θ (x(t), y(t)) vr west beach east beach swimmer current x y (0, 0) (1, 0) vs vr x (a) (b) FIGURE 3.2.8 Path of swimmer in Problem 28 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. y-directions. If are parametric equa-tions of the swimmer’s path, then .] 29. (a) Solve the DE in Problem 28 subject to y(1) 0. For convenience let (b) Determine the values of vs for which the swimmer will reach the point (0, 0) by examining in the cases k 1, k 1, and 0 k 1. 30. Old Man River Keeps Moving . . . Suppose the man in Problem 28 again enters the current at (1, 0) but this time decides to swim so that his velocity vector vs is always directed toward the west beach. Assume that the speed mi/h is a constant. Show that a mathe-matical model for the path of the swimmer in the river is now 31. The current speed vr of a straight river such as that in Problem 28 is usually not a constant. Rather, an approxi-mation to the current speed (measured in miles per hour) could be a function such as whose values are small at the shores (in this case, vr(0) 0 and vr(1) 0) and largest in the middle of the river. Solve the DE in Problem 30 subject to y(1) 0, where vs 2 mi/h and vr(x) is as given. When the swim-mer makes it across the river, how far will he have to walk along the beach to reach the point (0, 0)? 32. Raindrops Keep Falling . . . When a bottle of liquid refreshment was opened recently, the following factoid was found inside the bottle cap: The average velocity of a falling raindr op is 7 miles/hour. A quick search of the Internet found that meteorologist Jeff Haby offers the additional information that an “average” spherical raindrop has a radius of 0.04 in. and an approximate volume of 0.000000155 ft3. Use this data and, if need be, dig up other data and make other reason-able assumptions to determine whether “average velocity of . . . 7 mi/h” is consistent with the models in Problems 35 and 36 in Exercises 3.1 and Problem 15 in this exer-cise set. Also see Problem 36 in Exercises 1.3. 33. Time Drips By The clepsydra, or water clock, was a device that the ancient Egyptians, Greeks, Romans, and Chinese used to measure the passage of time by observ-ing the change in the height of water that was permitted to ﬂow out of a small hole in the bottom of a container or tank. (a) Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) 2 ft corresponds to water ﬁlled to the top of the tank, a hole in the bottom is circular with radius in., g 32 ft/s2, and c 0.6. Use the differential equation in Problem 12 to ﬁnd the height h(t) of the water. 1 32 0 x 1, vr(x) 30x(1 x), dy dx vr vs . \|vs\| vs lim x : 0y(x) k vr>vs. v (dx>dt, dy>dt) x x(t), y y(t) 3.2 NONLINEAR MODELS ● 105 (b) For the tank in part (a), how far up from its bottom should a mark be made on its side, as shown in Figure 3.2.9, that corresponds to the passage of one hour? Next determine where to place the marks corresponding to the passage of 2 hr, 3 hr, . . . , 12 hr. Explain why these marks are not evenly spaced. 2 1 hour 2 hours 1 FIGURE 3.2.9 Clepsydra in Problem 33 2 1 FIGURE 3.2.10 Clepsydra in Problem 34 35. Suppose that r f(h) deﬁnes the shape of a water clock for which the time marks are equally spaced. Use the differential equation in Problem 12 to ﬁnd f(h) and sketch a typical graph of h as a function of r. Assume that the cross-sectional area Ah of the hole is constant. [Hint: In this situation dhdt a, where a 0 is a constant.] 34. (a) Suppose that a glass tank has the shape of a cone with circular cross section as shown in Figure 3.2.10. As in part (a) of Problem 33, assume that h(0) 2 ft corresponds to water ﬁlled to the top of the tank, a hole in the bottom is circular with radius in., g 32 ft/s2, and c 0.6. Use the differential equa-tion in Problem 12 to ﬁnd the height h(t) of the water. (b) Can this water clock measure 12 time intervals of length equal to 1 hour? Explain using sound mathematics. 1 32 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 106 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS MODELING WITH SYSTEMS OF FIRST-ORDER DEs REVIEW MATERIAL ●Section 1.3 INTRODUCTION This section is similar to Section 1.3 in that we are just going to discuss cer-tain mathematical models, but instead of a single differential equation the models will be systems of ﬁrst-order differential equations. Although some of the models will be based on topics that we explored in the preceding two sections, we are not going to develop any general methods for solv-ing these systems. There are reasons for this: First, we do not possess the necessary mathematical tools for solving systems at this point. Second, some of the systems that we discuss—notably the systems of nonlinear ﬁrst-order DEs—simply cannot be solved analytically. We shall examine solution methods for systems of linear DEs in Chapters 4, 7, and 8. 3.3 Linear/Nonlinear Systems We have seen that a single differential equation can serve as a mathematical model for a single population in an environment. But if there are, say, two interacting and perhaps competing species living in the same environment (for example, rabbits and foxes), then a model for their populations x(t) and y(t) might be a system of two ﬁrst-order differential equations such as . (1) When g1 and g2 are linear in the variables x and y—that is, g1 and g2 have the forms , where the coefﬁcients ci could depend on t—then (1) is said to be a linear system. A system of differential equations that is not linear is said to be nonlinear. Radioactive Series In the discussion of radioactive decay in Sections 1.3 and 3.1 we assumed that the rate of decay was proportional to the number A(t) of nuclei of the substance present at time t. When a substance decays by radioactivity, it usually doesn’t just transmute in one step into a stable substance; rather, the ﬁrst substance decays into another radioactive substance, which in turn decays into a third substance, and so on. This process, called a radioactive decay series, con-tinues until a stable element is reached. For example, the uranium decay series is U-238 : Th-234 : : Pb-206, where Pb-206 is a stable isotope of lead. The half-lives of the various elements in a radioactive series can range from billions of years (4.5 109 years for U-238) to a fraction of a second. Suppose a radioactive series is described schematically by , where k1 l1 0 and k2 l2 0 are the decay constants for substances X and Y, respectively, and Z is a stable element. Suppose, too, that x(t), y(t), and z(t) denote amounts of substances X, Y, and Z, respectively, remaining at time t. The decay of element X is described by , whereas the rate at which the second element Y decays is the net rate , dy dt 1x 2y dx dt 1x X 1 : Y 2 : Z g1(t, x, y) c1 x c2 y f1(t) and g2(t, x, y) c3 x c4 y f2(t) dy dt g2(t, x, y) dx dt g1(t, x, y) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.3 MODELING WITH SYSTEMS OF FIRST-ORDER DEs ● 107 since Y is gaining atoms from the decay of X and at the same time losing atoms because of its own decay. Since Z is a stable element, it is simply gaining atoms from the decay of element Y: . In other words, a model of the radioactive decay series for three elements is the linear system of three ﬁrst-order differential equations (2) Mixtures Consider the two tanks shown in Figure 3.3.1. Let us suppose for the sake of discussion that tank A contains 50 gallons of water in which 25 pounds of salt is dissolved. Suppose tank B contains 50 gallons of pure water. Liquid is pumped into and out of the tanks as indicated in the ﬁgure; the mixture exchanged between the two tanks and the liquid pumped out of tank B are assumed to be well stirred. We wish to construct a mathematical model that describes the number of pounds x1(t) and x2(t) of salt in tanks A and B, respectively, at time t. dz dt 2y. dy dt 1x 2y dx dt 1x dz dt 2y mixture 3 gal/min mixture 4 gal/min B A pure water 3 gal/min mixture 1 gal/min FIGURE 3.3.1 Connected mixing tanks By an analysis similar to that on page 24 in Section 1.3 and Example 5 of Section 3.1 we see that the net rate of change of x1(t) for tank A is dx1 ––– dt (3 gal/min) (0 lb/gal) (1 gal/min) ( lb/gal) (4 gal/min) ( lb/gal) x1 x2. input rate of salt output rate of salt x2 ––– 50 1 ––– 50 x1 ––– 50 2 ––– 25 Similarly, for tank B the net rate of change of x2(t) is 2 25x1 2 25x2. dx2 dt 4 x1 50 3 x2 50 1 x2 50 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Thus we obtain the linear system (3) Observe that the foregoing system is accompanied by the initial conditions x1(0) 25, x2(0) 0. A Predator-Prey Model Suppose that two different species of animals interact within the same environment or ecosystem, and suppose further that the ﬁrst species eats only vegetation and the second eats only the ﬁrst species. In other words, one species is a predator, and the other is a prey. For example, wolves hunt grass-eating caribou, sharks devour little ﬁsh, and the snowy owl pursues an arctic rodent called the lemming. For the sake of discussion, let us imagine that the predators are foxes and the prey are rabbits. Let x(t) and y(t) denote the fox and rabbit populations, respectively, at time t. If there were no rabbits, then one might expect that the foxes, lacking an adequate food supply, would decline in number according to . (4) When rabbits are present in the environment, however, it seems reasonable that the number of encounters or interactions between these two species per unit time is jointly proportional to their populations x and y—that is, proportional to the product xy. Thus when rabbits are present, there is a supply of food, so foxes are added to the system at a rate bxy, b 0. Adding this last rate to (4) gives a model for the fox population: (5) On the other hand, if there were no foxes, then the rabbits would, with an added assumption of unlimited food supply, grow at a rate that is proportional to the num-ber of rabbits present at time t: . (6) But when foxes are present, a model for the rabbit population is (6) decreased by cxy, c 0—that is, decreased by the rate at which the rabbits are eaten during their encounters with the foxes: (7) Equations (5) and (7) constitute a system of nonlinear differential equations (8) where a, b, c, and d are positive constants. This famous system of equations is known as the Lotka-Volterra predator-prey model. Except for two constant solutions, x(t) 0, y(t) 0 and x(t) dc, y(t) ab, the nonlinear system (8) cannot be solved in terms of elementary functions. However, we can analyze such systems quantitatively and qualitatively. See Chapter 9, “Numerical Solutions of Ordinary Differential Equations,” and Chapter 10, “Plane Autonomous Systems.” dy dt dy cxy y(d cx), dx dt ax bxy x(a by) dy dt dy cxy. dy dt dy, d 0 dx dt ax bxy. dx dt ax, a 0 dx2 dt 2 25 x1 2 25 x2. dx1 dt 2 25 x1 1 50 x2 108 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS Chapters 10–15 are in the expanded version of this text, Differential Equations with Boundary-Value Problems. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.3 MODELING WITH SYSTEMS OF FIRST-ORDER DEs ● 109 FIGURE 3.3.2 Populations of predators (red) and prey (blue) in Example 1 t population x, y time predators prey EXAMPLE 1 Predator-Prey Model Suppose represents a predator-prey model. Because we are dealing with populations, we have x(t) 0, y(t) 0. Figure 3.3.2, obtained with the aid of a numerical solver, shows typical population curves of the predators and prey for this model superim-posed on the same coordinate axes. The initial conditions used were x(0) 4, y(0) 4. The curve in red represents the population x(t) of the predators (foxes), and the blue curve is the population y(t) of the prey (rabbits). Observe that the model seems to predict that both populations x(t) and y(t) are periodic in time. This makes intuitive sense because as the number of prey decreases, the predator population eventually decreases because of a diminished food supply; but attendant to a decrease in the number of predators is an increase in the number of prey; this in turn gives rise to an increased number of predators, which ultimately brings about another decrease in the number of prey. Competition Models Now suppose two different species of animals occupy the same ecosystem, not as predator and prey but rather as competitors for the same resources (such as food and living space) in the system. In the absence of the other, let us assume that the rate at which each population grows is given by , (9) respectively. Since the two species compete, another assumption might be that each of these rates is diminished simply by the inﬂuence, or existence, of the other population. Thus a model for the two populations is given by the linear system (10) , where a, b, c, and d are positive constants. On the other hand, we might assume, as we did in (5), that each growth rate in (9) should be reduced by a rate proportional to the number of interactions between the two species: (11) . Inspection shows that this nonlinear system is similar to the Lotka-Volterra predator-prey model. Finally, it might be more realistic to replace the rates in (9), which indicate that the population of each species in isolation grows exponentially, with rates indicating that each population grows logistically (that is, over a long time the population is bounded): . (12) dx dt a1x b1x2 and dy dt a2 y b2 y2 dy dt cy dxy dx dt ax bxy dy dt cy dx dx dt ax by dx dt ax and dy dt cy dy dt 4.5y 0.9xy dx dt 0.16x 0.08xy Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. When these new rates are decreased by rates proportional to the number of interac-tions, we obtain another nonlinear model: (13) , where all coefﬁcients are positive. The linear system (10) and the nonlinear systems (11) and (13) are, of course, called competition models. Networks An electrical network having more than one loop also gives rise to simultaneous differential equations. As shown in Figure 3.3.3, the current i1(t) splits in the directions shown at point B1, called a branch point of the network. By Kirchhoff’s first la we can write . (14) We can also apply Kirchhoff’s second law to each loop. For loop A1B1B2A2A1, summing the voltage drops across each part of the loop gives . (15) Similarly, for loop A1B1C1C2B2A2A1 we ﬁnd . (16) Using (14) to eliminate i1 in (15) and (16) yields two linear ﬁrst-order equations for the currents i2(t) and i3(t): (17) We leave it as an exercise (see Problem 14 in Exercises 3.3) to show that the sys-tem of differential equations describing the currents i1(t) and i2(t) in the network con-taining a resistor, an inductor, and a capacitor shown in Figure 3.3.4 is (18) RC di2 dt i2 i1 0. L di1 dt Ri2 E(t) L2 di3 dt R1i2 R1i3 E(t). L1 di2 dt (R1 R2)i2 R1i3 E(t) E(t) i1R1 L2 di3 dt E(t) i1R1 L1 di2 dt i2R2 i1(t) i2(t) i3(t) dy dt a2y b2y 2 c2xy y(a2 b2y c2x) dx dt a1x b1x2 c1xy x(a1 b1x c1y) 110 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS A1 L1 R1 R2 A2 B1 B2 C1 C2 i1 i2 i3 L2 E FIGURE 3.3.3 Network whose model is given in (17) FIGURE 3.3.4 Network whose model is given in (18) i1 L R C i2 i3 E EXERCISES 3.3 Answers to selected odd-numbered problems begin on page ANS-4. Radioactive Series 1. We have not discussed methods by which systems of ﬁrst-order differential equations can be solved. Nevertheless, systems such as (2) can be solved with no knowledge other than how to solve a single linear ﬁrst-order equation. Find a solution of (2) subject to the initial conditions x(0) x0, y(0) 0, z(0) 0. 2. In Problem 1 suppose that time is measured in days, that the decay constants are k1 0.138629 and k2 0.004951, and that x0 20. Use a graphing utility to obtain the graphs of the solutions x(t), y(t), and z(t) on the same set of coordinate axes. Use the graphs to approximate the half-lives of substances X and Y. 3. Use the graphs in Problem 2 to approximate the times when the amounts x(t) and y(t) are the same, the times when the amounts x(t) and z(t) are the same, and the times when the amounts y(t) and z(t) are the same. Why does the time that is determined when the amounts y(t) and z(t) are the same make intuitive sense? 4. Construct a mathematical model for a radioactive series of four elements W, X, Y, and Z, where Z is a stable element. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.3 MODELING WITH SYSTEMS OF FIRST-ORDER DEs ● 111 Mixtures 5. Consider two tanks A and B, with liquid being pumped in and out at the same rates, as described by the system of equations (3). What is the system of differential equations if, instead of pure water, a brine solution containing 2 pounds of salt per gallon is pumped into tank A? 6. Use the information given in Figure 3.3.5 to construct a mathematical model for the number of pounds of salt x1(t), x2(t), and x3(t) at time t in tanks A, B, and C, respectively. x2(t), and x3(t) at time t in tanks A, B, and C, respectively. Without solving the system, predict limiting values of x1(t), x2(t), and x3(t) as t : . mixture 5 gal/min mixture 6 gal/min mixture 4 gal/min pure water 4 gal/min B 100 gal C 100 gal A 100 gal mixture 2 gal/min mixture 1 gal/min FIGURE 3.3.5 Mixing tanks in Problem 6 7. Two very large tanks A and B are each partially filled with 100 gallons of brine. Initially, 100 pounds of salt is dissolved in the solution in tank A and 50 pounds of salt is dissolved in the solution in tank B. The system is closed in that the well-stirred liquid is pumped only between the tanks, as shown in Figure 3.3.6. mixture 2 gal/min mixture 3 gal/min B 100 gal A 100 gal FIGURE 3.3.6 Mixing tanks in Problem 7 (a) Use the information given in the ﬁgure to construct a mathematical model for the number of pounds of salt x1(t) and x2(t) at time t in tanks A and B, respectively. (b) Find a relationship between the variables x1(t) and x2(t) that holds at time t. Explain why this relationship makes intuitive sense. Use this rela-tionship to help ﬁnd the amount of salt in tank B at t 30 min. 8. Three large tanks contain brine, as shown in Figure 3.3.7. Use the information in the ﬁgure to construct a mathe-matical model for the number of pounds of salt x1(t), FIGURE 3.3.7 Mixing tanks in Problem 8 mixture 4 gal/min mixture 4 gal/min mixture 4 gal/min pure water 4 gal/min B 150 gal C 100 gal A 200 gal Predator-Prey Models 9. Consider the Lotka-Volterra predator-prey model deﬁned by , where the populations x(t) (predators) and y(t) (prey) are measured in thousands. Suppose x(0) 6 and y(0) 6. Use a numerical solver to graph x(t) and y(t). Use the graphs to approximate the time t 0 when the two populations are ﬁrst equal. Use the graphs to approximate the period of each population. Competition Models 10. Consider the competition model deﬁned by , where the populations x(t) and y(t) are measured in thousands and t in years. Use a numerical solver to analyze the populations over a long period of time for each of the following cases: (a) x(0) 1.5, y(0) 3.5 (b) x(0) 1, y(0) 1 (c) x(0) 2, y(0) 7 (d) x(0) 4.5, y(0) 0.5 11. Consider the competition model deﬁned by , where the populations x(t) and y(t) are measured in thousands and t in years. Use a numerical solver to dy dt y(1.7 0.1y 0.15x) dx dt x(1 0.1x 0.05y) dy dt y(1 0.1y 0.3x) dx dt x(2 0.4x 0.3y) dy dt 0.2y 0.025xy dx dt 0.1x 0.02xy Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. analyze the populations over a long period of time for each of the following cases: (a) x(0) 1, y(0) 1 (b) x(0) 4, y(0) 10 (c) x(0) 9, y(0) 4 (d) x(0) 5.5, y(0) 3.5 Networks 12. Show that a system of differential equations that describes the currents i2(t) and i3(t) in the electrical network shown in Figure 3.3.8 is R1 di2 dt R2 di3 dt 1 C i3 0. L di2 dt L di3 dt R1i2 E(t) 112 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS let s(t), i(t), and r(t) denote, in turn, the number of peo-ple in the community (measured in hundreds) who are susceptible to the disease but not yet infected with it, the number of people who are infected with the dis-ease, and the number of people who have recovered from the disease. Explain why the system of differen-tial equations where k1 (called the infection rate) and k2 (called the removal rate) are positive constants, is a reasonable mathematical model, commonly called a SIR model, for the spread of the epidemic throughout the commu-nity. Give plausible initial conditions associated with this system of equations. 16. (a) In Problem 15, explain why it is sufﬁcient to analyze only . (b) Suppose k1 0.2, k2 0.7, and n 10. Choose various values of i(0) i0, 0 i0 10. Use a numerical solver to determine what the model pre-dicts about the epidemic in the two cases s0 k2k1 and s0 k2k1. In the case of an epidemic, estimate the number of people who are eventually infected. Project Problems 17. Concentration of a Nutrient Suppose compartments A and B shown in Figure 3.3.10 are ﬁlled with ﬂuids and are separated by a permeable membrane. The ﬁgure is a compartmental representation of the exterior and interior of a cell. Suppose, too, that a nutrient necessary for cell growth passes through the membrane. A model di dt k2i k1si ds dt k1si dr dt k2i, di dt k2i k1si ds dt k1si R1 E i1 L i2 i3 C R2 FIGURE 3.3.8 Network in Problem 12 i1 i2 i3 R1 R2 R3 E L1 L2 FIGURE 3.3.9 Network in Problem 13 13. Determine a system of ﬁrst-order differential equations that describes the currents i2(t) and i3(t) in the electrical network shown in Figure 3.3.9. 14. Show that the linear system given in (18) describes the currents i1(t) and i2(t) in the network shown in Figure 3.3.4. [Hint: dqdt i3.] Additional Nonlinear Models 15. SIR Model Acommunicable disease is spread through-out a small community, with a ﬁxed population of n peo-ple, by contact between infected individuals and people who are susceptible to the disease. Suppose that everyone is initially susceptible to the disease and that no one leaves the community while the epidemic is spreading.At time t, FIGURE 3.3.10 Nutrient ﬂow through a membrane in Problem 17 B A membrane fluid at concentration x(t) fluid at concentration y(t) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 3 IN REVIEW ● 113 solver to obtain numerical solution curves of (3) subject to the initial conditions x1(0) 25, x2(0) 0. 20. Newton’s Law of Cooling /Warming As shown in Figure 3.3.11, a small metal bar is placed inside con-tainer A, and container A then is placed within a much larger container B. As the metal bar cools, the ambient temperature TA(t) of the medium within container A changes according to Newton’s law of cooling. As container A cools, the temperature of the medium in-side container B does not change signiﬁcantly and can be considered to be a constant TB. Construct a mathe-matical model for the temperatures T(t) and TA(t), where T(t) is the temperature of the metal bar inside container A. As in Problems 1 and 18, this model can be solved by using prior knowledge. Find a solution of the system subject to the initial conditions T(0) T0, TA(0) T1. FIGURE 3.3.11 Container within a container in Problem 20 TA (t) container A container B TB = constant metal bar for the concentrations x(t) and y(t) of the nutrient in compartments A and B, respectively, at time t is given by the linear system of differential equations , where VA and VB are the volumes of the compartments, and k 0 is a permeability factor. Let x(0) x0 and y(0) y0 denote the initial concentrations of the nutri-ent. Solely on the basis of the equations in the system and the assumption x0 y0 0, sketch, on the same set of coordinate axes, possible solution curves of the sys-tem. Explain your reasoning. Discuss the behavior of the solutions over a long period of time. 18. The system in Problem 17, like the system in (2), can be solved with no advanced knowledge. Solve for x(t) and y(t) and compare their graphs with your sketches in Problem 17. Determine the limiting values of x(t) and y(t) as t : . Explain why the answer to the last ques-tion makes intuitive sense. 19. Mixtures Solely on the basis of the physical descrip-tion of the mixture problem on page 107 and in Figure 3.3.1, discuss the nature of the functions x1(t) and x2(t). What is the behavior of each function over a long period of time? Sketch possible graphs of x1(t) and x2(t). Check your conjectures by using a numerical dy dt VB (x y) dx dt VA (y x) CHAPTER 3 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-4. Answer Problems 1 and 2 without referring back to the text. Fill in the blank or answer true or false. 1. If P(t) P0e0.15t gives the population in an environment at time t, then a differential equation satisﬁed by P(t) is . 2. If the rate of decay of a radioactive substance is proportional to the amount A(t) remaining at time t, then the half-life of the substance is necessarily T (ln 2)k. The rate of decay of the substance at time t T is one-half the rate of decay at t 0. 3. In March 1976 the world population reached 4 billion. At that time, a popular news magazine predicted that with an average yearly growth rate of 1.8%, the world population would be 8 billion in 45 years. How does this value compare with the value predicted by the model that assumes that the rate of increase in population is proportional to the population present at time t? 4. Air containing 0.06% carbon dioxide is pumped into a room whose volume is 8000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial con-centration of 0.2% carbon dioxide in the room, deter-mine the subsequent amount in the room at time t. What is the concentration of carbon dioxide at 10 minutes? What is the steady-state, or equilibrium, concentration of carbon dioxide? 5. Solve the differential equation of the tractrix. See Problem 28 in Exercises 1.3. Assume that the initial point on the y-axis in (0, 10) and that the length of the rope is x 10 ft. dy dx y 1s2 y2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6. Suppose a cell is suspended in a solution containing a solute of constant concentration Cs. Suppose further that the cell has constant volume V and that the area of itspermeable membrane is the constant A.By Fick’s lawthe rate of change of its mass m is directly proportional to the area A and the difference Cs C(t), where C(t) is the con-centration of the solute inside the cell at time t. Find C(t) if m VC(t) and C(0) C0. See Figure 3.R.1. 114 ● CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS Find the current i(t) if the resistance is 0.2 ohm, the impressed voltage is E(t) 4, and i(0) 0. Graph i(t). 10. A classical problem in the calculus of variations is to ﬁnd the shape of a curve such that a bead, under the inﬂuence of gravity, will slide from point A(0, 0) to point B(x1, y1) in the least time. See Figure 3.R.2. It can be shown that a nonlinear differential for the shape y(x) of the path is y[1 (y)2] k, where k is a constant. First solve for dx in terms of y and dy, and then use the substitution y k sin2u to obtain a parametric form of the solution. The curve turns out to be a cycloid. concentration C(t) concentration Cs molecules of solute diffusing through cell membrane FIGURE 3.R.1 Cell in Problem 6 7. Suppose that as a body cools, the temperature of the surrounding medium increases because it completely absorbs the heat being lost by the body. Let T(t) and Tm(t) be the temperatures of the body and the medium at time t, respectively. If the initial temperature of the body is T1 and the initial temperature of the medium is T2, then it can be shown in this case that Newton’s law of cooling is dTdt k(T Tm), k 0, where Tm T2 B(T1 T), B 0 is a constant. (a) The foregoing DE is autonomous. Use the phase portrait concept of Section 2.1 to determine the limiting value of the temperature T(t) as t : . What is the limiting value of Tm(t) as t : ? (b) Verify your answers in part (a) by actually solving the differential equation. (c) Discuss a physical interpretation of your answers in part (a). 8. According to Stefan’s law of radiation the absolute temperature T of a body cooling in a medium at constant absolute temperature Tm is given by , where k is a constant. Stefan’s law can be used over a greater temperature range than Newton’s law of cooling. (a) Solve the differential equation. (b) Show that when T Tm is small in comparison to Tm then Newton’s law of cooling approximates Stefan’s law. [Hint: Think binomial series of the right-hand side of the DE.] 9. An LR-series circuit has a variable inductor with the inductance deﬁned by . L(t) 1 1 10 t, 0, 0 t 10 t 10 dT dt k(T 4 T 4 m) FIGURE 3.R.2 Sliding bead in Problem 10 x y B(x1, y1) A(0, 0) bead mg 11. A model for the populations of two interacting species of animals is Solve for x and y in terms of t. 12. Initially, two large tanks A and B each hold 100 gallons of brine. The well-stirred liquid is pumped between the tanks as shown in Figure 3.R.3. Use the information given in the ﬁgure to construct a mathematical model for the number of pounds of salt x1(t) and x2(t) at time t in tanks A and B, respectively. dy dt k2xy. dx dt k1x( x) FIGURE 3.R.3 Mixing tanks in Problem 12 2 lb/gal 7 gal/min mixture 5 gal/min A 100 gal B 100 gal mixture 3 gal/min mixture 1 gal/min mixture 4 gal/min When all the curves in a family G(x, y, c1) 0 intersect orthogonally all the curves in another family H(x, y, c2) 0, the families are said to be orthogonal trajectories of each other. See Figure 3.R.4. If dydx f (x, y) is the differential equation of one family, then the differential equation for the Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. orthogonal trajectories of this family is dydx 1f (x, y). In Problems 13 and 14 ﬁnd the differential equation of the given family. Find the orthogonal trajectories of this family. Use a graphing utility to graph both families on the same set of coordinate axes. CHAPTER 3 IN REVIEW ● 115 FIGURE 3.R.4 Orthogonal trajectories tangents H(x, y, c2) = 0 G(x, y, c1) = 0 13. y x 1 c1ex 14. 15. Potassium-40 Decay One of the most abundant metals found throughout the Earth’s crust and oceans is potas-sium. Although potassium occurs naturally in the form of three isotopes, only the isotope potassium-40 (K-40) is radioactive. This isotope is a bit unusual in that it decays by two different nuclear reactions. Over time, by emitting a beta particle, a great percentage of an ini-tial amount of K-40 decays into the stable isotope cal-cium-40 (Ca-40), whereas by electron capture a smaller percentage of K-40 decays into the stable isotope y 1 x c1 The knowledge of how K-40 decays is the basis for the potassium-argon dating method. This method can be used to ﬁnd the age of very old igneous rocks. Fossils can sometimes be dated indirectly by dating the igneous rocks in the substrata in which the fossils are found. argon-40 (Ar-40). Because the rates at which the amounts C(t) of Ca-40 and A(t) of Ar-40 increase are proportional to the amount K(t) of potassium present, and the rate at which potassium decreases is also proportional to K(t) we obtain the system of linear ﬁrst-order equations where and are positive constants of proportionality. (a) From the foregoing system of differential equations ﬁnd if Then ﬁnd and if and (b) It is known that and Find the half-life of K-40. (c) Use your solutions for C(t) and to determine the percentage of an initial amount of K-40 that de-cays into Ca-40 and the percentage that decays into Ar-40 over a very long period of time. K0 A(t) l2 0.5874 1010. l1 4.7526 1010 A(0) 0. C(0) 0 A(t) C(t) K(0) K0. K(t) l2 l1 dK dt (l1 l2)K, dA dt l2K dC dt l1K Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Higher-Order Differential Equations 4.1 Preliminary Theory—Linear Equations 4.1.1 Initial-Value and Boundary-Value Problems 4.1.2 Homogeneous Equations 4.1.3 Nonhomogeneous Equations 4.2 Reduction of Order 4.3 Homogeneous Linear Equations with Constant Coefﬁcients 4.4 Undetermined Coefﬁcients—Superposition Approach 4.5 Undetermined Coefﬁcients—Annihilator Approach 4.6 Variation of Parameters 4.7 Cauchy-Euler Equation 4.8 Green’s Functions 4.8.1 Initial-Value Problems 4.8.2 Boundary-Value Problems 4.9 Solving Systems of Linear DEs by Elimination 4.10 Nonlinear Differential Equations Chapter 4 in Review We turn now to the solution of ordinary differential equation of order two or higher. In the ﬁrst seven sections of this chapter we examine the underlying theory and solution methods for certain kinds of linear equations. In the new, but optional, Section 4.8 we build on the material of Section 4.6 to construct Green’s functions for solving linear initial-value and boundary-value problems. The elimination method of solving systems of linear equations is introduced in Section 4.9 because this method simply uncouples a system into individual linear equations in each dependent variable. The chapter concludes with a brief examination of nonlinear higher-order equations in Section 4.10. 4 116 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 117 PRELIMINARY THEORY—LINEAR EQUATIONS REVIEW MATERIAL ●Reread the Remarks at the end of Section 1.1 ●Section 2.3 (especially page 57) INTRODUCTION In Chapter 2 we saw that we could solve a few ﬁrst-order differential equa-tions by recognizing them as separable, linear, exact, homogeneous, or perhaps Bernoulli equations. Even though the solutions of these equations were in the form of a one-parameter family, this family, with one exception, did not represent the general solution of the differential equation. Only in the case of linear ﬁrst-order differential equations were we able to obtain general solutions, by paying attention to certain continuity conditions imposed on the coefﬁcients. Recall that a general solution is a family of solutions deﬁned on some interval I that contains all solutions of the DE that are deﬁned on I. Because our primary goal in this chapter is to ﬁnd general solutions of linear higher-order DEs, we ﬁrst need to examine some of the theory of linear equations. 4.1 4.1.1 INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS Initial-Value Problem In Section 1.2 we deﬁned an initial-value problem for a general nth-order differential equation. For a linear differential equation an nth-order initial-value problem is Solve: Subject to: . (1) Recall that for a problem such as this one we seek a function deﬁned on some interval I, containing x0, that satisﬁes the differential equation and the n initial conditions speciﬁed at x0: y(x0) y0, y(x0) y1, . . . , y(n1)(x0) yn1. We have already seen that in the case of a second-order initial-value problem a solution curve must pass through the point (x0, y0) and have slope y1 at this point. Existence and Uniqueness In Section 1.2 we stated a theorem that gave conditions under which the existence and uniqueness of a solution of a ﬁrst-order initial-value problem were guaranteed. The theorem that follows gives sufﬁcient conditions for the existence of a unique solution of the problem in (1). y(x0) y0, y(x0) y1 , . . . , y(n1)(x0) yn1 an(x) d ny dxn an1(x) d n1y dxn1 a1(x) dy dx a0(x)y g(x) THEOREM 4.1.1 Existence of a Unique Solution Let an(x), an1(x), . . . , a1(x), a0(x) and g(x) be continuous on an interval I and let an(x) 0 for every x in this interval. If x x0 is any point in this interval, then a solution y(x) of the initial-value problem (1) exists on the interval and is unique. EXAMPLE 1 Unique Solution of an IVP The initial-value problem 3y 5y y 7y 0, y(1) 0, y(1) 0, y (1) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 118 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS possesses the trivial solution y 0. Because the third-order equation is linear with constant coefﬁcients, it follows that all the conditions of Theorem 4.1.1 are fulﬁlled. Hence y 0 is the only solution on any interval containing x 1. FIGURE 4.1.1 Solution curves of a BVP that pass through two points I solutions of the DE (b, y1) (a, y0) x y EXAMPLE 2 Unique Solution of an IVP You should verify that the function y 3e2x e2x 3x is a solution of the initial-value problem Now the differential equation is linear, the coefﬁcients as well as g(x) 12x are continuous, and a2(x) 1 0 on any interval I containing x 0. We conclude from Theorem 4.1.1 that the given function is the unique solution on I. The requirements in Theorem 4.1.1 that ai(x), i 0, 1, 2, . . . , n be continuous and an(x) 0 for every x in I are both important. Speciﬁcally, if an(x) 0 for some x in the interval, then the solution of a linear initial-value problem may not be unique or even exist. For example, you should verify that the function y cx2 x 3 is a solution of the initial-value problem on the interval ( , ) for any choice of the parameter c. In other words, there is no unique solution of the problem. Although most of the conditions of Theorem 4.1.1 are satisﬁed, the obvious difﬁculties are that a2(x) x2 is zero at x 0 and that the initial conditions are also imposed at x 0. Boundary-Value Problem Another type of problem consists of solving a lin-ear differential equation of order two or greater in which the dependent variable y or its derivatives are speciﬁed at different points. A problem such as Solve: Subject to: is called a boundary-value problem (BVP). The prescribed values y(a) y0 and y(b) y1 are called boundary conditions. A solution of the foregoing problem is a function satisfying the differential equation on some interval I, containing a and b, whose graph passes through the two points (a, y0) and (b, y1). See Figure 4.1.1. For a second-order differential equation other pairs of boundary conditions could be where y0 and y1 denote arbitrary constants. These three pairs of conditions are just special cases of the general boundary conditions The next example shows that even when the conditions of Theorem 4.1.1 are fulﬁlled, a boundary-value problem may have several solutions (as suggested in Figure 4.1.1), a unique solution, or no solution at all. 2y(b) 2y(b) 2. 1y(a) 1y(a) 1 y(a) y0, y(b) y1, y(a) y0, y(b) y1 y(a) y0, y(b) y1 y(a) y0, y(b) y1 a2(x) d 2y dx2 a1(x) dy dx a0(x)y g(x) x2y 2xy 2y 6, y(0) 3, y(0) 1 y 4y 12x, y(0) 4, y(0) 1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 119 FIGURE 4.1.2 Solution curves for BVP in part (a) of Example 3 x c2 = 0 c2 = 1 c2 = c2 = c2 = − (0, 0) ( /2, 0) 1 1 t π 1 2 1 4 1 2 EXAMPLE 3 A BVP Can Have Many, One, or No Solutions In Example 7 of Section 1.1 we saw that the two-parameter family of solutions of the differential equation x 16x 0 is (2) (a) Suppose we now wish to determine the solution of the equation that further satisﬁes the boundary conditions x(0) 0, x(p2) 0. Observe that the ﬁrst condition 0 c1 cos 0 c2 sin 0 implies that c1 0, so x c2 sin 4t. But when t p2, 0 c2 sin 2p is satisﬁed for any choice of c2, since sin 2p 0. Hence the boundary-value problem (3) has inﬁnitely many solutions. Figure 4.1.2 shows the graphs of some of the members of the one-parameter family x c2 sin 4t that pass through the two points (0, 0) and (p2, 0). (b) If the boundary-value problem in (3) is changed to , (4) then x(0) 0 still requires c1 0 in the solution (2). But applying x(p8) 0 to x c2 sin 4t demands that 0 c2 sin(p2) c21. Hence x 0 is a solution of this new boundary-value problem. Indeed, it can be proved that x 0 is the only solution of (4). (c) Finally, if we change the problem to , (5) we ﬁnd again from x(0) 0 that c1 0, but applying x(p2) 1 to x c2 sin 4t leads to the contradiction 1 c2 sin 2p c2 0 0. Hence the boundary-value problem (5) has no solution. 4.1.2 HOMOGENEOUS EQUATIONS A linear nth-order differential equation of the form (6) is said to be homogeneous, whereas an equation (7) with g(x) not identically zero, is said to be nonhomogeneous. For example, 2y 3y 5y 0 is a homogeneous linear second-order differential equation, whereas x3y 6y 10y ex is a nonhomogeneous linear third-order differen-tial equation. The word homogeneous in this context does not refer to coefﬁcients that are homogeneous functions, as in Section 2.5. We shall see that to solve a nonhomogeneous linear equation (7), we must ﬁrst be able to solve the associated homogeneous equation (6). To avoid needless repetition throughout the remainder of this text, we shall, as a matter of course, make the following important assumptions when an(x) dny dxn an1(x) dn1y dxn1 a1(x) dy dx a0(x)y g(x), an(x) dny dxn an1(x) dn1y dxn1 a1(x) dy dx a0(x)y 0 x 16x 0, x(0) 0, x 2 1 x 16x 0, x(0) 0, x 8 0 x 16x 0, x(0) 0, x 2 0 x c1 cos 4t c2 sin 4t. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 120 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS stating definitions and theorems about linear equations (1). On some common interval I, • the coefﬁcient functions ai(x), i 0, 1, 2, . . . , n and g(x) are continuous; • an(x) 0 for every x in the interval. Differential Operators In calculus differentiation is often denoted by the cap-ital letter D—that is, dydx Dy. The symbol D is called a differential operator because it transforms a differentiable function into another function. For example, D(cos 4x) 4 sin 4x and D(5x3 6x2) 15x2 12x. Higher-order derivatives can be expressed in terms of D in a natural manner: where y represents a sufﬁciently differentiable function. Polynomial expressions involving D, such as D 3, D2 3D 4, and 5x3D3 6x2D2 4xD 9, are also differential operators. In general, we deﬁne an nth-order differential opera-tor or polynomial operator to be L an(x)Dn an1(x)Dn1 a1(x)D a0(x). (8) As a consequence of two basic properties of differentiation, D(cf(x)) cDf(x), c is a constant, and D{f(x) g(x)} Df(x) Dg(x), the differential operator L possesses a linearity property; that is, L operating on a linear combination of two differentiable functions is the same as the linear combination of L operating on the individual func-tions. In symbols this means that L{a f (x) bg(x)} aL( f (x)) bL(g(x)), (9) where a and b are constants. Because of (9) we say that the nth-order differential operator L is a linear operator. Differential Equations Any linear differential equation can be expressed in terms of the D notation. For example, the differential equation y 5y 6y 5x 3 can be written as D2y 5Dy 6y 5x 3 or (D2 5D 6)y 5x 3. Using (8), we can write the linear nth-order differential equations (6) and (7) compactly as respectively. Superposition Principle In the next theorem we see that the sum, or super-position, of two or more solutions of a homogeneous linear differential equation is also a solution. L(y) 0 and L(y) g(x), d dx dy dx d 2y dx2 D(Dy) D2y and, in general, dny dxn Dny, Please remember these two assumptions. THEOREM 4.1.2 Superposition Principle—Homogeneous Equations Let y1, y2, . . . , yk be solutions of the homogeneous nth-order differential equation (6) on an interval I. Then the linear combination where the ci, i 1, 2, . . . , k are arbitrary constants, is also a solution on the interval. y c1y1(x) c2y2(x) ckyk(x), PROOF We prove the case k 2. Let L be the differential operator deﬁned in (8), and let y1(x) and y2(x) be solutions of the homogeneous equation L( y) 0. If we deﬁne y c1y1(x) c2y2(x), then by linearity of L we have L(y) L{c1y1(x) c2y2(x)} c1 L(y1) c2 L(y2) c1 0 c2 0 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 121 FIGURE 4.1.3 Set consisting of f1 and f2 is linearly independent on ( , ) f1 = x x y f 2 = \|x\| x y (a) (b) COROLLARIES TO THEOREM 4.1.2 (A) A constant multiple y c1y1(x) of a solution y1(x) of a homogeneous linear differential equation is also a solution. (B) A homogeneous linear differential equation always possesses the trivial solution y 0. EXAMPLE 4 Superposition—Homogeneous DE The functions y1 x2 and y2 x2 ln x are both solutions of the homogeneous linear equation x3y 2xy 4y 0 on the interval (0, ). By the superposition principle the linear combination is also a solution of the equation on the interval. The function y e7x is a solution of y 9y 14y 0. Because the differen-tial equation is linear and homogeneous, the constant multiple y ce7x is also a solution. For various values of c we see that y 9e7x, y 0, , . . . are all solutions of the equation. Linear Dependence and Linear Independence The next two concepts are basic to the study of linear differential equations. y 15e7x y c1x2 c2x2 ln x DEFINITION 4.1.1 Linear Dependence/Independence A set of functions f1(x), f2(x), . . . , fn(x) is said to be linearly dependent on an interval I if there exist constants c1, c2, . . . , cn, not all zero, such that for every x in the interval. If the set of functions is not linearly dependent on the interval, it is said to be linearly independent. c1 f1(x) c2 f2(x) cn fn(x) 0 In other words, a set of functions is linearly independent on an interval I if the only constants for which for every x in the interval are . It is easy to understand these deﬁnitions for a set consisting of two functions f1(x) and f2(x). If the set of functions is linearly dependent on an interval, then there exist constants c1 and c2 that are not both zero such that for every x in the interval, c1 f1(x) c2 f2(x) 0. Therefore if we assume that c1 0, it follows that f1(x) (c2c1) f2(x); that is, if a set of two functions is linearly dependent, then one function is simply a constant multiple of the other . Conversely, if f1(x) c2 f2(x) for some constant c2, then (1) f1(x) c2 f2(x) 0 for every x in the interval. Hence the set of functions is linearly dependent because at least one of the constants (namely, c1 1) is not zero. We conclude that a set of two functions f1(x) and f2(x) is linearly independent when neither function is a constant multiple of the other on the interval. For example, the set of functions f1(x) sin 2x, f2(x) sin x cos x is linearly dependent on ( , ) because f1(x) is a constant multiple of f2(x). Recall from the double-angle formula for the sine that sin 2x 2 sin x cos x. On the other hand, the set of functions f1(x) x, f2(x) x is linearly independent on ( , ). Inspection of Figure 4.1.3 should convince you that neither function is a constant multiple of the other on the interval. c1 c2 cn 0 c1 f1(x) c2 f2(x) cn fn(x) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 122 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS It follows from the preceding discussion that the quotient f2(x)f1(x) is not a con-stant on an interval on which the set f1(x), f2(x) is linearly independent. This little fact will be used in the next section. EXAMPLE 5 Linearly Dependent Set of Functions The set of functions f1(x) cos2x, f2(x) sin2x, f3(x) sec2x, f4(x) tan2x is linearly dependent on the interval (p2, p2) because when c1 c2 1, c3 1, c4 1. We used here cos2x sin2x 1 and 1 tan2x sec2x. A set of functions f1(x), f2(x), . . . , fn(x) is linearly dependent on an interval if at least one function can be expressed as a linear combination of the remaining functions. c1 cos2x c2 sin2x c3 sec2x c4 tan2x 0 EXAMPLE 6 Linearly Dependent Set of Functions The set of functions , f3(x) x 1, f4(x) x2 is linearly dependent on the interval (0, ) because f2 can be written as a linear combi-nation of f1, f3, and f4. Observe that for every x in the interval (0, ). Solutions of Differential Equations We are primarily interested in linearly independent functions or, more to the point, linearly independent solutions of a lin-ear differential equation. Although we could always appeal directly to Deﬁnition 4.1.1, it turns out that the question of whether the set of n solutions y1, y2, . . . , yn of a homogeneous linear nth-order differential equation (6) is linearly independent can be settled somewhat mechanically by using a determinant. f2(x) 1 f1(x) 5 f3(x) 0 f4(x) f1(x) 1x 5, f2(x) 1x 5x DEFINITION 4.1.2 Wronskian Suppose each of the functions f1(x), f2(x), . . . , fn(x) possesses at least n 1 derivatives. The determinant where the primes denote derivatives, is called the Wronskian of the functions. W( f1, f2, . . . , fn) f1 f 1 f1 (n1) f2 f 2 f2 (n1) fn f n fn (n1) , THEOREM 4.1.3 Criterion for Linearly Independent Solutions Let y1, y2, . . . , yn be n solutions of the homogeneous linear nth-order differential equation (6) on an interval I. Then the set of solutions is linearly independent on I if and only if W(y1, y2, . . . , yn) 0 for every x in the interval. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. It follows from Theorem 4.1.3 that when y1, y2, . . . , yn are n solutions of (6) on an interval I, the Wronskian W( y1, y2, . . . , yn) is either identically zero or never zero on the interval. A set of n linearly independent solutions of a homogeneous linear nth-order differential equation is given a special name. 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 123 DEFINITION 4.1.3 Fundamental Set of Solutions Any set y1, y2, . . . , yn of n linearly independent solutions of the homogeneous linear nth-order differential equation (6) on an interval I is said to be a funda-mental set of solutions on the interval. The basic question of whether a fundamental set of solutions exists for a linear equation is answered in the next theorem. THEOREM 4.1.4 Existence of a Fundamental Set There exists a fundamental set of solutions for the homogeneous linear nth-order differential equation (6) on an interval I. Analogous to the fact that any vector in three dimensions can be expressed as a linear combination of the linearly independent vectors i, j, k, any solution of an nth-order homogeneous linear differential equation on an interval I can be expressed as a linear combination of n linearly independent solutions on I. In other words, n linearly independent solutions y1, y2, . . . , yn are the basic building blocks for the general solution of the equation. THEOREM 4.1.5 General Solution—Homogeneous Equations Let y1, y2, . . . , yn be a fundamental set of solutions of the homogeneous linear nth-order differential equation (6) on an interval I. Then the general solution of the equation on the interval is where ci, i 1, 2, . . . , n are arbitrary constants. y c1y1(x) c2y2(x) cnyn(x), Theorem 4.1.5 states that if Y(x) is any solution of (6) on the interval, then con-stants C1, C2, . . . , Cn can always be found so that We will prove the case when n 2. PROOF Let Y be a solution and let y1 and y2 be linearly independent solutions of a2y a1y a0y 0 on an interval I. Suppose that x t is a point in I for which W(y1(t), y2(t)) 0. Suppose also that Y(t) k1 and Y(t) k2. If we now examine the equations it follows that we can determine C1 and C2 uniquely, provided that the determinant of the coefﬁcients satisﬁes y1(t) y1 (t) y2(t) y2 (t) 0. C1y 1(t) C2y 2(t) k2, C1y1(t) C2y2(t) k1 Y(x) C1y1(x) C2y2(x) Cnyn(x). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 124 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS But this determinant is simply the Wronskian evaluated at x t, and by assumption, W 0. If we deﬁne G(x) C1y1(x) C2y2(x), we observe that G(x) satisﬁes the differential equation since it is a superposition of two known solutions; G(x) satisﬁes the initial conditions and Y(x) satisﬁes the same linear equation and the same initial conditions. Because the solution of this linear initial-value problem is unique (Theorem 4.1.1), we have Y(x) G(x) or Y(x) C1y1(x) C2y2(x). G(t) C1y1(t) C2y2(t) k1 and G(t) C1y 1(t) C2y 2(t) k2; EXAMPLE 7 General Solution of a Homogeneous DE The functions y1 e3x and y2 e3x are both solutions of the homogeneous linear equation y 9y 0 on the interval ( , ). By inspection the solutions are lin-early independent on the x-axis. This fact can be corroborated by observing that the Wronskian for every x. We conclude that y1 and y2 form a fundamental set of solutions, and consequently, y c1e3x c2e3x is the general solution of the equation on the interval. W(e3x, e3x) e3x 3e3x e3x 3e3x 6 0 EXAMPLE 8 A Solution Obtained from a General Solution The function y 4sinh 3x 5e3x is a solution of the differential equation in Example 7. (Verify this.) In view of Theorem 4.1.5 we must be able to obtain this solution from the general solution y c1e3x c2e3x. Observe that if we choose c1 2 and c2 7, then y 2e3x 7e3x can be rewritten as The last expression is recognized as y 4 sinh 3x 5e3x. y 2e3x 2e3x 5e3x 4 e3x e3x 2 5e3x. EXAMPLE 9 General Solution of a Homogeneous DE The functions y1 ex, y2 e2x, and y3 e3x satisfy the third-order equation y 6y 11y 6y 0. Since for every real value of x, the functions y1, y2, and y3 form a fundamental set of solu-tions on ( , ). We conclude that y c1ex c2e2x c3e3x is the general solution of the differential equation on the interval. 4.1.3 NONHOMOGENEOUS EQUATIONS Any function yp, free of arbitrary parameters, that satisﬁes (7) is said to be a particular solution or particular integral of the equation. For example, it is a straightforward task to show that the constant function yp 3 is a particular solution of the nonhomogeneous equation y 9y 27. W(ex, e2x, e3x) p ex ex ex e2x 2e2x 4e2x e3x 3e3x 9e3x p 2e6x 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Now if y1, y2, . . . , yk are solutions of (6) on an interval I and yp is any particular solution of (7) on I, then the linear combination (10) is also a solution of the nonhomogeneous equation (7). If you think about it, this makes sense, because the linear combination c1y1(x) c2y2(x) ckyk(x) is trans-formed into 0 by the operator L anDn an1Dn1 a1D a0, whereas yp is transformed into g(x). If we use k n linearly independent solutions of the nth-order equation (6), then the expression in (10) becomes the general solution of (7). y c1y1(x) c2y2(x) ckyk(x) yp 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 125 THEOREM 4.1.6 General Solution—Nonhomogeneous Equations Let yp be any particular solution of the nonhomogeneous linear nth-order differen-tial equation (7) on an interval I, and let y1, y2, . . . , yn be a fundamental set of so-lutions of the associated homogeneous differential equation (6) on I. Then the general solution of the equation on the interval is where the ci, i 1, 2, . . . , n are arbitrary constants. y c1y1(x) c2y2(x) cn yn(x) yp, PROOF Let L be the differential operator deﬁned in (8) and let Y(x) and yp(x) be particular solutions of the nonhomogeneous equation L(y) g(x). If we deﬁne u(x) Y(x) yp(x), then by linearity of L we have L(u) L{Y(x) yp(x)} L(Y(x)) L(yp(x)) g(x) g(x) 0. This shows that u(x) is a solution of the homogeneous equation L(y) 0. Hence by Theorem 4.1.5, , and so or Complementary Function We see in Theorem 4.1.6 that the general solu-tion of a nonhomogeneous linear equation consists of the sum of two functions: The linear combination , which is the general solution of (6), is called the complementary function for equation (7). In other words, to solve a nonhomogeneous linear differential equation, we ﬁrst solve the associated homogeneous equation and then ﬁnd any particular solution of the nonhomogeneous equation. The general solution of the nonhomogeneous equation is then y complementary function any particular solution yc yp. yc(x) c1y1(x) c2y2(x) cnyn(x) y c1y1(x) c2y2(x) cnyn(x) yp(x) yc(x) yp(x). Y(x) c1y1(x) c2y2(x) cnyn(x) yp(x). Y(x) yp(x) c1y1(x) c2y2(x) cnyn(x) u(x) c1y1(x) c2y2(x) cnyn(x) EXAMPLE 10 General Solution of a Nonhomogeneous DE By substitution the function is readily shown to be a particular solu-tion of the nonhomogeneous equation (11) y 6y 11y 6y 3x. yp 11 12 1 2 x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 126 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS To write the general solution of (11), we must also be able to solve the associated homogeneous equation But in Example 9 we saw that the general solution of this latter equation on the in-terval ( , ) was yc c1ex c2e2x c3e3x. Hence the general solution of (11) on the interval is Another Superposition Principle The last theorem of this discussion will be useful in Section 4.4 when we consider a method for ﬁnding particular solutions of nonhomogeneous equations. y yc yp c1ex c2e2x c3e3x 11 12 1 2 x. y 6y 11y 6y 0. THEOREM 4.1.7 Superposition Principle—Nonhomogeneous Equations Let , , . . . , be k particular solutions of the nonhomogeneous linear nth-order differential equation (7) on an interval I corresponding, in turn, to k dis-tinct functions g1, g2, . . . , gk. That is, suppose denotes a particular solution of the corresponding differential equation (12) where i 1, 2, . . . , k. Then (13) is a particular solution of (14) g1(x) g2(x) gk(x). an(x)y(n) an1(x)y(n1) a1(x)y a0(x)y yp yp1(x) yp2(x) ypk(x) an(x)y(n) an1(x)y(n1) a1(x)y a0(x)y gi(x), ypi ypk yp2 yp1 PROOF We prove the case k 2. Let L be the differential operator deﬁned in (8) and let and be particular solutions of the nonhomogeneous equations L(y) g1(x) and L( y) g2(x), respectively. If we deﬁne , we want to show that yp is a particular solution of L( y) g1(x) g2(x). The result follows again by the linearity of the operator L: L(yp) L{yp1(x) yp2(x)} L( yp1(x)) L( yp2(x)) g1(x) g2(x). yp yp1(x) yp2(x) yp2(x) yp1(x) EXAMPLE 11 Superposition—Nonhomogeneous DE You should verify that It follows from (13) of Theorem 4.1.7 that the superposition of , and , is a solution of y 3y 4y 16x2 24x 8 2e2x 2xex ex. g1(x) g3(x) g2(x) y yp1 yp2 yp3 4x2 e2x xex, yp3 yp1, yp2 yp3 xex is a particular solution of y 3y 4y 2xex ex. yp2 e2x is a particular solution of y 3y 4y 2e2x, yp1 4x2 is a particular solution of y 3y 4y 16x2 24x 8, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Note If the are particular solutions of (12) for i 1, 2, . . . , k, then the linear combination where the ci are constants, is also a particular solution of (14) when the right-hand member of the equation is the linear combination Before we actually start solving homogeneous and nonhomogeneous linear differential equations, we need one additional bit of theory, which is presented in the next section. c1g1(x) c2g2(x) ckgk(x). yp c1yp1 c2yp2 ckypk, ypi 4.1 PRELIMINARY THEORY—LINEAR EQUATIONS ● 127 REMARKS This remark is a continuation of the brief discussion of dynamical systems given at the end of Section 1.3. A dynamical system whose rule or mathematical model is a linear nth-order differential equation is said to be an nth-order linear system. The n time-dependent functions y(t), y(t), . . . , y(n1)(t) are the state variables of the system. Recall that their val-ues at some time t give the state of the system. The function g is variously called the input function, forcing function, or excitation function. A solu-tion y(t) of the differential equation is said to be the output or response of the system. Under the conditions stated in Theorem 4.1.1, the output or response y(t) is uniquely determined by the input and the state of the system prescribed at a time t0—that is, by the initial conditions y(t0), y(t0), . . . , y(n1)(t0). For a dynamical system to be a linear system, it is necessary that the super-position principle (Theorem 4.1.7) holds in the system; that is, the response of the system to a superposition of inputs is a superposition of outputs. We have already examined some simple linear systems in Section 3.1 (linear ﬁrst-order equations); in Section 5.1 we examine linear systems in which the mathe-matical models are second-order differential equations. an(t)y(n) an1(t)y(n1) a1(t)y a0(t)y g(t) EXERCISES 4.1 Answers to selected odd-numbered problems begin on page ANS-4. 4.1.1 INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS In Problems 1–4 the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. 1. y c1ex c2ex, ( , ); y y 0, y(0) 0, y(0) 1 2. y c1e4x c2ex, ( , ); y 3y 4y 0, y(0) 1, y(0) 2 3. y c1x c2x ln x, (0, ); x2y xy y 0, y(1) 3, y(1) 1 4. y c1 c2 cos x c3 sin x, ( , ); y y 0, y(p) 0, y(p) 2, y (p) 1 5. Given that y c1 c2x2 is a two-parameter family of solutions of xy y 0 on the interval ( , ), show that constants c1 and c2 cannot be found so that a member of the family satisﬁes the initial conditions y(0) 0, y(0) 1. Explain why this does not violate Theorem 4.1.1. 6. Find two members of the family of solutions in Problem 5 that satisfy the initial conditions y(0) 0, y(0) 0. 7. Given that x(t) c1 cos vt c2 sin vt is the general solution of x v2x 0 on the interval ( , ), show that a solution satisfying the initial conditions x(0) x0, x(0) x1 is given by x(t) x0 cos t x1 sin t. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 128 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 8. Use the general solution of x v2x 0 given in Problem 7 to show that a solution satisfying the initial conditions x(t0) x0, x(t0) x1 is the solution given in Problem 7 shifted by an amount t0: In Problems 9 and 10 ﬁnd an interval centered about x 0 for which the given initial-value problem has a unique solution. 9. (x 2)y 3y x, y(0) 0, y(0) 1 10. y (tan x)y ex, y(0) 1, y(0) 0 11. (a) Use the family in Problem 1 to ﬁnd a solution of y y 0 that satisﬁes the boundary conditions y(0) 0, y(1) 1. (b) The DE in part (a) has the alternative general solu-tion y c3 cosh x c4 sinh x on ( , ). Use this family to ﬁnd a solution that satisﬁes the boundary conditions in part (a). (c) Show that the solutions in parts (a) and (b) are equivalent 12. Use the family in Problem 5 to ﬁnd a solution of xy y 0 that satisﬁes the boundary conditions y(0) 1, y(1) 6. In Problems 13 and 14 the given two-parameter family is a solution of the indicated differential equation on the interval ( , ). Determine whether a member of the family can be found that satisﬁes the boundary conditions. 13. y c1ex cos x c2ex sin x; y 2y 2y 0 (a) y(0) 1, y(p) 0 (b) y(0) 1, y(p) 1 (c) y(0) 1, y(p2) (d) y(0) 0, y(p) 0. 14. y c1x2 c2x4 3; x2y 5xy 8y 24 (a) y(1) 0, y(1) 4 (b) y(0) 1, y(1) 2 (c) y(0) 3, y(1) 0 (d) y(1) 3, y(2) 15 4.1.2 HOMOGENEOUS EQUATIONS In Problems 15–22 determine whether the given set of func-tions is linearly independent on the interval ( , ). 15. f1(x) x, f2(x) x2, f3(x) 4x 3x2 16. f1(x) 0, f2(x) x, f3(x) ex 17. f1(x) 5, f2(x) cos2x, f3(x) sin2x 18. f1(x) cos 2x, f2(x) 1, f3(x) cos2x 19. f1(x) x, f2(x) x 1, f3(x) x 3 20. f1(x) 2 x, f2(x) 2 x 1 x(t) x0 cos (t t0) x1 sin (t t0). 21. f1(x) 1 x, f2(x) x, f3(x) x2 22. f1(x) ex, f2(x) ex, f3(x) sinh x In Problems 23–30 verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution. 23. y y 12y 0; e3x, e4x, ( , ) 24. y 4y 0; cosh 2x, sinh 2x, ( , ) 25. y 2y 5y 0; ex cos 2x, ex sin 2x, ( , ) 26. 4y 4y y 0; ex/2, xex/2, ( , ) 27. x2y 6xy 12y 0; x3, x4, (0, ) 28. x2y xy y 0; cos(ln x), sin(ln x), (0, ) 29. x3y 6x2y 4xy 4y 0; x, x2, x2 ln x, (0, ) 30. y(4) y 0; 1, x, cos x, sin x, ( , ) 4.1.3 NONHOMOGENEOUS EQUATIONS In Problems 31–34 verify that the given two-parameter fam-ily of functions is the general solution of the nonhomoge-neous differential equation on the indicated interval. 31. y 7y 10y 24ex; y c1e2x c2e5x 6ex, ( , ) 32. y y sec x; y c1 cos x c2 sin x x sin x (cos x) ln(cos x), (p2, p2) 33. y 4y 4y 2e2x 4x 12; y c1e2x c2xe2x x2e2x x 2, ( , ) 34. 2x2y 5xy y x2 x; 35. (a) Verify that and are, respec-tively, particular solutions of and (b) Use part (a) to ﬁnd particular solutions of and 36. (a) By inspection find a particular solution of y 2y 10. (b) By inspection find a particular solution of y 2y 4x. (c) Find a particular solution of y 2y 4x 10. (d) Find a particular solution of y 2y 8x 5. y 6y 5y 10x2 6x 32 e2x. y 6y 5y 5x2 3x 16 9e2x y 6y 5y 5x2 3x 16. y 6y 5y 9e2x yp2 x2 3x yp1 3e2x y c1x1/2 c2x1 1 15x2 1 6x, (0, ) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Discussion Problems 37. Let n 1, 2, 3, . . . . Discuss how the observations Dnxn1 0 and Dnxn n! can be used to ﬁnd the gen-eral solutions of the given differential equations. (a) y 0 (b) y 0 (c) y(4) 0 (d) y 2 (e) y 6 (f) y(4) 24 38. Suppose that y1 ex and y2 ex are two solutions of a homogeneous linear differential equation. Explain why y3 cosh x and y4 sinh x are also solutions of the equation. 39. (a) Verify that y1 x3 and y2 x3 are linearly independent solutions of the differential equation x2y 4xy 6y 0 on the interval ( , ). (b) Show that W( y1, y2) 0 for every real number x. Does this result violate Theorem 4.1.3? Explain. (c) Verify that Y1 x3 and Y2 x2 are also linearly independent solutions of the differential equation in part (a) on the interval ( , ). (d) Find a solution of the differential equation satisfy-ing y(0) 0, y(0) 0. 4.2 REDUCTION OF ORDER ● 129 (e) By the superposition principle, Theorem 4.1.2, both linear combinations y c1y1 c2y2 and Y c1Y1 c2Y2 are solutions of the differential equation. Discuss whether one, both, or neither of the linear combinations is a general solution of the differential equation on the interval ( , ). 40. Is the set of functions f1(x) ex2, f2(x) ex3 lin-early dependent or linearly independent on ( , )? Discuss. 41. Suppose y1, y2, . . . , yk are k linearly independent solu-tions on ( , ) of a homogeneous linear nth-order differential equation with constant coefﬁcients. By Theorem 4.1.2 it follows that yk1 0 is also a solution of the differential equation. Is the set of solutions y1, y2, . . . , yk, yk1 linearly dependent or linearly inde-pendent on ( , )? Discuss. 42. Suppose that y1, y2, . . . , yk are k nontrivial solutions of a homogeneous linear nth-order differential equation with constant coefﬁcients and that k n 1. Is the set of solutions y1, y2, . . . , yk linearly dependent or linearly independent on ( , )? Discuss. REDUCTION OF ORDER REVIEW MATERIAL ●Section 2.5 (using a substitution) ●Section 4.1 INTRODUCTION In the preceding section we saw that the general solution of a homogeneous linear second-order differential equation (1) is a linear combination y c1y1 c2y2, where y1 and y2 are solutions that constitute a linearly inde-pendent set on some interval I. Beginning in the next section, we examine a method for determining these solutions when the coefﬁcients of the differential equation in (1) are constants. This method, which is a straightforward exercise in algebra, breaks down in a few cases and yields only a single solution y1 of the DE. It turns out that we can construct a second solution y2 of a homogeneous equa-tion (1) (even when the coefﬁcients in (1) are variable) provided that we know a nontrivial solution y1 of the DE. The basic idea described in this section is that equation (1) can be reduced to a linear first-o der DE by means of a substitution involving the known solution y1. A second solution y2 of (1) is apparent after this ﬁrst-order differential equation is solved. a2(x)y a1(x)y a0(x)y 0 4.2 Reduction of Order Suppose that y1 denotes a nontrivial solution of (1) and that y1 is deﬁned on an interval I. We seek a second solution y2 so that the set consist-ing of y1 and y2 is linearly independent on I. Recall from Section 4.1 that if y1 and y2 are linearly independent, then their quotient y2y1 is nonconstant on I—that is, y2(x)y1(x) u(x) or y2(x) u(x)y1(x). The function u(x) can be found by substituting y2(x) u(x)y1(x) into the given differential equation. This method is called reduction of order because we must solve a linear ﬁrst-order differential equation to ﬁnd u. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 130 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 1 A Second Solution by Reduction of Order Given that y1 ex is a solution of y y 0 on the interval ( , ), use reduction of order to ﬁnd a second solution y2. SOLUTION If y u(x)y1(x) u(x)ex, then the Product Rule gives and so Since ex 0, the last equation requires u 2u 0. If we make the substitution w u, this linear second-order equation in u becomes w 2w 0, which is a linear first-order equation in w. Using the integrating factor e2x, we can write . After integrating, we get w c1e2x or u c1e2x. Integrating again then yields Thus . (2) By picking c2 0 and c1 2, we obtain the desired second solution, y2 ex. Because W(ex, ex) 0 for every x, the solutions are linearly independent on ( , ). Since we have shown that y1 ex and y2 ex are linearly independent solu-tions of a linear second-order equation, the expression in (2) is actually the general solution of y y 0 on ( , ). General Case Suppose we divide by a2(x) to put equation (1) in the standard form (3) where P(x) and Q(x) are continuous on some interval I. Let us suppose further that y1(x) is a known solution of (3) on I and that y1(x) 0 for every x in the interval. If we deﬁne y u(x)y1(x), it follows that This implies that we must have (4) where we have let w u. Observe that the last equation in (4) is both linear and separable. Separating variables and integrating, we obtain . We solve the last equation for w, use w u, and integrate again: . u c1 eP dx y1 2 dx c2 ln wy1 2 P dx c or wy1 2 c1eP dx dw w 2 y 1 y1 dx P dx 0 y1u (2y 1 Py1)u 0 or y1w (2y 1 Py1)w 0, y Py Qy u[y1 Py1 Qy1] y1u (2y1 Py1)u 0. zero y uy 1 y1u, y uy 1 2y 1u y1u y P(x)y Q(x)y 0, y u(x)ex c1 2 ex c2ex u 1 2 c1e2x c2. d dx [e2xw] 0 y y ex(u 2u) 0. y uex exu, y uex 2exu exu , Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.2 REDUCTION OF ORDER ● 131 By choosing c1 1 and c2 0, we ﬁnd from y u(x)y1(x) that a second solution of equation (3) is (5) It makes a good review of differentiation to verify that the function y2(x) deﬁned in (5) satisﬁes equation (3) and that y1 and y2 are linearly independent on any interval on which y1(x) is not zero. y2 y1(x) eP(x) dx y1 2(x) dx. EXAMPLE 2 A Second Solution by Formula (5) The function y1 x2 is a solution of x2y 3xy 4y 0. Find the general solu-tion of the differential equation on the interval (0, ). SOLUTION From the standard form of the equation, we ﬁnd from (5) . The general solution on the interval (0, ) is given by y c1y1 c2y2; that is, y c1x2 c2x2 ln x. x2 dx x x2 ln x ; e3d x /x eln x3 x3 y2 x2 e3dx /x x4 dx y 3 x y 4 x2 y 0, REMARKS (i) The derivation and use of formula (5) have been illustrated here because this formula appears again in the next section and in Sections 4.7 and 6.3. We use (5) simply to save time in obtaining a desired result. Your instructor will tell you whether you should memorize (5) or whether you should know the ﬁrst princi-ples of reduction of order. (ii) Reduction of order can be used to ﬁnd the general solution of a nonhomo-geneous equation a2(x)y a1(x)y a0(x)y g(x) whenever a solution y1 of the associated homogeneous equation is known. See Problems 17–20 in Exercises 4.2. EXERCISES 4.2 Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1–16 the indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to ﬁnd a second solution y2(x). 1. y 4y 4y 0; y1 e2x 2. y 2y y 0; y1 xex 3. y 16y 0; y1 cos 4x 4. y 9y 0; y1 sin 3x 5. y y 0; y1 cosh x 6. y 25y 0; y1 e5x 7. 9y 12y 4y 0; y1 e2x/3 8. 6y y y 0; y1 ex/3 9. x2y 7xy 16y 0; y1 x4 10. x2y 2xy 6y 0; y1 x2 11. xy y 0; y1 ln x 12. 4x2y y 0; y1 x1/2 ln x 13. x2y xy 2y 0; y1 x sin(ln x) 14. x2y 3xy 5y 0; y1 x2 cos(ln x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 132 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 15. (1 2x x2)y 2(1 x)y 2y 0; y1 x 1 16. (1 x2)y 2xy 0; y1 1 In Problems 17–20 the indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to ﬁnd a second solution y2(x) of the homogeneous equation and a particular solution of the given nonhomogeneous equation. 17. y 4y 2; y1 e2x 18. y y 1; y1 1 19. y 3y 2y 5e3x; y1 ex 20. y 4y 3y x; y1 ex Discussion Problems 21. (a) Give a convincing demonstration that the second-order equation ay by cy 0, a, b, and c con-stants, always possesses at least one solution of the form , m1 a constant. (b) Explain why the differential equation in part (a) must then have a second solution either of the form y1 em1x or of the form , m1 and m2 constants. (c) Reexamine Problems 1–8. Can you explain why the statements in parts (a) and (b) above are not contradicted by the answers to Problems 3–5? 22. Verify that y1(x) x is a solution of xy xy y 0. Use reduction of order to ﬁnd a second solution y2(x) in the form of an inﬁnite series. Conjecture an interval of deﬁnition for y2(x). Computer Lab Assignments 23. (a) Verify that y1(x) ex is a solution of xy (x 10)y 10y 0. (b) Use (5) to ﬁnd a second solution y2(x). Use a CAS to carry out the required integration. (c) Explain, using Corollary (A) of Theorem 4.1.2, why the second solution can be written compactly as . y2(x) 10 n0 1 n! xn y2 xem1x y2 em2 x HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS REVIEW MATERIAL ●Review Problems 27–30 in Exercises 1.1 and Theorem 4.1.5 ●Review the algebra of solving polynomial equations (see the Student Resource Manual) INTRODUCTION As a means of motivating the discussion in this section, let us return to ﬁrst-order differential equations—more speciﬁcally, to homogeneous linear equations ay by 0, where the coefﬁcients a 0 and b are constants. This type of equation can be solved either by separation of variables or with the aid of an integrating factor, but there is another solution method, one that uses only algebra. Before illustrating this alternative method, we make one observation: Solving ay by 0 for y yields y ky, where k is a constant. This observation reveals the nature of the unknown solution y; the only nontrivial elementary function whose derivative is a constant multiple of itself is an exponential function emx. Now the new solution method: If we substi-tute y emx and y memx into ay by 0, we get Since emx is never zero for real values of x, the last equation is satisﬁed only when m is a solution or root of the ﬁrst-degree polynomial equation am b 0. For this single value of m, y emx is a solution of the DE. To illustrate, consider the constant-coefﬁcient equation 2y 5y 0. It is not necessary to go through the differentiation and substitution of y emx into the DE; we merely have to form the equation 2m 5 0 and solve it for m. From we conclude that is a solution of 2y 5y 0, and its general solution on the interval ( , ) is In this section we will see that the foregoing procedure can produce exponential solutions for homogeneous linear higher-order DEs, (1) where the coefﬁcients ai, i 0, 1, . . . , n are real constants and an 0. any(n) an1y(n1) a2y a1y a0y 0, y c1e5x/2. y e5x/2 m 5 2 amemx bemx 0 or emx (am b) 0. 4.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Auxiliary Equation We begin by considering the special case of the second-order equation (2) where a, b, and c are constants. If we try to ﬁnd a solution of the form y emx, then after substitution of y memx and y m2emx, equation (2) becomes As in the introduction we argue that because emx 0 for all x, it is apparent that the only way y emx can satisfy the differential equation (2) is when m is chosen as a root of the quadratic equation (3) This last equation is called the auxiliary equation of the differential equa-tion (2). Since the two roots of (3) are and there will be three forms of the general solution of (2) corresponding to the three cases: • m1 and m2 real and distinct (b2 4ac 0), • m1 and m2 real and equal (b2 4ac 0), and • m1 and m2 conjugate complex numbers (b2 4ac 0). We discuss each of these cases in turn. Case I: Distinct Real Roots Under the assumption that the auxiliary equation (3) has two unequal real roots m1 and m2, we ﬁnd two solutions, and We see that these functions are linearly independent on ( , ) and hence form a fun-damental set. It follows that the general solution of (2) on this interval is (4) Case II: Repeated Real Roots When m1 m2, we necessarily obtain only one exponential solution, . From the quadratic formula we ﬁnd that m1 b2a since the only way to have m1 m2 is to have b2 4ac 0. It follows from (5) in Section 4.2 that a second solution of the equation is (5) In (5) we have used the fact that ba 2m1. The general solution is then (6) Case III: Conjugate Complex Roots If m1 and m2 are complex, then we can write m1 a ib and m2 a ib, where a and b 0 are real and i2 1. Formally, there is no difference between this case and Case I, and hence However, in practice we prefer to work with real functions instead of complex exponentials. To this end we use Euler’s formula: where u is any real number. It follows from this formula that (7) eix cos x i sin x and eix cos x i sin x, ei cos i sin , y C1e(ai)x C2e(ai)x. y c1em1x c2xem1x. y2 em1x e2m1x e2m1x dx em1x dx xem1x. y1 em1x y c1em1x c2em2x. y2 em2x. y1 em1x m2 (b 1b2 4ac)2a, m1 (b 1b2 4ac)2a am2 bm c 0. am2emx bmemx cemx 0 or emx(am2 bm c) 0. ay by cy 0, 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS ● 133 A formal derivation of Euler’s formula can be obtained from the Maclaurin series by substituting x iu, using i2 1, i3 i, . . . , and then separating the series into real and imaginary parts. The plausibility thus established, we can adopt cos u i sin u as the definitio of eiu. ex n0 xn n! Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 134 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS where we have used cos(bx) cos bx and sin(bx) sin bx. Note that by ﬁrst adding and then subtracting the two equations in (7), we obtain, respectively, Since y C1e(aib)x C2e(aib)x is a solution of (2) for any choice of the constants C1 and C2, the choices C1 C2 1 and C1 1, C2 1 give, in turn, two solutions: But and Hence from Corollary (A) of Theorem 4.1.2 the last two results show that eax cos bx and eax sin bx are real solutions of (2). Moreover, these solutions form a fundamen-tal set on ( , ). Consequently, the general solution is (8) y c1eax cos x c2eax sin x eax(c1 cos x c2 sin x). y2 eax(eix eix) 2ieax sin x. y1 eax(eix eix) 2eax cos x y1 e(ai)x e(ai)x and y2 e(ai)x e(ai)x. eix eix 2 cos x and eix eix 2i sin x. FIGURE 4.3.1 Solution curve of IVP in Example 2 x y 4 5 _4 _2 2 _3 _1 1 3 4 _2 _3 2 1 _1 3 EXAMPLE 1 Second-Order DEs Solve the following differential equations. (a) 2y 5y 3y 0 (b) y 10y 25y 0 (c) y 4y 7y 0 SOLUTION We give the auxiliary equations, the roots, and the corresponding gen-eral solutions. (a) 2m2 5m 3 (2m 1)(m 3) 0, , m2 3 From (4), y c1ex/2 c2e3x. (b) m2 10m 25 (m 5)2 0, m1 m2 5 From (6), y c1e5x c2xe5x. (c) From (8) with 2, 23, y e2x(c1 cos 23x c2 sin 23x). m2 4m 7 0, m1 2 23i, m2 2 23i m1 1 2 EXAMPLE 2 An Initial-Value Problem Solve 4y 4y 17y 0, y(0) 1, y(0) 2. SOLUTION By the quadratic formula we ﬁnd that the roots of the auxiliary equation 4m2 4m 17 0 are . Thus from (8) we have y ex/2(c1 cos 2x c2 sin 2x). Applying the condition y(0) 1, we see from e0(c1 cos 0 c2 sin 0) 1 that c1 1. Differentiating y ex/2(cos 2x c2 sin 2x) and then using y(0) 2 gives 2c2 2 or c2 . Hence the solution of the IVP is y ex/2(cos 2x sin 2x). In Figure 4.3.1 we see that the solution is oscillatory, but y : 0 as x : . Two Equations Worth Knowing The two differential equations , y k2y 0 and y k2y 0 3 4 3 4 1 2 m1 1 2 2i and m2 1 2 2i Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. where k is real, are important in applied mathematics. For y k2y 0 the auxiliary equation m2 k2 0 has imaginary roots m1 ki and m2 ki. With a 0 and b k in (8) the general solution of the DE is seen to be (9) On the other hand, the auxiliary equation m2 k2 0 for y k2y 0 has distinct real roots m1 k and m2 k, and so by (4) the general solution of the DE is (10) Notice that if we choose in (10), we get the particu-lar solutions . Since cosh kx and sinh kx are linearly independent on any interval of the x-axis, an alterna-tive form for the general solution of y k2y 0 is (11) See Problems 41 and 42 in Exercises 4.3. Higher-Order Equations In general, to solve an nth-order differential equa-tion (1), where the ai, i 0, 1, . . . , n are real constants, we must solve an nth-degree polynomial equation (12) If all the roots of (12) are real and distinct, then the general solution of (1) is It is somewhat harder to summarize the analogues of Cases II and III because the roots of an auxiliary equation of degree greater than two can occur in many combi-nations. For example, a ﬁfth-degree equation could have ﬁve distinct real roots, or three distinct real and two complex roots, or one real and four complex roots, or ﬁve real but equal roots, or ﬁve real roots but two of them equal, and so on. When m1 is a root of multiplicity k of an nth-degree auxiliary equation (that is, k roots are equal to m1), it can be shown that the linearly independent solutions are and the general solution must contain the linear combination Finally, it should be remembered that when the coefﬁcients are real, complex roots of an auxiliary equation always appear in conjugate pairs. Thus, for example, a cubic polynomial equation can have at most two complex roots. c1em1x c2xem1x c3x2em1x ckxk1em1x. em1x, xem1x, x2em1x, . . . , xk1em1x y c1em1x c2em2x cnemnx. anmn an1mn1 a2m2 a1m a0 0. y c1 cosh kx c2 sinh kx. y 1 2 (ekx ekx) cosh kx and y 1 2 (ekx ekx) sinh kx c1 c2 1 2 and c1 1 2, c2 1 2 y c1ekx c2ekx. y c1 cos kx c2 sin kx. 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS ● 135 EXAMPLE 3 Third-Order DE Solve y 3y 4y 0. SOLUTION It should be apparent from inspection of m3 3m2 4 0 that one root is m1 1, so m 1 is a factor of m3 3m2 4. By division we ﬁnd so the other roots are m2 m3 2. Thus the general solution of the DE is y c1ex c2e2x c3xe2x. m3 3m2 4 (m 1)(m2 4m 4) (m 1)(m 2)2, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 136 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 4 Fourth-Order DE Solve SOLUTION The auxiliary equation m4 2m2 1 (m2 1)2 0 has roots m1 m3 i and m2 m4 i. Thus from Case II the solution is By Euler’s formula the grouping C1eix C2eix can be rewritten as after a relabeling of constants. Similarly, x(C3eix C4eix) can be expressed as x(c3 cos x c4 sin x). Hence the general solution is Example 4 illustrates a special case when the auxiliary equation has repeated complex roots. In general, if m1 a ib, b 0 is a complex root of multiplicity k of an auxiliary equation with real coefﬁcients, then its conjugate m2 a ib is also a root of multiplicity k. From the 2k complex-valued solutions we conclude, with the aid of Euler’s formula, that the general solution of the corre-sponding differential equation must then contain a linear combination of the 2k real linearly independent solutions In Example 4 we identify k 2, a 0, and b 1. Of course the most difﬁcult aspect of solving constant-coefﬁcient differential equa-tions is ﬁnding roots of auxiliary equations of degree greater than two. For example, to solve 3y 5y 10y 4y 0, we must solve 3m3 5m2 10m 4 0. Something we can try is to test the auxiliary equation for rational roots. Recall that if m1 pq is a rational root (expressed in lowest terms) of an auxiliary equation with integer coefﬁcients, then p is a factor of a0 and q is a factor of an. For our speciﬁc cubic auxiliary equation, all the factors of a0 4 and an 3 are p: 1, 2, 4 and q: 1, 3, so the possible rational roots are . Each of these numbers can then be tested—say, by synthetic division. In this way we discover both the root and the factorization The quadratic formula then yields the remaining roots m2 1 i and m3 1 i. Therefore the general solution of 3y 5y 10y 4y 0 is y c1ex/3 ex(c2 cos x c3 sin x). Use of Computers Finding roots or approximation of roots of auxiliary equa-tions is a routine problem with an appropriate calculator or computer software. Polynomial equations (in one variable) of degree less than ﬁve can be solved by means of algebraic formulas using the solve commands in Mathematica and Maple. For auxiliary equations of degree ﬁve or greater it might be necessary to resort to nu-merical commands such as NSolve and FindRoot in Mathematica. Because of their capability of solving polynomial equations, it is not surprising that these computer 23 23 23 23 3m3 5m2 10m 4 (m 1 3)(3m2 6m 12). m1 1 3 p>q: 1, 2, 4, 1 3, 2 3, 4 3 anmn a1m a0 0 eax sin x, xeax sin x, x2eax sin x, . . . , xk1eax sin x. eax cos x, xeax cos x, x2eax cos x, . . . , xk1eax cos x, e(ai)x, xe(ai)x, x2e(ai)x, . . . , xk1e(ai)x, e(ai)x, xe(ai)x, x2e(ai)x, . . . , xk1e(ai)x, y c1 cos x c2 sin x c3x cos x c4x sin x. c1 cos x c2 sin x y C1eix C2eix C3xeix C4xeix. d 4y dx4 2 d 2y dx2 y 0. There is more on this in the SRM. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. algebra systems are also able, by means of their dsolve commands, to provide explicit solutions of homogeneous linear constant-coefﬁcient differential equations. In the classic text Differential Equations by Ralph Palmer Agnew (used by the author as a student) the following statement is made: It is not reasonable to expect students in this course to have computing skill and equipment necessary for efficient solving of equations such a (13) Although it is debatable whether computing skills have improved in the intervening years, it is a certainty that technology has. If one has access to a computer algebra sys-tem, equation (13) could now be considered reasonable. After simpliﬁcation and some relabeling of output, Mathematica yields the (approximate) general solution Finally, if we are faced with an initial-value problem consisting of, say, a fourth-order equation, then to ﬁt the general solution of the DE to the four initial conditions, we must solve four linear equations in four unknowns (the c1, c2, c3, c4 in the general solution). Using a CAS to solve the system can save lots of time. See Problems 69 and 70 in Exercises 4.3 and Problem 41 in Chapter 4 in Review. c3e0.476478x cos(0.759081x) c4e0.476478x sin(0.759081x). y c1e0.728852x cos(0.618605x) c2e0.728852x sin(0.618605x) 4.317 d 4y dx4 2.179 d 3y dx3 1.416 d 2y dx2 1.295 dy dx 3.169y 0. 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS ● 137 McGraw-Hill, New York, 1960. EXERCISES 4.3 Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1–14 ﬁnd the general solution of the given second-order differential equation. 1. 4y y 0 2. y 36y 0 3. y y 6y 0 4. y 3y 2y 0 5. y 8y 16y 0 6. y 10y 25y 0 7. 12y 5y 2y 0 8. y 4y y 0 9. y 9y 0 10. 3y y 0 11. y 4y 5y 0 12. 2y 2y y 0 13. 3y 2y y 0 14. 2y 3y 4y 0 In Problems 15–28 ﬁnd the general solution of the given higher-order differential equation. 15. y 4y 5y 0 16. y y 0 17. y 5y 3y 9y 0 18. y 3y 4y 12y 0 19. d3u dt3 d 2u dt2 2u 0 20. 21. y 3y 3y y 0 22. y 6y 12y 8y 0 23. y(4) y y 0 24. y(4) 2y y 0 25. 26. 27. 28. In Problems 29–36 solve the given initial-value problem. 29. y 16y 0, y(0) 2, y(0) 2 30. d 2y d2 y 0, y(p>3) 0, y(p>3) 2 2 d 5x ds5 7 d 4x ds4 12 d 3x ds3 8 d 2x ds2 0 d 5u dr5 5 d 4u dr4 2 d 3u dr3 10 d 2u dr2 du dr 5u 0 d 4y dx4 7 d 2y dx2 18y 0 16 d 4y dx4 24 d 2y dx2 9y 0 d 3x dt3 d 2x dt2 4x 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 138 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 31. 32. 4y 4y 3y 0, y(0) 1, y(0) 5 33. y y 2y 0, y(0) y(0) 0 34. y 2y y 0, y(0) 5, y(0) 10 35. y 12y 36y 0, y(0) 0, y(0) 1, y (0) 7 36. y 2y 5y 6y 0, y(0) y(0) 0, y (0) 1 In Problems 37–40 solve the given boundary-value problem. 37. y 10y 25y 0, y(0) 1, y(1) 0 38. y 4y 0, y(0) 0, y(p) 0 39. 40. y 2y 2y 0, y(0) 1, y(p) 1 In Problems 41 and 42 solve the given problem ﬁrst using the form of the general solution given in (10). Solve again, this time using the form given in (11). 41. y 3y 0, y(0) 1, y(0) 5 42. y y 0, y(0) 1, y(1) 0 In Problems 43–48 each ﬁgure represents the graph of a particular solution of one of the following differential equations: (a) y 3y 4y 0 (b) y 4y 0 (c) y 2y y 0 (d) y y 0 (e) y 2y 2y 0 (f) y 3y 2y 0 Match a solution curve with one of the differential equations. Explain your reasoning. y y 0, y(0) 0, y(p>2) 0 d 2y dt2 4 dy dt 5y 0, y(1) 0, y(1) 2 In Problems 49–58 ﬁnd a homogeneous linear differential equa-tion with constant coefﬁcients whose general solution is given. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. y c1cos x c2sin x c3cos2 x c4sin2x y c1 c2x c3e8x y c1 c2e2xcos5x c3e2xsin5x y c1excosx c2exsinx y c1 cosh7x c2 sinh7x y c1 cos3x c2 sin3x y c1e10x c2xe10x y c1 c2e2x y c1e4x c2e3x y c1ex c2e5x x y x y FIGURE 4.3.2 Graph for Problem 43 FIGURE 4.3.3 Graph for Problem 44 43. 44. x y FIGURE 4.3.4 Graph for Problem 45 45. x y FIGURE 4.3.5 Graph for Problem 46 46. π x y π x y FIGURE 4.3.6 Graph for Problem 47 FIGURE 4.3.7 Graph for Problem 48 47. 48. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Discussion Problems 59. Two roots of a cubic auxiliary equation with real coefﬁ-cients are and What is the corre-sponding homogeneous linear differential equation? Discuss: Is your answer unique? 60. Find the general solution of if is one root of its auxiliary equation. 61. Find the general solution of y 6y y 34y 0 if it is known that y1 e4x cos x is one solution. 62. To solve y(4) y 0, we must ﬁnd the roots of m4 1 0. This is a trivial problem using a CAS but can also be done by hand working with complex numbers. Observe that m4 1 (m2 1)2 2m2. How does this help? Solve the differential equation. 63. Verify that is a particular solution of y(4) y 0. Reconcile this particular solu-tion with the general solution of the DE. 64. Consider the boundary-value problem y ly 0, y(0) 0, y(p2) 0. Discuss: Is it possible to determine values of l so that the problem possesses (a) trivial solutions? (b) nontrivial solutions? y sinh x 2 cos (x p>6) m1 1 2 2y 7y 4y 4y 0 m2 3 i. m1 1 2 4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH ● 139 Computer Lab Assignments In Problems 65–68 use a computer either as an aid in solving the auxiliary equation or as a means of directly obtaining the general solution of the given differential equation. If you use a CAS to obtain the general solution, simplify the output and, if necessary, write the solution in terms of real functions. 65. y 6y 2y y 0 66. 6.11y 8.59y 7.93y 0.778y 0 67. 3.15y(4) 5.34y 6.33y 2.03y 0 68. y(4) 2y y 2y 0 In Problems 69 and 70 use a CAS as an aid in solving the aux-iliary equation. Form the general solution of the differential equation. Then use a CAS as an aid in solving the system of equations for the coefﬁcients ci, i 1, 2, 3, 4 that results when the initial conditions are applied to the general solution. 69. 2y(4) 3y 16y 15y 4y 0, y(0) 2, y(0) 6, y (0) 3, y(0) 70. y(4) 3y 3y y 0, y(0) y(0) 0, y (0) y(0) 1 1 2 Note to the Instructor: In this section the method of undetermined coefﬁcients is developed from the viewpoint of the superposition principle for nonhomogeneous equations (Theorem 4.7.1). In Section 4.5 an entirely different approach will be presented, one utilizing the concept of differential annihilator operators. Take your pick. UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH REVIEW MATERIAL ●Review Theorems 4.1.6 and 4.1.7 (Section 4.1) INTRODUCTION To solve a nonhomogeneous linear differential equation (1) we must do two things: • ﬁnd the complementary function yc and • ﬁnd any particular solution yp of the nonhomogeneous equation (1). Then, as was discussed in Section 4.1, the general solution of (1) is y yc yp. The complemen-tary function yc is the general solution of the associated homogeneous DE of (1), that is, . In Section 4.3 we saw how to solve these kinds of equations when the coefﬁcients were constants. Our goal in the present section is to develop a method for obtaining particular solutions. an y(n) an1y(n1) a1 y a0 y 0 an y(n) an1 y(n1) a1 y a0y g(x), 4.4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 140 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS Method of Undetermined Coefficient The ﬁrst of two ways we shall consider for obtaining a particular solution yp for a nonhomogeneous linear DE is called the method of undetermined coefficients The underlying idea behind this method is a conjecture about the form of yp, an educated guess really, that is motivated by the kinds of functions that make up the input function g(x). The general method is limited to linear DEs such as (1) where • the coefﬁcients ai, i 0, 1, . . . , n are constants and • g(x) is a constant k, a polynomial function, an exponential function eax, a sine or cosine function sin bx or cos bx, or ﬁnite sums and products of these functions. Note Strictly speaking, g(x) k (constant) is a polynomial function. Since a constant function is probably not the ﬁrst thing that comes to mind when you think of polynomial functions, for emphasis we shall continue to use the redundancy “constant functions, polynomials, . . . . ” The following functions are some examples of the types of inputs g(x) that are appropriate for this discussion: That is, g(x) is a linear combination of functions of the type g(x) sin 3x 5x cos 2x, g(x) xex sin x (3x2 1)e4x. g(x) 10, g(x) x2 5x, g(x) 15x 6 8ex, P(x) an xn an1 xn1 a1x a0, P(x) eax, P(x) eax sin x, and P(x) eax cos x, where n is a nonnegative integer and a and b are real numbers. The method of undetermined coefﬁcients is not applicable to equations of form (1) when and so on. Differential equations in which the input g(x) is a function of this last kind will be considered in Section 4.6. The set of functions that consists of constants, polynomials, exponentials eax, sines, and cosines has the remarkable property that derivatives of their sums and products are again sums and products of constants, polynomials, exponen-tials eax, sines, and cosines. Because the linear combination of derivatives must be identical to g(x), it seems reasonable to assume that yp has the same form as g(x). The next two examples illustrate the basic method. an y(n) p an1 yp (n1) a1 yp a0 yp g(x) ln x, g(x) 1 x , g(x) tan x, g(x) sin1x, EXAMPLE 1 General Solution Using Undetermined Coefficient Solve (2) SOLUTION Step 1. We ﬁrst solve the associated homogeneous equation y 4y 2y 0. From the quadratic formula we ﬁnd that the roots of the auxil-iary equation m2 4m 2 0 are and . Hence the complementary function is Step 2. Now, because the function g(x) is a quadratic polynomial, let us assume a particular solution that is also in the form of a quadratic polynomial: yp Ax2 Bx C. yc c1e(216)x c2e(216)x. m2 2 16 m1 2 16 y 4y 2y 2x2 3x 6. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. We seek to determine specifi coefﬁcients A, B, and C for which yp is a solution of (2). Substituting yp and the derivatives into the given differential equation (2), we get Because the last equation is supposed to be an identity, the coefﬁcients of like powers of x must be equal: That is, Solving this system of equations leads to the values A 1, , and C 9. Thus a particular solution is Step 3. The general solution of the given equation is y yc yp c1e(216)x c2e(216)x x2 5 2 x 9. yp x2 5 2 x 9. B 5 2 2A 2, 8A 2B 3, 2A 4B 2C 6. equal 2A x2 8A 2B x 2A 4B 2C 2x2 3x 6 y p 4y p 2yp 2A 8Ax 4B 2Ax2 2Bx 2C 2x2 3x 6. y p 2Ax B and y p 2A 4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH ● 141 EXAMPLE 2 Particular Solution Using Undetermined Coefficient Find a particular solution of y y y 2 sin 3x. SOLUTION A natural ﬁrst guess for a particular solution would be A sin 3x. But because successive differentiations of sin 3x produce sin 3x and cos 3x, we are prompted instead to assume a particular solution that includes both of these terms: Differentiating yp and substituting the results into the differential equation gives, after regrouping, or From the resulting system of equations, we get and . A particular solution of the equation is As we mentioned, the form that we assume for the particular solution yp is an educated guess; it is not a blind guess. This educated guess must take into consider-ation not only the types of functions that make up g(x) but also, as we shall see in Example 4, the functions that make up the complementary function yc. yp 6 73 cos 3x 16 73 sin 3x. B 16 73 A 6 73 8A 3B 0, 3A 8B 2, equal 8A 3B cos 3x 3A 8B sin 3x 0 cos 3x 2 sin 3x. y p y p yp (8A 3B) cos 3x (3A 8B) sin 3x 2 sin 3x yp A cos 3x B sin 3x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 142 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS (4) y p 2y p 3yp 3Ax 2A 3B 3Cxe2x (2C 3E )e2x 4x 5 6xe2x. From this identity we obtain the four equations The last equation in this system results from the interpretation that the coefﬁcient of e2x in the right member of (4) is zero. Solving, we ﬁnd , , C 2, and . Consequently, Step 3. The general solution of the equation is In light of the superposition principle (Theorem 4.1.7) we can also approach Example 3 from the viewpoint of solving two simpler problems. You should verify that substituting and yields, in turn, . A particular solution of (3) is then . The next example illustrates that sometimes the “obvious” assumption for the form of yp is not a correct assumption. yp yp1 yp2 yp1 4 3 x 23 9 and yp2 2x 4 3e2x yp2 Cxe2x Ee2x into y 2y 3y 6xe2x yp1 Ax B into y 2y 3y 4x 5 y c1ex c2e3x 4 3 x 23 9 2x 4 3 e2x. yp 4 3 x 23 9 2xe2x 4 3 e2x. E 4 3 B 23 9 A 4 3 3A 4, 2A 3B 5, 3C 6, 2C 3E 0. EXAMPLE 3 Forming yp by Superposition Solve (3) SOLUTION Step 1. First, the solution of the associated homogeneous equation y 2y 3y 0 is found to be yc c1ex c2e3x. Step 2. Next, the presence of 4x 5 in g(x) suggests that the particular solution includes a linear polynomial. Furthermore, because the derivative of the product xe2x produces 2xe2x and e2x, we also assume that the particular solution includes both xe2x and e2x. In other words, g is the sum of two basic kinds of functions: Correspondingly, the superposition principle for nonhomogeneous equations (Theorem 4.1.7) suggests that we seek a particular solution where . Substituting into the given equation (3) and grouping like terms gives yp Ax B Cxe2x Ee2x yp1 Ax B and yp2 Cxe2x Ee2x yp yp1 yp2, g(x) g1(x) g2(x) polynomial exponentials. y 2y 3y 4x 5 6xe2x. EXAMPLE 4 A Glitch in the Method Find a particular solution of y 5y 4y 8ex. SOLUTION Differentiation of ex produces no new functions. Therefore proceeding as we did in the earlier examples, we can reasonably assume a particular solution of the form yp Aex. But substitution of this expression into the differential equation Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. yields the contradictory statement 0 8ex, so we have clearly made the wrong guess for yp. The difﬁculty here is apparent on examining the complementary function yc c1ex c2e4x. Observe that our assumption Aex is already present in yc. This means that ex is a solution of the associated homogeneous differential equation, and a constant multiple Aex when substituted into the differential equation necessarily produces zero. What then should be the form of yp? Inspired by Case II of Section 4.3, let’s see whether we can ﬁnd a particular solution of the form Substituting and into the differential equation and simplifying gives From the last equality we see that the value of A is now determined as A . Therefore a particular solution of the given equation is The difference in the procedures used in Examples 1–3 and in Example 4 suggests that we consider two cases. The first case reflects the situation in Examples 1–3. Case I No function in the assumed particular solution is a solution of the asso-ciated homogeneous differential equation. In Table 4.4.1 we illustrate some speciﬁc examples of g(x) in (1) along with the corresponding form of the particular solution. We are, of course, taking for granted that no function in the assumed particular solution yp is duplicated by a function in the complementary function yc. yp 8 3xex. 8 3 y p 5y p 4yp 3Aex 8ex. y p Axex 2Aex y p Axex Aex yp Axex. 4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH ● 143 TABLE 4.4.1 Trial Particular Solutions g(x) Form of yp 1. 1 (any constant) A 2. 5x 7 Ax B 3. 3x2 2 Ax2 Bx C 4. x3 x 1 Ax3 Bx2 Cx E 5. sin 4x A cos 4x B sin 4x 6. cos 4x A cos 4x B sin 4x 7. e5x Ae5x 8. (9x 2)e5x (Ax B)e5x 9. x2e5x (Ax2 Bx C)e5x 10. e3x sin 4x Ae3x cos 4x Be3x sin 4x 11. 5x2 sin 4x (Ax2 Bx C) cos 4x (Ex2 Fx G) sin 4x 12. xe3x cos 4x (Ax B)e3x cos 4x (Cx E)e3x sin 4x EXAMPLE 5 Forms of Particular Solutions—Case I Determine the form of a particular solution of (a) y 8y 25y 5x3ex 7ex (b) y 4y x cos x SOLUTION (a) We can write g(x) (5x3 7)ex. Using entry 9 in Table 4.4.1 as a model, we assume a particular solution of the form Note that there is no duplication between the terms in yp and the terms in the comple-mentary function yc e4x(c1 cos 3x c2 sin 3x). yp (Ax3 Bx2 Cx E)ex. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 144 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS (b) The function g(x) x cos x is similar to entry 11 in Table 4.4.1 except, of course, that we use a linear rather than a quadratic polynomial and cos x and sin x instead of cos 4x and sin 4x in the form of yp: Again observe that there is no duplication of terms between yp and yc c1 cos 2x c2 sin 2x. If g(x) consists of a sum of, say, m terms of the kind listed in the table, then (as in Example 3) the assumption for a particular solution yp consists of the sum of the trial forms corresponding to these terms: The foregoing sentence can be put another way. Form Rule for Case I The form of y p is a linear combination of all linearly independent functions that ar e generated by r epeated differentiations of g (x). yp yp1 yp2 ypm. yp1, yp2, . . . , ypm yp (Ax B) cos x (Cx E) sin x. EXAMPLE 6 Forming yp by Superposition—Case I Determine the form of a particular solution of SOLUTION Corresponding to 3x2 we assume Corresponding to 5 sin 2x we assume Corresponding to 7xe6x we assume The assumption for the particular solution is then No term in this assumption duplicates a term in yc c1e2x c2e7x. Case II A function in the assumed particular solution is also a solution of the associated homogeneous differential equation. The next example is similar to Example 4. yp yp1 yp2 yp3 Ax2 Bx C E cos 2x F sin 2x (Gx H)e6x. yp3 (Gx H)e6x. yp2 E cos 2x F sin 2x. yp1 Ax2 Bx C. y 9y 14y 3x2 5 sin 2x 7xe6x. EXAMPLE 7 Particular Solution—Case II Find a particular solution of y 2y y ex. SOLUTION The complementary function is yc c1ex c2xex. As in Example 4, the assumption yp Aex will fail, since it is apparent from yc that ex is a solution of the associated homogeneous equation y 2y y 0. Moreover, we will not be able to ﬁnd a particular solution of the form yp Axex, since the term xex is also duplicated in yc. We next try Substituting into the given differential equation yields 2Aex ex, so Thus a particular solution is yp 1 2x2ex. A 1 2. yp Ax2ex. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Suppose again that g(x) consists of m terms of the kind given in Table 4.4.1, and suppose further that the usual assumption for a particular solution is where the are the trial particular solution forms corresponding to these terms. Under the circumstances described in Case II, we can make up the following general rule. Multiplication Rule for Case II If any contains terms that duplicate terms in yc, then that must be multiplied by xn, where n is the smallest positive integer that eliminates that duplication. ypi ypi ypi , i 1, 2, . . . , m yp yp1 yp2 ypm, 4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH ● 145 EXAMPLE 8 An Initial-Value Problem Solve y y 4x 10 sin x, y(p) 0, y(p) 2. SOLUTION The solution of the associated homogeneous equation y y 0 is yc c1 cos x c2 sin x. Because g(x) 4x 10 sin x is the sum of a linear polynomial and a sine function, our normal assumption for yp, from entries 2 and 5 of Table 4.4.1, would be the sum of and : (5) But there is an obvious duplication of the terms cos x and sin x in this assumed form and two terms in the complementary function. This duplication can be eliminated by simply multiplying by x. Instead of (5) we now use (6) Differentiating this expression and substituting the results into the differential equation gives and so A 4, B 0, 2C 10, and 2E 0. The solutions of the system are immediate: A 4, B 0, C 5, and E 0. Therefore from (6) we obtain yp 4x 5x cos x. The general solution of the given equation is We now apply the prescribed initial conditions to the general solution of the equation. First, y(p) c1 cos p c2 sin p 4p 5p cos p 0 yields c1 9p, since cos p 1 and sin p 0. Next, from the derivative and we ﬁnd c2 7. The solution of the initial-value is then y 9 cos x 7 sin x 4x 5x cos x. y() 9 sin c2 cos 4 5 sin 5 cos 2 y 9 sin x c2 cos x 4 5x sin x 5 cos x y yc yp c1 cos x c2 sin x 4x 5x cos x. y p yp Ax B 2C sin x 2E cos x 4x 10 sin x, yp Ax B Cx cos x Ex sin x. yp2 yp Ax B C cos x E sin x. yp2 C cos x E sin x yp1 Ax B EXAMPLE 9 Using the Multiplication Rule Solve y 6y 9y 6x2 2 12e3x. SOLUTION The complementary function is yc c1e3x c2xe3x. And so, based on entries 3 and 7 of Table 4.4.1, the usual assumption for a particular solution would be yp Ax2 Bx C Ee3x. yp1 yp2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 146 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS Inspection of these functions shows that the one term in is duplicated in yc. If we multiply by x, we note that the term xe3x is still part of yc. But multiplying by x2 eliminates all duplications. Thus the operative form of a particular solution is Differentiating this last form, substituting into the differential equation, and collecting like terms gives yp Ax2 Bx C Ex2e3x. yp2 yp2 yp2 y p 6y p 9yp 9Ax2 (12A 9B)x 2A 6B 9C 2Ee3x 6x2 2 12e3x. It follows from this identity that A , B , C , and E 6. Hence the general solution y yc yp is y c1e3x c2xe3x x2 x 6x2e3x. 2 3 8 9 2 3 2 3 8 9 2 3 EXAMPLE 10 Third-Order DE—Case I Solve y y ex cos x. SOLUTION From the characteristic equation m3 m2 0 we ﬁnd m1 m2 0 and m3 1. Hence the complementary function of the equation is yc c1 c2x c3ex. With g(x) ex cos x, we see from entry 10 of Table 4.4.1 that we should assume that Because there are no functions in yp that duplicate functions in the complementary solution, we proceed in the usual manner. From we get 2A 4B 1 and 4A 2B 0. This system gives and , so a particular solution is . The general solution of the equation is y yc yp c1 c2x c3ex 1 10 ex cos x 1 5 ex sin x. yp 1 10 ex cos x 1 5 ex sin x B 1 5 A 1 10 y p y p (2A 4B)ex cos x (4A 2B)ex sin x ex cos x yp Aex cos x Bex sin x. EXAMPLE 11 Fourth-Order DE—Case II Determine the form of a particular solution of y(4) y 1 x2ex. SOLUTION Comparing yc c1 c2x c3x2 c4ex with our normal assumption for a particular solution we see that the duplications between yc and yp are eliminated when is multiplied by x3 and is multiplied by x. Thus the correct assumption for a particular solution is yp Ax3 Bx3ex Cx2ex Exex. yp2 yp1 yp A Bx2ex Cxex Eex, yp1 yp2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH ● 147 REMARKS (i) In Problems 27–36 in Exercises 4.4 you are asked to solve initial-value problems, and in Problems 37–40 you are asked to solve boundary-value problems. As illustrated in Example 8, be sure to apply the initial conditions or the boundary conditions to the general solution y yc yp. Students often make the mistake of applying these conditions only to the complementary function yc because it is that part of the solution that contains the constants c1, c2, . . . , cn. (ii) From the “Form Rule for Case I” on page 144 of this section you see why the method of undetermined coefﬁcients is not well suited to nonhomogeneous linear DEs when the input function g(x) is something other than one of the four basic types highlighted in color on page 140. For example, if P(x) is a polyno-mial, then continued differentiation of P(x)eax sin bx will generate an indepen-dent set containing only a finit number of functions—all of the same type, namely, a polynomial times eax sin bx or a polynomial times eax cos bx. On the other hand, repeated differentiation of input functions such as g(x) ln x or g(x) tan1x generates an independent set containing an infinit number of functions: derivatives of tan1x: 1 1 x2, 2x (1 x2)2, 2 6x2 (1 x2)3 , . . . . derivatives of ln x: 1 x , 1 x2 , 2 x3, . . . , EXERCISES 4.4 Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1–26 solve the given differential equation by undetermined coefﬁcients. 1. y 3y 2y 6 2. 4y 9y 15 3. y 10y 25y 30x 3 4. y y 6y 2x 5. y y y x2 2x 6. y 8y 20y 100x2 26xex 7. y 3y 48x2e3x 8. 4y 4y 3y cos 2x 9. y y 3 10. y 2y 2x 5 e2x 11. 12. y 16y 2e4x 13. y 4y 3 sin 2x 14. y 4y (x2 3) sin 2x 15. y y 2x sin x y y 1 4 y 3 ex/2 1 4 16. y 5y 2x3 4x2 x 6 17. y 2y 5y ex cos 2x 18. y 2y 2y e2x(cos x 3 sin x) 19. y 2y y sin x 3 cos 2x 20. y 2y 24y 16 (x 2)e4x 21. y 6y 3 cos x 22. y 2y 4y 8y 6xe2x 23. y 3y 3y y x 4ex 24. y y 4y 4y 5 ex e2x 25. y(4) 2y y (x 1)2 26. y(4) y 4x 2xex In Problems 27–36 solve the given initial-value problem. 27. y 4y 2, 28. 2y 3y 2y 14x2 4x 11, y(0) 0, y(0) 0 29. 5y y 6x, y(0) 0, y(0) 10 30. y 4y 4y (3 x)e2x, y(0) 2, y(0) 5 31. y 4y 5y 35e4x, y(0) 3, y(0) 1 y(p>8) 1 2, y(p>8) 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 148 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 32. y y cosh x, y(0) 2, y(0) 12 33. , x(0) 0, x(0) 0 34. , x(0) 0, x(0) 0 35. y 2y y 2 24ex 40e5x, 36. y 8y 2x 5 8e2x, y(0) 5, y(0) 3, y (0) 4 In Problems 37–40 solve the given boundary-value problem. 37. y y x2 1, y(0) 5, y(1) 0 38. y 2y 2y 2x 2, y(0) 0, y(p) p 39. y 3y 6x, y(0) 0, y(1) y(1) 0 40. y 3y 6x, y(0) y(0) 0, y(1) 0 In Problems 41 and 42 solve the given initial-value problem in which the input function g(x) is discontinuous. [Hint: Solve each problem on two intervals, and then ﬁnd a solu-tion so that y and y are continuous at x p2 (Problem 41) and at x p (Problem 42).] 41. y 4y g(x), y(0) 1, y(0) 2, where 42. y 2y 10y g(x), y(0) 0, y(0) 0, where Discussion Problems 43. Consider the differential equation ay by cy ekx, where a, b, c, and k are constants. The auxiliary equation of the associated homogeneous equation is am2 bm c 0. (a) If k is not a root of the auxiliary equation, show that we can ﬁnd a particular solution of the form yp Aekx, where A 1(ak2 bk c). (b) If k is a root of the auxiliary equation of multiplicity one, show that we can ﬁnd a particular solution of the form yp Axekx, where A 1(2ak b). Explain how we know that k b(2a). (c) If k is a root of the auxiliary equation of multiplicity two, show that we can ﬁnd a particular solution of the form y Ax2ekx, where A 1(2a). 44. Discuss how the method of this section can be used to ﬁnd a particular solution of y y sin x cos 2x. Carry out your idea. g(x) 20, 0 x 0, x g(x) sin x, 0 x >2 0, x >2 y (0) 9 2 y(0) 5 2, y(0) 1 2, d 2x dt 2 2x F0 cos t d 2x dt2 2x F0 sin t 45. Without solving, match a solution curve of y y f(x) shown in the ﬁgure with one of the following functions: (i) f(x) 1, (ii) f(x) ex, (iii) f(x) ex, (iv) f(x) sin 2x, (v) f(x) ex sin x, (vi) f(x) sin x. Brieﬂy discuss your reasoning. x y FIGURE 4.4.1 Solution curve FIGURE 4.4.3 Solution curve FIGURE 4.4.4 Solution curve FIGURE 4.4.2 Solution curve x y x y x y (a) (b) (c) (d) Computer Lab Assignments In Problems 46 and 47 ﬁnd a particular solution of the given differential equation. Use a CAS as an aid in carrying out differentiations, simpliﬁcations, and algebra. 46. y 4y 8y (2x2 3x)e2x cos 2x (10x2 x 1)e2x sin 2x 47. y(4) 2y y 2 cos x 3x sin x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.5 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH ● 149 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH REVIEW MATERIAL ●Review Theorems 4.1.6 and 4.1.7 (Section 4.1) INTRODUCTION We saw in Section 4.1 that an nth-order differential equation can be written (1) where Dky dkydxk, k 0, 1, . . . , n. When it suits our purpose, (1) is also written as L(y) g(x), where L denotes the linear nth-order differential, or polynomial, operator (2) Not only is the operator notation a helpful shorthand, but also on a very practical level the application of differential operators enables us to justify the somewhat mind-numbing rules for determining the form of particular solution yp that were presented in the preceding section. In this section there are no special rules; the form of yp follows almost automatically once we have found an appropriate linear differential operator that annihilates g(x) in (1). Before investigating how this is done, we need to examine two concepts. anDn an1Dn1 a1D a0. anDny an1Dn1y a1Dy a0y g(x), 4.5 Factoring Operators When the coefﬁcients ai, i 0, 1, . . . , n are real con-stants, a linear differential operator (1) can be factored whenever the characteristic polynomial anmn an1mn1 a1m a0 factors. In other words, if r1 is a root of the auxiliary equation then L (D r1) P(D), where the polynomial expression P(D) is a linear differential operator of order n 1. For example, if we treat D as an algebraic quantity, then the operator D2 5D 6 can be factored as (D 2)(D 3) or as (D 3)(D 2). Thus if a function y f(x) possesses a second derivative, then This illustrates a general property: Factors of a linear differential operator with constant coefficients commute A differential equation such as y 4y 4y 0 can be written as (D2 4D 4)y 0 or (D 2)(D 2)y 0 or (D 2)2y 0. Annihilator Operator If L is a linear differential operator with constant co-efﬁcients and f is a sufﬁciently differentiable function such that then L is said to be an annihilator of the function. For example, a constant function y k is annihilated by D, since Dk 0. The function y x is annihilated by the differential operator D2 since the ﬁrst and second derivatives of x are 1 and 0, respectively. Similarly, D3x2 0, and so on. L( f (x)) 0, (D2 5D 6)y (D 2)(D 3)y (D 3)(D 2)y. anmn an1mn1 a1m a0 0, The differential operator Dn annihilates each of the functions 1, x, x2, . . . , xn1. (3) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 150 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS As an immediate consequence of (3) and the fact that differentiation can be done term by term, a polynomial (4) can be annihilated by ﬁnding an operator that annihilates the highest power of x. The functions that are annihilated by a linear nth-order differential operator L are simply those functions that can be obtained from the general solution of the homogeneous differential equation L(y) 0. c0 c1x c2x2 cn1x n1 The differential operator (D a)n annihilates each of the functions eax, xeax, x2eax, . . . , xn1eax. (5) To see this, note that the auxiliary equation of the homogeneous equation (D a)ny 0 is (m a)n 0. Since a is a root of multiplicity n, the general solution is (6) y c1eax c2xeax cnxn1eax. EXAMPLE 1 Annihilator Operators Find a differential operator that annihilates the given function. (a) 1 5x2 8x3 (b) e3x (c) 4e2x 10xe2x SOLUTION (a) From (3) we know that D4x3 0, so it follows from (4) that (b) From (5), with a 3 and n 1, we see that (c) From (5) and (6), with a 2 and n 2, we have When a and b, b 0 are real numbers, the quadratic formula reveals that [m2 2am (a2 b2)]n 0 has complex roots a ib, a ib, both of multi-plicity n. From the discussion at the end of Section 4.3 we have the next result. (D 2)2(4e2x 10xe2x) 0. (D 3)e3x 0. D4(1 5x2 8x3) 0. The differential operator [D2 2aD (a2 b2)]n annihilates each of the functions (7) ex cos x, xex cos x, x2ex cos x, . . . , xn1ex cos x, ex sin x, xex sin x, x2ex sin x, . . . , xn1ex sin x. EXAMPLE 2 Annihilator Operator Find a differential operator that annihilates 5ex cos 2x 9ex sin 2x. SOLUTION Inspection of the functions ex cos 2x and ex sin 2x shows that a 1 and b 2. Hence from (7) we conclude that D2 2D 5 will annihilate each function. Since D2 2D 5 is a linear operator, it will annihilate any linear combination of these functions such as 5ex cos 2x 9ex sin 2x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. When a 0 and n 1, a special case of (7) is (8) For example, D2 16 will annihilate any linear combination of sin 4x and cos 4x. We are often interested in annihilating the sum of two or more functions. As we have just seen in Examples 1 and 2, if L is a linear differential operator such that L(y1) 0 and L(y2) 0, then L will annihilate the linear combination c1y1(x) c2y2(x). This is a direct consequence of Theorem 4.1.2. Let us now suppose that L1 and L2 are linear differential operators with constant coefﬁcients such that L1 annihilates y1(x) and L2 annihilates y2(x), but L1(y2) 0 and L2(y1) 0. Then the product of differential operators L1L2 annihilates the sum c1y1(x) c2y2(x). We can easily demonstrate this, using linearity and the fact that L1L2 L2L1: For example, we know from (3) that D2 annihilates 7 x and from (8) that D2 16 annihilates sin 4x. Therefore the product of operators D2(D2 16) will annihilate the linear combination 7 x 6 sin 4x. Note The differential operator that annihilates a function is not unique. We saw in part (b) of Example 1 that D 3 will annihilate e3x, but so will differential operators of higher order as long as D 3 is one of the factors of the op-erator. For example, (D 3)(D 1), (D 3)2, and D3(D 3) all annihilate e3x. (Verify this.) As a matter of course, when we seek a differential annihilator for a function y f(x), we want the operator of lowest possible order that does the job. Undetermined Coefficient This brings us to the point of the preceding dis-cussion. Suppose that L(y) g(x) is a linear differential equation with constant coefﬁcients and that the input g(x) consists of ﬁnite sums and products of the func-tions listed in (3), (5), and (7)—that is, g(x) is a linear combination of functions of the form where m is a nonnegative integer and a and b are real numbers. We now know that such a function g(x) can be annihilated by a differential operator L1 of lowest order, consisting of a product of the operators Dn, (D a)n, and (D2 2aD a2 b2)n. Applying L1 to both sides of the equation L(y) g(x) yields L1L(y) L1(g(x)) 0. By solving the homogeneous higher-order equation L1L(y) 0, we can discover the form of a particular solution yp for the original nonhomogeneous equation L(y) g(x). We then substitute this assumed form into L(y) g(x) to ﬁnd an explicit particular solution. This procedure for determining yp, called the method of undetermined coefficients is illustrated in the next several examples. Before proceeding, recall that the general solution of a nonhomogeneous linear differential equation L(y) g(x) is y yc yp, where yc is the comple-mentary function—that is, the general solution of the associated homogeneous equation L(y) 0. The general solution of each equation L(y) g(x) is defined on the interval ( , ). k (constant), xm, xmex, xmex cos x, and xmex sin x, L1L2(y1 y2) L1L2(y1) L1L2(y2) L2L1(y1) L1L2(y2) L2[L1(y1)] L1[L2(y2)] 0. zero zero (D2 2) cos x sin x 0. 4.5 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH ● 151 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 152 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 General Solution Using Undetermined Coefficient Solve (9) SOLUTION Step 1. First, we solve the homogeneous equation y 3y 2y 0. Then, from the auxiliary equation m2 3m 2 (m 1)(m 2) 0 we ﬁnd m1 1 and m2 2, and so the complementary function is yc c1ex c2e2x. Step 2. Now, since 4x2 is annihilated by the differential operator D3, we see that D3(D2 3D 2)y 4D3x2 is the same as D3(D2 3D 2)y 0. (10) The auxiliary equation of the ﬁfth-order equation in (10), m3(m2 3m 2) 0 or m3(m 1)(m 2) 0, has roots m1 m2 m3 0, m4 1, and m5 2. Thus its general solution must be y c1 c2x c3x2 . (11) The terms in the shaded box in (11) constitute the complementary function of the original equation (9). We can then argue that a particular solution yp of (9) should also satisfy equation (10). This means that the terms remaining in (11) must be the basic form of yp: (12) where, for convenience, we have replaced c1, c2, and c3 by A, B, and C, respectively. For (12) to be a particular solution of (9), it is necessary to ﬁnd specifi coefﬁcients A, B, and C. Differentiating (12), we have and substitution into (9) then gives Because the last equation is supposed to be an identity, the coefﬁcients of like powers of x must be equal: That is (13) Solving the equations in (13) gives A 7, B 6, and C 2. Thus yp 7 6x 2x2. Step 3. The general solution of the equation in (9) is y yc yp or y c1ex c2e2x 7 6x 2x2. 2C 4, 2B 6C 0, 2A 3B 2C 0. equal 2C x2 2B 6C x 2A 3B 2C 4x2 0x 0. y p 3y p 2yp 2C 3B 6Cx 2A 2Bx 2Cx2 4x2. y p B 2Cx, y p 2C, yp A Bx Cx2, c4ex c5e2x y 3y 2y 4x2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.5 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH ● 153 EXAMPLE 4 General Solution Using Undetermined Coefficient Solve (14) SOLUTION Step 1. The auxiliary equation for the associated homogeneous equa-tion y 3y 0 is m2 3m m(m 3) 0, so yc c1 c2e3x. Step 2. Now, since (D 3)e3x 0 and (D2 1) sin x 0, we apply the differen-tial operator (D 3)(D2 1) to both sides of (14): (15) The auxiliary equation of (15) is Thus y After excluding the linear combination of terms in the box that corresponds to yc, we arrive at the form of yp: Substituting yp in (14) and simplifying yield Equating coefﬁcients gives 3A 8, B 3C 0, and 3B C 4. We ﬁnd , , and , and consequently, Step 3. The general solution of (14) is then y c1 c2e3x 8 3 xe3x 6 5 cos x 2 5 sin x. yp 8 3 xe3x 6 5 cos x 2 5 sin x. C 2 5 B 6 5 A 8 3 y p 3y p 3Ae3x (B 3C) cos x (3B C) sin x 8e3x 4 sin x. yp Axe3x B cos x C sin x. c3xe3x c4 cos x c5 sin x. c1 c2e3x (m 3)(m2 1)(m2 3m) 0 or m(m 3)2(m2 1) 0. (D 3)(D2 1)(D2 3D)y 0. y 3y 8e3x 4 sin x. EXAMPLE 5 General Solution Using Undetermined Coefficient Solve (16) SOLUTION The complementary function is yc c1 cos x c2 sin x. Now by com-paring cos x and x cos x with the functions in the ﬁrst row of (7), we see that a 0 and n 1, and so (D2 1)2 is an annihilator for the right-hand member of the equa-tion in (16). Applying this operator to the differential equation gives Since i and i are both complex roots of multiplicity 3 of the auxiliary equation of the last differential equation, we conclude that y We substitute into (16) and simplify: x cos x cos x. y p yp 4 Ex cos x 4 Cx sin x (2B 2C) cos x (2A 2E) sin x yp Ax cos x Bx sin x Cx2 cos x Ex2 sin x c3x cos x c4x sin x c5x2 cos x c6x2 sin x. c1 cos x c2 sin x (D2 1)2 (D2 1)y 0 or (D2 1)3y 0. y y x cos x cos x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 154 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS y c1 c2x c3x2 c4x3 c5e2x c6xe2x c7x2e2x c8x3e2x c9x4e2x c10e5x. (20) Because the linear combination corresponds to the complemen-tary function of (19), the remaining terms in (20) give the form of a particular solu-tion of the differential equation: Summary of the Method For your convenience the method of undeter-mined coefﬁcients is summarized as follows. yp Ax Bx2 Cx3 Ex2e2x Fx3e2x Gx4e2x He5x. c1 c5e2x c6xe2x EXAMPLE 6 Form of a Particular Solution Determine the form of a particular solution for (17) SOLUTION The complementary function for the given equation is yc c1ex c2xex. Now from (7), with a 2, b 1, and n 1, we know that Applying the operator D2 4D 5 to (17) gives (18) Since the roots of the auxiliary equation of (18) are 2 i, 2 i, 1, and 1, we see from y that a particular solution of (17) can be found with the form yp Ae2x cos x Be2x sin x. c3e2x cos x c4e2x sin x c1ex c2xex (D2 4D 5)(D2 2D 1)y 0. (D2 4D 5)e2x cos x 0. y 2y y 10e2x cos x. Equating coefﬁcients gives the equations 4E 1, 4C 0, 2B 2C 1, and 2A 2E 0, from which we ﬁnd , C 0, and . Hence the general solution of (16) is . y c1 cos x c2 sin x 1 4 x cos x 1 2 x sin x 1 4 x2 sin x E 1 4 A 1 4, B 1 2 EXAMPLE 7 Form of a Particular Solution Determine the form of a particular solution for (19) SOLUTION Observe that Therefore D3(D 2)3(D 5) applied to (19) gives or The roots of the auxiliary equation for the last differential equation are easily seen to be 0, 0, 0, 0, 2, 2, 2, 2, 2, and 5. Hence D4(D 2)5(D 5)y 0. D3(D 2)3(D 5)(D3 4D2 4D)y 0 D3(5x2 6x) 0, (D 2)3x2e2x 0, and (D 5)e5x 0. y 4y 4y 5x2 6x 4x2e2x 3e5x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.5 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH ● 155 UNDETERMINED COEFFICIENTS—ANNIHILATOR APPROACH The differential equation L(y) g(x) has constant coefﬁcients, and the func-tion g(x) consists of ﬁnite sums and products of constants, polynomials, expo-nential functions eax, sines, and cosines. (i) Find the complementary solution yc for the homogeneous equation L(y) 0. (ii) Operate on both sides of the nonhomogeneous equation L(y) g(x) with a differential operator L1 that annihilates the function g(x). (iii) Find the general solution of the higher-order homogeneous differential equation L1L(y) 0. (iv) Delete from the solution in step (iii) all those terms that are duplicated in the complementary solution yc found in step (i). Form a linear combination yp of the terms that remain. This is the form of a particular solution of L(y) g(x). (v) Substitute yp found in step (iv) into L(y) g(x). Match coefﬁcients of the various functions on each side of the equality, and solve the resulting system of equations for the unknown coefﬁcients in yp. (vi) With the particular solution found in step (v), form the general solution y yc yp of the given differential equation. REMARKS The method of undetermined coefﬁcients is not applicable to linear differential equations with variable coefﬁcients nor is it applicable to linear equations with constant coefﬁcients when g(x) is a function such as and so on. Differential equations in which the input g(x) is a function of this last kind will be considered in the next section. g(x) ln x, g(x) 1 x, g(x) tan x, g(x) sin1 x, EXERCISES 4.5 Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1–10 write the given differential equation in the form L(y) g(x), where L is a linear differential opera-tor with constant coefﬁcients. If possible, factor L. 1. 9y 4y sin x 2. y 5y x2 2x 3. y 4y 12y x 6 4. 2y 3y 2y 1 5. y 10y 25y ex 6. y 4y ex cos 2x 7. y 2y 13y 10y xex 8. y 4y 3y x2 cos x 3x 9. y(4) 8y 4 10. y(4) 8y 16y (x3 2x)e4x In Problems 11–14 verify that the given differential operator annihilates the indicated functions. 11. D4; y 10x3 2x 12. 2D 1; y 4ex/2 13. (D 2)(D 5); y e2x 3e5x 14. D2 64; y 2 cos 8x 5 sin 8x In Problems 15–26 ﬁnd a linear differential operator that annihilates the given function. 15. 1 6x 2x3 16. x3(1 5x) 17. 1 7e2x 18. x 3xe6x 19. cos 2x 20. 1 sin x 21. 13x 9x2 sin 4x 22. 8x sin x 10 cos 5x 23. ex 2xex x2ex 24. (2 ex)2 25. 3 ex cos 2x 26. ex sin x e2x cos x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 156 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS In Problems 27–34 ﬁnd linearly independent functions that are annihilated by the given differential operator. 27. D5 28. D2 4D 29. (D 6)(2D 3) 30. D2 9D 36 31. D2 5 32. D2 6D 10 33. D3 10D2 25D 34. D2(D 5)(D 7) In Problems 35–64 solve the given differential equation by undetermined coefﬁcients. 35. y 9y 54 36. 2y 7y 5y 29 37. y y 3 38. y 2y y 10 39. y 4y 4y 2x 6 40. y 3y 4x 5 41. y y 8x2 42. y 2y y x3 4x 43. y y 12y e4x 44. y 2y 2y 5e6x 45. y 2y 3y 4ex 9 46. y 6y 8y 3e2x 2x 47. y 25y 6 sin x 48. y 4y 4 cos x 3 sin x 8 49. y 6y 9y xe4x 50. y 3y 10y x(ex 1) 51. y y x2ex 5 52. y 2y y x2ex 53. y 2y 5y ex sin x 54. y y 1 4 y ex(sin 3x cos 3x) 55. y 25y 20 sin 5x 56. y y 4 cos x sin x 57. y y y x sin x 58. y 4y cos2x 59. y 8y 6x2 9x 2 60. y y y y xex ex 7 61. y 3y 3y y ex x 16 62. 2y 3y 3y 2y (ex ex)2 63. y(4) 2y y ex 1 64. y(4) 4y 5x2 e2x In Problems 65–72 solve the given initial-value problem. 65. y 64y 16, y(0) 1, y(0) 0 66. y y x, y(0) 1, y(0) 0 67. y 5y x 2, y(0) 0, y(0) 2 68. y 5y 6y 10e2x, y(0) 1, y(0) 1 69. y y 8 cos 2x 4 sin x, 70. y 2y y xex 5, y(0) 2, y(0) 2, y (0) 1 71. y 4y 8y x3, y(0) 2, y(0) 4 72. y(4) y x ex, y(0) 0, y(0) 0, y (0) 0, y(0) 0 Discussion Problems 73. Suppose L is a linear differential operator that factors but has variable coefﬁcients. Do the factors of L com-mute? Defend your answer. y(p>2) 1, y(p>2) 0 VARIATION OF PARAMETERS REVIEW MATERIAL ●Basic integration formulas and techniques from calculus ●Review Section 2.3 INTRODUCTION We pointed out in the discussions in Sections 4.4 and 4.5 that the method of undetermined coefﬁcients has two inherent weaknesses that limit its wider application to linear equations: The DE must have constant coefﬁcients and the input function must be of the type listed in Table 4.4.1. In this section we examine a method for determining a particular solution of a nonhomogeneous linear DE that has, in theory, no such restrictions on it. This method, due to the eminent astronomer and mathematician Joseph Louis Lagrange (1736–1813), is known as varia-tion of parameters. Before examining this powerful method for higher-order equations we revisit the solution of lin-ear ﬁrst-order differential equations that have been put into standard form. The discussion under the ﬁrst heading in this section is optional and is intended to motivate the main discussion of this section that starts under the second heading. If pressed for time this motivational material could be assigned for reading. yp g(x) 4.6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Linear First-Order DEs Revisited In Section 2.3 we saw that the general so-lution of a linear ﬁrst-order differential equation can be found by ﬁrst rewriting it in the standard form (1) and assuming that are continuous on an common interval I. Using the in-tegrating factor method, the general solution of (1) on the interval I, was found to be The foregoing solution has the same form as that given in Theorem 4.1.6, namely, In this case is a solution of the associated homogeneous equation (2) and (3) is a particular solution of the nonhomogeneous equation (1). As a means of moti-vating a method for solving nonhomogeneous linear equations of higher-order we propose to rederive the particular solution (3) by a method known as variation of parameters. Suppose that is a known solution of the homogeneous equation (2), that is, (4) It is easily shown that is a solution of (4) and because the equation is linear, is its general solution. Variation of parameters consists of ﬁnding a par-ticular solution of (1) of the form In other words, we have replaced the parameter by a function Substituting into (1) and using the Product Rule gives 0 because of (4) 4 so By separating variables and integrating, we ﬁnd Hence the sought-after particular solution is From the fact that we see the last result is identical to (3). y1 eP(x)dx yp u1y1 y1 f(x) y1(x)dx. du1 f(x) y1(x)dx yields u1 f (x) y1(x) dx. u1: y1 du1 dx f (x). u1 dy1 dx P(x)y1 y1 du1 dx f (x) u1 dy1 dx y1 du1 dx P(x)u1y1 f(x) d dx [u1y1] P(x)u1y1 f(x) yp u1y1 u1. c1 yp u1(x)y1(x). c1y1(x) y1 eP(x)dx dy1 dx P(x)y1 0. y1 yp eP(x)dxeP(x)dxf (x) dx dy dx P(x)y 0 yc c1eP(x)dx y yc yp. y c1eP(x)dx eP(x)dxeP(x)dxf(x)dx. P(x) and f(x) dy dx P(x)y f(x) a1(x)y a0(x)y g(x) 4.6 VARIATION OF PARAMETERS ● 157 The basic procedure is that used in Section 4.2. See (4) of Section 2.3. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 158 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS Linear Second-Order DEs Next we consider the case of a linear second-order equation (5) although, as we shall see, variation of parameters extends to higher-order equations. The method again begins by putting (5) into the standard form (6) by dividing by the leading coefﬁcient In (6) we suppose that coefﬁcient func-tions are continuous on some common interval I. As we have already seen in Section 4.3, there is no difﬁculty in obtaining the complementary solution the general solution of the associated homogeneous equation of (6), when the coefﬁcients are constants. Analogous to the preceding dis-cussion, we now ask: Can the parameters and in can be replaced with func-tions and or “variable parameters,” so that (7) is a particular solution of (6)? To answer this question we substitute (7) into (6). Using the Product Rule to differentiate twice, we get Substituting (7) and the foregoing derivatives into (6) and grouping terms yields zero zero 4 4 y p u1y 1 y 1u 1 y1u 1 u 1y 1 u2y 2 y 2u 2 y2u 2 u 2y 2. y p u1y 1 y1u 1 u2y 2 y2u 2 yp y u1(x)y1(x) u2(x)y2(x) u2, u1 yc c2 c1 yc c1y1(x) c2y2(x), P(x), Q(x), and f(x) a2(x). y P(x)y Q(x)y f(x) a2(x)y a1(x)y a0(x)y g(x), (8) d dx [y1u 1 y2u 2] P[y1u 1 y2u 2] y 1u 1 y 2u 2 f (x). d dx [y1u 1] d dx [y2u 2] P[y1u 1 y2u 2] y 1u 1 y 2u 2 y2u 2 u 2y 2 P[y1u 1 y2u 2] y 1u 1 y 2u 2 y p P(x)y p Q(x)yp u1[y 1 Py 1 Qy1] u2[y 2 Py 2 Qy2] y1u 1 u 1y 1 Because we seek to determine two unknown functions u1 and u2, reason dictates that we need two equations. We can obtain these equations by making the further assump-tion that the functions u1 and u2 satisfy This assumption does not come out of the blue but is prompted by the ﬁrst two terms in (8), since if we demand that , then (8) reduces to . We now have our desired two equations, albeit two equations for determining the derivatives and By Cramer’s Rule, the solution of the system can be expressed in terms of determinants: , (9) where . (10) The functions u1 and u2 are found by integrating the results in (9). The determinant W is recognized as the Wronskian of y1 and y2. By linear independence of y1 and y2 on I, we know that W(y1(x), y2(x)) 0 for every x in the interval. W y1 y 1 y2 y 2, W1 0 f (x) y2 y 2, W2 y1 y 1 0 f (x) u 1 W1 W y2 f (x) W and u 2 W2 W y1 f (x) W y 1u 1 y 2u 2 f (x) y1u 1 y2u 2 0 u 2. u 1 y 1u 1 y 2u 2 f (x) y1u 1 y2u 2 0 y1u 1 y2u 2 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.6 VARIATION OF PARAMETERS ● 159 Summary of the Method Usually, it is not a good idea to memorize for-mulas in lieu of understanding a procedure. However, the foregoing procedure is too long and complicated to use each time we wish to solve a differential equation. In this case it is more efﬁcient to simply use the formulas in (9). Thus to solve a2y a1y a0y g(x), ﬁrst ﬁnd the complementary function yc c1y1 c2y2 and then compute the Wronskian W( y1(x), y2(x)). By dividing by a2, we put the equation into the standard form y Py Qy f(x) to determine f(x). We ﬁnd u1 and u2 by integrating and , where W1 and W2 are deﬁned as in (10). A particular solution is yp u1y1 u2y2. The general solution of the equation is then y yc yp. u 2 W2>W u 1 W1>W W1 0 (x 1)e2x xe2x 2xe2x e2x (x 1)xe4x, W2 e2x 2e2x 0 (x 1)e2x (x 1)e4x, EXAMPLE 1 General Solution Using Variation of Parameters Solve y 4y 4y (x 1)e2x. SOLUTION From the auxiliary equation m2 4m 4 (m 2)2 0 we have yc c1e2x c2xe2x. With the identiﬁcations y1 e2x and y2 xe2x, we next com-pute the Wronskian: Since the given differential equation is already in form (6) (that is, the coefﬁcient of y is 1), we identify f(x) (x 1)e2x. From (10) we obtain W(e2x, xe2x) e2x 2e2x xe2x 2xe2x e2x e4x. and so from (9) It follows that and . Hence and y yc yp c1e2x c2xe2x 1 6x3e2x 1 2x2e2x. yp 1 3x3 1 2x2e2x 1 2x2 xxe2x 1 6x3e2x 1 2x2e2x u2 1 2x2 x u1 1 3x3 1 2x2 u 1 (x 1)xe4x e4x x2 x, u 2 (x 1)e4x e4x x 1. EXAMPLE 2 General Solution Using Variation of Parameters Solve 4y 36y csc 3x. SOLUTION We ﬁrst put the equation in the standard form (6) by dividing by 4: Because the roots of the auxiliary equation m2 9 0 are m1 3i and m2 3i, the complementary function is yc c1 cos 3x c2 sin 3x. Using y1 cos 3x, y2 sin 3x, and , we obtain W1 0 1 4 csc 3x sin 3x 3 cos 3x 1 4, W2 cos 3x 3 sin 3x 0 1 4 csc 3x 1 4 cos 3x sin 3x. W(cos 3x, sin 3x) cos 3x 3 sin 3x sin 3x 3 cos 3x 3, f (x) 1 4 csc 3x y 9y 1 4 csc 3x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 160 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS Integrating gives lnsin 3x. Thus a particular solution is The general solution of the equation is yp 1 12 x cos 3x 1 36 (sin 3x) ln sin 3x . u1 1 12x and u2 1 36 u 1 W1 W 1 12 and u 2 W2 W 1 12 cos 3x sin 3x EXAMPLE 3 General Solution Using Variation of Parameters Solve SOLUTION The auxiliary equation m2 1 0 yields m1 1 and m2 1. Therefore yc c1ex c2ex. Now W(ex, ex) 2, and Since the foregoing integrals are nonelementary, we are forced to write and so (12) In Example 3 we can integrate on any interval [x0, x] that does not contain the origin. We will solve the equation in Example 3 by an alternative method in Section 4.8. Higher-Order Equations The method that we have just examined for non homogeneous second-order differential equations can be generalized to linear nth-order equations that have been put into the standard form (13) If yc c1y1 c2y2 cnyn is the complementary function for (13), then a particular solution is yp u1(x)y1(x) u2(x)y2(x) un(x)yn(x), y(n) Pn1(x)y(n1) P1(x)y P0(x)y f (x). y yc yp c1ex c2ex 1 2 ex x x0 et t dt 1 2 ex x x0 et t dt. yp 1 2 ex x x0 et t dt 1 2 ex x x0 et t dt, u 2 ex(1>x) 2 , u2 1 2 x x0 et t dt. u 1 ex(1>x) 2 , u1 1 2 x x0 et t dt, y y 1 x. (11) y ycyp c1 cos 3x c2 sin 3x 1 12 x cos 3x 1 36 (sin 3x) ln sin 3x . Equation (11) represents the general solution of the differential equation on, say, the interval (0, p6). Constants of Integration When computing the indeﬁnite integrals of and , we need not introduce any constants. This is because C1y1 C2y2 u1y1 u2y2. (c1 a1)y1 (c2 b1)y2 u1y1 u2y2 y yc yp c1y1 c2y2 (u1 a1)y1 (u2 b1)y2 u 2 u 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. where the , k 1, 2, . . . , n are determined by the n equations (14) The ﬁrst n 1 equations in this system, like in (8), are assumptions that are made to simplify the resulting equation after yp u1(x)y1(x) un(x)yn(x) is substituted in (13). In this case Cramer’s Rule gives where W is the Wronskian of y1, y2, . . . , yn and Wk is the determinant obtained by replacing the kth column of the Wronskian by the column consisting of the right-hand side of (14)—that is, the column consisting of (0, 0, . . . , f(x)). When n 2, we get (9). When n 3, the particular solution is yp u1y1 u2y2 u3y3, where y1, y2, and y3 constitute a linearly independent set of solutions of the associated homogeneous DE and u1, u2, u3 are determined from (15) u 1 W1 W , u 2 W2 W , u 3 W3 W , u k Wk W, k 1, 2, . . . , n, y1u 1 y2u 2 0 y1 (n1)u 1 y2 (n1)u 2 yn (n1)u n f (x). y 1u 1 y 2u 2 y nu n 0 y1u 1 y2u 2 ynu n 0 u k 4.6 VARIATION OF PARAMETERS ● 161 W1 p 0 0 f (x) y2 y 2 y 2 y3 y 3 y 3 p , W2 p y1 y 1 y 1 0 0 f (x) y3 y 3 y 3 p , W3 p y1 y 1 y 1 y2 y 2 y 2 0 0 f (x) p , and W p y1 y 1 y 1 y2 y 2 y 2 y3 y 3 y 3 p . See Problems 25–28 in Exercises 4.6. REMARKS (i) Variation of parameters has a distinct advantage over the method of undetermined coefﬁcients in that it will always yield a particular solution yp provided that the associated homogeneous equation can be solved. The pre-sent method is not limited to a function f(x) that is a combination of the four types listed on page 140. As we shall see in the next section, variation of parameters, unlike undetermined coefﬁcients, is applicable to linear DEs with variable coefﬁcients. (ii) In the problems that follow, do not hesitate to simplify the form of yp. Depending on how the antiderivatives of and are found, you might not obtain the same yp as given in the answer section. For example, in Problem 3 in Exercises 4.6 both yp sin x x cos x and yp sin x x cos x are valid answers. In either case the general solution y yc yp simpliﬁes to y c1 cos x c2 sin x x cos x. Why? 1 2 1 2 1 4 1 2 1 2 u 2 u 1 EXERCISES 4.6 Answers to selected odd-numbered problems begin on page ANS-6. In Problems 1–18 solve each differential equation by varia-tion of parameters. 1. y y sec x 2. y y tan x 3. y y sin x 4. y y sec u tan u 5. y y cos2x 6. y y sec2x 7. y y cosh x 8. y y sinh 2x 9. 10. y 9y 9x e3x y 4y e2x x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 162 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 11. 12. 13. y 3y 2y sin ex 14. y 2y y et arctan t 15. y 2y y et ln t 16. 17. 3y 6y 6y ex sec x 18. In Problems 19–22 solve each differential equation by variation of parameters, subject to the initial conditions y(0) 1, y(0) 0. 19. 4y y xex/2 20. 2y y y x 1 21. y 2y 8y 2e2x ex 22. y 4y 4y (12x2 6x)e2x In Problems 23 and 24 the indicated functions are known lin-early independent solutions of the associated homogeneous differential equation on (0, ). Find the general solution of the given nonhomogeneous equation. 23. ; y1 x1/2 cos x, y2 x1/2 sin x x2y xy (x2 1 4)y x3/2 4y 4y y ex/211 x2 2y 2y y 41x y 2y y ex 1 x2 y 3y 2y 1 1 ex 24. x2y xy y sec(ln x); y1 cos(ln x), y2 sin(ln x) In Problems 25–28 solve the given third-order differential equation by variation of parameters. 25. y y tan x 26. y 4y sec 2x 27. 28. Discussion Problems In Problems 29 and 30 discuss how the methods of unde-termined coefﬁcients and variation of parameters can be combined to solve the given differential equation. Carry out your ideas. 29. 3y 6y 30y 15 sin x ex tan 3x 30. y 2y y 4x2 3 x1ex 31. What are the intervals of deﬁnition of the general solu-tions in Problems 1, 7, 9, and 18? Discuss why the inter-val of deﬁnition of the general solution in Problem 24 is not (0, ). 32. Find the general solution of x4y x3y 4x2y 1 given that y1 x2 is a solution of the associated homo-geneous equation. y 3y 2y e2x 1 ex y 2y y 2y e4x CAUCHY-EULER EQUATION REVIEW MATERIAL ●Review the concept of the auxiliary equation in Section 4.3. INTRODUCTION The same relative ease with which we were able to ﬁnd explicit solutions of higher-order linear differential equations with constant coefﬁcients in the preceding sections does not, in general, carry over to linear equations with variable coefﬁcients. We shall see in Chapter 6 that when a linear DE has variable coefﬁcients, the best that we can usually expect is to ﬁnd a solution in the form of an inﬁnite series. However, the type of differential equation that we consider in this section is an exception to this rule; it is a linear equation with variable coefﬁcients whose general solution can always be expressed in terms of powers of x, sines, cosines, and logarithmic functions. Moreover, its method of solution is quite similar to that for constant-coefﬁcient equations in that an auxiliary equation must be solved. 4.7 Cauchy-Euler Equation A linear differential equation of the form anxn dny dxn an1xn1 dn1y dxn1 a1x dy dx a0y g(x), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. where the coefﬁcients an, an1, . . . , a0 are constants, is known as a Cauchy-Euler equation. The differential equation is named in honor of two of the most proliﬁc mathematicians of all time. Augustin-Louis Cauchy (French, 1789–1857) and Leonhard Euler (Swiss, 1707–1783). The observable characteristic of this type of equation is that the degree k n, n 1, . . . , 1, 0 of the monomial coefﬁcients xk matches the order k of differentiation dkydxk: As in Section 4.3, we start the discussion with a detailed examination of the forms of the general solutions of the homogeneous second-order equation . (1) The solution of higher-order equations follows analogously. Also, we can solve the nonhomogeneous equation ax2y bxy cy g(x) by variation of parameters, once we have determined the complementary function yc. Note The coefﬁcient ax2 of y is zero at x 0. Hence to guarantee that the fundamental results of Theorem 4.1.1 are applicable to the Cauchy-Euler equation, we focus our attention on ﬁnding the general solutions deﬁned on the interval (0, ). Method of Solution We try a solution of the form y xm, where m is to be determined. Analogous to what happened when we substituted emx into a linear equa-tion with constant coefﬁcients, when we substitute xm, each term of a Cauchy-Euler equation becomes a polynomial in m times xm, since ax2 d 2y dx2 bx dy dx cy 0 anx n an1x n1 . . . . dny –––– dx n d n1y –––––– dx n1 same same 4.7 CAUCHY-EULER EQUATION ● 163 akxk dky dxk akxkm(m 1)(m 2) (m k 1)xmk akm(m 1)(m 2) (m k 1)xm. For example, when we substitute y xm, the second-order equation becomes ax2 d 2y dx2 bx dy dx cy am(m 1)xm bmxm cxm (am(m 1) bm c)xm. Thus y xm is a solution of the differential equation whenever m is a solution of the auxiliary equation (2) There are three different cases to be considered, depending on whether the roots of this quadratic equation are real and distinct, real and equal, or complex. In the last case the roots appear as a conjugate pair. Case I: Distinct Real Roots Let m1 and m2 denote the real roots of (1) such that m1 m2. Then and form a fundamental set of solutions. Hence the general solution is (3) y c1xm1 c2xm2. y2 xm2 y1 xm1 am(m 1) bm c 0 or am2 (b a)m c 0. EXAMPLE 1 Distinct Roots Solve x2 d 2y dx2 2x dy dx 4y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 164 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS SOLUTION Rather than just memorizing equation (2), it is preferable to assume y xm as the solution a few times to understand the origin and the difference between this new form of the auxiliary equation and that obtained in Section 4.3. Differentiate twice, and substitute back into the differential equation: if m2 3m 4 0. Now (m 1)(m 4) 0 implies m1 1, m2 4, so y c1x1 c2x4. Case II: Repeated Real Roots If the roots of (2) are repeated (that is, m1 m2), then we obtain only one solution—namely, When the roots of the quadratic equation am2 (b a)m c 0 are equal, the discriminant of the coef-ﬁcients is necessarily zero. It follows from the quadratic formula that the root must be m1 (b a)2a. Now we can construct a second solution y2, using (5) of Section 4.2. We ﬁrst write the Cauchy-Euler equation in the standard form and make the identiﬁcations P(x) bax and Thus The general solution is then (4) y c1xm1 c2xm1 ln x. xm1 dx x xm1 ln x. ; 2m1 (b a)/a xm1 xb/a x(ba)/adx ; e(b / a)ln x eln xb/a xb / a xm1 xb/a x2m1dx y2 xm1 e(b/a)ln x x2m1 dx (b>ax) dx (b>a) ln x. d 2y dx2 b ax dy dx c ax2 y 0 y xm1. xm(m(m 1) 2m 4) xm(m2 3m 4) 0 x2 d 2y dx2 2x dy dx 4y x2 m(m 1)xm2 2x mxm1 4xm dy dx mxm1, d 2y dx2 m(m 1)xm2, EXAMPLE 2 Repeated Roots Solve SOLUTION The substitution y xm yields when 4m2 4m 1 0 or (2m 1)2 0. Since , it follows from (4) that the general solution is y c1x1/2 c2x1/2 ln x. For higher-order equations, if m1 is a root of multiplicity k, then it can be shown that xm1, xm1 ln x, xm1(ln x)2, . . . , xm1(ln x)k1 m1 1 2 4x2 d 2y dx2 8x dy dx y xm(4m(m 1) 8m 1) xm(4m2 4m 1) 0 4x2 d 2y dx2 8x dy dx y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. are k linearly independent solutions. Correspondingly, the general solution of the dif-ferential equation must then contain a linear combination of these k solutions. Case III: Conjugate Complex Roots If the roots of (2) are the conjugate pair m1 a ib, m2 a ib, where a and b 0 are real, then a solution is But when the roots of the auxiliary equation are complex, as in the case of equations with constant coefﬁcients, we wish to write the solution in terms of real functions only. We note the identity which, by Euler’s formula, is the same as xib cos(b ln x) i sin(b ln x). Similarly, xib cos(b ln x) i sin(b ln x). Adding and subtracting the last two results yields xib xib 2 cos(b ln x) and xib xib 2i sin(b ln x), respectively. From the fact that y C1xaib C2xaib is a solution for any values of the constants, we see, in turn, for C1 C2 1 and C1 1, C2 1 that or are also solutions. Since W(xa cos(b ln x), xa sin(b ln x)) bx2a1 0, b 0 on the interval (0, ), we conclude that constitute a fundamental set of real solutions of the differential equation. Hence the general solution is (5) y x[c1 cos( ln x) c2 sin( ln x)]. y1 x cos( ln x) and y2 x sin( ln x) y1 2x cos( ln x) and y2 2ix sin( ln x) y1 x(xi xi) and y2 x(xi xi) xi (eln x)i ei ln x, y C1xi C2xi. 4.7 CAUCHY-EULER EQUATION ● 165 x y _1 0 1 1 x y 25 50 75 10 5 100 (a) solution for 0 x 1 (b) solution for 0 x 100 FIGURE 4.7.1 Solution curve of IVP in Example 3 EXAMPLE 3 An Initial-Value Problem Solve SOLUTION The y term is missing in the given Cauchy-Euler equation; neverthe-less, the substitution y xm yields when 4m2 4m 17 0. From the quadratic formula we ﬁnd that the roots are and . With the identiﬁcations and b 2 we see from (5) that the general solution of the differential equation is By applying the initial conditions to the foregoing solution and using ln 1 0, we then ﬁnd, in turn, that c1 1 and c2 0. Hence the solution of the initial-value problem is y x1/2 cos(2 ln x). The graph of this function, obtained with the aid of computer software, is given in Figure 4.7.1. The particular solution is seen to be oscillatory and unbounded as . x : y(1) 1, y(1) 1 2 y x1/2[c1 cos(2 ln x) c2 sin(2 ln x)]. 1 2 m2 1 2 2i m1 1 2 2i 4x2y 17y xm(4m(m 1) 17) xm(4m2 4m 17) 0 4x2y 17y 0, y(1) 1, y(1) 1 2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 166 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS The next example illustrates the solution of a third-order Cauchy-Euler equation. EXAMPLE 4 Third-Order Equation Solve SOLUTION The ﬁrst three derivatives of y xm are so the given differential equation becomes dy dx mxm1, d 2y dx2 m(m 1)xm2, d 3y dx3 m(m 1)(m 2)xm3, x3 d 3y dx3 5x2 d 2y dx2 7x dy dx 8y 0. xm(m3 2m2 4m 8) xm(m 2)(m2 4) 0. xm(m(m 1)(m 2) 5m(m 1) 7m 8) x3 d 3y dx3 5x2 d 2y dx2 7x dy dx 8y x3m(m 1)(m 2)xm3 5x2m(m 1)xm2 7xmxm1 8xm In this case we see that y xm will be a solution of the differential equation for m1 2, m2 2i, and m3 2i. Hence the general solution is y c1x2 c2 cos(2 ln x) c3 sin(2 ln x). Nonhomogeneous Equations The method of undetermined coefﬁcients described in Sections 4.5 and 4.6 does not carry over, in general, to nonhomoge-neous linear differential equations with variable coefﬁcients. Consequently, in our next example the method of variation of parameters is employed. EXAMPLE 5 Variation of Parameters Solve x2y 3xy 3y 2x4ex. SOLUTION Since the equation is nonhomogeneous, we ﬁrst solve the associated homogeneous equation. From the auxiliary equation (m 1)(m 3) 0 we ﬁnd yc c1x c2x3. Now before using variation of parameters to ﬁnd a particular solution yp u1y1 u2y2, recall that the formulas and , where W1, W2, and W are the determinants deﬁned on page 158, were derived under the assump-tion that the differential equation has been put into the standard form y P(x)y Q(x)y f(x). Therefore we divide the given equation by x2, and from we make the identiﬁcation f(x) 2x2ex. Now with y1 x, y2 x3, and y 3 x y 3 x2 y 2x2ex u 2 W2>W u 1 W1>W W x 1 x3 3x2 2x3, W1 0 2x2ex x3 3x2 2x5ex, W2 x 1 0 2x2ex 2x3ex, we ﬁnd The integral of the last function is immediate, but in the case of we integrate by parts twice. The results are u1 x2ex 2xex 2ex and u2 ex. Hence yp u1y1 u2y2 is Finally, y yc yp c1x c2x3 2x2ex 2xex. yp (x2ex 2xex 2ex)x exx3 2x2ex 2xex. u 1 u 1 2x5ex 2x3 x2ex and u 2 2x3ex 2x3 ex. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Reduction to Constant Coefficient The similarities between the forms of solutions of Cauchy-Euler equations and solutions of linear equations with constant coefﬁcients are not just a coincidence. For example, when the roots of the auxiliary equations for ay by cy 0 and ax2y bxy cy 0 are distinct and real, the respective general solutions are (5) In view of the identity eln x x, x 0, the second solution given in (5) can be expressed in the same form as the ﬁrst solution: where t ln x. This last result illustrates the fact that any Cauchy-Euler equation can always be rewritten as a linear differential equation with constant coefﬁcients by means of the substitution x et. The idea is to solve the new differential equation in terms of the variable t, using the methods of the previous sections, and, once the general solution is obtained, resubstitute t ln x. This method, illustrated in the last example, requires the use of the Chain Rule of differentiation. y c1em1 ln x c2em2 ln x c1em1t c2em2t, y c1em1x c2em2x and y c1xm1 c2xm2, x 0. 4.7 CAUCHY-EULER EQUATION ● 167 EXAMPLE 6 Changing to Constant Coefficient Solve x2y xy y ln x. SOLUTION With the substitution x et or t ln x, it follows that Substituting in the given differential equation and simplifying yields Since this last equation has constant coefﬁcients, its auxiliary equation is m2 2m 1 0, or (m 1)2 0. Thus we obtain yc c1et c2tet. By undetermined coefﬁcients we try a particular solution of the form yp A Bt. This assumption leads to 2B A Bt t, so A 2 and B 1. Using y yc yp, we get By resubstituting et x and t ln x we see that the general solution of the original differential equation on the interval (0, ) is y c1x c2x ln x 2 ln x. Solutions For x < 0 In the preceding discussion we have solved Cauchy-Euler equations for One way of solving a Cauchy-Euler equation for is to change the independent variable by means of the substitution (which implies and using the Chain Rule: dy dx dy dt dt dx dy dt and d2y dx2 d dt dy dt dt dx d 2y dt2. t 0) t x x 0 x 0. y c1et c2tet 2 t. d 2y dt2 2 dy dt y t. 1 x d 2y dt2 1 x dy dt 1 x2 1 x2 d 2y dt2 dy dt. ; Product Rule and Chain Rule d 2y dx2 1 x d dx dy dt dy dt 1 x2 ; Chain Rule dy dx dy dt dt dx 1 x dy dt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 168 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS See Problems 37 and 38 in Exercises 4.7. A Different Form A second-order equation of the form (6) is also a Cauchy-Euler equation. Observe that (6) reduces to (1) when We can solve (6) as we did (1), namely, seeking solutions of and using Alternatively, we can reduce (6) to the familiar form (1) by means of the change of independent variable solving the reduced equation, and resubstituting. See Problems 39–42 in Exercises 4.7. t x x0, dy dx m(x x0)m1 and d2y dx2 m(m 1)(x x0)m2. y (x x0)m x0 0. a(x x0)2 d2y dx2 b(x x0)dy dx cy 0 EXERCISES 4.7 Answers to selected odd-numbered problems begin on page ANS-6. In Problems 1–18 solve the given differential equation. 1. x2y 2y 0 2. 4x2y y 0 3. xy y 0 4. xy 3y 0 5. x2y xy 4y 0 6. x2y 5xy 3y 0 7. x2y 3xy 2y 0 8. x2y 3xy 4y 0 9. 25x2y 25xy y 0 10. 4x2y 4xy y 0 11. x2y 5xy 4y 0 12. x2y 8xy 6y 0 13. 3x2y 6xy y 0 14. x2y 7xy 41y 0 15. x3y 6y 0 16. x3y xy y 0 17. xy(4) 6y 0 18. x4y(4) 6x3y 9x2y 3xy y 0 In Problems 19–24 solve the given differential equation by variation of parameters. 19. xy 4y x4 20. 2x2y 5xy y x2 x 21. x2y xy y 2x 22. x2y 2xy 2y x4ex 23. x2y xy y ln x 24. In Problems 25–30 solve the given initial-value problem. Use a graphing utility to graph the solution curve. 25. x2y 3xy 0, y(1) 0, y(1) 4 26. x2y 5xy 8y 0, y(2) 32, y(2) 0 x2y xy y 1 x 1 27. x2y xy y 0, y(1) 1, y(1) 2 28. x2y 3xy 4y 0, y(1) 5, y(1) 3 29. 30. In Problems 31–36 use the substitution x et to transform the given Cauchy-Euler equation to a differential equation with constant coefﬁcients. Solve the original equation by solving the new equation using the procedures in Sections 4.3–4.5. 31. x2y 9xy 20y 0 32. x2y 9xy 25y 0 33. x2y 10xy 8y x2 34. x2y 4xy 6y ln x2 35. x2y 3xy 13y 4 3x 36. x3y 3x2y 6xy 6y 3 ln x3 In Problems 37 and 38 use the substitution to solve the given initial-value problem on the interval 37. 4x2y y 0, y(1) 2, y(1) 4 38. x2y 4xy 6y 0, y(2) 8, y(2) 0 In Problems 39 and 40 use to solve the given differential equation. 39. 40. (x 1)2y (x 1)y 5y 0 (x 3)2 y 8(x 1)y 14y 0 y (x x0)m ( , 0). t x x2y 5xy 8y 8x6, y1 2 0, y1 2 0 xy y x, y(1) 1, y(1) 1 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.8 GREEN’S FUNCTIONS ● 169 In Problems 41 and 42 use the substitution to solve the given differential equation. 41. 42. Discussion Problems 43. Give the largest interval over which the general solution of Problem 42 is deﬁned. 44. Can a Cauchy-Euler differential equation of lowest order with real coefﬁcients be found if it is known that 2 and 1 i are roots of its auxiliary equation? Carry out your ideas. 45. The initial-conditions y(0) y0, y(0) y1 apply to each of the following differential equations: x2y 0, x2y 2xy 2y 0, x2y 4xy 6y 0. (x 4)2y 5(x 4)y 9y 0 (x 2)2y (x 2)y y 0 t x x0 For what values of y0 and y1 does each initial-value problem have a solution? 46. What are the x-intercepts of the solution curve shown in Figure 4.7.1? How many x-intercepts are there for ? Computer Lab Assignments In Problems 47–50 solve the given differential equation by using a CAS to ﬁnd the (approximate) roots of the auxiliary equation. 47. 2x3y 10.98x2y 8.5xy 1.3y 0 48. x3y 4x2y 5xy 9y 0 49. x4y(4) 6x3y 3x2y 3xy 4y 0 50. x4y(4) 6x3y 33x2y 105xy 169y 0 51. Solve x3y x2y 2xy 6y x2 by variation of parameters. Use a CAS as an aid in computing roots of the auxiliary equation and the determinants given in (15) of Section 4.6. 0 x 1 2 GREEN’S FUNCTIONS REVIEW MATERIAL ●See the Remarks at the end of Section 4.1 for the deﬁnitions of response, input, and output. ●Differential operators in Section 4.1 and Section 4.5 ●The method of variation of parameters in Section 4.6 INTRODUCTION We will see in Chapter 5 that the linear second-order differential equation (1) plays an important role in many applications. In the mathematical analysis of physical systems it is often desirable to express the response or output of (1) subject to either initial conditions or boundary conditions directly in terms of the forcing function or input In this manner the response of the system can quickly be analyzed for different forcing functions. To see how this is done, we start by examining solutions of initial-value problems in which the DE (1) has been put into the standard form (2) by dividing the equation by the lead coefﬁcient We also assume throughout this section that the coefﬁcient functions and are continuous on some common interval I. f(x) P(x), Q(x), a2(x). y P(x)y Q(x)y f(x) g(x). y(x) a2(x)d2y dx2 a1(x) dy dx a0(x)y g(x) 4.8 4.8.1 INITIAL-VALUE PROBLEMS Three Initial-Value Problems We will see as the discussion unfolds that the solution of the second order initial-value problem (3) y P(x)y Q(x)y f(x), y(x0) y0, y(x0) y1 y(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 170 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS can be expressed as the superposition of two solutions: (4) where is the solution of the associated homogeneous DE with nonhomogeneous initial conditions (5) and is the solution of the nonhomogeneous DE with homogeneous (that is, zero) initial conditions (6) In the case where the coefﬁcients P and Q are constants the solution of the IVP (5) presents no difﬁculties: We use the method of Section 4.3 to ﬁnd the general solution of the homogeneous DE and then use the given initial conditions to determine the two constants in that solution. So we will focus on the solution of the IVP (6). Because of the zero initial conditions, the solution of (6) could describe a physical system that is initially at rest and so is sometimes called a rest solution. Green’s Function If form a fundamental set of solutions on the interval I of the associated homogeneous form of (2), then a particular solution of the nonhomogeneous equation (2) on the interval I can be found by variation of parame-ters. Recall from (3) of Section 4.6, the form of this solution is (7) The variable coefﬁcients in (7) are deﬁned by (9) of Section 4.6: (8) The linear independence of on the interval I guarantees that the Wronskian for all x in I. If x and are numbers in I, then integrating the derivatives and in (8) on the interval and substitut-ing the results into (7) give (9) where From the properties of the deﬁnite integral, the two integrals in the second line of (9) can be rewritten as a single integral (10) The function in (10), (11) is called the Green’s function for the differential equation (2). Observe that a Green’s function (11) depends only on the fundamental solutions of the associated homogeneous differential equation for (2) and not on the forcing function Therefore all linear second-order differential equations (2) with the same left-hand side but with different forcing functions have the same the Green’s function. So an alternative title for (11) is the Green’s function for the second-order differential operator L D2 P(x)D Q(x). f(x). y1(x) and y2(x) G(x, t) y1(t)y2(x) y1(x)y2(t) W(t) G(x, t) yp(x) x x0 G(x, t) f(t) dt. W(t) W(y1(t), y2(t)) y1(t) y 1(t) y2(t) y 2(t) x x0 y1(x)y2(t) W(t) f(t) dt x x0 y1(t)y2(x) W(t) f(t) dt, yp(x) y1(x) x x0 y2(t)f(t) W(t) dt y2(x) x x0 y1(t)f(t) W(t) dt [x0, x] u 2(x) u 1(x) x0 W W(y1(x), y2(x)) 0 y1(x) and y2(x) u 1(x) y2(x)f(x) W , u 2(x) y1(x)f(x) W . u1(x) and u2(x) yp(x) u1(x)y1(x) u2(x)y2(x). y1(x) and y2(x) y P(x)y Q(x)y f(x), y(x0) 0, y(x0) 0. yp(x) y P(x)y Q(x)y 0, y(x0) y0, y(x0) y1 yh(x) y(x) yh(x) yp(x), Because y1(x) and y2(x) are constant with respect to the integration on t, we can move these functions inside the definite intergrals Important. Read this paragraph a second time. Here at least one of the numbers y0 or y1 is assumed to be nonzero. If both y0 and y1 are 0, then the solution of the IVP is y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.8 GREEN’S FUNCTIONS ● 171 EXAMPLE 1 Particular Solution Use (10) and (11) to ﬁnd a particular solution of SOLUTION The solutions of the associated homogeneous equation are and It follows from (11) that the Green’s function is (12) Thus from (10), a particular solution of the DE is (13) yp(x) x x0 sinh(x t) f(t) dt. G(x, t) etex exet 2 ext e(xt) 2 sinh(x t). W(y1(t), y2(t)) 2. y2 ex, y1 ex, y y 0 y y f(x). EXAMPLE 2 General Solutions Find the general solution of following nonhomogeneous differential equations. (a) (b) SOLUTION From Example 1, both DEs possess the same complementary function . Moreover, as pointed out in the paragraph preceding Example 1, the Green’s function for both differential equations is (12). (a) With the identiﬁcations and we see from (13) that a partic-ular solution of is Thus the general solu-tion of the given DE on any interval not containing the origin is (14) You should compare this solution with that found in Example 3 of Section 4.6. (b) With in (13), a particular solution of is The general solution is then (15) Now consider the special initial-value problem (6) with homogeneous initial conditions. One way of solving the problem when has already been illus-trated in Sections 4.4 and 4.6, that is, apply the initial conditions to the general solution of the nonhomogeneous DE. But there is no actual need to do this because we already have a solution of the IVP at hand; it is the function deﬁned in (10). y(x0) 0, y(x0) 0 f(x) 0 y c1ex c2ex x x0 sinh(x t) e2t dt. y yc yp x x0sinh(x t) e2t dt. yp(x) y y e2x f(x) e2x y c1e x c2ex x x0 sinh(x t) t dt. [x0, x] y yc yp yp(x) x x0 sinh(x t) t dt. y y 1>x f(t) 1>t f(x) 1>x yc c1ex c2ex y y e2x y y 1>x THEOREM 4.8.1 Solution of the IVP (6) The function deﬁned in (10) is the solution of the initial-value problem (6). yp(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 172 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS PROOF By construction we know that in (10) satisﬁes the nonhomogeneous DE. Next, because a deﬁnite integral has the property we have Finally, to show that we utilize the Leibniz formula for the derivative of an integral: 0 from (11) 2 Hence, y p(x0) x0 x0 y1(t)y 2(x0) y 1(x0)y2(t) W(t) f(t) dt 0. x x0 y1(t)y 2(x) y 1(x)y2(t) W(t) f(t) dt. y p(x) G(x, x)f(x) y p(x0) 0 yp(x0) x0 x0 G(x0, t) f(t) dt 0. a a 0 yp(x) EXAMPLE 3 Example 2 Revisited Solve the initial-value problems (a) (b) SOLUTION (a) With it follows from (14) of Example 2 and Theorem 4.8.1 that the solution of the initial-value problem is where (b) Identifying we see from (15) that the solution of the IVP is (16) In part (b) of Example 3, we can carry out the integration in (16), but bear in mind that x is held constant throughout the integration with respect to t: 1 3e2x 1 2ex 1 6ex. 1 2ex x 0 et dt 1 2ex x 0 e3t dt yp(x) x 0 sinh(x t) e2t dt x 0 ext e(xt) 2 e2t dt yp(x) x 0 sinh(x t) e2t dt. x0 0 and f(t) e2t, [1, x], x 0. yp(x) x 1 sinh(x t) t dt, x0 1 and f(t) 1>t, y y e2x, y(0) 0, y(0) 0 y y 1>x, y(1) 0, y(1) 0 This formula, usually discussed in advanced calculus, is given by d dx v(x) u(x) F(x, t)dt F(x, v(x))v(x) F(x, u(x))u(x) v(x) u(x) xF(x, t)dt. EXAMPLE 4 Using (10) and (11) Solve the initial-value problem SOLUTION We begin by constructing the Green’s function for the given differen-tial equation. y 4y x, y(0) 0, y(0) 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The two linearly independent solutions of are From (11), with we ﬁnd With the further identiﬁcations and in (10) we see that a solution of the given initial-value problem is If we wish to evaluate the integral, we ﬁrst write and then use integration by parts: or Initial-Value Problems—Continued Finally, we are now in a position to make use of Theorem 4.8.1 to ﬁnd the solution of the initial-value problem posed in (3). It is simply the function already given in (4). yp(x) 1 4x 1 8 sin2x. yp(x) 1 2sin2x [1 2t sin2t 1 4 cos2t] x 0 1 2 cos2x [1 2t cos2t 1 4sin2t] x 0 yp(x) 1 2sin2x x 0 t cos2t dt 1 2 cos2x x 0 t sin2t dt yp(x) 1 2 x 0 tsin2(x t)dt. f(t) t x0 0 G(x, t) cos2t sin2x cos2x sin2t 2 1 2 sin2(x t). W(cos2t, sin2t) 2, y2(x) sin2x. y1(x) cos 2x and y 4y 0 4.8 GREEN’S FUNCTIONS ● 173 Here we have used the trigonometric identity sin(2x 2t) sin 2x cos 2t cos 2x sin 2t THEOREM 4.8.2 Solution of the IVP (3) If is the solution of the initial-value problem (5) and is the solution (10) of the initial-value problem (6) on the interval I, then (17) is the solution of the initial-value problem (3). y(x) yh(x) yp(x) yp(x) yh(x) PROOF Because is a linear combination of the fundamental solutions, it follows from (10) of Section 4.1 that is a solution of the nonhomoge-neous DE. Moreover, since satisﬁes the initial-conditions in (5) and satisﬁes the initial conditions in (6), we have, Keeping in mind the absence of a forcing function in (5) and the presence of such a term in (6), we see from (17) that the response of a physical system described by the initial-value problem (3) can be separated into two different responses: (18) 3 3 response of system response of system due to initial conditions due to the forcing function f If you wish to peek ahead, the following initial-value problem represents a pure resonance situation for a driven spring/mass system. See pages 200–202. y(x0) y0, y(x0) y1 y(x) yh(x) yp(x) y(x) y(x0) y h(x0) y p(x0) y1 0 y1. y(x0) yh(x0) yp(x0) y0 0 y0 yp yh y yh yp yh(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 174 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 5 Using Theorem 4.8.2 Solve the initial-value problem SOLUTION We solve two initial-value problems. First, we solve By applying the initial con-ditions to the general solution of the homogeneous DE, we ﬁnd that and Therefore, Next we solve Since the left-hand side of the differential equation is the same as the DE in Example 4, the Green’s function is the same, namely, With we see from (10) that the solution of this second problem is Finally, in view of (17) in Theorem 4.8.2, the solution of the original IVP is (19) If desired, we can integrate the deﬁnite integral in (19) by using the trigonomet-ric identity with and (20) Hence, the solution (19) can be rewritten as or (21) Note that the physical signiﬁcance indicated in (18) is lost in (21) after combining like terms in the two parts of the solution The beauty of the solution given in (19) is that we can immediately write down the response of a system if the initial conditions remain the same, but the forcing function is changed. For example, if the problem in Example 5 is changed to we simply replace in the integral in (19) by t and the solution is then Because the forcing function f is isolated in the particular solution the solution in (17) is useful when f is piecewise deﬁned. The next example illustrates this idea. yp(x) x x0G(x, t) f(t) dt, 1 4x cos2x 9 8sin2x. ; see Example 4 cos 2x sin2x 1 2 x 0 tsin2(x t) dt y(x) yh(x) yp(x) sin2t y 4y x, y(0) 1, y(0) 2, y(x) yh(x) yp(x). y(x) cos2x 7 8 sin2x 1 4 x cos2x. y(x) yh(x) yp(x) cos2x sin2x 1 8sin2x 1 4x cos2x, 1 8sin2x 1 4xcos2x. 1 4[1 4sin(2x 4t) tcos2x] x 0 1 4 x 0 [cos(2x 4t) cos2x] dt yp(x) 1 2 x 0 sin2(x t)sin2tdt B 2t: A 2(x t) sinAsinB 1 2[cos(A B) cos (A B)] y(x) yh(x) yp(x) cos2x sin2x 1 2 x 0 sin2(x t)sin2t dt. yp(x) 1 2x 0 sin2(x t)sin2t dt. f(t) sin2t G(x, t) 1 2sin2(x t). y 4y sin2x, y(0) 0, y(0) 0. yh(x) cos2x sin2x. c2 1. c1 1 y(x) c1cos2x c2 sin2x y 4y 0, y(0) 1, y(0) 2. y 4y sin 2x, y(0) 1, y(0) 2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.8 GREEN’S FUNCTIONS ● 175 EXAMPLE 6 An Initial-Value Problem Solve the initial-value problem where the forcing function f is piecewise deﬁned: SOLUTION From (19), with replaced by we can write Because f is deﬁned in three pieces, we consider three cases in the evaluation of the deﬁnite integral. For for and ﬁnally for we can use the integration following Example 5: Hence is and so Putting all the pieces together we get The three parts of are shown in different colors in Figure 4.8.1. We next examine how a boundary value problem (BVP) can be solved using a different kind of Green’s function. y(x) y(x) cos 2x sin 2x, x 0 (1 1 4 x) cos2x 7 8 sin 2x, 0 x 2 (1 1 2)cos2x sin2x, x 2. y(x) yh(x) yp(x) cos2x sin 2x yp(x). yp(x) 0, x 0 1 8 sin2x1 4x cos2x, 0 x 2 1 2 cos2x, x 2. yp(x) 1 2pcos 2x. ; sin(2x 8p) sin 2x 1 16 sin (2x 8p)1 2 pcos 2x 1 16sin 2x ; using the integration in (20) 1 4[1 4sin (2x 4t) tcos 2x] 2p 0 1 2p 2p 0 sin 2(x t) sin 2t dt yp(x) 1 2 2p 0 sin 2(x t) sin 2t dt 1 2 x 2p sin 2(x t) 0 dt x 2p, 1 8sin2x 1 4x cos2x, ; using the integration in (20) yp(x) 1 2 x 0 sin2(x t) sin2t dt 0 x 2, yp(x) 1 2 x 0 sin2(x t) 0 dt 0, x 0, y(x) cos 2x sin 2x 1 2 x 0 sin2(x t) f(t) dt. f(t), sin2t f(x) 0, x 0 sin 2x, 0 x 2 0, x 2. y 4y f(x), y(0) 1, y(0) 2, _1 3p 2p p _p 1 y x FIGURE 4.8.1 Graph of y(x) in Example 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 176 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 4.8.2 BOUNDARY-VALUE PROBLEMS In contrast to a second-order IVP, in which and are speciﬁed at the same point, a BVP for a second-order DE involves conditions on and that are speciﬁed at two different points and Conditions such as are just special cases of the more general homogeneous boundary conditions: (22) (23) where are constants. Speciﬁcally, our goal is to ﬁnd a integral solution that is analogous to (10) for nonhomogeneous boundary-value problems of the form (24) In addition to the usual assumptions that and are continuous on we assume that the homogeneous problem possesses only the trivial solution This latter assumption is sufﬁcient to guarantee that a unique solution of (24) exists and is given by an integral where is a Green’s function. The starting point in the construction of is again the variation of parame-ters formulas (7) and (8). Another Green’s Function Suppose are linearly independent solutions on of the associated homogeneous form of the DE in (24) and that x is a number in the interval Unlike the construction of (9) where we started by integrating the derivatives in (8) over the same interval, we now integrate the ﬁrst equation in (8) on and the second equation in (8) on (25) The reason for integrating over different intervals will become clear shortly. From (25), a particular solution of the DE is here we used the minus sign in (25) to reverse the limits of integration $1 1 %1 1 & or (26) The right-hand side of (26) can be written compactly as a single integral (27) where the function is (28) G(x, t) y1(t)y2(x) W(t) , a t x y1(x)y2(t) W(t) , x t b. G(x, t) yp(x) b a G(x, t) f(t)dt, yp(x) x a y2(x)y1(t) W(t) f(t) dt b x y1(x)y2(t) W(t) f(t)dt. yp(x) y1(x) b x y2(t) f(t) W(t) dt y2(x) x a y1(t) f(t) W(t) dt yp(x) u1(x)y1(x) u2(x)y2(x) u 1(x) and u 2(x) u1(x) x b y2(t) f(t) W(t) dt and u2(x) x a y1(t) f(t) W(t) dt. [a, x]: [b, x] [a, b]. [a, b] y1(x) and y2(x) G(x, t) G(x, t) yp(x) b aG(x, t) f(t)dt, y 0. A2y(b) B2y(b) 0, A1y(a) B1y(a) 0 y P(x)y Q(x)y 0, [a, b], f(x) P(x), Q(x), A2y(b) B2y(b) 0. A1y(a) B1y(a) 0 y P(x)y Q(x)y f(x), yp(x) A1, A2, B1, and B2 A2y(b) B2y(b) 0, A1y(a) B1y(a) 0 y(a) 0, y(b) 0; y(a) 0, y(b) 0; y(a) 0, y(b) 0. x b. x a y(x) y(x) y(x) y(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The piecewise-deﬁned function (28) is called a Green’s function for the boundary-value problem (24). It can be proved that is a continuous function of x on the interval Now if the solutions and used in the construction of in (28) are chosen in such a manner that at satisﬁes and at satisﬁes then, wondrously, deﬁned in (27) satisﬁes both homogeneous boundary conditions in (24). To see this we will need (29) and (30) Before proceeding, observe in (25) that and In view of the second of these two properties we can show that satisﬁes (22) whenever satisﬁes the same boundary condition. From (29) and (30) we have 0 0 2 2 (1111)1111 0 from (22) Likewise, implies that whenever satisﬁes (23) so does yp(x): y2(x) u1(b) 0 u1(a)[A1y1(a) B1y 1(a)] 0. A1yp(a) B1y p(a) A1[u1(a)y1(a) u2(a)y2(a)] B1[u1(a)y 1(a) u2(a)y 2(a)] y1(x) yp(x) u2(a) 0. u1(b) 0 u1(x)y 1(x) u2(x)y 2(x). y p(x) u1(x)y 1(x) y1(x)u 1(x) u2(x)y 2(x) y2(x)u 2(x) yp(x) u1(x)y1(x) u2(x)y2(x) yp(x) A2y2(b) B2y 2(b) 0, y2(x) x b, A1y1(a) B1y 1(a) 0, y1(x) x a, G(x, t) y2(x) y1(x) [a, b]. G(x, t) 4.8 GREEN’S FUNCTIONS ● 177 The second line in (30) results from the fact that . See the discussion in Section 4.6 following (4). y1(x)u 1(x) y2(x)u 2(x) 0 0 0 2 2 (1111)1111 0 from (22) u2(b)[A2y2(b) B2y2(b)] 0. A2yp(b) B2y p(b) A2[u1(b)y1(b) u2(b)y2(b)] B2[u1(b)y 1(b) u2(b)y 2(b)] THEOREM 4.8.3 Solution of the BVP (24) Let and be linearly independent solutions of on and suppose and satisfy (22) and (23), respectively. Then the function deﬁned in (27) is a solution of the boundary-value problem (24). yp(x) y2(x) y1(x) [a, b], y P(x)y Q(x)y 0 y2(x) y1(x) EXAMPLE 7 Using Theorem 4.8.3 Solve the boundary-value problem SOLUTION The solutions of the associated homogeneous equation are and and satisﬁes whereas sat-isﬁes The Wronskian is and so from (28) we see that the Green’s function for the boundary-value problem is G(x, t) 1 2 cos 2t sin 2x, 0 t x 1 2 cos 2x sin 2t, x t p>2. W(y1, y2) 2, y(p>2) 0. y2(x) y(0) 0 y1(x) y2(x) sin 2x y1(x) cos 2x y 4y 0 y 4y 3, y(0) 0, y(p>2) 0. The boundary condition y(0) 0 is a special case of (22) with a 0, A1 0, and B1 1. The boundary condition y(p/2) 0 is a special case of (23) with b p/2, A2 1, B2 0. The next theorem summarizes these results. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 178 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS It follows from Theorem 4.8.3 that a solution of the BVP is (27) with the identiﬁca-tions or, after evaluating the deﬁnite integrals, Don’t infer from the preceding example that the demand that satisfy (22) and satisfy (23) uniquely determines these functions. As we see in the last example, there is a certain arbitrariness in the selection of these functions. y2(x) y1(x) yp(x) 3 4 3 4 cos 2x. 3 1 2 sin 2x x 0 cos 2t dt 3 1 2 cos 2x p>2 x sin 2t dt, yp(x) 3 p>2 0 G(x, t)dt f(t) 3: a 0, b p>2, and EXAMPLE 8 A Boundary-Value Problem Solve the boundary-value problem SOLUTION The differential equation is recognized as a Cauchy-Euler DE. From the auxiliary equation the general so-lution of the associated homogeneous equation is Applying to this solution implies By choosing we get and On the other hand, applied to the general solution shows or The choice now gives and so The Wronskian of these two functions is Hence the Green’s function for the boundary-value problem is In order to identify the correct forcing function f we must write the DE in standard form: From this equation we see that and so in (27) becomes Straightforward deﬁnite integration and algebraic simpliﬁcation yield the solution yp(x) 3x5 15x3 12x. 4(4x x3) x 1 (t t3) dt 4(x x3) 2 x (4t t3)dt. yp(x) 24 2 1 G(x, t) t3dt yp(x) f(t) 24t3 y 3 xy 3 x2y 24x3. G(x, t) (t t3)(4x x3) 6t3 , 1 t x (x x3)(4t t3) 6t3 , x t 2. W(y1(x), y2(x)) x x3 4x x3 1 3x2 4 3x2 6x3. y2(x) 4x x3. c1 4 c2 1 c1 4c2. 2c1 8c2 0 y(2) 0 y1 x x3. c1 1 c2 1 c1 c2 0 or c1 c2. y(1) 0 y c1x c2x3. m(m 1) 3m 3 (m 1)(m 3) 0 x2y 3xy 3y 24x5, y(1) 0, y(2) 0. Verify yp(x) that satisfies the differentia equation and the two boundary conditions. REMARKS We have barely scratched the surface of the elegant, albeit complicated, theory of Green’s functions. Green’s functions can also be constructed for linear sec-ond-order partial differential equations, but we leave coverage of the latter topic to an advanced course. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.8 GREEN’S FUNCTIONS ● 179 EXERCISES 4.8 Answers to selected odd-numbered problems begin on page ANS-6. 4.8.1 INITIAL-VALUE PROBLEMS In Problems 1–6 proceed as in Example 1 to ﬁnd a particular solution of the given differential equation in the integral form (10). 1. 2. 3. 4. 5. 6. In Problems 7–12 proceed as in Example 2 to ﬁnd the general solution of the given differential equation. Use the results obtained in Problems 1–6. Do not evaluate the integral that deﬁnes 7. 8. 9. 10. 11. 12. In Problems 13–18 proceed as in Example 3 to ﬁnd a solu-tion of the given initial-value problem. Evaluate the integral that deﬁnes 13. 14. 15. 16. 17. 18. In Problems 19–30 proceed as in Example 5 to ﬁnd a solu-tion of the given initial-value problem. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. x2y xy y x2, y(1) 4, y(1) 3 x2y 6y ln x, y(1) 1, y(1) 3 x2y 2xy 2y x ln x, y(1) 1, y(1) 0 x2y 2xy 2y x, y(1) 2, y(1) 1 y 3y 2y 1 1 ex, y(0) 0, y(0) 1 y 3y 2y sin ex, y(0) 1, y(0) 0 y y sec2x, y(p) 1 2, y(p) 1 y y csc x cot x, y(p>2) p>2, y(p>2) 1 y 6y 9y x, y(0) 1, y(0) 3 y 10y 25y e5x, y(0) 1, y(0) 1 y y 1, y(0) 10, y(0) 1 y 4y e2x, y(0) 1, y(0) 4 y y sec2x, y(p) 0, y(p) 0 y y csc x cot x, y(p>2) 0, y(p>2) 0 y 6y 9y x, y(0) 0, y(0) 0 y 10y 25y e5x, y(0) 0, y(0) 0 y y 1, y(0) 0, y(0) 0 y 4y e2x, y(0) 0, y(0) 0 yp(x). y 2y 2y cos2x y 9y x sin x 4y 4y y arctan x y 2y y ex y 3y 10y x2 y 16y xe2x yp(x). y 2y 2y f(x) y 9y f(x) 4y 4y y f(x) y 2y y f(x) y 3y 10y f(x) y 16y f(x) yp(x) In Problems 31–34 proceed as in Example 6 to ﬁnd a solu-tion of the initial-value problem with the given piecewise-deﬁned forcing function. 31. where 32. where 33. where 34. where 4.8.2 BOUNDARY-VALUE PROBLEMS In Problems 35 and 36, (a) use (27) and (28) to ﬁnd a solu-tion of the boundary-value problem. (b) Verify that the function satisﬁes the differential equations and both boundary-conditions. 35. 36. 37. In Problem 35 ﬁnd a solution of the BVP when 38. In Problem 36 ﬁnd a solution of the BVP when In Problems 39–44 proceed as in Examples 7 and 8 to ﬁnd a solution of the given boundary-value problem. 39. 40. 41. 42. 43. 44. Discussion Problems 45. Suppose the solution of the boundary-value problem y Py Qy f(x), y(a) 0, y(b) 0, x2y 4xy 6y x4, y(1) y(1) 0, y(3) 0 x2y xy 1, y(e1) 0, y(1) 0 y y e2x, y(0) 0, y(1) 0 y 2y 2y ex, y(0) 0, y(p>2) 0 y 9y 1, y(0) 0, y(p) 0 y y 1, y(0) 0, y(1) 0 f(x) x. f(x) 1. y f(x), y(0) 0, y(1) y(1) 0 y f(x), y(0) 0, y(1) 0 yp(x) f(x) 0, x 0 cos x, 0 x 4p 0, x 4p y y f(x), y(0) 0, y(0) 1, f(x) 0, x 0 10, 0 x 3p 0, x 3p y y f(x), y(0) 1, y(0) 1, f(x) 0, x 0 x, x 0 y y f(x), y(0) 3, y(0) 2, f(x) 1, x 0 1, x 0 y y f(x), y(0) 8, y(0) 2, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 180 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS is given by where and are solutions of the associated homogeneous differential equation chosen in the construction of so that and Prove that the solution of the boundary-value problem with nonhomogeneous DE and boundary conditions, is given by y(x) yp(x) B y1(b)y1(x) A y2(a)y2(x). y Py Qy f(x), y(a) A, y(b) B y2(b) 0. y1(a) 0 G(x, t) y2(x) y1(x) yp(x) b aG(x, t)f(t)dt a b, [Hint: In your proof, you will have to show that Reread the assumptions fol-lowing (24).] 46. Use the result in Problem 45 to solve y y 1, y(0) 5, y(1) 10. y1(b) 0 and y2(a) 0. SOLVING SYSTEMS OF LINEAR DEs BY ELIMINATION REVIEW MATERIAL ●Because the method of systematic elimination uncouples a system into distinct linear ODEs in each dependent variable, this section gives you an opportunity to practice what you learned in Sections 4.3, 4.4 (or 4.5), and 4.6. INTRODUCTION Simultaneous ordinary differential equations involve two or more equations that contain derivatives of two or more dependent variables—the unknown functions—with respect to a single independent variable. The method of systematic elimination for solving systems of differential equations with constant coefﬁcients is based on the algebraic principle of elimination of variables. We shall see that the analogue of multiplying an algebraic equation by a constant is operating on an ODE with some combination of derivatives. 4.9 Systematic Elimination The elimination of an unknown in a system of linear differential equations is expedited by rewriting each equation in the system in differ-ential operator notation. Recall from Section 4.1 that a single linear equation where the ai, i 0, 1, . . . , n are constants, can be written as If the nth-order differential operator factors into differential operators of lower order, then the factors commute. Now, for exam-ple, to rewrite the system in terms of the operator D, we ﬁrst bring all terms involving the dependent variables to one side and group the same variables: x y 4x 2y et x 2x y x 3y sin t anDn an1D(n1) a1D a0 (anDn an1D(n1) a1D a0)y g(t). any(n) an1y(n1) a1y a0y g(t), x 2x x y 3y sin t x 4x y 2y et is the same as (D2 2D 1)x (D2 3)y sin t (D 4)x (D 2)y et. Solution of a System A solution of a system of differential equations is a set of sufﬁciently differentiable functions x f1(t), y f2(t), z f3(t), and so on that satisﬁes each equation in the system on some common interval I. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Method of Solution Consider the simple system of linear ﬁrst-order equations (1) Operating on the ﬁrst equation in (1) by D while multiplying the second by 3 and then adding eliminates y from the system and gives D2x 6x 0. Since the roots of the auxiliary equation of the last DE are and , we obtain (2) Multiplying the first equation in (1) by 2 while operating on the second by D and then subtracting gives the differential equation for y, D2y 6y 0. It follows immediately that (3) Now (2) and (3) do not satisfy the system (1) for every choice of c1, c2, c3, and c4 because the system itself puts a constraint on the number of parameters in a solu-tion that can be chosen arbitrarily. To see this, observe that substituting x(t) and y(t) into the ﬁrst equation of the original system (1) gives, after simpliﬁcation, Since the latter expression is to be zero for all values of t, we must have and These two equations enable us to write c3 as a multiple of c1 and c4 as a multiple of c2: . (4) Hence we conclude that a solution of the system must be You are urged to substitute (2) and (3) into the second equation of (1) and verify that the same relationship (4) holds between the constants. x(t) c1e16t c2e16t, y(t) 16 3 c1e16t 16 3 c2e16t. c3 16 3 c1 and c4 16 3 c2 16c2 3c4 0. 16c1 3c3 0 16c1 3c3e16t 16c2 3c4e16t 0. y(t) c3e16t c4e16t. x(t) c1e16t c2e16t. m2 16 m1 16 dx dt 3y dy dt 2x or, equivalently, Dx 3y 0 2x Dy 0. 4.9 SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION ● 181 EXAMPLE 1 Solution by Elimination Solve (5) SOLUTION Operating on the ﬁrst equation by D 3 and on the second by D and then subtracting eliminates x from the system. It follows that the differential equation for y is Since the characteristic equation of this last differential equation is m2 m 6 (m 2)(m 3) 0, we obtain the solution (6) Eliminating y in a similar manner yields (D2 D 6)x 0, from which we ﬁnd (7) x(t) c3e2t c4e3t. y(t) c1e2t c2e3t. [(D 3)(D 2) 2D]y 0 or (D2 D 6)y 0. (D 3)x 2y 0. Dx (D 2 )y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 182 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS As we noted in the foregoing discussion, a solution of (5) does not contain four in-dependent constants. Substituting (6) and (7) into the ﬁrst equation of (5) gives From 4c1 2c3 0 and c2 3c4 0 we get c3 2c1 and . Accordingly, a solution of the system is Because we could just as easily solve for c3 and c4 in terms of c1 and c2, the solution in Example 1 can be written in the alternative form It sometimes pays to keep one’s eyes open when solving systems. Had we solved for x ﬁrst in Example 1, then y could be found, along with the relationship between the constants, using the last equation in the system (5). You should verify that substituting x(t) into yields Also note in the initial dis-cussion that the relationship given in (4) and the solution y(t) of (1) could also have been obtained by using x(t) in (2) and the ﬁrst equation of (1) in the form y 1 3 Dx 1 3 26c1e16t 1 3 26c2e16t. y 1 2 c3e2t 3c4e3t. y 1 2 (Dx 3x) x(t) c3e2t c4e3t, y(t) 1 2 c3e2t 3c4e3t. x(t) 2c1e2t 1 3 c2e3t, y(t) c1e2t c2e3t. c4 1 3 c2 (4c1 2c3)e2t (c2 3c4)e3t 0. This might save you some time. EXAMPLE 2 Solution by Elimination Solve (8) SOLUTION First we write the system in differential operator notation: (9) Then, by eliminating x, we obtain or Since the roots of the auxiliary equation m(m2 4) 0 are m1 0, m2 2i, and m3 2i, the complementary function is yc c1 c2 cos 2t c3 sin 2t. To deter-mine the particular solution yp, we use undetermined coefﬁcients by assuming that yp At3 Bt2 Ct. Therefore The last equality implies that 12A 1, 8B 2, and 6A 4C 0; hence , and . Thus (10) Eliminating y from the system (9) leads to It should be obvious that xc c4 cos 2t c5 sin 2t and that undetermined coefﬁ-cients can be applied to obtain a particular solution of the form xp At2 Bt C. In this case the usual differentiations and algebra yield and so (11) x xc xp c4 cos 2t c5 sin 2t 1 4 t2 1 8. xp 1 4 t2 1 8, [(D 4) D(D 1)]x t2 or (D2 4)x t2. y yc yp c1 c2 cos 2t c3 sin 2t 1 12 t3 1 4 t2 1 8 t. C 1 8 A 1 12, B 1 4 y p 4y p 12At2 8Bt 6A 4C t2 2t. y p 3At2 2Bt C, y p 6At 2B, y p 6A, (D3 4D)y t2 2t. [(D 1)D2 (D 4)D]y (D 1)t2 (D 4)0 (D 1)x Dy 0. (D 4)x D2y t2 x x y 0. x 4x y t2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Now c4 and c5 can be expressed in terms of c2 and c3 by substituting (10) and (11) into either equation of (8). By using the second equation, we ﬁnd, after com-bining terms, so c5 2c4 2c2 0 and 2c5 c4 2c3 0. Solving for c4 and c5 in terms of c2 and c3 gives c4 (4c2 2c3) and c5 (2c2 4c3). Finally, a solution of (8) is found to be y(t) c1 c2 cos 2t c3 sin 2t 1 12 t3 1 4 t2 1 8 t. x(t) 1 5 (4c2 2c3) cos 2t 1 5 (2c2 4c3) sin 2t 1 4 t2 1 8, 1 5 1 5 (c5 2c4 2c2) sin 2t (2c5 c4 2c3) cos 2t 0, 4.9 SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION ● 183 EXAMPLE 3 A Mixture Problem Revisited In (3) of Section 3.3 we saw that the system of linear ﬁrst-order differential equations is a model for the number of pounds of salt x1(t) and x2(t) in brine mixtures in tanks A and B, respectively, shown in Figure 3.3.1. At that time we were not able to solve the system. But now, in terms of differential operators, the foregoing system can be written as Operating on the ﬁrst equation by multiplying the second equation by adding, and then simplifying gives (625D2 100D 3)x1 0. From the auxiliary equation we see immediately that x1(t) c1et/25 c2e3t/25. We can now obtain x2(t) by using the ﬁrst DE of the system in the form In this manner we ﬁnd the solution of the system to be In the original discussion on page 108 we assumed that the initial conditions were x1(0) 25 and x2(0) 0. Applying these conditions to the solution yields c1 c2 25 and 2c1 2c2 0. Solving these equations simultaneously gives Finally, a solution of the initial-value problem is The graphs of both of these equations are given in Figure 4.9.1. Consistent with the fact that pure water is being pumped into tank A we see in the ﬁgure that x1(t) : 0 and x2(t) : 0 as t : . x1(t) 25 2 et/25 25 2 e3t/25, x2(t) 25et/25 25e3t/25. c1 c2 25 2 . x1(t) c1et/25 c2e3t/25, x2(t) 2c1et/25 2c2e3t/25. x2 50(D 2 25)x1. 625m2 100m 3 (25m 1)(25m 3) 0 1 50, D 2 25, 2 25 x1 D 2 25 x2 0. D 2 25 x1 1 50 x2 0 dx2 dt 2 25 x1 2 25 x2 dx1 dt 2 25 x1 1 50 x2 FIGURE 4.9.1 Pounds of salt in tanks A and B in Example 3 20 5 10 15 25 20 40 x1(t) x2(t) 60 time pounds of salt 80 100 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 184 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS EXERCISES 4.9 Answers to selected odd-numbered problems begin on page ANS-6. In Problems 1–20 solve the given system of differential equations by systematic elimination. 1. 2. 3. 4. 5. (D2 5)x 2y 0 2x (D2 2)y 0 6. (D 1)x (D 1)y 2 3x (D 2)y 1 7. 8. 9. Dx D2y e3t (D 1)x (D 1)y 4e3t 10. D2x Dy t (D 3)x (D 3)y 2 11. (D2 1)x y 0 (D 1)x Dy 0 12. (2D2 D 1)x (2D 1)y 1 (D 1)x Dy 1 13. 14. 15. (D 1)x (D2 1)y 1 (D2 1)x (D 1)y 2 16. D2x 2(D2 D)y sin t x Dy 0 17. Dx y 18. Dx z et Dy z (D 1)x Dy Dz 0 Dz x x 2y Dz et 19. 20. dz dt x y dz dt x y dy dt y z dy dt x z dx dt x z dx dt 6y d2x dt2 dx dt x y 0 dx dt dy dt et dx dt x dy dt 5et 2 dx dt 5x dy dt et dx dt dy dt x 4y d 2y dt2 4x et d 2x dt2 dy dt 5x d 2x dt2 4y et dy dt x 2 dy dt x t dx dt 4y 1 dx dt y t dy dt x 2y dy dt x dx dt 4x 7y dx dt 2x y In Problems 21 and 22 solve the given initial-value problem. 21. 22. x(1) 0, y(1) 1 x(0) 0, y(0) 0 Mathematical Models 23. Projectile Motion A projectile shot from a gun has weight w mg and velocity v tangent to its path of motion. Ignoring air resistance and all other forces acting on the projectile except its weight, determine a system of differential equations that describes its path of motion. See Figure 4.9.2. Solve the system. [Hint: Use Newton’s second law of motion in the x and y directions.] dy dt 3x 2y dy dt 4x y dx dt y 1 dx dt 5x y FIGURE 4.9.2 Path of projectile in Problem 23 y x mg v 24. Projectile Motion with Air Resistance Determine a system of differential equations that describes the path of motion in Problem 23 if air resistance is a retarding force k (of magnitude k) acting tangent to the path of the projectile but opposite to its motion. See Figure 4.9.3. Solve the system. [Hint: k is a multiple of velocity, say, bv.] FIGURE 4.9.3 Forces in Problem 24 k v θ Discussion Problems 25. Examine and discuss the following system: Computer Lab Assignments 26. Reexamine Figure 4.9.1 in Example 3. Then use a root-ﬁnding application to determine when tank B contains more salt than tank A. (D 1)x 2(D 1)y 1. Dx 2Dy t2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.10 NONLINEAR DIFFERENTIAL EQUATIONS ● 185 27. (a) Reread Problem 8 of Exercises 3.3. In that problem you were asked to show that the system of differen-tial equations is a model for the amounts of salt in the connected mixing tanks A, B, and C shown in Figure 3.3.7. Solve the system subject to x1(0) 15, x2(t) 10, x3(t) 5. dx3 dt 2 75 x2 1 25 x3 dx2 dt 1 50 x1 2 75 x2 dx1 dt 1 50 x1 (b) Use a CAS to graph x1(t), x2(t), and x3(t) in the same coordinate plane (as in Figure 4.9.1) on the interval [0, 200]. (c) Because only pure water is pumped into Tank A, it stands to reason that the salt will eventually be ﬂushed out of all three tanks. Use a root-ﬁnding application of a CAS to determine the time when the amount of salt in each tank is less than or equal to 0.5 pound. When will the amounts of salt x1(t), x2(t), and x3(t) be simultaneously less than or equal to 0.5 pound? NONLINEAR DIFFERENTIAL EQUATIONS REVIEW MATERIAL ●Sections 2.2 and 2.5 ●Section 4.2 ●A review of Taylor series from calculus is also recommended. INTRODUCTION The difﬁculties that surround higher-order nonlinear differential equations and the few methods that yield analytic solutions are examined next. Two of the solution methods considered in this section employ a change of variable to reduce a nonlinear second-order DE to a ﬁrst-order DE. In that sense these methods are analogous to the material in Section 4.2. 4.10 Some Differences There are several signiﬁcant differences between linear and nonlinear differential equations. We saw in Section 4.1 that homogeneous linear equations of order two or higher have the property that a linear combination of solu-tions is also a solution (Theorem 4.1.2). Nonlinear equations do not possess this property of superposability. See Problems 1 and 18 in Exercises 4.10. We can ﬁnd general solutions of linear ﬁrst-order DEs and higher-order equations with constant coefﬁcients. Even when we can solve a nonlinear ﬁrst-order differential equation in the form of a one-parameter family, this family does not, as a rule, represent a gen-eral solution. Stated another way, nonlinear ﬁrst-order DEs can possess singular solutions, whereas linear equations cannot. But the major difference between linear and nonlinear equations of order two or higher lies in the realm of solvability. Given a linear equation, there is a chance that we can ﬁnd some form of a solution that we can look at—an explicit solution or perhaps a solution in the form of an inﬁnite series (see Chapter 6). On the other hand, nonlinear higher-order differential equa-tions virtually defy solution by analytical methods. Although this might sound dis-heartening, there are still things that can be done. As was pointed out at the end of Section 1.3, we can always analyze a nonlinear DE qualitatively and numerically. Let us make it clear at the outset that nonlinear higher-order differential equations are important—dare we say even more important than linear equations?—because as we ﬁne-tune the mathematical model of, say, a physical system, we also increase the likelihood that this higher-resolution model will be nonlinear. We begin by illustrating an analytical method that occasionally enables us to ﬁnd explicit/implicit solutions of special kinds of nonlinear second-order differential equations. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 186 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS Reduction of Order Nonlinear second-order differential equations F(x, y, y ) 0, EXAMPLE 1 Dependent Variable y Is Missing Solve y 2x(y)2. SOLUTION If we let u y, then dudx y . After substituting, the second-order equation reduces to a ﬁrst-order equation with separable variables; the independent variable is x and the dependent variable is u: The constant of integration is written as for convenience. The reason should be obvious in the next few steps. Because u1 1y, it follows that and so Independent Variable Missing Next we show how to solve an equation y dx x2 c1 2 or y 1 c1 tan1 x c1 c2. dy dx 1 x2 c1 2, c1 2 u1 x2 c1 2. u2 du 2x dx du dx 2xu2 or du u2 2x dx EXAMPLE 2 Independent Variable x Is Missing Solve yy (y)2. technique for an equation of the form F(x, y, y ) 0. If u y, then the differential equation becomes F(x, u, u) 0. If we can solve this last equation for u, we can ﬁnd y by integration. Note that since we are solving a second-order equation, its solution will contain two arbitrary constants. that has the form F(y, y, y ) 0. Once more we let u y, but because the indepen-dent variable x is missing, we use this substitution to transform the differential equa-tion into one in which the independent variable is y and the dependent variable is u. To this end we use the Chain Rule to compute the second derivative of y: In this case the ﬁrst-order equation that we must now solve is Fy, u, u du dy 0. y du dx du dy dy dx u du dy. where the dependent variable y is missing, and F(y, y, y ) 0, where the indepen-dent variable x is missing, can sometimes be solved by using first-order methods. Each equation can be reduced to a first-order equation by means of the substitu-tion u y. Dependent Variable Missing The next example illustrates the substitution Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SOLUTION With the aid of u y, the Chain Rule shown above, and separation of variables, the given differential equation becomes Integrating the last equation then yields lnu lny c1, which, in turn, gives u c2y, where the constant has been relabeled as c2. We now resubstitute u dydx, sepa-rate variables once again, integrate, and relabel constants a second time: Use of Taylor Series In some instances a solution of a nonlinear initial-value problem, in which the initial conditions are speciﬁed at x0, can be approximated by a Taylor series centered at x0. dy y c2 dx or lny c2x c3 or y c4ec2x. ec1 yu du dy u2 or du u dy y . 4.10 NONLINEAR DIFFERENTIAL EQUATIONS ● 187 EXAMPLE 3 Taylor Series Solution of an IVP Let us assume that a solution of the initial-value problem (1) exists. If we further assume that the solution y(x) of the problem is analytic at 0, then y(x) possesses a Taylor series expansion centered at 0: (2) Note that the values of the ﬁrst and second terms in the series (2) are known since those values are the speciﬁed initial conditions y(0) 1, y(0) 1. Moreover, the differential equation itself deﬁnes the value of the second derivative at 0: y (0) 0 y(0) y(0)2 0 (1) (1)2 2. We can then ﬁnd expressions for the higher derivatives y, y(4), . . . by calculating the successive derivatives of the differential equation: (3) (4) (5) and so on. Now using y(0) 1 and y(0) 1, we ﬁnd from (3) that y(0) 4. From the values y(0) 1, y(0) 1, and y (0) 2 we ﬁnd y(4)(0) 8 from (4). With the additional information that y(0) 4, we then see from (5) that y(5)(0) 24. Hence from (2) the ﬁrst six terms of a series solution of the initial-value problem (1) are Use of a Numerical Solver Numerical methods, such as Euler’s method or the Runge-Kutta method, are developed solely for ﬁrst-order differential equations and then are extended to systems of ﬁrst-order equations. To analyze an nth-order initial-value problem numerically, we express the nth-order ODE as a system of n ﬁrst-order equa-tions. In brief, here is how it is done for a second-order initial-value problem: First, solve y(x) 1 x x2 2 3 x3 1 3 x4 1 5 x5 . y(5)(x) d dx (y 2yy 2(y)2) y 2yy 6yy y(4)(x) d dx (1 y 2yy) y 2yy 2(y)2 y(x) d dx (x y y2) 1 y 2yy y(x) y(0) y(0) 1! x y (0) 2! x2 y(0) 3! x3 y(4)(0) 4! x4 y(5)(0) 5! x5 . y x y y2, y(0) 1, y(0) 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 188 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS for y —that is, put the DE into normal form y f(x, y, y)—and then let y u. For example, if we substitute y u in (6) then y u and y(x0) u(x0), so the initial-value problem (6) becomes However, it should be noted that a commercial numerical solver might not require that you supply the system. Subject to: y(x0) y0, u(x0) u0. Solve: y u u f(x, y, u) d 2y dx2 f (x, y, y), y(x0) y0, y(x0) u0, FIGURE 4.10.2 Numerical solution curve for the IVP in (1) y 10 20 x FIGURE 4.10.1 Comparison of two approximate solutions in Example 1 y x Taylor polynomial solution curve generated by a numerical solver EXAMPLE 4 Graphical Analysis of Example 3 Following the foregoing procedure, we ﬁnd that the second-order initial-value prob-lem in Example 3 is equivalent to with initial conditions y(0) 1, u(0) 1. With the aid of a numerical solver we get the solution curve shown in blue in Figure 4.10.1. For comparison the graph of the ﬁfth-degree Taylor polynomial is shown in red. Although we do not know the interval of convergence of the Taylor series obtained in Example 3, the closeness of the two curves in a neighborhood of the ori-gin suggests that the power series may converge on the interval (1, 1). Qualitative Questions The blue numerical solution curve in Figure 4.10.1 raises some questions of a qualitative nature: Is the solution of the original initial-value problem oscillatory as ? The graph generated by a numerical solver on the larger interval shown in Figure 4.10.2 would seem to suggest that the answer is yes. But this single example—or even an assortment of examples—does not answer the basic ques-tion as to whether all solutions of the differential equation y x y y2 are oscilla-tory in nature. Also, what is happening to the solution curve in Figure 4.10.2 when x is near 1? What is the behavior of solutions of the differential equation as ? Are solutions bounded as ? Questions such as these are not easily answered, in general, for nonlinear second-order differential equations. But certain kinds of second-order equations lend themselves to a systematic qualitative analysis, and these, like their ﬁrst-order relatives encountered in Section 2.1, are the kind that have no explicit dependence on the independent variable. Second-order ODEs of the form equations free of the independent variable x, are called autonomous. The differen-tial equation in Example 2 is autonomous, and because of the presence of the x term on its right-hand side, the equation in Example 3 is nonautonomous. For an in-depth treatment of the topic of stability of autonomous second-order differential equations and autonomous systems of differential equations, refer to Chapter 10 in Differential Equations with Boundary-Value Problems. F(y, y, y ) 0 or d 2y dx2 f (y, y), x : x : x : T5(x) 1 x x2 2 3x3 1 3x4 1 5x5 du dx x y y2 dy dx u Some numerical solvers require only that a second-order differential equation be expressed in normal form y f(x, y, y). The translation of the single equation into a system of two equations is then built into the computer program, since the ﬁrst equation of the system is always y u and the second equation is u f(x, y, u). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.10 NONLINEAR DIFFERENTIAL EQUATIONS ● 189 EXERCISES 4.10 Answers to selected odd-numbered problems begin on page ANS-7. In Problems 1 and 2 verify that y1 and y2 are solutions of the given differential equation but that y c1y1 c2y2 is, in general, not a solution. 1. (y )2 y2; y1 ex, y2 cos x 2. In Problems 3–8 solve the given differential equation by using the substitution u y. 3. y (y)2 1 0 4. y 1 (y)2 5. x2y (y)2 0 6. (y 1)y (y)2 7. y 2y( y)3 0 8. y2y y In Problems 9 and 10 solve the given initial-value problem. 9. 10. 11. Consider the initial-value problem y yy 0, y(0) 1, y(0) 1. (a) Use the DE and a numerical solver to graph the solution curve. (b) Find an explicit solution of the IVP. Use a graphing utility to graph this solution. (c) Find an interval of deﬁnition for the solution in part (b). 12. Find two solutions of the initial-value problem Use a numerical solver to graph the solution curves. In Problems 13 and 14 show that the substitution u y leads to a Bernoulli equation. Solve this equation (see Section 2.5). 13. xy y (y)3 14. xy y x( y)2 In Problems 15–18 proceed as in Example 3 and obtain the ﬁrst six nonzero terms of a Taylor series solution, centered at 0, of the given initial-value problem. Use a numerical solver and a graphing utility to compare the solution curve with the graph of the Taylor polynomial. 15. y x y2, y(0) 1, y(0) 1 16. y y2 1, y(0) 2, y(0) 3 17. y x2 y2 2y, y(0) 1, y(0) 1 18. y ey, y(0) 0, y(0) 1 ( y )2 ( y)2 1, y 2 1 2, y 2 13 2 . y x(y)2 0, y(1) 4, y(1) 2 2yy 1, y(0) 2, y(0) 1 yy 1 2 ( y)2; y1 1, y2 x2 19. In calculus the curvature of a curve that is deﬁned by a function y f(x) is deﬁned as Find y f(x) for which k 1. [Hint: For simplicity, ignore constants of integration.] Discussion Problems 20. In Problem 1 we saw that cos x and ex were solutions of the nonlinear equation (y )2 y2 0. Verify that sin x and ex are also solutions. Without attempting to solve the differential equation, discuss how these explicit solutions can be found by using knowledge about linear equations. Without attempting to verify, discuss why the linear combinations y c1ex c2ex c3 cos x c4 sin x and y c2ex c4 sin x are not, in general, solutions, but the two special linear combinations y c1ex c2ex and y c3 cos x c4 sin x must satisfy the differential equation. 21. Discuss how the method of reduction of order con-sidered in this section can be applied to the third-order differential equation . Carry out your ideas and solve the equation. 22. Discuss how to ﬁnd an alternative two-parameter fam-ily of solutions for the nonlinear differential equation y 2x(y)2 in Example 1. [Hint: Suppose that is used as the constant of integration instead of .] Mathematical Models 23. Motion in a For ce Field A mathematical model for the position x(t) of a body moving rectilinearly on the x-axis in an inverse-square force ﬁeld is given by Suppose that at t 0 the body starts from rest from the position x x0, x0 0. Show that the velocity of the body at time t is given by v2 2k2(1x 1x0). Use the last expression and a CAS to carry out the inte-gration to express time t in terms of x. 24. A mathematical model for the position x(t) of a moving object is . Use a numerical solver to graphically investigate the so-lutions of the equation subject to x(0) 0, x(0) x1, d 2x dt2 sin x 0 d 2x dt2 k2 x2. c1 2 c1 2 y 11 (y )2 y [1 ( y)2]3/2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 190 ● CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS x1 0. Discuss the motion of the object for t 0 and for various choices of x1. Investigate the equation d 2x dt2 dx dt sin x 0 in the same manner. Give a possible physical interpreta-tion of the dxdt term. CHAPTER 4 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-7. Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false. 1. The only solution of the initial-value problem y x2y 0, y(0) 0, y(0) 0 is _. 2. For the method of undetermined coefﬁcients, the assumed form of the particular solution yp for y y 1 ex is _. 3. A constant multiple of a solution of a linear differential equation is also a solution. _ 4. If the set consisting of two functions f1 and f2 is linearly independent on an interval I, then the Wronskian W( f1, f2) 0 for all x in I. _ 5. If is a solution of a homogeneous linear second-order differential with constant coefﬁcients, then the general solution of the DE is _. 6. If is a solution of a homoge-neous fourth-order linear differential equation with con-stant coefﬁcients, then the roots of the auxiliary equation are _. 7. If is the general solution of a homogeneous second-order Cauchy-Euler equation, then the DE is _. 8. is particular solution of for _. 9. If is a particular solution of and is a particular solution of then a particular solution of is _. 10. If and are solutions of homogeneous linear differential equation, then necessarily is also a solution of the DE. _ 11. Give an interval over which the set of two functions f1(x) x2 and f2(x) x x is linearly independent. Then give an interval over which the set consisting of f1 and f2 is linearly dependent. 12. Without the aid of the Wronskian, determine whether the given set of functions is linearly independent or linearly dependent on the indicated interval. (a) f1(x) ln x, f2(x) ln x2, (0, ) (b) f1(x) xn, f2(x) xn1, n 1, 2, . . . , ( , ) y 5ex 10ex y2 ex y1 ex y y x2 x y y x2, yp2 x2 2 y y x yp1 x A y y 1 yp Ax2 y c1x2 c2x2 ln x, x 0, y 1 x 6x2 3ex y sin 5x (c) f1(x) x, f2(x) x 1, ( , ) (d) (e) f1(x) 0, f2(x) x, (5, 5) (f) f1(x) 2, f2(x) 2x, ( , ) (g) f1(x) x2, f2(x) 1 x2, f3(x) 2 x2, ( , ) (h) f1(x) xex1, f2(x) (4x 5)ex, f3(x) xex, ( , ) 13. Suppose m1 3, m2 5, and m3 1 are roots of multiplicity one, two, and three, respectively, of an aux-iliary equation. Write down the general solution of the corresponding homogeneous linear DE if it is (a) an equation with constant coefﬁcients, (b) a Cauchy-Euler equation. 14. Consider the differential equation ay by cy g(x), where a, b, and c are constants. Choose the input func-tions g(x) for which the method of undetermined coefﬁ-cients is applicable and the input functions for which the method of variation of parameters is applicable. (a) g(x) ex ln x (b) g(x) x3 cos x (c) (d) g(x) 2x2ex (e) g(x) sin2x (f) In Problems 15–30 use the procedures developed in this chapter to ﬁnd the general solution of each differential equation. 15. y 2y 2y 0 16. 2y 2y 3y 0 17. y 10y 25y 0 18. 2y 9y 12y 5y 0 19. 3y 10y 15y 4y 0 20. 2y(4) 3y 2y 6y 4y 0 21. y 3y 5y 4x3 2x 22. y 2y y x2ex g(x) ex sin x g(x) sin x ex f1(x) cosx 2, f2(x) sin x, ( , ) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 4 IN REVIEW ● 191 23. y 5y 6y 8 2 sin x 24. y y 6 25. y 2y 2y ex tan x 26. 27. 6x2y 5xy y 0 28. 2x3y 19x2y 39xy 9y 0 29. x2y 4xy 6y 2x4 x2 30. x2y xy y x3 31. Write down the form of the general solution y yc yp of the given differential equation in the two cases v a and v a. Do not determine the coefﬁcients in yp. (a) y v2y sin ax (b) y v2y eax 32. (a) Given that y sin x is a solution of y(4) 2y 11y 2y 10y 0, ﬁnd the general solution of the DE without the aid of a calculator or a computer. (b) Find a linear second-order differential equation with constant coefﬁcients for which y1 1 and y2 ex are solutions of the associated homoge-neous equation and is a particular solution of the nonhomogeneous equation. 33. (a) Write the general solution of the fourth-order DE y(4) 2y y 0 entirely in terms of hyperbolic functions. (b) Write down the form of a particular solution of y(4) 2y y sinh x. 34. Consider the differential equation x2y (x2 2x)y (x 2)y x3. Verify that y1 x is one solution of the associated homogeneous equation. Then show that the method of reduction of order discussed in Section 4.2 leads to a second solution y2 of the homogeneous equation as well as a particular solution yp of the nonhomogeneous equa-tion. Form the general solution of the DE on the interval (0, ). yp 1 2x2 x y y 2ex ex ex In Problems 35–40 solve the given differential equation sub-ject to the indicated conditions. 35. 36. y 2y y 0, y(1) 0, y(0) 0 37. y y x sin x, y(0) 2, y(0) 3 38. 39. yy 4x, y(1) 5, y(1) 2 40. 2y 3y2, y(0) 1, y(0) 1 41. (a) Use a CAS as an aid in ﬁnding the roots of the aux-iliary equation for 12y(4) 64y 59y 23y 12y 0. Give the general solution of the equation. (b) Solve the DE in part (a) subject to the initial condi-tions y(0) 1, y(0) 2, y (0) 5, y(0) 0. Use a CAS as an aid in solving the resulting systems of four equations in four unknowns. 42. Find a member of the family of solutions of whose graph is tangent to the x-axis at x 1. Use a graphing utility to graph the solution curve. In Problems 43–46 use systematic elimination to solve the given system. 43. 44. 45. 46. 5x (D 3)y cos 2t (D 2 )x (D 1)y sin 2t 3 x (D 4 ) y 7et (D 2)x y et dy dt 3x 4y 4t dx dt 2x y t 2 dx dt 2 dy dt y 3 dx dt dy dt 2x 2y 1 xy y 1x 0 y y sec3x, y(0) 1, y(0) 1 2 y 2y 2y 0, y (p>2) 0, y() 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 192 Modeling with Higher-Order Differential Equations We have seen that a single differential equation can serve as a mathematical model for diverse physical systems. For this reason we examine just one application, the motion of a mass attached to a spring, in great detail in Section 5.1. Except for terminology and physical interpretations of the four terms in the linear differential equation the mathematics of, say, an electrical series circuit is identical to that of a vibrating spring/mass system. Forms of this linear second-order equation appear in the analysis of problems in many different areas of science and engineering. In Section 5.1 we deal exclusively with initial-value problems, whereas in Section 5.2 we examine applications described by boundary-value problems. In Section 5.2 we also see how some boundary-value problems lead to the important concepts of eigenvalues and eigenfunctions. Section 5.3 begins with a discussion on the differences between linear and nonlinear springs; we then show how the simple pendulum and a suspended wire lead to nonlinear models. a d 2y dt2 b dy dt cy g(t), 5.1 Linear Models: Initial-Value Problems 5.1.1 Spring/Mass Systems: Free Undamped Motion 5.1.2 Spring/Mass Systems: Free Damped Motion 5.1.3 Spring/Mass Systems: Driven Motion 5.1.4 Series Circuit Analogue 5.2 Linear Models: Boundary-Value Problems 5.3 Nonlinear Models Chapter 5 in Review 5 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 193 5.1.1 SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION Hooke’s Law Suppose that a ﬂexible spring is suspended vertically from a rigid support and then a mass m is attached to its free end. The amount of stretch, or elonga-tion, of the spring will of course depend on the mass; masses with different weights stretch the spring by differing amounts. By Hooke’s law the spring itself exerts a restor-ing force F opposite to the direction of elongation and proportional to the amount of elongation s. Simply stated, F ks, where k is a constant of proportionality called the spring constant. The spring is essentially characterized by the number k. For example, if a mass weighing 10 pounds stretches a spring foot, then implies k 20 lb/ft. Necessarily then, a mass weighing, say, 8 pounds stretches the same spring only foot. Newton’s Second Law After a mass m is attached to a spring, it stretches the spring by an amount s and attains a position of equilibrium at which its weight W is balanced by the restoring force ks. Recall that weight is deﬁned by W mg, where mass is measured in slugs, kilograms, or grams and g 32 ft/s2, 9.8 m/s2, or 980 cm/s2, respectively. As indicated in Figure 5.1.1(b), the condition of equilibrium is mg ks or mg ks 0. If the mass is displaced by an amount x from its equilibrium position, the restoring force of the spring is then k(x s). Assuming that there are no retarding forces acting on the system and assuming that the mass vibrates free of other external forces—free motion—we can equate Newton’s second law with the net, or resultant, force of the restoring force and the weight: (1) The negative sign in (1) indicates that the restoring force of the spring acts opposite to the direction of motion. Furthermore, we adopt the convention that displacements measured below the equilibrium position x 0 are positive. See Figure 5.1.2. d 2x ––– dt 2 k(s x) mg kx mg ks kx. m zero 2 5 10 k 1 2 1 2 m (a) (b) (c) unstretched motion l equilibrium position mg −ks = 0 m l l + s s x FIGURE 5.1.1 Spring/mass system m x = 0 x < 0 x > 0 FIGURE 5.1.2 Direction below the equilibrium position is positive. LINEAR MODELS: INITIAL-VALUE PROBLEMS REVIEW MATERIAL ●Sections 4.1, 4.3, and 4.4 ●Problems 29–36 in Exercises 4.3 ●Problems 27–36 in Exercises 4.4 INTRODUCTION In this section we are going to consider several linear dynamical systems in which each mathematical model is a second-order differential equation with constant coefﬁcients along with initial conditions speciﬁed at a time that we shall take to be t 0: . Recall that the function g is the input, driving function, or forcing function of the system. A solution y(t) of the differential equation on an interval I containing t 0 that satisﬁes the initial conditions is called the output or response of the system. a d 2y dt2 b dy dt cy g(t), y(0) y0, y(0) y1 5.1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 194 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS DE of Free Undamped Motion By dividing (1) by the mass m, we obtain the second-order differential equation d2xdt2 (km)x 0, or , (2) where v2 km. Equation (2) is said to describe simple harmonic motion or free undamped motion. Two obvious initial conditions associated with (2) are x(0) x0 and x(0) x1, the initial displacement and initial velocity of the mass, respectively. For example, if x0 0, x1 0, the mass starts from a point below the equilibrium position with an imparted upward velocity. When x(0) 0, the mass is said to be released from rest. For example, if x0 0, x1 0, the mass is released from rest from a point x0 units above the equilibrium position. d 2x dt2 2x 0 Equation of Motion To solve equation (2), we note that the solutions of its auxiliary equation m2 v2 0 are the complex numbers m1 vi, m2 vi. Thus from (8) of Section 4.3 we ﬁnd the general solution of (2) to be . (3) The period of motion described by (3) is T 2pv. The number T represents the time (measured in seconds) it takes the mass to execute one cycle of motion. A cycle is one complete oscillation of the mass, that is, the mass m moving from, say, the lowest point below the equilibrium position to the point highest above the equilibrium position and then back to the lowest point. From a graphical viewpoint T 2pv seconds is the length of the time interval between two successive maxima (or minima) of x(t). Keep in mind that a maximum of x(t) is a positive displacement corresponding to the mass attaining its greatest distance below the equilibrium position, whereas a minimum of x(t) is negative displacement corresponding to the mass attaining its greatest height above the equilibrium position. We refer to either case as an extreme displacement of the mass. The frequency of motion is f 1T v2p and is the number of cycles completed each second. For example, if x(t) 2 cos 3pt 4 sin 3pt, then the period is T 2p3p 23 s, and the frequency is f 32 cycles/s. From a graphical view-point the graph of x(t) repeats every second, that is, , and cycles of the graph are completed each second (or, equivalently, three cycles of the graph are completed every 2 seconds). The number (measured in radians per sec-ond) is called the circular frequency of the system. Depending on which text you read, both f v2p and v are also referred to as the natural frequency of the system. Finally, when the initial conditions are used to determine the constants c1 and c2 in (3), we say that the resulting particular solution or response is the equation of motion. 1k>m 3 2 x(t 2 3) x(t) 2 3 x(t) c1 cos t c2 sin t EXAMPLE 1 Free Undamped Motion A mass weighing 2 pounds stretches a spring 6 inches. At t 0 the mass is released from a point 8 inches below the equilibrium position with an upward velocity of . Determine the equation of motion. SOLUTION Because we are using the engineering system of units, the measure-ments given in terms of inches must be converted into feet: ; . In addition, we must convert the units of weight given in pounds into units of mass. From m Wg we have slug. Also, from Hooke’s law, implies that the spring constant is k 4 lb/ft. Hence (1) gives . The initial displacement and initial velocity are , , where the neg-ative sign in the last condition is a consequence of the fact that the mass is given an initial velocity in the negative, or upward, direction. x (0) 4 3 x(0) 2 3 1 16 d 2x dt2 4x or d 2x dt2 64x 0 2 k 1 2 m 2 32 1 16 8 in. 2 3 ft 6 in. 1 2 ft 4 3 ft/s Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 195 Now v2 64 or v 8, so the general solution of the differential equation is . (4) Applying the initial conditions to x(t) and x(t) gives and . Thus the equation of motion is . (5) Alternative Forms of x(t) When c1 0 and c2 0, the actual amplitude A of free vibrations is not obvious from inspection of equation (3). For example, although the mass in Example 1 is initially displaced foot beyond the equilibrium position, the amplitude of vibrations is a number larger than . Hence it is often con-venient to convert a solution of form (3) to the simpler form , (6) where and f is a phase angle deﬁned by . (7) To verify this, we expand (6) by the addition formula for the sine function: . (8) It follows from Figure 5.1.3 that if f is deﬁned by , then (8) becomes . A c1 A cos t A c2 A sin t c1 cos t c2 sin t x(t) sin c1 1c1 2 c2 2 c1 A , cos c2 1c1 2 c2 2 c2 A A sin t cos cos t sin ( sin )cos t ( cos )sin t sin c1 A cos c2 A tan c1 c2 A 2c1 2 c2 2 x(t) A sin(t ) 2 3 2 3 x(t) 2 3 cos 8t 1 6 sin 8t c2 1 6 c1 2 3 x(t) c1 cos 8t c2 sin 8t c1 c2 φ c1 + c2 2 2 FIGURE 5.1.3 A relationship between c1 0, c2 0 and phase angle f EXAMPLE 2 Alternative Form of Solution (5) In view of the foregoing discussion we can write solution (5) in the alternative form x(t) A sin(8t f). Computation of the amplitude is straightforward, , but some care should be exercised in computing the phase angle f defined by (7). With and we find tan f 4, and a calculator then gives tan1(4) 1.326 rad. This is not the phase angle, since tan1(4) is located in the fourth quadrant and therefore con-tradicts the fact that sin f 0 and cos f 0 because c1 0 and c2 0. Hence we must take f to be the second-quadrant angle f p (1.326) 1.816 rad. Thus (5) is the same as . (9) The period of this function is T 2p8 p4 s. You should be aware that some instructors in science and engineering prefer that (3) be expressed as a shifted cosine function x(t) A cos(vt f), (6) x(t) 117 6 sin(8t 1.816) c2 1 6 c1 2 3 A 2(2 3)2 (1 6)2 217 36 0.69 ft Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 196 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS where In this case the radian measured angle is deﬁned in slightly different manner than in (7): . (7) For example, in Example 2 with and (7) indicates that tan Because sin and cos the angle lies in the fourth quadrant and so rounded to three decimal places rad. From (6) we obtain a second alternative form of solution (5): . Graphical Interpretation Figure 5.1.4(a) illustrates the mass in Example 2 going through approximately two complete cycles of motion. Reading left to right, the ﬁrst ﬁve positions (marked with black dots) correspond to the initial position of the mass below the equilibrium position , the mass passing through the equilibrium position for the ﬁrst time heading upward (x 0), the mass at its extreme displacement above the equilibrium position , the mass at the equilibrium position for the second time heading downward (x 0), and the mass at its extreme displace-ment below the equilibrium position . The black dots on the graph of (9), given in Figure 5.1.4(b), also agree with the ﬁve positions just given. Note, however, that in Figure 5.1.4(b) the positive direction in the tx-plane is the usual upward (x 1176) (x 1176) (x 2 3) x(t) 217 6 cos(8t (0.245)) or x(t) 217 6 cos(8t 0.245) f tan1(1 4) 0.245 f f 0 f 0 f 1 4. c2 1 6, c1 2 3 sin c2 A cos c1 A tan c2 c1 f A 2c2 1 c2 2. x = − 6 17 x = 6 17 x = 0 2 3 x = x = 0 x = 0 x negative x positive (a) (b) x t (0, ) 2 3 period 4 π amplitude A = 6 17 x = 0 x negative x positive FIGURE 5.1.4 Simple harmonic motion Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 197 direction and so is opposite to the positive direction indicated in Figure 5.1.4(a). Hence the solid blue graph representing the motion of the mass in Figure 5.1.4(b) is the reﬂection through the t-axis of the blue dashed curve in Figure 5.1.4(a). Form (6) is very useful because it is easy to find values of time for which the graph of x(t) crosses the positive t-axis (the line x 0). We observe that sin(vt f) 0 when vt f np, where n is a nonnegative integer. Systems with Variable Spring Constants In the model discussed above we assumed an ideal world—a world in which the physical characteristics of the spring do not change over time. In the nonideal world, however, it seems reasonable to ex-pect that when a spring/mass system is in motion for a long period, the spring will weaken; in other words, the “spring constant” will vary—or, more speciﬁcally, decay—with time. In one model for the aging spring the spring constant k in (1) is replaced by the decreasing function K(t) keat, k 0, a 0. The linear differential equation mx keatx 0 cannot be solved by the methods that were considered in Chapter 4. Nevertheless, we can obtain two linearly independent solu-tions using the methods in Chapter 6. See Problem 15 in Exercises 5.1, Example 4 in Section 6.4, and Problems 33 and 39 in Exercises 6.4. When a spring/mass system is subjected to an environment in which the temperature is rapidly decreasing, it might make sense to replace the constant k with K(t) kt, k 0, a function that increases with time. The resulting model, mx ktx 0, is a form of Airy’s differential equation. Like the equation for an aging spring, Airy’s equation can be solved by the methods of Chapter 6. See Problem 16 in Exercises 5.1, Example 5 in Section 6.2, and Problems 34, 35, and 40 in Exercise 6.4. 5.1.2 SPRING/MASS SYSTEMS: FREE DAMPED MOTION The concept of free harmonic motion is somewhat unrealistic, since the motion described by equation (1) assumes that there are no retarding forces acting on the mov-ing mass. Unless the mass is suspended in a perfect vacuum, there will be at least a re-sisting force due to the surrounding medium. As Figure 5.1.5 shows, the mass could be suspended in a viscous medium or connected to a dashpot damping device. DE of Free Damped Motion In the study of mechanics, damping forces acting on a body are considered to be proportional to a power of the instantaneous velocity. In particular, we shall assume throughout the subsequent discussion that this force is given by a constant multiple of dxdt. When no other external forces are impressed on the system, it follows from Newton’s second law that , (10) where b is a positive damping constant and the negative sign is a consequence of the fact that the damping force acts in a direction opposite to the motion. Dividing (10) by the mass m, we ﬁnd that the differential equation of free damped motion is or , (11) where . (12) 2 m, 2 k m d 2x dt2 2 dx dt 2x 0 d 2x dt 2 b m dx dt k mx 0 m d 2x dt2 kx dx dt m (a) (b) m FIGURE 5.1.5 Damping devices Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 198 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS The symbol 2l is used only for algebraic convenience because the auxiliary equation is m2 2lm v2 0, and the corresponding roots are then . We can now distinguish three possible cases depending on the algebraic sign of l2 v2. Since each solution contains the damping factor elt, l 0, the displace-ments of the mass become negligible as time t increases. Case I: L2 V2 0 In this situation the system is said to be overdamped because the damping coefﬁcient b is large when compared to the spring constant k. The corresponding solution of (11) is or . (13) This equation represents a smooth and nonoscillatory motion. Figure 5.1.6 shows two possible graphs of x(t). Case II: L2 V2 0 The system is said to be critically damped because any slight decrease in the damping force would result in oscillatory motion. The general solution of (11) is or . (14) Some graphs of typical motion are given in Figure 5.1.7. Notice that the motion is quite similar to that of an overdamped system. It is also apparent from (14) that the mass can pass through the equilibrium position at most one time. Case III: L2 V2 0 In this case the system is said to be underdamped, since the damping coefﬁcient is small in comparison to the spring constant. The roots m1 and m2 are now complex: . Thus the general solution of equation (11) is . (15) As indicated in Figure 5.1.8, the motion described by (15) is oscillatory; but because of the coefﬁcient elt, the amplitudes of vibration as . t : : 0 x(t) e t (c1 cos 12 2t c2 sin 12 2t) m1 12 2i, m2 12 2i x(t) e t(c1 c2t) x(t) c1em1t c2tem1t x(t) e t (c1e1 22t c2e1 22t) x(t) c1em1t c2em2t m1 2 2 2, m2 2 2 2 t x FIGURE 5.1.6 Motion of an overdamped system t x underdamped undamped t x FIGURE 5.1.7 Motion of a critically damped system FIGURE 5.1.8 Motion of an underdamped system EXAMPLE 3 Overdamped Motion It is readily veriﬁed that the solution of the initial-value problem is (16) The problem can be interpreted as representing the overdamped motion of a mass on a spring. The mass is initially released from a position 1 unit below the equilibrium position with a downward velocity of 1 ft/s. To graph x(t), we find the value of t for which the function has an extremum—that is, the value of time for which the first derivative (velocity) is zero. Differentiating (16) gives , so x(t) 0 implies that x(t) 5 3et 8 3e4t x(t) 5 3 et 2 3 e4t. d 2x dt2 5 dx dt 4x 0, x(0) 1, x(0) 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 199 or . It follows from the first derivative test, as well as our physical intuition, that x(0.157) 1.069 ft is actually a maximum. In other words, the mass attains an extreme displacement of 1.069 feet below the equilib-rium position. We should also check to see whether the graph crosses the t-axis—that is, whether the mass passes through the equilibrium position. This cannot happen in this instance because the equation x(t) 0, or , has the physically irrelevant solu-tion . The graph of x(t), along with some other pertinent data, is given in Figure 5.1.9. t 1 3 ln 2 5 0.305 e3t 2 5 t 1 3 ln 8 5 0.157 e3t 8 5 1 3 2 t x 5 3 x = − e−t e−4t 2 3 t x(t) 1 0.601 1.5 0.370 2 0.225 2.5 0.137 3 0.083 (a) (b) FIGURE 5.1.9 Overdamped system in Example 3 EXAMPLE 4 Critically Damped Motion A mass weighing 8 pounds stretches a spring 2 feet. Assuming that a damping force numerically equal to 2 times the instantaneous velocity acts on the system, determine the equation of motion if the mass is initially released from the equilibrium position with an upward velocity of 3 ft/s. SOLUTION From Hooke’s law we see that 8 k(2) gives k 4 lb/ft and that W mg gives slug. The differential equation of motion is then . (17) The auxiliary equation for (17) is m2 8m 16 (m 4)2 0, so m1 m2 4. Hence the system is critically damped, and . (18) Applying the initial conditions x(0) 0 and x(0) 3, we ﬁnd, in turn, that c1 0 and c2 3. Thus the equation of motion is . (19) To graph x(t), we proceed as in Example 3. From x(t) 3e4t(1 4t) we see that x(t) 0 when . The corresponding extreme displacement is . As shown in Figure 5.1.10, we interpret this value to mean that the mass reaches a maximum height of 0.276 foot above the equilibrium position. x(1 4) 3(1 4)e1 0.276 ft t 1 4 x(t) 3te4t x(t) c1e4t c2te4t 1 4 d 2x dt 2 4x 2 dx dt or d 2x dt2 8 dx dt 16x 0 m 8 32 1 4 EXAMPLE 5 Underdamped Motion A mass weighing 16 pounds is attached to a 5-foot-long spring. At equilibrium the spring measures 8.2 feet. If the mass is initially released from rest at a point 2 feet above the equilibrium position, ﬁnd the displacements x(t) if it is further known that the sur-rounding medium offers a resistance numerically equal to the instantaneous velocity. SOLUTION The elongation of the spring after the mass is attached is 8.2 5 3.2 ft, so it follows from Hooke’s law that 16 k(3.2) or k 5 lb/ft. In addition, slug, so the differential equation is given by . (20) Proceeding, we ﬁnd that the roots of m2 2m 10 0 are m1 1 3i and m2 1 3i, which then implies that the system is underdamped, and (21) x(t) et(c1 cos 3t c2 sin 3t). 1 2 d 2x dt2 5x dx dt or d 2x dt2 2 dx dt 10x 0 m 16 32 1 2 − 0.276 t x t = maximum height above equilibrium position 1 4 FIGURE 5.1.10 Critically damped system in Example 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 200 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Finally, the initial conditions x(0) 2 and x(0) 0 yield c1 2 and , so the equation of motion is . (22) Alternative Form of x(t) In a manner identical to the procedure used on page 195, we can write any solution in the alternative form , (23) where and the phase angle f is determined from the equations . The coefﬁcient Aelt is sometimes called the damped amplitude of vibrations. Because (23) is not a periodic function, the number is called the quasi period and is the quasi frequency. The quasi period is the time interval between two successive maxima of x(t). You should verify, for the equation of motion in Example 5, that and f 4.391. Therefore an equivalent form of (22) is . 5.1.3 SPRING/MASS SYSTEMS: DRIVEN MOTION DE of Driven Motion with Damping Suppose we now take into con-sideration an external force f(t) acting on a vibrating mass on a spring. For example, f(t) could represent a driving force causing an oscillatory vertical motion of the support of the spring. See Figure 5.1.11. The inclusion of f(t) in the formu-lation of Newton’s second law gives the differential equation of driven or forced motion: . (24) Dividing (24) by m gives , (25) where F(t) f(t)m and, as in the preceding section, 2l bm, v2 km. To solve the latter nonhomogeneous equation, we can use either the method of undetermined coefﬁcients or variation of parameters. d 2x dt2 2 dx dt 2x F(t) m d 2x dt2 kx dx dt f(t) x(t) 2110 3 et sin(3t 4.391) A 21103 12 2 2 2 12 2 sin c1 A, cos c2 A, tan c1 c2 A 1c1 2 c2 2 x(t) Ae t sin(12 2t ) x(t) e t (c1 cos 12 2t c2 sin 12 2t) x(t) et 2 cos 3t 2 3 sin 3t c2 2 3 m FIGURE 5.1.11 Oscillatory vertical motion of the support EXAMPLE 6 Interpretation of an Initial-Value Problem Interpret and solve the initial-value problem . (26) SOLUTION We can interpret the problem to represent a vibrational system consist-ing of a mass ( slug or kilogram) attached to a spring (k 2 lb/ft or N/m). m 1 5 1 5 d 2x dt2 1.2 dx dt 2x 5 cos 4t, x(0) 1 2, x(0) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 201 The mass is initially released from rest unit (foot or meter) below the equilibrium position. The motion is damped (b 1.2) and is being driven by an external peri-odic (T p2 s) force beginning at t 0. Intuitively, we would expect that even with damping, the system would remain in motion until such time as the forcing function was “turned off,” in which case the amplitudes would diminish. However, as the problem is given, f(t) 5 cos 4t will remain “on” forever. We ﬁrst multiply the differential equation in (26) by 5 and solve by the usual methods. Because m1 3 i, m2 3 i, it follows that xc(t) e3t(c1 cos t c2 sin t). Using the method of undetermined coefficients, we assume a particular solution of the form xp(t) A cos 4t B sin 4t. Differentiating xp(t) and substituting into the DE gives . The resulting system of equations yields and . It follows that . (27) When we set t 0 in the above equation, we obtain . By differentiating the expression and then setting t 0, we also ﬁnd that . Therefore the equation of motion is . (28) x(t) e3t 38 51 cos t 86 51 sin t 25 102 cos 4t 50 51 sin 4t c2 86 51 c1 38 51 x(t) e3t(c1 cos t c2 sin t) 25 102 cos 4t 50 51 sin 4t B 50 51 A 25 102 6A 24B 25, 24A 6B 0 x p 6x p 10xp (6A 24B) cos 4t (24A 6B) sin 4t 25 cos 4t dx2 dt2 6 dx dt 10x 0 1 2 t x steady state xp(t) transient _1 1 π /2 (a) (b) t x x(t)=transient + steady state _1 1 π /2 FIGURE 5.1.12 Graph of solution in (28) of Example 6 x 2π π x1=7 x1=3 x1=0 x1=_3 t FIGURE 5.1.13 Graph of solution in Example 7 for various initial velocities x1 EXAMPLE 7 Transient/Steady-State Solutions The solution of the initial-value problem , where x1 is constant, is given by Solution curves for selected values of the initial velocity x1 are shown in Figure 5.1.13. The graphs show that the inﬂuence of the transient term is negligible for about t 3p2. x(t) (x1 2) et sin t 2 sin t. transient steady-state d 2x dt2 2 dx dt 2x 4 cos t 2 sin t, x(0) 0, x(0) x1 Transient and Steady-State Terms When F is a periodic function, such as F(t) F0 sin gt or F(t) F0 cos gt, the general solution of (25) for l 0 is the sum of a nonperiodic function xc(t) and a periodic function xp(t). Moreover, xc(t) dies off as time increases—that is, . Thus for large values of time, the dis-placements of the mass are closely approximated by the particular solution xp(t). The complementary function xc(t) is said to be a transient term or transient solution, and the function xp(t), the part of the solution that remains after an interval of time, is called a steady-state term or steady-state solution. Note therefore that the effect of the initial conditions on a spring/mass system driven by F is transient. In the particular solution (28), is a transient term, and is a steady-state term. The graphs of these two terms and the solution (28) are given in Figures 5.1.12(a) and 5.1.12(b), respectively. 25 102 cos 4t 50 51 sin 4t xp(t) e3t (38 51 cos t 86 51 sin t) limt: xc (t) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 202 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS DE of Driven Motion without Damping With a periodic impressed force and no damping force, there is no transient term in the solution of a problem. Also, we shall see that a periodic impressed force with a frequency near or the same as the frequency of free undamped vibrations can cause a severe problem in any oscillatory mechanical system. x t FIGURE 5.1.14 Pure resonance EXAMPLE 8 Undamped Forced Motion Solve the initial-value problem , (29) where F0 is a constant and g v. SOLUTION The complementary function is xc(t) c1 cos vt c2 sin vt. To obtain a particular solution, we assume xp(t) A cos gt B sin gt so that . Equating coefﬁcients immediately gives A 0 and B F0(v2 g2). Therefore . Applying the given initial conditions to the general solution yields c1 0 and c2 gF0v(v2 g2). Thus the solution is . (30) Pure Resonance Although equation (30) is not deﬁned for g v, it is interesting to observe that its limiting value as can be obtained by applying L’Hôpital’s Rule. This limiting process is analogous to “tuning in” the frequency of the driving force (g2p) to the frequency of free vibrations (v2p). Intuitively, we expect that over a length of time we should be able to substantially increase the amplitudes of vi-bration. For g v we deﬁne the solution to be (31) As suspected, when , the displacements become large; in fact, when tn npv, n 1, 2, . . . . The phenomenon that we have just described is known as pure resonance. The graph given in Figure 5.1.14 shows typical motion in this case. In conclusion it should be noted that there is no actual need to use a limiting process on (30) to obtain the solution for g v. Alternatively, equation (31) follows by solving the initial-value problem directly by conventional methods. d 2x dt2 2x F0 sin t, x(0) 0, x(0) 0 x(tn) B t : F0 22 sin t F0 2 t cos t. F0 sin t t cos t 22 F0 lim : sin t t cos t 2 x(t) lim :F0 sin t sin t (2 2) F0lim : d d ( sin t sin t) d d (3 2) : x(t) F0 (2 2) ( sin t sin t), x(t) c1 cos t c2 sin t F0 2 2 sin t xp(t) F0 2 2 sin t x p 2xp A(2 2) cos t B(2 2) sin t F0 sin t d 2x dt2 2x F0 sin t, x(0) 0, x(0) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 203 If the displacements of a spring/mass system were actually described by a func-tion such as (31), the system would necessarily fail. Large oscillations of the mass would eventually force the spring beyond its elastic limit. One might argue too that the resonating model presented in Figure 5.1.14 is completely unrealistic because it ignores the retarding effects of ever-present damping forces. Although it is true that pure resonance cannot occur when the smallest amount of damping is taken into con-sideration, large and equally destructive amplitudes of vibration (although bounded as ) can occur. See Problem 43 in Exercises 5.1. 5.1.4 SERIES CIRCUIT ANALOGUE LRC-Series Circuits As was mentioned in the introduction to this chapter, many different physical systems can be described by a linear second-order differential equation similar to the differential equation of forced motion with damping: . (32) If i(t) denotes current in the LRC-series electrical circuit shown in Figure 5.1.15, then the voltage drops across the inductor, resistor, and capacitor are as shown in Figure 1.3.4. By Kirchhoff’s second law the sum of these voltages equals the voltage E(t) impressed on the circuit; that is, . (33) But the charge q(t) on the capacitor is related to the current i(t) by i dqdt, so (33) becomes the linear second-order differential equation . (34) The nomenclature used in the analysis of circuits is similar to that used to describe spring/mass systems. If E(t) 0, the electrical vibrations of the circuit are said to be free. Because the auxiliary equation for (34) is Lm2 Rm 1C 0, there will be three forms of the solution with R 0, depending on the value of the discriminant R2 4LC. We say that the circuit is , , and . In each of these three cases the general solution of (34) contains the factor eRt/2L, so as . In the underdamped case when q(0) q0, the charge on the capacitor oscillates as it decays; in other words, the capacitor is charging and dis-charging as . When E(t) 0 and R 0, the circuit is said to be undamped, and the electrical vibrations do not approach zero as t increases without bound; the response of the circuit is simple harmonic. t : t : q(t) : 0 underdamped if R2 4L/C 0 critically damped if R2 4L/C 0 overdamped if R2 4L/C 0 L d 2q dt2 R dq dt 1 C q E(t) L di dt Ri 1 C q E(t) m d 2x dt2 dx dt kx f(t) t : C L E R FIGURE 5.1.15 LRC-series circuit EXAMPLE 9 Underdamped Series Circuit Find the charge q(t) on the capacitor in an LRC-series circuit when L 0.25 henry (h), R 10 ohms (), C 0.001 farad (f), E(t) 0, q(0) q0 coulombs (C), and i(0) 0. SOLUTION Since 1C 1000, equation (34) becomes . 1 4 q 10q 1000q 0 or q 40q 4000q 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 204 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Solving this homogeneous equation in the usual manner, we ﬁnd that the circuit is underdamped and q(t) e20t(c1 cos 60t c2 sin 60t). Applying the initial condi-tions, we ﬁnd c1 q0 and . Thus . Using (23), we can write the foregoing solution as . When there is an impressed voltage E(t) on the circuit, the electrical vibrations are said to be forced. In the case when R 0, the complementary function qc(t) of (34) is called a transient solution. If E(t) is periodic or a constant, then the particu-lar solution qp(t) of (34) is a steady-state solution. q(t) q0110 3 e20t sin(60t 1.249) q(t) q0e20t cos 60t 1 3 sin 60t c2 1 3 q0 EXAMPLE 10 Steady-State Current Find the steady-state solution qp(t) and the steady-state current in an LRC-series circuit when the impressed voltage is E(t) E0 sin gt. SOLUTION The steady-state solution qp(t) is a particular solution of the differential equation . Using the method of undetermined coefﬁcients, we assume a particular solution of the form qp(t) A sin gt B cos gt. Substituting this expression into the differen-tial equation, simplifying, and equating coefﬁcients gives It is convenient to express A and B in terms of some new symbols. If If Therefore A E0X(gZ2) and B E0R(gZ2), so the steady-state charge is . Now the steady-state current is given by : . (35) The quantities X Lg 1Cg and deﬁned in Example 10 are called the reactance and impedance, respectively, of the circuit. Both the reactance and the impedance are measured in ohms. Z 1X2 R2 ip(t) E0 Z R Z sin t X Z cos t ip(t) q p(t) qp(t) E0X Z2 sin t E0R Z2 cos t Z 1X2 R2, then Z 2 L22 2L C 1 C22 R2. X L 1 C , then X2 L22 2L C 1 C22. A E0L 1 C L22 2L C 1 C2 2 R2 , B E0R L22 2L C 1 C2 2 R2 . L d 2q dt2 R dq dt 1 C q E0 sin t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 205 EXERCISES 5.1 Answers to selected odd-numbered problems begin on page ANS-7. 5.1.1 SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION 1. A mass weighing 4 pounds is attached to a spring whose spring constant is 16 lb/ft. What is the period of simple harmonic motion? 2. A 20-kilogram mass is attached to a spring. If the fre-quency of simple harmonic motion is 2p cycles/s, what is the spring constant k? What is the frequency of simple harmonic motion if the original mass is replaced with an 80-kilogram mass? 3. A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 3 inches above the equi-librium position. Find the equation of motion. 4. Determine the equation of motion if the mass in Problem 3 is initially released from the equilibrium position with a downward velocity of 2 ft/s. 5. A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the equilibrium position. (a) Find the position of the mass at the times t p12, p8, p6, p4, and 9p32 s. (b) What is the velocity of the mass when t 3p16 s? In which direction is the mass heading at this instant? (c) At what times does the mass pass through the equi-librium position? 6. A force of 400 newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Find the equation of motion. 7. Another spring whose constant is 20 N/m is suspended from the same rigid support but parallel to the spring/mass system in Problem 6. A mass of 20 kilo-grams is attached to the second spring, and both masses are initially released from the equilibrium position with an upward velocity of 10 m/s. (a) Which mass exhibits the greater amplitude of motion? (b) Which mass is moving faster at t p4 s? At p2 s? (c) At what times are the two masses in the same position? Where are the masses at these times? In which directions are the masses moving? 8. A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 2 ft/s. How many complete cycles will the mass have com-pleted at the end of 4p seconds? 9. A mass weighing 8 pounds is attached to a spring. When set in motion, the spring/mass system exhibits simple harmonic motion. (a) Determine the equation of motion if the spring con-stant is 1 lb/ft and the mass is initially released from a point 6 inches below the equilibrium position with a downward velocity of (b) Express the equation of motion in the form given in (6). (c) Express the equation of motion in the form given in (6). 10. A mass weighing 10 pounds stretches a spring foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point foot above the equilibrium position with a downward veloc-ity of . (a) Express the equation of motion in the form given in (6). (b) Express the equation of motion in the form given in (6) (c) Use one of the solutions obtained in parts (a) and (b) to determine the times the mass attains a displace-ment below the equilibrium position numerically equal to the amplitude of motion. 11. A mass weighing 64 pounds stretches a spring 0.32 foot. The mass is initially released from a point 8 inches above the equilibrium position with a downward veloc-ity of 5 ft/s. (a) Find the equation of motion. (b) What are the amplitude and period of motion? (c) How many complete cycles will the mass have completed at the end of 3p seconds? (d) At what time does the mass pass through the equilib-rium position heading downward for the second time? (e) At what times does the mass attain its extreme displacements on either side of the equilibrium position? (f) What is the position of the mass at t 3 s? (g) What is the instantaneous velocity at t 3 s? (h) What is the acceleration at t 3 s? (i) What is the instantaneous velocity at the times when the mass passes through the equilibrium position? (j) At what times is the mass 5 inches below the equi-librium position? (k) At what times is the mass 5 inches below the equi-librium position heading in the upward direction? 1 2 5 4 ft/s 1 3 1 4 3 2 ft/s. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 206 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 12. A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of . Find the times at which the mass is heading downward at a velocity of 3 ft/s. 13. Under some circumstances when two parallel springs, with constants k1 and k2, support a single mass, the effective spring constant of the system is given by k 4k1k2(k1 k2). A mass weighing 20 pounds stretches one spring 6 inches and another spring 2 inches. The springs are attached to a common rigid support and then to a metal plate. As shown in Figure 5.1.16, the mass is attached to the center of the plate in the double-spring arrangement. Determine the effective spring constant of this system. Find the equa-tion of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s. 13 ft/s 14. A certain mass stretches one spring foot and another spring foot. The two springs are attached to a common rigid support in the manner described in Problem 13 and Figure 5.1.16. The ﬁrst mass is set aside, a mass weigh-ing 8 pounds is attached to the double-spring arrange-ment, and the system is set in motion. If the period of motion is p15 second, determine how much the ﬁrst mass weighs. 15. Amodel of a spring/mass system is 4x e0.1tx 0. By inspection of the differential equation only, discuss the be-havior of the system over a long period of time. 16. A model of a spring/mass system is 4x tx 0. By inspection of the differential equation only, discuss the behavior of the system over a long period of time. 5.1.2 SPRING/MASS SYSTEMS: FREE DAMPED MOTION In Problems 17–20 the given ﬁgure represents the graph of an equation of motion for a damped spring/mass system. Use the graph to determine (a) whether the initial displacement is above or below the equilibrium position and (b) whether the mass is initially released from rest, heading downward, or heading upward. 1 2 1 3 20 lb k 1 2 k FIGURE 5.1.16 Double-spring system in Problem 13 t x t x t x t x FIGURE 5.1.17 Graph for Problem 17 FIGURE 5.1.19 Graph for Problem 19 FIGURE 5.1.18 Graph for Problem 18 FIGURE 5.1.20 Graph for Problem 20 18. 20. 21. A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equi-librium position. What is the position of the mass at this instant? 19. 17. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 207 22. A 4-foot spring measures 8 feet long after a mass weigh-ing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numeri-cally equal to times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant? 23. A 1-kilogram mass is attached to a spring whose constant is 16 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equa-tions of motion if (a) the mass is initially released from rest from a point 1 meter below the equilibrium position, and then (b) the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 12 m/s. 24. In parts (a) and (b) of Problem 23 determine whether the mass passes through the equilibrium position. In each case ﬁnd the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant? 25. A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force that is numerically equal to 0.4 times the instantaneous velocity. (a) Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position. (b) Express the equation of motion in the form given in (23). (c) Find the ﬁrst time at which the mass passes through the equilibrium position heading upward. 26. After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. This mass is removed and replaced with another mass that weighs 8 pounds. The entire system is placed in a medium that offers a damping force that is numerically equal to the instanta-neous velocity. (a) Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s. (b) Express the equation of motion in the form given in (23). (c) Find the times at which the mass passes through the equilibrium position heading downward. (d) Graph the equation of motion. 27. A mass weighing 10 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping 1 2 12 force numerically equal to b (b 0) times the instanta-neous velocity. Determine the values of the damping con-stant b so that the subsequent motion is (a) overdamped, (b) critically damped, and (c) underdamped. 28. A mass weighing 24 pounds stretches a spring 4 feet. The subsequent motion takes place in medium that offers a damping force numerically equal to b (b 0) times the instantaneous velocity. If the mass is initially released from the equilibrium position with an upward velocity of 2 ft/s, show that when the equa-tion of motion is . 5.1.3 SPRING/MASS SYSTEMS: DRIVEN MOTION 29. Amass weighing 16 pounds stretches a spring feet. The mass is initially released from rest from a point 2 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to the instantaneous velocity. Find the equation of motion if the mass is driven by an external force equal to f(t) 10 cos 3t. 30. A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of 5 ft/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 2 times the instantaneous velocity. (a) Find the equation of motion if the mass is driven by an external force equal to f(t) 12 cos 2t 3 sin 2t. (b) Graph the transient and steady-state solutions on the same coordinate axes. (c) Graph the equation of motion. 31. A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t 0, an external force equal to f(t) 8 sin 4t is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity. 32. In Problem 31 determine the equation of motion if the external force is f(t) et sin 4t. Analyze the displace-ments for . 33. When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equi-librium position. Starting at t 0, a force equal to f(t) 68e2t cos 4t is applied to the system. Find the equation of motion in the absence of damping. 34. In Problem 33 write the equation of motion in the form x(t) Asin(vt f) Be2tsin(4t u). What is the amplitude of vibrations after a very long time? t : 1 2 8 3 x(t) 3 1 2 18 e2t/3 sinh 2 3 12 18t 312 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 208 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 35. A mass m is attached to the end of a spring whose con-stant is k. After the mass reaches equilibrium, its support begins to oscillate vertically about a horizontal line L according to a formula h(t). The value of h represents the distance in feet measured from L. See Figure 5.1.21. (a) Determine the differential equation of motion if the entire system moves through a medium offer-ing a damping force that is numerically equal to b(dxdt). (b) Solve the differential equation in part (a) if the spring is stretched 4 feet by a mass weighing 16 pounds and b 2, h(t) 5 cos t, x(0) x(0) 0. (b) Evaluate . 40. Compare the result obtained in part (b) of Problem 39 with the solution obtained using variation of parameters when the external force is F0 cos vt. 41. (a) Show that x(t) given in part (a) of Problem 39 can be written in the form . (b) If we deﬁne , show that when is small an approximate solution is . When is small, the frequency g2p of the impressed force is close to the frequency v2p of free vibrations. When this occurs, the motion is as indicated in Figure 5.1.22. Oscillations of this kind are called beats and are due to the fact that the frequency of sin t is quite small in compari-son to the frequency of sin gt. The dashed curves, or envelope of the graph of x(t), are obtained from the graphs of (F0 2g) sin t. Use a graphing utility with various values of F0, , and g to verify the graph in Figure 5.1.22. x(t) F0 2 sin t sin t 1 2 ( ) x(t) 2F0 2 2 sin 1 2 ( )t sin 1 2 ( )t lim : F0 2 2 (cos t cos t) 36. A mass of 100 grams is attached to a spring whose constant is 1600 dynes/cm. After the mass reaches equi-librium, its support oscillates according to the formula h(t) sin 8t, where h represents displacement from its original position. See Problem 35 and Figure 5.1.21. (a) In the absence of damping, determine the equation of motion if the mass starts from rest from the equi-librium position. (b) At what times does the mass pass through the equi-librium position? (c) At what times does the mass attain its extreme displacements? (d) What are the maximum and minimum displace-ments? (e) Graph the equation of motion. In Problems 37 and 38 solve the given initial-value problem. 37. 38. 39. (a) Show that the solution of the initial-value problem is . x(t) F0 2 2 (cos t cos t) d 2x dt2 2x F0 cos t, x(0) 0, x(0) 0 d 2x dt2 9x 5 sin 3t, x(0) 2, x(0) 0 x(0) 1, x(0) 1 d 2x dt2 4x 5 sin 2t 3 cos 2t, L support h(t) FIGURE 5.1.21 Oscillating support in Problem 35 t x FIGURE 5.1.22 Beats phenomenon in Problem 41 Computer Lab Assignments 42. Can there be beats when a damping force is added to the model in part (a) of Problem 39? Defend your position with graphs obtained either from the explicit solution of the problem or from solution curves obtained using a numerical solver. 43. (a) Show that the general solution of d 2x dt2 2 dx dt 2x F0 sin t d2x dt2 2 dx dt 2x F0cos t, x(0) 0, x(0) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.1 LINEAR MODELS: INITIAL-VALUE PROBLEMS ● 209 is where and the phase angles f and u are, respectively, defined by sin f c1A, cos f c2A and , . (b) The solution in part (a) has the form x(t) xc(t) xp(t). Inspection shows that xc(t) is tran-sient, and hence for large values of time, the solution is approximated by xp(t) g(g) sin(gt u), where . Although the amplitude g(g) of xp(t) is bounded as show that the maximum oscillations will occur at the value . What is the maximum value of g? The number is said to be the resonance frequency of the system. (c) When F0 2, m 1, and k 4, g becomes . Construct a table of the values of g1 and g(g1) corre-sponding to the damping coefﬁcients b 2, b 1, , and . Use a graphing utility to obtain the graphs of g corresponding to these damp-ing coefﬁcients. Use the same coordinate axes. This family of graphs is called the resonance curve or frequency response curve of the system. What is g1 approaching as ? What is happening to the resonance curve as ? 44. Consider a driven undamped spring/mass system described by the initial-value problem . (a) For n 2, discuss why there is a single frequency g12p at which the system is in pure resonance. (b) For n 3, discuss why there are two frequencies g12p and g22p at which the system is in pure resonance. (c) Suppose v 1 and F0 1. Use a numerical solver to obtain the graph of the solution of the initial-value problem for n 2 and g g1 in part (a). Obtain the graph of the solution of the initial-value problem for n 3 corresponding, in turn, to g g1 and g g2 in part (b). d 2x dt2 2x F0 sinn t, x(0) 0, x(0) 0 : 0 : 0 1 4 3 4, 1 2 g() 2 1(4 2)2 22 12 2 2/2 1 12 2 2 t : , g(y) F0 1(2 2)2 4 2 2 cos 2 2 1(2 2)2 4 2 2 sin 2 1(2 2)2 4 2 2 A 1c1 2 c2 2 F0 1(2 2)2 4 22 sin(t ), x(t) Aelt sin2v2 l2t f 5.1.4 SERIES CIRCUIT ANALOGUE 45. Find the charge on the capacitor in an LRC-series circuit at t 0.01 s when L 0.05 h, R 2 , C 0.01 f, E(t) 0 V, q(0) 5 C, and i(0) 0 A. Determine the ﬁrst time at which the charge on the capacitor is equal to zero. 46. Find the charge on the capacitor in an LRC-series circuit when , R 20 , , E(t) 0 V, q(0) 4 C, and i(0) 0 A. Is the charge on the capaci-tor ever equal to zero? In Problems 47 and 48 ﬁnd the charge on the capacitor and the current in the given LRC-series circuit. Find the maxi-mum charge on the capacitor. 47. , R 10 , , E(t) 300 V, q(0) 0 C, i(0) 0 A 48. L 1 h, R 100 , C 0.0004 f, E(t) 30 V, q(0) 0 C, i(0) 2 A 49. Find the steady-state charge and the steady-state current in an LRC-series circuit when L 1 h, R 2 , C 0.25 f, and E(t) 50 cos t V. 50. Show that the amplitude of the steady-state current in the LRC-series circuit in Example 10 is given by E0Z, where Z is the impedance of the circuit. 51. Use Problem 50 to show that the steady-state current in an LRC-series circuit when , R 20 , C 0.001 f, and E(t) 100 sin 60t V, is given by ip(t) 4.160 sin(60t 0.588). 52. Find the steady-state current in an LRC-series circuit when , R 20 , C 0.001 f, and E(t) 100 sin 60t 200 cos 40t V. 53. Find the charge on the capacitor in an LRC-series circuit when , R 10 , C 0.01 f, E(t) 150 V, q(0) 1 C, and i(0) 0 A. What is the charge on the capacitor after a long time? 54. Show that if L, R, C, and E0 are constant, then the amplitude of the steady-state current in Example 10 is a maximum when . What is the maximum amplitude? 55. Show that if L, R, E0, and g are constant, then the amplitude of the steady-state current in Example 10 is a maximum when the capacitance is C 1Lg2. 56. Find the charge on the capacitor and the current in an LC-series circuit when L 0.1 h, C 0.1 f, E(t) 100 sin gt V, q(0) 0 C, and i(0) 0 A. 57. Find the charge on the capacitor and the current in an LC-series circuit when E(t) E0 cos gt V, q(0) q0 C, and i(0) i0 A. 58. In Problem 57 ﬁnd the current when the circuit is in resonance. 1> 1LC L 1 2 h L 1 2 h L 1 2 h C 1 30 f L 5 3 h C 1 300 f L 1 4 h Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 210 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Deflection of a Beam Many structures are constructed by using girders or beams, and these beams deﬂect or distort under their own weight or under the inﬂuence of some external force. As we shall now see, this deﬂection y(x) is governed by a rela-tively simple linear fourth-order differential equation. To begin, let us assume that a beam of length L is homogeneous and has uniform cross sections along its length. In the absence of any load on the beam (including its weight), a curve joining the centroids of all its cross sections is a straight line called the axis of symmetry. See Figure 5.2.1(a). If a load is applied to the beam in a verti-cal plane containing the axis of symmetry, the beam, as shown in Figure 5.2.1(b), undergoes a distortion, and the curve connecting the centroids of all cross sections is called the deflection curve or elastic curve. The deﬂection curve approximates the shape of the beam. Now suppose that the x-axis coincides with the axis of symmetry and that the deﬂection y(x), measured from this axis, is positive if downward. In the theory of elasticity it is shown that the bending moment M(x) at a point x along the beam is related to the load per unit length w(x) by the equation . (1) In addition, the bending moment M(x) is proportional to the curvature k of the elas-tic curve , (2) where E and I are constants; E is Young’s modulus of elasticity of the material of the beam, and I is the moment of inertia of a cross section of the beam (about an axis known as the neutral axis). The product EI is called the flexural rigidity of the beam. Now, from calculus, curvature is given by k y[1 (y)2]3/2. When the deﬂection y(x) is small, the slope y 0, and so [1 (y)2]3/2 1. If we let k y, equation (2) becomes M EI y. The second derivative of this last expression is . (3) Using the given result in (1) to replace d2Mdx2 in (3), we see that the deﬂection y(x) satisﬁes the fourth-order differential equation . (4) EI d 4y dx4 w(x) d 2M dx2 EI d 2 dx2 y EI d 4y dx4 M(x) EI d2M dx2 w(x) axis of symmetry deflection curve (a) (b) FIGURE 5.2.1 Deﬂection of a homogeneous beam LINEAR MODELS: BOUNDARY-VALUE PROBLEMS REVIEW MATERIAL ●Section 4.1 (page 117) ●Problems 37–40 in Exercises 4.3 ●Problems 37–40 in Exercises 4.4 INTRODUCTION The preceding section was devoted to systems in which a second-order math-ematical model was accompanied by initial conditions—that is, side conditions that are speciﬁed on the unknown function and its ﬁrst derivative at a single point. But often the mathematical descrip-tion of a physical system demands that we solve a linear differential equation subject to boundary conditions—that is, conditions speciﬁed on the unknown function, or on one of its derivatives, or even on a linear combination of the unknown function and one of its derivatives at two (or more) different points. 5.2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.2 LINEAR MODELS: BOUNDARY-VALUE PROBLEMS ● 211 Boundary conditions associated with equation (4) depend on how the ends of the beam are supported. A cantilever beam is embedded or clamped at one end and free at the other. A diving board, an outstretched arm, an airplane wing, and a bal-cony are common examples of such beams, but even trees, ﬂagpoles, skyscrapers, and the George Washington Monument can act as cantilever beams because they are embedded at one end and are subject to the bending force of the wind. For a cantilever beam the deﬂection y(x) must satisfy the following two conditions at the embedded end x 0: • y(0) 0 because there is no deﬂection, and • y(0) 0 because the deﬂection curve is tangent to the x-axis (in other words, the slope of the deﬂection curve is zero at this point). At x L the free-end conditions are • y(L) 0 because the bending moment is zero, and • y(L) 0 because the shear force is zero. The function F(x) dMdx EI d3ydx3 is called the shear force. If an end of a beam is simply supported or hinged (also called pin supported and fulcrum supported) then we must have y 0 and y 0 at that end. Table 5.2.1 summarizes the boundary conditions that are associated with (4). See Figure 5.2.2. x = 0 x = L (a) embedded at both ends (b) cantilever beam: embedded at the left end, free at the right end (c) simply supported at both ends x = 0 x = L x = 0 x = L FIGURE 5.2.2 Beams with various end conditions TABLE 5.2.1 Ends of the Beam Boundary Conditions embedded y 0, y 0 free y 0, y 0 simply supported or hinged y 0, y 0 EXAMPLE 1 An Embedded Beam A beam of length L is embedded at both ends. Find the deﬂection of the beam if a con-stant load w0 is uniformly distributed along its length—that is, w(x) w0, 0 x L. SOLUTION From (4) we see that the deﬂection y(x) satisﬁes . Because the beam is embedded at both its left end (x 0) and its right end (x L), there is no vertical deﬂection and the line of deﬂection is horizontal at these points. Thus the boundary conditions are . We can solve the nonhomogeneous differential equation in the usual manner (ﬁnd yc by observing that m 0 is root of multiplicity four of the auxiliary equation m4 0 and then ﬁnd a particular solution yp by undetermined coefﬁcients), or we can simply integrate the equation d4ydx4 w0EI four times in succession. Either way, we ﬁnd the general solution of the equation y yc yp to be . Now the conditions y(0) 0 and y(0) 0 give, in turn, c1 0 and c2 0, whereas the remaining conditions y(L) 0 and y(L) 0 applied to yield the simultaneous equations 2c3 L 3c4 L2 w0 6EI L3 0. c3 L2 c4 L3 w0 24EI L4 0 y(x) c3x2 c4x3 w0 24EI x4 y(x) c1 c2x c3x2 c4x3 w0 24EI x4 y(0) 0, y(0) 0, y(L) 0, y(L) 0 EI d 4y dx4 w0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 212 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Solving this system gives c3 w0L224EI and c4 w0L12EI. Thus the deﬂection is or . By choosing w0 24EI, and L 1, we obtain the deﬂection curve in Figure 5.2.3. Eigenvalues and Eigenfunctions Many applied problems demand that we solve a two-point boundary-value problem (BVP) involving a linear differential equation that contains a parameter l. We seek the values of l for which the boundary-value problem has nontrivial, that is, nonzero, solutions. y(x) w0 24EI x2 (x L)2 y(x) w0L2 24EI x2 w0L 12EI x3 w0 24EI x4 x y 1 0.5 FIGURE 5.2.3 Deﬂection curve for BVP in Example 1 EXAMPLE 2 Nontrivial Solutions of a BVP Solve the boundary-value problem . SOLUTION We shall consider three cases: l 0, l 0, and l 0. Case I: For l 0 the solution of y 0 is y c1x c2. The conditions y(0) 0 and y(L) 0 applied to this solution imply, in turn, c2 0 and c1 0. Hence for l 0 the only solution of the boundary-value problem is the trivial solution y 0. Case II: For l 0 it is convenient to write l a2, where a denotes a positive number. With this notation the roots of the auxiliary equation m2 a2 0 are m1 a and m2 a. Since the interval on which we are working is ﬁnite, we choose to write the general solution of y a2y 0 as y c1 cosh ax c2 sinh ax. Now y(0) is , and so y(0) 0 implies that c1 0. Thus y c2 sinh ax. The second condition, y(L) 0, demands that c2 sinh aL 0. For a 0, sinh aL 0; consequently, we are forced to choose c2 0. Again the only solution of the BVP is the trivial solu-tion y 0. Case III: For l 0 we write l a2, where a is a positive number. Because the auxiliary equation m2 a2 0 has complex roots m1 ia and m2 ia, the general solution of y a2y 0 is y c1 cos ax c2 sin ax. As before, y(0) 0 yields c1 0, and so y c2 sin ax. Now the last condition y(L) 0, or , is satisﬁed by choosing c2 0. But this means that y 0. If we require c2 0, then sin aL 0 is satisﬁed whenever aL is an integer multiple of p. . Therefore for any real nonzero c2, y c2 sin(npxL) is a solution of the problem for each n. Because the differential equation is homogeneous, any constant multiple of a solution is also a solution, so we may, if desired, simply take c2 1. In other words, for each number in the sequence . . . , 1 2 L2, 2 42 L2 , 3 92 L2 , L n or n L or n n 2 n L 2 , n 1, 2, 3, . . . c2 sin L 0 y(0) c1 cosh 0 c2 sinh 0 c1 1 c2 0 c1 y y 0, y(0) 0, y(L) 0 Note that we use hyperbolic functions here. Reread “Two Equations Worth Knowing” on pages 134–135. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.2 LINEAR MODELS: BOUNDARY-VALUE PROBLEMS ● 213 the corresponding function in the sequence . . . is a nontrivial solution of the problem for n 1, 2, 3, . . . , respectively. The numbers ln n2p2L2, n 1, 2, 3, . . . for which the boundary-value problem in Example 2 possesses nontrivial solutions are known as eigenvalues. The nontrivial solutions that depend on these values of ln, yn c2 sin(npxL) or simply yn sin(npxL), are called eigenfunctions. The graphs of the eigenfunctions for n 1, 2, 3, 4, 5 are shown in Figure 5.2.4. Note that each graph passes through the two points (0, 0) and (0, L). y ny 0, y(0) 0, y(L) 0 y1 sin L x, y2 sin2 L x, y3 sin3 L x, L (a) (b) P x = 0 x y x = L FIGURE 5.2.5 Elastic column buckling under a compressive force EXAMPLE 3 Example 2 Revisited It follows from Example 2 and the preceding disucussion that the boundary-value problem possesses only the trivial solution y 0 because 5 is not an eigenvalue. Buckling of a Thin Vertical Column In the eighteenth century Leonhard Euler was one of the ﬁrst mathematicians to study an eigenvalue problem in analyz-ing how a thin elastic column buckles under a compressive axial force. Consider a long, slender vertical column of uniform cross section and length L. Let y(x) denote the deﬂection of the column when a constant vertical compressive force, or load, P is applied to its top, as shown in Figure 5.2.5. By comparing bend-ing moments at any point along the column, we obtain , (5) where E is Young’s modulus of elasticity and I is the moment of inertia of a cross section about a vertical line through its centroid. EI d 2y dx2 Py or EI d 2y dx2 Py 0 y 5y 0, y(0) 0, y(L) 0 –1 1 n = 2 n = 1 n = 4 n = 4 n = 5 n = 3 y x L FIGURE 5.2.4 Graphs of eigenfunctions for n 1, 2, 3, 4, 5 yn sin(npx>L), EXAMPLE 4 The Euler Load Find the deﬂection of a thin vertical homogeneous column of length L subjected to a constant axial load P if the column is hinged at both ends. SOLUTION The boundary-value problem to be solved is . First note that y 0 is a perfectly good solution of this problem. This solution has a simple intuitive interpretation: If the load P is not great enough, there is no deﬂection. The question then is this: For what values of P will the column bend? In mathematical terms: For what values of P does the given boundary-value problem possess nontrivial solutions? By writing l PEI, we see that is identical to the problem in Example 2. From Case III of that discussion we see that the deflections are yn(x) c2 sin(npxL) corresponding to the eigenvalues ln PnEI n2p 2L2, n 1, 2, 3, . . . . Physically, this means that the column will buckle or deflect only when the compressive force is one of the values Pn n2p 2EIL2, n 1, 2, 3, . . . . These different forces are called critical y y 0, y(0) 0, y(L) 0 EI d 2y dx2 Py 0, y(0) 0, y(L) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 214 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS loads. The deflection corresponding to the smallest critical load P1 p 2EIL2, called the Euler load, is y1(x) c2 sin(pxL) and is known as the first buckling mode. The deﬂection curves in Example 4 corresponding to n 1, n 2, and n 3 are shown in Figure 5.2.6. Note that if the original column has some sort of physical restraint put on it at x L 2, then the smallest critical load will be P2 4p2EIL2, and the deﬂection curve will be as shown in Figure 5.2.6(b). If restraints are put on the column at x L 3 and at x 2L 3, then the column will not buckle until the critical load P3 9p2EIL2 is applied, and the deﬂection curve will be as shown in Figure 5.2.6(c). See Problem 23 in Exercises 5.2. Rotating String The simple linear second-order differential equation (6) occurs again and again as a mathematical model. In Section 5.1 we saw (6) in the forms d2xdt2 (km)x 0 and d2qdt2 (1LC)q 0 as models for, respec-tively, the simple harmonic motion of a spring/mass system and the simple harmonic response of a series circuit. It is apparent when the model for the deﬂection of a thin column in (5) is written as d2ydx2 (PEI)y 0 that it is the same as (6). We encounter the basic equation (6) one more time in this section: as a model that deﬁnes the deﬂection curve or the shape y(x) assumed by a rotating string. The physical situ-ation is analogous to when two people hold a jump rope and twirl it in a synchronous manner. See Figures 5.2.7(a) and 5.2.7(b). Suppose a string of length L with constant linear density r (mass per unit length) is stretched along the x-axis and ﬁxed at x 0 and x L. Suppose the string is then rotated about that axis at a constant angular speed v. Consider a portion of the string on the interval [x, x x], where x is small. If the magnitude T of the tension T, acting tangential to the string, is constant along the string, then the desired differen-tial equation can be obtained by equating two different formulations of the net force acting on the string on the interval [x, x x]. First, we see from Figure 5.2.7(c) that the net vertical force is . (7) When angles u1 and u2 (measured in radians) are small, we have sin u2 tan u2 and sin u1 tan u1. Moreover, since tan u2 and tan u1 are, in turn, slopes of the lines con-taining the vectors T2 and T1, we can also write . Thus (7) becomes . (8) Second, we can obtain a different form of this same net force using Newton’s second law, F ma. Here the mass of the string on the interval is m r x; the centripetal acceleration of a body rotating with angular speed v in a circle of radius r is a rv2. With x small we take r y. Thus the net vertical force is also approximated by , (9) where the minus sign comes from the fact that the acceleration points in the direction opposite to the positive y-direction. Now by equating (8) and (9), we have F ( x)y2 F T [ y(x x) y(x)] tan 2 y(x x) and tan 1 y(x) F T sin 2 T sin 1 y y 0 L L x (b) y x (c) y x L (a) y FIGURE 5.2.6 Deﬂection curves corresponding to compressive forces P1, P2, P3 (a) (b) (c) ω x = 0 x = L y(x) x x x + ∆x 1 θ 2 θ T2 T1 FIGURE 5.2.7 Rotating string and forces acting on it (10) y(x x) y(x) ––––––––––––––––– x T[y(x x) y(x)] (rx)yv2 T rv2y 0. or difference quotient For x close to zero the difference quotient in (10) is approximately the second derivative d2ydx2. Finally, we arrive at the model . (11) T d 2y dx2 2y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.2 LINEAR MODELS: BOUNDARY-VALUE PROBLEMS ● 215 EXERCISES 5.2 Answers to selected odd-numbered problems begin on page ANS-8. Deflection of a Bea In Problems 1–5 solve equation (4) subject to the appropriate boundary conditions. The beam is of length L, and w0 is a constant. 1. (a) The beam is embedded at its left end and free at its right end, and w(x) w0, 0 x L. (b) Use a graphing utility to graph the deﬂection curve when w0 24EI and L 1. 2. (a) The beam is simply supported at both ends, and w(x) w0, 0 x L. (b) Use a graphing utility to graph the deﬂection curve when w0 24EI and L 1. 3. (a) The beam is embedded at its left end and simply sup-ported at its right end, and w(x) w0, 0 x L. (b) Use a graphing utility to graph the deﬂection curve when w0 48EI and L 1. 4. (a) The beam is embedded at its left end and simply sup-ported at its right end, and w(x) w0 sin(pxL), 0 x L. (b) Use a graphing utility to graph the deﬂection curve when w0 2p3EI and L 1. (c) Use a root-ﬁnding application of a CAS (or a graphic calculator) to approximate the point in the graph in part (b) at which the maximum deﬂection occurs. What is the maximum deﬂection? 5. (a) The beam is simply supported at both ends, and w(x) w0x, 0 x L. (b) Use a graphing utility to graph the deﬂection curve when w0 36EI and L 1. (c) Use a root-ﬁnding application of a CAS (or a graphic calculator) to approximate the point in the graph in part (b) at which the maximum deﬂection occurs. What is the maximum deﬂection? 6. (a) Find the maximum deﬂection of the cantilever beam in Problem 1. (b) How does the maximum deﬂection of a beam that is half as long compare with the value in part (a)? (c) Find the maximum deﬂection of the simply sup-ported beam in Problem 2. (d) How does the maximum deﬂection of the simply supported beam in part (c) compare with the value of maximum deﬂection of the embedded beam in Example 1? 7. A cantilever beam of length L is embedded at its right end, and a horizontal tensile force of P pounds is applied to its free left end. When the origin is taken at its free end, as shown in Figure 5.2.8, the deﬂection y(x) of the beam can be shown to satisfy the differential equation . Find the deﬂection of the cantilever beam if w(x) w0x, 0 x L, and y(0) 0, y(L) 0. EIy Py w(x)x 2 x O P y L x w0x FIGURE 5.2.8 Deﬂection of cantilever beam in Problem 7 Since the string is anchored at its ends x 0 and x L, we expect that the solution y(x) of equation (11) should also satisfy the boundary conditions y(0) 0 and y(L) 0. REMARKS (i) Eigenvalues are not always easily found, as they were in Example 2; you might have to approximate roots of equations such as tan x x or cos x cosh x 1. See Problems 34–38 in Exercises 5.2. (ii) Boundary conditions applied to a general solution of a linear differential equation can lead to a homogeneous algebraic system of linear equations in which the unknowns are the coefﬁcients ci in the general solution. A homoge-neous algebraic system of linear equations is always consistent because it possesses at least a trivial solution. But a homogeneous system of n linear equations in n unknowns has a nontrivial solution if and only if the determi-nant of the coefﬁcients equals zero. You might need to use this last fact in Problems 19 and 20 in Exercises 5.2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 216 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 8. When a compressive instead of a tensile force is applied at the free end of the beam in Problem 7, the differential equation of the deﬂection is . Solve this equation if w(x) w0x, 0 x L, and y(0) 0, y(L) 0. Eigenvalues and Eigenfunctions In Problems 9–18 ﬁnd the eigenvalues and eigenfunctions for the given boundary-value problem. 9. y ly 0, y(0) 0, y(p) 0 10. y ly 0, y(0) 0, y(p4) 0 11. y ly 0, y(0) 0, y(L) 0 12. y ly 0, y(0) 0, y(p2) 0 13. y ly 0, y(0) 0, y(p) 0 14. y ly 0, y(p) 0, y(p) 0 15. y 2y (l 1)y 0, y(0) 0, y(5) 0 16. y (l 1)y 0, y(0) 0, y(1) 0 17. x2y xy ly 0, y(1) 0, y(ep) 0 18. x2y xy ly 0, y(e1) 0, y(1) 0 In Problems 19 and 20 ﬁnd the eigenvalues and eigenfunc-tions for the given boundary-value problem. Consider only the case l a4, a 0. 19. y(4) ly 0, y(0) 0, y(0) 0, y(1) 0, y(1) 0 20. y(4) ly 0, y(0) 0, y(0) 0, y(p) 0, y(p) 0 Buckling of a Thin Column 21. Consider Figure 5.2.6. Where should physical restraints be placed on the column if we want the critical load to be P4? Sketch the deﬂection curve corresponding to this load. 22. The critical loads of thin columns depend on the end conditions of the column. The value of the Euler load P1 in Example 4 was derived under the assumption that the column was hinged at both ends. Suppose that a thin vertical homogeneous column is embedded at its base (x 0) and free at its top (x L) and that a constant axial load P is applied to its free end. This load either causes a small deﬂection d as shown in Figure 5.2.9 or does not cause such a deﬂection. In either case the dif-ferential equation for the deﬂection y(x) is . EI d2y dx2 Py P EIy Py w(x)x 2 (a) What is the predicted deﬂection when d 0? (b) When d 0, show that the Euler load for this col-umn is one-fourth of the Euler load for the hinged column in Example 4. 23. As was mentioned in Problem 22, the differential equa-tion (5) that governs the deﬂection y(x) of a thin elastic column subject to a constant compressive axial force P is valid only when the ends of the column are hinged. In general, the differential equation governing the deﬂection of the column is given by . Assume that the column is uniform (EI is a constant) and that the ends of the column are hinged. Show that the solution of this fourth-order differential equation subject to the boundary conditions y(0) 0, y(0) 0, y(L) 0, y(L) 0 is equivalent to the analysis in Example 4. 24. Suppose that a uniform thin elastic column is hinged at the end x 0 and embedded at the end x L. (a) Use the fourth-order differential equation given in Problem 23 to ﬁnd the eigenvalues ln, the critical loads Pn, the Euler load P1, and the deﬂections yn(x). (b) Use a graphing utility to graph the ﬁrst buckling mode. Rotating String 25. Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string: . For constant T and r, deﬁne the critical speeds of angu-lar rotation vn as the values of v for which the boundary-value problem has nontrivial solutions. Find the critical speeds vn and the corresponding deﬂections yn(x). T d2y dx2 2y 0, y(0) 0, y(L) 0 d2 dx2 EI d2y dx2 P d2y dx2 0 y x = 0 x = L P δ x FIGURE 5.2.9 Deﬂection of vertical column in Problem 22 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.2 LINEAR MODELS: BOUNDARY-VALUE PROBLEMS ● 217 26. When the magnitude of tension T is not constant, then a model for the deﬂection curve or shape y(x) assumed by a rotating string is given by . Suppose that 1 x e and that T(x) x2. (a) If y(1) 0, y(e) 0, and rv2 0.25, show that the critical speeds of angular rotation are and the corresponding deﬂections are yn(x) c2x1/2 sin(np ln x), n 1, 2, 3, . . . . (b) Use a graphing utility to graph the deﬂection curves on the interval [1, e] for n 1, 2, 3. Choose c2 1. Miscellaneous Boundary-Value Problems 27. Temperature in a Spher e Consider two concentric spheres of radius r a and r b, a b. See Figure 5.2.10. The temperature u(r) in the region between the spheres is determined from the boundary-value problem , where u0 and u1 are constants. Solve for u(r). r d2u dr2 2 du dr 0, u(a) u0, u(b) u1 n 1 2 2(4n22 1)> d dx T(x) dy dx 2y 0 where u0 and u1 are constants. Show that . Discussion Problems 29. Simple Harmonic Motion The model mx kx 0 for simple harmonic motion, discussed in Section 5.1, can be related to Example 2 of this section. Consider a free undamped spring/mass system for which the spring constant is, say, k 10 lb/ft. Deter-mine those masses mn that can be attached to the spring so that when each mass is released at the equilibrium position at t 0 with a nonzero velocity v0, it will then pass through the equilibrium position at t 1 second. How many times will each mass mn pass through the equilibrium position in the time interval 0 t 1? 30. Damped Motion Assume that the model for the spring/mass system in Problem 29 is replaced by mx 2x kx 0. In other words, the system is free but is subjected to damping numerically equal to 2 times the instantaneous velocity. With the same initial conditions and spring constant as in Problem 29, investigate whether a mass m can be found that will pass through the equilibrium position at t 1 second. In Problems 31 and 32 determine whether it is possible to ﬁnd values y0 and y1 (Problem 31) and values of L 0 (Problem 32) so that the given boundary-value problem has (a) precisely one nontrivial solution, (b) more than one solution, (c) no solution, (d) the trivial solution. 31. y 16y 0, y(0) y0, y(p2) y1 32. y 16y 0, y(0) 1, y(L) 1 33. Consider the boundary-value problem (a) The type of boundary conditions speciﬁed are called periodic boundary conditions. Give a geometric interpretation of these conditions. (b) Find the eigenvalues and eigenfunctions of the problem. (c) Use a graphing utility to graph some of the eigen-functions. Verify your geometric interpretation of the boundary conditions given in part (a). 34. Show that the eigenvalues and eigenfunctions of the boundary-value problem are and yn sin anx, respectively, where an, n 1, 2, 3, . . . are the consecutive positive roots of the equation tan a a. n 2 n y y 0, y(0) 0, y(1) y(1) 0 y y 0, y() y(), y() y(). u(r) u0 ln(r>b) u1 ln(r>a) ln(a>b) u = u1 u = u0 FIGURE 5.2.10 Concentric spheres in Problem 27 28. Temperature in a Ring The temperature u(r) in the circular ring shown in Figure 5.2.11 is determined from the boundary-value problem , r d2u dr2 du dr 0, u(a) u0, u(b) u1 FIGURE 5.2.11 Circular ring in Problem 28 a u = u1 u = u0 b Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 218 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Computer Lab Assignments 35. Use a CAS to plot graphs to convince yourself that the equation tan a a in Problem 34 has an inﬁnite number of roots. Explain why the negative roots of the equation can be ignored. Explain why l 0 is not an eigenvalue even though a 0 is an obvious solution of the equation tan a a. 36. Use a root-ﬁnding application of a CAS to approximate the ﬁrst four eigenvalues l1, l2, l3, and l4 for the BVP in Problem 34. In Problems 37 and 38 ﬁnd the eigenvalues and eigenfunc-tions of the given boundary-value problem. Use a CAS to approximate the ﬁrst four eigenvalues l1, l2, l3, and l4. 37. 38. y(4) ly 0, y(0) 0, y(0) 0, y(1) 0, y(1) 0 [Hint: Consider only l a4, a 0.] y y 0, y(0) 0, y(1) 1 2 y(1) 0 Nonlinear Springs The mathematical model in (1) of Section 5.1 has the form , (1) where F(x) kx. Because x denotes the displacement of the mass from its equilibrium position, F(x) kx is Hooke’s law—that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force F(x) kx is naturally referred to as a linear spring. But springs are seldom per-fectly linear. Depending on how it is constructed and the material that is used, a spring can range from “mushy,” or soft, to “stiff,” or hard, so its restorative force may vary from something below to something above that given by the linear law. In the case of free motion, if we assume that a nonaging spring has some nonlinear characteristics, then it might be reasonable to assume that the restorative force of a spring—that is, F(x) in (1)—is proportional to, say, the cube of the displacement x of the mass beyond its equilibrium position or that F(x) is a linear combination of powers of the displace-ment such as that given by the nonlinear function F(x) kx k1x3. A spring whose mathematical model incorporates a nonlinear restorative force, such as , (2) is called a nonlinear spring. In addition, we examined mathematical models in which damping imparted to the motion was proportional to the instantaneous velocity dxdt and the restoring force of a spring was given by the linear function F(x) kx. But these were simply assumptions; in more realistic situations damping could be proportional to some power of the instantaneous velocity dxdt. The nonlinear differential equation (3) m d 2x dt2 dx dt dx dt kx 0 m d 2x dt2 kx3 0 or m d 2x dt2 kx k1x3 0 m d 2x dt2 F(x) 0 NONLINEAR MODELS REVIEW MATERIAL ●Section 4.10 INTRODUCTION In this section we examine some nonlinear higher-order mathematical models. We are able to solve some of these models using the substitution method (leading to reduction of the order of the DE) introduced on page 186. In some cases in which the model cannot be solved, we show how a nonlinear DE can be replaced by a linear DE through a process called linearization. 5.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.3 NONLINEAR MODELS ● 219 is one model of a free spring/mass system in which the damping force is proportional to the square of the velocity. One can then envision other kinds of models: linear damping and nonlinear restoring force, nonlinear damping and nonlinear restoring force, and so on. The point is that nonlinear characteristics of a physical system lead to a mathematical model that is nonlinear. Notice in (2) that both F(x) kx3 and F(x) kx k1x3 are odd functions of x. To see why a polynomial function containing only odd powers of x provides a reasonable model for the restoring force, let us express F as a power series centered at the equilibrium position x 0: When the displacements x are small, the values of xn are negligible for n suffi-ciently large. If we truncate the power series with, say, the fourth term, then F(x) c0 c1x c2x2 c3x3. For the force at x 0, , and for the force at x 0, to have the same magnitude but act in the opposite direction, we must have F(x) F(x). Because this means that F is an odd function, we must have c0 0 and c2 0, and so F(x) c1x c3x3. Had we used only the ﬁrst two terms in the series, the same argument yields the linear function F(x) c1x. A restoring force with mixed powers, such as F(x) c1x c2x2, and the corresponding vibrations are said to be unsymmetrical. In the next discussion we shall write c1 k and c3 k1. Hard and Soft Springs Let us take a closer look at the equation in (1) in F(x) c0 c1x c2x2 c3x3 F(x) c0 c1x c2x2 c3x3 F(x) c0 c1x c2x2 c3x3 . F linear spring hard spring soft spring x FIGURE 5.3.1 Hard and soft springs (a) hard spring (b) soft spring x x(0)= 2, x'(0)= _3 t x(0)= 2, x'(0)= _3 t x x(0)= 2, x'(0)= 0 x(0)= 2, x'(0)= 0 FIGURE 5.3.2 Numerical solution curves EXAMPLE 1 Comparison of Hard and Soft Springs The differential equations (4) and (5) are special cases of the second equation in (2) and are models of a hard spring and a soft spring, respectively. Figure 5.3.2(a) shows two solutions of (4) and Figure 5.3.2(b) shows two solutions of (5) obtained from a numerical solver. The curves shown in red are solutions that satisfy the initial conditions x(0) 2, x(0) 3; the two curves in blue are solutions that satisfy x(0) 2, x(0) 0. These solution curves certainly suggest that the motion of a mass on the hard spring is oscillatory, whereas motion of a mass on the soft spring appears to be nonoscillatory. But we must be careful about drawing conclusions based on a couple of numerical solution curves. A more complete picture of the nature of the solutions of both of these equations can be obtained from the qualitative analysis discussed in Chapter 10. d 2x dt2 x x3 0 d 2x dt2 x x3 0 the case in which the restoring force is given by F(x) kx k1x3, k 0. The spring is said to be hard if k1 0 and soft if k1 0. Graphs of three types of restoring forces are illustrated in Figure 5.3.1. The next example illustrates these two special cases of the differential equation md2xdt2 kx k1x3 0, m 0, k 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 220 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Nonlinear Pendulum Any object that swings back and forth is called a physical pendulum. The simple pendulum is a special case of the physical pendu-lum and consists of a rod of length l to which a mass m is attached at one end. In describing the motion of a simple pendulum in a vertical plane, we make the simpli-fying assumptions that the mass of the rod is negligible and that no external damping or driving forces act on the system. The displacement angle u of the pendulum, measured from the vertical as shown in Figure 5.3.3, is considered positive when measured to the right of OP and negative to the left of OP. Now recall the arc s of a circle of radius l is related to the central angle u by the formula s lu. Hence angu-lar acceleration is . From Newton’s second law we then have . From Figure 5.3.3 we see that the magnitude of the tangential component of the force due to the weight W is mg sin u. In direction this force is mg sin u because it points to the left for u 0 and to the right for u 0. We equate the two different versions of the tangential force to obtain ml d2udt2 mg sin u, or . (6) Linearization Because of the presence of sin u, the model in (6) is non-linear. In an attempt to understand the behavior of the solutions of nonlinear higher-order differential equations, one sometimes tries to simplify the problem by replacing nonlinear terms by certain approximations. For example, the Maclaurin series for sin u is given by so if we use the approximation sin u u u36, equation (6) becomes . Observe that this last equation is the same as the second nonlinear equation in (2) with m 1, k gl, and k1 g6l. However, if we assume that the displacements u are small enough to justify using the replacement sin u u, then (6) becomes . (7) See Problem 25 in Exercises 5.3. If we set v2 gl, we recognize (7) as the differen-tial equation (2) of Section 5.1 that is a model for the free undamped vibrations of a lin-ear spring/mass system. In other words, (7) is again the basic linear equation y ly 0 discussed on page 212 of Section 5.2. As a consequence we say that equa-tion (7) is a linearization of equation (6). Because the general solution of (7) is u(t) c1 cos vt c2 sin vt, this linearization suggests that for initial conditions amenable to small oscillations the motion of the pendulum described by (6) will be periodic. d 2 dt2 g l 0 d 2u dt 2 g l u g 6l u3 0 sin 3 3! 5 5! . . . d 2 dt2 g l sin 0 F ma ml d 2 dt2 a d 2s dt2 l d 2 dt2 O θ θ P W = mg mg cos θ mg sin θ l FIGURE 5.3.3 Simple pendulum t 2 (0) = , (0) = 2 (0) = , (0) = (a) (b) (0) , (0) (c) (0) , (0) 2 1 2 1 2 1 2 1 2 1 2 1 2 FIGURE 5.3.4 In Example 2, oscillating pendulum in (b); whirling pendulum in (c) EXAMPLE 2 Two Initial-Value Problems The graphs in Figure 5.3.4(a) were obtained with the aid of a numerical solver and represent approximate or numerical solution curves of (6) when v2 1. The blue curve depicts the solution of (6) that satisﬁes the initial conditions whereas the red curve is the solution of (6) that satisﬁes (0) 1 2, (0) 1 2, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.3 NONLINEAR MODELS ● 221 u(0) 2. The blue curve represents a periodic solution—the pendulum oscillating back and forth as shown in Figure 5.3.4(b) with an apparent amplitude A 1. The red curve shows that u increases without bound as time increases—the pendulum, starting from the same initial displacement, is given an initial velocity of magnitude great enough to send it over the top; in other words, the pendulum is whirling about its pivot as shown in Figure 5.3.4(c). In the absence of damping, the motion in each case is continued indeﬁnitely. Telephone Wires The ﬁrst-order differential equation dydx WT1 is equation (16) of Section 1.3. This differential equation, established with the aid of Figure 1.3.8 on page 26, serves as a mathematical model for the shape of a ﬂex-ible cable suspended between two vertical supports when the cable is carrying a vertical load. In Section 2.2 we solved this simple DE under the assumption that the vertical load carried by the cables of a suspension bridge was the weight of a horizontal roadbed distributed evenly along the x-axis. With W rx, r the weight per unit length of the roadbed, the shape of each cable between the vertical supports turned out to be parabolic. We are now in a position to determine the shape of a uni-form ﬂexible cable hanging only under its own weight, such as a wire strung between two telephone posts. The vertical load is now the wire itself, and so if r is the linear density of the wire (measured, say, in pounds per feet) and s is the length of the segment P1P2 in Figure 1.3.8 then W rs. Hence . (8) Since the arc length between points P1 and P2 is given by , (9) it follows from the fundamental theorem of calculus that the derivative of (9) is . (10) Differentiating (8) with respect to x and using (10) lead to the second-order equation . (11) In the example that follows we solve (11) and show that the curve assumed by the suspended cable is a catenary. Before proceeding, observe that the nonlinear second-order differential equation (11) is one of those equations having the form F(x, y, y) 0 discussed in Section 4.10. Recall that we have a chance of solving an equation of this type by reducing the order of the equation by means of the substitu-tion u y. d 2y dx2 T1 ds dx or d 2y dx2 T1 B1 dy dx 2 ds dx B1 dy dx 2 s x 0 B1 dy dx 2 dx dy dx s 1 (0) 1 2, EXAMPLE 3 A Solution of (11) From the position of the y-axis in Figure 1.3.8 it is apparent that initial conditions associated with the second differential equation in (11) are y(0) a and y(0) 0. If we substitute u y, then the equation in (11) becomes . Sepa-rating variables, we ﬁnd that . du 11 u2 T1 dx gives sinh1u T1 x c1 du dx 1 11 u2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 222 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS Now, y(0) 0 is equivalent to u(0) 0. Since sinh1 0 0, c1 0, so u sinh (rxT1). Finally, by integrating both sides of . Using y(0) a, cosh 0 1, the last equation implies that c2 a T1r. Thus we see that the shape of the hanging wire is given by . In Example 3, had we been clever enough at the start to choose a T1r, then the solution of the problem would have been simply the hyperbolic cosine y (T1r) cosh (rxT1). Rocket Motion In (12) of Section 1.3 we saw that the differential equation of a free-falling body of mass m near the surface of the Earth is given by , where s represents the distance from the surface of the Earth to the object and the positive direction is considered to be upward. In other words, the underlying assumption here is that the distance s to the object is small when compared with the radius R of the Earth; put yet another way, the distance y from the center of the Earth to the object is approximately the same as R. If, on the other hand, the distance y to the object, such as a rocket or a space probe, is large when compared to R, then we combine Newton’s second law of motion and his universal law of gravitation to derive a differential equation in the variable y. Suppose a rocket is launched vertically upward from the ground as shown in Figure 5.3.5. If the positive direction is upward and air resistance is ignored, then the differential equation of motion after fuel burnout is , (12) where k is a constant of proportionality, y is the distance from the center of the Earth to the rocket, M is the mass of the Earth, and m is the mass of the rocket. To determine the constant k, we use the fact that when y R, kMmR2 mg or k gR2M. Thus the last equation in (12) becomes . (13) See Problem 14 in Exercises 5.3. Variable Mass Notice in the preceding discussion that we described the motion of the rocket after it has burned all its fuel, when presumably its mass m is constant. Of course, during its powered ascent the total mass of the rocket varies as its fuel is being expended. We saw in (17) of Exercises 1.3 that the second law of motion, as originally advanced by Newton, states that when a body of mass m moves through a force ﬁeld with velocity v, the time rate of change of the momentum mv of the body is equal to applied or net force F acting on the body: . (14) If m is constant, then (14) yields the more familiar form F m dvdt ma, where a is acceleration. We use the form of Newton’s second law given in (14) in the next example, in which the mass m of the body is variable. F d dt (mv) d 2y dt2 g R2 y2 m d 2y dt2 k Mm y2 or d 2y dt2 k M y2 m d 2s dt2 mg or simply d 2s dt2 g y (T1>) cosh(x> T1) a T1> dy dx sinh T1 x, we get y T1 cosh T1 x c2 v0 y center of Earth R FIGURE 5.3.5 Distance to rocket is large compared to R. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.3 NONLINEAR MODELS ● 223 x(t) 5 lb upward force FIGURE 5.3.6 Chain pulled upward by a constant force in Example 4 EXAMPLE 4 Chain Pulled Upward by a Constant Force A uniform 10-foot-long chain is coiled loosely on the ground. One end of the chain is pulled vertically upward by means of constant force of 5 pounds. The chain weighs 1 pound per foot. Determine the height of the end above ground level at time t. See Figure 5.3.6. SOLUTION Let us suppose that x x(t) denotes the height of the end of the chain in the air at time t, v dxdt, and the positive direction is upward. For the portion of the chain that is in the air at time t we have the following variable quantities: Thus from (14) we have (15) Because v dxdt, the last equation becomes . (16) The nonlinear second-order differential equation (16) has the form F(x, x, x) 0, which is the second of the two forms considered in Section 4.10 that can possibly be solved by reduction of order. To solve (16), we revert back to (15) and use v x along with the Chain Rule. From the second equation in (15) can be rewritten as . (17) On inspection (17) might appear intractable, since it cannot be characterized as any of the ﬁrst-order equations that were solved in Chapter 2. However, by rewriting (17) in differential form M(x, v)dx N(x, v)dv 0, we observe that although the equation (18) is not exact, it can be transformed into an exact equation by multiplying it by an integrating factor. From (Mv Nx)N 1x we see from (13) of Section 2.4 that an integrating factor is When (18) is multiplied by m(x) x, the resulting equation is exact (verify). By identifying f x xv2 32x2 160x, f v x2v and then proceeding as in Section 2.4, we obtain . (19) Since we have assumed that all of the chain is on the ﬂoor initially, we have x(0) 0. This last condition applied to (19) yields c1 0. By solving the algebraic equation for v dxdt 0, we get another ﬁrst-order differential equation, . dx dt B160 64 3 x 1 2 x2v2 32 3 x3 80x2 0 1 2 x2v2 32 3 x3 80x2 c1 edx/x eln x x. (v2 32x 160)dx xv dv 0 xv dv dx v2 160 32x dv dt dv dx dx dt v dv dx x d 2x dt2 dx dt 2 32x 160 Product Rule v 160 32x. x ( v) 5 x or x ––– 32 d ––– dt dv ––– dt dx ––– dt net force: F 5 W 5 x. mass: m W>g x>32, weight: W (x ft) (1 lb/ft) x, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 224 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS The last equation can be solved by separation of variables. You should verify that . (20) This time the initial condition x(0) 0 implies that . Finally, by squaring both sides of (20) and solving for x, we arrive at the desired result, (21) The graph of (21) given in Figure 5.3.7 should not, on physical grounds, be taken at face value. See Problem 15 in Exercises 5.3. x(t) 15 2 15 2 1 4110 15 t 2 . c2 31108 3 32 160 64 3 x 1/2 t c2 0.5 1 8 7 6 5 4 3 2 1 t x 1.5 2 2.5 0 FIGURE 5.3.7 Graph of (21) in Example 4 EXERCISES 5.3 Answers to selected odd-numbered problems begin on page ANS-9. To the Instructor In addition to Problems 24 and 25, all or portions of Problems 1–6, 8–13, 15, 20, and 21 could serve as Computer Lab Assignments. Nonlinear Springs In Problems 1–4 the given differential equation is model of an undamped spring/mass system in which the restoring force F(x) in (1) is nonlinear. For each equation use a numerical solver to plot the solution curves that satisfy the given initial conditions. If the solutions appear to be periodic use the solu-tion curve to estimate the period T of oscillations. 1. 2. 3. 4. 5. In Problem 3, suppose the mass is released from the initial position x(0) 1 with an initial velocity x(0) x1. Use a numerical solver to estimate the smallest value of x1 at which the motion of the mass is nonperiodic. 6. In Problem 3, suppose the mass is released from an initial position x(0) x0 with the initial velocity x(0) 1. Use a numerical solver to estimate an interval a x0 b for which the motion is oscillatory. 7. Find a linearization of the differential equation in Problem 4. x(0) 1, x(0) 1; x(0) 3, x(0) 1 d 2x dt2 xe0.01x 0, x(0) 1, x(0) 1; x(0) 3 2, x(0) 1 d 2x dt2 2x x2 0, x(0) 1, x(0) 1; x(0) 2, x(0) 2 d 2x dt2 4x 16x3 0, x(0) 1, x(0) 1; x(0) 1 2, x(0) 1 d 2x dt2 x3 0, 8. Consider the model of an undamped nonlinear spring/mass system given by x 8x 6x3 x5 0. Use a numerical solver to discuss the nature of the oscillations of the system corresponding to the initial conditions: In Problems 9 and 10 the given differential equation is a model of a damped nonlinear spring/mass system. Predict the behavior of each system as . For each equation use a numerical solver to obtain the solution curves satisfying the given initial conditions. 9. 10. 11. The model mx kx k1x3 F0cos vt of an undamped periodically driven spring/mass system is called Duffing s differential equation. Consider the initial-value problem x x k1x3 5 cos t, x(0) 1, x(0) 0. Use a nu-merical solver to investigate the behavior of the system for values of k1 0 ranging from k1 0.01 to k1 100. State your conclusions. 12. (a) Find values of k1 0 for which the system in Problem 11 is oscillatory. (b) Consider the initial-value problem x x k1x3 , x(0) 0, x(0) 0. Find values for k1 0 for which the system is oscillatory. cos 3 2 t x(0) 0, x(0) 3 2; x(0) 1, x(0) 1 d 2x dt2 dx dt x x3 0, x(0) 3, x(0) 4; x(0) 0, x(0) 8 d 2x dt2 dx dt x x3 0, t : x(0) 2, x(0) 0; x(0) 12, x(0) 1. x(0) 12, x(0) 1; x(0) 2, x(0) 1 2; x(0) 1, x(0) 1; x(0) 2, x(0) 1 2; Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.3 NONLINEAR MODELS ● 225 Nonlinear Pendulum 13. Consider the model of the free damped nonlinear pendulum given by . Use a numerical solver to investigate whether the motion in the two cases l2 v2 0 and l2 v2 0 corre-sponds, respectively, to the overdamped and underdamped cases discussed in Section 5.1 for spring/mass systems. For For and Rocket Motion 14. (a) Use the substitution v dydt to solve (13) for v in terms of y. Assuming that the velocity of the rocket at burnout is v v0 and y R at that instant, show that the approximate value of the constant c of integration is . (b) Use the solution for v in part (a) to show that the escape velocity of the rocket is given by . [Hint: Take and assume v 0 for all time t.] (c) The result in part (b) holds for any body in the Solar System. Use the values g 32 ft/s2 and R 4000 mi to show that the escape velocity from the Earth is (approximately) v0 25,000 mi/h. (d) Find the escape velocity from the Moon if the acceleration of gravity is 0.165g and R 1080 mi. Variable Mass 15. (a) In Example 4, how much of the chain would you intuitively expect the constant 5-pound force to be able to lift? (b) What is the initial velocity of the chain? (c) Why is the time interval corresponding to x(t) ! 0 given in Figure 5.3.7 not the interval I of deﬁnition of the solution (21)? Determine the interval I. How much chain is actually lifted? Explain any difference between this answer and your prediction in part (a). (d) Why would you expect x(t) to be a periodic solution? 16. A uniform chain of length L, measured in feet, is held vertically so that the lower end just touches the ﬂoor. The chain weighs 2 lb/ft. The upper end that is held is released from rest at t 0 and the chain falls straight down. If x(t) denotes the length of the chain on the ﬂoor at time t, air resistance is ignored, and the positive direc-tion is taken to be downward, then . (a) Solve for v in terms of x. Solve for x in terms of t. Express v in terms of t. (L x)d2x dt2 dx dt 2 Lg y : v0 12gR c gR 1 2 v0 2 u(0) 4. l2 v2 0, use l 1 3, v1, u(0) 2, and u(0) 2. l2 v2 0, use l2, v1, u(0)1, d 2 dt2 2 d dt 2 sin 0 (b) Determine how long it takes for the chain to fall completely to the ground. (c) What velocity does the model in part (a) predict for the upper end of the chain as it hits the ground? Miscellaneous Mathematical Models 17. Pursuit Curve In a naval exercise a ship S1 is pursued by a submarine S2 as shown in Figure 5.3.8. Ship S1 departs point (0, 0) at t 0 and proceeds along a straight-line course (the y-axis) at a constant speed v1. The subma-rine S2 keeps ship S1 in visual contact, indicated by the straightdashedline L intheﬁgure,whiletravelingatacon-stant speed v2 along a curve C. Assume that ship S2 starts at the point (a, 0), a 0, at t 0 and that L is tangent to C. (a) Determine a mathematical model that describes the curve C. (b) Find an explicit solution of the differential equation. For convenience deﬁne r v1v2. (c) Determine whether the paths of S1 and S2 will ever intersect by considering the cases r 1, r 1, and r 1. [Hint: , where s is arc length measured along C.] dt dx dt ds ds dx S2 x y S1 L C FIGURE 5.3.8 Pursuit curve in Problem 17 18. Pursuit Curve In another naval exercise a destroyer S1 pursues a submerged submarine S2. Suppose that S1 at (9, 0) on the x-axis detects S2 at (0, 0) and that S2 simultaneously detects S1. The captain of the destroyer S1 assumes that the submarine will take immediate eva-sive action and conjectures that its likely new course is the straight line indicated in Figure 5.3.9. When S1 is at (3, 0), it changes from its straight-line course toward the origin to a pursuit curve C. Assume that the speed of the destroyer is, at all times, a constant 30 mi/h and that the submarine’s speed is a constant 15 mi/h. (a) Explain why the captain waits until S1 reaches (3, 0) before ordering a course change to C. (b) Using polar coordinates, ﬁnd an equation r f(u) for the curve C. (c) Let T denote the time, measured from the initial detection, at which the destroyer intercepts the sub-marine. Find an upper bound for T. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 226 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 19. The Ballistic Pendulum Historically, in order to main-tain quality control over munitions (bullets) produced by an assembly line, the manufacturer would use a bal-listic pendulum to determine the muzzle velocity of a gun, that is, the speed of a bullet as it leaves the barrel. Invented in 1742 by the English engineer Benjamin Robins, the ballistic pendulum is simply a plane pendu-lum consisting of a rod of negligible mass to which a block of wood of mass mw is attached. The system is set in motion by the impact of a bullet which is moving horizontally at the unknown velocity vb; at the time of the impact, which we take as t 0, the combined mass is mw mb, where mb is the mass of the bullet imbedded in the wood. In (7) of this section, we saw that in the case of small oscillations, the angular displacement of a plane pendulum shown in Figure 5.3.3 is given by the linear DE where corre-sponds to motion to the right of vertical. The velocity vb can be found by measuring the height h of the mass mw mb at the maximum displacement angle shown in Figure 5.3.10. Intuitively, the horizontal velocity V of the com-bined mass (wood plus bullet) after impact is only a fraction of the velocity vb of the bullet, that is, Now recall, a distance s traveled by a particle moving along a circular path is related to the radius l and central angle by the formula By differentiating the last formula with respect to time t, it follows that the angular velocity of the mass and its linear velocity v are related by . Thus the initial angular velocity at the time t at which the bullet impacts the wood block is related to V by or (a) Solve the initial-value problem d2u dt2 g l u 0, u(0) 0, u(0) v0. v0 mb mw mb vb l . V lv0 v0 v lv v s lu. u V mb mw mbvb. umax u 0 u (g>l)u 0, u(t) (b) Use the result from part (a) to show that (c) Use Figure 5.3.10 to express cos in terms of l and h. Then use the ﬁrst two terms of the Maclaurin series for cos to express in terms of l and h. Finally, show that vb is given (approximately) by (d) Use the result in part (c) to ﬁnd vb and and h 6 cm. mb 5 g, mw 1 kg, vb mw mb mb 22gh. umax u umax vb mw mb mb 2lg umax. S2 L x y S1 C θ (3, 0) (9, 0) FIGURE 5.3.9 Pursuit curve in Problem 18 V h l mb vb h max mw mbmw FIGURE 5.3.10 Ballistic pendulum in Problem 19 20. Relief Supplies As shown in Figure 5.3.11, a plane ﬂying horizontally at a constant speed v0 drops a relief supply pack to a person on the ground. Assume the ori-gin is the point where the supply pack is released and that the positive x-axis points forward and that posi-tive y-axis points downward. Under the assumption that the horizontal and vertical components of the air resistance are proportional to and , re-spectively, and if the position of the supply pack is given by then its velocity is . Equating components in the vector form of Newton’s second law of motion, gives (a) Solve both of the foregoing initial-value problems by means of the substitutions and separation of variables. [Hint: See the Remarks at the end of Section 3.2.] (b) Suppose the plane ﬁles at an altitude of 1000 ft and that its constant speed is 300 mi/h. Assume that the constant of proportionality for air resistance is and that the supply pack weighs 256 lb. Use a root-ﬁnding application of a CAS or a graphic k 0.0053 w dy>dt, u dx>dt, m d 2y dt 2 mg k dy dt 2 , y(0) 0, y(0) 0. m d 2x dt 2 mg k dx dt 2 , x(0) 0, x(0) v0 m dv dt mg k dx dt 2 i dy dt 2 j v(t) (dx>dt)i (dy>dt)j r(t) x(t)i y(t)j, (dy>dt)2 (dx>dt)2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.3 NONLINEAR MODELS ● 227 calculator to determine the horizontal distance the pack travels, measured from its point of release to the point where it hits the ground. 24. Pendulum Motion on the Moon-Continued Repeat the two parts of Problem 23 this time using the linear model (7). Computer Lab Assignments 25. Consider the initial-value problem for a nonlinear pendulum. Since we cannot solve the differential equation, we can ﬁnd no explicit solution of this problem. But suppose we wish to determine the ﬁrst time t1 0 for which the pendulum in Figure 5.3.3, starting from its initial position to the right, reaches the position OP—that is, the ﬁrst positive root of u(t) 0. In this problem and the next we examine several ways to proceed. (a) Approximate t1 by solving the linear problem (b) Use the method illustrated in Example 3 of Sec-tion 4.10 to ﬁnd the ﬁrst four nonzero terms of a Taylor series solution u(t) centered at 0 for the non-linear initial-value problem. Give the exact values of all coefﬁcients. (c) Use the ﬁrst two terms of the Taylor series in part (b) to approximate t1. (d) Use the ﬁrst three terms of the Taylor series in part (b) to approximate t1. (e) Use a root-ﬁnding application of a CAS or a graphic calculator and the ﬁrst four terms of the Taylor series in part (b) to approximate t1. (f) In this part of the problem you are led through the commands in Mathematica that enable you to approximate the root t1. The procedure is easily modiﬁed so that any root of u(t) 0 can be approximated. (If you do not have Mathematica, adapt the given pr ocedure by finding the cor e-sponding syntax for the CAS you have on hand. ) Precisely reproduce and then, in turn, execute each line in the given sequence of commands. sol NDSolve[{y[t] Sin[y[t]] 0, y Pi/12, y 1/3}, y, {t, 0, 5}]//Flatten solution y[t]/.sol Clear[y] y[t]: Evaluate[solution] y[t] gr1 Plot[y[t], {t, 0, 5}] root FindRoot[y[t] 0, {t, 1}] (g) Appropriately modify the syntax in part (f) and ﬁnd the next two positive roots of u(t) 0. d 2 dt2 0, (0) 12, (0) 1 3. d 2 dt2 sin 0, (0) 12, (0) 1 3 supply pack target FIGURE 5.3.11 Airplane drop in Problem 20 Discussion Problems 21. Discuss why the damping term in equation (3) is written as 22. (a) Experiment with a calculator to ﬁnd an interval 0 u u1, where u is measured in radians, for which you think sin u u is a fairly good estimate. Then use a graphing utility to plot the graphs of y x and y sin x on the same coordinate axes for 0 x p2. Do the graphs conﬁrm your observations with the calculator? (b) Use a numerical solver to plot the solution curves of the initial-value problems and for several values of u0 in the interval 0 u u1 found in part (a). Then plot solution curves of the initial-value problems for several values of u0 for which u0 u1. 23. Pendulum Motion on the Moon Does a pendulum of length l oscillate faster on the Earth or on the Moon? (a) Take l 3 and g 32 for the acceleration of grav-ity on Earth. Use a numerical solver to generate a numerical solution curve for the nonlinear model (6) subject to the initial conditions Repeat using the same values but use 0.165g for the acceleration of gravity on the Moon. (b) From the graphs in part (a), determine which pendu-lum oscillates faster. Which pendulum has the greater amplitude of motion? u(0) 2. u(0) 1, d 2 dt2 0, (0) 0, (0) 0 d 2 dt2 sin 0, (0) 0, (0) 0 dx dt dx dt instead of dx dt 2 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 228 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 26. Consider a pendulum that is released from rest from an initial displacement of u0 radians. Solving the linear model (7) subject to the initial conditions u(0) u0, u(0) 0 gives . The period of oscillations predicted by this model is given by the familiar formula . The inter-esting thing about this formula for T is that it does not depend on the magnitude of the initial displacement u0. In other words, the linear model predicts that the time it would take the pendulum to swing from an ini-tial displacement of, say, u0 p2 ( 90°) to p2 and back again would be exactly the same as the time it would take to cycle from, say, u0 p360 ( 0.5°) to p360. This is intuitively unreasonable; the actual period must depend on u0. If we assume that g 32 ft/s2 and l 32 ft, then the period of oscillation of the linear model is T 2p s. Let us compare this last number with the period T 2 1g/l 2 1l/g (t) 0 cos 1g/lt predicted by the nonlinear model when u0 p4. Using a numerical solver that is capable of generating hard data, approximate the solution of on the interval 0 t 2. As in Problem 25, if t1 denotes the ﬁrst time the pendulum reaches the position OP in Figure 5.3.3, then the period of the nonlinear pendulum is 4t1. Here is another way of solving the equation u(t) 0. Experiment with small step sizes and advance the time, starting at t 0 and ending at t 2. From your hard data observe the time t1 when u(t) changes, for the ﬁrst time, from positive to negative. Use the value t1 to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period esti-mated by T 2p. d 2 dt2 sin 0, (0) 4 , (0) 0 CHAPTER 5 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-9. Answer Problems 1–8 without referring back to the text. Fill in the blank or answer true/false. 1. If a mass weighing 10 pounds stretches a spring 2.5 feet, a mass weighing 32 pounds will stretch it feet. 2. The period of simple harmonic motion of mass weigh-ing 8 pounds attached to a spring whose constant is 6.25 lb/ft is seconds. 3. The differential equation of a spring/mass system is x 16x 0. If the mass is initially released from a point 1 meter above the equilibrium position with a downward velocity of 3 m/s, the amplitude of vibra-tions is meters. 4. Pure resonance cannot take place in the presence of a damping force. 5. In the presence of a damping force, the displacements of a mass on a spring will always approach zero as . 6. A mass on a spring whose motion is critically damped can possibly pass through the equilibrium position twice. 7. At critical damping any increase in damping will result in an system. 8. If simple harmonic motion is described by , the phase angle f is when the initial conditions are and x(0) 1. In Problems 9 and 10 the eigenvalues and eigenfunc-tions of the boundary-value problem y ly 0, y(0) 0, y(p) 0 are ln n2, n 0, 1, 2, . . . , and y cos nx, respectively. Fill in the blanks. x(0) 1 2 x (12>2)sin(2t f) t : 9. A solution of the BVP when l 8 is y because . 10. A solution of the BVP when l 36 is y because . 11. A free undamped spring/mass system oscillates with a period of 3 seconds. When 8 pounds are removed from the spring, the system has a period of 2 seconds. What was the weight of the original mass on the spring? 12. A mass weighing 12 pounds stretches a spring 2 feet. The mass is initially released from a point 1 foot below the equilibrium position with an upward velocity of 4 ft/s. (a) Find the equation of motion. (b) What are the amplitude, period, and frequency of the simple harmonic motion? (c) At what times does the mass return to the point 1 foot below the equilibrium position? (d) At what times does the mass pass through the equilibrium position moving upward? Moving downward? (e) What is the velocity of the mass at ? (f) At what times is the velocity zero? 13. A force of 2 pounds stretches a spring 1 foot. With one end held ﬁxed, a mass weighing 8 pounds is attached to the other end. The system lies on a table that imparts a frictional force numerically equal to times the instantaneous velocity. Initially, the mass is displaced 4 inches above the equilibrium position and released from rest. Find the equation of motion if the motion takes place along a horizontal straight line that is taken as the x-axis. 3 2 t 3p>16 s Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 5 IN REVIEW ● 229 14. A mass weighing 32 pounds stretches a spring 6 inches. The mass moves through a medium offering a damping force that is numerically equal to b times the instanta-neous velocity. Determine the values of b 0 for which the spring/mass system will exhibit oscillatory motion. 15. Aspring with constant k 2 is suspended in a liquid that offers a damping force numerically equal to 4 times the instantaneous velocity. If a mass m is suspended from the spring, determine the values of m for which the subse-quent free motion is nonoscillatory. 16. The vertical motion of a mass attached to a spring is described by the IVP . Determine the maximum vertical displacement of the mass. 17. A mass weighing 4 pounds stretches a spring 18 inches. A periodic force equal to f(t) cos gt sin gt is impressed on the system starting at t 0. In the absence of a damping force, for what value of g will the system be in a state of pure resonance? 18. Find a particular solution for x 2lx v2x A, where A is a constant force. 19. A mass weighing 4 pounds is suspended from a spring whose constant is 3 lb/ft. The entire system is immersed in a ﬂuid offering a damping force numerically equal to the instantaneous velocity. Beginning at t 0, an external force equal to f(t) et is impressed on the system. Determine the equation of motion if the mass is initially released from rest at a point 2 feet below the equilibrium position. 20. (a) Two springs are attached in series as shown in Figure 5.R.1. If the mass of each spring is ignored, show that the effective spring constant k of the system is deﬁned by 1k 1k1 1k2. (b) A mass weighing W pounds stretches a spring foot and stretches a different spring foot. The two springs are attached, and the mass is then attached to the double spring as shown in Figure 5.R.1. Assume that the motion is free and that there is no damping force present. Determine the equation of motion if the mass is initially released at a point 1 foot below the equilibrium position with a downward velocity of . (c) Show that the maximum speed of the mass is 2 3 23g 1. 2 3 ft/s 1 4 1 2 x(0) 4, x(0) 2 1 4 x x x 0, 21. A series circuit contains an inductance of L 1 h, a capacitance of C 104 f, and an electromotive force of E(t) 100 sin 50t V. Initially, the charge q and current i are zero. (a) Determine the charge q(t). (b) Determine the current i(t). (c) Find the times for which the charge on the capacitor is zero. 22. (a) Show that the current i(t) in an LRC-series circuit satisﬁes , where E(t) denotes the derivative of E(t). (b) Two initial conditions i(0) and i(0) can be speciﬁed for the DE in part (a). If i(0) i0 and q(0) q0, what is i(0)? 23. Consider the boundary-value problem . Show that except for the case l 0, there are two independent eigenfunctions corresponding to each eigenvalue. 24. A bead is constrained to slide along a frictionless rod of length L. The rod is rotating in a vertical plane with a constant angular velocity v about a pivot P ﬁxed at the midpoint of the rod, but the design of the pivot allows the bead to move along the entire length of the rod. Let r(t) denote the position of the bead relative to this rotat-ing coordinate system as shown in Figure 5.R.2. To apply Newton’s second law of motion to this rotating frame of reference, it is necessary to use the fact that the net force acting on the bead is the sum of the real forces (in this case, the force due to gravity) and the inertial forces (coriolis, transverse, and centrifugal). The math-ematics is a little complicated, so we just give the result-ing differential equation for r: . (a) Solve the foregoing DE subject to the initial conditions r(0) r0, r(0) v0. (b) Determine the initial conditions for which the bead exhibits simple harmonic motion. What is the min-imum length L of the rod for which it can accom-modate simple harmonic motion of the bead? (c) For initial conditions other than those obtained in part (b), the bead must eventually ﬂy off the rod. Explain using the solution r(t) in part (a). (d) Suppose v 1 rad/s. Use a graphing utility to graph the solution r(t) for the initial conditions r(0) 0, r(0) v0, where v0 is 0, 10, 15, 16, 16.1, and 17. m d 2r dt2 m2r mg sin t y y 0, y(0) y(2), y(0) y(2) L d 2i dt2 R di dt 1 C i E(t) k2 k1 FIGURE 5.R.1 Attached springs in Problem 20 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 230 ● CHAPTER 5 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS (e) Suppose the length of the rod is L 40 ft. For each pair of initial conditions in part (d), use a root-ﬁnding application to ﬁnd the total time that the bead stays on the rod. bead P r(t) t ω FIGURE 5.R.2 Rotating rod in Problem 24 rigid support frictionless surface x = 0 x(t) < 0 x(t) > 0 m m (a) equilibrium (b) motion FIGURE 5.R.3 Sliding spring/mass system in Problem 25 Figure 5.R.4. If the spring constants are and deter-mine a differential equation for the displacement of the freely sliding mass. x(t) k2, k1 25. Suppose a mass m lying on a ﬂat dry frictionless surface is attached to the free end of a spring whose constant is k. In Figure 5.R.3(a) the mass is shown at the equilib-rium position x 0, that is, the spring is neither stretched nor compressed. As shown in Figure 5.R.3(b), the displacement of the mass to the right of the equi-librium position is positive and negative to the left. Determine a differential equation for the displacement of the freely sliding mass. Discuss the difference be-tween the derivation of this DE and the analysis leading to (1) of Section 5.1. x(t) x(t) rigid support rigid support k1 k2 m FIGURE 5.R.4 Double spring system in Problem 26 27. Suppose the mass m in the spring/mass system in Problem 25 slides over a dry surface whose coefﬁcient of sliding friction is . If the retarding force of kinetic friction has the constant magnitude where mg is the weight of the mass, and acts opposite to the direction of motion, then it is known as coulomb friction. By using the signum function determine a piecewise-deﬁned differential equation for the displacement of the damped sliding mass. 28. For simplicity, let us assume in Problem 27 that and (a) Find the displacement of the mass if it is re-leased from rest from a point units to the right of the equilibrium position, that is, the initial condi-tions are When released, intuitively the motion of the mass will be to the left. Give a time interval over which this solution is deﬁned. Where is the mass at time t1? (b) For assume that the motion is now to the right. Using initial conditions at ﬁnd and give a time interval [t1, t2] over which this solution is deﬁned. Where is the mass at time t2? (c) For assume that the motion is now to the left. Using initial conditions at ﬁnd and give a time interval [t2, t3] over which this solution is deﬁned. Where is the mass at time t3? (d) Using initial conditions at t3, show that the model predicts that there is no furthere motion for (e) Graph the displacement on the interval [0, t3]. x(t) t t3. x(t) t2, t t2 x(t) t1, t t1 [0, t1] x(0) 5.5, x(0) 0. 5.5 x(t) fk 1. k 1, m 1, x(t) sgn(x) 1, x 0 (motion to left) 1, x 0 (motion to right) fk "mg, " 0 26. Suppose the mass m on the ﬂat, dry, frictionless surface in Problem 25 is attached to two springs as shown Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 231 231 6 Series Solutions of Linear Equations 6.1 Review of Power Series 6.2 Solutions About Ordinary Points 6.3 Solutions About Singular Points 6.4 Special Functions Chapter 6 in Review Up to this point in our study of differential equations we have primarily solved linear equations of order two (or higher) that have constant coefﬁcients. The only exception was the Cauchy-Euler equation in Section 4.7. In applications, higher-order linear equations with variable coefﬁcients are just as important as, if not more than, differential equations with constant coefﬁcients. As pointed out in Section 4.7, even a simple linear second-order equation with variable coefﬁcients such as does not possess solutions that are elementary functions. But this is not to say that we can’t ﬁnd two linearly independent solutions of we can. In Sections 6.2 and 6.4 we shall see that the functions that are solutions of this equation are deﬁned by inﬁnite series. In this chapter we shall study two inﬁnite-series methods for ﬁnding solutions of homogeneous linear second-order DEs where the variable coefﬁcients are, for the most part, simple polynomial functions. a2(x), a1(x), and a0(x) a2(x)y a1(x)y a0(x)y 0, y xy 0; y xy 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 232 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS x a a + R a − R divergence divergence absolute convergence series may converge or diverge at endpoints FIGURE 6.1.1 Absolute convergence within the interval of convergence and divergence outside of this interval REVIEW OF POWER SERIES REVIEW MATERIAL ●Inﬁnite series of constants, p-series, harmonic series, alternating harmonic series, geometric series, tests for convergence especially the ratio test ●Power series, Taylor series, Maclaurin series (See any calculus text) INTRODUCTION In Section 4.3 we saw that solving a homogeneous linear DE with constant coefﬁcients was essentially a problem in algebra. By ﬁnding the roots of the auxiliary equation, we could write a general solution of the DE as a linear combination of the elementary functions But as was pointed out in the introduction to Section 4.7, most linear higher-order DEs with variable coefﬁcients cannot be solved in terms of elementary functions. A usual course of action for equations of this sort is to assume a solution in the form of an inﬁnite series and proceed in a manner similar to the method of undetermined coefﬁcients (Section 4.4). In Section 6.2 we consider linear second-order DEs with variable coefﬁcients that possess solutions in the form of a power series, and so it is appropriate that we begin this chapter with a review of that topic. eax, xkeax, xkeaxcosbx, and xkeaxsinbx. 6.1 Power Series Recall from calculus that power series in is an inﬁnite series of the form Such a series also said to be a power series centered at a. For example, the power series is centered at a 1. In the next section we will be concerned principally with power series in x, in other words, power series that are centered at . For example, is a power series in x. Important Facts The following bulleted list summarizes some important facts about power series • Convergence A power series is convergent at a speciﬁed value of x if its sequence of partial sums converges, that is, exists. If the limit does not exist at x, then the series is said to be divergent. • Interval of Convergence Every power series has an interval of convergence. The interval of convergence is the set of all real numbers x for which the series converges. The center of the interval of convergence is the center a of the series. • Radius of Convergence The radius R of the interval of convergence of a power series is called its radius of convergence. If then a power series converges for and diverges for If the series converges only at its center a, then If the series converges for all x, then we write Recall, the absolute-value inequality is equivalent to the simultaneous inequality Apower series may or may not converge at the endpoints of this interval. • Absolute Convergence Within its interval of convergence a power series converges absolutely. In other words, if x is in the interval of convergence and is not an endpoint of the interval, then the series of absolute values converges. See Figure 6.1.1. n0 cn(x a)n a R and a R a R x a R. x a R R . R 0. x a R. x a R R 0, cn (x a)n lim N : N n0 lim N : SN(x) {SN(x)} n0cn(x a)n. n0 2nxn 1 2x 4x2 . . . a 0 n0(x 1)n n0 cn(x a)n c0 c1(x a) c2(x a)2 . . .. x a The index of summation need not start at n 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. • Ratio Test Convergence of power series can often be determined by the ratio test. Suppose for all n in and that If the series converges absolutely; if the series diverges; and if the test is inconclusive. The ratio test is always inconclusive at an endpoint a R. L 1 L 1 L 1, lim n: cn1(x a)n1 cn(x a)n x a lim n: cn1 cn L. n0 cn(x a)n, cn 0 6.1 REVIEW OF POWER SERIES ● 233 EXAMPLE 1 Interval of Convergence Find the interval and radius of convergence for SOLUTION The ratio test gives The series converges absolutely for or or 1 x 5. This last inequality deﬁnes the open interval of convergence. The series diverges for , that is, for x 5 or x 1. At the left endpoint x 1 of the open interval of convergence, the series of constants is convergent by the alternating series test. At the right endpoint x 5, the series is the divergent harmonic series. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. • A Power Series Defines a Functio A power series deﬁnes a function, that is, whose domain is the interval of convergence of the series. If the radius of convergence is R 0 or then f is continuous, differentiable, and integrable on the intervals (a R, a R) or , respectively. Moreover, f(x) and f(x)dx can be found by term-by-term differentiation and integration. Convergence at an endpoint may be either lost by differentiation or gained through integration. If is a power series in x, then the ﬁrst two derivatives are and Notice that the ﬁrst term in the ﬁrst derivative and the ﬁrst two terms in the second derivative are zero. We omit these zero terms and write . (1) Be sure you understand the two results given in (1); especially note where the index of summation starts in each series. These results are important and will be used in all examples in the next section. • Identity Property If 0, R 0, for all numbers x in some open interval, then for all n. • Analytic at a Point A function f is said to be analytic at a point a if it can be represented by a power series in x a with either a positive or an inﬁnite radius of convergence. In calculus it is seen that inﬁnitely cn 0 n0 cn(x a)n y n2 cnn(n 1)xn2 2c2 6c3x 12c4x2 . . . y n1 cnnxn1 c1 2c2x 3c3x2 4c4x3 . . . y n0 n(n 1)xn2. y n0 nxn1 y n1 cnxn c0 c1x c2x2 c3x3 . . . (, ) R , f (x) n0 cn(x a)n n1 (1>n) n1 ((1)n>n) x 3 2 x 3 2 1 2 x 3 1 lim n: (x 3)n1 2n1(n 1) (x 3)n 2nn x 3 lim n: n 1 2n 1 2 x 3 . n1 (x 3)n 2nn . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 234 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS differentiable functions such as ex and so on, can be represented by Taylor series or by a Maclaurin series . You might remember some of the following Maclaurin series representations. Interval Maclaurin Series of Convergence (2) These results can be used to obtain power series representations of other functions. For example, if we wish to ﬁnd the Maclaurin series representation of, say, we need only replace x in the Maclaurin series for Similarly, to obtain a Taylor series representation of centered at we replace x by in the Maclaurin series for ln(1 x): x 1 a 1 ln x ex2 1 x2 1! x4 2! x6 3! . . . n0 1 n! x2n. ex: ex2 (1, 1) 1 1 x 1 x x2 x3 . . . n0 xn (1, 1] ln(1 x) x x2 2 x3 3 x4 4 . . . n1 (1)n1 n xn (, ) sinh x x x3 3! x5 5! x7 7! . . . n0 1 (2n 1)!x2n1 (, ) cosh x 1 x2 2! x4 4! x6 6! . . . n0 1 (2n)!x2n [1, 1] tan1 x x x3 3 x5 5 x7 7 . . . n0 (1)n 2n 1x2n1 (, ) sin x x x3 3! x5 5! x7 7! . . . n0 (1)n (2n 1)!x2n1 (, ) cos x 1 x2 2! x4 4! x6 6! . . . n0 (1)n (2n)! x2n (, ) ex 1 x 1! x2 2! x3 3! . . . n0 1 n!xn n0 f (n)(0) n! xn f(0) f(0) 1! x f (0) 1! x2 . . . n0 f (n)(a) n! (x a)n f (a) f(a) 1! (x a) f (a) 1! (x a)2 . . . ln(1 x), cos x, ex, sinx, ln x ln(1 (x 1)) (x 1) (x 1)2 2 (x 1)3 3 (x 1)4 4 . . . n1 (1)n1 n (x 1)n. The interval of convergence for the power series representation of is the same as that of that is, But the interval of convergence of the Taylor series of is now this interval is shifted 1 unit to the right. • Arithmetic of Power Series Power series can be combined through the operations of addition, multiplication, and division. The procedures for powers series are similar to the way in which two polynomials are added, multiplied, and divided —that is, we add coefﬁcients of like powers of x, use the distributive law and collect like terms, and perform long division. (1, 1] (0, 2]; ln x (, ). ex, ex2 You can also verify that the interval of convergence is (0, 2] by using the ratio test. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.1 REVIEW OF POWER SERIES ● 235 EXAMPLE 2 Multiplication of Power Series Find a power series representation of SOLUTION We use the power series for and Since the power series of and both converge on the product series converges on the same interval. Problems involving multiplication or division of power series can be done with minimal fuss using a computer algebra system. Shifting the Summation Index For the three remaining sections of this chap-ter, it is crucial that you become adept at simplifying the sum of two or more power series, each series expressed in summation notation, to an expression with a single As the next example illustrates, combining two or more summations as a single summa-tion often requires a reindexing, that is, a shift in the index of summation. . (, ), sin x ex x x2 x3 3 x5 30 . . . . (1)x (1)x2 1 6 1 2 x3 1 6 1 6x4 1 120 1 12 1 24x5 . . . exsinx 1 x x2 2 x3 6 x4 24 . . .x x3 6 x5 120 x7 5040 . . . sinx: ex ex sin x. EXAMPLE 3 Addition of Power Series Write as one power series. SOLUTION In order to add the two series given in summation notation, it is neces-sary that both indices of summation start with the same number and that the powers of x in each series be “in phase,” in other words, if one series starts with a multiple of, say, x to the ﬁrst power, then we want the other series to start with the same power. Note that in the given problem, the first series starts with x0 whereas the second series starts with x1. By writing the ﬁrst term of the ﬁrst series outside of the summa-tion notation, (3) we see that both series on the right side start with the same power of x, namely, x1. Now to get the same summation index we are inspired by the exponents of x; we let in the ﬁrst series and at the same time let in the second series. For in we get and for in we get and so the right-hand side of (3) becomes (4) same same 2c2 (k 2)(k 1)ck2x k ck1x k. k1 k1 k 1, k n 1 n 0 k 1, k n 2 n 3 k n 1 k n 2 series starts with x for n 3 series starts with x for n 0 n(n 1)cnx n2 cnx n1 2 1c2x 0 n(n 1)cnx n2 cnxn1 n2 n0 n3 n0 n2 n(n 1)cnxn2 n0 cnxn1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Remember the summation index is a “dummy” variable; the fact that in one case and in the other should cause no confusion if you keep in mind that it is the value of the summation index that is important. In both cases k takes on the same successive values when n takes on the values for and for We are now in a position to add the series in (4) term-by-term: (5) If you are not totally convinced of the result in (5), then write out a few terms on both sides of the equality. A Preview The point of this section is to remind you of the salient facts about power series so that you are comfortable using power series in the next section to ﬁnd solutions of linear second-order DEs. In the last example in this section we tie up many of the concepts just discussed; it also gives a preview of the method that will used in Section 6.2. We purposely keep the example simple by solving a linear ﬁrst-order equation. Also suspend, for the sake of illustration, the fact that you already know how to solve the given equation by the integrating-factor method in Section 2.3. n2 n(n 1)cnxn2 n0 cnxn1 2c2 k1 [(k 2)(k 1)ck2 ck1]xk. k n 1. n 0, 1, 2, . . . k n 1 n 2, 3, 4, . . . k 1, 2, 3, . . . k n 1 k n 2 236 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS EXAMPLE 4 A Power Series Solution Find a power series solution of the differential equation SOLUTION We break down the solution into a sequence of steps. (i) First calculate the derivative of the assumed solution: (ii) Then substitute into the given DE: (iii) Now shift the indices of summation. When the indices of summation have the same starting point and the powers of x agree, combine the summations: y y n1 cnnxn1 n0 cnxn y y n1 cnnxn1 n0 cnxn. y and y ; see the first line in (1) y n1 cnnxn1 y y 0. y n0 cnxn 123 123 k n1 k n (iv) Because we want for all x in some interval, is an identity and so we must have ck1 1 k 1 ck, k 0, 1, 2, . . . . ck1(k 1) ck 0, or k0 [ck1(k 1) ck]xk 0 y y 0 k0 [ck1(k 1) ck]xk. k0 ck1(k 1)xk k0 ckxk Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (v) By letting k take on successive integer values starting with we ﬁnd and so on, where is arbitrary. (vi) Using the original assumed solution and the results in part (v) we obtain a formal power series solution It should be fairly obvious that the pattern of the coefﬁcients in part (v) is so that in summation notation we can write (8) From the ﬁrst power series representation in (2) the solution in (8) is recognized as Had you used the method of Section 2.3, you would have found that is a solution of on the interval This interval is also the interval of convergence of the power series in (8). (, ). y y 0 y cex y c0ex. y c0 k0 (1)k k! xk. ck c0(1)k>k!, k 0, 1, 2, . . . . c01 x 1 2x2 1 3 2x3 1 4 3 2x4 . . .. c0 c0x 1 2c0x2 c0 1 3 2x3 c0 1 4 3 2x4 . . . y c0 c1x c2x2 c3x3 c4x4 . . . c0 c4 1 4c2 1 4 1 3 2c0 1 4 3 2c0 c3 1 3c2 1 3 1 2c0 1 3 2c0 c2 1 2c1 1 2(c0) 1 2c0 c1 1 1c0 c0 k 0, 6.1 REVIEW OF POWER SERIES ● 237 If desired we could switch back to n as the index of summation. EXERCISES 6.1 Answers to selected odd-numbered problems begin on page ANS-9. In Problems 1–10 ﬁnd the interval and radius of convergence for the given power series. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. In Problems 11–16 use an appropriate series in (2) to ﬁnd the Maclaurin series of the given function. Write your answer in summation notation. 11. 12. 13. 14. x 1 x2 1 2 x xe3x ex>2 n0 (1)n 9n x2n1 k1 25k 52k x 3 k k0 3k(4x 5)k k1 1 k2 k(3x 1)k k0 k!(x 1)k k1 (1)k 10k (x 5)k n0 5n n!xn n1 2n n xn n1 1 n2 xn n1 (1)n n xn 15. 16. In Problems 17 and 18 use an appropriate series in (2) to ﬁnd the Taylor series of the given function centered at the indi-cated value of a. Write your answer in summation notation. 17. [Hint: Use periodicity.] 18. [Hint: ] In Problems 19 and 20 the given function is analytic at Use appropriate series in (2) and multiplication to ﬁnd the ﬁrst four nonzero terms of the Maclaurin series of the given function. 19. 20. In Problems 21 and 22 the given function is analytic at Use appropriate series in (2) and long division to ﬁnd the ﬁrst four nonzero terms of the Maclaurin series of the given function. 21. 22. tan x sec x a 0. excos x sin x cos x a 0. x 2[1 (x 2)>2] ln x; a 2 sinx, a 2p sin x2 ln(1 x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 238 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS In Problems 23 and 24 use a substitution to shift the summa-tion index so that the general term of given power series involves 23. 24. In Problems 25–30 proceed as in Example 3 to rewrite the given expression using a single power series whose general term involves 25. 26. 27. 28. 29. 30. n2 n(n 1)cnxn 2 n2 n(n 1)cnxn2 3 n1 ncnxn n2 n(n 1)cnxn2 2 n1 ncnxn n0 cnxn n2 n(n 1)cnxn2 n0 cnxn2 n1 2ncnxn1 n0 6cnxn1 n1 ncnxn1 3 n0 cnxn2 n1 ncnxn1 n0 cnxn xk. n3 (2n 1)cnxn3 n1 ncnxn2 xk. In Problems 31–34 verify by direct substitution that the given power series is a solution of the indicated differential equation. [Hint: For a power let 31. 32. 33. 34. In Problems 35–38 proceed as in Example 4 and ﬁnd a power series solution of the given linear ﬁrst-order differential equation. 35. 36. 37. 38. Discussion Problems 39. In Problem 19, ﬁnd an easier way than multiplying two power series to obtain the Maclaurin series representa-tion of 40. In Problem 21, what do you think is the interval of con-vergence for the Maclaurin series of sec x? sin x cos x. (1 x)y y 0 y xy 4y y 0 y 5y 0 y n0 cnxn y n0 (1)n 22n(n!)2x2n, xy y xy 0 y n1 (1)n1 n xn, (x 1)y y 0 y n0 (1)nx2n, (1 x2)y 2xy 0 y n0 (1)n n! x2n, y 2xy 0 k n 1.] x2n1 SOLUTIONS ABOUT ORDINARY POINTS REVIEW MATERIAL ●Power series, analytic at a point, shifting the index of summation in Section 6.1 INTRODUCTION At the end of the last section we illustrated how to obtain a power series solution of a linear ﬁrst-order differential equation. In this section we turn to the more important problem of ﬁnding power series solutions of linear second-order equations. More to the point, we are going to ﬁnd solutions of linear second-order equations in the form of power series whose center is a number that is an ordinary point of the DE. We begin with the definition of an ordinary point. x0 6.2 A Definitio If we divide the homogeneous linear second-order differential equation (1) by the lead coefﬁcient we obtain the standard form (2) We have the following deﬁnition. y P(x)y Q(x)y 0. a2(x) a2(x)y a1(x)y a0(x)y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.2 SOLUTIONS ABOUT ORDINARY POINTS ● 239 DEFINITION 6.2.1 Ordinary and Singular Points A point is said to be an ordinary point of the differential of the differential equation (1) if both coefﬁcients and in the standard form (2) are analytic at A point that is not an ordinary point of (1) is said to be a singular point of the DE. x0. Q(x) P(x) x x0 EXAMPLE 1 Ordinary Points (a) A homogeneous linear second-order differential equation with constant coefﬁcients, such as can have no singular points. In other words, every ﬁnite value of x is an ordinary point of such equations. (b) Every ﬁnite value of x is an ordinary point of the differential equation Speciﬁcally is an ordinary point of the DE, because we have already seen in (2) of Section 6.1 that both and are analytic at this point. The negation of the second sentence in Deﬁnition 6.2.1 stipulates that if at least one of the coefﬁcient functions in (2) fails to be analytic at then is a singular point. x0 x0, P(x) and Q(x) sin x ex x 0 y exy (sin x)y 0. y y 0 and y 3y 2y 0, EXAMPLE 2 Singular Points (a) The differential equation is already in standard form. The coefﬁcient functions are Now is analytic at every real number, and is analytic at every positive real number. However, since is discontinuous at it cannot be represented by a power series in x, that is, a power series centered at 0. We conclude that is a singular point of the DE. (b) By putting in the standard form , we see that fails to be analytic at . Hence is a singular point of the equation. Polynomial Coefficient We will primarily be interested in the case when the coefﬁcients in (1) are polynomial functions with no common factors. A polynomial function is analytic at any value of x, and a rational function is analytic except at points where its denominator is zero. Thus, in (2) both coefﬁcients P(x) a1(x) a2(x) and Q(x) a0(x) a2(x) a2(x), a1(x), and a0(x) x 0 x 0 P(x) 1/x y 1 x y y 0 xy y xy 0 x 0 x 0 Q(x) ln x Q(x) ln x P(x) x P(x) x and Q(x) ln x. y xy (lnx)y 0 For our purposes, ordinary points and singular points will always be ﬁnite points. It is possible for an ODE to have, say, a singular point at inﬁnity. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 240 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS are analytic except at those numbers for which It follows, then, that A number is an ordinary point of (1) if whereas is a singular point of (1) if a2(x0) 0. x x0 a2(x0) 0, x x0 a2(x) 0. EXAMPLE 3 Ordinary and Singular Points (a) The only singular points of the differential equation are the solutions of All other values of x are ordinary points. (b) Inspection of the Cauchy-Euler shows that it has a singular point at All other values of x are ordinary points. (c) Singular points need not be real numbers. The equation has singular points at the solutions of —namely, All other values of x, real or complex, are ordinary points. We state the following theorem about the existence of power series solutions without proof. x i. x2 1 0 (x2 1)y xy y 0 x 0. x2y y 0 a2(x) x2 0 at x 0 b x2 1 0 or x 1. (x2 1)y 2xy 6y 0 THEOREM 6.2.1 Existence of Power Series Solutions If is an ordinary point of the differential equation (1), we can always ﬁnd two linearly independent solutions in the form of a power series centered at that is, A power series solution converges at least on some interval deﬁned by , where R is the distance from to the closest singular point. x0 x x0 & R y n0 cn(x x0)n. x0, x x0 A solution of the form is said to be a solution about the ordinary point x0. The distance R in Theorem 6.2.1 is the minimum value or lower bound for the radius of convergence. y n0 cn(x x0)n EXAMPLE 4 Minimum Radius of Convergence Find the minimum radius of convergence of a power series solution of the second-order differential equation (a) about the ordinary point (b) about the ordinary point SOLUTION By the quadratic formula we see from that the singular points of the given differential equation are the complex numbers 1 2i. x2 2x 5 0 x 1. x 0, (x2 2x 5)y xy y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (a) Because is an ordinary point of the DE, Theorem 6.2.1 guarantees that we can ﬁnd two power series solutions centered at 0. That is, solutions that look like and, moreover, we know without actually ﬁnding these solutions that each series must converge at least for , where is the distance in the complex plane from either of the numbers (the point or (the point to the ordinary point 0 (the point See Figure 6.2.1. (b) Because is an ordinary point of the DE, Theorem 6.2.1 guarantees that we can ﬁnd two power series solutions that look like Each of power series converges at least for since the distance from each of the singular points to (the point is In part (a) of Example 4, one of the two power series solutions centered at 0 of the differential equation is valid on an interval much larger than in actual fact this solution is valid on the interval because it can be shown that one of the two solutions about 0 reduces to a polynomial. Note In the examples that follow as well as in the problems of Exercises 6.2 we will, for the sake of simplicity, ﬁnd only power series solutions about the ordinary point If it is necessary to ﬁnd a power series solutions of an ODE about an ordinary point we can simply make the change of variable in the equation (this translates to ﬁnd solutions of the new equation of the form , and then resubstitute Finding a Power Series Solution Finding a power series solution of a homo-geneous linear second-order ODE has been accurately described as “the method of undetermined series coefﬁcients” since the procedure is quite analogous to what we did in Section 4.4. In case you did not work through Example 4 of Section 6.1 here, in brief, is the idea. Substitute into the differential equation, combine series as we did in Example 3 of Section 6.1, and then equate the all coefﬁcients to the right-hand side of the equation to determine the coefﬁcients But because the right-hand side is zero, the last step requires, by the identity property in the bulleted list in Section 6.1, that all coefﬁcients of x must be equated to zero. No, this does not mean that all coefﬁcients are zero; this would not make sense, after all Theorem 6.2.1 guarantees that we can ﬁnd two solutions. We will see in Example 5 how the single assumption that leads to two sets of coefﬁ-cients so that we have two distinct power series and both expanded about the ordinary point The general solution of the differential equation is ; indeed, it can be shown that and . C2 c1 C1 c0 y C1y1(x) C2y2(x) x 0. y2 (x), y1(x) y n0cnxn c0 c1x c2x2 . . . cn. y n0cnxn t x x0. y n0cntn t 0), x x0 t x x0 x0 0, x 0. (, ) (15, 15); R 18 212. (1, 0)) 1 x 1 212 y n0cn(x 1)n. x 1 (0, 0)). (1, 2)) 1 2i (1, 2)) 1 2i R 25 x 25 y n0 cnxn x 0 6.2 SOLUTIONS ABOUT ORDINARY POINTS ● 241 FIGURE 6.2.1 Distance from singular points to the ordinary point 0 in Example 4 y x 1 1 + 2i 1 − 2i i 5 5 EXAMPLE 5 Power Series Solutions Solve SOLUTION Since there are no singular points, Theorem 6.2.1 guarantees two power series solutions centered at 0 that converge for Substituting and the second derivative (see (1) in Section 6.1) into the differential equation give (3) We have already added the last two series on the right-hand side of the equality in (3) by shifting the summation index. From the result given in (5) of Section 6.1 (4) y xy 2c2 k1 [(k 1)(k 2)ck2 ck1]xk 0. y xy n2 cnn(n 1)xn2 x n0 cnxn n2 cnn(n 1)xn2 n0 cnxn1. y n2 n(n 1)cnxn2 y n0 cnxn x . y xy 0. Before working through this example, we recommend that you reread Example 4 of Section 6.1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. At this point we invoke the identity property. Since (4) is identically zero, it is neces-sary that the coefﬁcient of each power of x be set equal to zero—that is, 2c2 0 (it is the coefﬁcient of x0), and (5) Now 2c2 0 obviously dictates that c2 0. But the expression in (5), called a recurrence relation, determines the ck in such a manner that we can choose a certain subset of the set of coefﬁcients to be nonzero. Since (k 1)(k 2) 0 for all val-ues of k, we can solve (5) for ck2 in terms of ck1: (6) This relation generates consecutive coefﬁcients of the assumed solution one at a time as we let k take on the successive integers indicated in (6): and so on. Now substituting the coefﬁcients just obtained into the original assumption ; c8 is zero k 9, c11 c8 10 11 0 k 8, c10 c7 9 10 1 3 4 6 7 9 10 c1 k 7, c9 c6 8 9 1 2 3 5 6 8 9 c0 ; c5 is zero k 6, c8 c5 7 8 0 k 5, c7 c4 6 7 1 3 4 6 7 c1 k 4, c6 c3 5 6 1 2 3 5 6 c0 ; c2 is zero k 3, c5 c2 4 5 0 k 2, c4 c1 3 4 k 1, c3 c0 2 3 ck2 ck1 (k 1)(k 2) , k 1, 2, 3, . . . . (k 1)(k 2)ck2 ck1 0, k 1, 2, 3, . . . . 242 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS y c0 c1x c2x2 c3x3 c4x4 c5x5 c6x6 c7x7 c8x8 c9x9 c10x10 c11x11 , we get c1 3 4 6 7 x7 0 c0 2 3 5 6 8 9 x9 c1 3 4 6 7 9 10 x10 0 . y c0 c1x 0 c0 2 3 x3 c1 3 4 x4 0 c0 2 3 5 6 x6 After grouping the terms containing c0 and the terms containing c1, we obtain y c0y1(x) c1y2(x), where y2(x) x 1 3 4 x4 1 3 4 6 7 x7 1 3 4 6 7 9 10 x10 x k1 (1)k 3 4 (3k)(3k 1) x3k1. y1(x) 1 1 2 3 x3 1 2 3 5 6 x6 1 2 3 5 6 8 9 x9 1 k1 (1)k 2 3 (3k 1)(3k) x3k Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Because the recursive use of (6) leaves c0 and c1 completely undetermined, they can be chosen arbitrarily. As was mentioned prior to this example, the linear combination y c0y1(x) c1y2(x) actually represents the general solution of the differential equation. Although we know from Theorem 6.2.1 that each series solu-tion converges for that is, on the interval . This fact can also be ver-iﬁed by the ratio test. The differential equation in Example 5 is called Airy’s equation and is named after the English mathematician and astronomer George Biddel Airy (1801–1892). Airy’s differential equation is encountered in the study of diffraction of light, diffrac-tion of radio waves around the surface of the Earth, aerodynamics, and the deﬂection of a uniform thin vertical column that bends under its own weight. Other common forms of Airy’s equation are y xy 0 and y 2xy 0. See Problem 41 in Exercises 6.4 for an application of the last equation. (, ) x , 6.2 SOLUTIONS ABOUT ORDINARY POINTS ● 243 EXAMPLE 6 Power Series Solution Solve (x2 1)y xy y 0. SOLUTION As we have already seen on page 240, the given differential equation has singular points at x i, and so a power series solution centered at 0 will converge at least for 1, where 1 is the distance in the complex plane from 0 to either i or i. The assumption and its ﬁrst two derivatives lead to y n0 cnxn x (x 2 1) n(n 1)cnx n2 x ncnx n1 cnx n n2 n1 n0 n(n 1)cnx n n(n 1)cnx n2 ncnx n cnx n n2 n2 n1 n0 2c2 c0 6c3x [k(k 1)ck (k 2)(k 1)ck2 kck ck]xk k2 2c2 c0 6c3x [(k 1)(k 1)ck (k 2)(k 1)ck2]x k 0. k2 n(n 1)cnx n2 ncnx n cnx n n4 n2 n2 2c2x 0 c0x 0 6c3x c1x c1x n(n 1)cnx n n2 kn kn2 kn kn From this identity we conclude that 2c2 c0 0, 6c3 0, and Thus Substituting k 2, 3, 4, . . . into the last formula gives c4 1 4 c2 1 2 4 c0 1 222! c0 ck2 1 k k 2 ck, k 2, 3, 4, . . . . c3 0 c2 1 2 c0 (k 1)(k 1)ck (k 2)(k 1)ck2 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. and so on. Therefore c10 7 10 c8 3 5 7 2 4 6 8 10 c0 1 3 5 7 255! c0, ; c7 is zero c9 6 9 c7 0, c8 5 8 c6 3 5 2 4 6 8 c0 1 3 5 244! c0 ; c5 is zero c7 4 7 c5 0 c6 3 6 c4 3 2 4 6 c0 1 3 233! c0 ; c3 is zero c5 2 5 c3 0 244 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS c5 c3 c2 4 5 c0 4 5 1 6 1 2 c0 30 c4 c2 c1 3 4 c0 2 3 4 c0 24 c3 c1 c0 2 3 c0 2 3 c0 6 c2 1 2 c0 c0 0, c1 0 c5 c3 c2 4 5 c1 4 5 6 c1 120 c4 c2 c1 3 4 c1 3 4 c1 12 c3 c1 c0 2 3 c1 2 3 c1 6 c2 1 2 c0 0 c0 0, c1 0 c0y1(x) c1y2(x). c01 1 2 x2 1 222! x4 1 3 233! x6 1 3 5 244! x8 1 3 5 7 255! x10 c1x y c0 c1x c2x2 c3x3 c4x4 c5x5 c6x6 c7x7 c8x8 c9x9 c10x10 The solutions are the polynomial y2(x) x and the power series y1(x) 1 1 2 x2 n2 (1)n11 3 5 2n 3 2nn! x2n, x 1. EXAMPLE 7 Three-Term Recurrence Relation If we seek a power series solution for the differential equation we obtain and the three-term recurrence relation It follows from these two results that all coefﬁcients cn, for n 3, are expressed in terms of both c0 and c1. To simplify life, we can ﬁrst choose c0 0, c1 0; this yields coefﬁcients for one solution expressed entirely in terms of c0. Next, if we choose c0 0, c1 0, then coefﬁcients for the other solution are expressed in terms of c1. Using in both cases, the recurrence relation for k 1, 2, 3, . . . gives c2 1 2 c0 ck2 ck ck1 (k 1)(k 2), k 1, 2, 3, . . . . c2 1 2 c0 y (1 x)y 0, y n0 cnxn Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. and so on. Finally, we see that the general solution of the equation is y c0y1(x) c1y2(x), where and Each series converges for all ﬁnite values of x. Nonpolynomial Coefficient The next example illustrates how to ﬁnd a power series solution about the ordinary point x0 0 of a differential equation when its coefﬁcients are not polynomials. In this example we see an application of the multiplication of two power series. y2(x) x 1 6 x3 1 12 x4 1 120 x5 . y1(x) 1 1 2 x2 1 6 x3 1 24 x4 1 30 x5 6.2 SOLUTIONS ABOUT ORDINARY POINTS ● 245 EXAMPLE 8 DE with Nonpolynomial Coefficient Solve y (cos x)y 0. SOLUTION We see that x 0 is an ordinary point of the equation because, as we have already seen, cos x is analytic at that point. Using the Maclaurin series for cos x given in (2) of Section 6.1, along with the usual assumption and the results in (1) of Section 6.1 we ﬁnd y n0 cnxn 2c2 c0 (6c3 c1)x 12c4 c2 1 2 c0x2 20c5 c3 1 2 c1x3 0. 2c2 6c3x 12c4x2 20c5x3 1 x2 2! x4 4! (c0 c1x c2x2 c3x3 ) y (cos x)y n2 n(n 1)cnxn2 1 x2 2! x4 4! x6 6! n0 cnxn It follows that 2c2 c0 0, 6c3 c1 0, 12c4 c2 1 2 c0 0, 20c5 c3 1 2 c1 0, and so on. This gives By group-ing terms, we arrive at the general solution y c0y1(x) c1y 2(x), where Because the differential equation has no ﬁnite singular points, both power series con-verge for Solution Curves The approximate graph of a power series solution can be obtained in several ways. We can always resort to graphing the terms in the sequence of partial sums of the series—in other words, the graphs of the polynomials For large values of N, SN(x) should give us an indi-cation of the behavior of y(x) near the ordinary point x 0. We can also obtain an ap-proximate or numerical solution curve by using a solver as we did in Section 4.10. For example, if you carefully scrutinize the series solutions of Airy’s equation in SN(x) N n0 cnxn. n0 cnxn y(x) x . y1(x) 1 1 2 x2 1 12 x4 and y2(x) x 1 6 x3 1 30 x5 . c5 1 30 c1, . . . . c4 1 12 c0, c3 1 6 c1, c2 1 2 c0, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Example 5, you should see that y1(x) and y2(x) are, in turn, the solutions of the initial-value problems (11) The speciﬁed initial conditions “pick out” the solutions y1(x) and y2(x) from y c0y1(x) c1y2(x), since it should be apparent from our basic series assumption that y(0) c0 and y(0) c1. Now if your numerical solver requires a system of equations, the substitution y u in y xy 0 gives y u xy, and so a system of two ﬁrst-order equations equivalent to Airy’s equation is (12) Initial conditions for the system in (12) are the two sets of initial conditions in (11) rewritten as y(0) 1, u(0) 0, and y(0) 0, u(0) 1. The graphs of y1(x) and y2(x) shown in Figure 6.2.2 were obtained with the aid of a numerical solver. The fact that the numerical solution curves appear to be oscillatory is consistent with the fact that Airy’s equation appeared in Section 5.1 (page 197) in the form mx ktx 0 as a model of a spring whose “spring constant” K(t) kt increases with time. u xy. y u y n0 cnxn y xy 0, y(0) 0, y(0) 1. y xy 0, y(0) 1, y(0) 0, 246 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS _2 2 4 6 10 8 1 2 3 x y1 _2 _1 _2 _3 2 4 6 10 8 1 x y2 (a) plot of y1(x) (b) plot of y2(x) FIGURE 6.2.2 Numerical solution curves for Airy’s DE REMARKS (i) In the problems that follow, do not expect to be able to write a solution in terms of summation notation in each case. Even though we can generate as many terms as desired in a series solution either through the use of a recurrence relation or, as in Example 8, by multiplication, it might not be possible to deduce any general term for the coefﬁcients cn. We might have to settle, as we did in Examples 7 and 8, for just writing out the ﬁrst few terms of the series. (ii) A point x0 is an ordinary point of a nonhomogeneous linear second-order DE y P(x)y Q(x)y f (x) if P(x), Q(x), and f(x) are analytic at x0. Moreover, Theorem 6.2.1 extends to such DEs; in other words, we can ﬁnd power series solutions of nonhomogeneous linear DEs in the same manner as in Examples 5–8. See Problem 26 in Exercises 6.2. y n0 cn(x x0)n y n0 cnxn EXERCISES 6.2 Answers to selected odd-numbered problems begin on page ANS-9. In Problems 1 and 2 without actually solving the given differential equation, ﬁnd the minimum radius of convergence of power series solutions about the ordinary point About the ordinary point 1. 2. In Problems 3–6 ﬁnd two power series solutions of the given differential equation about the ordinary point Compare the series solutions with the solutions of the differential equa-tions obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solutions. 3. 4. 5. 6. y 2y 0 y y 0 y y 0 y y 0 x 0. (x2 2x 10)y xy 4y 0 (x2 25)y 2xy y 0 x 1. x 0. In Problems 7–18 ﬁnd two power series solutions of the given differential equation about the ordinary point x 0. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. (x2 1)y xy y 0 (x2 2)y 3xy y 0 (x2 1)y 6y 0 y (x 1)y y 0 (x 2)y xy y 0 (x 1)y y 0 y 2xy 2y 0 y x2y xy 0 y xy 2y 0 y 2xy y 0 y x2y 0 y xy 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.3 SOLUTIONS ABOUT SINGULAR POINTS ● 247 In Problems 19–22 use the power series method to solve the given initial-value problem. 19. 20. 21. 22. In Problems 23 and 24 use the procedure in Example 8 to ﬁnd two power series solutions of the given differential equation about the ordinary point x 0. 23. 24. Discussion Problems 25. Without actually solving the differential equation ﬁnd the minimum radius of convergence of power series solutions about the ordi-nary point About the ordinary point 26. How can the power series method be used to solve the nonhomogeneous equation about the ordi-nary point Of ? Carry out your ideas by solving both DEs. 27. Is x 0 an ordinary or a singular point of the differen-tial equation ? Defend your answer with sound mathematics. [Hint: Use the Maclaurin series of and then examine (sin x)>x. sin x xy (sin x)y 0 y 4xy 4y ex x 0? y xy 1 x 1. x 0. (cos x)y y 5y 0, y exy y 0 y (sin x)y 0 (x2 1)y 2xy 0, y(0) 0, y(0) 1 y 2xy 8y 0, y(0) 3, y(0) 0 (x 1)y (2 x)y y 0, y(0) 2, y(0) 1 (x 1)y xy y 0, y(0) 2,y(0) 6 28. Is an ordinary point of the differential equation Computer Lab Assignments 29. (a) Find two power series solutions for y xy y 0 and express the solutions y1(x) and y2(x) in terms of summation notation. (b) Use a CAS to graph the partial sums SN(x) for y1(x). Use N 2, 3, 5, 6, 8, 10. Repeat using the partial sums SN(x) for y2(x). (c) Compare the graphs obtained in part (b) with the curve obtained by using a numerical solver. Use the initial-conditions y1(0) 1, y 1(0) 0, and y2(0) 0, y 2(0) 1. (d) Reexamine the solution y1(x) in part (a). Express this series as an elementary function. Then use (5) of Section 4.2 to ﬁnd a second solution of the equa-tion. Verify that this second solution is the same as the power series solution y2(x). 30. (a) Find one more nonzero term for each of the solu-tions y1(x) and y2(x) in Example 8. (b) Find a series solution y(x) of the initial-value problem y (cos x)y 0, y(0) 1, y(0) 1. (c) Use a CAS to graph the partial sums SN(x) for the solution y(x) in part (b). Use N 2, 3, 4, 5, 6, 7. (d) Compare the graphs obtained in part (c) with the curve obtained using a numerical solver for the initial-value problem in part (b). y 5xy 1xy 0? x 0 SOLUTIONS ABOUT SINGULAR POINTS REVIEW MATERIAL ●Section 4.2 (especially (5) of that section) ●The deﬁnition of a singular point in Deﬁnition 6.2.1 INTRODUCTION The two differential equations y xy 0 and xy y 0 are similar only in that they are both examples of simple linear second-order DEs with variable coefﬁcients. That is all they have in common. Since x 0 is an ordinary point of y xy 0, we saw in Section 6.2 that there was no problem in ﬁnding two distinct power series solutions centered at that point. In contrast, because x 0 is a singular point of xy y 0, ﬁnding two inﬁnite series—notice that we did not say power series—solutions of the equation about that point becomes a more difﬁcult task. The solution method that is discussed in this section does not always yield two inﬁnite series solutions. When only one solution is found, we can use the formula given in (5) of Section 4.2 to ﬁnd a second solution. 6.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. A Definitio A singular point x0 of a linear differential equation (1) is further classiﬁed as either regular or irregular. The classiﬁcation again depends on the functions P and Q in the standard form (2) y P(x)y Q(x)y 0. a2(x)y a1(x)y a0(x)y 0 248 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS DEFINITION 6.3.1 Regular and Irregular Singular Points A singular point x x0 is said to be a regular singular point of the differential equation (1) if the functions p(x) (x x0) P(x) and q(x) (x x0)2Q(x) are both analytic at x0. A singular point that is not regular is said to be an irregular singular point of the equation. The second sentence in Deﬁnition 6.3.1 indicates that if one or both of the func-tions p(x) (x x0) P(x) and q(x) (x x0)2Q(x) fail to be analytic at x0, then x0 is an irregular singular point. Polynomial Coefficient As in Section 6.2, we are mainly interested in linear equations (1) where the coefﬁcients a2(x), a1(x), and a0(x) are polynomials with no common factors. We have already seen that if a2(x0) 0, then x x0 is a singular point of (1), since at least one of the rational functions P(x) a1(x) a2(x) and Q(x) a0(x) a2(x) in the standard form (2) fails to be analytic at that point. But since a2(x) is a polynomial and x0 is one of its zeros, it follows from the Factor Theorem of algebra that x x0 is a factor of a2(x). This means that after a1(x) a2(x) and a0(x) a2(x) are reduced to lowest terms, the factor x x0 must remain, to some positive integer power, in one or both denominators. Now suppose that x x0 is a singular point of (1) but both the functions defined by the products p(x) (x x0) P(x) and q(x) (x x0)2Q(x) are analytic at x0. We are led to the conclu-sion that multiplying P(x) by x x0 and Q(x) by (x x0)2 has the effect (through cancellation) that x x0 no longer appears in either denominator. We can now determine whether x0 is regular by a quick visual check of denominators: If x x0 appears at most to the first power in the denominator of (x) and at most to the second power in the denominator of Q(x), then x x0 is a regular singular point. Moreover, observe that if x x0 is a regular singular point and we multiply (2) by (x x0)2, then the original DE can be put into the form (3) where p and q are analytic at x x0. (x x0)2y (x x0)p(x)y q(x)y 0, EXAMPLE 1 Classification of Singula Points It should be clear that x 2 and x 2 are singular points of After dividing the equation by (x2 4)2 (x 2)2(x 2)2 and reducing the coefﬁcients to lowest terms, we ﬁnd that We now test P(x) and Q(x) at each singular point. P(x) 3 (x 2)(x 2)2 and Q(x) 5 (x 2)2(x 2)2. (x2 4)2y 3(x 2)y 5y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. For x 2 to be a regular singular point, the factor x 2 can appear at most to the ﬁrst power in the denominator of P(x) and at most to the second power in the denom-inator of Q(x). A check of the denominators of P(x) and Q(x) shows that both these conditions are satisﬁed, so x 2 is a regular singular point. Alternatively, we are led to the same conclusion by noting that both rational functions are analytic at x 2. Now since the factor x (2) x 2 appears to the second power in the denominator of P(x), we can conclude immediately that x 2 is an irregular singular point of the equation. This also follows from the fact that is not analytic at x 2. In Example 1, notice that since x 2 is a regular singular point, the original equation can be written as As another example, we can see that x 0 is an irregular singular point of x3y 2xy 8y 0 by inspection of the denominators of P(x) 2 x2 and Q(x) 8 x3. On the other hand, x 0 is a regular singular point of xy 2xy 8y 0, since x 0 and (x 0)2 do not even appear in the respective denominators of P(x) 2 and Q(x) 8 x. For a singular point x x0 any nonnegative power of x x0 less than one (namely, zero) and any nonnegative power less than two (namely, zero and one) in the denominators of P(x) and Q(x), re-spectively, imply that x0 is a regular singular point. A singular point can be a complex number. You should verify that x 3i and x 3i are two regular singular points of (x2 9)y 3xy (1 x)y 0. Note Any second-order Cauchy-Euler equation ax2y bxy cy 0, where a, b, and c are real constants, has a regular singular point at x 0. You should verify that two solutions of the Cauchy-Euler equation x2y 3xy 4y 0 on the interval (0, ) are y1 x2 and y2 x2 ln x. If we attempted to ﬁnd a power series solution about the regular singular point x 0 (namely, ), we would succeed in obtaining only the polynomial solution y1 x2. The fact that we would not obtain the second so-lution is not surprising because ln x (and consequently y2 x2 ln x) is not analytic at x 0—that is, y2 does not possess a Taylor series expansion centered at x 0. Method of Frobenius To solve a differential equation (1) about a regular sin-gular point, we employ the following theorem due to the eminent German mathe-matician Ferdinand Georg Frobenius (1849–1917). y n0 cnxn (x 2)2y (x 2) y y 0. p(x) analytic at x 2 q(x) analytic at x 2 3 –––––––– (x 2)2 5 –––––––– (x 2)2 p(x) (x 2)P(x) 3 (x 2)(x 2) p(x) (x 2)P(x) 3 (x 2)2 and q(x) (x 2)2Q(x) 5 (x 2)2 6.3 SOLUTIONS ABOUT SINGULAR POINTS ● 249 THEOREM 6.3.1 Frobenius’ Theorem If x x0 is a regular singular point of the differential equation (1), then there exists at least one solution of the form (4) where the number r is a constant to be determined. The series will converge at least on some interval 0 x x0 R. y (x x0)r n0 cn(x x0)n n0 cn(x x0)nr, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Notice the words at least in the ﬁrst sentence of Theorem 6.3.1. This means that in contrast to Theorem 6.2.1, Theorem 6.3.1 gives us no assurance that two series solutions of the type indicated in (4) can be found. The method of Frobenius, ﬁnding series solutions about a regular singular point x0, is similar to the power-series method in the preceding section in that we substitute into the given differential equation and determine the unknown coefﬁcients cn by a recurrence rela-tion. However, we have an additional task in this procedure: Before determining the co-efﬁcients, we must ﬁnd the unknown exponent r. If r is found to be a number that is not a nonnegative integer, then the corresponding solution is not a power series. As we did in the discussion of solutions about ordinary points, we shall always assume, for the sake of simplicity in solving differential equations, that the regular singular point is x 0. y n0 cn(x x0)nr y n0 cn(x x0)nr 250 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS EXAMPLE 2 Two Series Solutions Because x 0 is a regular singular point of the differential equation (5) we try to ﬁnd a solution of the form Now so y n0 (n r)cnxnr1 and y n0 (n r)(n r 1)cnxnr2, y n0 cnxnr. 3xy y y 0, x rr(3r 2)c0x1 k0 [(k r 1)(3k 3r 1)ck1 ck]x k 0, 1444442444443 123 k n1 k n x rr(3r 2)c0x1 n1 (n r)(3n 3r 2)cnxn1 n0 cnxn n0 (n r)(3n 3r 2)cnxnr1 n0 cnxnr 3xy y y 3 n0 (n r)(n r 1)cn xnr1 n0 (n r)cnxnr1 n0 cnxnr which implies that r(3r 2)c0 0 and Because nothing is gained by taking c0 0, we must then have (6) and (7) When substituted in (7), the two values of r that satisfy the quadratic equation (6), and r2 0, give two different recurrence relations: (8) (9) r2 0, ck1 ck (k 1)(3k 1), k 0, 1, 2, . . . . r1 2 3, ck1 ck (3k 5)(k 1), k 0, 1, 2, . . . r1 2 3 ck1 ck (k r 1)(3k 3r 1), k 0, 1, 2, . . . . r(3r 2) 0 (k r 1)(3k 3r 1)ck1 ck 0, k 0, 1, 2, . . . . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. From (8) we ﬁnd From (9) we ﬁnd 6.3 SOLUTIONS ABOUT SINGULAR POINTS ● 251 cn c0 n!5 8 11 (3n 2). c4 c3 14 4 c0 4!5 8 11 14 c3 c2 11 3 c0 3!5 8 11 c2 c1 8 2 c0 2!5 8 c1 c0 5 1 cn c0 n!1 4 7 (3n 2). c4 c3 4 10 c0 4!1 4 7 10 c3 c2 3 7 c0 3!1 4 7 c2 c1 2 4 c0 2!1 4 c1 c0 1 1 Here we encounter something that did not happen when we obtained solutions about an ordinary point; we have what looks to be two different sets of coefﬁ-cients, but each set contains the same multiple c0. If we omit this term, the series solutions are (10) (11) By the ratio test it can be demonstrated that both (10) and (11) converge for all val-ues of x—that is, Also, it should be apparent from the form of these solutions that neither series is a constant multiple of the other, and therefore y1(x) and y2(x) are linearly independent on the entire x-axis. Hence by the superposition prin-ciple, y C1y1(x) C2y2(x) is another solution of (5). On any interval that does not contain the origin, such as (0, ), this linear combination represents the general solu-tion of the differential equation. Indicial Equation Equation (6) is called the indicial equation of the problem, and the values and r2 0 are called the indicial roots, or exponents, of the singularity x 0. In general, after substituting into the given dif-ferential equation and simplifying, the indicial equation is a quadratic equation in r that results from equating the total coefficient of the lowest power of x to ze o. We solve for the two values of r and substitute these values into a recurrence relation such as (7). Theorem 6.3.1 guarantees that at least one solution of the assumed series form can be found. It is possible to obtain the indicial equation in advance of substituting into the differential equation. If x 0 is a regular singular point of (1), then by Deﬁnition 6.3.1 both functions p(x) xP(x) and q(x) x2Q(x), where P and Q are deﬁned by the standard form (2), are analytic at x 0; that is, the power series expansions y n0 cnxnr y n0 cnxnr r1 2 3 x . y2(x) x01 n1 1 n!1 4 7 (3n 2) xn. y1(x) x2/31 n1 1 n!5 8 11 (3n 2) xn (12) p(x) xP(x) a0 a1x a2x2 and q(x) x2Q(x) b0 b1x b2x2 are valid on intervals that have a positive radius of convergence. By multiplying (2) by x2, we get the form given in (3): (13) x2y x[xP(x)]y [x2Q(x)]y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. After substituting and the two series in (12) into (13) and carrying out the multiplication of series, we ﬁnd the general indicial equation to be (14) where a0 and b0 are as deﬁned in (12). See Problems 13 and 14 in Exercises 6.3. r(r 1) a0r b0 0, y n0 cnxnr 252 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS EXAMPLE 3 Two Series Solutions Solve 2xy (1 x)y y 0. SOLUTION Substituting gives y n0 cnxnr 2xy (1 x)y y 2 (n r)(n r 1)cnx nr1 (n r)cnx nr1 n0 n0 (n r)(2n 2r 1)cnx nr1 (n r 1)cnx nr n0 n0 xr [r(2r 1)c0x1 [(k r 1)(2k 2r 1)ck1 (k r 1)ck]xk], k0 (n r)cnx nr cnx nr n0 n0 xr [r(2r 1)c0x1 (n r)(2n 2r 1)cnx n1 (n r 1)cnx n] n1 n0 kn1 kn which implies that (15) and (16) k 0, 1, 2, . . . . From (15) we see that the indicial roots are and r2 0. For we can divide by in (16) to obtain (17) whereas for r2 0, (16) becomes (18) From (17) we ﬁnd From (18) we ﬁnd ck1 ck 2k 1, k 0, 1, 2, . . . . ck1 ck 2(k 1), k 0, 1, 2, . . . , k 3 2 r1 1 2 r1 1 2 (k r 1)(2k 2r 1)ck1 (k r 1)ck 0, r(2r 1) 0 cn (1)nc0 2nn! . c4 c3 2 4 c0 24 4! c3 c2 2 3 c0 23 3! c2 c1 2 2 c0 22 2! c1 c0 2 1 cn (1)nc0 1 3 5 7 (2n 1) . c4 c3 7 c0 1 3 5 7 c3 c2 5 c0 1 3 5 c2 c1 3 c0 1 3 c1 c0 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Thus for the indicial root we obtain the solution where we have again omitted c0. The series converges for x 0; as given, the series is not deﬁned for negative values of x because of the presence of x1/2. For r2 0 a second solution is On the interval (0, ) the general solution is y C1y1(x) C2y2(x). y2(x) 1 n1 (1)n 1 3 5 7 (2n 1)xn, x . y1(x) x1/21 n1 (1)n 2nn! xn n0 (1)n 2nn! xn1/2 , r1 1 2 6.3 SOLUTIONS ABOUT SINGULAR POINTS ● 253 EXAMPLE 4 Only One Series Solution Solve xy y 0. SOLUTION From xP(x) 0, x2Q(x) x and the fact that 0 and x are their own power series centered at 0 we conclude that a0 0 and b0 0, so from (14) the indicial equation is r(r 1) 0. You should verify that the two recurrence relations corresponding to the indicial roots r1 1 and r2 0 yield exactly the same set of coefﬁcients. In other words, in this case the method of Frobenius produces only a single series solution Three Cases For the sake of discussion let us again suppose that x 0 is a regular singular point of equation (1) and that the indicial roots r1 and r2 of the singularity are real. When using the method of Frobenius, we distinguish three cases corresponding to the nature of the indicial roots r1 and r2. In the ﬁrst two cases the symbol r1 denotes the largest of two distinct roots, that is, r1 r2. In the last case r1 r2. Case I: If r1 and r2 are distinct and the difference r1 r2 is not a positive inte-ger, then there exist two linearly independent solutions of equation (1) of the form This is the case illustrated in Examples 2 and 3. Next we assume that the difference of the roots is N, where N is a positive integer. In this case the second solution may contain a logarithm. Case II: If r1 and r2 are distinct and the difference r1 r2 is a positive integer, then there exist two linearly independent solutions of equation (1) of the form (19) (20) where C is a constant that could be zero. Finally, in the last case, the case when r1 r2, a second solution will always contain a logarithm. The situation is analogous to the solution of a Cauchy-Euler equation when the roots of the auxiliary equation are equal. y2(x) Cy1(x) ln x n0 bnxnr2, b0 0, y1(x) n0 cnxnr1, c0 0, y1(x) n0 cn xnr1, c0 0, y2(x) n0 bn xnr2, b0 0. y1(x) n0 (1)n n!(n 1)! xn1 x 1 2 x2 1 12 x3 1 144 x4 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Case III: If r1 and r2 are equal, then there always exist two linearly indepen-dent solutions of equation (1) of the form (21) (22) Finding a Second Solution When the difference r1 r2 is a positive integer (Case II), we may or may not be able to ﬁnd two solutions having the form This is something that we do not know in advance but is determined after we have found the indicial roots and have carefully examined the recurrence relation that deﬁnes the coefﬁcients cn. We just may be lucky enough to ﬁnd two solutions that involve only powers of x, that is, (equation (19)) and (equation (20) with C 0). See Problem 31 in Exercises 6.3. On the other hand, in Example 4 we see that the difference of the in-dicial roots is a positive integer (r1 r2 1) and the method of Frobenius failed to give a second series solution. In this situation equation (20), with C 0, indicates what the second solution looks like. Finally, when the difference r1 r2 is a zero (Case III), the method of Frobenius fails to give a second series solution; the second solution (22) always contains a logarithm and can be shown to be equivalent to (20) with C 1. One way to obtain the second solution with the logarithmic term is to use the fact that (23) is also a solution of y P(x)y Q(x)y 0 whenever y1(x) is a known solution. We illustrate how to use (23) in the next example. y2(x) y1(x) e P(x)dx y2 1(x) dx y2(x) n0 bnxnr2 y1(x) n0 cnxnr1 y n0 cnxnr. y2(x) y1(x) ln x n1 bnxnr1. y1(x) n0 cnxnr1, c0 0, 254 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS EXAMPLE 5 Example 4 Revisited Using a CAS Find the general solution of xy y 0. SOLUTION From the known solution given in Example 4, we can construct a second solution y2(x) using formula (23). Those with the time, energy, and patience can carry out the drudgery of squaring a series, long division, and integration of the quotient by hand. But all these operations can be done with relative ease with the help of a CAS. We give the results: y1(x) ln x y1(x) 1 x 7 12 x 19 144 x2 , y1(x) 1 x ln x 7 12 x 19 144 x2 y1(x) 1 x2 1 x 7 12 19 72 x dx y1(x) dx x2 x3 5 12 x4 7 72 x5 y2(x) y1(x) e∫0dx [y1(x)]2 dx y1(x) dx x 1 2 x2 1 12 x3 1 144 x4 2 y1(x) x 1 2 x2 1 12 x3 1 144 x4 , ; after long division ; after integrating ; after squaring Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. or On the interval (0, ) the general solution is y C1y1(x) C2y2(x). Note that the ﬁnal form of y2 in Example 5 matches (20) with C 1; the series in the brackets corresponds to the summation in (20) with r2 0. y2(x) y1(x) ln x 1 1 2 x 1 2 x2 . 6.3 SOLUTIONS ABOUT SINGULAR POINTS ● 255 REMARKS (i) The three different forms of a linear second-order differential equation in (1), (2), and (3) were used to discuss various theoretical concepts. But on a practical level, when it comes to actually solving a differential equation using the method of Frobenius, it is advisable to work with the form of the DE given in (1). (ii) When the difference of indicial roots r1 r2 is a positive integer (r1 r2), it sometimes pays to iterate the recurrence relation using the smaller root r2 ﬁrst. See Problems 31 and 32 in Exercises 6.3. (iii) Because an indicial root r is a solution of a quadratic equation, it could be complex. We shall not, however, investigate this case. (iv) If x 0 is an irregular singular point, then we might not be able to ﬁnd any solution of the DE of form y n0 cnxnr. EXERCISES 6.3 Answers to selected odd-numbered problems begin on page ANS-10. In Problems 1–10 determine the singular points of the given differential equation. Classify each singular point as regular or irregular. 1. x3y 4x2y 3y 0 2. x(x 3)2y y 0 3. (x2 9)2y (x 3)y 2y 0 4. 5. (x3 4x)y 2xy 6y 0 6. x2(x 5)2y 4xy (x2 25)y 0 7. (x2 x 6)y (x 3)y (x 2)y 0 8. x(x2 1)2y y 0 9. x3(x2 25)(x 2)2y 3x(x 2)y 7(x 5)y 0 10. (x3 2x2 3x)2y x(x 3)2y (x 1)y 0 In Problems 11 and 12 put the given differential equation into form (3) for each regular singular point of the equation. Identify the functions p(x) and q(x). 11. (x2 1)y 5(x 1)y (x2 x)y 0 12. xy (x 3)y 7x2y 0 y 1 x y 1 (x 1)3 y 0 In Problems 13 and 14, x 0 is a regular singular point of the given differential equation. Use the general form of the indi-cial equation in (14) to ﬁnd the indicial roots of the singu-larity. Without solving, discuss the number of series solutions you would expect to ﬁnd using the method of Frobenius. 13. 14. xy y 10y 0 In Problems 15–24, x 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about x 0. Form the general solution on (0, ). 15. 2xy y 2y 0 16. 2xy 5y xy 0 17. 18. 2x2y xy (x2 1)y 0 19. 3xy (2 x)y y 0 20. 21. 2xy (3 2x)y y 0 x2y (x 2 9)y 0 4xy 1 2y y 0 x2y (5 3 x x2)y 1 3 y 0 ; after multiplying out Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 256 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS 22. 23. 9x2y 9x2y 2y 0 24. 2x2y 3xy (2x 1)y 0 In Problems 25–30, x 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of Frobenius to obtain at least one series solution about x 0. Use (23) where necessary and a CAS, if instructed, to ﬁnd a second solution. Form the general solution on (0, ). 25. xy 2y xy 0 26. 27. xy xy y 0 28. 29. xy (1 x)y y 0 30. xy y y 0 In Problems 31 and 32, x 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the recur-rence relation found by the method of Frobenius ﬁrst with the larger root r1. How many solutions did you ﬁnd? Next use the recurrence relation with the smaller root r2. How many solutions did you ﬁnd? 31. xy (x 6)y 3y 0 32. x(x 1)y 3y 2y 0 33. (a) The differential equation x4y y 0 has an irregular singular point at x 0. Show that the sub-stitution t 1 x yields the DE which now has a regular singular point at t 0. (b) Use the method of this section to ﬁnd two series solutions of the second equation in part (a) about the regular singular point t 0. (c) Express each series solution of the original equation in terms of elementary functions. Mathematical Model 34. Buckling of a Tapered Column In Example 4 of Section 5.2 we saw that when a constant vertical compressive force or load P was applied to a thin column of uniform cross section, the deﬂection y(x) was a solution of the boundary-value problem (24) The assumption here is that the column is hinged at both ends. The column will buckle or deﬂect only when the compressive force is a critical load Pn. EI d 2y dx2 Py 0, y(0) 0, y(L) 0. d 2y dt2 2 t dy dt y 0, y 3 x y 2y 0 x2y xy (x2 1 4)y 0 x2y xy (x2 4 9)y 0 (a) In this problem let us assume that the column is of length L, is hinged at both ends, has circular cross sections, and is tapered as shown in Figure 6.3.1(a). If the column, a truncated cone, has a linear taper y cx as shown in cross section in Figure 6.3.1(b), the moment of inertia of a cross section with respect to an axis perpendicular to the xy-plane is where r y and y cx. Hence we can write I(x) I0(x b)4, where Sub-stituting I(x) into the differential equation in (24), we see that the deﬂection in this case is determined from the BVP where Pb4 EI0. Use the results of Problem 33 to ﬁnd the critical loads Pn for the tapered column. Use an appropriate identity to express the buckling modes yn(x) as a single function. (b) Use a CAS to plot the graph of the ﬁrst buckling mode y1(x) corresponding to the Euler load P1 when b 11 and a 1. x4 d 2y dx2 y 0, y(a) 0, y(b) 0, I0 I(b) 1 4(cb)4. I 1 4r4, x = a y P x = b y = cx b − a = L L (a) (b) x FIGURE 6.3.1 Tapered column in Problem 34 Discussion Problems 35. Discuss how you would deﬁne a regular singular point for the linear third-order differential equation 36. Each of the differential equations has an irregular singular point at x 0. Determine whether the method of Frobenius yields a series solu-tion of each differential equation about x 0. Discuss and explain your ﬁndings. 37. We have seen that x 0 is a regular singular point of any Cauchy-Euler equation ax2y bxy cy 0. Are the indicial equation (14) for a Cauchy-Euler equa-tion and its auxiliary equation related? Discuss. x3y y 0 and x2y (3x 1)y y 0 a3(x)y a2(x)y a1(x)y a0(x)y 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.4 SPECIAL FUNCTIONS ● 257 SPECIAL FUNCTIONS REVIEW MATERIAL ●Sections 6.2 and 6.3 INTRODUCTION In the Remarks at the end of Section 2.3 we mentioned the branch of mathemat-ics called special functions. Perhaps a better title for this ﬁeld of applied mathematics might be named functions because many of the functions studied bear proper names: Bessel functions, Legendre functions, Airy functions, Chebyshev polynomials, Hermite polynomials, Jacobi polyno-mials, Laguerre polynomials, Gauss’ hypergeometric function, Mathieu functions, and so on. Historically, special functions were often the by-product of necessity: Someone needed a solution of a very specialized differential equation that arose from an attempt to solve a physical problem. In effect, a special function was determined or deﬁned by the differential equation and many properties of the function could be discerned from the series form of the solution. In this section we use the methods of Sections 6.2 and 6.3 to ﬁnd solutions of two differential equations (1) (2) that arise in advanced studies of applied mathematics, physics, and engineering. They are called, respectively, Bessel’s equation of order , named after the German mathematician and astronomer Friedrich Wilhelm Bessel (1784–1846), and Legendre’s equation of order n, named after the French mathematician Adrien-Marie Legendre (1752–1833). When we solve (1) we shall assume that whereas in (2) we shall consider only the case when n in a nonnegative integer. 0, (1 x2)y 2xy n(n 1)y 0 x 2y xy (x2 2)y 0 6.4 Solution of Bessel’s Equation Because is a regular singular point of Bessel’s equation, we know that there exists at least one solution of the form Substituting the last expression into (1) gives y n0 cnxnr. x 0 (3) c0(r2 2)xr xr n1 cn[(n r)2 2]xn xr n0 cnxn2. c0(r2 r r 2)xr xr n1 cn[(n r)(n r 1) (n r) 2]xn xr n0 cnxn2 x2y xy (x2 2)y n0 cn(n r)(n r 1)xnr n0 cn(n r)xnr n0 cnxnr2 2 n0 cnxnr From (3) we see that the indicial equation is r2 2 0, so the indicial roots are r1 and r2 . When r1 , (3) becomes xn cnn(n 2n)xn xn cnx n2 n1 n0 xn[(1 2n)c1x [(k 2)(k 2 2n)ck2 ck]x k2] 0. k0 xn[(1 2n)c1x cnn(n 2n)x n cnx n2] n2 n0 k n 2 k n Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Therefore by the usual argument we can write (1 2)c1 0 and or (4) The choice c1 0 in (4) implies that so for k 0, 2, 4, . . . we ﬁnd, after letting k 2 2n, n 1, 2, 3, . . . , that (5) Thus (6) It is standard practice to choose c0 to be a speciﬁc value, namely, where (1 ) is the gamma function. See Appendix I. Since this latter function possesses the convenient property (1 ) (), we can reduce the indicated product in the denominator of (6) to one term. For example, Hence we can write (6) as for n 0, 1, 2, . . . . Bessel Functions of the First Kind Using the coefﬁcients c2n just obtained c2n (1)n 22n n!(1 )(2 ) (n )(1 ) (1)n 22n n!(1 n) (1 2) (2 )(2 ) (2 )(1 )(1 ). (1 1) (1 )(1 ) c0 1 2(1 ), c2n (1)nc0 22nn!(1 )(2 ) (n ), n 1, 2, 3, . . . . c6 c4 22 3(3 ) c0 26 1 2 3(1 )(2 )(3 ) c4 c2 22 2(2 ) c0 24 1 2(1 )(2 ) c2 c0 22 1 (1 ) c2n c2n2 22n(n ). c3 c5 c7 0, ck2 ck (k 2)(k 2 2), k 0, 1, 2, . . . . (k 2)(k 2 2)ck2 ck 0 258 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS When we replace x by \|x\|, the series given in (7) and (8) converge for 0 \|x\| . and r , a series solution of (1) is This solution is usually denoted by J(x): (7) If 0, the series converges at least on the interval [0, ). Also, for the second exponent r2 we obtain, in exactly the same manner, (8) The functions J(x) and J(x) are called Bessel functions of the first kin of order and , respectively. Depending on the value of , (8) may contain negative powers of x and hence converges on (0, ). J (x) n0 (1)n n!(1 n) x 2 2n . J(x) n0 (1)n n!(1 n) x 2 2n . y n0c2n x2n. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Now some care must be taken in writing the general solution of (1). When 0, it is apparent that (7) and (8) are the same. If 0 and r1 r2 () 2 is not a positive integer, it follows from Case I of Section 6.3 that J (x) and J (x) are linearly independent solutions of (1) on (0, ), and so the general solution on the interval is y c1J (x) c2J (x). But we also know from Case II of Section 6.3 that when r1 r2 2 is a positive integer, a second series solution of (1) may exist. In this second case we distinguish two possibilities. When m positive integer, J m(x) deﬁned by (8) and Jm(x) are not linearly independent solutions. It can be shown that J m is a constant multiple of Jm (see Property (i) on page 262). In addition, r1 r2 2 can be a positive integer when is half an odd positive integer. It can be shown in this latter event that J (x) and J (x) are linearly independent. In other words, the general solution of (1) on (0, ) is (9) The graphs of y J0(x) and y J1(x) are given in Figure 6.4.1. y c1J (x) c2J (x), integer. 6.4 SPECIAL FUNCTIONS ● 259 EXAMPLE 1 Bessel’s Equation of Order By identifying we can see from (9) that the general solution of the equation on (0, ) is Bessel Functions of the Second Kind If integer, the function deﬁned by the linear combination (10) and the function J(x) are linearly independent solutions of (1). Thus another form of the general solution of (1) is y c1J(x) c2Y(x), provided that integer. As m an integer, (10) has the indeterminate form 0 0. However, it can be shown by L’Hôpital’s Rule that exists. Moreover, the function and Jm(x) are linearly independent solutions of x2y xy (x2 m2)y 0. Hence for any value of the general solution of (1) on (0, ) can be written as (11) Y(x) is called the Bessel function of the second kindof order . Figure 6.4.2 shows the graphs of Y0(x) and Y1(x). y c1J (x) c2Y (x). Ym(x) lim :m Y (x) lim :m Y (x) : m, Y (x) cos J (x) J (x) sin y c1J1/2(x) c2J1/2(x). x2y xy (x2 1 4)y 0 2 1 4 and 1 2, 1 2 2 4 6 8 _0.4 0.2 0.4 0.6 0.8 1 _0.2 x y J1 J0 FIGURE 6.4.1 Bessel functions of the ﬁrst kind for n 0, 1, 2, 3, 4 2 4 6 8 1 _3 _2.5 _2 _1.5 _1 _0.5 0.5 x y Y0 Y1 FIGURE 6.4.2 Bessel functions of the second kind for n 0, 1, 2, 3, 4 EXAMPLE 2 Bessel’s Equation of Order 3 By identifying 2 9 and 3, we see from (11) that the general solution of the equation x2y xy (x2 9)y 0 on (0, ) is y c1J3(x) c2Y3(x). DES Solvable in Terms of Bessel Functions Sometimes it is possible to transform a differential equation into equation (1) by means of a change of variable. We can then express the solution of the original equation in terms of Bessel func-tions. For example, if we let t x, 0, in (12) then by the Chain Rule, dy dx dy dt dt dx dy dt and d 2y dx2 d dt dy dx dt dx 2 d 2y dt2 . x2y xy (a2x2 2)y 0, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Accordingly, (12) becomes 260 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS t 2 2 d 2y dt 2 t dy dt (t2 2)y 0 or t2 d 2y dt2 t dy dt (t2 2)y 0. The last equation is Bessel’s equation of order with solution y c1J(t) c2Y(t). By resubstituting t x in the last expression, we ﬁnd that the general solution of (12) is (13) Equation (12), called the parametric Bessel equation of order , and its general solution (13) are very important in the study of certain boundary-value problems involving partial differential equations that are expressed in cylindrical coordinates. Modified Bessel Functions Another equation that bears a resemblance to (1) is the modified Bessel equation of orde , (14) This DE can be solved in the manner just illustrated for (12). This time if we let where then (14) becomes Because solutions of the last DE are J(t) and Y(t), complex-valued solutions of (14) are J(ix) and Y(ix). A real-valued solution, called the modified Bessel func-tion of the first kin of order , is deﬁned in terms of J(ix): (15) See Problem 21 in Exercises 6.4. Analogous to (10), the modified Bessel function of the second kind of order integer is deﬁned to be (16) and for integer n, Because I and K are linearly independent on the interval (0, ) for any value of v, the general solution of (14) on that interval is (17) The graphs of and are given in Figure 6.4.3 and the graphs of and are given in Figure 6.4.4. Unlike the Bessel functions of the ﬁrst and second kinds, the modiﬁed Bessel functions of the ﬁrst and second kind are not oscillatory. Figures 6.4.3 and 6.4.4 also illustrate the fact that the modiﬁed Bessel functions and have no real zeros in the interval Also notice that the modiﬁed Bessel functions of the second kind like the Bessel functions of the second kind become unbounded as A change of variable in (14) gives us the parametric form of the modiﬁed Bessel equation of order The general solution of the last equation on the interval is y c1I(ax) c2K(ax). (0, ) x2y xy (a2x2 n2)y 0. : x : 0. Yn(x) Kn(x) (0, ). Kn(x), n 0, 1, 2, . . . In(x) y K2(x) y K0(x), y K1(x), y I2(x) y I0(x), y I1(x), y c1I(x) c2K(x). Kn(x) lim :n K(x). K(x) 2 I(x) I(x) sin , I(x) i J (ix). t2 d 2y dt2 t dy dt (t2 2)y 0. i2 1, t ix, x2y xy (x2 2)y 0. y c1J (x) c2Y (x). 1 2 3 1 1.5 2 2.5 3 0.5 x y I0 I1 I2 FIGURE 6.4.3 Modiﬁed Bessel functions of the ﬁrst kind for n 0, 1, 2 1 2 3 1 1.5 2 2.5 3 0.5 x y K2 K1 K0 FIGURE 6.4.4 Modiﬁed Bessel functions of the second kind for n 0, 1, 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Yet another equation, important because many DEs ﬁt into its form by appro-priate choices of the parameters, is (18) Although we shall not supply the details, the general solution of (18), (19) can be found by means of a change in both the independent and the dependent variables: If p is not an integer, then Yp in (19) can be replaced by Jp. z bxc, y(x) z b a/c w(z). y xac1Jp(bxc) c2Yp(bxc), y 1 2a x y b2c2x2c2 a2 p2c2 x2 y 0, p 0. 6.4 SPECIAL FUNCTIONS ● 261 EXAMPLE 3 Using (18) Find the general solution of xy 3y 9y 0 on (0, ). SOLUTION By writing the given DE as we can make the following identiﬁcations with (18): The ﬁrst and third equations imply that a 1 and With these values the second and fourth equations are satisﬁed by taking b 6 and p 2. From (19) we ﬁnd that the general solution of the given DE on the interval (0, ) is y x1[c1J2(6x1/2) c2Y2(6x1/2)]. c 1 2. 1 2a 3, b2c2 9, 2c 2 1, and a2 p2c2 0. y 3 x y 9 x y 0, EXAMPLE 4 The Aging Spring Revisited Recall that in Section 5.1 we saw that one mathematical model for the free undamped motion of a mass on an aging spring is given by mx ketx 0, 0. We are now in a position to ﬁnd the general solution of the equation. It is left as a problem to show that the change of variables transforms the differential equation of the aging spring into The last equation is recognized as (1) with 0 and where the symbols x and s play the roles of y and x, respectively. The general solution of the new equation is x c1J0(s) c2Y0(s). If we resubstitute s, then the general solution of mx ketx 0 is seen to be See Problems 33 and 39 in Exercises 6.4. The other model that was discussed in Section 5.1 of a spring whose character-istics change with time was mx ktx 0. By dividing through by m, we see that the equation is Airy’s equation y 2xy 0. See Example 5 in Section 6.2. The general solution of Airy’s differential equation can also be written in terms of Bessel functions. See Problems 34, 35, and 40 in Exercises 6.4. x k mtx 0 x(t) c1J0 2 B k m e t/2 c2Y0 2 B k m e t/2. s2 d 2x ds2 s dx ds s2x 0. s 2 B k m et/2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Properties We list below a few of the more useful properties of Bessel functions of order m, m 0, 1, 2, . . .: (i) (ii) (iii) (iv) Note that Property (ii) indicates that Jm(x) is an even function if m is an even integer and an odd function if m is an odd integer. The graphs of Y0(x) and Y1(x) in Figure 6.4.2 illustrate Property (iv), namely, Ym(x) is unbounded at the origin. This last fact is not obvious from (10). The solutions of the Bessel equation of order 0 can be obtained by using the solutions y1(x) in (21) and y2(x) in (22) of Section 6.3. It can be shown that (21) of Section 6.3 is y1(x) J0(x), whereas (22) of that section is The Bessel function of the second kind of order 0, Y0(x), is then deﬁned to be the linear combination for x 0. That is, where 0.57721566 . . . is Euler’s constant. Because of the presence of the logarithmic term, it is apparent that Y0(x) is discontinuous at x 0. Numerical Values The ﬁrst ﬁve nonnegative zeros of J0(x), J1(x), Y0(x), and Y1(x) are given in Table 6.4.1. Some additional function values of these four functions are given in Table 6.4.2. Y0(x) 2 J0(x) ln x 2 2 k1 (1)k (k!)21 1 2 1 k x 2 2k , Y0(x) 2 ( ln 2)y1(x) 2 y2(x) y2(x) J0(x)ln x k1 (1)k (k!)2 1 1 2 1 k x 2 2k . lim x:0 Ym(x) . Jm(0) 0, 1, m 0 m 0, Jm(x) (1)mJm(x), J m(x) (1)mJm(x), 262 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS TABLE 6.4.2 Numerical Values of J0, J1, Y0, and Y1 x J0(x) J1(x) Y0(x) Y1(x) 0 1.0000 0.0000 — — 1 0.7652 0.4401 0.0883 0.7812 2 0.2239 0.5767 0.5104 0.1070 3 0.2601 0.3391 0.3769 0.3247 4 0.3971 0.0660 0.0169 0.3979 5 0.1776 0.3276 0.3085 0.1479 6 0.1506 0.2767 0.2882 0.1750 7 0.3001 0.0047 0.0259 0.3027 8 0.1717 0.2346 0.2235 0.1581 9 0.0903 0.2453 0.2499 0.1043 10 0.2459 0.0435 0.0557 0.2490 11 0.1712 0.1768 0.1688 0.1637 12 0.0477 0.2234 0.2252 0.0571 13 0.2069 0.0703 0.0782 0.2101 14 0.1711 0.1334 0.1272 0.1666 15 0.0142 0.2051 0.2055 0.0211 TABLE 6.4.1 Zeros of J0, J1, Y0, and Y1 J0(x) J1(x) Y0(x) Y1(x) 2.4048 0.0000 0.8936 2.1971 5.5201 3.8317 3.9577 5.4297 8.6537 7.0156 7.0861 8.5960 11.7915 10.1735 10.2223 11.7492 14.9309 13.3237 13.3611 14.8974 Differential Recurrence Relation Recurrence formulas that relate Bessel functions of different orders are important in theory and in applications. In the next example we derive a differential recurrence relation. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.4 SPECIAL FUNCTIONS ● 263 EXAMPLE 5 Derivation Using the Series Definitio Derive the formula SOLUTION It follows from (7) that The result in Example 5 can be written in an alternative form. Dividing by x gives This last expression is recognized as a linear ﬁrst-order differential equation in J(x). Multiplying both sides of the equality by the integrating factor x then yields (20) It can be shown in a similar manner that (21) See Problem 27 in Exercises 6.4. The differential recurrence relations (20) and (21) are also valid for the Bessel function of the second kind Y(x). Observe that when 0, it follows from (20) that (22) An application of these results is given in Problem 39 of Exercises 6.4. Bessel Functions of Half-Integral Order When the order is half an odd in-teger, that is, Bessel functions of the ﬁrst and second kinds can be expressed in terms of the elementary functions and powers of x. Let’s consider the case when From (7) J1/2(x) n0 (1)n n!(1 1 2 n) x 2 2n1/2 . 1 2. cosx, sinx, 1 2, 3 2, 5 2, . . . , J 0(x) J1(x) and Y 0(x) Y1(x). d dx [xJ(x)] xJ1(x). d dx [xJ(x)] xJ1(x). J n(x) x Jn(x) Jn1(x). xJ (x) J(x) xJ1(x) xJv(x) ( ) 2n n0 k n 1 (1)n(2n ) ––––––––––––––– n! (1 n) x – 2 L J(x) x ( ) 2n1 n1 (1)n ––––––––––––––––––––– (n 1)! (1 n) x – 2 L ( ) 2n n0 (1)n ––––––––––––––– n! (1 n) x – 2 L 2 ( ) 2n n0 (1)nn ––––––––––––––– n! (1 n) x – 2 L J(x) x J(x) xJ1(x). ( ) 2k1 k0 (1)k ––––––––––––––– k! (2 k) x – 2 L xJ (x) J(x) xJ1(x). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In view of the property (1 ) () and the fact that the values of for n 0, n 1, n 2, and n 3 are, respectively, In general, Hence From (2) of Section 6.1 you should recognize that the inﬁnite series in the last line is the Maclaurin series for and so we have shown that (23) We leave it as an exercise to show that (24) See Figure 6.4.5 and Problems 31, 32, and 38 in Exercises 6.4. If n is an integer, then is half an odd integer. Because and we see from (10) that For we have, in turn, and In view of (23) and (24) these results are the same as (25) and (26) Spherical Bessel Functions Bessel functions of half-integral order are used to deﬁne two more important functions: and (27) The function is called the spherical Bessel function of the first kin and is the spherical Bessel function of the second kind. For example, for the expressions in (27) become and . y0(x) B p 2xY1>2(x) B p 2x B 2 px cosx cosx x j0(x) B p 2xJ1>2(x) B p 2x B 2 px sinx sinx x n 0 yn(x) jn(x) yn(x) B p 2xYn1>2(x). jn(x) B p 2xJn1>2(x) Y1>2(x) B 2 px sin x. Y1>2(x) B 2 px cosx Y1>2(x) J1>2(x). Y1>2(x) J1>2(x) n 0 and n 1 J(n1>2)(x). (1)n1 Yn1>2(x) sin(n 1 2)p cosnp (1)n, cos(n 1 2) p 0 n n 1 2 J1/2(x) B 2 x cos x. J1/2(x) B 2 x sin x. sin x, J1/2(x) n0 (1)n n!(2n 1)! 22n1n! 1 x 2 2n1/2 B 2 x n0 (1)n (2n 1)! x2n1. (1 1 2 n) (2n 1)! 22n1n! 1. ( 9 2) (1 7 2) 7 2( 7 2) 7 5 26 2! 1 7 6 5! 26 6 2! 1 7! 273! 1. ( 7 2) (1 5 2) 5 2( 5 2) 5 3 23 1 5 4 3 2 1 234 2 1 5! 252! 1 ( 5 2) (1 3 2) 3 2( 3 2) 3 22 1 (3 2) (1 1 2) 1 2( 1 2) 1 2 1 (1 1 2 n) (1 2) 1 264 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS 2 4 6 8 10 12 14 0 0.5 1 −0.5 x y J -1/2 J1/2 FIGURE 6.4.5 Bessel functions of order (red) and order 1 2 1 2 (blue) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. It is apparent from (27) and Figure 6.4.2 for the spherical Bessel of the second kind becomes unbounded as . Spherical Bessel functions arise in the solution of a special partial differential equation expressed in spherical coordinates. See Problem 54 in Exercises 6.4 and Problem 13 in Exercises 13.3. x : 0 yn(x) n 0 6.4 SPECIAL FUNCTIONS ● 265 Solution of Legendre’ s Equation Since x 0 is an ordinary point of Legendre’s equation (2), we substitute the series shift summation in-dices, and combine series to get y k0 ckxk, j2 [( j 2)( j 1)cj2 (n j)(n j 1)cj]x j 0 (1 x2)y 2xy n(n 1)y [n(n 1)c0 2c2] [(n 1)(n 2)c1 6c3]x which implies that or (28) If we let j take on the values 2, 3, 4, . . . , the recurrence relation (28) yields and so on. Thus for at least we obtain two linearly independent power series solutions: (29) Notice that if n is an even integer, the ﬁrst series terminates, whereas y2(x) is an inﬁnite series. For example, if n 4, then Similarly, when n is an odd integer, the series for y2(x) terminates with xn; that is, when n is a nonnegative integer , we obtain an nth-degr ee polynomial solution of Legendre’s equation. y1(x) c01 4 5 2! x2 2 4 5 7 4! x4 c01 10x2 35 3 x4. (n 5)(n 3)(n 1)(n 2)(n 4)(n 6) 7! x7 . y2(x) c1x (n 1)(n 2) 3! x3 (n 3)(n 1)(n 2)(n 4) 5! x5 (n 4)(n 2)n(n 1)(n 3)(n 5) 6! x6 y1(x) c01 n(n 1) 2! x2 (n 2)n(n 1)(n 3) 4! x4 x 1 c7 (n 5)(n 6) 7 6 c5 (n 5)(n 3)(n 1)(n 2)(n 4)(n 6) 7! c1 c6 (n 4)(n 5) 6 5 c4 (n 4)(n 2)n(n 1)(n 3)(n 5) 6! c0 c5 (n 3)(n 4) 5 4 c3 (n 3)(n 1)(n 2)(n 4) 5! c1 c4 (n 2)(n 3) 4 3 c2 (n 2)n(n 1)(n 3) 4! c0 cj2 (n j)(n j 1) ( j 2)( j 1) cj, j 2, 3, 4, . . . . c3 (n 1)(n 2) 3! c1 c2 n(n 1) 2! c0 ( j 2)( j 1)cj2 (n j)(n j 1)cj 0 (n 1)(n 2)c1 6c3 0 n(n 1)c0 2c2 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 266 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS Because we know that a constant multiple of a solution of Legendre’s equation is also a solution, it is traditional to choose speciﬁc values for c0 or c1, depending on whether n is an even or odd positive integer, respectively. For n 0 we choose c0 1, and for n 2, 4, 6, . . . whereas for n 1 we choose c1 1, and for n 3, 5, 7, . . . For example, when n 4, we have Legendre Polynomials These speciﬁc nth-degree polynomial solutions are called Legendre polynomials and are denoted by Pn(x). From the series for y1(x) and y2(x) and from the above choices of c0 and c1 we ﬁnd that the ﬁrst several Legendre polynomials are (30) Remember, P0(x), P1(x), P2(x), P3(x), . . . are, in turn, particular solutions of the differential equations (31) The graphs, on the interval [1, 1], of the six Legendre polynomials in (30) are given in Figure 6.4.6. Properties You are encouraged to verify the following properties using the Legendre polynomials in (30). (i) (ii) (iii) (iv) (v) Property (i) indicates, as is apparent in Figure 6.4.6, that Pn(x) is an even or odd function according to whether n is even or odd. Recurrence Relation Recurrence relations that relate Legendre polynomials of different degrees are also important in some aspects of their applications. We state, without proof, the three-term recurrence relation (32) which is valid for k 1, 2, 3, . . . . In (30) we listed the ﬁrst six Legendre polynomials. If, say, we wish to ﬁnd P6(x), we can use (32) with k 5. This relation expresses P6(x) in terms of the known P4(x) and P5(x). See Problem 45 in Exercises 6.4. (k 1)Pk1(x) (2k 1)xPk(x) kPk1(x) 0, P n(0) 0, n even Pn(0) 0, n odd Pn(1) (1)n Pn(1) 1 Pn(x) (1)nPn(x) n 0: n 1: n 2: n 3: (1 x2)y 2xy 0, (1 x2)y 2xy 2y 0, (1 x2)y 2xy 6y 0, (1 x2)y 2xy 12y 0, P0(x) 1, P1(x) x, P2(x) 1 2 (3x2 1), P3(x) 1 2 (5x3 3x), P4(x) 1 8 (35x4 30x2 3), P5(x) 1 8 (63x5 70x3 15x). y1(x) (1)4/2 1 3 2 4 1 10x2 35 3 x4 1 8 (35x4 30x2 3). c1 (1)(n1)/2 1 3 n 2 4 (n 1). c0 (1)n /2 1 3 (n 1) 2 4 n , x y 1 -1 -1 -0.5 0.5 1 -0.5 0.5 P 1 P 0 P 2 FIGURE 6.4.6 Legendre polynomials for n 0, 1, 2, 3, 4, 5 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Another formula, although not a recurrence relation, can generate the Legendre polynomials by differentiation. Rodrigues’ formula for these polynomials is (33) See Problem 48 in Exercises 6.4. Pn(x) 1 2nn! d n dxn (x2 1)n, n 0, 1, 2, . . . . 6.4 SPECIAL FUNCTIONS ● 267 REMARKS Although we have assumed that the parameter n in Legendre’s differential equa-tion (1 x2)y 2xy n(n 1)y 0, represented a nonnegative integer, in a more general setting n can represent any real number. Any solution of Legendre’s equation is called a Legendre function. If n is not a nonnegative integer, then both Legendre functions y1(x) and y2(x) given in (29) are inﬁnite series convergent on the open interval (1, 1) and divergent (unbounded) at x 1. If n is a nonnegative integer, then as we have just seen one of the Legendre functions in (29) is a polynomial and the other is an inﬁnite series convergent for 1 x 1. You should be aware of the fact that Legendre’s equation possesses solutions that are bounded on the closed interval [1, 1] only in the case when n 0, 1, 2, . . . . More to the point, the only Legendre functions that are bounded on the closed interval [1, 1] are the Legendre poly-nomials Pn(x) or constant multiples of these polynomials. See Problem 47 in Exercises 6.4 and Problem 24 in Chapter 6 in Review. EXERCISES 6.4 Answers to selected odd-numbered problems begin on page ANS-11. Bessel’s Equation In Problems 1–6 use (1) to ﬁnd the general solution of the given differential equation on (0, ). 1. 2. x2y xy (x2 1)y 0 3. 4x2y 4xy (4x2 25)y 0 4. 16x2y 16xy (16x2 1)y 0 5. xy y xy 0 6. In Problems 7–10 use (12) to ﬁnd the general solution of the given differential equation on (0, ). 7. x2y xy (9x2 4)y 0 8. x2y xy 36x2 1 4y 0 d dx [xy] x 4 xy 0 x2y xy x2 1 9y 0 9. 10. x2y xy (2x2 64)y 0 In Problems 11 and 12 use the indicated change of variable to ﬁnd the general solution of the given differential equation on (0, ). 11. x2y 2xy 2x2y 0; y x1/2v(x) 12. In Problems 13–20 use (18) to ﬁnd the general solution of the given differential equation on (0, ). 13. xy 2y 4y 0 14. xy 3y xy 0 15. xy y xy 0 16. xy 5y xy 0 17. x2y (x2 2)y 0 18. 4x2y (16x2 1)y 0 x2y (2x2 2 1 4)y 0; y 1x v(x) x2y xy 25x2 4 9y 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 268 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS 19. xy 3y x3y 0 20. 9x2y 9xy (x6 36)y 0 21. Use the series in (7) to verify that I(x) i J(ix) is a real function. 22. Assume that b in equation (18) can be pure imaginary, that is, b i, 0, i2 1. Use this assumption to express the general solution of the given differential equation in terms the modiﬁed Bessel functions In and Kn. (a) y x2y 0 (b) xy y 7x3y 0 In Problems 23–26 ﬁrst use (18) to express the general solu-tion of the given differential equation in terms of Bessel func-tions. Then use (23) and (24) to express the general solution in terms of elementary functions. 23. y y 0 24. x2y 4xy (x2 2)y 0 25. 16x2y 32xy (x4 12)y 0 26. 4x2y 4xy (16x2 3)y 0 27. (a) Proceed as in Example 5 to show that xJ (x) J(x) xJ1(x). [Hint: Write 2n 2(n ) .] (b) Use the result in part (a) to derive (21). 28. Use the formula obtained in Example 5 along with part (a) of Problem 27 to derive the recurrence relation 2J(x) xJ1(x) xJ1(x). In Problems 29 and 30 use (20) or (21) to obtain the given result. 29. 30. J 0(x) J1(x) J1(x) 31. Proceed as on page 264 to derive the elementary form of J1/2(x) given in (24). 32. Use the recurrence relation in Problem 28 along with (23) and (24) to express J3/2(x), J3/2(x), J5/2(x) and J5/2(x) in terms of sin x, cos x, and powers of x. 33. Use the change of variables to show that the differential equation of the aging spring mx ketx 0, 0, becomes s2 d 2x ds2 s dx ds s2x 0. s 2 B k m et /2 x 0 rJ0(r) dr xJ1(x) 34. Show that is a solution of Airy’s differential equation y 2xy 0, x 0, whenever w is a solution of Bessel’s equation of order that is, t 0. [Hint: After differentiating, substituting, and simplifying, then let ] 35. (a) Use the result of Problem 34 to express the general solution of Airy’s differential equation for x 0 in terms of Bessel functions. (b) Verify the results in part (a) using (18). 36. Use the Table 6.4.1 to ﬁnd the ﬁrst three positive eigen-values and corresponding eigenfunctions of the boundary-value problem [Hint: By identifying 2, the DE is the parametric Bessel equation of order zero.] 37. (a) Use (18) to show that the general solution of the differential equation xy y 0 on the interval (0, ) is (b) Verify by direct substitution that is a particular solution of the DE in the case 1. Computer Lab Assignments 38. Use a CAS to graph J3/2(x), J3/2(x), J5/2(x), and J5/2(x). 39. (a) Use the general solution given in Example 4 to solve the IVP Also use and along with Table 6.4.1 or a CAS to evaluate coefﬁcients. (b) Use a CAS to graph the solution obtained in part (a) for 0 t . 40. (a) Use the general solution obtained in Problem 35 to solve the IVP Use a CAS to evaluate coefﬁcients. (b) Use a CAS to graph the solution obtained in part (a) for 0 t 200. 41. Column Bending Under Its Own Weight A uniform thin column of length L, positioned vertically with one 4x tx 0, x(0.1) 1, x(0.1) 1 2. Y 0(x) Y1(x) J 0(x) J1(x) 4x e0.1tx 0, x(0) 1, x(0) 1 2. y 1xJ1(21x) y c11xJ1(21x) c21xY1(21x). y(x), y(x) bounded as x : 0, y(2) 0. xy y xy 0, t 2 3x3/2. t2w tw (t2 1 9)w 0, 1 3, y x1/2w(2 3x3/2) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.4 SPECIAL FUNCTIONS ● 269 end embedded in the ground, will deﬂect, or bend away, from the vertical under the inﬂuence of its own weight when its length or height exceeds a certain critical value. It can be shown that the angular deﬂection (x) of the column from the vertical at a point P(x) is a solution of the boundary-value problem: where E is Young’s modulus, I is the cross-sectional moment of inertia, is the constant linear density, and x is the distance along the column measured from its base. See Figure 6.4.7. The column will bend only for those values of L for which the boundary-value problem has a nontrivial solution. (a) Restate the boundary-value problem by making the change of variables t L x. Then use the results of a problem earlier in this exercise set to express the general solution of the differential equation in terms of Bessel functions. (b) Use the general solution found in part (a) to ﬁnd a solution of the BVP and an equation which deﬁnes the critical length L, that is, the smallest value of L for which the column will start to bend. (c) With the aid of a CAS, ﬁnd the critical length L of a solid steel rod of radius r 0.05 in., g 0.28 A lb/in., E 2.6 107 lb/in.2, A r2, and I 1 4r4. EI d 2 dx2 g(L x) 0, (0) 0, (L) 0, Use the information in Problem 37 to ﬁnd a solu-tion of if it is known that is not zero at x 0. (b) Use Table 6.4.1 to ﬁnd the Euler load P1 for the column. (c) Use a CAS to graph the ﬁrst buckling mode y1(x) corresponding to the Euler load P1. For simplicity assume that c1 1 and L 1. 43. Pendulum of Varying Length For the simple pendu-lum described on page 220 of Section 5.3, suppose that the rod holding the mass m at one end is replaced by a ﬂexible wire or string and that the wire is strung over a pulley at the point of support O in Figure 5.3.3. In this manner, while it is in motion in a vertical plane, the mass m can be raised or lowered. In other words, the length l(t) of the pendulum varies with time. Under the same assumptions leading to equation (6) in Sec-tion 5.3, it can be shown that the differential equation for the displacement angle is now (a) If l increases at constant rate v and if l(0) l0, show that a linearization of the foregoing DE is (34) (b) Make the change of variables x (l0 vt) v and show that (34) becomes (c) Use part (b) and (18) to express the general solution of equation (34) in terms of Bessel functions. (d) Use the general solution obtained in part (c) to solve the initial-value problem consisting of equation (34) and the initial conditions (0) 0, (0) 0. [Hints: To simplify calculations, use a further change of variable Also, recall that (20) holds for both J1(u) and Y1(u). Finally, the identity will be helpful.] 2 u J1(u)Y2(u) J2(u)Y1(u) u 2 v 1g(l0 vt) 2 B g v x1/ 2. d 2 dx 2 2 x d dx g vx 0. (l0 vt) 2v g 0. l 2l g sin 0. 1xY1(21x) M x L d 2y dx2 Py 0, y(0) 0, y(L) 0 x = 0 x θ P(x) ground FIGURE 6.4.7 Beam in Problem 41 42. Buckling of a Thin Vertical Column In Example 4 of Section 5.2 we saw that when a constant vertical compressive force, or load, P was applied to a thin column of uniform cross section and hinged at both ends, the deﬂection y(x) is a solution of the BVP: (a) If the bending stiffness factor EI is proportional to x, then EI(x) kx, where k is a constant of proportionality. If EI(L) kL M is the maximum stiffness factor, then k M L and so EI(x) Mx L. EI d 2y dx2 Py 0, y(0) 0, y(L) 0. See Mathematical Methods in Physical Sciences, Mary Boas, John Wiley & Sons, Inc., 1966. Also see the article by Borelli, Coleman, and Hobson in Mathematics Magazine, vol. 58, no. 2, March 1985. (problem continues on page 270) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 270 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS (e) Use a CAS to graph the solution (t) of the IVP in part (d) when l0 1 ft, 0 radian, and Experiment with the graph using dif-ferent time intervals such as [0, 10], [0, 30], and so on. (f) What do the graphs indicate about the displacement angle (t) as the length l of the wire increases with time? Legendre’s Equation 44. (a) Use the explicit solutions y1(x) and y2(x) of Legendre’s equation given in (29) and the appropri-ate choice of c0 and c1 to ﬁnd the Legendre polyno-mials P6(x) and P7(x). (b) Write the differential equations for which P6(x) and P7(x) are particular solutions. 45. Use the recurrence relation (32) and P0(x) 1, P1(x) x, to generate the next six Legendre polynomials. 46. Show that the differential equation can be transformed into Legendre’s equation by means of the substitution x cos . 47. Find the ﬁrst three positive values of for which the problem has nontrivial solutions. Computer Lab Assignments 48. For purposes of this problem ignore the list of Legendre polynomials given on page 266 and the graphs given in Figure 6.4.3. Use Rodrigues’ formula (33) to generate the Legendre polynomials P1(x), P2(x), . . . , P7(x). Use a CAS to carry out the differentiations and simpliﬁcations. 49. Use a CAS to graph P1(x), P2(x), . . . , P7(x) on the interval [1, 1]. 50. Use a root-ﬁnding application to ﬁnd the zeros of P1(x), P2(x), . . . , P7(x). If the Legendre polynomials are built-in functions of your CAS, ﬁnd zeros of Legendre polynomials of higher degree. Form a con-jecture about the location of the zeros of any Legendre polynomial Pn(x), and then investigate to see whether it is true. Miscellaneous Differential Equations 51. The differential equation y 2xy 2ay 0 y(0) 0, y(x), y(x) bounded on [1,1] (1 x2)y 2xy y 0, sin d 2y d 2 cos dy d n(n 1)(sin )y 0 v 1 60 ft/s. 1 10 is known as Hermite’s equation of order after the French mathematician Charles Hermite (1822–1901). Show that the general solution of the equation is where are power series solutions centered at the ordinary point 0. 52. (a) When is a nonnegative integer, Hermite’s differential equation always possesses a polynomial solution of degree n. Use given in Problem 51, to ﬁnd polynomial solutions for and Then use to ﬁnd polynomial solutions for and (b) A Hermite polynomial is deﬁned to be the nth degree polynomial solution of Hermite’s equa-tion multiplied by an appropriate constant so that the coefﬁcient of in is Use the polyno-mial solutions in part (a) to show that the ﬁrst six Hermite polynomials are 53. The differential equation , where is a parameter, is known as Chebyshev’s equation after the Russian mathematician Pafnuty Chebyshev (1821–1894). When is a nonnega-tive integer, Chebyshev’s differential equation always possesses a polynomial solution of degree n. Find a ﬁfth degree polynomial solution of this differential equation. 54. If n is an integer, use the substitution to show that the general solution of the differential equation on the interval is where are the spherical Bessel func-tions of the ﬁrst and second kind deﬁned in (27). jn(ax) and yn(ax) R(x) c1 jn(ax) c2 yn(ax), (0, ) x2R 2xR [a2x2 n(n 1)]R 0 R(x) (ax)1>2Z(x) a n a (1 x2)y xy a2y 0 H5(x) 32x5 160x3 120x. H4(x) 16x4 48x2 12 H3(x) 8x3 12x H2(x) 4x2 2 H1(x) 2x H0(x) 1 2n. Hn(x) xn Hn(x) n 5. n 1, n 3, y2(x) n 4. n 0, n 2, y1(x), a n y2(x) x k1 (1)k 2k(a1)(a3) . . . (a 2k1) (2k 1)! x2k1 y1(x) 1 k1 (1)k 2ka(a 2) . . . (a 2k 2) (2k)! x 2k y(x) c0y1(x) c1y2(x), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 6 IN REVIEW ● 271 CHAPTER 6 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-11. In Problems 1 and 2 answer true or false without referring back to the text. 1. The general solution of x2y xy (x2 1)y 0 is y c1J1(x) c2J1(x). 2. Because x 0 is an irregular singular point of x3y xy y 0, the DE possesses no solution that is analytic at x 0. 3. Both power series solutions of y ln(x 1)y y 0 centered at the ordinary point x 0 are guaranteed to converge for all x in which one of the following intervals? (a) (, ) (b) (1, ) (c) (d) [1, 1] 4. x 0 is an ordinary point of a certain linear differential equation. After the assumed solution is substituted into the DE, the following algebraic system is obtained by equating the coefﬁcients of x0, x1, x2, and x3 to zero: Bearing in mind that c0 and c1 are arbitrary, write down the ﬁrst ﬁve terms of two power series solutions of the differential equation. 5. Suppose the power series is known to converge at 2 and diverge at 13. Discuss whether the series converges at 7, 0, 7, 10, and 11. Possible answers are does, does not, might. 6. Use the Maclaurin series for sin x and cos x along with long division to ﬁnd the ﬁrst three nonzero terms of a power series in x for the function In Problems 7 and 8 construct a linear second-order differen-tial equation that has the given properties. 7. A regular singular point at x 1 and an irregular singular point at x 0 8. Regular singular points at x 1 and at x 3 In Problems 9–14 use an appropriate inﬁnite series method about x 0 to ﬁnd two solutions of the given differential equation. f (x) sin x cos x. k0 ck(x 4)k 20c5 8c4 c3 2 3c2 0. 12c4 6c3 c2 1 3c1 0 6c3 4c2 c1 0 2c2 2c1 c0 0 y n0 cnxn [1 2, 1 2] 9. 2xy y y 0 10. y xy y 0 11. (x 1)y 3y 0 12. y x2y xy 0 13. xy (x 2)y 2y 0 14. (cos x)y y 0 In Problems 15 and 16 solve the given initial-value problem. 15. y xy 2y 0, y(0) 3, y(0) 2 16. (x 2)y 3y 0, y(0) 0, y(0) 1 17. Without actually solving the differential equation (1 2 sin x)y xy 0, ﬁnd a lower bound for the radius of convergence of power series solutions about the ordinary point x 0. 18. Even though x 0 is an ordinary point of the differen-tial equation, explain why it is not a good idea to try to ﬁnd a solution of the IVP of the form Using power series, ﬁnd a better way to solve the problem. In Problems 19 and 20 investigate whether x 0 is an ordi-nary point, singular point, or irregular singular point of the given differential equation. [Hint: Recall the Maclaurin series for cos x and ex.] 19. xy (1 cos x)y x2y 0 20. (ex 1 x)y xy 0 21. Note that x 0 is an ordinary point of the differential equation y x2y 2xy 5 2x 10x3. Use the assumption to ﬁnd the general solution y yc yp that consists of three power series centered at x 0. 22. The ﬁrst-order differential equation dy dx x2 y2 cannot be solved in terms of elementary functions. However, a solution can be expressed in terms of Bessel functions. (a) Show that the substitution leads to the equation u x2u 0. (b) Use (18) in Section 6.4 to ﬁnd the general solution of u x2u 0. (c) Use (20) and (21) in Section 6.4 in the forms and J (x) xJ(x) J1(x) J (x) xJ(x) J1(x) y 1 u du dx y n0 cnxn y n0 cnxn. y xy y 0, y(1) 6, y(1) 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 272 ● CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS as an aid to show that a one-parameter family of solutions of dy dx x2 y2 is given by 23. (a) Use (10) of Section 6.4 and Problem 32 of Exer-cises 6.4 to show that (b) Use (15) of Section 6.4 to show that (c) Use (16) of Section 6.4 and part (b) to show that 24. (a) From (30) and (31) of Section 6.4 we know that when n 0, Legendre’s differential equation (1 x2)y 2xy 0 has the polynomial solu-tion y P0(x) 1. Use (5) of Section 4.2 to show K1/2(x) B 2x ex. I1/2(x) B 2 x sinh x and I1/2(x) B 2 x cosh x. Y3/2(x) B 2 x cos x x sin x y x J3/4( 1 2x2) cJ3/4( 1 2x2) cJ1/4( 1 2x2) J1/4( 1 2x2) . that a second Legendre function satisfying the DE for 1 x 1 is (b) We also know from (30) and (31) of Section 6.4 that when n 1, Legendre’s differential equation (1 x2)y 2xy 2y 0 possesses the polyno-mial solution y P1(x) x. Use (5) of Section 4.2 to show that a second Legendre function satisfying the DE for 1 x 1 is (c) Use a graphing utility to graph the logarithmic Legendre functions given in parts (a) and (b). 25. (a) Use binomial series to formally show that (b) Use the result obtained in part (a) to show that Pn(1) 1 and Pn(1) (1)n. See Properties (ii) and (iii) on page 266. (1 2xt t2)1/2 n0 Pn(x)tn. y x 2 ln 1 x 1 x 1. y 1 2 ln 1 x 1 x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 273 7.1 Deﬁnition of the Laplace Transform 7.2 Inverse Transforms and Transforms of Derivatives 7.2.1 Inverse Transforms 7.2.2 Transforms of Derivatives 7.3 Operational Properties I 7.3.1 Translation on the s-Axis 7.3.2 Translation on the t-Axis 7.4 Operational Properties II 7.4.1 Derivatives of a Transform 7.4.2 Transforms of Integrals 7.4.3 Transform of a Periodic Function 7.5 The Dirac Delta Function 7.6 Systems of Linear Differential Equations Chapter 7 in Review In the linear mathematical models for a physical system such as a spring/mass system or a series electrical circuit, the right-hand member, or input, of the differential equations is a driving function and represents either an external force f(t) or an impressed voltage E(t). In Section 5.1 we considered problems in which the functions f and E were continuous. However, discontinuous driving functions are not uncommon. For example, the impressed voltage on a circuit could be piecewise continuous and periodic, such as the “sawtooth” function shown on the left. Solving the differential equation of the circuit in this case is difﬁcult using the techniques of Chapter 4. The Laplace transform studied in this chapter is an invaluable tool that simpliﬁes the solution of problems such as these. m d 2x dt2 b dx dt kx f(t) or Ld 2q dt2 Rdq dt 1 C q E(t) The Laplace Transform 7 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 274 ● CHAPTER 7 THE LAPLACE TRANSFORM Integral Transform If f(x, y) is a function of two variables, then a deﬁnite in-tegral of f with respect to one of the variables leads to a function of the other variable. For example, by holding y constant, we see that . Similarly, a deﬁ-nite integral such as transforms a function f of the variable t into a function F of the variable s. We are particularly interested in an integral transform, where the interval of integration is the unbounded interval [0, ). If f(t) is deﬁned for t 0, then the improper integral is deﬁned as a limit: . (1) If the limit in (1) exists, then we say that the integral exists or is convergent; if the limit does not exist, the integral does not exist and is divergent. The limit in (1) will, in general, exist for only certain values of the variable s. A Definitio The function K(s, t) in (1) is called the kernel of the transform. 0 K(s, t) f(t) dt lim b : b 0 K(s, t) f(t) dt 0 K(s, t) f(t) dt b a K(s, t) f(t) dt 2 1 2xy2 dx 3y2 DEFINITION OF THE LAPLACE TRANSFORM REVIEW MATERIAL ●Improper integrals with inﬁnite limits of integration ●Integration by parts and partial fraction decomposition INTRODUCTION In elementary calculus you learned that differentiation and integration are transforms; this means, roughly speaking, that these operations transform a function into another function. For example, the function f(x) x2 is transformed, in turn, into a linear function and a family of cubic polynomial functions by the operations of differentiation and integration: and Moreover, these two transforms possess the linearity property that the transform of a linear combination of functions is a linear combination of the transforms. For a and b constants and provided that each derivative and integral exists. In this section we will examine a special type of integral transform called the Laplace transform. In addition to possessing the linearity property the Laplace transform has many other interesting properties that make it very useful in solving linear initial-value problems. [ f (x) g(x)] dx f (x) dx g(x) dx d dx [ f (x) g(x)] f (x) g(x) x2 dx 1 3 x3 c. d dx x2 2x 7.1 We will assume throughout that s is a real variable. The choice K(s, t) e st as the kernel gives us an especially important integral transform. DEFINITION 7.1.1 Laplace Transform Let f be a function deﬁned for t 0. Then the integral (2) is said to be the Laplace transform of f, provided that the integral converges. The Laplace transform is named in honor of the French mathematician and astronomer Pierre-Simon Marquis de Laplace (1749–1827). { f (t)} 0 e st f (t) dt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.1 DEFINITION OF THE LAPLACE TRANSFORM ● 275 When the deﬁning integral (2) converges, the result is a function of s. In general discussion we shall use a lowercase letter to denote the function being transformed and the corresponding capital letter to denote its Laplace transform—for example, . As the next four examples show, the domain of the function F(s) depends on the function f(t). {f (t) } F(s), {g(t)} G(s), {y(t)} Y(s) EXAMPLE 1 Applying Definition 7.1. Evaluate . SOLUTION From (2), provided that s 0. In other words, when s 0, the exponent sb is negative, and as . The integral diverges for s 0. The use of the limit sign becomes somewhat tedious, so we shall adopt the notation 0 as a shorthand for writing limb: ( ) b 0. For example, . At the upper limit, it is understood that we mean as for s 0. t : e st : 0 {1} 0 e st (1) dt e st s 0 1 s, s 0 b : e sb : 0 lim b : e st s 0 b lim b : e sb 1 s 1 s {1} 0 e st(1) dt lim b : b 0 e st dt {1} EXAMPLE 2 Applying Definition 7.1. Evaluate . SOLUTION From Deﬁnition 7.1.1 we have . Integrating by parts and using s 0, along with the result from Example 1, we obtain . {t} te st s 0 1 s 0 e st dt 1 s {1} 1 s 1 s 1 s2 lim t : te st 0, {t} 0 e st t dt {t} EXAMPLE 3 Applying Definition 7.1. Evaluate (a) (b) SOLUTION In each case we use Deﬁnition 7.1.1. (a) The last result is valid for s 3 because in order to have limt : e (s3)t 0 we must require that s 3 0 or s 3. 1 s 3. e (s3)t s 3 0 {e 3t} 0 e 3t e st dt 0 e (s3)t dt {e5t} {e 3t} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 276 ● CHAPTER 7 THE LAPLACE TRANSFORM (b) In contrast to part (a), this result is valid for s 5 because limt : e (s 5)t 0 demands s 5 0 or s 5. 1 s 5. e (s 5)t s 5 0 {e5t} 0 e5t e st dt 0 e (s 5)t dt EXAMPLE 4 Applying Definition 7.1. Evaluate . SOLUTION From Deﬁnition 7.1.1 and two applications of integration by parts we obtain At this point we have an equation with on both sides of the equality. Solving for that quantity yields the result . Is a Linear Transform For a linear combination of functions we can write whenever both integrals converge for s c. Hence it follows that . (3) Because of the property given in (3), is said to be a linear transform. { f (t) g(t)} { f (t) } {g(t)} F(s) G(s) 0 e st [ f (t) g(t)] dt 0 e st f (t) dt 0 e st g(t) dt {sin 2t} 2 s2 4, s 0 {sin 2t} lim e st cos 2t 0, s 0 t: Laplace transform of sin 2t e st sin 2t –––––––––––– s 2 – s 2 – s {sin 2t} e st sin 2t dt {sin 2t}. e st cos 2t dt 0 0 0 e st cos 2t –––––––––––– s 2 – s 2 – s 2 –– s2 4 –– s2 [ e st sin 2t dt] 0 0 e st cos 2t dt, s 0 0 {sin 2t} EXAMPLE 5 Linearity of the Laplace Transform In this example we use the results of the preceding examples to illustrate the linear-ity of the Laplace transform. (a) From Examples 1 and 2 we have for s 0, . (b) From Examples 3 and 4 we have for s 5, {4e5t 10 sin 2t} 4 {e5t} 10 {sin2t} 4 s 5 20 s2 4. {1 5t} {1} 5 {t} 1 s 5 s2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.1 DEFINITION OF THE LAPLACE TRANSFORM ● 277 t f(t) b t1 t3 t2 a FIGURE 7.1.1 Piecewise continuous function (c) From Examples 1, 2, and 3 we have for s 0, We state the generalization of some of the preceding examples by means of the next theorem. From this point on we shall also refrain from stating any restrictions on s; it is understood that s is sufﬁciently restricted to guarantee the convergence of the appropriate Laplace transform. 20 s 3 7 s2 9 s. {20e 3t 7t 9} 20 {e 3t} 7 {t} 9 {1} THEOREM 7.1.1 Transforms of Some Basic Functions (a) (b) (c) (d) (e) (f) (g) {cosh kt} s s2 k2 {sinh kt} k s2 k2 {cos kt} s s2 k2 {sin kt} k s2 k2 {eat} 1 s a {tn} n! sn1, n 1, 2, 3, . . . {1} 1 s This result in (b) of Theorem 7.1.1 can be formally justiﬁed for n a positive integer using intergration by parts to ﬁrst show that . Then for n 1, 2, and 3, we have, respectively, If we carry on in this manner, you should be convinced that Sufficient Conditions for Existence of {f(t)} The integral that deﬁnes the Laplace transform does not have to converge. For example, neither nor exists. Sufﬁcient conditions guaranteeing the existence of are that f be piecewise continuous on [0, ) and that f be of exponential order for t T. Recall that a function f is piecewise continuous on [0, ) if, in any interval 0 a t b, there are at most a ﬁnite number of points tk, k 1, 2, . . . , n (tk 1 tk) at which f has ﬁnite discontinuities and is continuous on each open interval (tk 1, tk). See Figure 7.1.1. The concept of exponential order is deﬁned in the following manner. DEFINITION 7.1.2 Exponential Order A function f is said to be of exponential order if there exist constants c, M 0, and T 0 such that f (t) Mect for all t T. {f (t) } {et2} {1>t} {t n} n . . . 3 2 1 sn1 n! sn1. {t3} 3 s {t2} 3 s 2 1 s3 3 2 1 s4 {t2} 2 s {t} 2 s 1 s2 2 1 s3 {t} 1 s {1} 1 s 1 s 1 s2 {t n} n s {t n 1} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 278 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.1.2 f is of exponential order f ( t ) t T Me ct ( c > 0) f(t) e t 2 t f(t) c e ct FIGURE 7.1.4 is not of exponential order et2 t e −t 2 cos t e t (a) (b) (c) t e t t 2 e t f(t) f(t) f(t) t FIGURE 7.1.3 Three functions of exponential order A positive integral power of t is always of exponential order, since, for c 0, is equivalent to showing that is ﬁnite for n 1, 2, 3, . . . . The result follows from n applications of L’Hôpital’s rule. A function such as is not of exponential order since, as shown in Figure 7.1.4, grows faster than any positive linear power of e for . This can also be seen from as . t : et2 ect et2 ct et(t c) : t > c > 0 et2 f(t) et2 limt : tn>ect tn Mect or tn ect M for t T If f is an increasing function, then the condition f(t) Mect, t T, simply states that the graph of f on the interval (T, ) does not grow faster than the graph of the exponential function Mect, where c is a positive constant. See Figure 7.1.2. The functions f (t) t, f (t) e t, and f (t) 2 cos t are all of exponential order because for c 1, M 1, T 0 we have, respectively, for t 0 . A comparison of the graphs on the interval [0, ) is given in Figure 7.1.3. t et, e t et, and 2 cos t 2et THEOREM 7.1.2 Sufficient Conditions fo Existence If f is piecewise continuous on [0, ) and of exponential order, then exists for s c. {f(t)} PROOF By the additive interval property of deﬁnite integrals we can write . The integral I1 exists because it can be written as a sum of integrals over intervals on which e st f(t) is continuous. Now since f is of exponential order, there exist constants c, M 0, T 0 so that f(t) Mect for t T. We can then write for s c. Since converges, the integral converges by the comparison test for improper integrals. This, in turn, implies that I2 exists T e st f (t) dt T Me (s c)t dt I2 T e st f (t) dt M T e stect dt M T e (s c)t dt M e (s c)T s c { f(t) } T 0 e st f(t) dt T e st f(t) dt I1 I2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.1 DEFINITION OF THE LAPLACE TRANSFORM ● 279 for s c. The existence of I1 and I2 implies that exists for s c. {f (t) } 0 e st f (t) dt t y 3 2 FIGURE 7.1.5 Piecewise continuous function in Example 6 EXAMPLE 6 Transform of a Piecewise Continuous Function Evaluate {f(t)} where SOLUTION The function f, shown in Figure 7.1.5, is piecewise continuous and of exponential order for t 0. Since f is deﬁned in two pieces, {f(t)} is expressed as the sum of two integrals: We conclude this section with an additional bit of theory related to the types of functions of s that we will, generally, be working with. The next theorem indicates that not every arbitrary function of s is a Laplace transform of a piecewise continuous func-tion of exponential order. 2e 3s s , s 0. 0 2e st s 3 {f (t) } 0 e st f (t) dt 3 0 e st(0) dt 3 e st(2) dt f (t) 0, 0 t 3 2, t 3. THEOREM 7.1.3 Behavior of F(s) as If f is piecewise continuous on [0, ) and of exponential order and F(s) {f(t)}, then lim s:F(s) 0. s : PROOF Since f is of exponential order, there exist constants g, M1 0, and T 0 so that f(t) M1egt for t T. Also, since f is piecewise continuous for 0 t T, it is necessarily bounded on the interval; that is, f(t) M2 M2e0t. If M denotes the maximum of the set {M1, M2} and c denotes the maximum of {0, g}, then for s c. As , we have , and so F(s) { f (t)} : 0. F(s) : 0 s : F(s) 0 e st f (t) dt M 0 e stect dt M 0 e (s c)t dt M s c REMARKS (i) Throughout this chapter we shall be concerned primarily with functions that are both piecewise continuous and of exponential order. We note, however, that these two conditions are sufﬁcient but not necessary for the existence of a Laplace transform. The function f (t) t 1/2 is not piecewise continuous on the interval [0, ), but its Laplace transform exists. The function is not of exponential order, but it can be shown that its Laplace transform exists. See Problems 43 and 54 in Exercises 7.1. (ii) As a consequence of Theorem 7.1.3 we can say that functions of s such as F1(s) 1 and F2(s) s(s 1) are not the Laplace transforms of piecewise continuous functions of exponential order, since F1(s) 0 and F2(s) 0 as . But you should not conclude from this that F1(s) and F2(s) are not Laplace transforms. There are other kinds of functions. s : : / : / f(t) 2tet2 cos et2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 280 ● CHAPTER 7 THE LAPLACE TRANSFORM EXERCISES 7.1 Answers to selected odd-numbered problems begin on page ANS-11. In Problems 1–18 use Deﬁnition 7.1.1 to ﬁnd {f(t)}. 1. 2. 3. 4. 5. 6. 7. f (t) 0, cos t, 0 t >2 t >2 f (t) sin t, 0, 0 t t f (t) 2t 1, 0, 0 t 1 t 1 f (t) t, 1, 0 t 1 t 1 f (t) 4, 0, 0 t 2 t 2 f (t) 1, 1, 0 t 1 t 1 9. FIGURE 7.1.7 Graph for Problem 8 FIGURE 7.1.8 Graph for Problem 9 FIGURE 7.1.9 Graph for Problem 10 t f(t) (2, 2) 1 1 FIGURE 7.1.6 Graph for Problem 7 t f(t) (2, 2) 1 1 t f(t) 1 1 f(t) a c b t 8. 10. 23. f(t) t2 6t 3 24. f(t) 4t2 16t 9 25. f(t) (t 1)3 26. f(t) (2t 1)3 27. f(t) 1 e4t 28. f(t) t2 e 9t 5 29. f(t) (1 e2t)2 30. f(t) (et e t)2 31. f(t) 4t2 5 sin 3t 32. f(t) cos 5t sin 2t 33. f(t) sinh kt 34. f(t) cosh kt 35. f(t) et sinh t 36. f(t) e t cosh t In Problems 37–40 ﬁnd {f(t)} by ﬁrst using a trigono-metric identity. 37. f(t) sin 2t cos 2t 38. f(t) cos2t 39. f(t) sin(4t 5) 40. 41. We have encountered the gamma function in our study of Bessel functions in Section 6.4 (page 258). One deﬁnition of this function is given by the improper integral Use this deﬁnition to show that 42. Use Problem 41 and a change of variables to obtain the generalization of the result in Theorem 7.1.1(b). In Problems 43–46 use Problems 41 and 42 and the fact that to find the Laplace transform of the given function. 43. f(t) t 1/2 44. f(t) t1/2 45. f(t) t3/2 46. f(t) 2t1/2 8t5/2 Discussion Problems 47. Make up a function F(t) that is of exponential order but where f(t) F(t) is not of exponential order. Make up a function f that is not of exponential order but whose Laplace transform exists. 48. Suppose that for s c1 and that for s c2. When does 49. Figure 7.1.4 suggests, but does not prove, that the func-tion is not of exponential order. How does the observation that for and t sufﬁciently large, show that for any c? 50. Use part (c) of Theorem 7.1.1 to show that {e(aib)t} , where a and b are real s a ib (s a)2 b2 et 2 Mect M 0 t2 ln M ct, f (t) et 2 {f1(t) f2(t)} F1(s) F2(s)? {f2(t)} F2(s) {f1(t)} F1(s) (1 2) 1p {t} ( 1) s1 , 1, (a 1) a(a). () 0 t a 1e t dt, a 0. (a) f (t) 10 cost 6 11. f(t) et7 12. f(t) e 2t 5 13. f(t) te4t 14. f(t) t2e 2t 15. f(t) e t sin t 16. f(t) et cos t 17. f(t) t cos t 18. f(t) t sin t In Problems 19–36 use Theorem 7.1.1 to ﬁnd {f (t)}. 19. f(t) 2t4 20. f(t) t5 21. f(t) 4t 10 22. f(t) 7t 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES ● 281 and i2 1. Show how Euler’s formula (page 133) can then be used to deduce the results . 51. Under what conditions is a linear function f(x) mx b, m 0, a linear transform? 52. Explain why the function is not piecewise continuous on [0, ). 53. Show that the function does not possess a Laplace transform. [Hint: Write as two im-proper integrals: Show that I1 diverges.] {1>t 2} 1 0 e st t 2 dt 1 e st t 2 dt I1 I2. {1>t2} f(t) 1>t2 f(t) t, 0 t 2 4, 2 t 5 1>(t 5), t 5 {eat sin bt} b (s a)2 b2 {eat cos bt} s a (s a)2 b2 54. Show that the Laplace transform exists. [Hint: Start with integration by parts.] 55. If and is a constant, show that . This result is known as the change of scale theor em. 56. Use the given Laplace transform and the result in Problem 55 to find the indicated Laplace transform. Assume that a and k are positive constants. (a) (b) (c) (d) {sin t sinh t} 2s s4 4; {sin kt sinh kt} {1 cos t} 1 s(s2 1); {1 cos kt} {sin t} 1 s2 1; {sin kt} {et} 1 s 1; {eat} {f(at)} 1 a F s a a 0 {f(t)} F(s) {2tet2 coset2} Transform Inverse Transform e 3t 1 1 s 3 {e 3t} 1 s 3 t 1 1 s2 {t} 1 s2 1 1 1 s {1} 1 s 7.2.1 INVERSE TRANSFORMS The Inverse Problem If F(s) represents the Laplace transform of a function f(t), that is, , we then say f(t) is the inverse Laplace transform of F(s) and write . For example, from Examples 1, 2, and 3 of Section 7.1 we have, respectively, f(t) 1{F(s)} {f(t)} F(s) INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES REVIEW MATERIAL ●Partial fraction decomposition ●See the Student Resource Manual INTRODUCTION In this section we take a few small steps into an investigation of how the Laplace transform can be used to solve certain types of equations for an unknown function. We begin the discussion with the concept of the inverse Laplace transform or, more precisely, the inverse of a Laplace transform F(s). After some important preliminary background material on the Laplace transform of derivatives f(t), f (t), . . . , we then illustrate how both the Laplace transform and the in-verse Laplace transform come into play in solving some simple ordinary differential equations. 7.2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 282 ● CHAPTER 7 THE LAPLACE TRANSFORM We shall see shortly that in the application of the Laplace transform to equa-tions we are not able to determine an unknown function f(t) directly; rather, we are able to solve for the Laplace transform F(s) of f(t); but from that knowledge we ascertain f by computing . The idea is simply this: Suppose is a Laplace transform; ﬁnd a function f(t) such that We shall show how to solve this problem in Example 2. For future reference the analogue of Theorem 7.1.1 for the inverse transform is presented as our next theorem. {f(t)} F(s). F(s) 2s 6 s2 4 f (t) 1{F(s)} In evaluating inverse transforms, it often happens that a function of s under con-sideration does not match exactly the form of a Laplace transform F(s) given in a table. It may be necessary to “ﬁx up” the function of s by multiplying and dividing by an appropriate constant. EXAMPLE 1 Applying Theorem 7.2.1 Evaluate (a) (b) . SOLUTION (a) To match the form given in part (b) of Theorem 7.2.1, we identify n 1 5 or n 4 and then multiply and divide by 4!: . (b) To match the form given in part (d) of Theorem 7.2.1, we identify k2 7, so . We ﬁx up the expression by multiplying and dividing by : . 1 is a Linear T ransform The inverse Laplace transform is also a linear transform; that is, for constants a and b , (1) where F and G are the transforms of some functions f and g. Like (3) of Section 7.1, (1) extends to any ﬁnite linear combination of Laplace transforms. 1{F(s) G(s)} 1{F(s)} 1{G(s)} 1 1 s2 7 1 17 1 17 s2 7 1 17 sin17t 17 k 17 1 1 s5 1 4! 1 4! s5 1 24 t4 1 1 s2 7 1 1 s5 THEOREM 7.2.1 Some Inverse Transforms (a) (b) (c) (d) (e) (f) (g) cosh kt 1 s s2 k2 sinh kt 1 k s2 k2 cos kt 1 s s2 k2 sin kt 1 k s2 k2 eat 1 1 s a tn 1 n! sn1, n 1, 2, 3, . . . 1 1 1 s Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES ● 283 (2) termwise division parts (e) and (d) of Theorem 7.2.1 with k 2 linearity and fixing up constants 2s 6 ––––––––– s2 4 6 – 2 1{ } 1{ 2 cos 2t 3 sin 2t. } 2 1{ 1{ } 2s ––––––– s2 4 6 ––––––– s2 4 } 2 ––––––– s2 4 s ––––––– s2 4 Partial Fractions Partial fractions play an important role in ﬁnding inverse Laplace transforms. The decomposition of a rational expression into component frac-tions can be done quickly by means of a single command on most computer algebra systems. Indeed, some CASs have packages that implement Laplace transform and inverse Laplace transform commands. But for those of you without access to such software, we will review in this and subsequent sections some of the basic algebra in the important cases in which the denominator of a Laplace transform F(s) contains distinct linear factors, repeated linear factors, and quadratic polynomials with no real factors. Although we shall examine each of these cases as this chapter develops, it still might be a good idea for you to consult either a calculus text or a current precal-culus text for a more comprehensive review of this theory. The following example illustrates partial fraction decomposition in the case when the denominator of F(s) is factorable into distinct linear factors. EXAMPLE 2 Termwise Division and Linearity Evaluate . SOLUTION We ﬁrst rewrite the given function of s as two expressions by means of termwise division and then use (1): 1 2s 6 s2 4 EXAMPLE 3 Partial Fractions: Distinct Linear Factors Evaluate . SOLUTION There exist unique real constants A, B, and C so that Since the denominators are identical, the numerators are identical: . (3) By comparing coefﬁcients of powers of s on both sides of the equality, we know that (3) is equivalent to a system of three equations in the three unknowns A, B, and C. However, there is a shortcut for determining these unknowns. If we set s 1, s 2, and s 4 in (3), we obtain, respectively, , and so , , and . Hence the partial fraction decomposition is , (4) s2 6s 9 (s 1)(s 2)(s 4) 16>5 s 1 25>6 s 2 1>30 s 4 C 1 30 B 25 6 A 16 5 16 A( 1)(5), 25 B(1)(6), and 1 C( 5)( 6) s2 6s 9 A(s 2)(s 4) B(s 1)(s 4) C(s 1)(s 2) A(s 2)(s 4) B(s 1)(s 4) C(s 1)(s 2) (s 1)(s 2)(s 4) . s2 6s 9 (s 1)(s 2)(s 4) A s 1 B s 2 C s 4 1 s2 6s 9 (s 1)(s 2)(s 4) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 284 ● CHAPTER 7 THE LAPLACE TRANSFORM and thus, from the linearity of 1 and part (c) of Theorem 7.2.1, 1 s2 6s 9 (s 1)(s 2)(s 4) 16 5 1 1 s 1 25 6 1 1 s 2 1 30 1 1 s 4 . (5) 7.2.2 TRANSFORMS OF DERIVATIVES Transform a Derivative As was pointed out in the introduction to this chap-ter, our immediate goal is to use the Laplace transform to solve differential equations. To that end we need to evaluate quantities such as and . For example, if f is continuous for t 0, then integration by parts gives or (6) Here we have assumed that as . Similarly, with the aid of (6), or (7) In like manner it can be shown that (8) The recursive nature of the Laplace transform of the derivatives of a function f should be apparent from the results in (6), (7), and (8). The next theorem gives the Laplace transform of the nth derivative of f. The proof is omitted. { f (t)} s3F(s) s2f (0) sf(0) f (0). { f (t)} s2F(s) sf (0) f(0). ; from (6) s[sF(s) f (0)] f (0) f(0) s{ f(t)} { f (t)} 0 e st f (t) dt e st f (t) 0 s 0 e st f (t) dt t : e st f(t) : 0 { f(t)} sF(s) f (0). f (0) s{ f (t)} { f(t)} 0 e st f (t) dt e st f (t)0 s 0 e st f (t) dt {d2y>dt2} {dy>dt} 16 5 et 25 6 e2t 1 30 e 4t Solving Linear ODEs It is apparent from the general result given in THEOREM 7.2.2 Transform of a Derivative If f, f, . . . , f (n 1) are continuous on [0, ) and are of exponential order and if f (n)(t) is piecewise continuous on [0, ), then where . F(s) { f(t)} { f (n)(t)} snF(s) sn 1f(0) sn 2f(0) f (n 1)(0), Theorem 7.2.2 that depends on and the n 1 derivatives of y(t) evaluated at This property makes the Laplace transform ideally suited for solving linear initial-value problems in which the differential equation has con-stant coefficients Such a differential equation is simply a linear combination of terms y, y, y, . . . , y(n): y(0) y0, y(0) y1, . . . , y(n 1)(0) yn 1, an d ny dtn an 1 d n 1y dtn 1 a0y g(t), t 0. Y(s) {y(t) } {dny>dtn} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES ● 285 where the ai, i 0, 1, . . . , n and y0, y1, . . . , yn 1 are constants. By the linearity prop-erty the Laplace transform of this linear combination is a linear combination of Laplace transforms: (9) From Theorem 7.2.2, (9) becomes , (10) where G(s). In other words, The Laplace transform of a linear differential equation with constant coefficient becomes an algebraic equation in Y(s). If we solve the general transformed equation (10) for the symbol Y(s), we ﬁrst obtain P(s)Y(s) Q(s) G(s) and then write , (11) where is a polynomial in s of degree less than or equal to consisting of the various products of the coefﬁcients ai, . . . , n and the prescribed initial conditions y0, y1, . . . , yn 1, and G(s) is the Laplace transform of g(t). Typically, we put the two terms in (11) over the least common denominator and then decompose the expression into two or more partial fractions. Finally, the solution y(t) of the original initial-value problem is , where the inverse transform is done term by term. The procedure is summarized in the diagram in Figure 7.2.1. y(t) 1{Y(s)} i 1, n 1 P(s) ansn an 1sn 1 a0, Q(s) Y(s) Q(s) P(s) G(s) P(s) {y(t)} Y(s) and {g(t)} an 1[sn 1Y(s) sn 2y(0) y(n 2)(0)] a0Y(s) G(s) an [snY(s) sn 1y(0) y(n 1)(0)] an d ny dtn an 1 d n 1y dt n 1 a0 {y} {g(t)}. The polynomial P(s) is the same as the nth-degree auxiliary polynomial in (12) in Section 4.3 with the usual symbol m replaced by s. The next example illustrates the foregoing method of solving DEs, as well as partial fraction decomposition in the case when the denominator of Y(s) contains a quadratic polynomial with no real factors. Apply Laplace transform Apply inverse Laplace transform Find unknown y(t) that satisfies DE and initial conditions Transformed DE becomes an algebraic equation in Y(s) Solve transformed equation for Y(s) Solution y(t) of original IVP −1 FIGURE 7.2.1 Steps in solving an IVP by the Laplace transform EXAMPLE 4 Solving a First-Order IVP Use the Laplace transform to solve the initial-value problem . SOLUTION We ﬁrst take the transform of each member of the differential equation: . (12) dy dt 3{y} 13{sin 2t} dy dt 3y 13 sin 2t, y(0) 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 286 ● CHAPTER 7 THE LAPLACE TRANSFORM From (6), , and from part (d) of Theorem 7.1.1, , so (12) is the same as . Solving the last equation for Y(s), we get . (13) Since the quadratic polynomial s2 4 does not factor using real numbers, its assumed numerator in the partial fraction decomposition is a linear polynomial in s: . Putting the right-hand side of the equality over a common denominator and equating numerators gives 6s2 50 A(s2 4) (Bs C)(s 3). Setting s 3 then immediately yields A 8. Since the denominator has no more real zeros, we equate the coefﬁcients of s2 and s: 6 A B and 0 3B C. Using the value of A in the ﬁrst equation gives B 2, and then using this last value in the second equation gives C 6. Thus . We are not quite ﬁnished because the last rational expression still has to be written as two fractions. This was done by termwise division in Example 2. From (2) of that example, . It follows from parts (c), (d), and (e) of Theorem 7.2.1 that the solution of the initial-value problem is y(t) 8e 3t 2 cos 2t 3 sin 2t. y(t) 8 1 1 s 3 2 1 s s2 4 3 1 2 s2 4 Y(s) 6s2 50 (s 3)(s2 4) 8 s 3 2s 6 s2 4 6s2 50 (s 3)(s2 4) A s 3 Bs C s2 4 Y(s) 6 s 3 26 (s 3)(s2 4) 6s2 50 (s 3)(s2 4) sY(s) 6 3Y(s) 26 s2 4 or (s 3)Y(s) 6 26 s2 4 {sin 2t} 2>(s2 4) {dy>dt} sY(s) y(0) sY(s) 6 EXAMPLE 5 Solving a Second-Order IVP Solve y 3y 2y e 4t, y(0) 1, y(0) 5. SOLUTION Proceeding as in Example 4, we transform the DE. We take the sum of the transforms of each term, use (6) and (7), use the given initial conditions, use (c) of Theorem 7.1.1, and then solve for Y(s): . (14) The details of the partial fraction decomposition of Y(s) have already been carried out in Example 3. In view of the results in (4) and (5) we have the solution of the initial-value problem . y(t) 1{Y(s)} 16 5 et 25 6 e2t 1 30 e 4t Y(s) s 2 s2 3s 2 1 (s2 3s 2)(s 4) s2 6s 9 (s 1)(s 2)(s 4) (s2 3s 2)Y(s) s 2 1 s 4 s2Y(s) sy(0) y(0) 3[sY(s) y(0)] 2Y(s) 1 s 4 d2y dt2 3 dy dt 2{y} {e 4t} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. REMARKS (i) The inverse Laplace transform of a function F(s) may not be unique; in other words, it is possible that and yet f1 f2. For our purposes this is not anything to be concerned about. If f1 and f2 are piecewise continuous on [0, ) and of exponential order, then f1 and f2 are essentially the same. See Problem 44 in Exercises 7.2. However, if f1 and f2 are continuous on [0, ) and , then f1 f2 on the interval. (ii) This remark is for those of you who will be required to do partial fraction decompositions by hand. There is another way of determining the coefﬁcients in a partial fraction decomposition in the special case when is a rational function of s and the denominator of F is a product of distinct linear factors. Let us illustrate by reexamining Example 3. Suppose we multiply both sides of the assumed decomposition (15) by, say, s 1, simplify, and then set s 1. Since the coefﬁcients of B and C on the right-hand side of the equality are zero, we get . Written another way, , where we have shaded, or covered up, the factor that canceled when the left-hand side was multiplied by s 1. Now to obtain B and C, we simply evaluate the left-hand side of (15) while covering up, in turn, s 2 and s 4: s2 6s 9 –––––––––––––––––––––– (s 1)(s 2)(s 4) 1 – – – 30 s 4 C. and s2 6s 9 –––––––––––––––––––––– (s 1)(s 2)(s 4) 25 – – – 6 s2 B s2 6s 9 (s 1) (s 2)(s 4) s1 16 5 A s2 6s 9 (s 2)(s 4) s1 A or A 16 5 s2 6s 9 (s 1)(s 2)(s 4) A s 1 B s 2 C s 4 { f(t)} F(s) { f1(t)} { f2(t)} { f1(t)} { f2(t)} 7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES ● 287 Examples 4 and 5 illustrate the basic procedure for using the Laplace transform to solve a linear initial-value problem, but these examples may appear to demonstrate a method that is not much better than the approach to such problems outlined in Sections 2.3 and 4.3–4.6. Don’t draw any negative conclusions from only two examples. Yes, there is a lot of algebra inherent in the use of the Laplace transform, but observe that we do not have to use variation of parameters or worry about the cases and algebra in the method of undetermined coefﬁcients. Moreover, since the method incorporates the prescribed initial conditions directly into the solution, there is no need for the separate operation of applying the initial conditions to the general solution of the DE to ﬁnd speciﬁc constants in a particular solution of the IVP. The Laplace transform has many operational properties. In the sections that fol-low we will examine some of these properties and see how they enable us to solve problems of greater complexity. y c1y1 c2y2 cn yn yp Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 288 ● CHAPTER 7 THE LAPLACE TRANSFORM The desired decomposition (15) is given in (4). This special technique for determining coefﬁcients is naturally known as the cover-up method. (iii) In this remark we continue our introduction to the terminology of dynamical systems. Because of (9) and (10) the Laplace trans-form is well adapted to linear dynamical systems. The polynomial in (11) is the total coefﬁcient of Y(s) in (10) and is simply the left-hand side of the DE with the derivatives dkydtk replaced by powers sk, k 0, 1, . . . , n. It is usual practice to call the recipro-cal of P(s)—namely, W(s) 1P(s)—the transfer function of the system and write (11) as . (16) In this manner we have separated, in an additive sense, the effects on the response that are due to the initial conditions (that is, W(s)Q(s)) from those due to the input function g (that is, W(s)G(s)). See (13) and (14). Hence the response y(t) of the system is a superposition of two responses: . If the input is g(t) 0, then the solution of the problem is . This solution is called the zero-input response of the system. On the other hand, the function is the output due to the input g(t). Now if the initial state of the system is the zero state (all the initial conditions are zero), then Q(s) 0, and so the only solution of the initial-value problem is y1(t). The latter solution is called the zero-state response of the system. Both y0(t) and y1(t) are particular solutions: y0(t) is a solution of the IVP consisting of the associated ho-mogeneous equation with the given initial conditions, and y1(t) is a solution of the IVP consisting of the nonhomogeneous equation with zero initial conditions. In Example 5 we see from (14) that the transfer function is W(s) 1(s2 3s 2), the zero-input response is , and the zero-state response is . Verify that the sum of y0(t) and y1(t) is the solution y(t) in Example 5 and that y0(0) 1, , whereas y1(0) 0, . y 1(0) 0 y 0(0) 5 y1(t) 1 1 (s 1)(s 2)(s 4) 1 5 et 1 6 e2t 1 30 e 4t y0(t) 1 s 2 (s 1)(s 2) 3et 4e2t y1(t) 1{W(s)G(s)} y0(t) 1{W(s)Q(s)} y(t) 1{W(s)Q(s)} 1{W(s)G(s)} y0(t) y1(t) Y(s) W(s)Q(s) W(s)G(s) P(s) ansn an 1sn 1 a0 EXERCISES 7.2 Answers to selected odd-numbered problems begin on page ANS-11. 7.2.1 INVERSE TRANSFORMS In Problems 1–30 use appropriate algebra and Theorem 7.2.1 to ﬁnd the given inverse Laplace transform. 1. 2. 3. 4. 5. 6. 1 (s 2)2 s3 1 (s 1)3 s4 1 2 s 1 s3 2 1 1 s2 48 s5 1 1 s4 1 1 s3 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 1 s 1 s2 2 1 2s 6 s2 9 1 1 4s2 1 1 4s 4s2 1 1 10s s2 16 1 5 s2 49 1 1 5s 2 1 1 4s 1 1 4 s 6 s5 1 s 8 1 1 s2 1 s 1 s 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 289 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 7.2.2 TRANSFORMS OF DERIVATIVES In Problems 31–40 use the Laplace transform to solve the given initial-value problem. 31. 32. 33. y 6y e4t, y(0) 2 34. y y 2 cos 5t, y(0) 0 35. y 5y 4y 0, y(0) 1, y(0) 0 36. y 4y 6e3t 3e t, y(0) 1, y(0) 1 37. 38. y 9y et, y(0) 0, y(0) 0 y y 22 sin 22t, y(0) 10, y(0) 0 2 dy dt y 0, y(0) 3 dy dt y 1, y(0) 0 1 6s 3 s4 5s2 4 1 1 (s2 1)(s2 4) 1 1 s4 9 1 2s 4 (s2 s)(s2 1) 1 s (s 2)(s2 4) 1 1 s3 5s 1 s2 1 s(s 1)(s 1)(s 2) 1 s (s 2)(s 3)(s 6) 1 s 3 s 13s 13 1 0.9s (s 0.1)(s 0.2) 1 1 s2 s 20 1 s s2 2s 3 1 s 1 s2 4s 1 1 s2 3s 39. 2y 3y 3y 2y e t, y(0) 0, y(0) 0, y(0) 1 40. y 2y y 2y sin 3t, y(0) 0, y(0) 0, y(0) 1 The inverse forms of the results in Problem 50 in Exercises 7.1 are In Problems 41 and 42 use the Laplace transform and these inverses to solve the given initial-value problem. 41. y y e 3t cos 2t, y(0) 0 42. y 2y 5y 0, y(0) 1, y(0) 3 Discussion Problems 43. (a) With a slight change in notation the transform in (6) is the same as With f(t) teat, discuss how this result in conjunc-tion with (c) of Theorem 7.1.1 can be used to evalu-ate . (b) Proceed as in part (a), but this time discuss how to use (7) with f (t) t sin kt in conjunction with (d) and (e) of Theorem 7.1.1 to evaluate . 44. Make up two functions f1 and f2 that have the same Laplace transform. Do not think profound thoughts. 45. Reread (iii) in the Remarks on page 288. Find the zero-input and the zero-state response for the IVP in Problem 36. 46. Suppose f(t) is a function for which f(t) is piecewise continuous and of exponential order c. Use results in this section and Section 7.1 to justify , where F(s) { f (t)}. Verify this result with f(t) cos kt. f (0) lim s: sF(s) {t sin kt} {teat} { f(t)} s{ f (t)} f (0). 1 b (s a)2 b2 eat sin bt. 1 s a (s a)2 b2 eat cos bt OPERATIONAL PROPERTIES I REVIEW MATERIAL ●Keep practicing partial fraction decomposition ●Completion of the square INTRODUCTION It is not convenient to use Deﬁnition 7.1.1 each time we wish to ﬁnd the Laplace transform of a function f(t). For example, the integration by parts involved in evaluating, say, is formidable, to say the least. In this section and the next we present several labor-saving operational properties of the Laplace transform that enable us to build up a more extensive list of transforms (see the table in Appendix III) without having to resort to the basic deﬁnition and integration. {et t2 sin 3t} 7.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 290 ● CHAPTER 7 THE LAPLACE TRANSFORM 7.3.1 TRANSLATION ON THE s-AXIS A Translation Evaluating transforms such as and is straightforward provided that we know (and we do) and . In general, if we know the Laplace transform of a function f, , it is pos-sible to compute the Laplace transform of an exponential multiple of f, that is, with no additional effort other than translating, or shifting, the transform F(s) to This result is known as the first translation theorem or firs shifting theorem. F(s a). {eat f (t) }, { f (t)} F(s) {cos 4t} {t3} {e 2t cos 4t} {e5tt3} PROOF The proof is immediate, since by Deﬁnition 7.1.1 . If we consider s a real variable, then the graph of F(s a) is the graph of F(s) shifted on the s-axis by the amount a. If a 0, the graph of F(s) is shifted a units to the right, whereas if a 0, the graph is shifted a units to the left. See Figure 7.3.1. For emphasis it is sometimes useful to use the symbolism , where means that in the Laplace transform F(s) of f(t) we replace the symbol s wherever it appears by s a. s : s a {eat f (t)} { f (t)} s:s a {eat f (t)} 0 e steat f (t) dt 0 e (s a)t f (t) dt F(s a) s F ( s ) s = a , a > 0 F F( s −a) FIGURE 7.3.1 Shift on s-axis THEOREM 7.3.1 First Translation Theorem If and a is any real number, then . {eat f(t)} F(s a) {f(t)} F(s) EXAMPLE 1 Using the First Translation Theorem Evaluate (a) (b) . SOLUTION The results follow from Theorems 7.1.1 and 7.3.1. (a) (b) Inverse Form of Theorem 7.3.1 To compute the inverse of F(s a), we must recognize F(s), ﬁnd f(t) by taking the inverse Laplace transform of F(s), and then multiply f(t) by the exponential function eat. This procedure can be summarized symbolically in the following manner: , (1) where . The ﬁrst part of the next example illustrates partial fraction decomposition in the case when the denominator of Y(s) contains repeated linear factors. f(t) 1{F(s)} 1{F(s a)} 1{F(s)s:s a} eat f (t) {e 2t cos 4t} {cos 4t} s:s ( 2) s s2 16 s:s2 s 2 (s 2)2 16 {e5tt3} {t3} s: s 5 3! s4 s:s 5 6 (s 5)4 {e 2t cos 4t} {e5tt3} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 291 EXAMPLE 2 Partial Fractions: Repeated Linear Factors Evaluate (a) (b) . SOLUTION (a) A repeated linear factor is a term (s a)n, where a is a real number and n is a positive integer 2. Recall that if (s a)n appears in the denominator of a rational expression, then the assumed decomposition contains n partial fractions with constant numerators and denominators s a, (s a)2, . . . , (s a)n. Hence with a 3 and n 2 we write . By putting the two terms on the right-hand side over a common denominator, we obtain the numerator 2s 5 A(s 3) B, and this identity yields A 2 and B 11. Therefore (2) and (3) Now 1(s 3)2 is F(s) 1s2 shifted three units to the right. Since , it follows from (1) that . Finally, (3) is . (4) (b) To start, observe that the quadratic polynomial s2 4s 6 has no real zeros and so has no real linear factors. In this situation we complete the square: . (5) Our goal here is to recognize the expression on the right-hand side as some Laplace transform F(s) in which s has been replaced throughout by s 2. What we are trying to do is analogous to working part (b) of Example 1 backwards. The denom-inator in (5) is already in the correct form—that is, s2 2 with s replaced by s 2. However, we must ﬁx up the numerator by manipulating the constants: . Now by termwise division, the linearity of 1, parts (e) and (d) of Theorem 7.2.1, and ﬁnally (1), (6) (7) 1 2 e 2t cos 12t 12 3 e 2t sin 12t. 1 2 1 s s2 2s:s2 2 312 1 12 s2 2s:s2 1 s>2 5> 3 s2 4s 6 1 2 1 s 2 (s 2)2 2 2 3 1 1 (s 2)2 2 s> 2 5> 3 (s 2)2 2 1 2(s 2) 2 3 (s 2)2 2 1 2 s 2 (s 2)2 2 2 3 1 (s 2)2 2 1 2s 5 3 1 2 (s 2) 5 3 2 2 1 2 (s 2) 2 3 s>2 5>3 s2 4s 6 s>2 5>3 (s 2)2 2 1 2s 5 (s 3)2 2e3t 11e3tt 1 1 (s 3)2 1 1 s2 s:s 3 e3tt 1{1>s2} t 1 2s 5 (s 3)2 2 1 1 s 3 11 1 1 (s 3)2. 2s 5 (s 3)2 2 s 3 11 (s 3)2 2s 5 (s 3)2 A s 3 B (s 3)2 1 s>2 5>3 s2 4s 6 1 2s 5 (s 3)2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 292 ● CHAPTER 7 THE LAPLACE TRANSFORM EXAMPLE 3 An Initial-Value Problem Solve y 6y 9y t2e3t, y(0) 2, y(0) 17. SOLUTION Before transforming the DE, note that its right-hand side is similar to the function in part (a) of Example 1. After using linearity, Theorem 7.3.1, and the initial conditions, we simplify and then solve for : . The ﬁrst term on the right-hand side was already decomposed into individual partial fractions in (2) in part (a) of Example 2: . Thus . (8) From the inverse form (1) of Theorem 7.3.1, the last two terms in (8) are . Thus (8) is y(t) 2e3t 11te3t 1 12t4e3t. 1 1 s2s:s 3 te3t and 1 4! s5s:s 3 t 4e3t y(t) 2 1 1 s 3 11 1 1 (s 3)2 2 4! 1 4! (s 3)5 Y(s) 2 s 3 11 (s 3)2 2 (s 3)5 Y(s) 2s 5 (s 3)2 2 (s 3)5 (s 3)2Y(s) 2s 5 2 (s 3)3 (s2 6s 9)Y(s) 2s 5 2 (s 3)3 s2Y(s) sy(0) y(0) 6[sY(s) y(0)] 9Y(s) 2 (s 3)3 {y} 6{y} 9{y} {t2e3t} Y(s) { f (t)} EXAMPLE 4 An Initial-Value Problem Solve y 4y 6y 1 e t, y(0) 0, y(0) 0. SOLUTION Since the quadratic term in the denominator does not factor into real linear factors, the partial fraction decomposition for Y(s) is found to be . Moreover, in preparation for taking the inverse transform we already manipulated the last term into the necessary form in part (b) of Example 2. So in view of the results in (6) and (7) we have the solution Y(s) 1>6 s 1>3 s 1 s> 2 5> 3 s2 4s 6 Y(s) 2s 1 s(s 1)(s2 4s 6) (s2 4s 6)Y(s) 2s 1 s(s 1) s2Y(s) sy(0) y(0) 4[sY(s) y(0)] 6Y(s) 1 s 1 s 1 {y} 4{y} 6{y} {1} {e t} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 293 . 1 6 1 3 e t 1 2 e 2t cos 12t 12 3 e 2t sin 12t y(t) 1 6 1 1 s 1 3 1 1 s 1 1 2 1 s 2 (s 2)2 2 2 312 1 12 (s 2)2 2 7.3.2 TRANSLATION ON THE t-AXIS Unit Step Function In engineering, one frequently encounters functions that are either “off” or “on.” For example, an external force acting on a mechanical system or a voltage impressed on a circuit can be turned off after a period of time. It is convenient, then, to deﬁne a special function that is the number 0 (off) up to a certain time t a and then the number 1 (on) after that time. This function is called the unit step function or the Heaviside function, named after the English polymath Oliver Heaviside (1850–1925). Notice that we deﬁne only on the nonnegative t-axis, since this is all that we are concerned with in the study of the Laplace transform. In a broader sense for t a. The graph of is given in Figure 7.3.2. In the case when a 0, we take When a function f deﬁned for t 0 is multiplied by , the unit step function “turns off” a portion of the graph of that function. For example, consider the function f(t) 2t 3. To “turn off” the portion of the graph of f for 0 t 1, we simply form the product (2t 3) . See Figure 7.3.3. In general, the graph of f(t) is 0 (off) for 0 t a and is the portion of the graph of f (on) for t a. The unit step function can also be used to write piecewise-deﬁned functions in a compact form. For example, if we consider 0 t 2, 2 t 3, and t 3 and the corresponding values of and , it should be apparent that the piecewise-deﬁned function shown in Figure 7.3.4 is the same as . Also, a general piecewise-deﬁned function of the type (9) is the same as . (10) Similarly, a function of the type (11) can be written (12) f (t) g(t)[(t a) (t b)]. f(t) 0, g(t), 0, 0 t a a t b t b f(t) g(t) g(t) (t a) h(t) (t a) f(t) g(t), h(t), 0 t a t a f(t) 2 3(t 2) (t 3) (t 3) (t 2) (t a) (t 1) (t a) (t) 1 for t 0. (t a) (t a) 0 (t a) FIGURE 7.3.2 Graph of unit step function t 1 a FIGURE 7.3.3 Function is f(t) (2t 3) (t 1) 1 y t FIGURE 7.3.4 Function is f(t) 2 3(t 2) (t 3) −1 2 t f(t) DEFINITION 7.3.1 Unit Step Function The unit step function is deﬁned to be (t a) 0, 1, 0 t a t a. (t a) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 294 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.3.5 Function f in Example 5 100 5 f(t) t FIGURE 7.3.6 Shift on t-axis (a) f(t), t 0 (b) f(t a) (t a) t f(t) t f(t) a zero for 0 t a one for t a {f(t a) (t a)} estf(t a) (t a) dt estf(t a) (t a) dt estf(t a) dt. a 0 a a Now if we let v t a, dv dt in the last integral, then PROOF By the additive interval property of integrals, can be written as two integrals: (t a) dt 0 e st f (t a) . { f (t a) (t a)} 0 e s(va) f (v) dv e as 0 e sv f (v) dv e as{ f (t)} We often wish to ﬁnd the Laplace transform of just a unit step function. This can be from either Deﬁnition 7.1.1 or Theorem 7.3.2. If we identify f(t) 1 in Theorem 7.3.2, then f(t a) 1, , and so . (14) {(t a)} e as s F(s) {1} 1>s EXAMPLE 5 A Piecewise-Defined Functio Express in terms of unit step functions. Graph. SOLUTION The graph of f is given in Figure 7.3.5. Now from (9) and (10) with a 5, g(t) 20t, and h(t) 0 we get . Consider a general function y f(t) deﬁned for t 0. The piecewise-deﬁned function (13) plays a signiﬁcant role in the discussion that follows. As shown in Figure 7.3.6, for a 0 the graph of the function coincides with the graph of y for (which is the entire graph of shifted a units to the right on the t-axis), but is identically zero for We saw in Theorem 7.3.1 that an exponential multiple of f(t) results in a transla-tion of the transform F(s) on the s-axis. As a consequence of the next theorem we see that whenever F(s) is multiplied by an exponential function e as, a 0, the inverse transform of the product e as F(s) is the function f shifted along the t-axis in the man-ner illustrated in Figure 7.3.6(b). This result, presented next in its direct transform version, is called the second translation theorem or second shifting theorem. 0 t a. y f(t), t 0 t a f(t a) y f(t a) (t a) f(t a) (t a) 0, f(t a), 0 t a t a f (t) 20t 20t (t 5) f (t) 20t, 0, 0 t 5 t 5 THEOREM 7.3.2 Second Translation Theorem If and a 0, then . { f(t a) (t a)} e asF(s) F(s) { f(t)} EXAMPLE 6 Figure 7.3.4 Revisited Find the Laplace transform the function f in Figure 7.3.4. SOLUTION We use f expressed in terms of the unit step function f(t) 2 3 (t 2) (t 3) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 295 and the result given in (14): Inverse Form of Theorem 7.3.2 If , the inverse form of f(t) 1{F(s)} 2 s 3 e 2s s e 3s s . { f(t)} 2{1} 3{(t 2)} {(t 3)} EXAMPLE 7 Using Formula (15) Evaluate . SOLUTION (a) With the three identiﬁcations a 2, F(s) 1(s 4), and 1{F(s)} e4t, we have from (15) . (b) With a p2, F(s) s(s2 9), and , (15) yields . The last expression can be simpliﬁed somewhat by using the addition formula for the cosine. Verify that the result is the same as Alternative Form of Theorem 7.3.2 We are frequently confronted with the problem of ﬁnding the Laplace transform of a product of a function g and a unit step function where the function g lacks the precise shifted form in Theorem 7.3.2. To ﬁnd the Laplace transform of , it is possible to ﬁx up g(t) into the required form by algebraic manipulations. For example, if we wanted to use Theorem 7.3.2 to ﬁnd the Laplace transform of , we would have to force into the form You should work through the details and verify that is an identity. Therefore where each term on the right-hand side can now be evaluated by Theorem 7.3.2. But since these manipulations are time consuming and often not obvious, it is simpler to devise an alternative version of Theorem 7.3.2. Using Deﬁnition 7.1.1, the deﬁnition of , and the substitution u t a, we obtain . That is, . (16) {g(t)(t a)} e as {g(t a)} {g(t) (t a)} a e st g(t) dt 0 e s(ua) g(u a) du (t a) {t2(t 2)} {(t 2)2 (t 2) 4(t 2) (t 2) 4(t 2)}, t2 (t 2)2 4(t 2) 4 f(t 2). g(t) t2 t2(t 2) f(t a) g(t)(t a) f(t a) (t a) sin 3t t 2. 1 s s2 9 e s/2 cos 3t 2 t 2 1{F(s)} cos 3t 1 1 s 4 e 2s e4(t 2) (t 2) (b) 1 s s2 9 e s/2 (a) 1 1 s 4 e 2s EXAMPLE 8 Second Translation Theorem—Alternative Form Evaluate . SOLUTION With g(t) cos t and a p, then g(t p) cos(t p) cos t by the addition formula for the cosine function. Hence by (16), {cos t (t )} e s {cos t} s s2 1 e s. {cos t (t )} Theorem 7.3.2, a 0, is . (15) 1{e asF(s)} f(t a) (t a) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 296 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.3.7 Graph of function (18) in Example 9 _2 1 2 3 4 5 _1 t y 2 π π 3π (18) 5e t, 5e t 3 2 e (t ) 3 2 sin t 3 2 cos t, 0 t t . ; trigonometric identities 5e t 3 2 [e (t ) sin t cos t] (t ) y(t) 5e t 3 2 e (t ) (t ) 3 2 sin(t ) (t ) 3 2 cos(t ) (t ) We obtained the graph of (18) shown in Figure 7.3.7 by using a graphing utility. Beams In Section 5.2 we saw that the static deﬂection y(x) of a uniform beam of length L carrying load w(x) per unit length is found from the linear fourth-order differential equation (19) where E is Young’s modulus of elasticity and I is a moment of inertia of a cross section of the beam. The Laplace transform is particularly useful in solving (19) when w(x) is piecewise-deﬁned. However, to use the Laplace transform, we must tacitly assume that y(x) and w(x) are deﬁned on (0, ) rather than on (0, L). Note, too, that the next example is a boundary-value problem rather than an initial-value problem. EI d4y dx4 w(x), FIGURE 7.3.8 Embedded beam with variable load in Example 10 wall x y L w(x) 1 1 s 1 e s e (t ) (t ), 1 1 s2 1 e s sin(t ) (t ), and . Thus the inverse of (17) is 1 s s2 1 e s cos(t ) (t ) EXAMPLE 9 An Initial-Value Problem Solve SOLUTION The function f can be written as , so by linear-ity, the results of Example 7, and the usual partial fractions, we have . (17) Now proceeding as we did in Example 7, it follows from (15) with a p that the inverses of the terms inside the brackets are Y(s) 5 s 1 3 2 1 s 1 e s 1 s2 1 e s s s2 1 e s (s 1)Y(s) 5 3s s2 1 e s sY(s) y(0) Y(s) 3 s s2 1 e s {y} {y} 3{cos t (t )} f(t) 3 cos t (t ) y y f(t), y(0) 5, where f(t) 0, 3 cos t, 0 t t . EXAMPLE 10 A Boundary-Value Problem A beam of length L is embedded at both ends, as shown in Figure 7.3.8. Find the deﬂection of the beam when the load is given by w(x) w01 2 L x, 0, 0 x L>2 L>2 x L. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 297 SOLUTION Recall that because the beam is embedded at both ends, the boundary conditions are y(0) 0, y(0) 0, y(L) 0, y(L) 0. Now by (10) we can express w(x) in terms of the unit step function: Transforming (19) with respect to the variable x gives or If we let c1 y(0) and c2 y(0), then , and consequently Y(s) c1 s3 c2 s4 2w0 EIL L>2 s5 1 s6 1 s6 e Ls/2 s4Y(s) sy(0) y(0) 2w0 EIL L>2 s 1 s2 1 s2 e Ls/2 . EIs4Y(s) s3y(0) s2y(0) sy(0) y(0) 2w0 L L>2 s 1 s2 1 s2 e Ls/2 2w0 L L 2 x x L 2 x L 2 . w(x) w01 2 L x w01 2 L x x L 2 c1 2 x2 c2 6 x3 w0 60 EIL 5L 2 x4 x5 x L 2 5 x L 2 . y(x) c1 2! 1 2! s3 c2 3! 1 3! s4 2w0 EIL L>2 4! 1 4! s5 1 5! 1 5! s6 1 5! 1 5! s6 e Ls/ 2 Applying the conditions y(L) 0 and y(L) 0 to the last result yields a system of equations for c1 and c2: Solving, we ﬁnd c1 23w0L2(960EI) and c2 9w0L(40EI). Thus the deﬂec-tion is given by y(x) 23w0L2 1920EI x2 3w0L 80EI x3 w0 60EIL 5L 2 x4 x5 x L 2 5 x L 2 . c1 L c2 L2 2 85w0L3 960EI 0. c1 L2 2 c2 L3 6 49w0L4 1920EI 0 EXERCISES 7.3 Answers to selected odd-numbered problems begin on page ANS-12. 7.3.1 TRANSLATION ON THE s-AXIS In Problems 1–20 ﬁnd either F(s) or f(t), as indicated. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. e3t 9 4t 10 sin t 2 {(1 et 3e 4t) cos 5t} {e 2t cos 4t} {et sin 3t} {e2t(t 1)2} {t(et e2t)2} {t10e 7t} {t3e 2t} {te 6t} {te10t} 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 1 (s 1)2 (s 2)4 1 2s 1 s2(s 1)3 1 5s (s 2)2 1 s (s 1)2 1 2s 5 s2 6s 34 1 s s2 4s 5 1 1 s2 2s 5 1 1 s2 6s 10 1 1 (s 1)4 1 1 (s 2)3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 298 ● CHAPTER 7 THE LAPLACE TRANSFORM In Problems 21–30 use the Laplace transform to solve the given initial-value problem. 21. y 4y e 4t, y(0) 2 22. y y 1 tet, y(0) 0 23. y 2y y 0, y(0) 1, y(0) 1 24. y 4y 4y t3e2t, y(0) 0, y(0) 0 25. y 6y 9y t, y(0) 0, y(0) 1 26. y 4y 4y t3, y(0) 1, y(0) 0 27. y 6y 13y 0, y(0) 0, y(0) 3 28. 2y 20y 51y 0, y(0) 2, y(0) 0 29. y y et cos t, y(0) 0, y(0) 0 30. y 2y 5y 1 t, y(0) 0, y(0) 4 In Problems 31 and 32 use the Laplace transform and the procedure outlined in Example 10 to solve the given boundary-value problem. 31. y 2y y 0, y(0) 2, y(1) 2 32. y 8y 20y 0, y(0) 0, y(p) 0 33. A 4-pound weight stretches a spring 2 feet. The weight is released from rest 18 inches above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to times the instantaneous velocity. Use the Laplace transform to ﬁnd the equation of motion x(t). 34. Recall that the differential equation for the instanta-neous charge q(t) on the capacitor in an LRC-series circuit is given by . (20) See Section 5.1. Use the Laplace transform to ﬁnd q(t) when L 1 h, R 20 , C 0.005 f, E(t) 150 V, t 0, q(0) 0, and i(0) 0. What is the current i(t)? 35. Consider a battery of constant voltage E0 that charges the capacitor shown in Figure 7.3.9. Divide equa-tion (20) by L and deﬁne 2l RL and v2 1LC. Use the Laplace transform to show that the solution q(t) of q 2lq v2q E0L subject to q(0) 0, i(0) 0 is q(t) E0C 1 e t (cosh 12 2t 12 2 sinh 12 2t) , , E0C[1 e t (1 t)], , E0C 1 e t (cos 12 2t 12 2 sin 12 2t) , . L d 2q dt2 R dq dt 1 C q E(t) 7 8 36. Use the Laplace transform to find the charge q(t) in an RC series circuit when q(0) 0 and E(t) E0e kt, k 0. Consider two cases: k 1RC and k 1RC. 7.3.2 TRANSLATION ON THE t-AXIS In Problems 37–48 ﬁnd either F(s) or f(t), as indicated. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. In Problems 49–54 match the given graph with one of the functions in (a)–(f). The graph of f(t) is given in Figure 7.3.10. (a) (b) (c) (d) (e) (f) f (t a) (t a) f (t a) (t b) f (t) (t a) f(t) (t b) f (t) f (t) (t b) f (t) (t a) f (t b) (t b) f (t) f (t) (t a) 1 e 2s s2(s 1) 1 e s s(s 1) 1 se s/2 s2 4 1 e s s2 1 1 (1 e 2s)2 s 2 1 e 2s s3 sin t t 2 {cos 2t (t )} {(3t 1)(t 1)} {t (t 2)} {e2 t (t 2)} {(t 1)(t 1)} FIGURE 7.3.9 Series circuit in Problem 35 E0 R C L FIGURE 7.3.10 Graph for Problems 49–54 t f(t) a b 49. FIGURE 7.3.11 Graph for Problem 49 t f(t) a b Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.3 OPERATIONAL PROPERTIES I ● 299 FIGURE 7.3.12 Graph for Problem 50 t f(t) a b FIGURE 7.3.13 Graph for Problem 51 t f(t) a b FIGURE 7.3.14 Graph for Problem 52 t f(t) a b FIGURE 7.3.15 Graph for Problem 53 t f(t) a b FIGURE 7.3.16 Graph for Problem 54 t f(t) a b 50. 51. 52. 53. 54. In Problems 55–62 write each function in terms of unit step functions. Find the Laplace transform of the given function. 55. 56. 57. f (t) 0, t2, 0 t 1 t 1 f (t) 1, 0, 1, 0 t 4 4 t 5 t 5 f (t) 2, 2, 0 t 3 t 3 58. 59. 60. f (t) sin t, 0, 0 t 2 t 2 f (t) t, 0, 0 t 2 t 2 f (t) 0, sin t, 0 t 3 >2 t 3 >2 61. 62. FIGURE 7.3.18 Graph for Problem 62 3 2 1 staircase function t f(t) 1 2 3 4 FIGURE 7.3.17 Graph for Problem 61 1 rectangular pulse t b a f(t) In Problems 63–70 use the Laplace transform to solve the given initial-value problem. 63. y y f(t), y(0) 0, where f(t) 64. y y f(t), y(0) 0, where 65. y 2y f(t), y(0) 0, where 66. where 67. , y(0) 1, y(0) 0 68. , y(0) 0, y(0) 1 69. where 70. y 4y 3y 1 (t 2) (t 4) (t 6), y(0) 0, y(0) 0 f (t) 0, 1, 0, 0 t t 2 t 2 y y f(t), y(0) 0, y(0) 1, y 5y 6y (t 1) y 4y sin t (t 2 ) f(t) 1, 0, 0 t 1 t 1 y 4y f (t), y(0) 0, y(0) 1, f(t) t, 0, 0 t 1 t 1 f (t) 1, 1, 0 t 1 t 1 0, 5, 0 t 1 t 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 300 ● CHAPTER 7 THE LAPLACE TRANSFORM 71. Suppose a 32-pound weight stretches a spring 2 feet. If the weight is released from rest at the equilibrium position, ﬁnd the equation of motion x(t) if an impressed force f(t) 20t acts on the system for 0 t 5 and is then removed (see Example 5). Ignore any damping forces. Use a graphing utility to graph x(t) on the interval [0, 10]. 72. Solve Problem 71 if the impressed force f(t) sin t acts on the system for 0 t 2p and is then removed. In Problems 73 and 74 use the Laplace transform to ﬁnd the charge q(t) on the capacitor in an RC-series circuit subject to the given conditions. 73. q(0) 0, R 2.5 , C 0.08 f, E(t) given in Figure 7.3.19 t E(t) 3 5 FIGURE 7.3.19 E(t) in Problem 73 t E(t) 1.5 30 30et FIGURE 7.3.20 E(t) in Problem 74 74. q(0) q0, R 10 , C 0.1 f, E(t) given in Figure 7.3.20 75. (a) Use the Laplace transform to find the current i(t) in a single-loop LR-series circuit when i(0) 0, L 1 h, R 10 , and E(t) is as given in Figure 7.3.21. (b) Use a computer graphing program to graph i(t) for 0 t 6. Use the graph to estimate imax and imin, the maximum and minimum values of the current. FIGURE 7.3.21 E(t) in Problem 75 FIGURE 7.3.22 E(t) in Problem 76 /2 1 −1 t E(t) 3 /2 π sin t, 0 ≤t < 3 /2 π π π t 3 1 E(t) E0 76. (a) Use the Laplace transform to ﬁnd the charge q(t) on the capacitor in an RC-series circuit when q(0) 0, R 50 , C 0.01 f, and E(t) is as given in Figure 7.3.22. (b) Assume that E0 100 V. Use a computer graphing program to graph q(t) for 0 t 6. Use the graph to estimate qmax, the maximum value of the charge. 77. A cantilever beam is embedded at its left end and free at its right end. Use the Laplace transform to ﬁnd the deﬂection y(x) when the load is given by 78. Solve Problem 77 when the load is given by 79. Find the deﬂection y(x) of a cantilever beam embedded at its left end and free at its right end when the load is as given in Example 10. 80. A beam is embedded at its left end and simply supported at its right end. Find the deﬂection y(x) when the load is as given in Problem 77. Mathematical Model 81. Cake Inside an Oven Reread Example 4 in Sec-tion 3.1 on the cooling of a cake that is taken out of an oven. (a) Devise a mathematical model for the temperature of a cake while it is inside the oven based on the fol-lowing assumptions: At t 0 the cake mixture is at the room temperature of 70°; the oven is not pre-heated, so at t 0, when the cake mixture is placed into the oven, the temperature inside the oven is also 70°; the temperature of the oven increases linearly until t 4 minutes, when the desired temperature of 300° is attained; the oven temperature is a con-stant 300° for t 4. (b) Use the Laplace transform to solve the initial-value problem in part (a). w(x) 0, w0, 0, 0 x L>3 L> 3 x 2L> 3 2L> 3 x L. w(x) w0, 0, 0 x L> 2 L> 2 x L. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 301 Discussion Problems 82. Discuss how you would ﬁx up each of the following functions so that Theorem 7.3.2 could be used directly to ﬁnd the given Laplace transform. Check your answers using (16) of this section. (a) (b) (c) (d) 83. (a) Assume that Theorem 7.3.1 holds when the sym-bol a is replaced by ki, where k is a real number {(t2 3t)(t 2)} {cos t (t )} {et (t 5)} {(2t 1)(t 1)} and i2 1. Show that can be used to deduce (b) Now use the Laplace transform to solve the initial-value problem x v2x cos vt, x(0) 0, x(0) 0. {t sin kt} 2ks (s2 k2)2. {t cos kt} s2 k2 (s2 k2)2 {tekti} 7.4.1 DERIVATIVES OF A TRANSFORM Multiplying a Function by tn The Laplace transform of the product of a function f(t) with t can be found by differentiating the Laplace transform of f(t). To motivate this result, let us assume that exists and that it is possible to interchange the order of differentiation and integration. Then ; that is, . We can use the last result to ﬁnd the Laplace transform of t2f(t): {tf(t)} d ds { f (t)} d ds F(s) d ds 0 e st f (t) dt 0 s [e st f (t)] dt 0 e st tf (t) dt {tf (t)} F(s) { f (t)} . {t2 f (t)} {t tf(t)} d ds {tf (t)} d ds d ds {f (t)} d 2 ds2 { f (t) } The preceding two cases suggest the general result for . {tn f(t)} OPERATIONAL PROPERTIES II REVIEW MATERIAL ●Deﬁnition 7.1.1 ●Theorems 7.3.1 and 7.3.2 INTRODUCTION In this section we develop several more operational properties of the Laplace transform. Speciﬁcally, we shall see how to ﬁnd the transform of a function f(t) that is multiplied by a monomial t n, the transform of a special type of integral, and the transform of a pe-riodic function. The last two transform properties allow us to solve some equations that we have not encountered up to this point: Volterra integral equations, integrodifferential equations, and or-dinary differential equations in which the input function is a periodic piecewise-deﬁned function. 7.4 THEOREM 7.4.1 Derivatives of Transforms If and n 1, 2, 3, . . . , then . {tn f(t)} ( 1)n dn dsn F(s) F(s) { f (t)} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 302 ● CHAPTER 7 THE LAPLACE TRANSFORM EXAMPLE 1 Using Theorem 7.4.1 Evaluate . SOLUTION With f(t) sin kt, F(s) k(s2 k2), and n 1, Theorem 7.4.1 gives . If we want to evaluate and , all we need do, in turn, is take the negative of the derivative with respect to s of the result in Example 1 and then take the negative of the derivative with respect to s of . Note To ﬁnd transforms of functions tneat we can use either Theorem 7.3.1 or Theorem 7.4.1. For example, Theorem 7.3.1: . Theorem 7.4.1: . {te3t} d ds {e3t} d ds 1 s 3 (s 3) 2 1 (s 3)2 {te3t} {t}s :s 3 1 s2s:s 3 1 (s 3)2 {t2 sin kt} {t3 sin kt} {t2 sin kt} {t sin kt} d ds {sin kt} d ds k s2 k2 2ks (s2 k2)2 {t sin kt} EXAMPLE 2 An Initial-Value Problem Solve x 16x cos 4t, x(0) 0, x(0) 1. SOLUTION The initial-value problem could describe the forced, undamped, and resonant motion of a mass on a spring. The mass starts with an initial velocity of 1 ft/s in the downward direction from the equilibrium position. Transforming the differential equation gives . Now we just saw in Example 1 that , (1) and so with the identiﬁcation k 4 in (1) and in part (d) of Theorem 7.2.1, we obtain 7.4.2 TRANSFORMS OF INTEGRALS Convolution If functions f and g are piecewise continuous on the interval 1 4 sin 4t 1 8 t sin 4t. x(t) 1 4 1 4 s2 16 1 8 1 8s (s2 16)2 1 2ks (s2 k2)2 t sin kt (s2 16) X(s) 1 s s2 16 or X(s) 1 s2 16 s (s2 16)2 [0, ), then a special product, denoted by f g, is deﬁned by the integral (2) and is called the convolution of f and g. The convolution f g is a function of t. For example, . (3) et sin t t 0 e sin(t ) d 1 2 ( sin t cos t et) f g t 0 f () g(t ) d Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 303 It is left as an exercise to show that that is, f g g f. This means that the convolution of two functions is commutative. It is not true that the integral of a product of functions is the product of the integrals. However, it is true that the Laplace transform of the special product (2) is the product of the Laplace transform of f and g. This means that it is possible to ﬁnd the Laplace transform of the convolution of two functions without actually evaluat-ing the integral as we did in (3). The result that follows is known as the convolution theorem. t 0 f() g(t ) d t 0 f(t ) g() d; PROOF Let and . Proceeding formally, we have Holding t ﬁxed, we let t t b, dt db, so that In the tt-plane we are integrating over the shaded region in Figure 7.4.1. Since f and g are piecewise continuous on [0, ) and of exponential order, it is possible to inter-change the order of integration: F(s)G(s) 0 f () d e stg(t ) dt. 0 f () d 0 e s()g() d. 0 0 e s() f ()g() d d F(s)G(s) 0 e s f () d 0 e s g() d G(s) {g(t)} 0 e sg() d F(s) { f(t)} 0 e s f() d F(s) G(s) 0 e st dt t 0 f ()g(t ) d 0 e st t 0 f () g(t ) ddt {f g}. . t 0 e sin(t ) d {et} {sin t} 1 s 1 1 s2 1 1 (s 1)(s2 1) FIGURE 7.4.1 Changing order of integration from t ﬁrst to t ﬁrst t τ τ τ = t : 0 to t t: to ∞ τ THEOREM 7.4.2 Convolution Theorem If f(t) and g(t) are piecewise continuous on [0, ) and of exponential order, then . { f g} { f (t)} {g(t)} F(s)G(s) EXAMPLE 3 Transform of a Convolution Evaluate SOLUTION With f(t) et and g(t) sin t, the convolution theorem states that the Laplace transform of the convolution of f and g is the product of their Laplace transforms: t 0 e sin(t ) d. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 304 ● CHAPTER 7 THE LAPLACE TRANSFORM Inverse Form of Theorem 7.4.2 The convolution theorem is sometimes useful in ﬁnding the inverse Laplace transform of the product of two Laplace transforms. From Theorem 7.4.2 we have (4) Many of the results in the table of Laplace transforms in Appendix III can be derived using (4). For example, in the next example we obtain entry 25 of the table: . (5) {sin kt kt cos kt} 2k3 (s2 k2)2 1{F(s)G(s)} f g. EXAMPLE 4 Inverse Transform as a Convolution Evaluate . SOLUTION Let so that . In this case (4) gives . (6) With the aid of the product-to-sum trigonometric identity and the substitutions A kt and B k(t t) we can carry out the integration in (6): Multiplying both sides by 2k3 gives the inverse form of (5). Transform of an Integral When g(t) 1 and , the convolution theorem implies that the Laplace transform of the integral of f is . (7) The inverse form of (7), , (8) can be used in lieu of partial fractions when sn is a factor of the denominator and is easy to integrate. For example, we know for f(t) sin t that and so by (8) and so on. 1 1 s3(s2 1) 1 1>s2(s2 1) s t 0 ( sin ) d 1 2 t2 1 cos t 1 1 s2(s2 1) 1 1>s(s2 1) s t 0 (1 cos ) d t sin t 1 1 s(s2 1) 1 1>(s2 1) s t 0 sin d 1 cos t F(s) 1>(s2 1), f(t) 1{F(s)} t 0 f() d 1 F(s) s t 0 f() d F(s) s {g(t)} G(s) 1>s sin kt kt cos kt 2k3 . 1 2k2 1 2k sin k(2 t) cos kt t 0 1 1 (s2 k2)2 1 2k2 t 0 [cos k(2 t) cos kt] d sin A sin B 1 2 [cos(A B) cos(A B)] 1 1 (s2 k2)2 1 k2 t 0 sin k sin k(t ) d f(t) g(t) 1 k 1 k s2 k2 1 k sin kt F(s) G(s) 1 s2 k2 1 1 (s2 k2)2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 305 Volterra Integral Equation The convolution theorem and the result in (7) are useful in solving other types of equations in which an unknown function appears under an integral sign. In the next example we solve a Volterra integral equation for f(t), . (9) The functions g(t) and h(t) are known. Notice that the integral in (9) has the convo-lution form (2) with the symbol h playing the part of g. f(t) g(t) t 0 f() h(t ) d FIGURE 7.4.2 LRC-series circuit C L E R EXAMPLE 5 An Integral Equation Solve . SOLUTION In the integral we identify h(t t) et t so that h(t) et. We take the Laplace transform of each term; in particular, by Theorem 7.4.2 the transform of the integral is the product of and : . After solving the last equation for F(s) and carrying out the partial fraction decomposition, we find . The inverse transform then gives Series Circuits In a single-loop or series circuit, Kirchhoff’s second law states that the sum of the voltage drops across an inductor, resistor, and capacitor is equal to the impressed voltage E(t). Now it is known that the voltage drops across an in-ductor, resistor, and capacitor are, respectively, , where i(t) is the current and L, R, and C are constants. It follows that the current in a circuit, such as that shown in Figure 7.4.2, is governed by the integrodifferential equation . (10) L di dt Ri(t) 1 C t 0 i() d E(t) L di dt, Ri(t), and 1 C t 0 i() d 3t2 t3 1 2e t. f(t) 3 1 2! s3 1 3! s4 1 1 s 2 1 1 s 1 F(s) 6 s3 6 s4 1 s 2 s 1 F(s) 3 2 s3 1 s 1 F(s) 1 s 1 {et} 1>(s 1) { f(t)} F(s) f(t) 3t2 e t t 0 f() et d for f(t) EXAMPLE 6 An Integrodifferential Equation Determine the current i(t) in a single-loop LRC-series circuit when L 0.1 h, R 2 , C 0.1 f, i(0) 0, and the impressed voltage is . E(t) 120t 120t (t 1) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 306 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.4.3 Graph of current i(t) in Example 6 1 0.5 2 1.5 2.5 20 10 _30 _20 _10 t i SOLUTION With the given data equation (10) becomes Now by (7), I(s)s, where . Thus the Laplace transform of the integrodifferential equation is . ; by (16) of Section 7.3 Multiplying this equation by 10s, using s2 20s 100 (s 10)2, and then solving for I(s) gives . By partial fractions, From the inverse form of the second translation theorem, (15) of Section 7.3, we ﬁnally obtain Written as a piecewise-deﬁned function, the current is Using this last expression and a CAS, we graph i(t) on each of the two intervals and then combine the graphs. Note in Figure 7.4.3 that even though the input E(t) is discontinuous, the output or response i(t) is a continuous function. Post Script—Green’s Functions Redux By applying the Laplace transform to the initial-value problem where a and b are constants, we ﬁnd that the transform of y(t) is where . By rewriting the foregoing transform as the product we can use the inverse form of the convolution theorem (4) to write the solution of the IVP as (11) y(t) t 0 g(t ) f()d, Y(s) 1 s2 as bF(s) F(s) {f(t)} Y(s) F(s) s2 as b, y ay by f(t), y(0) 0, y(0) 0, i(t) 12 12e 10t 120te 10t, 12e 10t 12e 10(t 1) 120te 10t 1080(t 1)e 10(t 1), 0 t 1 t 1. 120te 10t 1080(t 1)e 10(t 1) (t 1). i(t) 12[1 (t 1)] 12[e 10t e 10(t 1)(t 1)] 1>100 s 10 e s 1>10 (s 10)2 e s 1 (s 10)2 e s . I(s) 1200 1>100 s 1>100 s 10 1>10 (s 10)2 1>100 s e s I(s) 1200 1 s(s 10)2 1 s(s 10)2 e s 1 (s 10)2 e s 0.1sI(s) 2I(s) 10 I(s) s 120 1 s2 1 s2 e s 1 s e s I(s) {i(t)} {t 0 i() d} 0.1 di dt 2i 10 t 0 i() d 120t 120t (t 1). Optional material if Section 4.8 was covered. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 307 where and On the other hand, we know from (10) of Section 4.8 that the solution of the IVP is also given by (12) where is the Green’s function for the differential equation. By comparing (11) and (12) we see that the Green’s function for the differential equation is related to by (13) For example, for the initial-value problem we ﬁnd Thus from (13) we see that the Green’s function for the DE is See Example 4 in Section 4.8. 7.4.3 TRANSFORM OF A PERIODIC FUNCTION Periodic Function If a periodic function has period T, T 0, then G(t, ) g(t ) 1 2 sin 2(t ). y 4y f(t) 1 1 s2 4 1 2 sin 2t g(t). y 4y f(t), y(0) 0, y(0) 0 G(t, ) g(t ). 1 1 s2 as b g(t) G(t, t) y(t) t 0 G(t, ) f() d, 1{F(s)} f(t). 1 1 s2 as b g(t) In Example 4 of Section 4.8, the roles of the symbols x and t are played by t and in this discussion. THEOREM 7.4.3 Transform of a Periodic Function If f(t) is piecewise continuous on [0, ), of exponential order, and periodic with period T, then { f (t)} 1 1 e sT T 0 e st f (t) dt. PROOF Write the Laplace transform of f as two integrals: . When we let t u T, the last integral becomes . Therefore . Solving the equation in the last line for proves the theorem. { f(t)} { f(t)} T 0 e st f(t) dt e sT { f(t)} T e st f (t) dt 0 e s(uT ) f (u T ) du e sT 0 e su f (u) du e sT { f (t)} { f(t)} T 0 e st f(t) dt T e st f(t) dt EXAMPLE 7 Transform of a Periodic Function Find the Laplace transform of the periodic function shown in Figure 7.4.4. SOLUTION The function E(t) is called a square wave and has period T 2. For 0 t 2, E(t) can be deﬁned by E(t) 1, 0, 0 t 1 1 t 2 t E(t) 1 4 3 2 1 FIGURE 7.4.4 Square wave in Example 7 f (t T ) f (t). The next theorem shows that the Laplace transform of a periodic function can be obtained by integration over one period. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 308 ● CHAPTER 7 THE LAPLACE TRANSFORM From we can then rewrite (16) as 1 s(s R>L) L>R s L>R s R>L . 1 R 1 s e s s e 2s s e 3s s 1 R 1 s R>L 1 s R>L e s e 2s s R>L e 3s s R>L I(s) 1 R 1 s 1 s R>L(1 e s e 2s e 3s ) By applying the form of the second translation theorem to each term of both series, we obtain 1 R (e Rt/L e R(t 1)/L (t 1) e R(t 2)/L (t 2) e R(t 3)/L (t 3) ) i(t) 1 R (1 (t 1) (t 2) (t 3) ) or, equivalently, To interpret the solution, let us assume for the sake of illustration that R 1, L 1, and 0 t 4. In this case i(t) 1 R (1 e Rt/L) 1 R n1 ( 1)n (1 e R(t n)/L) (t n). ; i(t) 1 e t (1 et 1) (t 1) (1 e (t 2)) (t 2) (1 e (t 3)) (t 3) and outside the interval by E(t 2) E(t). Now from Theorem 7.4.3 . (14) 1 s(1 e s) ; 1 e 2s (1 e s)(1 e s) 1 1 e 2s 1 e s s {E(t)} 1 1 e 2s 2 0 e st E(t) dt 1 1 e 2s 1 0 e st 1dt 2 1 e st 0 dt 1 1 x 1 x x2 x3 becomes 1 1 e s 1 e s e 2s e 3s . EXAMPLE 8 A Periodic Impressed Voltage The differential equation for the current i(t) in a single-loop LR-series circuit is . (15) Determine the current i(t) when i(0) 0 and E(t) is the square wave function shown in Figure 7.4.4. SOLUTION If we use the result in (14) of the preceding example, the Laplace trans-form of the DE is . (16) To ﬁnd the inverse Laplace transform of the last function, we ﬁrst make use of geo-metric series. With the identiﬁcation x e s, s 0, the geometric series LsI(s) RI(s) 1 s(1 e s) or I(s) 1>L s(s R>L) 1 1 e s L di dt Ri E(t) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 309 in other words, The graph of i(t) for 0 t 4, given in Figure 7.4.5, was obtained with the help of a CAS. i(t) 1 e t, e t e (t 1), 1 e t e (t 1) e (t 2), e t e (t 1) e (t 2) e (t 3), 0 t 1 1 t 2 2 t 3 3 t 4. 2 1 3 4 2 1.5 1 0.5 t i FIGURE 7.4.5 Graph of current i(t) in Example 8 EXERCISES 7.4 Answers to selected odd-numbered problems begin on page ANS-12. 7.4.1 DERIVATIVES OF A TRANSFORM In Problems 1–8 use Theorem 7.4.1 to evaluate the given Laplace transform. 1. 2. 3. 4. 5. 6. 7. 8. In Problems 9–14 use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. 9. y y t sin t, y(0) 0 10. y y tet sin t, y(0) 0 11. y 9y cos 3t, y(0) 2, y(0) 5 12. y y sin t, y(0) 1, y(0) 1 13. y 16y f(t), y(0) 0, y(0) 1, where 14. y y f(t), y(0) 1, y(0) 0, where In Problems 15 and 16 use a graphing utility to graph the indicated solution. 15. y(t) of Problem 13 for 0 t 2p 16. y(t) of Problem 14 for 0 t 3p In some instances the Laplace transform can be used to solve linear differential equations with variable monomial coefﬁ-cients. In Problems 17 and 18 use Theorem 7.4.1 to reduce the given differential equation to a linear ﬁrst-order DE in the transformed function . Solve the ﬁrst-order DE for Y(s) and then ﬁnd . 17. ty y 2t2, y(0) 0 18. 2y ty 2y 10, y(0) y(0) 0 y(t) 1{Y(s)} Y(s) {y(t)} f(t) 1, sin t, 0 t >2 t >2 f(t) cos 4t, 0, 0 t t {te 3t cos 3t} {te2t sin 6t} {t2 cos t} {t2 sinh t} {t sinh 3t} {t cos 2t} {t3et} {te 10t} 7.4.2 TRANSFORMS OF INTEGRALS In Problems 19–30 use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. In Problems 31–34 use (8) to evaluate the given inverse transform. 31. 32. 33. 34. 35. The table in Appendix III does not contain an entry for . (a) Use (4) along with the results in (5) to evaluate this inverse transform. Use a CAS as an aid in evaluating the convolution integral. (b) Reexamine your answer to part (a). Could you have obtained the result in a different manner? 36. Use the Laplace transform and the results of Problem 35 to solve the initial-value problem . Use a graphing utility to graph the solution. y y sin t t sin t, y(0) 0, y(0) 0 1 8k3s (s2 k2)3 1 1 s(s a)2 1 1 s3(s 1) 1 1 s2(s 1) 1 1 s(s 1) t t 0 e d t t 0 sin d t 0 sin cos (t ) d t 0 et d t 0 sin d t 0 e cos d t 0 cos d t 0 e d {e2t sin t} {e t et cos t} {t2 tet} {1 t3} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 310 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.4.7 Graph for Problem 50 FIGURE 7.4.8 Graph for Problem 51 FIGURE 7.4.9 Graph for Problem 52 FIGURE 7.4.10 Graph for Problem 53 1 square wave t 2a a f(t) 3a 4a sawtooth function t 2b b a f(t) 3b 4b 1 triangular wave t 2 f(t) 3 4 1 1 full-wave rectification of sin t t f(t) 4 3 2 π π π π 50. 51. 52. 53. FIGURE 7.4.11 Graph for Problem 54 4 3 2 π π π π 1 half-wave rectification of sin t t f(t) 54. In Problems 55 and 56 solve equation (15) subject to i(0) 0 with E(t) as given. Use a graphing utility to graph the solution for 0 t 4 in the case when L 1 and R 1. 55. E(t) is the meander function in Problem 49 with amplitude 1 and a 1. 56. E(t) is the sawtooth function in Problem 51 with amplitude 1 and b 1. In Problems 37–46 use the Laplace transform to solve the given integral equation or integrodifferential equation. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. In Problems 47 and 48 solve equation (10) subject to i(0) 0 with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for 0 t 3. 47. L 0.1 h, R 3 , C 0.05 f, 48. L 0.005 h, R 1 , C 0.02 f, 7.4.3 TRANSFORM OF A PERIODIC FUNCTION In Problems 49–54 use Theorem 7.4.3 to ﬁnd the Laplace transform of the given periodic function. E(t) 100[t (t 1)(t 1)] E(t) 100[(t 1) (t 2)] dy dt 6y(t) 9 t 0 y() d 1, y(0) 0 y(t) 1 sin t t 0 y() d, y(0) 0 t 2 f (t) t 0 (e e ) f (t ) d f (t) 1 t 8 3 t 0 ( t)3 f () d f (t) cos t t 0 e f (t ) d f (t) t 0 f () d 1 f (t) 2 t 0 f () cos (t ) d 4e t sin t f (t) tet t 0 f (t ) d f (t) 2t 4 t 0 sin f (t ) d f (t) t 0 (t ) f () d t 49. FIGURE 7.4.6 Graph for Problem 49 1 meander function t 2a a f(t) 3a 4a 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.4 OPERATIONAL PROPERTIES II ● 311 In Problems 57 and 58 solve the model for a driven spring/ mass system with damping where the driving function f is as speciﬁed. Use a graphing utility to graph x(t) for the indicated values of t. 57. , b 1, k 5, f is the meander function in Problem 49 with amplitude 10, and a p, 0 t 2p. 58. m 1, b 2, k 1, f is the square wave in Problem 50 with amplitude 5, and a p, 0 t 4p. Discussion Problems 59. Discuss how Theorem 7.4.1 can be used to ﬁnd . 60. In Section 6.4 we saw that ty y ty 0 is Bessel’s equation of order n 0. In view of (22) of that section and Table 6.4.1 a solution of the initial-value problem ty y ty 0, y(0) 1, y(0) 0, is y J0(t). Use this result and the procedure outlined in the instructions to Problems 17 and 18 to show that . [Hint: You might need to use Problem 46 in Exercises 7.2.] 61. (a) Laguerre’s differential equation ty (1 t)y ny 0 is known to possess polynomial solutions when n is a nonnegative integer. These solutions are naturally called Laguerre polynomials and are denoted by Ln(t). Find y Ln(t), for n 0, 1, 2, 3, 4 if it is known that Ln(0) 1. (b) Show that , where and y Ln(t) is a polynomial solution of the DE in part (a). Conclude that . This last relation for generating the Laguerre poly-nomials is the analogue of Rodrigues’ formula for the Legendre polynomials. See (33) in Section 6.4. 62. The Laplace transform exists, but without ﬁnd-ing it solve the initial-value problem 63. Solve the integral equation f (t) et et t 0 e t f () d. y(0) 0, y(0) 0. y y e t2, {e t2} Ln(t) et n! dn dtn tne t, n 0, 1, 2, . . . Y(s) {y} et n! dn dtn tne t Y(s) {J0(t)} 1 1s2 1 1ln s 3 s 1 m 1 2 m d 2x dt2 dx dt kx f (t), x(0) 0, x(0) 0, 64. (a) Show that the square wave function E(t) given in Figure 7.4.4 can be written (b) Obtain (14) of this section by taking the Laplace transform of each term in the series in part (a). 65. Use the Laplace transform as an aide in evaluating the improper integral 66. If we assume that exists and , then Use this result to ﬁnd the Laplace transform of the given function. The symbols a and k are positive constants. (a) (b) 67. Transform of the Logarithm Because has an inﬁnite discontinuity at it might be assumed that does not exist; however, this is incorrect. The point of this problem to guide you through the formal steps leading to the Laplace transform of (a) Use integration by parts to show that (b) If , use Theorem 7.4.1 with to show that part (a) becomes Find an explicit solution of the foregoing dif-ferential equation. (c) Finally, the integral deﬁnition of Euler’s constant (sometimes called the Euler-Mascheroni constant) is , where . . . . Use in the solution in part (b) to show that Computer Lab Assignments 68. In this problem you are led through the commands in Mathematica that enable you to obtain the symbolic Laplace transform of a differential equation and the so-lution of the initial-value problem by ﬁnding the inverse transform. In Mathematica the Laplace transform of a function y(t) is obtained using LaplaceTransform [y[t], t, s]. In line two of the syntax we replace LaplaceTransform [y[t], t, s] by the symbol Y. (If you {ln t} s ln s s , s 0. Y(1) 0.5772156649 0 e t ln t dt Y(s) sdY ds Y 1 s. n 1 {ln t} Y(s) {ln t} s {t ln t} 1 s. f(t) ln t, t 0. {ln t} t 0 f(t) lnt f(t) 2(1 cos kt) t f(t) sin at t f(t) t s F(u)du. {f(t)} F(s) {f(t)>t} 0 te 2t sin 4t dt. E(t) k0 ( 1)k (t k). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 312 ● CHAPTER 7 THE LAPLACE TRANSFORM do not have Mathematica, then adapt the given pr oce-dure by finding the cor esponding syntax for the CAS you have on hand.) Consider the initial-value problem . Load the Laplace transform package. Precisely reproduce and then, in turn, execute each line in the following sequence of commands. Either copy the output by hand or print out the results. diffequat y[t] 6y[t] 9y[t] t Sin[t] transformdeq LaplaceTransform [diffequat, t, s] /. {y 2, y 1, LaplaceTransform [y[t], t, s] Y} soln Solve[transformdeq, Y]//Flatten Y Y/.soln InverseLaplaceTransform[Y, s, t] y 6y 9y t sin t, y(0) 2, y(0) 1 69. Appropriately modify the procedure of Problem 68 to ﬁnd a solution of . 70. The charge q(t) on a capacitor in an LC-series circuit is given by . Appropriately modify the procedure of Problem 68 to ﬁnd q(t). Graph your solution. q(0) 0, q(0) 0 d 2q dt2 q 1 4(t ) 6(t 3 ), y(0) 0, y(0) 0, y(0) 1 y 3y 4y 0, Unit Impulse Mechanical systems are often acted on by an external force (or electromotive force in an electrical circuit) of large magnitude that acts only for a very short period of time. For example, a vibrating airplane wing could be struck by lightning, a mass on a spring could be given a sharp blow by a ball peen hammer, and a ball (baseball, golf ball, tennis ball) could be sent soaring when struck violently by some kind of club (baseball bat, golf club, tennis racket). See Figure 7.5.1. The graph of the piecewise-deﬁned function (1) a 0, t0 0, shown in Figure 7.5.2(a), could serve as a model for such a force. For a small value of a, da (t t0) is essentially a constant function of large magnitude that is “on” for just a very short period of time, around t0. The behavior of da(t t0) as is illustrated in Figure 7.5.2(b). The function da(t t0) is called a unit impulse, because it possesses the integration property . Dirac Delta Function In practice it is convenient to work with another 0 a(t t0) dt 1 a : 0 a(t t0) 0, 1 2a, 0, 0 t t0 a t0 a t t0 a t t0 a, FIGURE 7.5.1 A golf club applies a force of large magnitude on the ball for a very short period of time THE DIRAC DELTA FUNCTION INTRODUCTION In the last paragraph on page 279, we indicated that as an immediate conse-quence of Theorem 7.1.3, F(s) 1 cannot be the Laplace transform of a function f that is piecewise continuous on [0, ) and of exponential order. In the discussion that follows we are going to intro-duce a function that is very different from the kinds that you have studied in previous courses. We shall see that there does indeed exist a function—or, more precisely, a generalized function—whose Laplace transform is F(s) 1. 7.5 type of unit impulse, a “function” that approximates da(t t0) and is defined by the limit (2) (t t0) lim a : 0a(t t0). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.5 THE DIRAC DELTA FUNCTION ● 313 The latter expression, which is not a function at all, can be characterized by the two properties . The unit impulse d(t t0) is called the Dirac delta function. It is possible to obtain the Laplace transform of the Dirac delta function by the for-mal assumption that . {(t t0)} lima : 0 {a(t t0)} (i) (t t0) , 0, t t0 t t0 and (ii) 0 (t t0) dt 1 FIGURE 7.5.2 Unit impulse (b) behavior of a as a 0 t t0 y t t0 −a 2a 1/2a t0 y t0 + a (a) graph of a(t t0) EXAMPLE 1 Two Initial-Value Problems Solve y y 4 d(t 2p) subject to (a) y(0) 1, y(0) 0 (b) y(0) 0, y(0) 0. The two initial-value problems could serve as models for describing the motion of a mass on a spring moving in a medium in which damping is negligible. At t 2p the mass is given a sharp blow. In (a) the mass is released from rest 1 unit below the equilibrium position. In (b) the mass is at rest in the equilibrium position. SOLUTION (a) From (3) the Laplace transform of the differential equation is . Using the inverse form of the second translation theorem, we ﬁnd . Since sin(t 2p) sin t, the foregoing solution can be written as (5) y(t) cos t, 0 t 2 cos t 4 sin t, t 2 . y(t) cos t 4 sin (t 2 ) (t 2 ) s2Y(s) s Y(s) 4e 2 s or Y(s) s s2 1 4e 2 s s2 1 THEOREM 7.5.1 Transform of the Dirac Delta Function For t0 0, (3) {(t t0)} e st0. PROOF To begin, we can write da(t t0 ) in terms of the unit step function by virtue of (11) and (12) of Section 7.3: By linearity and (14) of Section 7.3 the Laplace transform of this last expression is (4) Since (4) has the indeterminate form 00 as , we apply L’Hôpital’s Rule: . Now when t0 0, it seems plausible to conclude from (3) that The last result emphasizes the fact that d(t) is not the usual type of function that we have been considering, since we expect from Theorem 7.1.3 that {f(t)} : 0 as s : . {(t)} 1. {(t t0)} lim a : 0 {a(t t0)} e st0 lim a : 0 esa e sa 2sa e st0 a : 0 {a(t t0)} 1 2a e s(t0 a) s e s(t0a) s e st0 esa e sa 2sa . a(t t0) 1 2a [(t (t0 a)) (t (t0 a))]. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 314 ● CHAPTER 7 THE LAPLACE TRANSFORM In Figure 7.5.3 we see from the graph of (5) that the mass is exhibiting simple harmonic motion until it is struck at t 2p. The inﬂuence of the unit impulse is to increase the amplitude of vibration to for t 2p. (b) In this case the transform of the equation is simply and so (6) The graph of (6) in Figure 7.5.4 shows, as we would expect from the initial conditions that the mass exhibits no motion until it is struck at t 2p. REMARKS (i) If d(t t0) were a function in the usual sense, then property (i) on page 313 would imply rather than . Because the Dirac delta function did not “behave” like an ordinary function, even though its users produced correct results, it was met initially with great scorn by mathe-maticians. However, in the 1940s Dirac’s controversial function was put on a rigorous footing by the French mathematician Laurent Schwartz in his book La Théorie de distribution, and this, in turn, led to an entirely new branch of mathematics known as the theory of distributions or generalized functions. In this theory (2) is not an accepted deﬁnition of d(t t0), nor does one speak of a function whose values are either or 0. Although we shall not pursue this topic any further, sufﬁce it to say that the Dirac delta function is best character-ized by its effect on other functions. If f is a continuous function, then (7) can be taken as the definitio of d(t t0). This result is known as the sifting property, since d(t t0) has the effect of sifting the value f (t0) out of the set of values of f on [0, ). Note that property (ii) (with f(t) 1) and (3) (with f(t) e st) are consistent with (7). (ii) In (iii) in the Remarks at the end of Section 7.2 we indicated that the trans-fer function of a general linear nth-order differential equation with constant coefﬁcients is W(s) 1P(s), where . The transfer function is the Laplace transform of function w(t), called the weight function of a linear system. But w(t) can also be characterized in terms of the discussion at hand. For simplicity let us consider a second-order linear system in which the input is a unit impulse at t 0: . Applying the Laplace transform and using shows that the trans-form of the response y in this case is the transfer function {(t)} 1 a2y a1y a0y (t), y(0) 0, y(0) 0 P(s) ansn an 1sn 1 a0 0 f(t) (t t0) dt f(t0) 0 (t t0) dt 1 0 (t t0) dt 0 0, 0 t 2 4 sin t, t 2 . y(t) 4 sin (t 2 ) (t 2 ) Y(s) 4e 2 s s2 1, 117 FIGURE 7.5.4 No motion until mass is struck at t 2p in part (b) of Example 1 FIGURE 7.5.3 Mass is struck at t 2p in part (a) of Example 1 t y 1 −1 2 4 π π t y 1 −1 2 4 π π . Y(s) 1 a2s2 a1s a0 1 P(s) W(s) and so y 1 1 P(s) w(t) From this we can see, in general, that the weight function y w(t) of an nth-order linear system is the zero-state response of the system to a unit impulse. For this reason w(t) is also called the impulse response of the system. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.6 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS ● 315 Coupled Springs Two masses m1 and m2 are connected to two springs A and B of negligible mass having spring constants k1 and k2, respectively. In turn the two springs are attached as shown in Figure 7.6.1 on page 316. Let x1(t) and x2(t) denote the vertical displacements of the masses from their equilibrium positions. When the system is in motion, spring B is subject to both an elongation and a compression; hence its net elongation is x2 x1. Therefore it follows from Hooke’s law that springs A and B exert forces k1x1 and k2(x2 x1), respectively, on m1. If no exter-nal force is impressed on the system and if no damping force is present, then the net force on m1 is k1x1 k2(x2 x1). By Newton’s second law we can write . m1 d 2x1 dt2 k 1x1 k2(x2 x1) EXERCISES 7.5 Answers to selected odd-numbered problems begin on page ANS-13. In Problems 1–12 use the Laplace transform to solve the given initial-value problem. 1. y 3y d(t 2), y(0) 0 2. y y d(t 1), y(0) 2 3. y y d(t 2p), y(0) 0, y(0) 1 4. y 16y d(t 2p), y(0) 0, y(0) 0 5. 6. y y d(t 2p) d(t 4p), y(0) 1, y(0) 0 7. y 2y d(t 1), y(0) 0, y(0) 1 8. y 2y 1 d(t 2), y(0) 0, y(0) 1 9. y 4y 5y d(t 2p), y(0) 0, y(0) 0 10. y 2y y d(t 1), y(0) 0, y(0) 0 11. y 4y 13y d(t p) d(t 3p), y(0) 1, y(0) 0 12. y 7y 6y et d(t 2) d(t 4), y(0) 0, y(0) 0 13. A uniform beam of length L carries a concentrated load w0 at . The beam is embedded at its left end and x 1 2L y(0) 0, y(0) 0 y y (t 1 2 ) (t 3 2 ), is free at its right end. Use the Laplace transform to determine the deﬂection y(x) from where y(0) 0, y(0) 0, y(L) 0, and y(L) 0. 14. Solve the differential equation in Problem 13 subject to y(0) 0, y(0) 0, y(L) 0, y(L) 0. In this case the beam is embedded at both ends. See Figure 7.5.5. EI d 4y dx4 w0 x 1 2L, FIGURE 7.5.5 Beam in Problem 14 x y L w0 Discussion Problems 15. Someone tells you that the solutions of the two IVPs are exactly the same. Do you agree or disagree? Defend your answer. y 2y 10y 0, y 2y 10y (t), y(0) 0, y(0) 1 y(0) 0, y(0) 0 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS REVIEW MATERIAL ●Solving systems of two equations in two unknowns INTRODUCTION When initial conditions are speciﬁed, the Laplace transform of each equation in a system of linear differential equations with constant coefﬁcients reduces the system of DEs to a set of simultaneous algebraic equations in the transformed functions. We solve the system of algebraic equations for each of the transformed functions and then ﬁnd the inverse Laplace trans-forms in the usual manner. 7.6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 316 ● CHAPTER 7 THE LAPLACE TRANSFORM 5 2.5 10 7.5 15 12.5 _0.4 0.2 0.4 _0.2 t x1 (a) plot of x1(t) (b) plot of x2(t) 5 2.5 10 7.5 15 12.5 _0.4 0.2 0.4 _0.2 t x2 FIGURE 7.6.2 Displacements of the two masses in Example 1 Similarly, the net force exerted on mass m2 is due solely to the net elongation of B; that is, k2(x2 x1). Hence we have . In other words, the motion of the coupled system is represented by the system of simultaneous second-order differential equations (1) In the next example we solve (1) under the assumptions that k1 6, k2 4, m1 1, m2 1, and that the masses start from their equilibrium positions with opposite unit velocities. m2x 2 k2(x2 x1). m1x 1 k1x1 k2(x2 x1) m2 d 2x2 dt2 k2(x2 x1) m2 k1 k2 k1 k (x2 − x1) 2 k (x2 − x1) 2 x2 x1 = 0 x2 = 0 x1 x1 A m1 B m1 m2 m2 (a) equilibrium (b) motion (c) forces m1 FIGURE 7.6.1 Coupled spring/mass system EXAMPLE 1 Coupled Springs Solve (2) subject to SOLUTION The Laplace transform of each equation is where . The preceding system is the same as (3) Solving (3) for X1(s) and using partial fractions on the result yields and therefore Substituting the expression for X1(s) into the ﬁrst equation of (3) gives and 12 5 sin 12t 13 10 sin 213t. x2(t) 2 512 1 12 s2 2 3 5112 1 112 s2 12 X2(s) s2 6 (s2 2)(s2 12) 2> 5 s2 2 3> 5 s2 12 12 10 sin 12t 13 5 sin 213t. x1(t) 1 512 1 12 s2 2 6 5112 1 112 s2 12 X1(s) s2 (s2 2)(s2 12) 1>5 s2 2 6>5 s2 12, 4 X1(s) (s2 4) X2(s) 1. (s2 10) X1(s) 4X2(s) 1 X1(s) {x1(t)} and X2(s) {x2(t)} 4X1(s) s2X2(s) sx2(0) x2 (0) 4X2(s) 0, s2X1(s) sx1(0) x1 (0) 10X1(s) 4X2(s) 0 x1(0) 0, x 1(0) 1, x2(0) 0, x 2(0) 1. 4x 1 x 2 4x2 0 x 1 10x1 4x2 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.6 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS ● 317 Finally, the solution to the given system (2) is (4) The graphs of x1 and x2 in Figure 7.6.2 reveal the complicated oscillatory motion of each mass. Networks In (18) of Section 3.3 we saw the currents i1(t) and i2(t) in the network shown in Figure 7.6.3, containing an inductor, a resistor, and a capacitor, were governed by the system of ﬁrst-order differential equations (5) We solve this system by the Laplace transform in the next example. RC di2 dt i2 i1 0. L di1 dt Ri2 E(t) x2(t) 12 5 sin 12t 13 10 sin 2 13t. x1(t) 12 10 sin 12t 13 5 sin 2 13t FIGURE 7.6.3 Electrical network R i1 L i2 i3 C E EXAMPLE 2 An Electrical Network Solve the system in (5) under the conditions E(t) 60 V, L 1 h, R 50 , C 10 4 f, and the currents i1 and i2 are initially zero. SOLUTION We must solve subject to i1(0) 0, i2(0) 0. Applying the Laplace transform to each equation of the system and simplifying gives where and . Solving the system for I1 and I2 and decomposing the results into partial fractions gives Taking the inverse Laplace transform, we ﬁnd the currents to be i2(t) 6 5 6 5 e 100t 120te 100t. i1(t) 6 5 6 5 e 100t 60te 100t I2(s) 12,000 s(s 100)2 6>5 s 6>5 s 100 120 (s 100)2. I1(s) 60s 12,000 s(s 100)2 6>5 s 6>5 s 100 60 (s 100)2 I2(s) {i2(t)} I1(s) {i1(t)} 200I1(s) (s 200)I2(s) 0, sI1(s) 50I2(s) 60 s 50(10 4) di2 dt i2 i1 0 di1 dt 50i2 60 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 318 ● CHAPTER 7 THE LAPLACE TRANSFORM Note that both i1(t) and i2(t) in Example 2 tend toward the value as t : . Furthermore, since the current through the capacitor is i3(t) i1(t) i2(t) 60te 100t, we observe that as . Double Pendulum Consider the double-pendulum system consisting of a pendulum attached to a pendulum shown in Figure 7.6.4. We assume that the system oscillates in a vertical plane under the inﬂuence of gravity, that the mass of each rod is negligible, and that no damping forces act on the system. Figure 7.6.4 also shows that the displacement angle u1 is measured (in radians) from a vertical line extending downward from the pivot of the system and that u2 is measured from a vertical line extending downward from the center of mass m1. The positive direction is to the right; the negative direction is to the left. As we might expect from the analysis lead-ing to equation (6) of Section 5.3, the system of differential equations describing the motion is nonlinear: t : i3(t) : 0 E>R 6 5 FIGURE 7.6.4 Double pendulum 1 θ 2 θ l1 m1 m2 l2 (6) m2l2 22 m2l1l21 cos (1 2) m2l1l2(1 )2 sin (1 2) m2l2g sin 2 0. (m1 m2)l1 21 m2l1l22 cos (1 2) m2l1l2(2 )2 sin (1 2) (m1 m2)l1g sin 1 0 But if the displacements u1(t) and u2(t) are assumed to be small, then the approximations cos(u1 u2) 1, sin(u1 u2) 0, sin u1 u1, sin u2 u2 enable us to replace system (6) by the linearization (7) m2l2 22 m2l1l21 m2l2g2 0. (m1 m2)l1 21 m2l1l22 (m1 m2)l1g1 0 (a) t 0 (b) t 1.4 (c) t 2.5 (d) t 8.5 FIGURE 7.6.5 Positions of masses on double pendulum at various times in Example 3 EXAMPLE 3 Double Pendulum It is left as an exercise to ﬁll in the details of using the Laplace transform to solve system (7) when m1 3, m2 1, l1 l2 16, u1(0) 1, u2(0) 1, , and . You should ﬁnd that (8) With the aid of a CAS the positions of the two masses at t 0 and at subsequent times are shown in Figure 7.6.5. See Problem 21 in Exercises 7.6. 2(t) 1 2 cos 2 13 t 3 2 cos 2t. 1(t) 1 4 cos 2 13 t 3 4 cos 2t 2(0) 0 1(0) 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7.6 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS ● 319 EXERCISES 7.6 Answers to selected odd-numbered problems begin on page ANS-13. In Problems 1–12 use the Laplace transform to solve the given system of differential equations. 1. 2. x(0) 0, y(0) 1 x(0) 1, y(0) 1 3. 4. x(0) 1, y(0) 2 x(0) 0, y(0) 0 5. x(0) 0, y(0) 0 6. x(0) 0, y(0) 1 7. 8. x(0) 0, x(0) 2, x(0) 1, x(0) 0, y(0) 0, y(0) 1 y(0) 1, y(0) 5 9. 10. x(0) 8, x(0) 0, x(0) 0, y(0) 0, y(0) 0, y(0) 0 y(0) 0, y(0) 0 11. x(0) 0, x(0) 2, y(0) 0 12. x(0) 0, y(0) 1 2 dy dt 3x y (t 1) dx dt 4x 2y 2(t 1) d 2x dt2 3y te t d 2x dt2 3 dy dt 3y 0 dx dt 2x 2 d 3y dt3 0 d 2x dt2 d 2y dt2 4t dx dt 4x d 3y dt3 6 sin t d 2x dt2 d 2y dt2 t2 d 2y dt2 dy dt 4 dx dt 0 d 2y dt2 y x 0 d 2x dt2 dx dt dy dt 0 d 2x dt2 x y 0 dx dt dy dt 2y 0 dx dt x dy dt y 0 dx dt dy dt 3x 3y 2 2 dx dt dy dt 2x 1 dx dt x dy dt y et dy dt 5x y dx dt 3x dy dt 1 dx dt x 2y dy dt 8x t dy dt 2x dx dt 2y et dx dt x y 13. Solve system (1) when k1 3, k2 2, m1 1, m2 1 and x1(0) 0, , x2(0) 1, . 14. Derive the system of differential equations describing the straight-line vertical motion of the coupled springs shown in Figure 7.6.6. Use the Laplace transform to solve the system when k1 1, k2 1, k3 1, m1 1, m2 1 and x1(0) 0, , x2(0) 0, . x 2(0) 1 x 1(0) 1 x 2(0) 0 x 1(0) 1 k m2 k2 3 x2 = 0 m1 k1 x1 = 0 FIGURE 7.6.6 Coupled springs in Problem 14 15. (a) Show that the system of differential equations for the currents i2(t) and i3(t) in the electrical network shown in Figure 7.6.7 is (b) Solve the system in part (a) if R 5 , L1 0.01 h, L2 0.0125 h, E 100 V, i2(0) 0, and i3(0) 0. (c) Determine the current i1(t). L2 di3 dt Ri2 Ri3 E(t). L1 di2 dt Ri2 Ri3 E(t) FIGURE 7.6.7 Network in Problem 15 L1 R E i1 i2 i3 L2 16. (a) In Problem 12 in Exercises 3.3 you were asked to show that the currents i2(t) and i3(t) in the electrical network shown in Figure 7.6.8 satisfy R1 di2 dt R2 di3 dt 1 C i3 0. L di2 dt L di3 dt R1i2 E(t) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 320 ● CHAPTER 7 THE LAPLACE TRANSFORM Solve the system if R1 10 , R2 5 , L 1 h, C 0.2 f, i2(0) 0, and i3(0) 0. (b) Determine the current i1(t). E(t) 120, 0, 0 t 2 t 2, Computer Lab Assignments 21. (a) Use the Laplace transform and the information given in Example 3 to obtain the solution (8) of the system given in (7). (b) Use a graphing utility to graph u1(t) and u2(t) in the tu-plane. Which mass has extreme displacements of greater magnitude? Use the graphs to estimate the ﬁrst time that each mass passes through its equilib-rium position. Discuss whether the motion of the pendulums is periodic. (c) Graph u1(t) and u2(t) in the u1u2-plane as parametric equations. The curve deﬁned by these parametric equations is called a Lissajous curve. (d) The positions of the masses at t 0 are given in Figure 7.6.5(a). Note that we have used 1 radian 57.3°. Use a calculator or a table application in a CAS to construct a table of values of the angles u1 and u2 for t 1, 2, . . . , 10 s. Then plot the positions of the two masses at these times. (e) Use a CAS to ﬁnd the ﬁrst time that u1(t) u2(t) and compute the corresponding angular value. Plot the positions of the two masses at these times. (f) Utilize the CAS to draw appropriate lines to sim-ulate the pendulum rods, as in Figure 7.6.5. Use the animation capability of your CAS to make a “movie” of the motion of the double pendulum from t 0 to t 10 using a time increment of 0.1. [Hint: Express the coordinates (x1(t), y1(t)) and (x2(t), y2(t)) of the masses m1 and m2, respec-tively, in terms of u1(t) and u2(t).] FIGURE 7.6.8 Network in Problem 16 R1 E i1 L i2 i3 C R2 17. Solve the system given in (17) of Section 3.3 when R1 6 , R2 5 , L1 1 h, L2 1 h, E(t) 50 sin t V, i2(0) 0, and i3(0) 0. 18. Solve (5) when E 60 V, , R 50 , C 10 4 f, i1(0) 0, and i2(0) 0. 19. Solve (5) when E 60 V, L 2 h, R 50 , C 10 4 f, i1(0) 0, and i2(0) 0. 20. (a) Show that the system of differential equations for the charge on the capacitor q(t) and the current i3(t) in the electrical network shown in Figure 7.6.9 is (b) Find the charge on the capacitor when L 1 h, R1 1 , R2 1 , C 1 f, i3(0) 0, and q(0) 0. E(t) 0, 50e t, 0 t 1 t 1, L di3 dt R2i3 1 C q 0. R1 dq dt 1 C q R1i3 E(t) L 1 2 h FIGURE 7.6.9 Network in Problem 20 R1 E i1 i2 i3 L C R2 CHAPTER 7 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-13. In Problems 1 and 2 use the deﬁnition of the Laplace transform to ﬁnd . 1. 2. f (t) 0, 1, 0, 0 t 2 2 t 4 t 4 f (t) t, 2 t, 0 t 1 t 1 { f (t)} In Problems 3–24 ﬁll in the blanks or answer true or false. 3. If f is not piecewise continuous on [0, ), then will not exist. _ 4. The function f(t) (et)10 is not of exponential order. _ 5. F(s) s2(s2 4) is not the Laplace transform of a function that is piecewise continuous and of exponential order. _ { f (t)} Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. CHAPTER 7 IN REVIEW ● 321 6. If and , then . _ 7. _ 8. _ 9. _ 10. _ 11. _ 12. _ 13. _ 14. _ 15. _ 16. _ 17. _ 18. _ 19. _ 20. _ 21. exists for s _. 22. If , then _. 23. If and k 0, then _. 24. _ whereas _. In Problems 25–28 use the unit step function to ﬁnd an equation for each graph in terms of the function y f (t), whose graph is given in Figure 7.R.1. {eatt 0 f () d} {t 0 ea f () d} {eatf (t k)(t k)} { f(t)} F(s) {te8t f (t)} { f (t)} F(s) {e 5t} 1 1 L2s2 n2 2 1 s s2 2 e s 1 e 5s s2 1 s s2 10s 29 1 1 s2 5 1 1 (s 5)3 1 1 3s 1 1 20 s6 {sin 2t (t )} {t sin 2t} {e 3t sin 2t} {sin 2t} {te 7t} {e 7t} 1{F(s)G(s)} f (t)g(t) {g(t)} G(s) { f (t)} F(s) In Problems 29–32 express f in terms of unit step functions. Find and . {etf (t)} { f (t)} FIGURE 7.R.2 Graph for Problem 25 t0 t y t0 t y y = f(t) FIGURE 7.R.1 Graph for Problems 25–28 25. FIGURE 7.R.3 Graph for Problem 26 FIGURE 7.R.4 Graph for Problem 27 FIGURE 7.R.5 Graph for Problem 28 t0 t y t0 t y t0 t y t1 26. 27. 28. 29. 30. 31. 32. FIGURE 7.R.6 Graph for Problem 29 FIGURE 7.R.7 Graph for Problem 30 FIGURE 7.R.8 Graph for Problem 31 FIGURE 7.R.9 Graph for Problem 32 1 1 2 3 4 t f(t) 2 1 −1 t f(t) π π π 3 π π y = sin 3 t, ≤t ≤ 1 2 3 2 1 t f(t) (3, 3) 1 2 1 t f(t) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In Problems 33–40 use the Laplace transform to solve the given equation. 33. y 2y y et, y(0) 0, y(0) 5 34. y 8y 20y tet, y(0) 0, y(0) 0 35. y 6y 5y t t (t 2), y(0) 1, y(0) 0 36. y 5y f (t), where 37. , where is given in Figure 7.R.10. f(t) y 2y f(t), y(0) 1 f (t) t2, 0, 0 t 1 t 1, y(0) 1 45. A uniform cantilever beam of length L is embedded at its left end (x 0) and free at its right end. Find the deflection y(x) if the load per unit length is given by . 46. When a uniform beam is supported by an elastic foundation, the differential equation for its deﬂection y(x) is , where k is the modulus of the foundation and ky is the restoring force of the foundation that acts in the direction opposite to that of the load w(x). See Figure 7.R.11. For algebraic convenience suppose that the differential equa-tion is written as , where a (k4EI)1/4. Assume L p and a 1. Find the deﬂection y(x) of a beam that is supported on an elastic foundation when (a) the beam is simply supported at both ends and a con-stant load w0 is uniformly distributed along its length, (b) the beam is embedded at both ends and w(x) is a concentrated load w0 applied at x p2. [Hint: In both parts of this problem use entries 35 and 36 in the table of Laplace transforms in Appendix III.] d 4y dx4 4a4y w(x) EI EI d 4y dx4 ky w(x) w(x) 2w0 L L 2 x x L 2 x L 2 FIGURE 7.R.11 Beam on elastic foundation in Problem 46 0 x y L w(x) elastic foundation 47. (a) Suppose two identical pendulums are coupled by means of a spring with constant k. See Figure 7.R.12. Under the same assumptions made in the discussion preceding Example 3 in Section 7.6, it can be shown that when the displacement angles u1(t) and u2(t) are small, the system of linear differential equations describing the motion is Use the Laplace transform to solve the system when u1(0) u0, u1 (0) 0, u2(0) c0, u2 (0) 0, where u0 and c0 constants. For convenience let v2 gl, K km. (b) Use the solution in part (a) to discuss the motion of the coupled pendulums in the special case when 2 g l2 k m (1 2). 1 g l1 k m (1 2) 322 ● CHAPTER 7 THE LAPLACE TRANSFORM FIGURE 7.R.10 Graph for Problem 37 1 2 3 1 t f(t) 38. , where 39. 40. In Problems 41 and 42 use the Laplace transform to solve each system. 41. x y t 42. x y e2t 4x y 0 2x y e2t x(0) 1, y(0) 2 x(0) 0, y(0) 0, x(0) 0, y(0) 0 43. The current i(t) in an RC-series circuit can be deter-mined from the integral equation , where E(t) is the impressed voltage. Determine i(t) when R 10 , C 0.5 f, and E(t) 2(t2 t). 44. A series circuit contains an inductor, a resistor, and a capacitor for which , R 10 , and C 0.01 f, respectively. The voltage is applied to the circuit. Determine the instantaneous charge q(t) on the capacitor for t 0 if q(0) 0 and q(0) 0. E(t) 10, 0, 0 t 5 t 5 L 1 2 h Ri 1 C t 0 i() d E(t) t 0 f () f (t ) d 6t3 y(t) cos t t 0 y() cos(t ) d, y(0) 1 f(t) 12 k0 ( 1)k (t k). y 5y 4y f(t), y(0) 0, y(0) 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. the initial conditions are u1(0) u0, u1 (0) 0, u2(0) u0, u2 (0) 0. When the initial conditions are u1(0) u0, u1 (0) 0, u2(0) u0, u2 (0) 0. CHAPTER 7 IN REVIEW ● 323 FIGURE 7.R.12 Coupled pendulums in Problem 47 1 θ θ2 m l l m (f) Show that each successive oscillation is shorter than the preceding one. (g) Predict the long-term behavior of the system. 49. Range of a Projectile—No Air Resistance (a) A pro-jectile, such as the canon ball shown in Fig-ure 7.R.13, has weight and initial velocity v0 that is tangent to its path of motion. If air resis-tance and all other forces except its weight are ig-nored, we saw in Problem 23 of Exercises 4.9 that motion of the projectile is describe by the system of linear differential equations Use the Laplace transform to solve this system sub-ject to the initial conditions , where is con-stant and is the constant angle of elevation shown in Figure 7.R.13. The solutions and are para-metric equations of the trajectory of the projectile. (b) Use in part (a) to eliminate the parameter t in . Use the resulting equation for y to show that the horizontal range R of the projectile is given by (c) From the formula in part (b), we see that R is a max-imum when sin or when Show that the same range—less than the maximum—can be attained by ﬁring the gun at either of two comple-mentary angles and The only difference is that the smaller angle results in a low trajectory whereas the larger angle gives a high trajectory. (d) Suppose and Use part (b) to ﬁnd the horizontal range of the pro-jectile. Find the time when the projectile hits the ground. (e) Use the parametric equations and in part (a) along with the numerical data in part (d) to plot the ballistic curve of the projectile. Repeat with and Superimpose both curves on the same coordinate system. v0 300 ft/s. u 52 y(t) x(t) v0 300 ft/s. g 32 ft/s2, u 38, p>2 u. u u p>4. 2u 1 R v2 0 g sin 2u. y(t) x(t) y(t) x(t) u v0 v0 y(0) 0, y(0) v0 sin u, x(0) 0, x(0) v0 cos u m d 2y dt 2 mg. m d 2x dt 2 0 w mg 2F>v2 48. Coulomb Friction Revisited In Problem 27 in Chap-ter 5 in Review we examined a spring/mass system in which a mass m slides over a dry horizontal surface whose coefﬁcient of kinetic friction is a constant . The constant retarding force of the dry sur-face that acts opposite to the direction of motion is called Coulomb friction after the French physicist Charles-Augustin de Coulomb (1736–1806). You were asked to show that the piecewise-deﬁned differen-tial equation for the displacement of the mass is given by (a) Suppose that the mass is released from rest from a point and that there are no other external forces. Then the differential equations describing the motion of the mass m are , and so on, where , and . Show that the times correspond to . (b) Explain why, in general, the initial displacement must satisfy (c) Explain why the interval is ap-propriately called the “dead zone” of the system. (d) Use the Laplace transform and the concept of the meander function to solve for the displacement (e) Show that in the case and that on the interval your solution agrees with parts (a) and (b) of Problem 28 in Chapter 5 in Review. [0, 2p) x0 5.5 m 1, k 1, fk 1, x(t) for t 0. F>v2 x F>v2 v2 & x0 & F. x(t) 0 T, 3T>2, . . . 0, T>2, T 2p>v g 32 v2 k>m, F fk>m g, x v2x F, T t 3T>2 x v2x F, T>2 t T x v2x F, 0 t T>2 x(0) x0 0 md2x dt2 kx fk, x 0 (motion to left) fk, x 0 (motion to right). x(t) fk mg FIGURE 7.R.13 Projectile in Problem 49 x y R Horizontal Range v0 θ Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 50. Range of a Pr ojectile—With Air Resistance (a) Now suppose that air resistance is a retarding force tan-gent to the path but acts opposite to the motion. If we take air resistance to be proportional to the ve-locity of the projectile, then we saw in Problem 24 of Exercises 4.9 that motion of the projectile is de-scribe by the system of linear differential equations where Use the Laplace transform to solve this system subject to the initial conditions sin , where and are constant. u v0 v0 u cos u, y(0) 0, y(0) v0 x(0) v0 x(0) 0, b 0. m d 2y dt 2 mg b dy dt, m d 2x dt 2 b dx dt 324 ● CHAPTER 7 THE LAPLACE TRANSFORM (b) Suppose slug, and Use a CAS to ﬁnd the time when the projectile hits the ground and then compute its corresponding horizontal range. (c) Repeat part (c) using the complementary angle and compare the range with that found in part (b). Does the property in part (c) of Problem 49 hold? (d) Use the parametric equations and in part (a) along with the numerical data in part (b) to plot the ballistic curve of the projectile. Repeat with the same numerical data in part (b) but take Superimpose both curves on the same coordinate system. Compare these curves with those obtained in part (e) of Problem 49. u 52. y(t) x(t) u 52 v0 300 ft/s. u 38, g 32 ft/s2, b 0.02, m 1 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 325 8.1 Preliminary Theory—Linear Systems 8.2 Homogeneous Linear Systems 8.2.1 Distinct Real Eigenvalues 8.2.2 Repeated Eigenvalues 8.2.3 Complex Eigenvalues 8.3 Nonhomogeneous Linear Systems 8.3.1 Undetermined Coefﬁcients 8.3.2 Variation of Parameters 8.4 Matrix Exponential Chapter 8 in Review We encountered systems of ordinary differential equations in Sections 3.3, 4.9, and 7.6 and were able to solve some of these systems by means of either systematic elimination or by the Laplace transform. In this chapter we are going to concentrate only on systems of linear first-o der differential equations. Although most of the systems that are considered could be solved using elimination or the Laplace transform, we are going to develop a general theory for these kinds of systems and in the case of systems with constant coefﬁcients, a method of solution that utilizes some basic concepts from the algebra of matrices. We will see that this general theory and solution procedure is similar to that of linear higher-order differential equations considered in Chapter 4. This material is fundamental to the analysis of systems of nonlinear ﬁrst-order equations in Chapter 10. Systems of Linear First-Order Differential Equations 8 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 326 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS Linear Systems When each of the functions g1, g2, . . . , gn in (2) is linear in the dependent variables x1, x2, . . . , xn, we get the normal form of a ﬁrst-order system of linear equations: We refer to a system of the form given in (3) simply as a linear system. We assume that the coefﬁcients aij as well as the functions fi are continuous on a common interval I. When fi(t) 0, i 1, 2, . . . , n, the linear system (3) is said to be homogeneous; otherwise, it is nonhomogeneous. Matrix Form of a Linear System If X, A(t), and F(t) denote the respective matrices x1(t) x2(t) xn(t) X ( ) , a11(t) a21(t) an1(t) a1n(t) a2n(t) ann(t) a12(t) a22(t) an2(t) . . . . . . . . . A(t) ( ) , f1(t) f2(t) fn(t) F(t) ( ) , . . . . . . . . . . . . a11(t)x1 a12(t)x2 . . . a1n(t)xn f1(t) a21(t)x1 a22(t)x2 . . . a2n(t)xn f2(t) an1(t)x1 an2(t)x2 . . . ann(t)xn fn(t). dx1 ––– dt dx2 ––– dt dxn ––– dt . . . . . . (3) PRELIMINARY THEORY—LINEAR SYSTEMS REVIEW MATERIAL ●Matrix notation and properties are used extensively throughout this chapter. It is imperative that you review either Appendix II or a linear algebra text if you unfamiliar with these concepts. INTRODUCTION Recall that in Section 4.9 we illustrated how to solve systems of n linear differential equations in n unknowns of the form (1) where the Pij were polynomials of various degrees in the differential operator D. In this chapter we conﬁne our study to systems of ﬁrst-order DEs that are special cases of systems that have the normal form A system such as (2) of n ﬁrst-order equations is called a first-orde system. g1(t, x1, x2, . . . , xn) g2(t, x1, x2, . . . , xn) gn(t, x1, x2, . . . , xn). dx1 ––– dt dx2 ––– dt dxn ––– dt . . . . . . P11(D)x1 P12(D)x2 . . . P1n(D)xn b1(t) P21(D)x1 P22(D)x2 . . . P2n(D)xn b2(t) Pn1(D)x1 Pn2(D)x2 . . . Pnn(D)xn bn(t), . . . . . . 8.1 (2) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. then the system of linear ﬁrst-order differential equations (3) can be written as or simply (4) If the system is homogeneous, its matrix form is then (5) X AX. X AX F. ( d –– dt x1 x2 xn ) a11(t) a21(t) an1(t) a1n(t) a2n(t) ann(t) a12(t) a22(t) an2(t) . . . . . . . . . ( ( x1 x2 xn ) ( ) f1(t) f2(t) fn(t) ) . . . . . . . . . . . . . . . 8.1 PRELIMINARY THEORY—LINEAR SYSTEMS ● 327 EXAMPLE 1 Systems Written in Matrix Notation (a) If , then the matrix form of the homogeneous system (b) If , then the matrix form of the nonhomogeneous system dx dt 6x y z t dy dt 8x 7y z 10t dz dt 2x 9y z 6t is X 6 8 2 1 7 9 1 1 1X t 10t 6t. X x y z dx dt 3x 4y dy dt 5x 7y is X 3 5 4 7X. X x y DEFINITION 8.1.1 Solution Vector A solution vector on an interval I is any column matrix whose entries are differentiable functions satisfying the system (4) on the interval. x1(t) x2(t) xn(t) X ( ) . . . A solution vector of (4) is, of course, equivalent to n scalar equations x1 f1(t), x2 f2(t), . . . , xn fn(t) and can be interpreted geometrically as a set of parametric equations of a space curve. In the important case n 2 the equations x1 f1(t), x2 f2(t) represent a curve in the x1x2-plane. It is common practice to call a curve in the plane a trajectory and to call the x1x2-plane the phase plane. We will come back to these concepts and illustrate them in the next section. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 328 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 2 Verification of Solution Verify that on the interval (, ) are solutions of . (6) SOLUTION From and we see that and Much of the theory of systems of n linear ﬁrst-order differential equations is similar to that of linear nth-order differential equations. Initial-Value Problem Let t0 denote a point on an interval I and where the gi, i 1, 2, . . . , n are given constants. Then the problem (7) is an initial-value problem on the interval. Subject to: X(t0) X0 Solve: X A(t)X F(t) x1(t0) x2(t0) xn(t0) X(t0) ( and ) 1 2 n X0 ( ) , . . . . . . AX2 1 5 3 3 3e6t 5e6t 3e6t 15e6t 15e6t 15e6t 18e6t 30e6t X 2. AX1 1 5 3 3 e2t e2t e2t 3e2t 5e2t 3e2t 2e2t 2e2t X 1, X 2 18e6t 30e6t X 1 2e2t 2e2t X 1 5 3 3X X1 1 1e2t e2t e2t and X2 3 5e6t 3e6t 5e6t THEOREM 8.1.1 Existence of a Unique Solution Let the entries of the matrices A(t) and F(t) be functions continuous on a common interval I that contains the point t0. Then there exists a unique solution of the initial-value problem (7) on the interval. Homogeneous Systems In the next several deﬁnitions and theorems we are concerned only with homogeneous systems. Without stating it, we shall always assume that the aij and the fi are continuous functions of t on some common interval I. Superposition Principle The following result is a superposition principle for solutions of linear systems. THEOREM 8.1.2 Superposition Principle Let X1, X2, . . . , Xk be a set of solution vectors of the homogeneous system (5) on an interval I. Then the linear combination where the ci, i 1, 2, . . . , k are arbitrary constants, is also a solution on the interval. X c1X1 c2X2 ckXk, Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. It follows from Theorem 8.1.2 that a constant multiple of any solution vector of a homogeneous system of linear ﬁrst-order differential equations is also a solution. 8.1 PRELIMINARY THEORY—LINEAR SYSTEMS ● 329 EXAMPLE 3 Using the Superposition Principle You should practice by verifying that the two vectors are solutions of the system (8) By the superposition principle the linear combination is yet another solution of the system. Linear Dependence and Linear Independence We are primarily inter-ested in linearly independent solutions of the homogeneous system (5). X c1X1 c2X2 c1 cos t 1 2 cos t 1 2 sin t cos t sin t c2 0 et 0 X 1 1 2 0 1 0 1 0 1 X. X1 cos t 1 2 cos t 1 2 sin t cos t sin t and X2 0 et 0 DEFINITION 8.1.2 Linear Dependence/Independence Let X1, X2, . . . , Xk be a set of solution vectors of the homogeneous system (5) on an interval I. We say that the set is linearly dependent on the interval if there exist constants c1, c2, . . . , ck, not all zero, such that for every t in the interval. If the set of vectors is not linearly dependent on the interval, it is said to be linearly independent. c1X1 c2X2 ckXk 0 The case when k 2 should be clear; two solution vectors X1 and X2 are linearly dependent if one is a constant multiple of the other, and conversely. For k 2 a set of solution vectors is linearly dependent if we can express at least one solution vector as a linear combination of the remaining vectors. Wronskian As in our earlier consideration of the theory of a single ordi-nary differential equation, we can introduce the concept of the Wronskian determinant as a test for linear independence. We state the following theorem without proof. THEOREM 8.1.3 Criterion for Linearly Independent Solutions Let X1 ( x11 x21 xn1 x12 x22 xn2 ) , X2 ( . . . , ) , x1n x2n xnn Xn ( ) . . . . . . . . . (continues on page 330) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 330 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS be n solution vectors of the homogeneous system (5) on an interval I. Then the set of solution vectors is linearly independent on I if and only if the Wronskian (9) for every t in the interval. W(X1, X2, . . . , Xn) x11 x21 xn1 x1n x2n xnn x12 x22 xn2 . . . . . . . . . . . . . . . 0 It can be shown that if X1, X2, . . . , Xn are solution vectors of (5), then for every t in I either or Thus if we can show that W 0 for some t0 in I, then W 0 for every t, and hence the solutions are linearly independent on the interval. Notice that, unlike our deﬁnition of the Wronskian in Section 4.1, here the deﬁnition of the determinant (9) does not involve differentiation. W(X1, X2, . . . , Xn) 0. W(X1, X2, . . . , Xn) 0 EXAMPLE 4 Linearly Independent Solutions In Example 2 we saw that and are solutions of system (6). Clearly, X1 and X2 are linearly independent on the interval (, ), since neither vector is a constant multiple of the other. In addition, we have for all real values of t. W(X1, X2) e2t e2t 3e6t 5e6t 8e4t 0 X2 3 5e6t X1 1 1e2t DEFINITION 8.1.3 Fundamental Set of Solutions Any set of n linearly independent solution vectors of the homogeneous system (5) on an interval I is said to be a fundamental set of solutions on the interval. X1, X2, . . . , Xn THEOREM 8.1.4 Existence of a Fundamental Set There exists a fundamental set of solutions for the homogeneous system (5) on an interval I. The next two theorems are the linear system equivalents of Theorems 4.1.5 and 4.1.6. THEOREM 8.1.5 General Solution—Homogeneous Systems Let be a fundamental set of solutions of the homogeneous system (5) on an interval I. Then the general solution of the system on the interval is where the ci, i 1, 2, . . . , n are arbitrary constants. X c1X1 c2X2 cnXn, X1, X2, . . . , Xn Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.1 PRELIMINARY THEORY—LINEAR SYSTEMS ● 331 EXAMPLE 5 General Solution of System (6) From Example 2 we know that and are linearly independent solutions of (6) on (, ). Hence X1 and X2 form a fundamental set of solutions on the interval. The general solution of the system on the interval is then (10) X c1X1 c2X2 c1 1 1e2t c2 3 5e6t. X2 3 5e6t X1 1 1e2t EXAMPLE 6 General Solution of System (8) The vectors are solutions of the system (8) in Example 3 (see Problem 16 in Exercises 8.1). Now for all real values of t. We conclude that X1, X2, and X3 form a fundamental set of solutions on (, ). Thus the general solution of the system on the interval is the linear combination X c1X1 c2X2 c3X3; that is, Nonhomogeneous Systems For nonhomogeneous systems a particular solution Xp on an interval I is any vector, free of arbitrary parameters, whose entries are functions that satisfy the system (4). X c1 cos t 1 2cos t 1 2sin t cos t sin t c2 0 1 0 et c3 sin t 1 2sin t 1 2cos t sin t cos t . W(X1, X2, X3) p cos t 1 2cos t 1 2sin t cos t sin t 0 et 0 sin t 1 2sin t 1 2cos t sin t cos t p et 0 X1 cos t 1 2cos t 1 2 sin t cos t sin t , X2 0 1 0 et, X3 sin t 1 2sin t 1 2cos t sin t cos t THEOREM 8.1.6 General Solution—Nonhomogeneous Systems Let Xp be a given solution of the nonhomogeneous system (4) on an interval I and let denote the general solution on the same interval of the associated homo-geneous system (5). Then the general solution of the nonhomogeneous system on the interval is The general solution Xc of the associated homogeneous system (5) is called the complementary function of the nonhomogeneous system (4). X Xc Xp. Xc c1X1 c2X2 cnXn Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 332 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 7 General Solution—Nonhomogeneous System The vector is a particular solution of the nonhomogeneous system (11) on the interval (, ). (Verify this.) The complementary function of (11) on the same interval, or the general solution of , was seen in (10) of Example 5 to be . Hence by Theorem 8.1.6 is the general solution of (11) on (, ). X Xc Xp c1 1 1e2t c2 3 5e6t 3t 4 5t 6 Xc c1 1 1e2t c2 3 5e6t X 1 5 3 3X X 1 5 3 3X 12t 11 3 Xp 3t 4 5t 6 EXERCISES 8.1 Answers to selected odd-numbered problems begin on page ANS-14. In Problems 1–6 write the linear system in matrix form. 1. 2. 3. 4. 5. 6. In Problems 7–10 write the given system without the use of matrices. 7. X 4 1 2 3X 1 1et dz dt y 6z et dy dt 5x 9z 4etcos 2t dx dt 3x 4y etsin 2t dz dt x y z t2 t 2 dy dt 2x y z 3t2 dx dt x y z t 1 dz dt x z dz dt 10x 4y 3z dy dt x 2z dy dt 6x y dx dt x y dx dt 3x 4y 9z dy dt 5x dy dt 4x 8y dx dt 4x 7y dx dt 3x 5y 8. 9. 10. In Problems 11–16 verify that the vector X is a solution of the given system. 11. 12. 13. 14. X 2 1 1 0X; X 1 3et 4 4tet X 1 1 1 4 1X; X 1 2e3t/2 dy dt 2x 4y; X 5 cos t 3 cos t sin tet dx dt 2x 5y dy dt 4x 7y; X 1 2e5t dx dt 3x 4y d dt x y 3 1 7 1 x y 4 8sin t t 4 2t 1e4t d dt x y z 1 3 2 1 4 5 2 1 6 x y z 1 2 2 et 3 1 1 t X 7 4 0 5 1 2 9 1 3 X 0 2 1 e5t 8 0 3 e2t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 15. 16. In Problems 17–20 the given vectors are solutions of a system X AX. Determine whether the vectors form a fundamental set on the interval (, ). 17. 18. 19. 20. In Problems 21–24 verify that the vector Xp is a particular solution of the given system. 21. dy dt 3x 2y 4t 18; Xp 2 1t 5 1 dx dt x 4y 2t 7 X1 1 6 13 , X2 1 2 1 e4t, X3 2 3 2 e3t X3 3 6 12 t 2 4 4 X1 1 2 4 t 1 2 2 , X2 1 2 4 , X1 1 1et, X2 2 6et 8 8tet X1 1 1e2t, X2 1 1e6t X 1 1 2 0 1 0 1 0 1 X; X sin t 1 2 sin t 1 2 cos t sin t cos t X 1 6 1 2 1 2 1 0 1 X; X 1 6 13 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 333 22. 23. 24. 25. Prove that the general solution of on the interval (, ) is 26. Prove that the general solution of on the interval (, ) is 1 0t2 2 4t 1 0. X c1 1 1 12e12t c2 1 1 12e12t X 1 1 1 1X 1 1t2 4 6t 1 5 X c1 6 1 5 et c2 3 1 1 e2t c3 2 1 1 e3t. X 0 1 1 6 0 1 0 1 0 X X 1 4 6 2 2 1 3 0 0 X 1 4 3 sin 3t; Xp sin 3t 0 cos 3t X 2 3 1 4X 1 7et; Xp 1 1et 1 1tet X 2 1 1 1X 5 2; Xp 1 3 HOMOGENEOUS LINEAR SYSTEMS REVIEW MATERIAL ●Section II.3 of Appendix II ●Also the Student Resource Manual INTRODUCTION We saw in Example 5 of Section 8.1 that the general solution of the homogeneous system is . Because the solution vectors X1 and X2 have the form , i 1, 2, Xi k1 k2e it X c1X1 c2X2 c1 1 1e2t c2 3 5e6t X 1 5 3 3X 8.2 (continues on page 334) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 334 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS where k1, k2, l1, and l2 are constants, we are prompted to ask whether we can always ﬁnd a solution of the form (1) for the general homogeneous linear ﬁrst-order system (2) where A is an n n matrix of constants. X AX, X ( k1 k2 kn) e lt Kelt . . . THEOREM 8.2.1 General Solution—Homogeneous Systems Let l1, l2, . . . , ln be n distinct real eigenvalues of the coefﬁcient matrix A of the homogeneous system (2) and let K1, K2, . . . , Kn be the corresponding eigenvec-tors. Then the general solution of (2) on the interval (, ) is given by X c1K1e 1t c2K2e 2t cnKne nt. Eigenvalues and Eigenvectors If (1) is to be a solution vector of the homoge-neous linear system (2), then X Klelt, so the system becomes Klelt AKelt. After dividing out elt and rearranging, we obtain AK lK or AK lK 0. Since K IK, the last equation is the same as (3) The matrix equation (3) is equivalent to the simultaneous algebraic equations Thus to ﬁnd a nontrivial solution X of (2), we must ﬁrst ﬁnd a nontrivial solution of the foregoing system; in other words, we must ﬁnd a nontrivial vector K that satisﬁes (3). But for (3) to have solutions other than the obvious solution , we must have This polynomial equation in l is called the characteristic equation of the matrix A; its solutions are the eigenvalues of A. A solution K 0 of (3) corresponding to an eigenvalue l is called an eigenvector of A. A solution of the homogeneous system (2) is then X Kelt. In the discussion that follows we examine three cases: real and distinct eigen-values (that is, no eigenvalues are equal), repeated eigenvalues, and, ﬁnally, complex eigenvalues. 8.2.1 DISTINCT REAL EIGENVALUES When the n n matrix A possesses n distinct real eigenvalues l1, l2, . . . , ln, then a set of n linearly independent eigenvectors K1, K2, . . . , Kn can always be found, and is a fundamental set of solutions of (2) on the interval (, ). X1 K1e 1t, X2 K2e 2t, . . . , Xn Kne nt det(A I) 0. k1 k2 kn 0 (a11 l)k1 a12k2 . . . a1nkn 0 a2nkn 0 a21k1 (a22 l)k2 . . . an1k1 an2k2 . . . (ann l)kn 0. . . . . . . (A I)K 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 335 _1 _2 _3 2 3 1 _1 _2 _3 2 3 1 1 2 3 4 5 6 t x (a) graph of x et 3e4t (b) graph of y et 2e4t (c) trajectory define by x et 3e4t, y et 2e4t in the phase plane _2 _4 _6 2 4 6 t y _2 _4 _6 _8 _10 12.5 15 10 5 7.5 2.5 2 4 x y FIGURE 8.2.1 A solution from (5) yields three different curves in three different planes EXAMPLE 1 Distinct Eigenvalues Solve (4) SOLUTION We ﬁrst ﬁnd the eigenvalues and eigenvectors of the matrix of coefﬁcients. From the characteristic equation we see that the eigenvalues are l1 1 and l2 4. Now for l1 1, (3) is equivalent to Thus k1 k2. When k2 1, the related eigenvector is For 2 4 we have so therefore with k2 2 the corresponding eigenvector is Since the matrix of coefﬁcients A is a 2 2 matrix and since we have found two lin-early independent solutions of (4), we conclude that the general solution of the system is (5) Phase Portrait You should keep ﬁrmly in mind that writing a solution of a sys-tem of linear ﬁrst-order differential equations in terms of matrices is simply an alternative to the method that we employed in Section 4.9, that is, listing the individ-ual functions and the relationship between the constants. If we add the vectors on the right-hand side of (5) and then equate the entries with the corresponding entries in the vector on the left-hand side, we obtain the more familiar statement As was pointed out in Section 8.1, we can interpret these equations as parametric equations of curves in the xy-plane or phase plane. Each curve, corresponding to speciﬁc choices for c1 and c2, is called a trajectory. For the choice of constants c1 c2 1 in the solution (5) we see in Figure 8.2.1 the graph of x(t) in the tx-plane, the graph of y(t) in the ty-plane, and the trajectory consisting of the points x c1et 3c2e4t, y c1et 2c2e4t. X c1X1 c2X2 c1 1 1et c2 3 2e4t. X1 1 1et and X2 3 2e4t, K2 3 2. k1 3 2 k2; 2k1 3k2 0 2k1 3k2 0 K1 1 1. 2k1 2k2 0. 3k1 3k2 0 det(A I) 2 2 3 1 2 3 4 ( 1)( 4) 0 dy dt 2x y. dx dt 2x 3y Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. (x(t), y(t)) in the phase plane. A collection of representative trajectories in the phase plane, as shown in Figure 8.2.2, is said to be a phase portrait of the given linear system. What appears to be two red lines in Figure 8.2.2 are actually four red half-lines deﬁned parametrically in the ﬁrst, second, third, and fourth quadrants by the solutions X2, X1, X2, and X1, respectively. For example, the Cartesian equations , and y x, x 0, of the half-lines in the ﬁrst and fourth quadrants were obtained by eliminating the parameter t in the solutions x 3e4t, y 2e4t, and x et, y et, respectively. Moreover, each eigenvector can be visualized as a two-dimensional vector lying along one of these half-lines. The eigenvector lies along in the ﬁrst quadrant, and lies along y x in the fourth quadrant. Each vector starts at the origin; K2 terminates at the point (2, 3), and K1 terminates at (1, 1). The origin is not only a constant solution x 0, y 0 of every 2 2 homoge-neous linear system X AX, but also an important point in the qualitative study of such systems. If we think in physical terms, the arrowheads on each trajectory in Figure 8.2.2 indicate the direction that a particle with coordinates (x(t), y(t)) on that trajectory at time t moves as time increases. Observe that the arrowheads, with the exception of only those on the half-lines in the second and fourth quadrants, indicate that a particle moves away from the origin as time t increases. If we imagine time ranging from to , then inspection of the solution x c1et 3c2e4t, y c1et 2c2e4t, c1 0, c2 0 shows that a trajectory, or moving particle, “starts” asymptotic to one of the half-lines deﬁned by X1 or X1 (since e4t is negli-gible for ) and “ﬁnishes” asymptotic to one of the half-lines deﬁned by X2 and X2 (since et is negligible for ). We note in passing that Figure 8.2.2 represents a phase portrait that is typical of all 2 2 homogeneous linear systems X AX with real eigenvalues of opposite signs. See Problem 17 in Exercises 8.2. Moreover, phase portraits in the two cases when distinct real eigenvalues have the same algebraic sign are typical of all such 2 2 linear systems; the only difference is that the arrowheads indicate that a parti-cle moves away from the origin on any trajectory as when both l1 and l2 are positive and moves toward the origin on any trajectory when both l1 and l2 are neg-ative. Consequently, we call the origin a repeller in the case l1 0, l2 0 and an attractor in the case l1 0, l2 0. See Problem 18 in Exercises 8.2. The origin in Figure 8.2.2 is neither a repeller nor an attractor. Investigation of the remaining case when l 0 is an eigenvalue of a 2 2 homogeneous linear system is left as an exercise. See Problem 49 in Exercises 8.2. t : t : t : K1 1 1 y 2 3 x K2 3 2 y 2 3 x, x 0 336 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS x y X1 X2 FIGURE 8.2.2 A phase portrait of system (4) EXAMPLE 2 Distinct Eigenvalues Solve (6) SOLUTION Using the cofactors of the third row, we ﬁnd and so the eigenvalues are l1 3, l2 4, and l3 5. det(A I) p 4 1 0 1 5 1 1 1 3 p ( 3)( 4)( 5) 0, dz dt y 3 z. dy dt x 5 y z dx dt 4x y z Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. For l1 3 Gauss-Jordan elimination gives Therefore k1 k3 and k2 0. The choice k3 1 gives an eigenvector and corre-sponding solution vector (7) Similarly, for l2 4 implies that k1 10k3 and k2 k3. Choosing k3 1, we get a second eigenvector and solution vector (8) Finally, when l3 5, the augmented matrices yield (9) The general solution of (6) is a linear combination of the solution vectors in (7), (8), and (9): Use of Computers Software packages such as MATLAB, Mathematica, Maple, and DERIVE can be real time savers in ﬁnding eigenvalues and eigenvectors of a matrix A. 8.2.2 REPEATED EIGENVALUES Of course, not all of the n eigenvalues l1, l2, . . . , ln of an n n matrix A need be distinct; that is, some of the eigenvalues may be repeated. For example, the charac-teristic equation of the coefﬁcient matrix in the system (10) X 3 2 18 9X X c1 1 0 1e3t c2 10 1 1e4t c3 1 8 1e5t. K3 1 8 1 , X3 1 8 1 e5t. (A 5I0) ( 9 1 0 1 1 8 0 0 0 1 0 1 ) ( 1 0 0 1 8 0 0 0 0 0 1 0 ) row operations K2 10 1 1 , X2 10 1 1 e4t. (A 4I0) ( 0 1 0 1 1 1 0 0 0 1 9 1 ) ( 1 0 0 10 1 0 0 0 0 0 1 0 ) row operations K1 1 0 1 , X1 1 0 1 e3t. (A 3I0) ( 1 1 0 1 1 0 0 0 0 1 8 1 ) ( 1 0 0 1 0 0 0 0 0 0 1 0 ) . row operations 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 337 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. is readily shown to be (l 3)2 0, and therefore l1 l2 3 is a root of multi-plicity two. For this value we ﬁnd the single eigenvector (11) is one solution of (10). But since we are obviously interested in forming the general solution of the system, we need to pursue the question of ﬁnding a second solution. In general, if m is a positive integer and (l l1)m is a factor of the characteristic equation while (l l1)m1 is not a factor, then l1 is said to be an eigenvalue of multiplicity m. The next three examples illustrate the following cases: (i) For some n n matrices A it may be possible to ﬁnd m linearly inde-pendent eigenvectors K1, K2, . . . , Km corresponding to an eigenvalue 1 of multiplicity m n. In this case the general solution of the system contains the linear combination (ii) If there is only one eigenvector corresponding to the eigenvalue l1 of multiplicity m, then m linearly independent solutions of the form where Kij are column vectors, can always be found. Eigenvalue of Multiplicity T wo We begin by considering eigenvalues of multiplicity two. In the ﬁrst example we illustrate a matrix for which we can ﬁnd two distinct eigenvectors corresponding to a double eigenvalue. X1 K11e l1t X2 K21te l1t K22e l1t Xm Km1 e l1t Km2 e l1t . . . Kmme l1t, tm1 –––––––– (m 1)! tm2 –––––––– (m 2)! . . . c1K1e 1t c2K2e 1t cmKme 1t. K1 3 1, so X1 3 1e3t 338 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Repeated Eigenvalues Solve SOLUTION Expanding the determinant in the characteristic equation yields (l 1)2(l 5) 0. We see that l1 l2 1 and l3 5. For l1 1 Gauss-Jordan elimination immediately gives (A I0) ( 2 2 2 2 2 2 0 0 0 2 2 2 ) ( 1 0 0 1 0 0 0 0 0 1 0 0 ). row operations det(A I) p 1 2 2 2 1 2 2 2 1 p 0 X 1 2 2 2 1 2 2 2 1 X. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The ﬁrst row of the last matrix means k1 k2 k3 0 or k1 k2 k3. The choices k2 1, k3 0 and k2 1, k3 1 yield, in turn, k1 1 and k1 0. Thus two eigenvectors corresponding to l1 1 are Since neither eigenvector is a constant multiple of the other, we have found two linearly independent solutions, corresponding to the same eigenvalue. Lastly, for l3 5 the reduction implies that k1 k3 and k2 k3. Picking k3 1 gives k1 1, k2 1; thus a third eigenvector is We conclude that the general solution of the system is The matrix of coefﬁcients A in Example 3 is a special kind of matrix known as a symmetric matrix. An n n matrix A is said to be symmetric if its transpose AT (where the rows and columns are interchanged) is the same as A—that is, if AT A. It can be proved that if the matrix A in the system X AX is symmetric and has real entries, then we can always ﬁnd n linearly independent eigen-vectors K1, K2, . . . , Kn, and the general solution of such a system is as given in Theorem 8.2.1. As illustrated in Example 3, this result holds even when some of the eigenvalues are repeated. Second Solution Now suppose that l1 is an eigenvalue of multiplicity two and that there is only one eigenvector associated with this value. A second solution can be found of the form , (12) where and ) K ( k1 k2 kn ) . P ( p1 p2 pn . . . . . . X2 Kte 1t Pe 1t X c1 1 1 0 et c2 0 1 1 et c3 1 1 1 e5t. K3 1 1 1 . (A 5I0) ( 4 2 2 2 2 4 0 0 0 2 4 2 ) ( 1 0 0 1 1 0 0 0 0 0 1 0 ) row operations X1 1 1 0 et and X2 0 1 1 et, K1 1 1 0 and K2 0 1 1 . 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 339 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. To see this, we substitute (12) into the system X AX and simplify: Since this last equation is to hold for all values of t, we must have (13) and (14) Equation (13) simply states that K must be an eigenvector of A associated with l1. By solving (13), we ﬁnd one solution . To ﬁnd the second solution X2, we need only solve the additional system (14) for the vector P. X1 Ke 1t (A 1I)P K. (A 1I)K 0 (AK 1K)te 1t (AP 1P K)e 1t 0. 340 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS x y X1 FIGURE 8.2.3 A phase portrait of system (10) EXAMPLE 4 Repeated Eigenvalues Find the general solution of the system given in (10). SOLUTION From (11) we know that l1 3 and that one solution is . Identifying , we ﬁnd from (14) that we must now solve . Since this system is obviously equivalent to one equation, we have an inﬁnite number of choices for p1 and p2. For example, by choosing p1 1, we ﬁnd . However, for simplicity we shall choose so that p2 0. Hence . Thus from (12) we ﬁnd . The general solution of (10) is then X c1X1 c2X2 or By assigning various values to c1 and c2 in the solution in Example 4, we can plot trajectories of the system in (10). A phase portrait of (10) is given in Figure 8.2.3. The solutions X1 and X1 determine two half-lines and , respectively, shown in red in the ﬁgure. Because the single eigenvalue is negative and as on every trajectory, we have as . This is why the arrowheads in Figure 8.2.3 indicate that a particle on any trajectory moves toward the origin as time increases and why the origin is an attractor in this case. Moreover, a moving particle or trajectory , approaches (0, 0) tangentially to one of the half-lines as . In contrast, when the repeated eigen-value is positive, the situation is reversed and the origin is a repeller. See Problem 21 in Exercises 8.2. Analogous to Figure 8.2.2, Figure 8.2.3 is typical of all 2 2 homogeneous linear systems X AX that have two repeated negative eigenvalues. See Problem 32 in Exercises 8.2. Eigenvalue of Multiplicity Three When the coefﬁcient matrix A has only one eigenvector associated with an eigenvalue l1 of multiplicity three, we can ﬁnd a t : y c1e3t c2te3t, c2 0 x 3c1e3t c2(3te3t 1 2e3t), t : (x(t), y(t)) : (0, 0) t : e3t : 0 y 1 3x, x 0 y 1 3x, x 0 X c1 3 1e3t c2 3 1te3t 1 2 0e3t. X2 3 1te3t 1 2 0e3t P 1 2 0 p1 1 2 p2 1 6 (A 3I)P K or 6p1 18p2 3 2p1 6p2 1 K 3 1 and P p1 p2 X1 3 1e3t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. second solution of the form (12) and a third solution of the form , (15) where By substituting (15) into the system X AX, we ﬁnd that the column vectors K, P, and Q must satisfy (16) (17) and (18) Of course, the solutions of (16) and (17) can be used in forming the solutions X1 and X2. (A 1I)Q P. (A 1I)P K (A 1I)K 0 and ) , K ( k1 k2 kn . . . ) , P ( p1 p2 pn . . . ) . Q ( q1 q2 qn . . . X3 K t2 2 e 1t Pte 1t Qe 1t 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 341 EXAMPLE 5 Repeated Eigenvalues Solve . SOLUTION The characteristic equation (l 2)3 0 shows that l1 2 is an eigenvalue of multiplicity three. By solving (A 2I)K 0, we ﬁnd the single eigenvector We next solve the systems (A 2I)P K and (A 2I)Q P in succession and ﬁnd that Using (12) and (15), we see that the general solution of the system is . X c1 1 0 0 e2t c2 1 0 0 te2t 0 1 0 e2t c3 1 0 0 t2 2 e2t 0 1 0 te2t 0 6 5 1 5 e2t P 0 1 0 and Q 0 6 5 1 5 . K 1 0 0 . X 2 0 0 1 2 0 6 5 2 X REMARKS When an eigenvalue l1 has multiplicity m, either we can ﬁnd m linearly independent eigenvectors or the number of corresponding eigenvectors is less than m. Hence the two cases listed on page 338 are not all the possibilities under which a repeated eigenvalue can occur. It can happen, say, that a 5 5 matrix has an eigenvalue of multiplicity ﬁve and there exist three corresponding lin-early independent eigenvectors. See Problems 31 and 50 in Exercises 8.2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.2.3 COMPLEX EIGENVALUES If l1 a bi and l2 a bi, b 0, i2 1 are complex eigenvalues of the coefﬁcient matrix A, we can then certainly expect their corresponding eigenvectors to also have complex entries. For example, the characteristic equation of the system (19) is From the quadratic formula we ﬁnd l1 5 2i, l2 5 2i. Now for l1 5 2i we must solve Since k2 (1 2i)k1,† the choice k1 1 gives the following eigenvector and corresponding solution vector: In like manner, for l2 5 2i we ﬁnd We can verify by means of the Wronskian that these solution vectors are linearly independent, and so the general solution of (19) is (20) Note that the entries in K2 corresponding to l2 are the conjugates of the entries in K1 corresponding to l1. The conjugate of l1 is, of course, l2. We write this as and . We have illustrated the following general result. K2 K1 2 1 X c1 1 1 2ie(52i )t c2 1 1 2ie(52i )t. K2 1 1 2i, X2 1 1 2ie(52i)t. K1 1 1 2i, X1 1 1 2ie(52i)t. 5k1 (1 2i) k2 0. (1 2i)k1 k2 0 det(A I) 6 5 1 4 2 10 29 0. dx dt 6x y dy dt 5x 4y 342 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS When the characteristic equation has real coefﬁcients, complex eigenvalues always appear in conjugate pairs. †Note that the second equation is simply (1 2i) times the ﬁrst. THEOREM 8.2.2 Solutions Corresponding to a Complex Eigenvalue Let A be the coefﬁcient matrix having real entries of the homogeneous system (2), and let K1 be an eigenvector corresponding to the complex eigenvalue l1 a ib, a and b real. Then are solutions of (2). K1e 1t and K1e 1t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. It is desirable and relatively easy to rewrite a solution such as (20) in terms of real functions. To this end we ﬁrst use Euler’s formula to write Then, after we multiply complex numbers, collect terms, and replace c1 c2 by C1 and (c1 c2)i by C2, (20) becomes (21) where and It is now important to realize that the vectors X1 and X2 in (21) constitute a linearly independent set of real solutions of the original system. Consequently, we are justi-ﬁed in ignoring the relationship between C1, C2 and c1, c2, and we can regard C1 and C2 as completely arbitrary and real. In other words, the linear combination (21) is an alternative general solution of (19). Moreover, with the real form given in (21) we are able to obtain a phase portrait of the system in (19). From (21) we ﬁnd x(t) and y(t) to be By plotting the trajectories (x(t), y(t)) for various values of C1 and C2, we obtain the phase portrait of (19) shown in Figure 8.2.4. Because the real part of l1 is 5 0, as . This is why the arrowheads in Figure 8.2.4 point away from the origin; a particle on any trajectory spirals away from the origin as . The origin is a repeller. The process by which we obtained the real solutions in (21) can be generalized. Let K1 be an eigenvector of the coefﬁcient matrix A (with real entries) corresponding to the complex eigenvalue l1 a ib. Then the solution vectors in Theorem 8.2.2 can be written as By the superposition principle, Theorem 8.1.2, the following vectors are also solutions: Both and are real numbers for any complex number z a ib. Therefore, the entries in the column vectors and are real numbers. By deﬁning (22) we are led to the following theorem. B1 1 2 (K1 K1) and B2 i 2 (K1 K1), 1 2i(K1 K1) 1 2(K1 K1) 1 2i(z z) b 1 2(z z) a X2 i 2(K1e 1t K1e 1t) i 2(K1 K1)et cos t 1 2(K1 K1)et sin t. X1 1 2(K1e 1t K1e 1t) 1 2(K1 K1)et cos t i 2(K1 K1)et sin t K1e 1t K1eteit K1et(cos t i sin t). K1e 1t K1eteit K1et(cos t i sin t) t : t : e5t : y (C1 2C2)e5t cos 2t (2C1 C2)e5t sin 2t. x C1e5t cos 2t C2e5t sin 2t X2 0 2cos 2t 1 1sin 2te5t. X1 1 1cos 2t 0 2sin 2te5t X C1X1 C2X2, e(52i )t e5te2ti e5t(cos 2t i sin 2t). e(52i )t e5te2ti e5t(cos 2t i sin 2t) 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 343 FIGURE 8.2.4 A phase portrait of system (19) x y Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The matrices B1 and B2 in (22) are often denoted by (24) since these vectors are, respectively, the real and imaginary parts of the eigenvector K1. For example, (21) follows from (23) with B1 Re(K1) 1 1 and B2 Im(K1) 0 2. K1 1 1 2i 1 1 i 0 2, B1 Re(K1) and B2 Im(K1) 344 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS THEOREM 8.2.3 Real Solutions Corresponding to a Complex Eigenvalue Let l1 a ib be a complex eigenvalue of the coefﬁcient matrix A in the homogeneous system (2) and let B1 and B2 denote the column vectors deﬁned in (22). Then (23) are linearly independent solutions of (2) on (, ). X2 [B2 cos t B1 sin t]et X1 [B1 cos t B2 sin t]et EXAMPLE 6 Complex Eigenvalues Solve the initial-value problem (25) SOLUTION First we obtain the eigenvalues from The eigenvalues are l1 2i and . For l1 the system gives k1 (2 2i)k2. By choosing k2 1, we get Now from (24) we form Since a 0, it follows from (23) that the general solution of the system is (26) c1 2 cos 2t 2 sin 2t cos 2t c2 2 cos 2t 2 sin 2t sin 2t . X c1 2 1cos 2t 2 0sin 2t c2 2 0cos 2t 2 1sin 2t B1 Re(K1) 2 1 and B2 Im(K1) 2 0. K1 2 2i 1 2 1 i 2 0. k1 (2 2i) k2 0 (2 2i ) k1 8k2 0 2 1 2i det(A I) 2 1 8 2 2 4 0. X 2 1 8 2X, X(0) 2 1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. X X. (29) 0 0 k1 m1 k2 m1 k2 m2 0 0 k2 m1 k2 m2 1 0 0 0 0 1 0 0 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 345 Some graphs of the curves or trajectories deﬁned by solution (26) of the system are illustrated in the phase portrait in Figure 8.2.5. Now the initial condition or, equivalently, x(0) 2 and y(0) 1 yields the algebraic system 2c1 2c2 2, c1 1, whose solution is c1 1, c2 0. Thus the solution to the problem is . The speciﬁc trajectory deﬁned parametrically by the particular solution x 2 cos 2t 2 sin 2t, y cos 2t is the red curve in Figure 8.2.5. Note that this curve passes through (2, 1). X 2 cos 2t 2 sin 2t cos 2t X(0) 2 1 FIGURE 8.2.5 A phase portrait of (25) in Example 6 x y (2, _1) REMARKS In this section we have examined exclusively homogeneous ﬁrst-order systems of linear equations in normal form X AX. But often the mathematical model of a dynamical physical system is a homogeneous second-order system whose normal form is X AX. For example, the model for the coupled springs in (1) of Section 7.6, (27) can be written as where Since M is nonsingular, we can solve for X as X AX, where A M1K. Thus (27) is equivalent to (28) The methods of this section can be used to solve such a system in two ways: • First, the original system (27) can be transformed into a ﬁrst-order system by means of substitutions. If we let and , then and and so (27) is equivalent to a system of four linear ﬁrst-order DEs: or By ﬁnding the eigenvalues and eigenvectors of the coefﬁcient matrix A in (29), we see that the solution of this ﬁrst-order system gives the complete state of the physical system—the positions of the masses relative to the equilibrium positions (x1 and x2) as well as the velocities of the masses (x3 and x4) at time t. See Problem 48(a) in Exercises 8.2. x 4 k2 m2 x1 k2 m2 x2 x 3 k1 m1 k2 m1x1 k2 m1 x2 x 2 x4 x 1 x3 x 4 x 2 x 3 x 1 x 2 x4 x 1 x3 X k1 m1 k2 m1 k2 m2 k2 m1 k2 m2 X. M m1 0 0 m2, K k1 k2 k2 k2 k2, and X x1(t) x2(t). MX KX, m2x 2 k2(x2 x1), m1x 1 k1x1 k2(x2 x1) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 346 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS • Second, because (27) describes free undamped motion, it can be argued that real-valued solutions of the second-order system (28) will have the form , (30) where V is a column matrix of constants. Substituting either of the functions in (30) into X AX yields (A v2I)V 0. (Verify.) By identiﬁcation with (3) of this section we conclude that l v2 represents an eigenvalue and V a corresponding eigenvector of A. It can be shown that the eigenvalues , i 1, 2 of A are negative, and so is a real number and represents a (circular) frequency of vibration (see (4) of Section 7.6). By superposition of solutions the general solution of (28) is then (31) where V1 and V2 are, in turn, real eigenvectors of A corresponding to l1 and l2. The result given in (31) generalizes. If are distinct negative eigenvalues and V1, V2, . . . , Vn are corresponding real eigenvectors of the n n coefﬁcient matrix A, then the homogeneous second-order system X AX has the general solution (32) where ai and bi represent arbitrary constants. See Problem 48(b) in Exercises 8.2. X n i1 (ai cos it bi sin it)Vi, 1 2, 2 2, . . . , n 2 (c1 cos 1t c2 sin 1t)V1 (c3 cos 2t c4 sin 2t)V2, X c1V1 cos 1t c2V1 sin 1t c3V2 cos 2t c4V2 sin 2t i 1 i i i 2 X V cos t and X V sin t EXERCISES 8.2 Answers to selected odd-numbered problems begin on page ANS-14. 8.2.1 DISTINCT REAL EIGENVALUES In Problems 1–12 ﬁnd the general solution of the given system. 1. 2. 3. 4. 5. 6. 7. 8. dz dt 5y 2z dz dt y z dy dt 5x 10y 4z dy dt 2y dx dt 2x 7y dx dt x y z X 6 3 2 1X X 10 8 5 12X dy dt 3 4x 2y dy dt 5 2x 2y dx dt 5 2x 2y dx dt 4x 2y dy dt x 3y dy dt 4x 3y dx dt 2x 2y dx dt x 2y 9. 10. 11. 12. In Problems 13 and 14 solve the given initial-value problem. 13. 14. X 1 0 1 1 2 1 4 0 1 X, X(0) 1 3 0 X 1 2 1 0 1 2X, X(0) 3 5 X 1 4 0 4 1 0 2 2 6 X X 1 3 4 1 8 1 3 2 1 4 0 3 1 2 X X 1 0 1 0 1 0 1 0 1 X X 1 1 0 1 2 3 0 1 1 X Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Computer Lab Assignments In Problems 15 and 16 use a CAS or linear algebra software as an aid in ﬁnding the general solution of the given system. 15. 16. 17. (a) Use computer software to obtain the phase portrait of the system in Problem 5. If possible, include arrowheads as in Figure 8.2.2. Also include four half-lines in your phase portrait. (b) Obtain the Cartesian equations of each of the four half-lines in part (a). (c) Draw the eigenvectors on your phase portrait of the system. 18. Find phase portraits for the systems in Problems 2 and 4. For each system ﬁnd any half-line trajectories and include these lines in your phase portrait. 8.2.2 REPEATED EIGENVALUES In Problems 19–28 ﬁnd the general solution of the given system. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. X 4 0 0 1 4 0 0 1 4 X X 1 2 0 0 2 1 0 1 0 X X 1 0 0 0 3 1 0 1 1 X X 5 1 0 4 0 2 0 2 5 X dz dt 4x 2y 3z dz dt x y z dy dt 2x 2z dy dt x y z dx dt 3x 2y 4z dx dt 3x y z X 12 4 9 0X X 1 3 3 5X dy dt 5x 4y dy dt 9x 3y dx dt 6x 5y dx dt 3x y X 1 0 1 0 2.8 0 5.1 2 1 0 2 0 3 3.1 0 1.8 1 0 4 1.5 0 3 0 0 1 X X 0.9 0.7 1.1 2.1 6.5 1.7 3.2 4.2 3.4 X 8.2 HOMOGENEOUS LINEAR SYSTEMS ● 347 In Problems 29 and 30 solve the given initial-value problem. 29. 30. 31. Show that the 5 5 matrix has an eigenvalue l1 of multiplicity 5. Show that three linearly independent eigenvectors corresponding to l1 can be found. Computer Lab Assignments 32. Find phase portraits for the systems in Problems 20 and 21. For each system ﬁnd any half-line trajectories and include these lines in your phase portrait. 8.2.3 COMPLEX EIGENVALUES In Problems 33–44 ﬁnd the general solution of the given system. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. X 4 0 4 0 6 0 1 0 4 X X 1 1 1 1 1 0 2 0 1 X dz dt 4x 3z dz dt y dy dt 3x 6z dy dt z dx dt 2x y 2z dx dt z X 1 1 8 3X X 4 5 5 4X dy dt 2x 6y dy dt 2x 3y dx dt 4x 5y dx dt 5x y dy dt 2x y dy dt 5x 2y dx dt x y dx dt 6x y A 2 0 0 0 0 1 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 1 2 X 0 0 1 0 1 0 1 0 0 X, X(0) 1 2 5 X 2 1 4 6X, X(0) 1 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 43. 44. In Problems 45 and 46 solve the given initial-value problem. 45. 46. Computer Lab Assignments 47. Find phase portraits for the systems in Problems 36, 37, and 38. 48. (a) Solve (2) of Section 7.6 using the ﬁrst method outlined in the Remarks (page 345)—that is, express (2) of Section 7.6 as a ﬁrst-order system of four lin-ear equations. Use a CAS or linear algebra software as an aid in ﬁnding eigenvalues and eigenvectors of a 4 4 matrix. Then apply the initial conditions to your general solution to obtain (4) of Section 7.6. (b) Solve (2) of Section 7.6 using the second method out-lined in the Remarks—that is, express (2) of Sec-tion 7.6 as a second-order system of two linear equa-tions. Assume solutions of the form X V sin vt X 6 5 1 4X, X(0) 2 8 X 1 1 1 12 2 1 14 3 2 X, X(0) 4 6 7 X 2 1 1 4 2 0 4 0 2 X X 2 5 0 5 6 0 1 4 2 X 348 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS and X V cos vt. Find the eigenvalues and eigen-vectors of a 2 2 matrix. As in part (a), obtain (4) of Section 7.6. Discussion Problems 49. Solve each of the following linear systems. (a) (b) Find a phase portrait of each system. What is the geo-metric signiﬁcance of the line y x in each portrait? 50. Consider the 5 5 matrix given in Problem 31. Solve the system X AX without the aid of matrix methods, but write the general solution using matrix notation. Use the general solution as a basis for a discussion of how the system can be solved using the matrix methods of this section. Carry out your ideas. 51. Obtain a Cartesian equation of the curve deﬁned parametrically by the solution of the linear system in Example 6. Identify the curve passing through (2, 1) in Figure 8.2.5. [Hint: Compute x2, y2, and xy.] 52. Examine your phase portraits in Problem 47. Under what conditions will the phase portrait of a 2 2 homogeneous linear system with complex eigenvalues consist of a family of closed curves? consist of a family of spirals? Under what conditions is the origin (0, 0) a repeller? An attractor? X 1 1 1 1X X 1 1 1 1X 8.3.1 UNDETERMINED COEFFICIENTS The Assumptions As in Section 4.4, the method of undetermined coefﬁcients consists of making an educated guess about the form of a particular solution vector Xp; the guess is motivated by the types of functions that make up the entries of the NONHOMOGENEOUS LINEAR SYSTEMS REVIEW MATERIAL ●Section 4.4 (Undetermined Coefﬁcients) ●Section 4.6 (Variation of Parameters) INTRODUCTION In Section 8.1 we saw that the general solution of a nonhomogeneous linear system X AX F(t) on an interval I is X Xc Xp, where is the complementary function or general solution of the associated homogeneous linear system X AX and Xp is any particular solution of the nonhomogeneous system. In Section 8.2 we saw how to obtain Xc when the coefﬁcient matrix A was an n n matrix of constants. In the present section we consider two methods for obtaining Xp. The methods of undetermined coefficient and variation of parameters used in Chapter 4 to ﬁnd particular solutions of nonhomogeneous linear ODEs can both be adapted to the solution of nonhomogeneous linear systems X AX F(t). Of the two methods, variation of parameters is the more powerful technique. However, there are instances when the method of undetermined coefﬁcients provides a quick means of ﬁnding a particular solution. Xc c1X1 c2X2 cnXn 8.3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. column matrix F(t). Not surprisingly, the matrix version of undetermined coefﬁcients is applicable to X AX F(t) only when the entries of A are constants and the entries of F(t) are constants, polynomials, exponential functions, sines and cosines, or ﬁnite sums and products of these functions. 8.3 NONHOMOGENEOUS LINEAR SYSTEMS ● 349 EXAMPLE 1 Undetermined Coefficient Solve the system on (, ). SOLUTION We ﬁrst solve the associated homogeneous system The characteristic equation of the coefﬁcient matrix A, yields the complex eigenvalues l1 i and . By the procedures of Section 8.2 we ﬁnd Now since F(t) is a constant vector, we assume a constant particular solution vector . Substituting this latter assumption into the original system and equat-ing entries leads to Solving this algebraic system gives a1 14 and b1 11, and so a particular solution is . The general solution of the original system of DEs on the interval (, ) is then X Xc Xp or X c1 cos t sin t cos t c2 cos t sin t sin t 14 11. Xp 14 11 0 a1 b1 3. 0 a1 2b1 8 Xp a1 b1 Xc c1 cos t sin t cos t c2 cos t sin t sin t . 2 1 i det(A I) 1 1 2 1 2 1 0, X 1 1 2 1X. X 1 1 2 1X 8 3 EXAMPLE 2 Undetermined Coefficient Solve the system on (, ). SOLUTION The eigenvalues and corresponding eigenvectors of the associated homogeneous system are found to be l1 2, l2 7, , and . Hence the complementary function is Xc c1 1 4e2t c2 1 1e7t. K2 1 1 K1 1 4 X 6 4 1 3X X 6 4 1 3X 6t 10t 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Now because F(t) can be written , we shall try to ﬁnd a particular solution of the system that possesses the same form: Substituting this last assumption into the given system yields or From the last identity we obtain four algebraic equations in four unknowns Solving the ﬁrst two equations simultaneously yields a2 2 and b2 6. We then substitute these values into the last two equations and solve for a1 and b1. The results are . It follows, therefore, that a particular solution vector is . The general solution of the system on (, ) is X Xc Xp or . X c1 1 4e2t c2 1 1e7t 2 6t 4 7 10 7 Xp 2 6t 4 7 10 7 a1 4 7, b1 10 7 6a2 b2 6 0 4a2 3b2 10 0 and 6a1 b1 a2 0 4a1 3b1 b2 4 0. 0 0 (6a2 b2 6)t 6a1 b1 a2 (4a2 3b2 10)t 4a1 3b1 b2 4. a2 b2 6 4 1 3 a2 b2t a1 b1 6 10t 0 4 Xp a2 b2t a1 b1. F(t) 6 10t 0 4 350 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Form of Xp Determine the form of a particular solution vector Xp for the system SOLUTION Because F(t) can be written in matrix terms as a natural assumption for a particular solution would be Xp a3 b3et a2 b2t a1 b1. F(t) 2 1et 0 5t 1 7 dy dt x y et 5t 7. dx dt 5x 3y 2et 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.3 NONHOMOGENEOUS LINEAR SYSTEMS ● 351 REMARKS The method of undetermined coefﬁcients for linear systems is not as straightfor-ward as the last three examples would seem to indicate. In Section 4.4 the form of a particular solution yp was predicated on prior knowledge of the comple-mentary function yc. The same is true for the formation of Xp. But there are fur-ther difﬁculties: The special rules governing the form of yp in Section 4.4 do not quite carry to the formation of Xp. For example, if F(t) is a constant vector, as in Example 1, and l 0 is an eigenvalue of multiplicity one, then Xc contains a constant vector. Under the Multiplication Rule on page 145 we would ordinarily try a particular solution of the form . This is not the proper assumption for linear systems; it should be . Similarly, in Example 3, if we replace et in F(t) by e2t (l 2 is an eigenvalue), then the correct form of the particular solution vector is Rather than delving into these difﬁculties, we turn instead to the method of variation of parameters. Xp a4 b4te2t a3 b3e2t a2 b2t a1 b1. Xp a2 b2t a1 b1 Xp a1 b1t 8.3.2 VARIATION OF PARAMETERS A Fundamental Matrix If X1, X2, . . . , Xn is a fundamental set of solutions of the homogeneous system X AX on an interval I, then its general solution on the in-terval is the linear combination (1) The last matrix in (1) is recognized as the product of an n n matrix with an n 1 matrix. In other words, the general solution (1) can be written as the product , (2) where C is an n 1 column vector of arbitrary constants c1, c2, . . . , cn and the n n matrix, whose columns consist of the entries of the solution vectors of the system X AX, is called a fundamental matrix of the system on the interval. x11 x21 xn1 (t) ( ) , x1n x2n xnn x12 x22 xn2 . . . . . . . . . . . . . . . X (t)C x11 x21 xn1 x12 x22 xn2 X c1( ) c2( ) . . . cn( . . . . . . x1n x2n xnn c1x11 c2x12 . . . cnx1n c1x21 c2x22 . . . cnx2n c1xn1 c2xn2 . . . cnxnn ) ( ) . . . . . . . X c1X1 c2X2 cnXn or Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In the discussion that follows we need to use two properties of a fundamental matrix: • A fundamental matrix is nonsingular. • If is a fundamental matrix of the system X AX, then . (3) A reexamination of (9) of Theorem 8.1.3 shows that det is the same as the Wronskian W(X1, X2, . . . , Xn). Hence the linear independence of the columns of on the interval I guarantees that det for every t in the interval. Since is nonsingular, the multiplicative inverse exists for every t in the inter-val. The result given in (3) follows immediately from the fact that every column of is a solution vector of X AX. Variation of Parameters Analogous to the procedure in Section 4.6 we ask whether it is possible to replace the matrix of constants C in (2) by a column matrix of functions (4) is a particular solution of the nonhomogeneous system . (5) By the Product Rule the derivative of the last expression in (4) is . (6) Note that the order of the products in (6) is very important. Since U(t) is a column matrix, the products and are not deﬁned. Substituting (4) and (6) into (5) gives (7) Now if we use (3) to replace , (7) becomes or (8) Multiplying both sides of equation (8) by gives . Since , we conclude that a particular solution of (5) is . (9) To calculate the indeﬁnite integral of the column matrix in (9), we inte-grate each entry. Thus the general solution of the system (5) is X Xc Xp or . (10) Note that it is not necessary to use a constant of integration in the evaluation of for the same reasons stated in the discussion of variation of parame-ters in Section 4.6. 1(t)F(t) dt X (t)C (t)1(t)F(t) dt 1(t)F(t) Xp (t)1(t)F(t) dt Xp (t)U(t) U(t) 1(t)F(t) and so U(t) 1(t)F(t) dt 1(t) (t)U(t) F(t). (t)U(t) A(t)U(t) A(t)U(t) F(t) (t) (t)U(t) (t)U(t) A(t)U(t) F(t). U(t)(t) U(t)(t) X p (t)U(t) (t)U(t) X AX F(t) u1(t) u2(t) un(t) U(t) ( Xp (t)U(t) so ) . . . (t) 1(t) (t) (t) 0 (t) (t) (t) A(t) (t) (t) 352 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.3 NONHOMOGENEOUS LINEAR SYSTEMS ● 353 EXAMPLE 4 Variation of Parameters Solve the system (11) on (, ). SOLUTION We ﬁrst solve the associated homogeneous system . (12) The characteristic equation of the coefﬁcient matrix is , so the eigenvalues are l1 2 and l2 5. By the usual method we ﬁnd that the eigenvectors corresponding to l1 and l2 are, respectively, and . The solution vectors of the homogeneous system (12) are then . The entries in X1 form the ﬁrst column of , and the entries in X2 form the second column of . Hence . From (9) we obtain the particular solution Hence from (10) the general solution of (11) on the interval is . c1 1 1e2t c2 1 2e5t 6 5 3 5 t 27 50 21 50 1 4 1 2 et X e2t e2t e5t 2e5t c1 c2 6 5t 27 50 1 4et 3 5t 21 50 1 2et 6 5t 27 50 1 4et 3 5t 21 50 1 2et. e2t e2t e5t 2e5t te2t 1 2e2t 1 3et 1 5te5t 1 25e5t 1 12e4t e2t e2t e5t 2e5t 2te2t 1 3et te5t 1 3e4t dt Xp (t)1(t)F(t) dt e2t e2t e5t 2e5t 2 3e2t 1 3e5t 1 3e2t 1 3e5t 3t et dt (t) e2t e2t e5t 2e5t and 1(t) 2 3e2t 1 3e5t 1 3e2t 1 3e5t (t) (t) X1 1 1e2t e2t e2t and X2 1 2e5t e5t 2e5t K2 1 2 K1 1 1 det(A I) 3 2 1 4 ( 2)( 5) 0 X 3 2 1 4X X 3 2 1 4X 3t et Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Initial-Value Problem The general solution of (5) on an interval can be writ-ten in the alternative manner , (13) where t and t0 are points in the interval. This last form is useful in solving (5) subject to an initial condition X(t0) X0, because the limits of integration are chosen so that the particular solution vanishes at t t0. Substituting t t0 into (13) yields from which we get . Substituting this last result into (13) gives the following solution of the initial-value problem: . (14) X (t)1(t0)X0 (t) t t0 1(s)F(s) ds C 1(t0)X0 X0 (t0)C X (t)C (t) t t0 1(s)F(s) ds 354 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS FIGURE 8.3.1 Network in Problem 10 R1 R2 L1 L2 i1 i2 i3 E 8.3.2 VARIATION OF PARAMETERS In Problems 11–30 use variation of parameters to solve the given system. 11. 12. 13. X 3 3 4 5 1X 1 1et/2 dy dt 3x 2y 4t dx dt 2x y dy dt 2x 2y 1 dx dt 3x 3y 4 EXERCISES 8.3 Answers to selected odd-numbered problems begin on page ANS-15. 8.3.1 UNDETERMINED COEFFICIENTS In Problems 1–8 use the method of undetermined coefﬁ-cients to solve the given system. 1. 2. 3. 4. 5. 6. 7. 8. 9. Solve subject to . X(0) 4 5 X 1 3 2 4X 3 3 X 0 0 5 0 5 0 5 0 0X 5 10 40 X 1 0 0 1 2 0 1 3 5X 1 1 2e4t X 1 1 5 1X sin t 2 cos t X 4 9 1 3 6X 3 10et X 1 4 4 1X 4t 9e6t t e6t X 1 3 3 1X 2t2 t 5 dy dt x 11y 6 dx dt 5x 9y 2 dy dt x 2y 5 dx dt 2x 3y 7 10. (a) The system of differential equations for the currents i2(t) and i3(t) in the electrical network shown in Figure 8.3.1 is . Use the method of undetermined coefﬁcients to solve the system if R1 2 , R2 3 , L1 1 h, L2 1 h, E 60 V, i2(0) 0, and i3(0) 0. (b) Determine the current i1(t). d dt i2 i3 R1>L1 R1>L2 R1>L1 (R1 R2)>L2 i2 i3 E>L1 E>L2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. In Problems 31 and 32 use (14) to solve the given initial-value problem. 31. 32. X 1 1 1 1X 1>t 1>t, X(1) 2 1 X 3 1 1 3X 4e2t 4e4t, X(0) 1 1 X 3 1 1 1 1 1 1 1 1X 0 t 2et X 1 1 0 1 1 0 0 0 3X et e2t te3t X 1 1 2 1X tan t 1 X 1 1 2 2 1X csc t sec tet X 0 1 1 0X 1 cot t X 0 1 1 0X 0 sec t tan t X 2 8 2 6X 1 3 e2t t X 1 1 1 1X cos t sin tet X 1 1 1 1X 3 3et X 0 1 1 0X sec t 0 X 3 2 2 1X 1 1 X 3 2 2 1X 2et et X 1 1 8 1X et tet X 1 1 8 1X 12 12t X 0 1 2 3X 2 e3t X 0 1 2 3X 1 1et X 2 4 1 2X sin 2t 2 cos 2te2t 8.3 NONHOMOGENEOUS LINEAR SYSTEMS ● 355 33. The system of differential equations for the currents i1(t) and i2(t) in the electrical network shown in Figure 8.3.2 is . Use variation of parameters to solve the system if R1 8 , R2 3 , L1 1 h, L2 1 h, E(t) 100 sin t V, i1(0) 0, and i2(0) 0. d dt i1 i2 (R1 R2)>L2 R2>L1 R2>L2 R2>L1 i1 i2 E>L2 0 FIGURE 8.3.2 Network in Problem 33 i1 i2 i3 R1 R2 E L1 L2 Discussion Problems 34. If y1 and y2 are linearly independent solutions of the associated homogeneous DE for y P(x)y Q(x)y f (x), show in the case of a nonhomogeneous linear second-order DE that (9) reduces to the form of variation of parameters discussed in Section 4.6. Computer Lab Assignments 35. Solving a nonhomogeneous linear system X AX F(t) by variation of parameters when A is a 3 3 (or larger) matrix is almost an impossible task to do by hand. Consider the system (a) Use a CAS or linear algebra software to ﬁnd the eigenvalues and eigenvectors of the coefﬁcient matrix. (b) Form a fundamental matrix and use the com-puter to ﬁnd . (c) Use the computer to carry out the computations of: where C is a column matrix of constants c1, c2, c3, and c4. (d) Rewrite the computer output for the general solu-tion of the system in the form X Xc Xp, where Xc c1X1 c2X2 c3X3 c4X4. (t)C, and (t)C 1(t)F(t) dt, 1(t)F(t), 1(t)F(t) dt, (t)1(t)F(t) dt, 1(t) (t) X 2 1 0 0 2 3 0 0 2 0 4 2 1 3 2 1 X tet et e2t 1 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Homogeneous Systems We shall now see that it is possible to deﬁne a ma-trix exponential eAt so that (1) is a solution of the homogeneous system X AX. Here A is an n n matrix of constants, and C is an n 1 column matrix of arbitrary constants. Note in (1) that the matrix C post multiplies eAt because we want eAt to be an n n matrix. While the complete development of the meaning and theory of the matrix exponential would require a thorough knowledge of matrix algebra, one way of deﬁning eAt is inspired by the power series representation of the scalar exponential function eat: (2) The series in (2) converges for all t. Using this series, with 1 replaced by the identity matrix I and the constant a replaced by an n n matrix A of constants, we arrive at a deﬁnition for the n n matrix eAt. 1 at a2 t2 2! k t k k! k0 k tk k!. eat 1 at (at)2 2! (at)k k! X eAtC 356 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS MATRIX EXPONENTIAL REVIEW MATERIAL ●Appendix II.1 (Deﬁnitions II.10 and II.11) INTRODUCTION Matrices can be used in an entirely different manner to solve a system of linear ﬁrst-order differential equations. Recall that the simple linear ﬁrst-order differential equation x ax, where a is constant, has the general solution x ceat, where c is a constant. It seems natural then to ask whether we can deﬁne a matrix exponential function eAt, where A is a matrix of constants, so that a solution of the linear system X AX is eAt. 8.4 DEFINITION 8.4.1 Matrix Exponential For any n n matrix A, . (3) eAt I At A2 t2 2! Ak tk k! k0 Ak tk k! It can be shown that the series given in (3) converges to an n n matrix for every value of t. Also, A2 AA, A3 A(A2), and so on. EXAMPLE 1 Matrix Exponential Using (3) Compute for the matrix SOLUTION From the various powers A2 22 0 0 32, A3 23 0 0 33, A4 24 0 0 34, . . . , An 2n 0 0 3n, . . . , A 2 0 0 3. eAt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.4 MATRIX EXPONENTIAL ● 357 we see from (3) that In view of (2) and the identiﬁcations and the power series in the ﬁrst and second rows of the last matrix represent, respectively, and so we have . The matrix in Example 1 is an example of a diagonal matrix. In general, an matrix A is a diagonal matrix if all its entries off the main diagonal are zero, that is, . Hence if A is any diagonal matrix it follows from Example 1 that . eAt ea11t 0 o 0 0 ea22t o 0 . . . . . . . . . 0 0 o eannt n n A a11 0 o 0 0 a22 o 0 . . . . . . . . . 0 0 o ann n n 2 2 eAt e2t 0 0 e3t e2t and e3t a 3, a 2 1 2t 22 t2 2! . . . 0 0 1 3t 32 t2 2! . . .. 1 0 0 1 2 0 0 3t 22 0 0 32 t2 2! . . . 2n 0 0 3n t n n! . . . eAt I At A 2 2! t2 . . . AI At A2 t2 2! AeAt. d dt eAt d dt I At A2 t2 2! Ak tk k! A A2t 1 2! A3t2 Because of (4), we can now prove that (1) is a solution of X AX for every n 1 vector C of constants: eAt is a Fundamental Matrix If we denote the matrix exponential eAt by X d dt eAtC AeAtC A(eAtC) AX. the symbol (t), then (4) is equivalent to the matrix differential equation (t) A(t) (see (3) of Section 8.3). In addition, it follows immediately from Derivative of eAt The derivative of the matrix exponential is analogous to the differentiation property of the scalar exponential . To justify , (4) we differentiate (3) term by term: d dt eAt AeAt d dt eat aeat Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Deﬁnition 8.4.1 that (0) eA0 I, and so det (0) 0. It turns out that these two properties are sufﬁcient for us to conclude that (t) is a fundamental matrix of the system X AX. Nonhomogeneous Systems We saw in (4) of Section 2.3 that the general 358 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS solution of the single linear ﬁrst-order differential equation x ax f (t), where a is a constant, can be expressed as . For a nonhomogeneous system of linear ﬁrst-order differential equations it can be shown that the general solution of X AX F(t), where A is an n n matrix of constants, is . (5) Since the matrix exponential eAt is a fundamental matrix, it is always nonsingular and eAs (eAs)1. In practice, eAs can be obtained from eAt by simply replacing t by s. Computation of eAt The deﬁnition of eAt given in (3) can, of course, always be used to compute eAt. However, the practical utility of (3) is limited by the fact that the entries in eAt are power series in t. With a natural desire to work with simple and familiar things, we then try to recognize whether these series deﬁne a closed-form function. Fortunately, there are many alternative ways of computing eAt; the follow-ing discussion shows how the Laplace transform can be used. Use of the Laplace T ransform We saw in (5) that X eAt is a solution of X Xc Xp eAtC eAt t t0 eAsF(s) ds x xc xp ceat eat t t0 easf (s) ds EXAMPLE 2 Matrix Exponential Using (7) Use the Laplace transform to compute eAt for . SOLUTION First we compute the matrix sI A and ﬁnd its inverse: (sI A)1 s 1 2 1 s 2 1 s 2 s(s 1) 2 s(s 1) 1 s(s 1) s 1 s(s 1) . sI A s 1 2 1 s 2, A 1 2 1 2 X AX. Indeed, since eA0 I, X eAt is a solution of the initial-value problem . (6) If , then the Laplace transform of (6) is . Multiplying the last equation by (sI A)1 implies that x(s) (sI A)1 I (sI A)1. In other words, (7) eAt 1{(sI A)1}. {eAt} (sI A)1 or sx(s) X(0) Ax(s) or (sI A)x(s) I x(s) {X(t)} {eAt} X AX, X(0) I Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Then we decompose the entries of the last matrix into partial fractions: . (8) It follows from (7) that the inverse Laplace transform of (8) gives the desired result, . Use of Computers For those who are willing to momentarily trade under-standing for speed of solution, eAt can be computed with the aid of computer software. See Problems 27 and 28 in Exercises 8.4. eAt 2 et 2 2et 1 et 1 2et (sI A)1 2 s 1 s 1 2 s 2 s 1 1 s 1 s 1 1 s 2 s 1 8.4 MATRIX EXPONENTIAL ● 359 EXERCISES 8.4 Answers to selected odd-numbered problems begin on page ANS-16. In Problems 1 and 2 use (3) to compute eAt and eAt. 1. 2. In Problems 3 and 4 use (3) to compute eAt. 3. 4. In Problems 5–8 use (1) to ﬁnd the general solution of the given system. 5. 6. 7. 8. In Problems 9–12 use (5) to ﬁnd the general solution of the given system. 9. 10. 11. 12. X 0 1 1 0X cosh t sinh t X 0 1 1 0X 1 1 X 1 0 0 2X t e4t X 1 0 0 2X 3 1 X 0 3 5 0 0 1 0 0 0X X 1 1 2 1 1 2 1 1 2X X 0 1 1 0X X 1 0 0 2X A 0 3 5 0 0 1 0 0 0 A 1 1 2 1 1 2 1 1 2 A 0 1 1 0 A 1 0 0 2 13. Solve the system in Problem 7 subject to the initial condition . 14. Solve the system in Problem 9 subject to the initial condition . In Problems 15–18 use the method of Example 2 to com-pute eAt for the coefﬁcient matrix. Use (1) to ﬁnd the general solution of the given system. 15. 16. 17. 18. Let P denote a matrix whose columns are eigenvectors K1, K2, . . . , Kn corresponding to distinct eigenvalues l1, l2, . . . , ln of an n n matrix A. Then it can be shown that A PDP1, where D is a diagonal matrix deﬁned by (9) In Problems 19 and 20 verify the foregoing result for the given matrix. 19. 20. A 2 1 1 2 A 2 3 1 6 l1 0 0 D ( ) . 0 0 ln 0 l2 0 . . . . . . . . . . . . . . . . . . X 0 2 1 2X X 5 1 9 1X X 4 1 2 1X X 4 4 3 4X X(0) 4 3 X(0) 1 4 6 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 21. Suppose A PDP1, where D is deﬁned as in (9). Use (3) to show that eAt PeDtP1. 22. If D is deﬁned as in (9), then ﬁnd eDt. In Problems 23 and 24 use the results of Problems 19–22 to solve the given system. 23. 24. Discussion Problems 25. Reread the discussion leading to the result given in (7). Does the matrix sI A always have an inverse? Discuss. 26. A matrix A is said to be nilpotent if there exists some positive integer m such that Am 0. Verify that is nilpotent. Discuss why it is relatively easy to compute eAt when A is nilpotent. Compute eAt and then use (1) to solve the system X AX. A 1 1 1 1 0 1 1 1 1 X 2 1 1 2X X 2 3 1 6X 360 ● CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS Computer Lab Assignments 27. (a) Use (1) to ﬁnd the general solution of . Use a CAS to ﬁnd eAt. Then use the computer to ﬁnd eigenvalues and eigenvectors of the coefﬁcient matrix and form the general solution in the manner of Section 8.2. Finally, reconcile the two forms of the general solu-tion of the system. (b) Use (1) to ﬁnd the general solution of . Use a CAS to ﬁnd eAt. In the case of complex output, utilize the software to do the simpliﬁcation; for example, in Mathematica, if m MatrixExp[A t] has complex entries, then try the command Simplify[ComplexExpand[m]]. 28. Use (1) to ﬁnd the general solution of . Use MATLAB or a CAS to ﬁnd eAt. X 4 0 1 0 0 5 0 3 6 0 1 0 0 4 0 2 X X 3 2 1 1X A 4 3 2 3 X 4 3 2 3X CHAPTER 8 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-16. In Problems 1 and 2 ﬁll in the blanks. 1. The vector is a solution of for k _. 2. The vector is solution of the initial-value problem for c1 _ and c2 ____. 3. Consider the linear system . Without attempting to solve the system, determine which one of the vectors K1 0 1 1, K2 1 1 1, K3 3 1 1, K4 6 2 5 X 4 1 1 6 3 4 6 2 3X X 1 6 10 3X, X(0) 2 0 X c1 1 1e9t c2 5 3e7t X 1 2 4 1X 8 1 X k 4 5 is an eigenvector of the coefﬁcient matrix. What is the solution of the system corresponding to this eigenvector? 4. Consider the linear system X AX of two differential equations, where A is a real coefﬁcient matrix. What is the general solution of the system if it is known that l1 1 2i is an eigenvalue and is a corre-sponding eigenvector? In Problems 5–14 solve the given linear system. 5. 6. 7. 8. 9. 10. X 0 1 2 2 1 2 1 2 1X X 1 0 4 1 1 3 1 3 1X X 2 2 5 4X X 1 2 2 1X dy dt 2x 4y dy dt x dx dt 4x 2y dx dt 2x y K1 1 i Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 11. 12. 13. 14. 15. (a) Consider the linear system X AX of three ﬁrst-order differential equations, where the coefﬁcient matrix is A 5 3 5 3 5 5 3 3 3 X 3 1 1 1X 2 1e2t X 1 2 1 1X 1 cot t X 1 1 2 2 1X 0 et tan t X 2 0 8 4X 2 16t CHAPTER 8 IN REVIEW ● 361 and l 2 is known to be an eigenvalue of multi-plicity two. Find two different solutions of the sys-tem corresponding to this eigenvalue without using a special formula (such as (12) of Section 8.2). (b) Use the procedure of part (a) to solve . 16. Verify that is a solution of the linear system for arbitrary constants c1 and c2. By hand, draw a phase portrait of the system. X 1 0 0 1X X c1 c2et X 1 1 1 1 1 1 1 1 1X Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 362 Numerical Solutions of Ordinary Differential Equations 9.1 Euler Methods and Error Analysis 9.2 Runge-Kutta Methods 9.3 Multistep Methods 9.4 Higher-Order Equations and Systems 9.5 Second-Order Boundary-Value Problems Chapter 9 in Review Even if it can be shown that a solution of a differential equation exists, we might not be able to exhibit it in explicit or implicit form. In many instances we have to be content with an approximation of the solution. If a solution exists, it represents a set of points in the Cartesian plane. In this chapter we continue to explore the basic idea introduced in Section 2.6, that is, using the differential equation to construct an algorithm to approximate the y-coordinates of points on the actual solution curve. Our concentration in this chapter is primarily on ﬁrst-order initial-value problems We saw in Section 4.10 that numerical procedures developed for ﬁrst-order DEs extend in a natural way to systems of ﬁrst-order equations. Because of this extension, we are able to approximate solutions of a higher-order equation by rewriting it as a system of ﬁrst-order DEs. Chapter 9 concludes with a method for approximating solutions of linear second-order boundary-value problems. dy>dx f(x, y), y(x0) y0. 9 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. A Comparison In Problem 4 in Exercises 2.6 you were asked to use Euler’s method to obtain the approximate value of y(1.5) for the solution of the initial-value problem y 2xy, y(1) 1. You should have obtained the analytic solution and results similar to those given in Tables 9.1.1 and 9.1.2. y ex21 9.1 EULER METHODS AND ERROR ANALYSIS ● 363 TABLE 9.1.1 Euler’s Method with h 0.1 Actual Abs. % Rel. xn yn value error error 1.00 1.0000 1.0000 0.0000 0.00 1.10 1.2000 1.2337 0.0337 2.73 1.20 1.4640 1.5527 0.0887 5.71 1.30 1.8154 1.9937 0.1784 8.95 1.40 2.2874 2.6117 0.3244 12.42 1.50 2.9278 3.4903 0.5625 16.12 TABLE 9.1.2 Euler’s Method with h 0.05 Actual Abs. % Rel. xn yn value error error 1.00 1.0000 1.0000 0.0000 0.00 1.05 1.1000 1.1079 0.0079 0.72 1.10 1.2155 1.2337 0.0182 1.47 1.15 1.3492 1.3806 0.0314 2.27 1.20 1.5044 1.5527 0.0483 3.11 1.25 1.6849 1.7551 0.0702 4.00 1.30 1.8955 1.9937 0.0982 4.93 1.35 2.1419 2.2762 0.1343 5.90 1.40 2.4311 2.6117 0.1806 6.92 1.45 2.7714 3.0117 0.2403 7.98 1.50 3.1733 3.4903 0.3171 9.08 EULER METHODS AND ERROR ANALYSIS REVIEW MATERIAL ●Section 2.6 INTRODUCTION In Chapter 2 we examined one of the simplest numerical methods for approximating solutions of ﬁrst-order initial-value problems y f(x, y), y(x0) y0. Recall that the backbone of Euler’s method is the formula (1) where f is the function obtained from the differential equation y f(x, y). The recursive use of (1) for n 0, 1, 2, . . . yields the y-coordinates y1, y2, y3, . . . of points on successive “tangent lines” to the solution curve at x1, x2, x3, . . . or xn x0 nh, where h is a constant and is the size of the step between xn and xn1. The values y1, y2, y3, . . . approximate the values of a solution y(x) of the IVP at x1, x2, x3, . . . . But whatever advantage (1) has in its simplicity is lost in the crudeness of its approximations. yn1 yn hf (xn, yn), 9.1 In this case, with a step size h 0.1 a 16% relative error in the calculation of the approximation to y(1.5) is totally unacceptable. At the expense of doubling the number of calculations, some improvement in accuracy is obtained by halving the step size to h 0.05. Errors in Numerical Methods In choosing and using a numerical method for the solution of an initial-value problem, we must be aware of the various sources of errors. For some kinds of computation the accumulation of errors might reduce the accuracy of an approximation to the point of making the computation useless. On the other hand, depending on the use to which a numerical solution may be put, extreme accuracy might not be worth the added expense and complication. One source of error that is always present in calculations is round-off error. This error results from the fact that any calculator or computer can represent numbers using only a ﬁnite number of digits. Suppose, for the sake of illustration, that we have Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 364 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS a calculator that uses base 10 arithmetic and carries four digits, so that is repre-sented in the calculator as 0.3333 and is represented as 0.1111. If we use this calculator to compute for x 0.3334, we obtain With the help of a little algebra, however, we see that so when . This exam-ple shows that the effects of round-off error can be quite serious unless some care is taken. One way to reduce the effect of round-off error is to minimize the number of calculations. Another technique on a computer is to use double-precision arithmetic to check the results. In general, round-off error is unpredictable and difﬁcult to ana-lyze, and we will neglect it in the error analysis that follows. We will concentrate on investigating the error introduced by using a formula or algorithm to approximate the values of the solution. Truncation Errors for Euler’s Method In the sequence of values y1, y2, y3, . . . generated from (1), usually the value of y1 will not agree with the actual solu-tion at x1—namely, y(x1)—because the algorithm gives only a straight-line approximation to the solution. See Figure 2.6.2. The error is called the local truncation error, formula error, or discretization error. It occurs at each step; that is, if we assume that yn is accurate, then yn1 will contain local truncation error. To derive a formula for the local truncation error for Euler’s method, we use Taylor’s formula with remainder. If a function y(x) possesses k 1 derivatives that are continuous on an open interval containing a and x, then where c is some point between a and x. Setting k 1, a xn, and x xn1 xn h, we get or Euler’s method (1) is the last formula without the last term; hence the local truncation error in yn1 is The value of c is usually unknown (it exists theoretically), so the exact error cannot be calculated, but an upper bound on the absolute value of the error is In discussing errors that arise from the use of numerical methods, it is helpful to use the notation O(hn). To deﬁne this concept, we let e(h) denote the error in a numerical calculation depending on h. Then e(h) is said to be of order hn, denoted by O(hn), if there exist a constant C and a positive integer n such that e(h) Chn for h sufﬁciently small. Thus the local truncation error for Euler’s method is O(h2). We note that, in general, if e(h) in a numerical method is of order hn and h is halved, the new error is approximately C(h2)n Chn2n; that is, the error is reduced by a factor of 12n. where M max xn x xn1 y(x). Mh2> 2!, y(c) h2 2!, where xn c xn1. yn1 y(xn1) yn hf(xn, yn) y(c) . h2 –– 2! y(xn1) y(xn) y(xn) h 1! y(c) h2 2! y(x) y(a) y(a) x a 1! y(k)(a) (x a)k k! y(k1)(c) (x a)k1 (k 1)! , x 0.3334, (x2 1 9)(x 1 3) 0.3334 0.3333 0.6667 x2 1 9 x 1 3 (x 1 3)(x 1 3) x 1 3 x 1 3, (0.3334)2 0.1111 0.3334 0.3333 0.1112 0.1111 0.3334 0.3333 1. (x2 1 9)(x 1 3) 1 9 1 3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.1 EULER METHODS AND ERROR ANALYSIS ● 365 EXAMPLE 1 Bound for Local Truncation Errors Find a bound for the local truncation errors for Euler’s method applied to y 2xy, y(1) 1. SOLUTION From the solution we get , so the local truncation error is where c is between xn and xn h. In particular, for h 0.1 we can get an upper bound on the local truncation error for y1 by replacing c by 1.1: From Table 9.1.1 we see that the error after the ﬁrst step is 0.0337, less than the value given by the bound. Similarly, we can get a bound for the local truncation error for any of the ﬁve steps given in Table 9.1.1 by replacing c by 1.5 (this value of c gives the largest value of y(c) for any of the steps and may be too generous for the ﬁrst few steps). Doing this gives (2) as an upper bound for the local truncation error in each step. Note that if h is halved to 0.05 in Example 1, then the error bound is 0.0480, about one-fourth as much as shown in (2). This is expected because the local trunca-tion error for Euler’s method is O(h2). In the above analysis we assumed that the value of yn was exact in the calcula-tion of yn1, but it is not because it contains local truncation errors from previous steps. The total error in yn1 is an accumulation of the errors in each of the previous steps. This total error is called the global truncation error. A complete analysis of the global truncation error is beyond the scope of this text, but it can be shown that the global truncation error for Euler’s method is O(h). We expect that, for Euler’s method, if the step size is halved the error will be approximately halved as well. This is borne out in Tables 9.1.1 and 9.1.2 where the absolute error at x 1.50 with h 0.1 is 0.5625 and with h 0.05 is 0.3171, approximately half as large. In general it can be shown that if a method for the numerical solution of a differential equation has local truncation error O(ha1), then the global truncation error is O(ha). For the remainder of this section and in the subsequent sections we study meth-ods that give signiﬁcantly greater accuracy than does Euler’s method. Improved Euler’s Method The numerical method deﬁned by the formula (3) where (4) is commonly known as the improved Euler’s method. To compute yn1 for n 0, 1, 2, . . . from (3), we must, at each step, ﬁrst use Euler’s method (4) to obtain an initial estimate . For example, with n 0, (4) gives , and then, knowing this value, we use (3) to get , where y1 y0 h f (x0, y0) f (x1, y 1 ) 2 y 1 y0 hf(x0, y0) y n1 y n1 yn h f(xn, yn), yn1 yn h f (xn, yn) f (xn1, y n1) 2 , [2 (4)(1.5)2]e((1.5)21) (0.1)2 2 0.1920 [2 (4)(1.1)2]e((1.1)21) (0.1)2 2 0.0422. y(c) h2 2 (2 4c2)e(c21) h2 2 , y (2 4x2)ex21 y ex21 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 366 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS x1 x0 h. These equations can be readily visualized. In Figure 9.1.1, observe that m0 f(x0, y0) and are slopes of the solid straight lines shown passing through the points (x0, y0) and , respectively. By taking an average of these slopes, that is, , we obtain the slope of the parallel dashed skew lines. With the ﬁrst step, rather than advancing along the line through (x0, y0) with slope f(x0, y0) to the point with y-coordinate obtained by Euler’s method, we advance instead along the red dashed line through (x0, y0) with slope mave until we reach x1. It seems plausible from inspection of the ﬁgure that y1 is an improvement over . In general, the improved Euler’s method is an example of a predictor-corrector method. The value of given by (4) predicts a value of y(xn), whereas the value of yn1 deﬁned by formula (3) corrects this estimate. y n1 y 1 y 1 mave f (x0, y0) f (x1, y1 ) 2 (x1, y 1 ) m1 f (x1, y 1 ) ( x 1 , y 1 ) ( x 1 , y 1 ) 0 1 m ave x y x0 x1 h (x0, y0) (x1, ) (x1, ) m0 = f(x0, y0) m1 = f(x1, y 1) (x1, y(x1)) solution curve f(x0, y0) + f(x1, y 1) 2 mave = FIGURE 9.1.1 Slope of red dashed line is the average of m0 and m1 TABLE 9.1.4 Improved Euler’s Method with h 0.05 Actual Abs. % Rel. xn yn value error error 1.00 1.0000 1.0000 0.0000 0.00 1.05 1.1077 1.1079 0.0002 0.02 1.10 1.2332 1.2337 0.0004 0.04 1.15 1.3798 1.3806 0.0008 0.06 1.20 1.5514 1.5527 0.0013 0.08 1.25 1.7531 1.7551 0.0020 0.11 1.30 1.9909 1.9937 0.0029 0.14 1.35 2.2721 2.2762 0.0041 0.18 1.40 2.6060 2.6117 0.0057 0.22 1.45 3.0038 3.0117 0.0079 0.26 1.50 3.4795 3.4904 0.0108 0.31 TABLE 9.1.3 Improved Euler’s Method with h 0.1 Actual Abs. % Rel. xn yn value error error 1.00 1.0000 1.0000 0.0000 0.00 1.10 1.2320 1.2337 0.0017 0.14 1.20 1.5479 1.5527 0.0048 0.31 1.30 1.9832 1.9937 0.0106 0.53 1.40 2.5908 2.6117 0.0209 0.80 1.50 3.4509 3.4904 0.0394 1.13 A brief word of caution is in order here. We cannot compute all the values of ﬁrst and then substitute these values into formula (3). In other words, we cannot use the data in Table 9.1.1 to help construct the values in Table 9.1.3. Why not? Truncation Errors for the Improved Euler’ s Method The local trunca-tion error for the improved Euler’s method is O(h3). The derivation of this result is similar to the derivation of the local truncation error for Euler’s method. Since the y n EXAMPLE 2 Improved Euler’s Method Use the improved Euler’s method to obtain the approximate value of y(1.5) for the solution of the initial-value problem y 2xy, y(1) 1. Compare the results for h 0.1 and h 0.05. SOLUTION With x0 1, y0 1, f(xn, yn) 2xnyn, n 0, and h 0.1, we ﬁrst compute (4): We use this last value in (3) along with x1 1 h 1 0.1 1.1: The comparative values of the calculations for h 0.1 and h 0.05 are given in Tables 9.1.3 and 9.1.4, respectively. y1 y0 (0.1) 2x0y0 2x 1y 1 2 1 (0.1) 2(1)(1) 2(1.1)(1.2) 2 1.232. y 1 y0 (0.1)(2x0y0) 1 (0.1)2(1)(1) 1.2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. local truncation error for the improved Euler’s method is O(h3), the global truncation error is O(h2). This can be seen in Example 2; when the step size is halved from h 0.1 to h 0.05, the absolute error at x 1.50 is reduced from 0.0394 to 0.0108, a reduction of approximately (1 2) 2 1 4. 9.1 EULER METHODS AND ERROR ANALYSIS ● 367 EXERCISES 9.1 Answers to selected odd-numbered problems begin on page ANS-16. In Problems 1–10 use the improved Euler’s method to obtain a four-decimal approximation of the indicated value. First use h 0.1 and then use h 0.05. 1. y 2x 3y 1, y(1) 5; y(1.5) 2. y 4x 2y, y(0) 2; y(0.5) 3. y 1 y2, y(0) 0; y(0.5) 4. y x2 y2, y(0) 1; y(0.5) 5. y ey, y(0) 0; y(0.5) 6. y x y2, y(0) 0; y(0.5) 7. y (x y)2, y(0) 0.5; y(0.5) 8. 9. 10. y y y2, y(0) 0.5; y(0.5) 11. Consider the initial-value problem y (x y 1)2, y(0) 2. Use the improved Euler’s method with h 0.1 and h 0.05 to obtain approximate values of the solution at x 0.5. At each step compare the approximate value with the actual value of the analytic solution. 12. Although it might not be obvious from the differential equation, its solution could “behave badly” near a point x at which we wish to approximate y(x). Numerical pro-cedures may give widely differing results near this point. Let y(x) be the solution of the initial-value prob-lem y x2 y3, y(1) 1. (a) Use a numerical solver to graph the solution on the interval [1, 1.4]. (b) Using the step size h 0.1, compare the results obtained from Euler’s method with the results from the improved Euler’s method in the approximation of y(1.4). 13. Consider the initial-value problem y 2y, y(0) 1. The analytic solution is y e2x. (a) Approximate y(0.1) using one step and Euler’s method. (b) Find a bound for the local truncation error in y1. (c) Compare the error in y1 with your error bound. (d) Approximate y(0.1) using two steps and Euler’s method. y xy2 y x, y(1) 1; y(1.5) y xy 1y, y(0) 1; y(0.5) (e) Verify that the global truncation error for Euler’s method is O(h) by comparing the errors in parts (a) and (d). 14. Repeat Problem 13 using the improved Euler’s method. Its global truncation error is O(h2). 15. Repeat Problem 13 using the initial-value problem y x 2y, y(0) 1. The analytic solution is 16. Repeat Problem 15 using the improved Euler’s method. Its global truncation error is O(h2). 17. Consider the initial-value problem y 2x 3y 1, y(1) 5. The analytic solution is (a) Find a formula involving c and h for the local trunca-tion error in the nth step if Euler’s method is used. (b) Find a bound for the local truncation error in each step if h 0.1 is used to approximate y(1.5). (c) Approximate y(1.5) using h 0.1 and h 0.05 with Euler’s method. See Problem 1 in Exercises 2.6. (d) Calculate the errors in part (c) and verify that the global truncation error of Euler’s method is O(h). 18. Repeat Problem 17 using the improved Euler’s method, which has a global truncation error O(h2). See Problem 1. You might need to keep more than four decimal places to see the effect of reducing the order of the error. 19. Repeat Problem 17 for the initial-value problem y ey, y(0) 0. The analytic solution is y(x) ln(x 1). Approximate y(0.5). See Problem 5 in Exercises 2.6. 20. Repeat Problem 19 using the improved Euler’s method, which has global truncation error O(h2). See Problem 5. You might need to keep more than four decimal places to see the effect of reducing the order of error. Discussion Problems 21. Answer the question “Why not?” that follows the three sentences after Example 2 on page 366. y(x) 1 9 2 3x 38 9 e3(x1). y 1 2x 1 4 5 4e2x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 368 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS Runge-Kutta Methods Fundamentally, all Runge-Kutta methods are gener-alizations of the basic Euler formula (1) of Section 9.1 in that the slope function f is replaced by a weighted average of slopes over the interval xn x xn1. That is, (1) Here the weights wi, i 1, 2, . . . , m, are constants that generally satisfy w1 w2 wm 1, and each ki, i 1, 2, . . . , m, is the function f evalu-ated at a selected point (x, y) for which xn x xn1. We shall see that the ki are defined recursively. The number m is called the order of the method. Observe that by taking m 1, w1 1, and k1 f(xn, yn), we get the familiar Euler formula yn1 yn h f (xn, yn). Hence Euler’s method is said to be a first-order Runge-Kutta method. The average in (1) is not formed willy-nilly, but parameters are chosen so that (1) agrees with a Taylor polynomial of degree m. As we saw in the preceding section, if a function y(x) possesses k 1 derivatives that are continuous on an open interval containing a and x, then we can write where c is some number between a and x. If we replace a by xn and x by xn1 xn h, then the foregoing formula becomes where c is now some number between xn and xn1. When y(x) is a solution of y f (x, y) in the case k 1 and the remainder is small, we see that a Taylor polynomial y(xn1) y(xn) hy(xn) of degree one agrees with the approximation formula of Euler’s method yn1 yn hy n yn h f (xn, yn). 1 2 h2y(c) y(xn1) y(xn h) y(xn) hy(xn) h2 2! y(xn) hk1 (k 1)! y(k1)(c), y(x) y(a) y(a) x a 1! y(a) (x a)2 2! y(k1)(c) (x a)k1 (k 1)! , weighted average yn1 yn h (w1k1 w2k2 … wmkm). RUNGE-KUTTA METHODS REVIEW MATERIAL ●Section 2.6 (see page 78) INTRODUCTION Probably one of the more popular as well as most accurate numerical proce-dures used in obtaining approximate solutions to a ﬁrst-order initial-value problem y f(x, y), y(x0) y0 is the fourth-order Runge-Kutta method. As the name suggests, there are Runge-Kutta methods of different orders. 9.2 A Second-Order Runge-Kutta Method To further illustrate (1), we con-sider now a second-order Runge-Kutta procedure. This consists of ﬁnding con-stants or parameters w1, w2, a, and b so that the formula (2) where k2 f(xn h, yn hk1), k1 f(xn, yn) yn1 yn h(w1k1 w2k2), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. agrees with a Taylor polynomial of degree two. For our purposes it sufﬁces to say that this can be done whenever the constants satisfy (3) This is an algebraic system of three equations in four unknowns and has inﬁnitely many solutions: (4) where w2 0. For example, the choice yields , and so (2) becomes where Since xn h xn1 and yn hk1 yn hf(xn, yn), the foregoing result is rec-ognized to be the improved Euler’s method that is summarized in (3) and (4) of Section 9.1. In view of the fact that w2 0 can be chosen arbitrarily in (4), there are many possible second-order Runge-Kutta methods. See Problem 2 in Exercises 9.2. We shall skip any discussion of third-order methods in order to come to the prin-cipal point of discussion in this section. k1 f (xn, yn) and k2 f (xn h, yn hk1). yn1 yn h 2 (k1 k2), w1 1 2, 1, and 1 w2 1 2 w1 1 w2, 1 2w2 , and 1 2w2 , w1 w2 1, w2 1 2, and w2 1 2. 9.2 RUNGE-KUTTA METHODS ● 369 A Fourth-Order Runge-Kutta Method A fourth-order Runge-Kutta procedure consists of ﬁnding parameters so that the formula (5) where agrees with a Taylor polynomial of degree four. This results in a system of 11 equa-tions in 13 unknowns. The most commonly used set of values for the parameters yields the following result: (6) While other fourth-order formulas are easily derived, the algorithm summarized in (6) is so widely used and recognized as a valuable computational tool it is often referred to as the fourth-order Runge-Kutta method or the classical Runge-Kutta method. It is (6) that we have in mind, hereafter, when we use the abbreviation the RK4 method. You are advised to look carefully at the formulas in (6); note that k2 depends on k1, k3 depends on k2, and k4 depends on k3. Also, k2 and k3 involve approximations to the slope at the midpoint of the interval deﬁned by xn x xn1. xn 1 2 h k4 f (xn h, yn hk3). k3 f (xn 1 2h, yn 1 2hk2) k2 f (xn 1 2h, yn 1 2hk1) k1 f (xn, yn) yn1 yn h 6 (k1 2k2 2k3 k4), k4 f (xn 3h, yn 4hk1 5hk2 6hk3), k3 f (xn 2h, yn 2hk1 3hk2) k2 f (xn 1h, yn 1hk1) k1 f (xn, yn) yn1 yn h(w1k1 w2k2 w3k3 w4k4), Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 370 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS TABLE 9.2.1 RK4 Method with h 0.1 Actual Abs. % Rel. xn yn value error error 1.00 1.0000 1.0000 0.0000 0.00 1.10 1.2337 1.2337 0.0000 0.00 1.20 1.5527 1.5527 0.0000 0.00 1.30 1.9937 1.9937 0.0000 0.00 1.40 2.6116 2.6117 0.0001 0.00 1.50 3.4902 3.4904 0.0001 0.00 Truncation Errors for the RK4 Method In Section 9.1 we saw that global truncation errors for Euler’s method and for the improved Euler’s method are, re-spectively, O(h) and O(h2). Because the ﬁrst equation in (6) agrees with a Taylor polynomial of degree four, the local truncation error for this method is y(5)(c) h55! or O(h5), and the global truncation error is thus O(h4). It is now obvious why Euler’s method, the improved Euler’s method, and (6) are first-, second-, and fourth-order Runge-Kutta methods, respectively. EXAMPLE 1 RK4 Method Use the RK4 method with h 0.1 to obtain an approximation to y(1.5) for the solu-tion of y 2xy, y(1) 1. SOLUTION For the sake of illustration let us compute the case when n 0. From (6) we ﬁnd and therefore The remaining calculations are summarized in Table 9.2.1, whose entries are rounded to four decimal places. Inspection of Table 9.2.1 shows why the fourth-order Runge-Kutta method is so popular. If four-decimal-place accuracy is all that we desire, there is no need to use a smaller step size. Table 9.2.2 compares the results of applying Euler’s, the improved Euler’s, and the fourth-order Runge-Kutta methods to the initial-value problem y 2xy, y(1) 1. (See Tables 9.1.1–9.1.4.) 1 0.1 6 (2 2(2.31) 2(2.34255) 2.715361) 1.23367435. y1 y0 0.1 6 (k1 2k2 2k3 k4) 2(x0 0.1)(y0 0.234255) 2.715361 k4 f(x0 (0.1), y0 (0.1)2.34255) 2(x0 1 2(0.1))(y0 1 2(0.231)) 2.34255 k3 f (x0 1 2(0.1), y0 1 2(0.1)2.31) 2(x0 1 2(0.1))(y0 1 2(0.2)) 2.31 k2 f (x0 1 2(0.1), y0 1 2(0.1)2) k1 f (x0, y0) 2x0y0 2 TABLE 9.2.2 y 2xy, y(1) 1 Comparison of numerical methods with h 0.1 Comparison of numerical methods with h 0.05 Improved Actual Improved Actual xn Euler Euler RK4 value xn Euler Euler RK4 value 1.00 1.0000 1.0000 1.0000 1.0000 1.00 1.0000 1.0000 1.0000 1.0000 1.10 1.2000 1.2320 1.2337 1.2337 1.05 1.1000 1.1077 1.1079 1.1079 1.20 1.4640 1.5479 1.5527 1.5527 1.10 1.2155 1.2332 1.2337 1.2337 1.30 1.8154 1.9832 1.9937 1.9937 1.15 1.3492 1.3798 1.3806 1.3806 1.40 2.2874 2.5908 2.6116 2.6117 1.20 1.5044 1.5514 1.5527 1.5527 1.50 2.9278 3.4509 3.4902 3.4904 1.25 1.6849 1.7531 1.7551 1.7551 1.30 1.8955 1.9909 1.9937 1.9937 1.35 2.1419 2.2721 2.2762 2.2762 1.40 2.4311 2.6060 2.6117 2.6117 1.45 2.7714 3.0038 3.0117 3.0117 1.50 3.1733 3.4795 3.4903 3.4904 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.2 RUNGE-KUTTA METHODS ● 371 TABLE 9.2.3 RK4 Method h Approx. Error 0.1 3.49021064 1.32321089 104 0.05 3.49033382 9.13776090 106 The Runge-Kutta method of order four used in RKF45 is not the same as that given in (6). EXAMPLE 2 Bound for Local Truncation Errors Find a bound for the local truncation errors for the RK4 method applied to y 2xy, y(1) 1. SOLUTION By computing the ﬁfth derivative of the known solution we get (7) Thus with c 1.5, (7) yields a bound of 0.00028 on the local truncation error for each of the ﬁve steps when h 0.1. Note that in Table 9.2.1 the error in y1 is much less than this bound. Table 9.2.3 gives the approximations to the solution of the initial-value problem at x 1.5 that are obtained from the RK4 method. By computing the value of the an-alytic solution at x 1.5, we can ﬁnd the error in these approximations. Because the method is so accurate, many decimal places must be used in the numerical solution to see the effect of halving the step size. Note that when h is halved, from h 0.1 to h 0.05, the error is divided by a factor of about 24 16, as expected. y(5)(c) h5 5! (120c 160c3 32c5)ec21 h5 5!. y(x) ex21, EXERCISES 9.2 Answers to selected odd-numbered problems begin on page ANS-17. 1. Use the RK4 method with h 0.1 to approximate y(0.5), where y(x) is the solution of the initial-value problem y (x y 1)2, y(0) 2. Compare this approximate value with the actual value obtained in Problem 11 in Exercises 9.1. 2. Assume that in (4). Use the resulting second-order Runge-Kutta method to approximate y(0.5), where y(x) is the solution of the initial-value problem in Problem 1. Compare this approximate value with the ap-proximate value obtained in Problem 11 in Exercises 9.1. In Problems 3–12 use the RK4 method with h 0.1 to obtain a four-decimal approximation of the indicated value. 3. y 2x 3y 1, y(1) 5; y(1.5) 4. y 4x 2y, y(0) 2; y(0.5) 5. y 1 y2, y(0) 0; y(0.5) w2 3 4 6. y x2 y2, y(0) 1; y(0.5) 7. y ey, y(0) 0; y(0.5) 8. y x y2, y(0) 0; y(0.5) 9. y (x y)2, y(0) 0.5; y(0.5) 10. 11. 12. y y y2, y(0) 0.5; y(0.5) 13. If air resistance is proportional to the square of the instan-taneous velocity, then the velocity v of a mass m dropped from a given height is determined from Let v(0) 0, k 0.125, m 5 slugs, and g 32 ft/s2. m dv dt mg kv2, k 0. y xy2 y x, y(1) 1; y(1.5) y xy 1y, y(0) 1; y(0.5) Adaptive Methods We have seen that the accuracy of a numerical method for approximating solutions of differential equations can be improved by decreasing the step size h. Of course, this enhanced accuracy is usually obtained at a cost— namely, increased computation time and greater possibility of round-off error. In general, over the interval of approximation there may be subintervals where a rela-tively large step size sufﬁces and other subintervals where a smaller step is necessary to keep the truncation error within a desired limit. Numerical methods that use a vari-able step size are called adaptive methods. One of the more popular of the adaptive routines is the Runge-Kutta-Fehlberg method. Because Fehlberg employed two Runge-Kutta methods of differing orders, a fourth- and a ﬁfth-order method, this al-gorithm is frequently denoted as the RKF45 method. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 372 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS (a) Use the RK4 method with h 1 to approximate the velocity v(5). (b) Use a numerical solver to graph the solution of the IVP on the interval [0, 6]. (c) Use separation of variables to solve the IVP and then ﬁnd the actual value v(5). 14. A mathematical model for the area A (in cm2) that a colony of bacteria (B. dendroides) occupies is given by Suppose that the initial area is 0.24 cm2. (a) Use the RK4 method with h 0.5 to complete the following table: dA dt A(2.128 0.0432A). 18. Consider the initial-value problem y 2x 3y 1, y(1) 5. The analytic solution is (a) Find a formula involving c and h for the local trunca-tion error in the nth step if the RK4 method is used. (b) Find a bound for the local truncation error in each step if h 0.1 is used to approximate y(1.5). (c) Approximate y(1.5) using the RK4 method with h 0.1 and h 0.05. See Problem 3. You will need to carry more than six decimal places to see the effect of reducing the step size. 19. Repeat Problem 18 for the initial-value problem y ey, y(0) 0. The analytic solution is y(x) ln(x 1). Approximate y(0.5). See Problem 7. Discussion Problems 20. A count of the number of evaluations of the function f used in solving the initial-value problem y f(x, y), y(x0) y0 is used as a measure of the computational complexity of a numerical method. Determine the num-ber of evaluations of f required for each step of Euler’s, the improved Euler’s, and the RK4 methods. By consid-ering some speciﬁc examples, compare the accuracy of these methods when used with comparable computa-tional complexities. Computer Lab Assignments 21. The RK4 method for solving an initial-value problem over an interval [a, b] results in a ﬁnite set of points that are supposed to approximate points on the graph of the exact solution. To expand this set of discrete points to an approximate solution deﬁned at all points on the interval [a, b], we can use an interpolating function. This is a function, supported by most computer algebra systems, that agrees with the given data exactly and as-sumes a smooth transition between data points. These interpolating functions may be polynomials or sets of polynomials joined together smoothly. In Mathematica the command yInterpolation[data] can be used to obtain an interpolating function through the points data {{x0, y0}, {x1, y1}, . . . , {xn, yn}}. The inter-polating function y[x] can now be treated like any other function built into the computer algebra system. (a) Find the analytic solution of the initial-value prob-lem y y 10 sin 3x; y(0) 0 on the interval [0, 2]. Graph this solution and ﬁnd its positive roots. (b) Use the RK4 method with h 0.1 to approximate a solution of the initial-value problem in part (a). Obtain an interpolating function and graph it. Find the positive roots of the interpolating function of the interval [0, 2]. y(x) 1 9 2 3x 38 9 e3(x1). t (days) 1 2 3 4 5 A (observed) 2.78 13.53 36.30 47.50 49.40 A (approximated) (b) Use a numerical solver to graph the solution of the initial-value problem. Estimate the values A(1), A(2), A(3), A(4), and A(5) from the graph. (c) Use separation of variables to solve the initial-value problem and compute the actual values A(1), A(2), A(3), A(4), and A(5). 15. Consider the initial-value problem y x2 y3, y(1) 1. See Problem 12 in Exercises 9.1. (a) Compare the results obtained from using the RK4 method over the interval [1, 1.4] with step sizes h 0.1 and h 0.05. (b) Use a numerical solver to graph the solution of the initial-value problem on the interval [1, 1.4]. 16. Consider the initial-value problem y 2y, y(0) 1. The analytic solution is y(x) e2x. (a) Approximate y(0.1) using one step and the RK4 method. (b) Find a bound for the local truncation error in y1. (c) Compare the error in y1 with your error bound. (d) Approximate y(0.1) using two steps and the RK4 method. (e) Verify that the global truncation error for the RK4 method is O(h4) by comparing the errors in parts (a) and (d). 17. Repeat Problem 16 using the initial-value problem y 2y x, y(0) 1. The analytic solution is y(x) 1 2x 1 4 5 4e2x. See V. A. Kostitzin, Mathematical Biology (London: Harrap, 1939). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.3 MULTISTEP METHODS ● 373 MULTISTEP METHODS REVIEW MATERIAL ●Sections 9.1 and 9.2 INTRODUCTION Euler’s method, the improved Euler’s method, and the Runge-Kutta methods are examples of single-step or starting methods. In these methods each successive value yn1 is computed based only on information about the immediately preceding value yn. On the other hand, multistep or continuing methods use the values from several computed steps to obtain the value of yn1. There are a large number of multistep method formulas for approximating solutions of DEs, but since it is not our intention to survey the vast ﬁeld of numerical procedures, we will consider only one such method here. 9.3 Adams-Bashforth-Moulton Method The multistep method that is dis-cussed in this section is called the fourth-order Adams-Bashforth-Moulton method. Like the improved Euler’s method it is a predictor-corrector method—that is, one formula is used to predict a value , which in turn is used to obtain a cor-rected value yn1. The predictor in this method is the Adams-Bashforth formula (1) for n 3. The value of is then substituted into the Adams-Moulton corrector (2) Notice that formula (1) requires that we know the values of y0, y1, y2, and y3 to obtain y4. The value of y0 is, of course, the given initial condition. The local trunca-tion error of the Adams-Bashforth-Moulton method is O(h5), the values of y1, y2, and y3 are generally computed by a method with the same error property, such as the fourth-order Runge-Kutta method. y n1 f (xn1, y n1). yn1 yn h 24(9y n1 19y n 5y n1 y n2) y n1 y n3 f (xn3, yn3) y n2 f (xn2, yn2) y n1 f (xn1, yn1) y n f (xn, yn) y n1 yn h 24 (55y n 59y n1 37y n2 9y n3), y n1 EXAMPLE 1 Adams-Bashforth-Moulton Method Use the Adams-Bashforth-Moulton method with h 0.2 to obtain an approximation to y(0.8) for the solution of SOLUTION With a step size of h 0.2, y(0.8) will be approximated by y4. To get started, we use the RK4 method with x0 0, y0 1, and h 0.2 to obtain y1 1.02140000, y2 1.09181796, y3 1.22210646. y x y 1, y(0) 1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 374 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS Now with the identiﬁcations x0 0, x1 0.2, x2 0.4, x3 0.6, and f (x, y) x y 1, we ﬁnd With the foregoing values the predictor (1) then gives To use the corrector (2), we ﬁrst need Finally, (2) yields You should verify that the actual value of y(0.8) in Example 1 is y(0.8) 1.42554093. See Problem 1 in Exercises 9.3. y4 y3 0.2 24 (9y 4 19y 3 5y 2 y 1) 1.42552788. y 4 f(x4, y 4) 0.8 1.42535975 1 1.22535975. y 4 y3 0.2 24 (55y 3 59y 2 37y 1 9y 0) 1.42535975. y 3 f (x3, y3) (0.6) (1.22210646) 1 0.82210646. y 2 f (x2, y2) (0.4) (1.09181796) 1 0.49181796 y 1 f (x1, y1) (0.2) (1.02140000) 1 0.22140000 y 0 f (x0, y0) (0) (1) 1 0 Stability of Numerical Methods An important consideration in using nu-merical methods to approximate the solution of an initial-value problem is the stabil-ity of the method. Simply stated, a numerical method is stable if small changes in the initial condition result in only small changes in the computed solution. A numerical method is said to be unstable if it is not stable. The reason that stability considera-tions are important is that in each step after the ﬁrst step of a numerical technique we are essentially starting over again with a new initial-value problem, where the initial condition is the approximate solution value computed in the preceding step. Because of the presence of round-off error, this value will almost certainly vary at least slightly from the true value of the solution. Besides round-off error, another common source of error occurs in the initial condition itself; in physical applications the data are often obtained by imprecise measurements. One possible method for detecting instability in the numerical solution of a spe-ciﬁc initial-value problem is to compare the approximate solutions obtained when decreasing step sizes are used. If the numerical method is unstable, the error may actually increase with smaller step sizes. Another way of checking stability is to observe what happens to solutions when the initial condition is slightly perturbed (for example, change y(0) 1 to y(0) 0.999). For a more detailed and precise discussion of stability, consult a numerical analysis text. In general, all of the methods that we have discussed in this chapter have good stability characteristics. Advantages and Disadvantages of Multistep Methods Many consid-erations enter into the choice of a method to solve a differential equation numerically. Single-step methods, particularly the RK4 method, are often chosen because of their accuracy and the fact that they are easy to program. However, a major drawback is that the right-hand side of the differential equation must be eval-uated many times at each step. For instance, the RK4 method requires four function evaluations for each step. On the other hand, if the function evaluations in the previous step have been calculated and stored, a multistep method requires only one new function evaluation for each step. This can lead to great savings in time and expense. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. As an example, solving y f(x, y), y(x0) y0 numerically using n steps by the fourth-order Runge-Kutta method requires 4n function evaluations. The Adams-Bashforth multistep method requires 16 function evaluations for the Runge-Kutta fourth-order starter and n 4 for the n Adams-Bashforth steps, giving a total of n 12 function evaluations for this method. In general the Adams-Bashforth multi-step method requires slightly more than a quarter of the number of function evalua-tions required for the RK4 method. If the evaluation of f(x, y) is complicated, the multistep method will be more efﬁcient. Another issue that is involved with multistep methods is how many times the Adams-Moulton corrector formula should be repeated in each step. Each time the corrector is used, another function evaluation is done, and so the accuracy is increased at the expense of losing an advantage of the multistep method. In prac-tice, the corrector is calculated once, and if the value of yn1 is changed by a large amount, the entire problem is restarted using a smaller step size. This is often the basis of the variable step size methods, whose discussion is beyond the scope of this text. 9.4 HIGHER-ORDER EQUATIONS AND SYSTEMS ● 375 EXERCISES 9.3 Answers to selected odd-numbered problems begin on page ANS-17. 1. Find the analytic solution of the initial-value problem in Example 1. Compare the actual values of y(0.2), y(0.4), y(0.6), and y(0.8) with the approximations y1, y2, y3, and y4. 2. Write a computer program to implement the Adams-Bashforth-Moulton method. In Problems 3 and 4 use the Adams-Bashforth-Moulton method to approximate y(0.8), where y(x) is the solution of the given initial-value problem. Use h 0.2 and the RK4 method to compute y1, y2, and y3. 3. y 2x 3y 1, y(0) 1 4. y 4x 2y, y(0) 2 In Problems 5–8 use the Adams-Bashforth-Moulton method to approximate y(1.0), where y(x) is the solution of the given initial-value problem. First use h 0.2 and then use h 0.1. Use the RK4 method to compute y1, y2, and y3. 5. y 1 y2, y(0) 0 6. y y cos x, y(0) 1 7. y (x y)2, y(0) 0 8. y xy 1y, y(0) 1 HIGHER-ORDER EQUATIONS AND SYSTEMS REVIEW MATERIAL ●Section 1.1 (normal form of a second-order DE) ●Section 4.10 (second-order DE written as a system of ﬁrst-order DEs) INTRODUCTION So far, we have focused on numerical techniques that can be used to approx-imate the solution of a ﬁrst-order initial-value problem y f(x, y), y(x0) y0. In order to approxi-mate the solution of a second-order initial-value problem, we must express a second-order DE as a system of two ﬁrst-order DEs. To do this, we begin by writing the second-order DE in normal form by solving for y in terms of x, y, and y. 9.4 Second-Order IVPs A second-order initial-value problem (1) y f(x, y, y), y(x0) y0, y(x0) u0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 376 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS x y 2 1 2 1 0.2 Euler’s method approximate y(0.2) (a) Euler’s method (red) and the RK4 method (blue) (b) RK4 method x y 10 20 5 15 2 1 RK4 method FIGURE 9.4.1 Numerical solution curves generated by different methods can be expressed as an initial-value problem for a system of first-order differen-tial equations. If we let y u, the differential equation in (1) becomes the system (2) Since y(x0) u(x0), the corresponding initial conditions for (2) are then y(x0) y0, u(x0) u0. The system (2) can now be solved numerically by simply applying a par-ticular numerical method to each ﬁrst-order differential equation in the system. For example, Euler’s method applied to the system (2) would be (3) whereas the fourth-order Runge-Kutta method, or RK4 method, would be (4) where In general, we can express every nth-order differential equation y(n) f(x, y, y, . . . , y(n1)) as a system of n first-order equations using the substitutions y u1, y u2, y u3, . . . , y (n1) un. k4 f (xn h, yn hm3, un hk3). m4 un hk3 k3 f (xn 1 2h, yn 1 2hm2, un 1 2hk2) m3 un 1 2hk2 k2 f (xn 1 2h, yn 1 2hm1, un 1 2hk1) m2 un 1 2hk1 k1 f (xn, yn, un) m1 un un1 un h 6 (k1 2k2 2k3 k4) yn1 yn h 6 (m1 2m2 2m3 m4) un1 un h f (xn, yn, un), yn1 yn hun u f(x, y, u). y u EXAMPLE 1 Euler’s Method Use Euler’s method to obtain the approximate value of y(0.2), where y(x) is the solution of the initial-value problem (5) SOLUTION In terms of the substitution y u, the equation is equivalent to the system . Thus from (3) we obtain Using the step size h 0.1 and y0 1, u0 2, we ﬁnd In other words, y(0.2) 1.39 and y(0.2) 1.761. With the aid of the graphing feature of a numerical solver, in Figure 9.4.1(a) we compare the solution curve of (5) generated by Euler’s method (h 0.1) on the u2 u1 (0.1)[x1u1 y1] 1.9 (0.1)[(0.1)(1.9) 1.2] 1.761. y2 y1 (0.1)u1 1.2 (0.1)(1.9) 1.39 u1 u0 (0.1) [x0u0 y0] 2 (0.1)[(0)(2) 1] 1.9 y1 y0 (0.1)u0 1 (0.1)2 1.2 un1 un h[xnun yn]. yn1 yn hun u xu y y u y xy y 0, y(0) 1, y(0) 2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.4 HIGHER-ORDER EQUATIONS AND SYSTEMS ● 377 interval [0, 3] with the solution curve generated by the RK4 method (h 0.1). From Figure 9.4.1(b) it appears that the solution y(x) of (4) has the property that and . If desired, we can use the method of Section 6.2 to obtain two power series solutions of the differential equation in (5). But unless this method reveals that the DE possesses an elementary solution, we will still only be able to approximate y(0.2) using a partial sum. Reinspection of the inﬁnite series solutions of Airy’s differential equation y xy 0, given on page 242, does not reveal the oscillatory behavior of the solutions y1(x) and y2(x) exhibited in the graphs in Figure 6.2.2. Those graphs were obtained from a numerical solver using the RK4 method with a step size of h 0.1. Systems Reduced to First-Order Systems Using a procedure similar to that just discussed for second-order equations, we can often reduce a system of higher-order differential equations to a system of ﬁrst-order equations by ﬁrst solv-ing for the highest-order derivative of each dependent variable and then making appropriate substitutions for the lower-order derivatives. x : y(x) : 0 EXAMPLE 2 A System Rewritten as a First-Order System Write as a system of ﬁrst-order differential equations. SOLUTION Write the system as and then eliminate y by multiplying the second equation by 2 and subtracting. This gives . Since the second equation of the system already expresses the highest-order derivative of y in terms of the remaining functions, we are now in a position to introduce new vari-ables. If we let x u and y v, the expressions for x and y become, respectively, The original system can then be written in the form It might not always be possible to carry out the reductions illustrated in Example 2. Numerical Solution of a System The solution of a system of the form g1(t, x1, x2, . . . , xn) g2(t, x1, x2, . . . , xn) gn(t, x1, x2, . . . , xn) . . . . . . dx1 ––– dt dx2 ––– dt dxn ––– dt v 2x 2y 3t2. u 9x 4y u et 6t2 y v x u v y 2x 2y 3t2. u x 9x 4y u et 6t2 x 9x 4y x et 6t2 y 3t2 2x 2y x 2y et 5x x 2x y 2y 3t2 x x 5x 2y et Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 378 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS can be approximated by a version of Euler’s, the Runge-Kutta, or the Adams-Bashforth-Moulton method adapted to the system. For instance, the RK4 method applied to the system (6) looks like this: (7) yn1 yn h 6 (k1 2k2 2k3 k4), xn1 xn h 6 (m1 2m2 2m3 m4) x(t0) x0, y(t0) y0, y g(t, x, y) x f(t, x, y) where (8) k4 g(tn h, xn hm3, yn hk3). m4 f(tn h, xn hm3, yn hk3) k3 g(tn 1 2 h, xn 1 2 hm2, yn 1 2 hk2) m3 f(tn 1 2h, xn 1 2 hm2, yn 1 2 hk2) k2 g(tn 1 2 h, xn 1 2 hm1, yn 1 2 hk1) m2 f(tn 1 2 h, xn 1 2 hm1, yn 1 2 hk1) k1 g(tn, xn, yn) m1 f (tn, xn, yn) EXAMPLE 3 RK4 Method Consider the initial-value problem Use the RK4 method to approximate x(0.6) and y(0.6). Compare the results for h 0.2 and h 0.1. SOLUTION We illustrate the computations of x1 and y1 with step size h 0.2. With the identiﬁcations f(t, x, y) 2x 4y, g(t, x, y) x 6y, t0 0, x0 1, and y0 6 we see from (8) that k4 g(t0 h, x0 hm3, y0 hk3) g(0.2, 9.608, 19.416) 106.888. m4 f (t0 h, x0 hm3, y0 hk3) f (0.2, 9.608, 19.416) 96.88 k3 g(t0 1 2h, x0 1 2hm2, y0 1 2hk2) g(0.1, 3.12, 11.7) 67.08 m3 f (t0 1 2h, x0 1 2hm2, y0 1 2hk2) f (0.1, 3.12, 11.7) 53.04 k2 g(t0 1 2h, x0 1 2hm1, y0 1 2hk1) g(0.1, 1.2, 9.7) 57 m2 f (t0 1 2h, x0 1 2hm1, y0 1 2hk1) f (0.1, 1.2, 9.7) 41.2 k1 g(t0, x0, y0) g(0, 1, 6) 1(1) 6(6) 37 m1 f (t0, x0, y0) f (0, 1, 6) 2(1) 4(6) 22 x(0) 1, y(0) 6. y x 6y x 2x 4y Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.4 HIGHER-ORDER EQUATIONS AND SYSTEMS ● 379 Therefore from (7) we get where, as usual, the computed values of x1 and y1 are rounded to four decimal places. These numbers give us the approximation x1 x(0.2) and y1 y(0.2). The subsequent values, obtained with the aid of a computer, are summarized in Tables 9.4.1 and 9.4.2. You should verify that the solution of the initial-value problem in Example 3 is given by x(t) (26t 1)e4t, y(t) (13t 6)e4t. From these equations we see that the actual values x(0.6) 160.9384 and y(0.6) 152.1198 compare favorably with the entries in the last line of Table 9.4.2. The graph of the solution in a neighborhood of t 0 is shown in Figure 9.4.2; the graph was obtained from a numerical solver using the RK4 method with h 0.1. In conclusion, we state Euler’s method for the general system (6): yn1 yn hg(tn, xn, yn). xn1 xn h f(tn, xn, yn) 6 0.2 6 (37 2(57) 2(67.08) 106.888) 19.0683, y1 y0 0.2 6 (k1 2k2 2k3 k4) 1 0.2 6 (22 2(41.2) 2(53.04) 96.88) 9.2453 x1 x0 0.2 6 (m1 2m2 2m3 m4) FIGURE 9.4.3 Network in Problem 6 i1 i2 i3 R R L L R E t x, y _1 1 y(t) x(t) FIGURE 9.4.2 Numerical solution curves for IVP in Example 3 TABLE 9.4.1 h 0.2 tn xn yn 0.00 1.0000 6.0000 0.20 9.2453 19.0683 0.40 46.0327 55.1203 0.60 158.9430 150.8192 TABLE 9.4.2 h 0.1 tn xn yn 0.00 1.0000 6.0000 0.10 2.3840 10.8883 0.20 9.3379 19.1332 0.30 22.5541 32.8539 0.40 46.5103 55.4420 0.50 88.5729 93.3006 0.60 160.7563 152.0025 EXERCISES 9.4 Answers to selected odd-numbered problems begin on page ANS-17. 1. Use Euler’s method to approximate y(0.2), where y(x) is the solution of the initial-value problem Use h 0.1. Find the analytic solution of the problem, and compare the actual value of y(0.2) with y2. 2. Use Euler’s method to approximate y(1.2), where y(x) is the solution of the initial-value problem where x 0. Use h 0.1. Find the analytic solution of the problem, and compare the actual value of y(1.2) with y2. In Problems 3 and 4 repeat the indicated problem using the RK4 method. First use h 0.2 and then use h 0.1. 3. Problem 1 4. Problem 2 5. Use the RK4 method to approximate y(0.2), where y(x) is the solution of the initial-value problem First use h 0.2 and then use h 0.1. y 2y 2y et cos t, y(0) 1, y(0) 2. x2y 2xy 2y 0, y(1) 4, y(1) 9, y 4y 4y 0, y(0) 2, y(0) 1. 6. When E 100 V, R 10 , and L 1 h, the sys-tem of differential equations for the currents i1(t) and i3(t) in the electrical network given in Figure 9.4.3 is where i1(0) 0 and i3(0) 0. Use the RK4 method to approximate i1(t) and i3(t) at t 0.1, 0.2, 0.3, 0.4, and 0.5. Use h 0.1. Use a numerical solver to graph the solution for 0 t 5. Use the graphs to predict the behavior of i1(t) and i3(t) as . t : di3 dt 10i1 20i3, di1 dt 20i1 10i3 100 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 380 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS In Problems 7–12 use the Runge-Kutta method to approxi-mate x(0.2) and y(0.2). First use h 0.2 and then use h 0.1. Use a numerical solver and h 0.1 to graph the so-lution in a neighborhood of t 0. 7. x 2x y 8. x x 2y y x y 4x 3y x(0) 6, y(0) 2 x(0) 1, y(0) 1 9. x y t 10. x 6x y 6t y x t y 4x 3y 10t 4 x(0) 3, y(0) 5 x(0) 0.5, y(0) 0.2 11. x 4x y 7t 12. x y 4t x y 2y 3t x y y 6t2 10 x(0) 1, y(0) 2 x(0) 3, y(0) 1 SECOND-ORDER BOUNDARY-VALUE PROBLEMS REVIEW MATERIAL ●Section 4.1 (page 118) ●Exercises 4.3 (Problems 37–40) ●Exercises 4.4 (Problems 37–40) ●Section 5.2 INTRODUCTION We just saw in Section 9.4 how to approximate the solution of a second-order initial-value problem y f (x, y, y), y(x0) y0, y(x0) u0. In this section we are going to examine two methods for approximating a solution of a second-order boundary-value problem y f (x, y, y), y(a) a, y(b) b. Unlike the procedures that are used with second-order initial-value problems, the methods of second-order boundary-value problems do not require writing the second-order DE as a system of first-order DEs. 9.5 Finite Difference Approximations The Taylor series expansion, centered at a point a, of a function y(x) is If we set h x a, then the preceding line is the same as For the subsequent discussion it is convenient then to rewrite this last expression in two alternative forms: (1) and (2) If h is small, we can ignore terms involving h4, h5, . . . since these values are neg-ligible. Indeed, if we ignore all terms involving h2 and higher, then solving (1) and (2), in turn, for y(x) yields the following approximations for the ﬁrst derivative: (3) (4) y(x) 1 h [y(x) y(x h)]. y(x) 1 h [y(x h) y(x)] y(x h) y(x) y(x)h y(x) h2 2 y (x) h3 6 . y(x h) y(x) y(x)h y(x) h2 2 y (x) h3 6 y(x) y(a) y(a) h 1! y(a) h2 2! y (a) h3 3! . y(x) y(a) y(a) x a 1! y(a) (x a)2 2! y (a) (x a)3 3! . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.5 SECOND-ORDER BOUNDARY-VALUE PROBLEMS ● 381 Subtracting (1) and (2) also gives (5) On the other hand, if we ignore terms involving h3 and higher, then by adding (1) and (2), we obtain an approximation for the second derivative y(x): (6) The right-hand sides of (3), (4), (5), and (6) are called difference quotients. The expressions and are called finite differences. Speciﬁcally, y(x h) y(x) is called a forward difference, y(x) y(x h) is a backward difference, and both y(x h) y(x h) and y(x h) 2y(x) y(x h) are called central differences. The results given in (5) and (6) are referred to as central difference approximations for the derivatives y and y. Finite Difference Method Consider now a linear second-order boundary-value problem (7) Suppose represents a regular partition of the interval [a, b], that is, xi a ih, where i 0, 1, 2, . . . , n and h (b a)n. The points are called interior mesh points of the interval [a, b]. If we let and if y and y in (7) are replaced by the central difference approximations (5) and (6), we get or, after simplifying, (8) The last equation, known as a finite difference equation, is an approximation to the differential equation. It enables us to approximate the solution y(x) of (7) at the interior mesh points x1, x2, . . . , xn1 of the interval [a, b]. By letting i take on the values 1, 2, . . . , n 1 in (8), we obtain n 1 equations in the n 1 unknowns y1, y2, . . . , yn1. Bear in mind that we know y0 and yn, since these are the prescribed boundary conditions y0 y(x0) y(a) a and yn y(xn) y(b) b. In Example 1 we consider a boundary-value problem for which we can compare the approximate values that we ﬁnd with the actual values of an explicit solution. 1 h 2 Piyi1 (2 h2Qi)yi 1 h 2 Piyi1 h2 fi. yi1 2 yi yi1 h2 Pi yi1 yi1 2 h Qiyi fi yi y(xi), Pi P(xi), Qi Q(xi), and fi f(xi) x1 a h, x2 a 2h, . . . , xn1 a (n 1)h a x0 x1 x2 xn1 xn b y P(x)y Q(x)y f(x), y(a) , y(b) . y(x h) 2y(x) y(x h) y(x h) y(x), y(x) y(x h), y(x h) y(x h), y(x) 1 h2 [y(x h) 2y(x) y(x h)]. y(x) 1 2h [ y(x h) y(x h)]. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 382 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS EXAMPLE 1 Using the Finite Difference Method Use the difference equation (8) with n 4 to approximate the solution of the boundary-value problem y 4y 0, y(0) 0, y(1) 5. SOLUTION To use (8), we identify P(x) 0, Q(x) 4, f(x) 0, and . Hence the difference equation is (9) Now the interior points are , so for i 1, 2, and 3, (9) yields the following system for the corresponding y1, y2, and y3: With the boundary conditions y0 0 and y4 5 the foregoing system becomes 2.25y1 y2 0 y1 2.25y2 y3 0 y2 2.25y3 5. Solving the system gives y1 0.7256, y2 1.6327, and y3 2.9479. Now the general solution of the given differential equation is y c1 cosh 2x c2 sinh 2x. The condition y(0) 0 implies that c1 0. The other boundary condition gives c2. In this way we see that a solution of the boundary-value problem is y(x) (5 sinh 2x)sinh 2. Thus the actual values (rounded to four decimal places) of this solution at the interior points are as follows: y(0.25) 0.7184, y(0.5) 1.6201, and y(0.75) 2.9354. The accuracy of the approximations in Example 1 can be improved by using a smaller value of h. Of course, the trade-off here is that a smaller value of h necessitates solving a larger system of equations. It is left as an exercise to show that with , approximations to y(0.25), y(0.5), and y(0.75) are 0.7202, 1.6233, and 2.9386, respec-tively. See Problem 11 in Exercises 9.5. h 1 8 y4 2.25y3 y2 0. y3 2.25y2 y1 0 y2 2.25y1 y0 0 x1 0 1 4, x2 0 2 4, x3 0 3 4 yi1 2.25yi yi1 0. h (1 0)>4 1 4 EXAMPLE 2 Using the Finite Difference Method Use the difference equation (8) with n 10 to approximate the solution of SOLUTION In this case we identify P(x) 3, Q(x) 2, f(x) 4x2, and h (2 1)10 0.1, and so (8) becomes (10) Now the interior points are x1 1.1, x2 1.2, x3 1.3, x4 1.4, x5 1.5, x6 1.6, x7 1.7, x8 1.8, and x9 1.9. For i 1, 2, . . . , 9 and y0 1, y10 6, (10) gives a system of nine equations and nine unknowns: 1.15y5 1.98y4 0.85y3 0.0784 1.15y4 1.98y3 0.85y2 0.0676 1.15y3 1.98y2 0.85y1 0.0576 1.15y2 1.98y1 0.8016 1.15yi1 1.98yi 0.85yi1 0.04xi 2. y 3y 2y 4x2, y(1) 1, y(2) 6. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. We can solve this large system using Gaussian elimination or, with relative ease, by means of a computer algebra system. The result is found to be y1 2.4047, y2 3.4432, y3 4.2010, y4 4.7469, y5 5.1359, y6 5.4124, y7 5.6117, y8 5.7620, and y9 5.8855. Shooting Method Another way of approximating a solution of a boundary- 1.98y9 0.85y8 6.7556. 1.15y9 1.98y8 0.85y7 0.1296 1.15y8 1.98y7 0.85y6 0.1156 1.15y7 1.98y6 0.85y5 0.1024 1.15y6 1.98y5 0.85y4 0.0900 9.5 SECOND-ORDER BOUNDARY-VALUE PROBLEMS ● 383 REMARKS The approximation method using ﬁnite differences can be extended to boundary-value problems in which the ﬁrst derivative is speciﬁed at a boundary—for example, a problem such as y f(x, y, y), y(a) a, y(b) b. See Problem 13 in Exercises 9.5. EXERCISES 9.5 Answers to selected odd-numbered problems begin on page ANS-17. In Problems 1–10 use the ﬁnite difference method and the indicated value of n to approximate the solution of the given boundary-value problem. 1. y 9y 0, y(0) 4, y(2) 1; n 4 2. y y x2, y(0) 0, y(1) 0; n 4 3. y 2y y 5x, y(0) 0, y(1) 0; n 5 4. y 10y 25y 1, y(0) 1, y(1) 0; n 5 5. y 4y 4y (x 1)e2x, y(0) 3, y(1) 0; n 6 6. y 5y 41x, y(1) 1, y(2) 1; n 6 7. x2y 3xy 3y 0, y(1) 5, y(2) 0; n 8 8. x2y xy y ln x, y(1) 0, y(2) 2; n 8 9. y (1 x)y xy x, y(0) 0, y(1) 2; n 10 10. y xy y x, y(0) 1, y(1) 0; n 10 11. Rework Example 1 using n 8. 12. The electrostatic potential u between two concentric spheres of radius r 1 and r 4 is determined from d2u dr2 2 r du dr 0, u(1) 50, u(4) 100. value problem y f(x, y, y), y(a) a, y(b) b is called the shooting method. The starting point in this method is the replacement of the second-order boundary-value problem by a second-order initial-value problem (11) The number m1 in (11) is simply a guess for the unknown slope of the solution curve at the known point (a, y(a)). We then apply one of the step-by-step numerical techniques to the second-order equation in (11) to ﬁnd an approximation b1 for the value of y(b). If b1 agrees with the given value y(b) b to some preassigned tolerance, we stop; other-wise, the calculations are repeated, starting with a different guess y(a) m2 to obtain a second approximation b2 for y(b). This method can be continued in a trial-and-error manner, or the subsequent slopes m3, m4, . . . can be adjusted in some systematic way; linear interpolation is particularly successful when the differential equation in (11) is linear. The procedure is analogous to shooting (the “aim” is the choice of the initial slope) at a target until the bull’s-eye y(b) is hit. See Problem 14 in Exercises 9.5. Of course, underlying the use of these numerical methods is the assumption, which we know is not always warranted, that a solution of the boundary-value prob-lem exists. y f(x, y, y), y(a) a, y(a) m1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 384 ● CHAPTER 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS Use the method of this section with n 6 to approxi-mate the solution of this boundary-value problem. 13. Consider the boundary-value problem y xy 0, y(0) 1, y(1) 1. (a) Find the difference equation corresponding to the differential equation. Show that for i 0, 1, 2, . . . , n 1 the difference equation yields n equations in n 1 unknows y1, y0, y1, y2, . . . , yn1. Here y1 and y0 are unknowns, since y1 represents an approximation to y at the exterior point x h and y0 is not speciﬁed at x 0. (b) Use the central difference approximation (5) to show that y1 y1 2h. Use this equation to elim-inate y1 from the system in part (a). (c) Use n 5 and the system of equations found in parts (a) and (b) to approximate the solution of the original boundary-value problem. Computer Lab Assignments 14. Consider the boundary-value problem y y sin (xy), y(0) 1, y(1) 1.5. Use the shooting method to approx-imate the solution of this problem. (The approximation can be obtained using a numerical technique—say, the RK4 method with h 0.1; or, even better, if you have access to a CAS such as Mathematica or Maple, the NDSolve function can be used.) CHAPTER 9 IN REVIEW Answers to selected odd-numbered problems begin on page ANS-17. In Problems 1–4 construct a table comparing the indicated values of y(x) using Euler’s method, the improved Euler’s method, and the RK4 method. Compute to four rounded dec-imal places. First use h 0.1 and then use h 0.05. 1. y 2 ln xy, y(1) 2; y(1.1), y(1.2), y(1.3), y(1.4), y(1.5) 2. y sin x2 cos y2, y(0) 0; y(0.1), y(0.2), y(0.3), y(0.4), y(0.5) 3. y(0.6), y(0.7), y(0.8), y(0.9), y(1.0) 4. y xy y2, y(1) 1; y(1.1), y(1.2), y(1.3), y(1.4), y(1.5) 5. Use Euler’s method to approximate y(0.2), where y(x) is the solution of the initial-value problem y (2x 1)y 1, y(0) 3, y(0) 1. First use one step with h 0.2 and then repeat the calculations using two steps with h 0.1. y 1x y, y(0.5) 0.5; 6. Use the Adams-Bashforth-Moulton method to approxi-mate y(0.4), where y(x) is the solution of the initial-value problem y 4x 2y, y(0) 2. Use h 0.1 and the RK4 method to compute y1, y2, and y3. 7. Use Euler’s method with h 0.1 to approximate x(0.2) and y(0.2), where x(t), y(t) is the solution of the initial-value problem 8. Use the ﬁnite difference method with n 10 to approximate the solution of the boundary-value problem y 6.55(1 x)y 1, y(0) 0, y(1) 0. x(0) 1, y(0) 2. y x y x x y Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Euler’s integral deﬁnition of the gamma function is (1) Convergence of the integral requires that x 1 1 or x 0. The recurrence relation (2) which we saw in Section 6.4, can be obtained from (1) with integration by parts. Now when and thus (2) gives , and so on. In this manner, it is seen that when n is a positive integer, (n 1) n!. For this reason, the gamma function is often called the generalized factorial function. Although the integral form (1) does not converge for x 0, it can be shown by means of alternative deﬁnitions that the gamma function is deﬁned for all real and complex numbers except x n, n 0, 1, 2, . . . . As a consequence, (2) is actually valid for x n. The graph of (x), considered as a function of a real variable x, is as given in Figure I.1. Observe that the nonpositive integers correspond to vertical asymptotes of the graph. In Problems 31 and 32 of Exercises 6.4 we used the fact that This result can be derived from (1) by setting : (3) When we let t u2, (3) can be written as But so Switching to polar coordinates u r cos u, v r sin u enables us to evaluate the double integral: Hence (4) [(1 2)] 2 or (1 2) 1 . 4 0 0 e(u2v2 ) du dv 4 /2 0 0 er 2 r dr d . [(1 2)] 2 2 0 eu2 du2 0 ev2 dv 4 0 0 e(u2v2) du dv. 0 eu2 du 0 ev2 dv, (1 2) 2 0 eu2 du. (1 2) 0 t1/2et dt. x 1 2 (1 2) 1 . (4) 3(3) 3 2 1 (3) 2(2) 2 1 (2) 1(1) 1 x 1, (1) 0 et dt 1, (x 1) x(x), (x) 0 tx1et dt. FIGURE I.1 Graph of for x neither 0 nor a negative integer (x) APP-1 Γ(x) x Appendix I Gamma Function Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. EXAMPLE 1 Value of Evaluate SOLUTION In view of (2) and (4), it follows that, with Therefore (1 2) 2(1 2) 21 . (1 2) 1 2(1 2). x 1 2, (1 2). (1 2) EXERCISES FOR APPENDIX I Answers to selected odd-numbered problems begin on page ANS-31. 1. Evaluate. (a) (5) (b) (7) (c) (d) 2. Use (1) and the fact that to evaluate [Hint: Let t x5.] 3. Use (1) and the fact that to evaluate 4. Evaluate [Hint: Let t ln x.] 1 0 x3 ln 1 x 3 dx. 0 x4ex3 dx. (5 3) 0.89 0 x5ex 5 dx. (6 5) 0.92 (5 2) (3 2) 5. Use the fact that to show that (x) is unbounded as 6. Use (1) to derive (2) for x 0. 7. A deﬁnition of the gamma function due to Carl Friedrich Gauss that is valid for all real numbers, except is given by Use this deﬁnition to show that . (x 1) x(x) (x) lim n: n!nx x(x 1)(x 2) . . . (x n). x 0, 1, 2, . . . , x : 0. (x) 1 0 t x1et dt APP-2 ● APPENDIX I GAMMA FUNCTION Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. II.1 BASIC DEFINITIONS AND THEORY APP-3 Appendix II Matrices DEFINITION II.1 Matrix A matrix A is any rectangular array of numbers or functions: (1) a11 a21 am1 a1n a2n amn a12 a22 am2 . . . . . . . . . A ( . ) . . . . . . If a matrix has m rows and n columns, we say that its size is m by n (written m n). An n n matrix is called a square matrix of order n. The entry in the ith row and jth column of an m n matrix A is written aij. An m n matrix A is then abbreviated as A (aij)mn or simply A (aij). A 1 1 matrix is sximply one constant or function. DEFINITION II.2 Equality of Matrices Two m n matrices A and B are equal if aij bij for each i and j. DEFINITION II.3 Column Matrix A column matrix X is any matrix having n rows and one column: b11 b21 bn1 X ( ) (bi1)n1. . . . A column matrix is also called a column vector or simply a vector. DEFINITION II.4 Multiples of Matrices A multiple of a matrix A is deﬁned to be where k is a constant or a function. ka11 ka21 kam1 ka1n ka2n kamn ka12 ka22 kam2 . . . . . . . . . kA ( ) (kaij)mn, . . . . . . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APP-4 ● APPENDIX II MATRICES EXAMPLE 1 Multiples of Matrices (a) (b) We note in passing that, for any matrix A, the product kA is the same as Ak. For example, e3t 2 5 2e3t 5e3t 2 5 e3t. et 1 2 4 et 2et 4et 5 2 4 1 5 3 1 6 10 20 1 15 5 30 DEFINITION II.5 Addition of Matrices The sum of two m n matrices A and B is deﬁned to be the matrix A B (aij bij)mn. In other words, when adding two matrices of the same size, we add the correspond-ing entries. EXAMPLE 2 Matrix Addition The sum of and is A B 2 4 1 7 3 (8) 0 9 4 3 6 5 6 1 10 (1) 5 2 6 6 5 9 7 11 5 9 3. B 4 7 8 9 3 5 1 1 2 A 2 1 3 0 4 6 6 10 5 EXAMPLE 3 A Matrix Written as a Sum of Column Matrices The single matrix can be written as the sum of three column vectors: The difference of two m n matrices is deﬁned in the usual manner: A B A (B), where B (1)B. 3t2 2et t2 7t 5t 3t2 t2 0 0 7t 5t 2et 0 0 3 1 0t 2 0 7 5t 2 0 0et. 3t2 2et t2 7t 5t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-5 q q DEFINITION II.6 Multiplication of Matrices Let A be a matrix having m rows and n columns and B be a matrix having n rows and p columns. We deﬁne the product AB to be the m p matrix ( aikbkj)mp . k1 n a11b11 a12b21 a21b11 a22b21 am1b11 am2b21 a1nbn1 a2nbn1 amnbn1 ( ) . . . . . . . . . . . . a11b1p a12b2p a21b1p a22b2p am1b1p am2b2p a1nbnp a2nbnp amnbnp . . . . . . . . . . . . . . . . . . . . . a11 a21 am1 a1n a2n amn a12 a22 am2 . . . . . . . . . AB ( ) ( ) . . . . . . b11 b21 bn1 b1p b2p bnp b12 b22 bn2 . . . . . . . . . . . . . . . Note carefully in Deﬁnition II.6 that the product AB C is deﬁned only when the number of columns in the matrix A is the same as the number of rows in B. The size of the product can be determined from Also, you might recognize that the entries in, say, the ith row of the ﬁnal matrix AB are formed by using the component deﬁnition of the inner, or dot, product of the ith row of A with each of the columns of B. AmnBnp Cmp. EXAMPLE 4 Multiplication of Matrices (a) For and (b) For and In general, matrix multiplication is not co mmutative; that is, AB BA. Observe in part (a) of Example 4 that whereas in part (b) the product BA is not deﬁned, since Deﬁnition II.6 requires that the ﬁrst matrix (in this case B) have the same number of columns as the second matrix has rows. We are particularly interested in the product of a square matrix and a column vector. BA 30 53 48 82, AB 5 (4) 8 2 5 (3) 8 0 1 (4) 0 2 1 (3) 0 0 2 (4) 7 2 2 (3) 7 0 4 15 4 3 6 6. B 4 3 2 0, A 5 8 1 0 2 7 AB 4 9 7 6 4 (2) 7 8 3 9 5 6 3 (2) 5 8 78 48 57 34. B 9 2 6 8, A 4 7 3 5 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APP-6 ● APPENDIX II MATRICES EXAMPLE 5 Multiplication of Matrices (a) (b) Multiplicative Identity For a given positive integer n the n n matrix is called the multiplicative identity matrix. It follows from Deﬁnition II.6 that for any n n matrix A. Also, it is readily veriﬁed that, if X is an n 1 column matrix, then IX X. Zero Matrix A matrix consisting of all zero entries is called a zero matrix and is denoted by 0. For example, and so on. If A and 0 are m n matrices, then Associative Law Although we shall not prove it, matrix multiplication is associative. If A is an m p matrix, B a p r matrix, and C an r n matrix, then is an m n matrix. Distributive Law If all products are deﬁned, multiplication is distributive over addition: Determinant of a Matrix Associated with every square matrix A of constants is a number called the determinant of the matrix, which is denoted by det A. A(B C) AB AC and (B C)A BA CA. A(BC) (AB)C A 0 0 A A. 0 0 0, 0 0 0 0 0, 0 0 0 0 0 0 0, AI IA A. 1 0 0 0 1 0 0 0 0 . . . . . . . . . I ( ) . . . 0 0 1 . . . 4 2 3 8 x y 4x 2y 3x 8y 2 1 3 0 4 5 1 7 9 3 6 4 2 (3) (1) 6 3 4 0 (3) 4 6 5 4 1 (3) (7) 6 9 4 0 44 9 EXAMPLE 6 Determinant of a Square Matrix For we expand det A by cofactors of the ﬁrst row: 3(20 2) 6(8 1) 2(4 5) 18. det A p 3 6 2 2 5 1 1 2 4 p 3 5 1 2 4 6 2 1 1 4 2 2 5 1 2 A 3 6 2 2 5 1 1 2 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-7 It can be proved that a determinant det A can be expanded by cofactors using any row or column. If det A has a row (or a column) containing many zero entries, then wisdom dictates that we expand the determinant by that row (or column). DEFINITION II.7 Transpose of a Matrix The transpose of the m n matrix (1) is the n m matrix AT given by In other words, the rows of a matrix A become the columns of its transpose AT. a11 a12 a1n am1 am2 amn a21 a22 a2n . . . . . . . . . AT ( . ) . . . . . . EXAMPLE 7 Transpose of a Matrix (a) The transpose of is (b) If then XT (5 0 3). X 5 0 3, AT 3 2 1 6 5 2 2 1 4. A 3 6 2 2 5 1 1 2 4 DEFINITION II.8 Multiplicative Inverse of a Matrix Let A be an n n matrix. If there exists an n n matrix B such that where I is the multiplicative identity, then B is said to be the multiplicative inverse of A and is denoted by B A1. AB BA I, DEFINITION II.9 Nonsingular/Singular Matrices Let A be an n n matrix. If det A 0, then A is said to be nonsingular. If det A 0, then A is said to be singular. The following theorem gives a necessary and sufﬁcient condition for a square matrix to have a multiplicative inverse. THEOREM II.1 Nonsingularity Implies A Has an Inverse An n n matrix A has a multiplicative inverse A1 if and only if A is nonsingular. The following theorem gives one way of ﬁnding the multiplicative inverse for a nonsingular matrix. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APP-8 ● APPENDIX II MATRICES THEOREM II.2 A Formula for the Inverse of a Matrix Let A be an n n nonsingular matrix and let Cij (1)ijMij, where Mij is the determinant of the (n 1) (n 1) matrix obtained by deleting the ith row and jth column from A. Then (2) A1 1 det A (Cij)T. Each Cij in Theorem II.2 is simply the cofactor (signed minor) of the corresponding entry aij in A. Note that the transpose is utilized in formula (2). For future reference we observe in the case of a 2 2 nonsingular matrix that C11 a22, C12 a21, C21 a12, and C22 a11. Thus (3) For a 3 3 nonsingular matrix and so on. Carrying out the transposition gives (4) A1 1 det A C11 C21 C31 C12 C22 C32 C13 C23 C33. C11 a 22 a 23 a32 a33, C12 a 21 a 23 a31 a33, C13 a 21 a 22 a31 a32, A a11 a12 a13 a 21 a 22 a 23 a31 a32 a33, A1 1 det A a 22 a 21 a12 a11 T 1 det A a 22 a12 a 21 a11. A a11 a12 a 21 a 22 EXAMPLE 9 Inverse of a 3 3 Matrix Find the multiplicative inverse for A 2 2 0 2 1 1 3 0 1. EXAMPLE 8 Inverse of a 2 2 Matrix Find the multiplicative inverse for SOLUTION Since det A 10 8 2 0, A is nonsingular. It follows from Theorem II.1 that A1 exists. From (3) we ﬁnd Not every square matrix has a multiplicative inverse. The matrix is singular, since det A 0. Hence A1 does not exist. A 2 2 3 3 A1 1 2 10 4 2 1 5 2 1 1 2. A 1 4 2 10. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SOLUTION Since det A 12 0, the given matrix is nonsingular. The cofactors corresponding to the entries in each row of det A are If follows from (4) that You are urged to verify that A1A AA1 I. Formula (2) presents obvious difﬁculties for nonsingular matrices larger than 3 3. For example, to apply (2) to a 4 4 matrix, we would have to calculate sixteen 3 3 determinants. In the case of a large matrix there are more efﬁcient ways of ﬁnding A1. The curious reader is referred to any text in linear algebra. Since our goal is to apply the concept of a matrix to systems of linear ﬁrst-order differential equations, we need the following deﬁnitions. A1 1 12 1 2 2 5 2 2 3 6 6 1 12 1 6 1 6 5 12 1 6 1 6 1 4 1 2 1 2. C31 2 0 1 1 2 C32 2 0 2 1 2 C33 2 2 2 1 6. C21 2 0 0 1 2 C22 2 0 3 1 2 C23 2 2 3 0 6 C11 1 1 0 1 1 C12 2 1 3 1 5 C13 2 1 3 0 3 APPENDIX II MATRICES ● APP-9 Strictly speaking, a determinant is a number, but it is sometimes convenient to refer to a determinant as if it were an array. DEFINITION II.10 Derivative of a Matrix of Functions If A(t) (aij(t))mn is a matrix whose entries are functions differentiable on a common interval, then dA dt d dt ai j mn . DEFINITION II.11 Integral of a Matrix of Functions If A(t) (aij(t))mn is a matrix whose entries are functions continuous on a common interval containing t and t0, then t t 0 A(s) ds t t 0 ai j(s) ds mn . EXAMPLE 10 Derivative/Integral of a Matrix If X(t) sin 2t e3t 8t 1, then X(t) d dt sin 2t d dt e3t d dt (8t 1) 2 cos 2t 3e3t 8 To differentiate (integrate) a matrix of functions, we simply differentiate (integrate) each entry. The derivative of a matrix is also denoted by A(t). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. and II.2 GAUSSIAN AND GAUSS-JORDAN ELIMINATION Matrices are an invaluable aid in solving algebraic systems of n linear equations in n variables or unknowns, (5) If A denotes the matrix of coefﬁcients in (5), we know that Cramer’s rule could be used to solve the system whenever det A 0. However, that rule requires a herculean effort if A is larger than 3 3. The procedure that we shall now consider has the distinct advantage of being not only an efﬁcient way of handling large systems, but also a means of solving consistent systems (5) in which det A 0 and a means of solving m linear equations in n unknowns. a11x1 a12x2 a1nxn b1 a21x1 a22x2 a2nxn b2 M M an1x1 an2x2 annxn bn. t 0 X(s) ds t 0 sin 2s ds t 0 e3s ds t 0 (8s 1) ds 1 2 cos 2t 1 2 1 3 e3t 1 3 4t2 t . APP-10 ● APPENDIX II MATRICES DEFINITION II.12 Augmented Matrix The augmented matrix of the system (5) is the n (n 1) matrix a11 a21 an1 a1n a2n ann a12 a22 an2 . . . . . . . . . ( . . . b1 b2 bn) . . . . If B is the column matrix of the bi, i 1, 2, . . . , n, the augmented matrix of (5) is denoted by Elementary Row Operations Recall from algebra that we can transform an algebraic system of equations into an equivalent system (that is, one having the same solution) by multiplying an equation by a nonzero constant, interchanging the posi-tions of any two equations in a system, and adding a nonzero constant multiple of an equation to another equation. These operations on equations in a system are, in turn, equivalent to elementary row operations on an augmented matrix: (i) Multiply a row by a nonzero constant. (ii) Interchange any two rows. (iii) Add a nonzero constant multiple of one row to any other row. Elimination Methods To solve a system such as (5) using an augmented ma-trix, we use either Gaussian elimination or the Gauss-Jordan elimination method. In the former method, we carry out a succession of elementary row operations until we arrive at an augmented matrix in row-echelon form: (i) The ﬁrst nonzero entry in a nonzero row is 1. (ii) In consecutive nonzero rows the ﬁrst entry 1 in the lower row appears to the right of the ﬁrst 1 in the higher row. (iii) Rows consisting of all 0’s are at the bottom of the matrix. (AB). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-11 In the Gauss-Jordan method the row operations are continued until we obtain an aug-mented matrix that is in reduced row-echelon form. A reduced row-echelon matrix has the same three properties listed above in addition to the following one: (iv) A column containing a ﬁrst entry 1 has 0’s everywhere else. We can always interchange equations so that the ﬁrst equation contains the variable x1. EXAMPLE 11 Row-Echelon/Reduced Row-Echelon Form (a) The augmented matrices are in row-echelon form. You should verify that the three criteria are satisﬁed. (b) The augmented matrices are in reduced row-echelon form. Note that the remaining entries in the columns containing a leading entry 1 are all 0’s. Note that in Gaussian elimination we stop once we have obtained an augmented matrix in row-echelon form. In other words, by using different sequences of row operations we may arrive at different row-echelon forms. This method then requires the use of back-substitution. In Gauss-Jordan elimination we stop when we have obtained the augmented matrix in reduced row-echelon form. Any sequence of row operations will lead to the same augmented matrix in reduced row-echelon form. This method does not require back-substitution; the solution of the system will be apparent by inspection of the ﬁnal matrix. In terms of the equations of the original system, our goal in both methods is simply to make the coefﬁcient of x1 in the ﬁrst equation equal to 1 and then use multiples of that equation to eliminate x1 from other equations. The process is repeated on the other variables. To keep track of the row operations on an augmented matrix, we utilize the following notation: Symbol Meaning Rij Interchange rows i and j cRi Multiply the ith row by the nonzero constant c cRi Rj Multiply the ith row by c and add to the jth row 1 0 0 0 1 0 0 0 0 7 p 1 0 and 0 0 0 0 1 0 6 0 0 1 6 4 1 0 0 5 1 0 0 0 0 2 p 1 0 and 0 0 0 0 1 0 6 0 2 1 2 4 EXAMPLE 12 Solution by Elimination Solve using (a) Gaussian elimination and (b) Gauss-Jordan elimination. 5x1 7 x2 4 x3 9 x1 2 x2 x3 1 2x1 6 x2 x3 7 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SOLUTION (a) Using row operations on the augmented matrix of the system, we obtain APP-12 ● APPENDIX II MATRICES ( 2 1 5 1 1 4 7 1 9 6 2 7 ) ( 1 2 5 1 1 4 2 6 7 R12 1 7 9 ) ( 1 0 0 1 3 1 2 2 3 1 9 14 ) 2R1 R2 5R1 R3 ( 1 0 0 1 1 1 14 2 1 3 ) ( 1 0 0 1 2 1 0 3R2 R3 1 ) ( 1 0 0 1 1 2 1 0 1 5 ). R2 R3 1 _ 2 2 __ 11 3 _ 2 9 _ 2 3 _ 2 11 __ 2 55 __ 2 9 _ 2 3 _ 2 9 _ 2 The last matrix is in row-echelon form and represents the system Substituting x3 5 into the second equation then gives x2 3. Substituting both these values back into the ﬁrst equation ﬁnally yields x1 10. (b) We start with the last matrix above. Since the ﬁrst entries in the second and third rows are 1’s, we must, in turn, make the remaining entries in the second and third columns 0’s: The last matrix is now in reduced row-echelon form. Because of what the matrix means in terms of equations, it is evident that the solution of the system is x1 10, x2 3, x3 5. ( 1 0 0 1 1 1 5 2 1 0 ) ( 1 0 0 4 1 0 1 0 2R2 R1 10 5) ( 1 0 0 0 0 1 0 1 0 10 3 5 ) . 4R3 R1 R3 R2 3 _ 2 3 _ 2 9 _ 2 3 _ 2 9 _ 2 x3 5. x2 3 2 x3 9 2 x1 2x2 x3 1 EXAMPLE 13 Gauss-Jordan Elimination Solve SOLUTION We solve the system using Gauss-Jordan elimination: In this case, the last matrix in reduced row-echelon form implies that the original system of three equations in three unknowns is really equivalent to two equations in three unknowns. Since only z is common to both equations (the nonzero rows), we R2 R3 1 __ 11 1 __ 11 ( 1 4 2 2 3 7 7 5 19 3 1 5 ) ( 1 0 0 2 11 11 3 11 11 7 33 33 ) 4R1 R2 2R1 R3 ( 1 0 0 2 1 1 7 3 3 3 1 1 ) ( 1 0 0 1 1 0 0 1 0 3R2 R1 R2 R3 1 3 0 ). 2x 5 y 7z 19. 4x y 3z 5 x 3 y 2z 7 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-13 can assign its values arbitrarily. If we let z t, where t represents any real number, then we see that the system has inﬁnitely many solutions: x 2 t, y 3 t, z t. Geometrically, these equations are the parametric equations for the line of intersection of the planes x 0y z 2 and 0x y z 3. Using Row Operations to Find an Inverse Because of the number of deter-minants that must be evaluated, formula (2) in Theorem II.2 is seldom used to ﬁnd the inverse when the matrix A is large. In the case of 3 3 or larger matrices the method described in the next theorem is a particularly efﬁcient means for ﬁnding A1. ( 2 2 5 1 4 6 0 3 5 1 0 0 0 1 0 0 0 1 ) ( 1 2 5 4 6 0 3 5 0 0 0 1 0 0 0 1 ) R1 2R1 R2 5R1 R3 1 _ 2 1 _ 2 1 _ 2 ( 1 0 0 5 0 3 5 1 0 1 0 0 0 1 ) 1 _ 2 1 _ 2 17 __ 2 5 _ 2 THEOREM II.3 Finding A1 Using Elementary Row Operations If an n n matrix A can be transformed into the n n identity I by a sequence of elementary row operations, then A is nonsingular. The same sequence of operations that transforms A into the identity I will also trans-form I into A1. It is convenient to carry out these row operations on A and I simultaneously by means of an n 2n matrix obtained by augmenting A with the identity I as shown here: The procedure for ﬁnding A1 is outlined in the following diagram: ( A I ) (I A1). Perform row operations on A until I is obtained. This means that A is nonsingular. By simultaneously applying the same row operations to I, we get A1. (A I) ( a11 a21 an1 a1n a2n ann a12 a22 an2 . . . . . . . . . . . . . . . . . . . . . 1 1 0 0 0 0 . . . . . . 0 0 1 . . .) . EXAMPLE 14 Inverse by Elementary Row Operations Find the multiplicative inverse for SOLUTION We shall use the same notation as we did when we reduced an augmented matrix to reduced row-echelon form: A 2 0 1 2 3 4 5 5 6. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Because I appears to the left of the vertical line, we conclude that the matrix to the right of the line is If row reduction of (AI) leads to the situation where the matrix B contains a row of zeros, then necessarily A is singular. Since fur-ther reduction of B always yields another matrix with a row of zeros, we can never transform A into I. II.3 THE EIGENVALUE PROBLEM Gauss-Jordan elimination can be used to ﬁnd the eigenvectors of a square matrix. (A I) (B C), row operations A1 2 5 3 8 17 10 5 10 6. ( 1 0 0 1 0 1 0 5 0 10 0 0 6 ) 30R3 R3 R1 R3 R2 1 _ 2 1 _ 3 1 _ 3 1 _ 2 5 _ 3 ( 1 0 0 0 0 1 0 1 0 2 8 5 5 17 10 3 10 6 ) . 5 _ 3 1 _ 3 ( 1 0 0 0 1 1 0 0 0 0 ) R2 R3 R2 R3 1 _ 5 1 _ 3 1 _ 2 1 _ 2 5 _ 3 17 __ 10 1 _ 3 1 _ 2 1 _ 3 1 _ 5 0 0 0 ) 1 _ 2 1 _ 3 1 _ 3 1 _ 5 ( 1 0 0 0 1 0 1 _ 2 5 _ 3 1 _ 3 1 __ 30 1 _ 6 APP-14 ● APPENDIX II MATRICES EXAMPLE 15 Eigenvector of a Matrix Verify that is an eigenvector of the matrix A 0 1 3 2 3 3 2 1 1. K 1 1 1 DEFINITION II.13 Eigenvalues and Eigenvectors Let A be an n n matrix. A number l is said to be an eigenvalue of A if there exists a nonzero solution vector K of the linear system (6) The solution vector K is said to be an eigenvector corresponding to the eigenvalue l. AK K. The word eigenvalue is a combination of German and English terms adapted from the German word eigenwert, which, translated literally, is “proper value.” Eigenvalues and eigenvectors are also called characteristic values and character-istic vectors, respectively. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-15 SOLUTION By carrying out the multiplication AK, we see that We see from the preceding line and Deﬁnition II.13 that l 2 is an eigenvalue of A. Using properties of matrix algebra, we can write (6) in the alternative form (7) where I is the multiplicative identity. If we let then (7) is the same as (8) Although an obvious solution of (8) is k1 0, k2 0, . . . , kn 0, we are seeking only nontrivial solutions. It is known that a homogeneous system of n linear equa-tions in n unknowns (that is, bi 0, i 1, 2, . . . , n in (5)) has a nontrivial solution if and only if the determinant of the coefﬁcient matrix is equal to zero. Thus to ﬁnd a nonzero solution K for (7), we must have (9) Inspection of (8) shows that the expansion of det(A lI) by cofactors results in an nth-degree polynomial in l. The equation (9) is called the characteristic equation of A. Thus the eigenvalues of A are the roots of the characteristic equation. To ﬁnd an eigenvector corresponding to an eigenvalue l, we simply solve the system of equations (A lI)K 0 by applying Gauss-Jordan elimination to the augmented matrix (A I0). det(A I) 0. (a11 l)k1 a21k1 an1k1 a12k2 (a22 l)k2 an2k2 (ann l)kn 0. a1nk n 0 a2nk n 0 . . . . . . . . . . . . . . . K k1 k2 M kn , (A I)K 0, AK ( 0 2 2 1 1 1 3 3 1 1 3 1 1 1 1 2 2 2 ) ( ) ( ) (2) ( ) (2)K. eigenvalue EXAMPLE 16 Eigenvalues/Eigenvectors Find the eigenvalues and eigenvectors of SOLUTION To expand the determinant in the characteristic equation, we use the cofactors of the second row: From l3 l2 12l l(l 4)(l 3) 0 we see that the eigenvalues are l1 0, l2 4, and l3 3. To ﬁnd the eigenvectors, we must now reduce three times corresponding to the three distinct eigenvalues. (A I0) det(A I) p 1 2 1 6 1 0 1 2 1 p 3 2 12 0. A 1 2 1 6 1 0 1 2 1. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. For l1 0 we have Thus we see that and Choosing k3 13, we get the eigenvector For l2 4, K1 1 6 13. k2 6 13 k3. k1 1 13 k3 6 __ 13 6 __ 13 1 __ 13 6R1 R2 R1 R3 R2 1 __ 13 (A 0I 0) ( 1 6 1 2 1 2 1 0 1 0 0 0 ) ( 1 0 0 1 6 0 0 0 0 2 13 0 ) 2R2 R1 ( 1 0 0 2 1 0 1 0 0 0 0 ) ( 1 0 0 0 0 0 0 0 1 0 ) . APP-16 ● APPENDIX II MATRICES R2 R3 2R2 R1 R2 R3 ( 1 0 0 2 1 1 3 2 2 0 0 0 ) ( 1 0 0 1 2 0 0 0 0 0 1 0 ) 6R1 R2 5R1 R3 ( 1 0 0 3 18 16 0 0 0 2 9 8 ) 1 _ 9 1 _ 8 R3 R31 (A 4I 0) ( 5 6 1 2 3 2 1 0 3 0 0 0 ) ( 1 6 5 3 0 1 0 0 0 2 3 2 ) implies that k1 k3 and k2 2k3. Choosing k3 1 then yields the second eigenvector Finally, for l3 3 Gauss-Jordan elimination gives so k1 k3 and The choice of k3 2 leads to the third eigenvector: When an n n matrix A possesses n distinct eigenvalues l1, l2, . . . , ln, it can be proved that a set of n linearly independent† eigenvectors K1, K2, . . . , Kn can be found. However, when the characteristic equation has repeated roots, it may not be possible to ﬁnd n linearly independent eigenvectors for A. K3 2 3 2. k2 3 2 k3. (A 3I 0) ( 2 6 1 1 0 4 0 0 0 2 4 2 ) ( 1 0 0 1 0 0 0 0 0 1 0 ) , row operations 3 _ 2 K2 1 2 1. Of course, k3 could be chosen as any nonzero number. In other words, a nonzero constant multiple of an eigenvector is also an eigenvector. †Linear independence of column vectors is deﬁned in exactly the same manner as for functions. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-17 EXAMPLE 17 Eigenvalues/Eigenvectors Find the eigenvalues and eigenvectors of SOLUTION From the characteristic equation we see that l1 l2 5 is an eigenvalue of multiplicity two. In the case of a 2 2 matrix there is no need to use Gauss-Jordan elimination. To ﬁnd the eigenvector(s) corresponding to l1 5, we resort to the system in its equivalent form It is apparent from this system that k1 2k2. Thus if we choose k2 1, we ﬁnd the single eigenvector K1 2 1. k1 2k2 0. 2k1 4k2 0 (A 5I0) det(A I) 3 4 1 7 ( 5)2 0 A 3 4 1 7. EXAMPLE 18 Eigenvalues/Eigenvectors Find the eigenvalues and eigenvectors of . SOLUTION The characteristic equation shows that l1 11 and that l2 l3 8 is an eigenvalue of multiplicity two. For l1 11 Gauss-Jordan elimination gives Hence k1 k3 and k2 k3. If k3 1, then Now for l2 8 we have (A 8I 0) ( 1 1 1 1 1 1 0 0 0 1 1 1 ) ( 1 0 0 1 0 0 0 0 0 1 0 0 ) . row operations K1 1 1 1. (A 11I 0) ( 2 1 1 1 1 2 0 0 0 1 2 1 ) ( 1 0 0 1 1 0 0 0 0 0 1 0 ) . row operations det(A I) p 9 1 1 1 9 1 1 1 9 p ( 11)( 8)2 0 A 9 1 1 1 9 1 1 1 9 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In the equation k1 k2 k3 0, we are free to select two of the variables arbitrar-ily. Choosing, on the one hand, k2 1, k3 0 and, on the other, k2 0, k3 1, we obtain two linearly independent eigenvectors K2 1 1 0 and K3 1 0 1. APP-18 ● APPENDIX II MATRICES EXERCISES FOR APPENDIX II Answers to selected odd-numbered problems begin on page ANS-31. II.1 BASIC DEFINITIONS AND THEORY 1. If and ﬁnd (a) A B (b) B A (c) 2A 3B 2. If and ﬁnd (a) A B (b) B A (c) 2(A B) 3. If and ﬁnd (a) AB (b) BA (c) A2 AA (d) B2 BB 4. If and ﬁnd (a) AB (b) BA 5. If and ﬁnd (a) BC (b) A(BC) (c) C(BA) (d) A(B C) 6. If and ﬁnd (a) AB (b) BA (c) (BA)C (d) (AB)C 7. If and ﬁnd (a) ATA (b) BTB (c) A BT B (2 4 5), A 4 8 10 C 1 2 4 0 1 1 3 2 1, A (5 6 7), B 3 4 1, C 0 2 3 4, A 1 2 2 4, B 6 3 2 1, B 4 6 3 1 3 2, A 1 4 5 10 8 12 B 1 6 3 2, A 2 3 5 4 B 3 1 0 2 4 2, A 2 0 4 1 7 3 B 2 6 8 10, A 4 5 6 9 8. If and ﬁnd (a) A BT (b) 2AT BT (c) AT(A B) 9. If and ﬁnd (a) (AB)T (b) BTAT 10. If and ﬁnd (a) AT BT (b) (A B)T In Problems 11–14 write the given sum as a single column matrix. 11. 12. 13. 14. In Problems 15–22 determine whether the given matrix is singular or nonsingular. If it is nonsingular, ﬁnd A1 using Theorem II.2. 15. 16. 17. 18. 19. 20. A 3 2 1 4 1 0 2 5 1 A 2 1 0 1 2 1 1 2 1 A 7 10 2 2 A 4 8 3 5 A 2 5 1 4 A 3 6 2 4 1 3 4 2 5 1 0 4 2 t 2t 1 t t 1 4 2 8 6 2 3 1 4 2 5 1 6 2 3 7 2 3t 2 t 1 (t 1) 1 t 3 2 3t 4 5t 4 1 2 2 2 8 3 2 3 B 3 11 7 2, A 5 9 4 6 B 5 10 2 5, A 3 4 8 1 B 2 3 5 7, A 1 2 2 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APPENDIX II MATRICES ● APP-19 21. 22. In Problems 23 and 24 show that the given matrix is nonsingular for every real value of t. Find A1(t) using Theorem II.2. 23. 24. In Problems 25–28 ﬁnd dXdt. 25. 26. 27. 28. 29. Let Find (a) (b) (c) 30. Let and Find (a) (b) (c) (d) (e) A(t)B(t) (f) (g) II.2 GAUSSIAN AND GAUSS-JORDAN ELIMINATION In Problems 31–38 solve the given system of equations by either Gaussian elimination or Gauss-Jordan elimination. 31. x y 2z 14 32. 5x 2y 4z 10 2x y z 0 x y z 9 6x 3y 4z 1 4x 3y 3z 1 33. y z 5 34. 3x y z 4 5x 4y 16z 10 4x 2y z 7 x y 5z 7 x y 3z 6 t 1 A(s)B(s) ds d dtA(t)B(t) 2 1 B(t) dt 1 0 A(t) dt dB dt dA dt B(t) 6t 2 1>t 4t. A(t) 1 t2 1 3t t2 t t 0 A(s) ds 2 0 A(t) dt dA dt A(t) e4t cos t 2t 3t2 1. X 5te2t t sin 3t X 2 1 1e2t 4 2 1e3t X 1 2 sin 2t 4 cos 2t 3 sin 2t 5 cos 2t X 5et 2et 7et A(t) 2et sin t 2et cos t et cos t et sin t A(t) 2et e4t 4et 3e4t A 4 1 1 6 2 3 2 1 2 A 2 1 1 1 2 3 3 2 4 35. 2x y z 4 36. x 2z 8 10x 2y 2z 1 x 2y 2z 4 6x 2y 4z 8 2x 5y 6z 6 37. x1 x2 x3 x4 1 38. 2x1 x2 x3 0 x1 x2 x3 x4 3 x1 3x2 x3 0 x1 x2 x3 x4 3 7x1 x2 3x3 0 4x1 x2 2x3 x4 0 In Problems 39 and 40 use Gauss-Jordan elimination to demonstrate that the given system of equations has no solution. 39. x 2y 4z 2 40. x1 x2 x3 3x4 1 2x 4y 3z 1 x2 x3 4x4 0 x 2y z 7 x1 2x2 2x3 x4 6 4x1 7x2 7x3 9 In Problems 41–46 use Theorem II.3 to ﬁnd A1 for the given matrix or show that no inverse exists. 41. 42. 43. 44. 45. 46. II.3 THE EIGENVALUE PROBLEM In Problems 47–54 ﬁnd the eigenvalues and eigenvectors of the given matrix. 47. 48. 49. 50. 51. 52. 53. 54. 1 6 0 0 2 1 0 1 2 0 4 0 1 4 0 0 0 2 3 0 0 0 2 0 4 0 1 5 1 0 0 5 9 5 1 0 1 1 1 4 1 8 1 16 0 2 1 2 1 1 2 7 8 A 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 A 1 2 3 1 1 0 2 1 2 1 3 0 1 1 2 1 A 1 2 3 0 1 4 0 0 8 A 1 3 0 1 2 1 0 1 2 A 2 4 2 4 2 2 8 10 6 A 4 2 3 2 1 0 1 2 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. In Problems 55 and 56 show that the given matrix has complex eigenvalues. Find the eigenvectors of the matrix. 55. 56. Miscellaneous Problems 57. If A(t) is a 2 2 matrix of differentiable functions and X(t) is a 2 1 column matrix of differentiable func-tions, prove the product rule 58. Derive formula (3). [Hint: Find a matrix for which AB I. Solve for b11, b12, b21, and b22. Then show that BA I.] B b11 b12 b21 b22 d dt [A(t)X(t)] A(t)X(t) A(t)X(t). 2 1 0 5 2 4 0 1 2 1 2 5 1 APP-20 ● APPENDIX II MATRICES 59. If A is nonsingular and AB AC, show that B C. 60. If A and B are nonsingular, show that (AB)1 B1A1. 61. Let A and B be n n matrices. In general, is 62. A square matrix A is said to be a diagonal matrix if all its entries off the main diagonal are zero—that is, aij 0, i j. The entries aii on the main diagonal may or may not be zero. The multiplicative identity matrix I is an example of a diagonal matrix. (a) Find the inverse of the 2 2 diagonal matrix when a11 0, a22 0. (b) Find the inverse of a 3 3 diagonal matrix A whose main diagonal entries aii are all nonzero. (c) In general, what is the inverse of an n n diagonal matrix A whose main diagonal entries aii are all nonzero? A a11 0 0 a22 (A B)2 A2 2AB B2? Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APP-21 f (t) 1. 1 2. t 3. tn n a positive integer 4. t1/2 5. t1/2 6. ta 7. sin kt 8. cos kt 9. sin2 kt 10. cos2 kt 11. eat 12. sinh kt 13. cosh kt 14. sinh2kt 15. cosh2kt 16. teat 17. tn eat n a positive integer n! (s a)n1, 1 (s a)2 s2 2k2 s(s2 4k2) 2k2 s(s2 4k2) s s2 k2 k s2 k2 1 s a s2 2k2 s(s2 4k2) 2k2 s(s2 4k2) s s2 k2 k s2 k2 ( 1) s1 , a 1 1 2s3/2 B s n! sn1, 1 s2 1 s { f (t)} F(s) Appendix III Laplace Transforms Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. APP-22 ● APPENDIX III LAPLACE TRANSFORMS f (t) 18. eat sin kt 19. eat cos kt 20. eat sinh kt 21. eat cosh kt 22. t sin kt 23. t cos kt 24. sin kt kt cos kt 25. sin kt kt cos kt 26. t sinh kt 27. t cosh kt 28. 29. 30. 1 cos kt 31. kt sin kt 32. 33. 34. sin kt sinh kt 35. sin kt cosh kt 36. cos kt sinh kt 37. cos kt cosh kt s3 s4 4k4 k(s2 2k2) s4 4k4 k(s2 2k2) s4 4k4 2k2s s4 4k4 s (s2 a2)(s2 b2) cos bt cos at a2 b2 1 (s2 a2)(s2 b2) a sin bt b sin at ab (a2 b2) k3 s2(s2 k2) k2 s(s2 k2) s (s a)(s b) aeat bebt a b 1 (s a)(s b) eat ebt a b s2 k2 (s2 k2)2 2ks (s2 k2)2 2k3 (s2 k2)2 2ks2 (s2 k2)2 s2 k2 (s2 k2)2 2ks (s2 k2)2 s a (s a)2 k2 k (s a)2 k2 s a (s a)2 k2 k (s a)2 k2 { f (t)} F(s) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. f (t) 38. J0(kt) 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. eat f (t) F(s a) 51. 52. easF(s) 53. eas 54. f (n)(t) 55. tn f(t) 56. F(s)G(s) 57. d(t) 1 58. d(t t0) est0 t 0 f ( )g(t ) d (1)n dn dsn F(s) snF(s) s(n1) f (0) f (n1)(0) { g(t a)} g(t)(t a) f (t a)(t a) eas s (t a) erfc a 2 1t bea1s s(1s b) eabeb2terfc b 1t a 2 1t ea1s 1s(1s b) eabeb2t erfc b 1t a 2 1t ea1s s1s 2 B t ea2/4t a erfc a 21t ea1s s erfc a 21t ea1s a 21t3 ea2/4t ea 1s 1s 1 1t ea2/4t 1 2 arctan a b s 1 2 arctan a b s sin at cos bt t arctan a s sin at t ln s2 k2 s2 2(1 cosh kt) t ln s2 k2 s2 2(1 cos kt) t ln s a s b ebt eat t 1 1s2 k2 { f (t)} F(s) APPENDIX III LAPLACE TRANSFORMS ● APP-23 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-1 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS EXERCISES 1.1 (PAGE 10) 1. linear, second order 3. linear, fourth order 5. nonlinear, second order 7. linear, third order 9. linear in x but nonlinear in y 15. domain of function is [2, ); largest interval of deﬁnition for solution is (2, ) 17. domain of function is the set of real numbers except x 2 and x 2; largest intervals of deﬁnition for solution are (, 2), (2, 2), or (2, ) 19. deﬁned on (, ln 2) or on (ln 2, ) 27. m 2 29. m 2, m 3 31. m 0, m 1 33. y 2 35. no constant solutions EXERCISES 1.2 (PAGE 17) 1. y 1(1 4ex) 3. y 1(x2 1); (1, ) 5. y 1(x2 1); (, ) 7. x cos t 8 sin t 9. 11. 13. y 5ex1 15. y 0, y x3 17. half-planes deﬁned by either y 0 or y 0 19. half-planes deﬁned by either x 0 or x 0 21. the regions deﬁned by y 2, y 2, or 2 y 2 23. any region not containing (0, 0) 25. yes 27. no 29. (a) y cx (b) any rectangular region not touching the y-axis (c) No, the function is not differentiable at x 0. 31. (b) y 1(1 x) on (, 1); y 1(x 1) on (1, ); (c) y 0 on (, ) 39. 41. y 0 43. no solution EXERCISES 1.3 (PAGE 28) 1. 3. 7. dx dt kx(1000 x) dP dt k1P k2P2 dP dt kP r; dP dt kP r y 3sin 2x y 3 2 ex 1 2 ex x 13 4 cos t 1 4 sin t X et 1 et 2 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. CHAPTER 1 IN REVIEW (PAGE 33) 1. 3. y k2y 0 5. y 2y y 0 7. (a), (d) 9. (b) 11. (b) 13. y c1 and y c2ex, c1 and c2 constants 15. y x2 y2 17. (a) The domain is the set of all real numbers. (b) either (, 0) or (0, ) 19. For x0 1 the interval is (, 0), and for x0 2 the interval is (0, ). 21. (c) 23. (, ) 25. (0, ) 31. 33. 35. y0 3, y1 0 37. EXERCISES 2.1 (PAGE 43) 21. 0 is asymptotically stable (attractor); 3 is unstable (repeller). 23. 2 is semi-stable. 25. 2 is unstable (repeller); 0 is semi-stable; 2 is asymptotically stable (attractor). 27. 1 is asymptotically stable (attractor); 0 is unstable (repeller). 39. 0 P0 hk 41. 1mg>k dP dt k(P 200 10t) y 3 2 e3x3 9 2 ex1 2x. y 1 2 e3x 1 2 ex 2x y x2, x2, x 0 x 0 dy dx 10y dy dx x 1x2 y2 y dx dt kx r, k 0 dA dt k(M A), k 0 d 2r dt2 gR2 r 2 0 mdv dt vdm dt kv mg R m d 2x dt2 kx m dv dt mg kv2 L di dt Ri E(t) dh dt c 450 1h dA dt 7 600 t A 6 dA dt 1 100 A 0; A(0) 50 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-2 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 2 EXERCISES 2.2 (PAGE 51) 1. 3. 5. y cx4 7. 3e2y 2e3x c 9. 11. 4 cos y 2x sin 2x c 13. (ex 1)2 2(ey 1)1 c 15. S cekr 17. 19. ( y 3)5 ex c(x 4)5 ey 21. 23. 25. 27. 29. 31. 33. 35. (a) 37. y 1 and y 1 are singular solutions of Problem 21; of Problem 22 39. y 1 41. 45. 47. 49. 57. y(x) (4hL2)x2 a EXERCISES 2.3 (PAGE 61) 1. y ce5x, (, ) 3. ; is transient 5. ; is transient 7. y x1 ln x cx1, (0, ); solution is transient 9. y cx x cos x, (0, ) 11. ; is transient 13. ; is transient 15. x 2y6 cy4, (0, ) 17. y sin x c cos x, (p2, p2) 19. (x 1)exy x2 c, (1, ); solution is transient 21. (sec u tan u)r u cos u c, (p2, p2) 23. y e3x cx1e3x, (0, ); solution is transient 25. 27. 29. i E R i0 E ReRt>L; (, ) y x1ex (2 e)x1; (0, ) y 1 5x 1 25 76 25e5x; (, ) cx2ex y 1 2 x2ex cx2ex, (0, ) cx4 y 1 7 x3 1 5 x cx4, (0, ) cex3 y 1 3 cex3, (, ) cex y 1 4 e3x cex, (, ) y 221xe1x e1x 4 y [1 c(1 2x)]2 y tan x sec x c y 1 1 10 tan ( 1 10 x) y 0 y 2, y 2, y 2 3 e4x1 3 e4x1 y ln(2 ex); (, ln2) y 2x2 x 1; , 1 15 2 . y ex 4 e-t2dt y 1 2 x 13 2 11 x2 y e(11/x) x x tan (4t 3 4 ) y sin (1 2 x2 c) P cet 1 cet 1 3 x3 ln x 1 9 x3 1 2 y2 2y ln y c y 1 3 e3x c y 1 5 cos 5x c 31. 33. 35. 37. 39. 41. 43. 53. E(t) E0e(t4)/RC EXERCISES 2.4 (PAGE 69) 1. 3. 5. x2y2 3x 4y c 7. not exact 9. 11. not exact 13. xy 2xex 2ex 2x3 c 15. x3y3 tan1 3x c 17. 19. t4y 5t3 ty y3 c 21. 23. 4ty t2 5t 3y2 y 8 25. y2 sin x x3y x2 y ln y y 0 27. k 10 29. x2y2 cos x c 31. x2y2 x3 c 33. 3x2y3 y4 c 35. 37. 39. (c) 45. (a) (b) 12.7 ft/s EXERCISES 2.5 (PAGE 74) 1. 3. 5. 7. ln(x2 y2) 2 tan1(yx) c 9. 4x y(lny c)2 11. y3 3x3 lnx 8x3 13. lnx ey/x 1 15. y3 1 cx3 17. 19. et/y ct 21. 23. y x 1 tan(x c) y3 9 5 x1 49 5 x6 y3 x 1 3 ce3x x y lnx cy (x y)lnx y y c(x y) y x lnx cx v(x) 8 B x 3 9 x2 y2(x) x2 1x4 x3 4 y1(x) x2 1x4 x3 4 ey 2 (x2 4) 20 2ye3x 10 3 e3x x c 1 3 x3 x2y xy2 y 4 3 lncos x cos x sin y c xy3 y2 cos x 1 2 x2 c 5 2 x2 4xy 2y4 c x2 x 3 2 y2 7y c y ex21 1 2 1 ex2 (erf(x) erf(1)) y 2x 1 4e2x, 4x2 ln x (1 4e2)x2, 0 x 1 x 1 y 1 2 3 2 ex2, (1 2 e 3 2)ex2, 0 x 1 x 1 y 1 2 (1 e2x), 1 2 (e6 1)e2x, 0 x 3 x 3 y 2 3ecos x; (, ) (x 1)y xlnx x 21; (0, ) y 2x 1 5>x; (0, ) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-3 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 3 25. 2y 2x sin 2(x y) c 27. 4( y 2x 3) (x c)2 29. 35. (b) EXERCISES 2.6 (PAGE 79) 1. y2 2.9800, y4 3.1151 3. y10 2.5937, y20 2.6533; y ex 5. y5 0.4198, y10 0.4124 7. y5 0.5639, y10 0.5565 9. y5 1.2194, y10 1.2696 13. Euler: y10 3.8191, y20 5.9363 RK4: y10 42.9931, y20 84.0132 CHAPTER 2 IN REVIEW (PAGE 80) 1. Ak, a repeller for k 0, an attractor for k 0 3. true 5. 7. true 9. 11. 13. 15. semi-stable for n even and unstable for n odd; semi-stable for n even and asymptotically stable for n odd. 19. 2x sin 2x 2 ln( y2 1) c 21. (6x 1)y3 3x3 c 23. 25. 27. y csc x, (p, 2p) 29. (b) EXERCISES 3.1 (PAGE 90) 1. 7.9 yr; 10 yr 3. 760; approximately 11 persons/yr 5. 11 h 7. 136.5 h 9. I(15) 0.00098I0 or approximately 0.1% of I0 11. 15,600 years 13. T(1) 36.67° F; approximately 3.06 min 15. approximately 82.1 s; approximately 145.7 s 17. 390° 19. about 1.6 hours prior to the discovery of the body 21. A(t) 200 170et/50 y 1 4 (x 2 1y0 x0)2, (x0 2 1y0, ) y 1 4 c(x2 4)4 Q ct1 1 25 t4 (1 5 ln t) dy dx ( y 1)2 ( y 3)2 dy dx (sinx)y x y c1eex d3y dx3 x siny y 2 x (1 4 x cx3)1 cot(x y) csc(x y) x 12 1 23. A(t) 1000 1000et/100 25. 27. 64.38 lb 29. 31. 33. 35. (a) (b) as (c) 39. (a) (c) 41. (a) 43. (a) As . (b) x(t) rk (rk)ekt; (ln 2)k 47. (c) 1.988 ft EXERCISES 3.2 (PAGE 100) 1. (a) N 2000 (b) 3. 1,000,000; 5.29 mo 5. (b) (c) For 0 P0 1, time of extinction is . 7. ; time of extinction is 9. 29.3 g; as ; 0 g of A and 30 g of B 11. (a) (b) or 30.36 min 13. (a) approximately 858.65 s or 14.31 min (b) 243 s or 4.05 min 576 110 s h(t) 1H 4Ah Aw t 2 ; I is 0 t 1HAw 4Ah t : X : 60 t 2 13 tan1 5 13 tan1 2P0 5 13 P(t) 5 2 13 2 tan 13 2 t tan1 2P0 5 13 t 1 3 ln 4(P0 1) P0 4 P(t) 4(P0 1) (P0 4)e3t (P0 1) (P0 4)e3t N(t) 2000 et 1999 et; N(10) 1834 t : , x(t) : r>k P(t) P0 e(k1k2)t 331 3 seconds v(t) g 4k k t r0 gr0 4k r0 k t r0 3 m k v0 mg k s(t) mg k t m k v0 mg k ekt/m t : v : mg k v(t) mg k v0 mg k ekt /m i(t) 60 60et/10, 0 t 20 60(e2 1)et/10, t 20 q(t) 1 100 1 100e50t; i(t) 1 2e50t i(t) 3 5 3 5 e500t; i : 3 5 as t : A(t) 1000 10t 1 10(100 t)2; 100 min Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-4 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS 15. (a) (b) (c) , where c2 (mk)ln cosh c1 17. (a) , where r is the weight density of water (b) (c) 19. (a) W 0 and W 2 (b) W(x) 2 sech2(x c1) (c) W(x) 2 sech2x 21. (a) (b) approximately 724 months (b) approximately 12,839 and 28,630,966 EXERCISES 3.3 (PAGE 110) 1. 3. 5, 20, 147 days. The time when y(t) and z(t) are the same makes sense because most of A and half of B are gone, so half of C should have been formed. 5. 7. (a) (b) x1(t) x2(t) 150; x2(30) 47.4 lb 13. 15. i(0) i0, s(0) n i0, r(0) 0 L2 di3 dt R1i2 (R1 R3) i3 E(t) L1 di2 dt (R1 R2)i2 R1 i3 E(t) dx2 dt 2 x1 100 t 3 x2 100 t dx1 dt 3 x2 100 t 2 x1 100 t dx2 dt 2 25 x1 2 25 x2 dx1 dt 6 2 25 x1 1 50 x2 z(t) x0 1 2 2 1 e1t 1 2 1 e2 t y(t) x01 2 1 (e1t e2t) x(t) x0 e1t P(t) 1 (0.001350t 100.01)100 B mg V k v(t) B mg V k tanh 1kmg k V m t c1 m dv dt mg kv2 V s(t) m k ln cosh B kg m t c1 c2 B mg k where c1 tanh1 B k mg v0 v(t) B mg k tanh B kg m t c1 CHAPTER 3 IN REVIEW (PAGE 113) 1. dPdt 0.15P 3. P(45) 8.99 billion 5. 7. (a) (b) 9. 11. 13. x y 1 c2ey 15. (a) (b) (c) EXERCISES 4.1 (PAGE 127) 1. 3. y 3x 4x ln x 9. (, 2) 11. (a) (b) 13. (a) y ex cos x ex sin x (b) no solution (c) y ex cos x ep/2ex sin x (d) y c2ex sin x, where c2 is arbitrary 15. dependent 17. dependent 19. dependent 21. independent 23. The functions satisfy the DE and are linearly independent on the interval since W(e3x, e4x) 7ex 0; y c1e3x c2e4x. 25. The functions satisfy the DE and are linearly independent on the interval since W(ex cos 2x, ex sin 2x) 2e2x 0; y c1ex cos 2x c2ex sin 2x. 27. The functions satisfy the DE and are linearly independent on the interval since W(x3, x4) x6 0; y c1x3 c2x4. 29. The functions satisfy the DE and are linearly independent on the interval since W(x, x2, x2 ln x) 9x6 0; y c1x c2x2 c3x2 ln x. 35. (b) yp x2 3x 3e2x; yp 2x2 6x 1 3 e2x y sinh x sinh 1 y e e2 1 (ex ex) y 1 2 ex 1 2 ex 89%, 11% 1.3 109 years A(t) l2 l1 l2 K0[1 e(l1l2)t] C(t) l1 l1 l2 K0[1 e(l1l2)t], K(t) K0e(l1l2)t, x(t) ac1eak1t 1 c1eak1t , y(t) c2(1 c1eak1t)k2 /k1 i(t) 4t 1 5 t2, 0 t 10 20, t 10 T(t) BT1 T2 1 B T1 T2 1 B ek(1B)t BT1 T2 1 B , BT1 T2 1 B x 10 ln 10 1100 y2 y 1100 y2 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 4 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-5 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 4 EXERCISES 4.2 (PAGE 131) 1. y2 xe2x 3. y2 sin 4x 5. y2 sinh x 7. y2 xe2x/3 9. y2 x4 lnx 11. y2 1 13. y2 x cos(ln x) 15. y2 x2 x 2 17. 19. EXERCISES 4.3 (PAGE 137) 1. y c1 c2ex/4 3. y c1e3x c2e2x 5. y c1e4x c2xe4x 7. y c1e2x/3 c2ex/4 9. y c1cos 3x c2sin 3x 11. y e2x(c1cos x c2sin x) 13. 15. y c1 c2ex c3e5x 17. y c1ex c2e3x c3xe3x 19. u c1et et (c2cos t c3sin t) 21. y c1ex c2xex c3x2ex 23. 25. 27. u c1er c2rer c3er c4rer c5e5r 29. 31. 33. y 0 35. 37. y e5x xe5x 39. y 0 41. 49. y 6y 5y 0 51. y 2y 0 53. y 9y 0 55. y 2y 2y 0 57. y 8y 0 EXERCISES 4.4 (PAGE 147) 1. y c1ex c2e2x 3 3. 5. 7. 9. y c1 c2ex 3x 11. 13. 15. 17. 19. 12 25 sin 2x 9 25 cos 2x y c1ex c2xex 1 2 cos x y c1ex cos 2x c2ex sin 2x 1 4 xex sin 2x y c1 cos x c2 sin x 1 2 x2 cos x 1 2 x sin x y c1 cos 2x c2 sin 2x 3 4 x cos 2x y c1ex/2 c2xex/2 12 1 2 x2ex/2 y c1 cos 13x c2 sin 13x (4x2 4x 4 3)e3x y c1e2x c2xe2x x2 4x 7 2 y c1e5x c2xe5x 6 5 x 3 5 y cosh 13x 5 13 sinh13x y 1 2 1 5 13 e13x 1 2 1 5 13 e13x; y 5 36 5 36 e6x 1 6 xe6x y 1 3 e(t1) 1 3 e5(t1) y 2 cos 4x 1 2 sin 4x c3 x cos 1 2 13 x c4 x sin 1 2 13 x y c1 cos 1 2 13 x c2 sin 1 2 13 x y c1 c2 x ex/2 (c3 cos 1 2 13 x c4 sin 1 2 13 x) y ex /3(c1 cos 1 3 12 x c2 sin 1 3 12 x) y2 e2x, yp 5 2 e3x y2 e2x, yp 1 2 21. 23. 25. y c1 cos x c2 sin x c3x cos x c4x sin x x2 2x 3 27. 29. y 200 200ex/5 3x2 30x 31. y 10e2x cos x 9e2x sin x 7e4x 33. 35. 37. y 6 cos x 6(cot 1) sin x x2 1 39. 41. EXERCISES 4.5 (PAGE 155) 1. (3D 2)(3D 2)y sin x 3. (D 6)(D 2)y x 6 5. D(D 5)2y ex 7. (D 1)(D 2)(D 5)y xex 9. D(D 2)(D2 2D 4)y 4 15. D4 17. D(D 2) 19. D2 4 21. D3(D2 16) 23. (D 1)(D 1)3 25. D(D2 2D 5) 27. 1, x, x2, x3, x4 29. e6x, e3x/2 31. 33. 1, e5x, xe5x 35. y c1e3x c2e3x 6 37. y c1 c2ex 3x 39. 41. 43. 45. y c1ex c2e3x ex 3 47. 49. 51. 53. 55. y c1cos 5x c2sin 5x 2x cos 5x 57. sin x 2 cos x x cos x 59. 61. 63. 65. 67. 69. 71. y 2e2x cos 2x 3 64 e2x sin 2x 1 8 x3 3 16 x2 3 32 x y cos x 11 3 sin x 8 3 cos 2x 2x cos x y 41 125 41 125 e5x 1 10 x2 9 25 x y 5 8 e8x 5 8 e8x 1 4 y c1 c2x c3ex c4xex 1 2 x2ex 1 2 x2 y c1ex c2xex c3x2ex 1 6 x3ex x 13 y c1 c2x c3e8x 11 256 x2 7 32 x3 1 16 x4 y ex/2c1 cos 13 2 x c2 sin 13 2 x y ex (c1cos 2x c2sin 2x) 1 3 ex sin x y c1ex c2ex 1 6 x3ex 1 4 x2ex 1 4 xex 5 y c1e3x c2xe3x 1 49 xe4x 2 343 e4x y c1 cos 5x c2 sin 5x 1 4 sin x y c1e3x c2e4x 1 7 xe4x y c1 c2x c3ex 2 3 x4 8 3 x3 8x2 y c1e2x c2xe2x 1 2 x 1 cos 15x, sin 15x y cos 2x 5 6 sin 2x 1 3 sin x, 2 3 cos 2x 5 6 sin 2x, 0 x >2 x >2 y 4 sin 13x sin 13 13 cos 13 2x y 11 11ex 9xex 2x 12x2ex 1 2 e5x x F0 22 sin t F0 2 t cos t y 12 sin 2x 1 2 y c1ex c2xex c3x2ex x 3 2 3 x3ex y c1 c2x c3e6x 1 4 x2 6 37 cos x 1 37 sin x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-6 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 4 EXERCISES 4.6 (PAGE 161) 1. 3. 5. 7. 9. 11. y c1ex c2e2x (ex e2x) ln(1 ex) 13. y c1e2x c2ex e2x sin ex 15. 17. 19. 21. 23. y c1x1/2 cos x c2x1/2 sin x x1/2 25. 27. EXERCISES 4.7 (PAGE 168) 1. y c1x1 c2x2 3. y c1 c2 ln x 5. y c1 cos(2 ln x) c2 sin(2 ln x) 7. 9. 11. y c1x2 c2x2 ln x 13. 15. 17. y c1 c2x c3x2 c4x3 19. 21. y c1x c2x ln x x(ln x)2 23. y c1x1 c2x ln x 25. y 2 2x2 27. y cos(ln x) 2 sin(ln x) 29. 31. y c1x10 c2x2 33. 35. 37. y 2(x)1/2 5(x)1/2 ln(x), x 0 39. 41. y c1cos[ln(x 2)] c2sin[ln(x 2)] y c1(x 3)2 c2(x 3)7 y x2[c1 cos(3 ln x) c2 sin(3 ln x)] 4 13 3 10 x y c1x1 c2x8 1 30 x2 y 3 4 ln x 1 4 x2 y c1 c2x5 1 5 x5 ln x y c1x3 c2 cos(12 ln x) c3 sin(12 ln x) y x1/2[c1 cos(1 6 13 ln x) c2 sin(1 6 13 ln x)] y c1 cos (1 5 ln x) c2 sin (1 5 ln x) y c1x(216) c2x(216) y c1ex c2ex c3e2x 1 30 e4x sin x lnsec x tan x y c1 c2 cos x c3 sin x ln cos x y 4 9 e4x 25 36 e2x 1 4 e2x 1 9 ex y 1 4 ex/2 3 4 ex/2 1 8 x2ex/2 1 4 xex/2 1 3 ex cos x lncos x y c1ex sin x c2ex cos x 1 3 xex sin x y c1et c2tet 1 2 t2et ln t 3 4 t2et x0 0 y c1e2x c2e2x 1 4 e2x ln x e2x x x0 e4t t dt, y c1ex c2ex 1 2 x sinh x y c1 cos x c2 sin x 1 2 1 6 cos 2x y c1 cos x c2 sin x 1 2 x cos x y c1 cos x c2 sin x x sin x cos x lncos x EXERCISES 4.8 (PAGE 179) 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. where yp(x) 33. where 35. 37. 39. 41. 43. EXERCISES 4.9 (PAGE 184) 1. x c1et c2tet y (c1 c2)et c2tet 3. x c1 cos t c2 sin t t 1 y c1 sin t c2 cos t t 1 yp(x) 1 2(lnx)2 1 2 lnx yp(x) excos x exsin x ex yp(x) sin(x 1) sin1 sinx sin1 1 yp(x) 1 2x2 1 2x yp(x) (x 1) x 0 tf(t)dt x 1 x (t 1)f(t)dt yp(x) 0, x 0 10 10cos x, 0 x 3 20cos x, x 3 y cosx sinx yp(x), 1 cosh x, x 0 1 cosh x, x 0 y(x) 5ex 3ex yp(x), y 46 45x3 1 20x2 1 36 1 6ln x y 4x 2x2 x lnx y (cos1 2)ex (1 sin1 cos1)e2x e2xsinex y xsinx cosxlnsinx y e5x 6xe5x 1 2x2e5x y 25 16e2x 9 16e2x 1 4xe2x yp(x) cosx p 2 sinx x sinx cosx lnsinx yp(x) 1 2x2e5x yp(x) 1 4xe2x 1 16e2x 1 16e2x y c1cos3x c2sin3x 1 3 x x0 sin3(x t)(t sint)dt y c1ex c2xex x x0 (x t)e(xt)etdt y c1e4x c2e4x 1 4 x x0 sinh4(x t)te2tdt yp(x) 1 3 x x0 sin3(x t)f(t)dt yp(x) x x0 (x t)e(xt)f(t)dt yp(x) 1 4 x x0 sinh 4(x t)f(t)dt Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-7 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 5 5. 7. 9. 11. 13. 15. 17. 19. x 6c1et 3c2e2t 2c3e3t y c1et c2e2t c3e3t z 5c1et c2e2t c3e3t 21. x e3t3 te3t3 y e3t3 2te3t3 23. mx 0 my mg; x c1t c2 EXERCISES 4.10 (PAGE 189) 3. 5. 7. 9. 11. 13. 15. 17. 19. y 11 x2 y 1 x 1 2 x2 2 3 x3 1 4 x4 7 60 x5 y 1 x 1 2 x2 1 2 x3 1 6 x4 1 10 x5 y 1 c1 11 c2 1x2 c2 y tan (1 4 1 2 x), 1 2 x 3 2 y 2 3 (x 1)3>2 4 3 1 3 y3 c1y x c2 y 1 c2 1 ln c1x 1 1 c1 x c2 y ln cos (c1 x) c2 y 1 2 gt2 c3t c4 ( 1 2 13c2 1 2 c3)et/2 cos 1 2 13t z c1et (1 2 c2 1 2 13c3)et/2 sin 1 2 13t ( 1 2 13c2 1 2 c3)et/2 cos 1 2 13t y c1et (1 2 c2 1 2 13c3)et/2 sin 1 2 13t x c1et c2et/2 sin 1 2 13t c3et/2 cos 1 2 13t y (c1 c2 2) (c2 1)t c4et 1 2 t2 x c1 c2t c3et c4et 1 2 t2 y 3 4 c1e4t c2 5et x c1e4t 4 3 et (1 2 13c2 3 2 c3)et/2 sin 1 2 13t y (3 2 c2 1 2 13c3)et/2 cos 1 2 13t x c1et c2et/2 cos 1 2 13t c3et/2 sin 1 2 13t y c1 c2 sin t c3 cos t 4 15 e3t x c1 c2 cos t c3 sin t 17 15 e3t y c1e2t c2e2t c3 sin 2t c4 cos 2t 1 5 et x c1e2t c2e2t c3 sin 2t c4 cos 2t 1 5 et y c1 sin t c2 cos t c3 sin 16t c4 cos 16t x 1 2 c1 sin t 1 2 c2 cos t 2c3 sin 16t 2c4 cos 16t CHAPTER 4 IN REVIEW (PAGE 190) 1. y 0 3. false 5. 7. 9. 11. (, 0); (0, ) 13. y c1e3x c2e5x c3xe5x c4ex c5xex c6x2ex; y c1x3 c2x5 c3x5 ln x c4x c5xln x c6x(ln x)2 15. 17. y c1 c2e5x c3xe5x 19. 21. 23. 25. 27. y c1x1/3 c2x1/2 29. y c1x2 c2x3 x4 x2 ln x 31. (a) ; (b) 33. (a) y c1cosh x c2sinh x c3x cosh x c4x sinh x (b) yp Ax2 cosh x Bx2 sinh x 35. y exp cos x 37. 39. y x2 4 43. y c1et c2e2t 3 45. x c1et c2e5t tet y c1et 3c2e5t tet 2et EXERCISES 5.1 (PAGE 205) 1. 3. 5. (a) (b) 4 ft/s; downward (c) t (2n 1) 16 , n 0, 1, 2, . . . x ( 4) 1 2; x (9 32) 12 4 x ( 12) 1 4; x ( 8) 1 2; x ( 6) 1 4; x(t) 1 4 cos 4 16 t 12 8 x c1et 3 2 c2e2t 5 2 y 13 4 ex 5 4 ex x 1 2 sin x y c1ex c2ex Axex , y c1ex c2ex Aex , ; Bx sin x, y c1 cos x c2sin x Ax cos x B sin x, y c1 cos x c2sin x A cos x ex cos x ln sec x tan x y ex (c1 cos x c2 sin x) y c1 c2e2x c3e3x 1 5 sin x 1 5 cos x 4 3 x 46 125 x 222 625 y e3x/2 (c2 cos 1 2 111x c3 sin 1 2 111x) 4 5 x3 36 25 x2 y c1ex /3 e3x/2 (c2 cos 1 2 17x c3 sin 1 2 17x) y c1e(113)x c2e(113)x yp x2 x 2 x2y 3xy 4y 0 y c1cos5x c2sin5x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-8 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 5 7. (a) the 20-kg mass (b) the 20-kg mass; the 50-kg mass (c) t np, n 0, 1, 2, . . . ; at the equilibrium position; the 50-kg mass is moving upward whereas the 20-kg mass is moving upward when n is even and downward when n is odd. 9. (a) (b) (c) 11. (a) (b) (c) 15 cycles (d) 0.721 s (e) (f) x(3) 0.597 ft (g) x (3) 5.814 ft/s (h) x(3) 59.702 ft/s2 (i) (j) (k) 13. 17. (a) above (b) heading upward 19. (a) below (b) heading upward 21. that is, the weight is approximately 0.14 ft below the equilibrium position. 23. (a) (b) 25. (a) (b) (c) t 1.294 s 27. (a) (b) (c) 29. 31. 33. 2e2t sin 4t x(t) 1 2 cos 4t 9 4 sin 4t 1 2 e2t cos 4t x(t) 1 4 e4t te4t 1 4 cos 4t 10 3 (cos 3t sin 3t) x(t) et/2 4 3 cos 247 2 t 64 3 147 sin 247 2 t 0 5 2 5 2 5 2 x(t) 25 2 e2t sin (4t 4.249) x(t) e2t (cos 4t 1 2 sin 4t) x(t) 2 3 e2t 5 3 e8t x(t) 4 3 e2t 1 3 e8t 1 4 s; 1 2 s, x (1 2) e2; 120 lb/ft; x(t) 23 12 sin 813 t 0.3545 n 5 , n 0, 1, 2, . . . 0.1451 n 5 ; 0.3545 n 5 , n 0, 1, 2, . . . 81 3 ft/s (2n 1) 20 0.0927, n 0, 1, 2, . . . 5 6 ft; 5 5 6 sin (10t 0.927) x(t) 2 3 cos 10t 1 2 sin 10t x(t) 213 4 cos(2t 0.983) x(t) 213 4 sin(2t 0.588) x(t) 1 2 cos 2t 3 4 sin 2t 35. (a) where 2l bm and v2 km (b) 37. 39. (b) 45. 4.568 C; 0.0509 s 47. q(t) 10 10e3t(cos 3t sin 3t) i(t) 60e3t sin 3t; 10.432 C 49. 53. 57. EXERCISES 5.2 (PAGE 215) 1. (a) 3. (a) 5. (a) (c) x 0.51933, ymax 0.234799 7. 9. ln n2, n 1, 2, 3, . . . ; y sin nx 11. y cos (2n 1)x 2L n (2n 1)22 4L2 , n 1, 2, 3, . . . ; w0 2P x2 w0EI P2 w0EI P2 sinh B P EI L w0L 1EI P 1P sinh B P EI x cosh B P EI L y(x) w0EI P2 cosh B P EI x y(x) w0 360EI (7L4x 10L2x3 3x5) y(x) w0 48EI (3L2x2 5Lx3 2x4) y(x) w0 24EI (6L2x2 4Lx3 x4) E0C 1 2LC sin t i(t) i0 cos t 1LC 1 1LC q0 E0C 1 2LC sin t 1LC 1LCi0 sin t 1LC E0C 1 2LC cos t q(t) q0 E0C 1 2LC cos t 1LC q(t) 1 2 e10t (cos 10t sin 10t) 3 2 ; 3 2 C ip 100 13 cos t 150 13 sin t qp 100 13 sin t 150 13 cos t F0 2 t sin t x(t) cos 2t 1 8 sin 2t 3 4 t sin 2t 5 4 t cos 2t 32 13 sin t x(t) e2t (56 13 cos 2t 72 13 sin 2t) 56 13 cos t d 2x dt2 2 dx dt 2x 2h(t), m d 2x dt2 k(x h) dx dt or Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-9 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 6 13. ln n2, n 0, 1, 2, . . . ; y cos nx 15. 17. ln n2, n 1, 2, 3, . . . ; y sin(n ln x) 19. ln n4p4, n 1, 2, 3, . . . ; y sin npx 21. x L4, x L2, x 3 L4 25. 27. EXERCISE 5.3 (PAGE 224) 7. 15. (a) 5 ft (b) (c) ; 7.5 ft 17. (a) . When t 0, x a, y 0, dydx 0. (b) When r 1, When r 1, (c) The paths intersect when r 1. 19. (a) (b) use at (c) use cos (d) CHAPTER 5 IN REVIEW (PAGE 228) 1. 8 ft 3. 5. False; there could be an impressed force driving the system. 7. overdamped 9. y 0 since l 8 is not an eigenvalue 11. 14.4 lb 13. 15. 0 m 2 17. 19. x(t) e4t (26 17 cos 2 12 t 28 17 12 sin 2 12 t) 8 17 et 8 3 13 x(t) 2 3e2t 1 3 e4t 5 4 m vb 21,797 cm/s umax 1 1 2 u2 max umax, sin 1g>l t 1 u(t) v0B l g sinB g l t y(x) 1 2 1 2a (x2 a2) 1 a ln a x ar 1 r2 y(x) a 2 1 1 r x a 1r 1 1 r x a 1r xy r 11 (y )2 0 t 3 8 110 4 110 ft/s d 2x dt2 x 0 u(r) u0 u1 b a ab r u1b u0a b a n n 1T L 1 , n 1, 2, 3, . . . ; y sin nx L n n22 25 , n 1, 2, 3, . . . ; y ex sin nx 5 21. (a) (b) (c) 25. 27. EXERCISES 6.1 (PAGE 237) 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 35. 37. EXERCISES 6.2 (PAGE 246) 1. 5; 4 3. 5. 7. 1 10 9 7 6 4 3 x10 y2(x) c1 x 1 4 3 x4 1 7 6 4 3 x7 1 9 8 6 5 3 2 x9 y1(x) c0 1 1 3 2 x3 1 6 5 3 2 x6 y2(x) c1 x 1 2! x2 1 3! x3 1 4! x4 . . . y1(x) c0 y2(x) c1 x 1 3! x3 1 5! x5 1 7! x7 . . . y1(x) c0 1 1 2! x2 1 4! x4 1 6! x6 . . . y c0 k0 1 k! x2 2 k y c0 k0 1 k! (5x)k c0 2c2 k1 [(k 2)(k 1)ck2 (2k 1)ck]x k 2c1 k1 [2(k 1)ck1 6ck1]xk k0 [(k 1)ck1 ck]x k k3 (k 2)ck2xk 1 1 2x2 5 24x4 61 720 x6 . . ., (p>2, p>2) x 2 3x3 2 15 x5 4 315x7 . . . n0 (1)n (2n 1)!(x 2p)2n1 n1 1 n xn n0 (1)n 2n1 xn n0 (1)n n!2n x n (75 32, 75 32), R 75 32 [0, 2 3], R 1 3 (5, 15), R 10 [1 2, 1 2), R 1 2 (1,1], R 1 mx fk sgn(x ) kx 0 m d 2x dt2 kx 0 t n 50, n 0, 1, 2, . . . i(t) 2 3 cos 100t 2 3 cos 50t q(t) 1 150 sin 100t 1 75 sin 50t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-10 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 6 9. 11. 13. 15. 17. 19. 8x 2ex 21. y(x) 3 12x2 4x4 23. EXERCISES 6.3 (PAGE 255) 1. x 0, irregular singular point 3. x 3, regular singular point; x 3, irregular singular point 5. x 0, 2i, 2i, regular singular points 7. x 3, 2, regular singular points 9. x 0, irregular singular point; x 5, 5, 2, regular singular points 11. 13. 15. C2 1 2x 2 x2 23 3 3! x3 23 9 7 5 3! x3 y(x) C1x3/2 1 2 5 x 22 7 5 2 x2 r1 3 2, r2 0 r1 1 3, r2 1 for x 1: p(x) 5(x 1) x 1 , q(x) x2 x for x 1: p(x) 5, q(x) x(x 1)2 x 1 y2(x) c1 [x 1 12 x4 1 180 x6 ] y1(x) c0 [1 1 6 x3 1 120 x5 ] y(x) 2 1 1 2! x2 1 3! x3 1 4! x4 6x y2(x) c1x 1 6 x3 14 2 5! x5 34 14 4 7! x7 y1(x) c0 1 1 4 x2 7 4 4! x4 23 7 8 6! x6 y2(x) c1 [x 1 2 x2 1 2 x3 1 4 x4 ] y1(x) c0 [1 1 2 x2 1 6 x3 1 6 x4 ] y1(x) c0; y2(x) c1 n1 1 n xn 82 52 22 10! x10 y2(x) c1 x 22 4! x4 52 22 7! x7 y1(x) c0 1 1 3! x3 42 6! x6 72 42 9! x9 y2(x) c1 x 1 3! x3 5 5! x5 45 7! x7 y1(x) c0 1 1 2! x2 3 4! x4 21 6! x6 17. 19. 21. 23. 25. r1 0, r2 1 27. r1 1, r2 0 29. r1 r2 0 where y1(x) n0 1 n! xn ex 1 3 3! x3 1 4 4! x4 y(x) C1y(x) C2 y1(x) ln x y1(x)x 1 4 x2 1 12 x3 1 72 x4 ] y(x) C1x C2 [ x ln x 1 1 2 x2 1 x [C1 sinh x C2 cosh x] C1x1 n0 1 (2n 1)! x2n1 C2x1 n0 1 (2n)! x2n y(x) C1 n0 1 (2n 1)! x2n C2x1 n0 1 (2n)! x2n C2x1/3[1 1 2 x 1 5 x2 7 120 x3 ] y(x) C1x2/3 [1 1 2 x 5 28 x2 1 21 x3 ] r1 2 3, r2 1 3 C2 1 1 3 x 1 6 x2 1 6 x3 23 4 11 9 7 x3 y(x) C1x5/2 1 2 2 7 x 22 3 9 7 x2 r1 5 2, r2 0 C2 1 1 2 x 1 5 2 x2 1 8 5 2 x3 1 33 3! x3 y(x) C1x1/3 1 1 3 x 1 32 2 x2 r1 1 3, r2 0 23 17 9 3! x3 c2 1 2x 22 9 2 x2 23 31 23 15 3! x3 y(x) c1x7/8 1 2 15 x 22 23 15 2 x2 r1 7 8, r2 0 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-11 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 7 33. (b) (c) EXERCISES 6.4 (PAGE 267) 1. y c1J1/3(x) c2J1/3(x) 3. y c1J5/2(x) c2J5/2(x) 5. y c1J0(x) c2Y0(x) 7. y c1J2(3x) c2Y2(3x) 9. y c1J2/3(5x) c2J2/3(5x) 11. y c1x1/2J1/2(ax) c2x1/2J1/2(ax) 13. y x1/2 [c1J1(4x1/2) c2Y1(4x1/2)] 15. y x [c1J1(x) c2Y1(x)] 17. y x1/2 c1J3/2(x) c2Y3/2(x) 19. 23. y x1/2 [c1J1/2(x) c2J1/2(x)] C1 sin x C2 cos x 25. 35. 45. P2(x), P3(x), P4(x), and P5(x) are given in the text, , 47. l1 2, l2 12, l3 30 53. CHAPTER 6 IN REVIEW (PAGE 271) 1. False 3. 7. x2(x 1)y y y 0 9. 11. 13. r1 3, r2 0 15. 17. 1 6 2 [x 1 2 x3 1 8 x5 1 48 x7 ] y(x) 3[1 x2 1 3 x4 1 15 x6 ] y2(x) C2 [1 x 1 2 x2] y1(x) C1x3 [1 1 4 x 1 20 x2 1 120 x3 ] y2(x) c1 [x 1 2 x3 1 4 x4 ] y1(x) c0 [1 3 2 x2 1 2 x3 5 8 x4 ] y2(x) C2 [1 x 1 6 x2 1 90 x3 ] y1(x) C1x1/2 [1 1 3 x 1 30 x2 1 630 x3 ] r1 1 2, r2 0 [1 2, 1 2] y x 4x3 16 5 x 5 P7(x) 1 16 (429x7 693x5 315x3 35x) P6(x) 1 16 (231x6 315x4 105x2 5) y c1x1/2J1/3(2 3 ax3/2) c2x1/2J1/3(2 3 ax3/2) C1x3/2sin(1 8 x2) C2 x3/2 cos(1 8 x2) y x1/2 [c1J1/2(1 8 x2) c2J1/2(1 8 x2)] y x1[c1J1/2(1 2 x2) c2J1/2(1 2 x2)] y C1x sin 1 x C2x cos 1 x y2(t) t1 n0 (1)n (2n)! (1 t)2n cos (1 t) t y1(t) n0 (1)n (2n 1)! (1 t)2n sin (1 t) 1 t 19. x 0 is an ordinary point 21. EXERCISES 7.1 (PAGE 280) 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. Use sinh and linearity to show that 35. 37. 39. 43. 45. EXERCISES 7.2 (PAGE 288) 1. 3. t 2t4 5. 7. t 1 e2t 9. 11. 13. 15. 2 cos 3t 2 sin 3t 17. 19. 21. 0.3e0.1t 0.6e0.2t 23. 25. 27. 4 3et cos t 3 sin t 1 5 1 5 cos 15 t 1 2 e2t e3t 1 2 e6t 3 4 e3t 1 4 et 1 3 1 3 e3t cos t 2 5 7 sin 7t 1 4 et/4 1 3t 3 2 t 2 1 6 t 3 1 2 t2 31p 4s5>2 1p s1>2 4 cos 5 (sin 5)s s2 16 2 s2 16 1 2(s 2) 1 2s {sinh kt} k s2 k2. kt ekt ekt 2 8 s3 15 s2 9 1 s 2 s 2 1 s 4 1 s 1 s 4 6 s4 6 s3 3 s2 1 s 2 s3 6 s2 3 s 4 s2 10 s 48 s5 s2 1 (s2 1)2 1 s2 2s 2 1 (s 4)2 e7 s 1 1 s 1 s2 1 s2 es 1 s es 1 s2 es 1 es s2 1 1 s2 1 s2 es 2 s es 1 s 1 32 2! x6 1 33 3! x9 1 4 7 10 x10 5 2 x2 1 3 x3 c1 x 1 4 x4 1 4 7 x7 y(x) c0 1 1 3 x3 1 32 2! x6 1 33 3! x9 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-12 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 7 29. 31. y 1 et 33. 35. 37. 39. 41. EXERCISES 7.3 (PAGE 297) 1. 3. 5. 7. 9. 11. 13. e3t sin t 15. e2t cos t 2e2t sin t 17. et tet 19. 21. y te4t 2e4t 23. y et 2tet 25. 27. 29. 31. y (e 1)tet (e 1)et 33. 37. 39. 41. 43. 45. 47. 49. (c) 51. (f) 53. (a) 55. 57. 59. 61. 63. 65. 67. 69. [1 cos(t 2)] (t 2) y sin t 1 cos(t ) 1 3 sin (t 2) (t 2) y cos 2t 1 6 sin 2(t 2) (t 2) 1 2 (t 1) (t 1) 1 4 e2(t1) (t 1) y 1 4 1 2 t 1 4 e2t 1 4 (t 1) y [5 5e(t1)] (t 1) f (t) (t a) (t b); {f (t)} eas s ebs s f (t) t t (t 2); {f (t)} 1 s2 e2s s2 2 e2s s f (t) t2 (t 1); {f (t)} 2 es s3 2 es s2 es s f (t) 2 4(t 3); {f (t)} 2 s 4 s e3s (t 1) e(t1) (t 1) sin t (t ) 1 2 (t 2)2 (t 2) s s2 4 es e2s s2 2 e2s s es s2 x(t) 3 2 e7t/2cos 115 2 t 7 115 10 e7t/2 sin 115 2 t y 1 2 1 2 et cos t 1 2 et sin t y 3 2 e3t sin 2t y 1 9 t 2 27 2 27 e3t 10 9 te3t 5 t 5et 4tet 3 2 t2 et 1 2 t2 e2t s s2 25 s 1 (s 1)2 25 3 s 4 (s 4)2 25 3 (s 1)2 9 1 (s 2)2 2 (s 3)2 1 (s 4)2 6 (s 2)4 1 (s 10)2 y 1 4 et 1 4 e3t cos 2t 1 4 e3t sin 2t y 8 9 et /2 1 9 e2t 5 18 et 1 2 et y 10 cos t 2 sin t 12 sin 12 t y 4 3 et 1 3 e4t y 1 10 e4t 19 10 e6t 1 3 sin t 1 6 sin 2t 71. 73. 75. (a) (b) imax 0.1 at t 1.7, imin 0.1 at t 4.7 77. 79. 81. (a) k[T 70 57.5t (230 57.5t)(t 4)] EXERCISES 7.4 (PAGE 309) 1. 3. 5. 7. 9. 11. 13. 17. 19. 21. 23. 25. 27. 29. 31. et 1 3s2 1 s2(s2 1)2 1 s2(s 1) s 1 s[(s 1)2 1] 1 s(s 1) s 1 (s 1)[(s 1)2 1] 6 s5 y 2 3 t 3 c1t 2 1 8 (t ) sin 4(t )(t ) y 1 4 sin 4t 1 8 t sin 4t y 2 cos 3t 5 3 sin 3t 1 6 t sin 3t y 1 2 et 1 2 cos t 1 2 t cos t 1 2 t sin t 12s 24 [(s 2)2 36]2 6s2 2 (s2 1)3 s2 4 (s2 4)2 1 (s 10)2 dT dt w0 60EIL 5L 2 x4 x5 x L 2 5 x L 2 y(x) w0L2 48EI x2 w0L 24EI x3 w0 24EI x L 2 4 x L 2 y(x) w0L2 16EI x2 w0L 12EI x3 w0 24EI x4 1 101 sint 3 2 t 3 2 10 101 cost 3 2 t 3 2 10 101 e10(t3/2) t 3 2 i(t) 1 101 e10t 1 101 cos t 10 101 sin t q(t) 2 5 (t 3) 2 5 e5(t3) (t 3) 25 4 cos 4(t 5) (t 5) 5 16 sin 4(t 5) (t 5) 25 4 (t 5) x(t) 5 4 t 5 16 sin 4t 5 4 (t 5) (t 5) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-13 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 7 33. 37. f(t) sin t 39. 41. f(t) et 43. 45. 47. 49. 51. 53. 55. 57. EXERCISES 7.5 (PAGE 315) 1. 3. 5. 7. 9. 11. 13. EXERCISES 7.6 (PAGE 319) 1. 3. 5. 7. 9. y 2 3! t3 1 4! t 4 x 8 2 3! t3 1 4! t 4 y 1 2 t 3 4 12 sin 12 t y 8 3 e3t 5 2 e2t 1 6 x 1 2 t 3 4 12 sin 12 t x 2e3t 5 2 e2t 1 2 y 2 cos 3t 7 3 sin 3t y 1 3 e2t 2 3 et x cos 3t 5 3 sin 3t x 1 3 e2t 1 3 et y(x) P0 EI L 4 x2 1 6 x3, 0 x L 2 P0L2 4EI 1 2 x L 12, L 2 x L 1 3 e2(t3) sin 3(t 3) (t 3) 1 3 e2(t) sin 3(t ) (t ) y e2t cos 3t 2 3 e2t sin 3t y e2(t2) sin t (t 2) y 1 2 1 2 e2t [1 2 1 2 e2(t1)] (t 1) y cos t (t 2) cos t (t 3 2 ) y sin t sin t (t 2) y e3(t2) (t 2) 1 3 e(tn) sin 3(t n) 4 n1 (1)n [1 e(tn) cos 3(t n) x(t) 2(1 et cos 3t 1 3 et sin 3t) 2 R n1 (1)n (1 eR(tn)/L)(t n) i(t) 1 R (1 eRt/L) coth (s>2) s2 1 a s 1 bs 1 ebs 1 1 eas s(1 eas) 100e10(t2) e20(t2) i(t) 100e10(t1) e20(t1) y(t) sin t 1 2 t sin t f (t) 3 8 e2t 1 8 e2t 1 2 cos 2t 1 4 sin 2t f (t) 1 8 et 1 8 et 3 4 tet 1 4 t 2et et 1 2 t 2 t 1 11. 13. 15. (b) (c) i1 20 20e900t 17. 19. CHAPTER 7 IN REVIEW (PAGE 320) 1. 3. false 5. true 7. 9. 11. 13. 15. 17. 19. 21. 5 23. ek(sa)F(s a) 25. 27. 29. ; ; 31. ; ; 33. 35. 9 100 e5(t2) (t 2) 1 5 (t 2) (t 2) 1 4 e(t2) (t 2) y 6 25 1 5 t 3 2 et 13 50 e5t 4 25 (t 2) y 5tet 1 2 t2 et {et f (t)} 2 s 1 1 (s 1)2 e2(s1) {f (t)} 2 s 1 s2 e2s f (t) 2 (t 2) (t 2) 1 s 1 e4(s1) {etf (t)} 1 (s 1)2 1 (s 1)2 e(s1) {f (t)} 1 s2 1 s2 es 1 s e4s f (t) t (t 1)(t 1) (t 4) f (t t0)(t t0) f (t)(t t0) cos (t 1)(t 1) sin (t 1)(t 1) e5t cos 2t 5 2 e5t sin 2t 1 2 t 2 e5t 1 6 t 5 4s (s2 4)2 2 s2 4 1 s 7 1 s2 2 s2 es i2 6 5 6 5 e100t cosh 50 12 t 612 5 e100t sinh 5012 t i1 6 5 6 5 e100t cosh 5012 t 912 10 e100t sinh 5012 t i3 30 13 e2t 250 1469 e15t 280 113 cos t 810 113 sin t i2 20 13 e2t 375 1469 e15t 145 113 cos t 85 113 sin t i3 80 9 80 9 e900t i2 100 9 100 9 e900t x2 2 5 sin t 16 15 sin 16 t 4 5 cos t 1 5 cos 16 t x1 1 5 sin t 216 15 sin 16 t 2 5 cos t 2 5 cos 16 t y 1 3 1 3 et 1 3 tet x 1 2 t2 t 1 et Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-14 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 8 37. 39. 41. 43. i(t) 9 2t 9et/5 45. 47. (a) 49. (a) (b) and use the double-angle formula for sin (d) approximately 2729 ft; approximately 11.54 s EXERCISES 8.1 (PAGE 332) 1. , where 3. 5. 7. 9. 17. Yes; W(X1, X2) 2e8t 0 implies that X1 and X2 are linearly independent on (, ). dz dt 2x 5y 6z 2et t dy dt 3x 4y z 2et t dx dt x y 2z et 3t dy dt x 3y et dx dt 4x 2y et where X x y z X 1 2 1 1 1 1 1 1 1X 0 3t2 t2 t 0 t 1 0 2, X 3 6 10 4 1 4 9 0 3X, where X x y z X x y X 3 5 4 8X 2u y(x) g 2v2 0 cos2 u x2 sin u cos u x; solve y(x) 0 x(t) (v0 cos u) t, y(t) 1 2 gt2 (v0 sin u)t 2(t) 0 0 2 cos t 0 0 2 cos 12 2K t 1(t) 0 0 2 cos t 0 0 2 cos 12 2K t 1 5 x L 2 5 x L 2 y(x) w0 12EIL 1 5 x5 L 2 x4 L2 2 x3 L3 4 x2 y t 9 4 e2t 1 4 e2t x 1 4 9 8 e2t 1 8 e2t y 1 t 1 2 t 2 1 4 1 2 (t 3) 1 4 e2(t3) 21 4 1 2 (t 2) 1 4 e2(t2) y(t) e2t 1 4 1 2 (t 1) 1 4 e2(t1) 19. No; W(X1, X2, X3) 0 for every t. The solution vectors are linearly dependent on (, ). Note that X3 2X1 X2. EXERCISES 8.2 (PAGE 346) 1. 3. 5. 7. 9. 11. 13. 19. 21. 23. 25. 27. c3 0 1 1 t2 2 et 0 1 0tet 1 2 0 0et X c1 0 1 1et c2 0 1 1tet 0 1 0et c3 2 0 1te5t 1 2 1 2 1e5t X c1 4 5 2 c2 2 0 1e5t X c1 1 1 1et c2 1 1 0e2t c3 1 0 1e2t X c1 1 1e2t c2 1 1te2t 1 3 0e2t X c1 1 3 c2 1 3t 1 4 1 4 X 3 1 1et/2 2 0 1et/2 X c1 4 0 1et c2 12 6 5et/2 c3 4 2 1e3t/2 X c1 1 0 1et c2 1 4 3e3t c3 1 1 3e2t X c1 1 0 0et c2 2 3 1e2t c3 1 0 2et X c1 5 2e8t c2 1 4e10t X c1 2 1e3t c2 2 5et X c1 1 2e5t c2 1 1et Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-15 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 8 29. 31. Corresponding to the eigenvalue l1 2 of multiplicity ﬁve, the eigenvectors are 33. 35. 37. 39. 41. 43. 45. EXERCISES 8.3 (PAGE 354) 1. 3. 1 4 1 4t 2 3 4 X c1 1 1e2t c2 1 1e4t 1 4 3 4 t2 X c1 1 1et c2 3 1et 1 3 6 5 cos 5t sin 5t sin 5t sin 5t X 25 7 6et cos 5t 5 sin 5t cos 5t cos 5t c3 3 cos 3t 4 sin 3t 5 sin 3t 0 e2t X c1 28 5 25e2t c2 4 cos 3t 3 sin 3t 5 cos 3t 0 e2t X c1 0 2 1et c2 sin t cos t cos tet c3 cos t sin t sin tet X c1 1 0 0 c2 cos t cos t sin t c3 sin t sin t cos t X c1 5 cos 3t 4 cos 3t 3sin3t c2 5sin3t 4 sin 3t 3 cos 3t X c1 cos t cos t sin te4t c2 sin t sin t cos te4t X c1 cos t 2 cos t sin te4t c2 sin t 2 sin t cos te4t K1 1 0 0 0 0 , K2 0 0 1 0 0 , K3 0 0 0 1 0 . X 7 2 1e4t 13 2t 1 t 1e4t 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 4 29 83 69sin t i1 i2 2 1 3e2t 6 29 3 1e12t 4 29 19 42cos t X 2 2te2t 1 1e2t 2 2te4t 2 0e4t 1 4 e2t 1 2 te2t et 1 4 e2t 1 2 te2t 1 2 t2e3t X c1 1 1 0 c2 1 1 0e2t c3 0 0 1e3t cos t 1 2 sin tet ln sin t 2 cos t sin tet ln cos t X c1 2 sin t cos tet c2 2 cos t sin tet 3 sin t 3 2 cos ttet sin t sin t tan t sin t cos t ln cos t X c1 cos t sin t c2 sin t cos t cos t sin tt X c1 cos t sin tet c2 sin t cos tet cos t sin ttet sin t cos t ln cos t X c1 cos t sin t c2 sin t cos t cos t sin tt X c1 1 1et c2 t 1 2 tet 1 2 2et X c1 4 1e3t c2 2 1e3t 12 0t 4 3 4 3 X c1 2 1et c2 1 1e2t 3 3et 4 2tet X c1 2 1et/2 c2 10 3e3t/2 13 2 13 4tet/2 15 2 9 4et/2 X c1 1 1 c2 3 2et 11 11t 15 10 X 13 1 1et 2 4 6e2t 9 6 X c1 1 0 0et c2 1 1 0e2t c3 1 2 2e5t 3 2 7 2 2e4t X c1 1 3e3t c2 1 9e7t 55 36 19 4 et Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-16 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 9 EXERCISES 8.4 (PAGE 359) 1. 3. 5. 7. 9. 11. 13. 15. 17. 23. CHAPTER 8 IN REVIEW (PAGE 360) 1. 5. 7. X c1 cos 2t sin 2tet c2 sin 2t cos 2tet X c1 1 1et c2 1 1tet 0 1et k 1 3 X c3 1 1e3t c4 1 3e5t X c1 3 2 e3t 1 2 e5t 3 2 e3t 3 2 e5t c2 1 2 e3t 1 2 e5t 1 2 e3t 3 2 e5t or X c1 1 3t t e2t c2 9t 1 3te2t eAt e2t 3te2t te2t 9te2t e2t 3te2t; X c3 3 2e2t c4 1 2e2t X c1 3 2 e2t 1 2 e2t e2t e2t c2 3 4 e2t 3 4 e2t 1 2 e2t 3 2 e2t or eAt 3 2 e2t 1 2 e2t e2t e2t 3 4 e2t 3 4 e2t 1 2 e2t 3 2 e2t; X t 1 t 2t 4 t t 1 2t 6 t t 2t 1 X c1 cosh t sinh t c2 sinh t cosh t 1 1 X c3 1 0et c4 0 1e2t 3 1 2 X c1 t 1 t 2t c2 t t 1 2t c3 t t 2t 1 X c1 1 0et c2 0 1e2t eAt t 1 t 2t t t 1 2t t t 2t 1 eAt et 0 0 e2t; eAt et 0 0 e2t 9. 11. 13. 15. (b) EXERCISES 9.1 (PAGE 367) 1. for h 0.1, y5 2.0801; for h 0.05, y10 2.0592 3. for h 0.1, y5 0.5470; for h 0.05, y10 0.5465 5. for h 0.1, y5 0.4053; for h 0.05, y10 0.4054 7. for h 0.1, y5 0.5503; for h 0.05, y10 0.5495 9. for h 0.1, y5 1.3260; for h 0.05, y10 1.3315 11. for h 0.1, y5 3.8254; for h 0.05, y10 3.8840; at x 0.5 the actual value is y(0.5) 3.9082 13. (a) y1 1.2 (b) (c) Actual value is y(0.1) 1.2214. Error is 0.0214. (d) If h 0.05, y2 1.21. (e) Error with h 0.1 is 0.0214. Error with h 0.05 is 0.0114. 15. (a) y1 0.8 (b) for 0 c 0.1. (c) Actual value is y(0.1) 0.8234. Error is 0.0234. (d) If h 0.05, y2 0.8125. (e) Error with h 0.1 is 0.0234. Error with h 0.05 is 0.0109. 17. (a) Error is 19h2e3(c1). (b) (c) If h 0.1, y5 1.8207. If h 0.05, y10 1.9424. (d) Error with h 0.1 is 0.2325. Error with h 0.05 is 0.1109. y(c) h2 2 19(0.1)2(1) 0.19 y(c) h2 2 5e2c (0.1)2 2 0.025e2c 0.025 0.0244 y(c) h2 2 4e2c (0.1)2 2 0.02e2c 0.02e0.2 X c1 1 1 0 c2 1 0 1 c3 1 1 1e3t sin t sin t cos t ln csc t cot t X c1 cos t cos t sin t c2 sin t sin t cos t 1 1 X c1 1 0e2t c2 4 1e4t 16 4t 11 1 X c1 2 3 1e2t c2 0 1 1e4t c3 7 12 16e3t Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 19. (a) Error is . (b) (c) If h 0.1, y5 0.4198. If h 0.05, y10 0.4124. (d) Error with h 0.1 is 0.0143. Error with h 0.05 is 0.0069. EXERCISES 9.2 (PAGE 371) 1. y5 3.9078; actual value is y(0.5) 3.9082 3. y5 2.0533 5. y5 0.5463 7. y5 0.4055 9. y5 0.5493 11. y5 1.3333 13. (a) 35.7130 (c) 15. (a) for h 0.1, y4 903.0282; for h 0.05, y8 1.1 1015 17. (a) y1 0.82341667 (b) (c) Actual value is y(0.1) 0.8234134413. Error is 3.225 106 3.333 106. (d) If h 0.05, y2 0.82341363. (e) Error with h 0.1 is 3.225 106. Error with h 0.05 is 1.854 107. 19. (a) (b) (c) From calculation with h 0.1, y5 0.40546517. From calculation with h 0.05, y10 0.40546511. EXERCISES 9.3 (PAGE 375) 1. y(x) x ex; actual values are y(0.2) 1.0214, y(0.4) 1.0918, y(0.6) 1.2221, y(0.8) 1.4255; approximations are given in Example 1. 3. y4 0.7232 5. for h 0.2, y5 1.5569; for h 0.1, y10 1.5576 7. for h 0.2, y5 0.2385; for h 0.1, y10 0.2384 EXERCISES 9.4 (PAGE 379) 1. y(x) 2e2x 5xe2x; y(0.2) 1.4918, y2 1.6800 3. y1 1.4928, y2 1.4919 5. y1 1.4640, y2 1.4640 7. x1 8.3055, y1 3.4199; x2 8.3055, y2 3.4199 24 (c 1)5 h5 5! 24 (0.1)5 5! 2.0000 106 y(5) (c) h5 5! 24 (c 1)5 h5 5! 3.333 106 y(5)(c) h5 5! 40e2c h5 5! 40e2(0) (0.1)5 5! v(t) B mg k tanh B kg m t; v(5) 35.7678 y(c) h2 2 (1) (0.1)2 2 0.005 1 (c 1)2 h2 2 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ● ANS-17 9. x1 3.9123, y1 4.2857; x2 3.9123, y2 4.2857 11. x1 0.4179, y1 2.1824; x2 0.4173, y2 2.1821 EXERCISES 9.5 (PAGE 383) 1. y1 5.6774, y2 2.5807, y3 6.3226 3. y1 0.2259, y2 0.3356, y3 0.3308, y4 0.2167 5. y1 3.3751, y2 3.6306, y3 3.6448, y4 3.2355, y5 2.1411 7. y1 3.8842, y2 2.9640, y3 2.2064, y4 1.5826, y5 1.0681, y6 0.6430, y7 0.2913 9. y1 0.2660, y2 0.5097, y3 0.7357, y4 0.9471, y5 1.1465, y6 1.3353, y7 1.5149, y8 1.6855, y9 1.8474 11. y1 0.3492, y2 0.7202, y3 1.1363, y4 1.6233, y5 2.2118, y6 2.9386, y7 3.8490 13. (c) y0 2.2755, y1 2.0755, y2 1.8589, y3 1.6126, y4 1.3275 CHAPTER 9 IN REVIEW (PAGE 384) 1. Comparison of numerical methods with h 0.1: Comparison of numerical methods with h 0.05: 3. Comparison of numerical methods with h 0.1: Improved xn Euler Euler RK4 0.60 0.6000 0.6048 0.6049 0.70 0.7095 0.7191 0.7194 0.80 0.8283 0.8427 0.8431 0.90 0.9559 0.9752 0.9757 1.00 1.0921 1.1163 1.1169 Improved xn Euler Euler RK4 1.10 2.1469 2.1554 2.1556 1.20 2.3272 2.3450 2.3454 1.30 2.5409 2.5689 2.5695 1.40 2.7883 2.8269 2.8278 1.50 3.0690 3.1187 3.1197 Improved xn Euler Euler RK4 1.10 2.1386 2.1549 2.1556 1.20 2.3097 2.3439 2.3454 1.30 2.5136 2.5672 2.5695 1.40 2.7504 2.8246 2.8278 1.50 3.0201 3.1157 3.1197 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • CHAPTER 9 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Comparison of numerical methods with h 0.05: 5. h 0.2: y(0.2) 3.2; h 0.1: y(0.2) 3.23 7. x(0.2) 1.62, y(0.2) 1.84 EXERCISES FOR APPENDIX I (PAGE APP-2) 1. (a) 24 (b) 720 (c) (d) 3. 0.297 EXERCISES FOR APPENDIX II (PAGE APP-18) 1. (a) (b) (c) 3. (a) (b) (c) (d) 5. (a) (b) (c) (d) 7. (a) 180 (b) (c) 9. (a) (b) 11. 13. 15. singular 17. nonsingular; 19. nonsingular; A1 1 2 0 2 4 1 2 3 1 2 5 A1 1 4 5 3 8 4 38 2 14 1 7 10 38 75 7 10 38 75 6 12 5 4 8 10 8 16 20 10 20 25 4 8 5 10 0 0 0 0 3 6 8 16 9 3 24 8 19 3 6 22 19 30 18 31 32 4 27 1 11 17 6 22 2 12 28 12 6 14 1 19 2 2 11 1 81 15 41 3 Improved xn Euler Euler RK4 0.60 0.6024 0.6049 0.6049 0.70 0.7144 0.7193 0.7194 0.80 0.8356 0.8430 0.8431 0.90 0.9657 0.9755 0.9757 1.00 1.1044 1.1168 1.1169 21. nonsingular; 23. 25. 27. 29. (a) (b) (c) 31. x 3, y 1, z 5 33. x 2 4t, y 5 t, z t 35. 37. 41. 43. 45. 47. 49. 51. 53. 55. K1 1 3i 5 , K2 1 3i 5 1 3i, 2 3i, K1 2 1 0, K2 0 0 1 1 2 3 2, K1 9 45 25, K2 1 1 1, K3 1 9 1 1 0, 2 4, 3 4, 1 2 4, K1 1 4 1 6, 2 1, K1 2 7, K2 1 1 A1 1 2 1 0 1 2 2 3 1 3 1 3 1 1 6 1 3 1 3 1 2 7 6 4 3 1 3 1 2 A1 5 2 1 6 2 1 3 1 1 A1 0 0 1 3 2 3 1 3 2 3 1 3 2 3 0 x1 1, x2 0, x3 2, x4 0 x 1 2, y 3 2, z 7 2 1 4e4t 1 4 t2 (1/) sin t t3 t 1 4e8 1 4 4 0 6 4e4t 2 sin t 6t dX dt 4 1 1e2t 12 2 1e3t dX dt 5et 2et 7et A1(t) 1 2e3t 3e4t 4et e4t 2et A1 1 9 2 13 8 2 5 1 1 7 5 ANS-18 ● ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS • APPENDIXS Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. I-1 A Absolute convergence of a power series, 232 Absolute error, 78 Acceleration due to gravity, 26, 193 Adams-Bashforth-Moulton method, 373 Adams-Bashforth predictor, 373 Adams-Moulton corrector, 373 Adaptive numerical method, 371 Addition: of matrices, APP-4 of power series, 234 Aging spring, 197, 261, 268 Agnew, Ralph Palmer, 33, 137 Air resistance: proportional to square of velocity, 30, 45, 102 proportional to velocity, 26, 45, 92 Airy, George Biddel, 243 Airy’s differential equation: deﬁnition of, 197, 243 numerical solution curves, 246 power series solutions, 241–243 solution in terms of Bessel functions, 261, 268 Algebra of matrices, APP-3 Algebraic equations, methods for solving, APP-10 Alternative form of second translation theorem, 295 Ambient temperature, 22 Amperes (A), 25 Amplitude: damped, 200 of free vibrations, 195 Analytic at a point, 233 Annihilator approach to method of undetermined coefﬁcients, 149 Annihilator differential operator, 149 Approaches to the study of differential equations: analytical, 27 numerical, 27 qualitative, 27 Archimedes’ principle, 30, 102 Arithmetic of power series, 234 Associated homogeneous differential equation, 119 Associated homogeneous system, 331, 348 Asymptotically stable critical point, 42 Attractor, 42, 336 Augmented matrix: deﬁnition of, APP-10 elementary row operations on, APP-10 in reduced row-echelon form, APP-11 in row-echelon form, APP-10 Autonomous differential equation: ﬁrst-order, 38 second-order, 188 translation property of, 42 Auxiliary equation: for Cauchy-Euler equations, 163 for linear equations with constant coefﬁcients, 133 roots of, 133, 163–165 Axis of symmetry, 210 B Backward difference, 381 Ballistic pendulum, 226 Beams: cantilever, 211 deﬂection curve of, 210 embedded, 211 free, 211 simply supported, 211 static deﬂection of, 296 supported on an elastic foundation, 322 Beats, 208 Bernoulli’s differential equation, 73 Bessel, Friedrich Wilhelm, 257 Bessel functions: aging spring and, 261, 268 differential equations solvable in terms of, 259–261 differential recurrence relations for, 262–263 of the ﬁrst kind, 258 graphs of, 259, 260, 264 of half-integral order, 263–264 modiﬁed of the ﬁrst kind, 260 modiﬁed of the second kind, 260 numerical values of, 262 of order n, 258 of order , 264 of order , 264 properties of, 262 recurrence relation for, 268 of the second kind, 258, 259 spherical, 264 zeros of, 262 Bessel’s differential equation: general solution of, 259 modiﬁed of order n, 260 of order n, 257 parametric of order n, 259–260 solution of, 257 1 2 1 2 Boundary conditions: deﬁnition of, 17, 118 periodic, 217 Boundary-value problem: deﬁnition of, 17, 118 numerical methods for ODEs, 381, 383 for an ordinary differential equations, 17, 118 shooting method for, 383 Branch point, 110 Buckling modes, 214 Buckling of a tapered column, 256 Buckling of a thin vertical column, 269 Buoyant force, 30 BVP, 17, 118 C Calculation of order hn, 364 Cantilever beam, 211 Capacitance, 25 Carbon dating, 85 Carrying capacity, 95 Catenary, 221 Cauchy, Augustin-Louis, 163 Cauchy-Euler differential equation: auxiliary equation for, 163 deﬁnition of, 162–163 general solution of, 163, 164, 165 method of solution for, 163 reduction to constant coefﬁcients, 167 Center of a power series, 232 Central difference, 381 Central difference approximations, 381 Chain pulled up by a constant force, 223 Change of scale theorem, 281 Characteristic equation of a matrix, 334, APP-15 Characteristic values, APP-14 Characteristic vectors, APP-14 Chebyshev, Pafnuty, 270 Chebyshev’s differential equation, 270 Chemical reactions: ﬁrst-order, 23 second-order, 23, 46, 98–99 Circuits, differential equations of, 25, 88–89 Circular frequency, 194 Clamped end of a beam, 211 Classiﬁcation of ordinary differential equations: by linearity, 4 by order, 3 by type, 2 Index INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. I-2 ● INDEX Closed form solution, 9 Clepsydra, 105 Coefﬁcient matrix, 326–327 Cofactor, APP-6 Column bending under its own weight, 268–269 Column matrix, 327, APP-1 Competition models, 109–110 Competition term, 96 Competitive interactions, 96, 109, 414 Complementary error function, 59 Complementary function: for a homogeneous linear differential equation, 125 for a homogeneous linear system, 331, 348 Concentration of a nutrient in a cell, 112 Continuing method, 373 Continuous compound interest, 22, 90 Convergent improper integral, 274 Convergent power series, 232 Convolution of two functions, 302 Convolution theorem, inverse form of, 304 Convolution theorem, Laplace transform, 303 Cooling/Warming, Newton’s Law of, 22–23, 86–87, 91 Coulomb, Charles Augustin de, 323 Coulomb friction, 230, 323 Coulombs (C), 25 Coupled pendulums, 323 Coupled springs, 315–316 Cover-up method, 288 Criterion for an exact differential, 64 Critical loads, 213–214 Critical point of an autonomous ﬁrst-order differential equation: asymptotically stable, 42 deﬁnition of, 38 isolated, 45 semi-stable, 42 unstable, 42 Critical speeds, 216–217 Critically damped series circuit, 203 Critically damped spring/mass system, 198 Curvature, 189, 210 Cycloid, 114 D Damped amplitude, 200 Damped motion, 197 Damped nonlinear pendulum, 225 Damping constant, 197 Damping factor, 198 Daphnia, 96 DE, 2 Dead sea scrolls, 86 Dead zone, 323 Decay, radioactive, 22, 85–86, 115 Decay constant, 85 Deﬁnition, interval of, 5 Deﬂection of a beam, 210–211, 296 Deﬂection curve, 210 Density-dependent hypothesis, 95 Derivative notation, 3 Derivatives of a Laplace transform, 301 Determinant of a square matrix: deﬁnition of, APP-6 expansion by cofactors, APP-6 Diagonal matrix, 357, APP-20 Difference equation replacement for an ODE, 381 Difference quotients, 381 Differences, ﬁnite, 381 Differential, exact, 64 Differential equation: autonomous, 38 Bernoulli, 73 Bessel, 257 Cauchy-Euler, 162 Chebyshev, 270 deﬁnition of, 2 exact, 64 families of solutions for, 7–8 ﬁrst order, 3, 35 Hermite, 270 homogeneous linear, 119 with homogeneous coefﬁcients, 71 Laguerre, 311 Legendre, 257 linear, 4, 54 modiﬁed Bessel, 260 nonautonomous, 38 nonhomogeneous linear, 119 nonlinear, 4 normal form of, 4 notation for, 3 order of, 3 ordinary, 2 parametric Bessel, 260 parametric modiﬁed Bessel, 260 partial, 2 Riccati, 75 separable, 46 solution of, 5–6, 8 standard form of, 54 systems of, 9, 106, 180, 365, 375–377, 385 type, 2 Differential equations as mathematical models, 20–21 Differential equations solvable in terms of Bessel functions, 259–261 Differential form of a ﬁrst-order equation, 3, 64 Differential of a function of two variables, 63 Differential operator, 120 Differential recurrence relation, 262–263 Differentiation notation, 3 Differentiation of a power series, 233 Dirac delta function: deﬁnition of, 313 Laplace transform of, 313 Direction ﬁeld of a ﬁrst-order differential equation: for an autonomous ﬁrst-order differential equation, 42 deﬁnition of, 36 method of isoclines for, 38, 44 nullclines for, 44 Discontinuous coefﬁcients, 59 Discretization error, 364 Distributions, theory of, 314 Divergent improper integral, 274 Divergent power series, 232 Domain: of a function, 6 of a solution, 6 Doomsday equation, 103 Dot notation, 3 Double cosine series, 490 Double eigenvalues, 496 Double pendulum, 318 Double spring systems, 206, 315–316, 319 Draining of a tank, 24, 101 Driven motion, 200 Driving function, 61, 193 Drosophila, 96 Dufﬁng’s differential equation, 224 Dynamical system, 28 E Effective spring constant, 206 Eigenfunctions of a boundary-value problem, 192, 213 Eigenvalues of a boundary-value problem, 192, 213 Eigenvalues of a matrix: complex, 342–344 deﬁnition of, 334, APP-14 distinct real, 334 of multiplicity m, 338 of multiplicity three, 340 of multiplicity two, 338, APP-17 repeated, 337 Eigenvectors of a matrix, 334, APP-14 Elastic curve, 210 Electrical networks, 110, 317 Electrical series circuits, analogy with spring/mass systems, 203 Electrical vibrations: forced, 204 free, 203 Elementary functions, 10 Elementary row operations: deﬁnition of, APP-10 notation for, APP-11 Elimination methods: for systems of algebraic equations, APP-10 for systems of ordinary differential equations, 180 Embedded end of a beam, 211 Emigration model, 98 INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. INDEX ● I-3 Environmental carrying capacity, 75 Equality of matrices, APP-3 Equation of motion, 194 Equilibrium point, 38 Equilibrium position, 193, 196 Equilibrium solution, 38 Error: absolute, 78 analysis, 363 discretization, 364 formula, 364 global truncation, 365 local truncation, 364, 366–367 percentage relative, 78 relative, 78 round off, 363–364 Error function, 59 Escape velocity, 225 Euler, Leonhard, 163 Euler load, 214 Euler’s constant, 262, 311 Euler’s formula, 133 Euler’s method: for ﬁrst-order differential equations, 76–77, 363 improved, 365 for second-order differential equations, 376 for systems, 379 Evaporating raindrop, 93 Evaporation, 102 Exact differential: criterion for, 64 deﬁnition of, 64 Exact differential equation: deﬁnition of, 64 method of solution, 65 Excitation function, 127 Existence, interval of, 5 Existence and uniqueness of a solution, 15–16, 117, 328 Explicit solution, 6 Exponential growth and decay, 84 Exponential matrix: computation of, 358 deﬁnition of, 356 derivative of, 357 Exponential order, 277 Exponents of a singularity, 251 Extreme displacement, 194 F Factorial function, APP-1 Falling body, 25, 26, 30 Falling chain, 70, 75 Falling raindrop, 33, 93, 105 Family of solutions, 7 Farads (f), 25 Fick’s law, 114 Finite difference approximations, 380–381 Finite difference equation, 381 Finite differences: backward, 381 central, 381 deﬁnition of, 381 forward, 381 First buckling mode, 214 First translation theorem: form of, 290 inverse form of, 290 First-order chemical reaction, 23, 84 First-order differential equations: applications of, 22–25, 83–84, 95 methods for solving, 46, 54, 63, 71 First-order initial-value problem, 13–14 First-order Runge-Kutta method, 368 First-order system of differential equations deﬁnition of, 326 linear system, 326 Flexural rigidity, 210 Folia of Descartes, 12 Forced electrical vibrations, 203–204 Forced motion of a spring/mass system, 200, 202 Forcing function, 127, 169, 193 Forgetfulness, 32 Formula error, 364 Forward difference, 381 Fourth-order Runge-Kutta method: for ﬁrst-order differential equations, 78, 369 for second-order differential equations, 376 for systems of ﬁrst-order equations, 378 truncation errors for, 370 Free electrical vibrations, 203 Free motion of a spring/mass system: damped, 197 undamped, 193–194 Freely falling body, 25 Frequency: circular, 194 of simple harmonic motion, 194 natural, 194 Frequency response curve, 209 Fresnel sine integral, 63 Frobenius, Ferdinand Georg, 249 Frobenius, method of, 250 Frobenius’ theorem, 249 Fulcrum supported ends of a beam, 211 Full-wave rectiﬁcation of sine function, 310 Functions deﬁned by integrals, 59–60 Fundamental matrix, 351, 357–358 Fundamental set of solutions: existence of, 123 of a linear differential equation, 123 of a linear system, 330 G g (acceleration due to gravity), 25, 193 Galileo Galilei, 26 Gamma function, 258, 280, APP-1 Gauss’ hypergeometric function, 257 Gaussian elimination, 383, APP-10 Gauss-Jordan elimination, 337, 338, APP-10 General form of a differential equation, 3 General solution: of Bessel’s differential equation, 259, 260 of a Cauchy-Euler differential equation, 163–165 of a differential equation, 10, 123, 125 of a homogeneous linear differential equation, 123 of a nonhomogeneous linear differential equation, 125 of a homogeneous system of linear differential equations, 330, 334 of a linear ﬁrst-order differential equation, 57 of the modiﬁed Bessel’s differential equation, 260 of a nonhomogeneous system of linear differential equations, 331, 348 Generalized factorial function, APP-1 Generalized functions, 314 Global truncation error, 365 Gompertz, Benjamin, 98 Gompertz differential equation, 98 Gospel of Judas, 86 Green’s function: for a boundary-value problem, 176–177 for an initial-value problem, 170 relationship to Laplace transform, 306–307 for a second-order differential operator, 170 Growth and decay, 84 Growth constant, 85 H Half-life: of carbon-14, 86 deﬁnition of, 85 of plutonium-239, 85 of potassium-40, 115 of radium-226, 85 of uranium-238, 85 Half-wave rectiﬁcation of sine function, 310 Hard spring, 219 Harvesting of a ﬁshery, model of, 98, 100 Heart pacemaker, model for, 63, 94 Heaviside, Oliver, 293 Heaviside function, 293 Henries (h), 25 Hermite, Charles, 270 Hermite polynomials, 270 Hermite’s differential equation, 270 Higher-order differential equations, 118, 135, 192 Hinged ends of a beam, 211 Hole through the Earth, 31 INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. I-4 ● INDEX Homogeneous differential equation: linear, 60, 119 with homogeneous coefﬁcients, 71 Homogeneous function of degree a, 71 Homogeneous systems: of algebraic equations, APP-15 of linear ﬁrst-order differential equations, 326 Hooke’s law, 31, 193 I IC, 13 Identity matrix, APP-6 Identity property of power series, 233 Immigration model, 98, 103 Impedance, 204 Implicit solution of an ODE, 6 Improved Euler method, 365–366 Impulse response, 314 Indicial equation, 251 Indicial roots, 251 Inductance, 25 Inﬂection, points of, 45, 97 Inhibition term, 96 Initial condition(s): for an ordinary differential equation, 13, 117 for a system of linear ﬁrst-order differential equations, 328 Initial-value problem: deﬁnition of, 13, 117 ﬁrst-order, 13, 362 geometric interpretation of, 14 for a linear system, 328 nth-order, 13, 117 second-order, 14, 375 Input, 61, 127, 169, 193 Integral curve, 7 Integral of a differential equation, 7 Integral equation, 305 Integral, Laplace transform of an, 304 Integral transform: deﬁnition of, 274 inverse of, 281 kernel of, 274 Laplace, 274 Integrating factor(s): for a linear ﬁrst-order differential equation, 55 for a nonexact ﬁrst-order differential equation, 67–68 Integration of a power series, 233 Integrodifferential equation, 305 Interactions, number of, 23 Interest compounded continuously, 90 Interior mesh points, 381 Interpolating function, 372 Interval: of convergence, 232 of deﬁnition, 5 of existence, 5 of existence and uniqueness, 16 of validity, 5 Inverse Laplace transform: deﬁnition of, 281 linearity of, 282 Inverse matrix: deﬁnition of, APP-7 by elementary row operations, APP-13 formula for, APP-8 Irregular singular point, 248 Isoclines, 38, 44 Isolated critical point, 45 IVP, 13 K Kernel of an integral transform, 274 Kinetic friction, 230 Kirchhoff’s ﬁrst law, 110 Kirchhoff’s second law, 25, 110 L Laguerre polynomials, 311 Laguerre’s differential equation, 311 Laplace, Pierre-Simon Marquis de, 274 Laplace transform: behavior as s : , 279 change of scale theorem for, 281 convolution theorem for, 303 deﬁnition of, 274 of a derivative, 284 derivatives of, 301 of Dirac delta function, 313 existence, sufﬁcient conditions for, 277–278 of an integral, 304 inverse of, 281 of a linear initial-value problem, 284–285 linearity of, 276 of a periodic function, 307 of systems of linear differential equations, 315 tables of, 277, APP-21 translation theorems for, 290, 294 of unit step function, 294 Lascaux cave paintings, dating of, 90 Law of mass action, 98 Leaking tanks, 24, 29–30, 101, 105 Least-squares line, 103 Legendre, Adrien-Marie, 257 Legendre function, 267 Legendre polynomials: ﬁrst six, 266 graphs of, 266 properties of, 266 recurrence relation for, 266 Rodrigues’ formula for, 267 Legendre’s differential equation: of order n, 257 solution of, 265–266 Leibniz notation, 3 Leibniz’s formula for differentiation of an integral, 172 Level curves, 49 Level of resolution of a mathematical model, 21 Libby, Willard, 85 Lineal element, 36 Linear dependence: of functions, 121 of solution vectors, 329 Linear differential operator, 120, 149–151, 170 Linear independence: of eigenvectors, APP-16 of functions, 121 of solutions, 121 of solution vectors, 329 and the Wronskian, 122, 329–330 Linear operator, 120 Linear ordinary differential equation: applications of, 84, 193, 210 associated homogeneous equation, 119 auxiliary equation for, 133 boundary-value problem for, 117 complementary function for, 125 deﬁnition of, 4 ﬁrst order, 54 fundamental set of solutions for, 123 general solution of, 57, 123, 125 homogeneous, 60, 119 initial-value problem for, 117 nonhomogeneous, 60, 119 particular solution of, 124 solution of, 56, 112–113, 139, 149, 157–158, 162–165, 241, 249 standard forms for, 54, 130, 157, 158, 160 superposition principles for, 120, 126 Linear regression, 103 Linear second-order boundary-value problem, 381 Linear spring, 218 Linear system, 127, 326 Linear systems of algebraic equations, APP-10 Linear systems of differential equations: deﬁnition of, 106, 326 homogeneous, 326, 333 matrix form of, 326 method for solving, 333, 348 nonhomogeneous, 326, 348 Linear transform, 276 Linearity property: of differentiation, 274 of integration, 274 of the inverse Laplace transform, 282 of the Laplace transform, 276 Linearization: of a differential equation, 220 of a function of one variable at a point, 76 INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. INDEX ● I-5 Lissajous curve, 320 Local truncation error, 364, 366, 370 Logistic curve, 96 Logistic differential equation, 75, 96 Logistic function, 96 Losing a solution, 48 Lotka-Volterra, equations of: competition model, 109 predator-prey model, 108 LR-series circuit, differential equation of, 30, 88 LRC-series circuit, differential equation of, 25, 203 M Malthus, Thomas, 21 Maple, 60, 384 Mass action, law of, 98 Mathematica, 60, 136–137, 337, 360, 384 Mathematical model(s): absolute temperature of a cooling body, 114 aging spring, 197 ballistic pendulum, 226 bobbing motion of a ﬂoating barrel, 30 box sliding down an inclined plane, 94–95 buckling of a thin vertical column, 213, 216 cables of a suspension bridge, 26–27 carbon dating, 85 chain pulled upward by a constant force, 223 chemical reactions, 23, 98, 101 competition models, 109–110 concentration of a nutrient in a cell, 114 constant harvest, 93, 98 continuous compound interest, 90 cooling cup of coffee, 91 cooling/warming, 22, 86 coupled pendulums, 318, 322–323 coupled springs, 316 deﬁnition of, 20–21 deﬂection of beams, 210–211 doomsday for a population, 103 double pendulum, 318 double spring, 206, 229, 230 draining a tank, 24, 29 dropping supplies from a plane, 226–225 drug infusion, 32 evaporating raindrop, 93 evaporation, 102 extinction of a population, 102 falling body (with air resistance), 26 falling body (with no air resistance), 25–26 forgetfulness, 32 ﬂuctuating population, 32, 93 ﬂuid ﬂow around a circular cylinder, 388 growth and decay, 84 hard spring, 219 harvesting ﬁsheries, 98 heart pacemaker, 94 hole through the Earth, 31 immigration, 98, 103 leaking tanks, 101 learning theory, 32 least time, 114 LR-series circuit, 30, 88, 92 LRC-series circuit, 25, 203 memorization, 94 mixtures, 24, 87, 107 networks, 110, 354–355 nutrient ﬂow through a membrane, 112 pendulum motion on the Earth, 220, 410 pendulum motion on the Moon, 227 population growth, 21 potassium-40 decay, 115 predator-prey, 108, 412–413 pursuit curves, 225 radioactive decay, 22 radioactive decay series, 106 raindrops, 33, 93, 105 range of a projectile, 323, 324 RC-series circuit, 30, 89, 92 reﬂecting surface, 32 resonance, 202 restocking ﬁsheries, 98 rocket motion, 31, 222 rotating ﬂuid, 32–33 rotating rod containing a sliding bead, 229–230 rotating string, 214 skydiving, 30, 93, 104 soft spring, 219, 406 solar collector, 102 spread of a disease, 23 spring/mass systems, 193–203, 316, 319 suspended cables, 26–27 snowplow problem, 33 swimming a river, 104, 105 temperature in a circular ring, 217 temperature in a sphere, 217 temperature in a thin rod, 217 terminal velocity, 45, 92 time of death, 91 tractrix, 32 tsunami, shape of, 102 U.S. population, 100 variable mass, 31, 222–223 water clock, 105 wire hanging under its own weight, 221 Mathieu functions, 257 Matrices: addition of, APP-4 associative law of, APP-6 augmented, APP-10 characteristic equation of, 334, APP-15 column, APP-3 deﬁnition of, APP-3 derivative of, APP-9 determinant of, APP-6 diagonal, APP-20 difference of, APP-4 distributive law for, APP-6 eigenvalue of, 334, APP-14 eigenvector of, 334, APP-14 elementary row operations on, APP-10 entry of, APP-3 equality of, APP-3 exponential, 356 fundamental, 351 integral of, APP-9 inverse of, APP-8, APP-13 multiples of, APP-3 multiplication of, APP-5 multiplicative identity, APP-6 multiplicative inverse, APP-7 nilpotent, 360 nonsingular, APP-7 product of, APP-5 reduced row-echelon form of, APP-11 row-echelon form of, APP-10 singular, APP-7 size, APP-3 square, APP-3 sum of, APP-4 symmetric, 339 transpose of, APP-7 vector, APP-3 zero, APP-6 Matrix. See Matrices. Matrix exponential: computation of, 356, 358 deﬁnition of, 356 derivative of, 357 as a fundamental matrix, 357–358 Matrix form of a linear system, 326–327 Meander function, 310 Memorization, mathematical model for, 32 Method of Frobenius, 249–250 Method of isoclines, 38 Method of undetermined coefﬁcients, 140, 151 Minor, APP-8 Mixtures: multiple tanks, 107, 111 single tank, 24, 87–88 Modeling process, steps in, 21 Modiﬁed Bessel equation: of order n, 260 general solution of, 260 parametric form of, 260 Modiﬁed Bessel functions: of the ﬁrst kind, 260 graphs of, 260 of the second kind, 260 Movie, 320 Multiplication: of matrices, APP-5 of power series, 234–235 Multiplicative identity, APP-6 Multiplicative inverse, APP-7 INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. I-6 ● INDEX Multiplicity of eigenvalues, 338, 340, APP-17 Multistep numerical method: advantages of, 374–375 deﬁnition of, 373 disadvantages of, 374–375 N Named functions, 257 Natural frequency of free undamped motion, 194 Networks, 110 Newton, Isaac, 25 Newton’s dot notation for differentiation, 3 Newton’s ﬁrst law of motion, 25 Newton’s law of cooling/warming: with constant ambient temperature, 22–23, 86–87, 91 with variable ambient temperature, 29, 91 Newton’s second law of motion, 25, 222 Newton’s second law of motion as the rate of change of momentum, 31, 222 Newton’s universal law of gravitation, 31 Nilpotent matrix, 360 Nonelementary integral, 51, 59 Nonhomogeneous linear differential equation, 60, 119 Nonhomogeneous systems of linear ﬁrst-order differential equations: deﬁnition of, 326 general solution of, 330, 331, 333–344 particular solution of, 331, 348–352 Nonlinear damping, 218–219 Nonlinear ordinary differential equation: deﬁnition of, 4 solvable by ﬁrst-order methods, 186 Taylor series solution of, 187 Nonlinear pendulum, 220, 410 Nonlinear spring: deﬁnition of, 218 hard, 218–219 soft, 218–219 Nonlinear system of differential equations, 106 Nonsingular matrix, APP-7 Normal form: of a linear system, 326 of an ordinary differential equation, 3 of a system of ﬁrst-order equations, 326 Notation for derivatives, 3 n-parameter family of solutions, 7 nth-order differential operator, 120 nth-order initial-value problem, 13 Nullcline, 44 Numerical methods: Adams-Bashforth-Moulton method, 373 adaptive methods, 371 applied to higher-order equations, 188, 375–376 applied to systems, 375–378 continuing, 373 errors in, 364 Euler’s method, 76–77, 363, 379 ﬁnite difference method, 381 improved Euler’s method, 365–366 multistep, 373 predictor-corrector method, 366, 373 RK4 method, 78, 369 RKF45 method, 371 shooting method, 383 single-step, 373 stability of, 374 starting, 373 truncation errors in, 364, 370 Numerical solution curve, 79 Numerical solver, 78–79, 187–188 Nutrient ﬂow through a membrane, 112 O ODE, 2 Ohms ( ), 25 Ohm’s Law, 89 One-dimensional phase portrait, 39 One-parameter family of solutions, 7 Order, exponential, 277 Order of a differential equation, 3 Order of a Runge-Kutta method, 368 Ordinary differential equation, 2 Ordinary point of a linear second-order differential equation: deﬁnition of, 239 solution about, 240 Orthogonal trajectories, 114–115 Output, 61, 127, 169, 193 Overdamped series circuit, 203 Overdamped spring/mass system, 198 P Parametric form of Bessel equation of order n, 259–260 Parametric form of modiﬁed Bessel equation of order n, 260 Partial differential equation, 2 Partial fractions, 283 Partial integral, 124 Particular solution: deﬁnition of, 7 of a linear differential equation, 124 of a system of linear differential equations, 331, 345 PDE, 2 Pendulum: ballistic, 226 double, 318 free damped, 225 linear, 220 nonlinear, 220, 225 period of, 228 physical, 220 simple, 220 spring-coupled, 322–323 Pendulum motion on the Moon, 227 Percentage relative error, 78 Period of simple harmonic motion, 194 Periodic boundary conditions, 217 Periodic function, Laplace transform of, 307 Phase angle, 195–196 Phase line, 39 Phase plane, 327, 335 Phase portrait(s): for ﬁrst-order equations, 39 for systems of two linear ﬁrst-order differential equations, 335–336 Physical pendulum, 220 Piecewise-continuous functions, 277 Pin supported ends of a beam, 211 Points of inﬂection, 45 Polynomial differential operator, 120 Population growth, 21 Population models: birth and death, 93 doomsday, 103 extinction, 103 ﬂuctuating, 93 harvesting, 45, 93, 98, 100 immigration, 98, 103 logistic, 45, 95–97, 100 Malthusian, 21–22 restocking, 98 Potassium-argon dating method, 115 Potassium-40 decay, 115 Power series: absolute convergence of, 232 arithmetic of, 234 center, 232 convergence of, 232 deﬁnes a function, 233 deﬁnition of, 232 differentiation of, 233 divergence of, 232 identity property of, 233 integration of, 233 interval of convergence, 232 Maclaurin, 234 radius of convergence, 232 ratio test for, 233 represents a continuous function, 233 represents an analytic function, 233 review of, 232 solutions of differential equations, 236, 240, 241 Taylor, 234 Power series solutions: existence of, 240 method of ﬁnding, 241 solution curves of, 245–246 Predator-prey model, 108 Predictor-corrector method, 366, 373 Prime notation, 3 Projectile motion, 184 INDEX Copyright 2012 Cengage Learning. 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INDEX ● I-7 Proportional quantities, 22 Pure resonance, 202 Pursuit curve, 225 Q Qualitative analysis of a ﬁrst-order DE, 36–42 Quasi frequency, 200 Quasi period, 200 R Radioactive decay, 22, 84–85, 106, 115 Radioactive decay series, 62, 106 Radius of convergence of a power series, 232 Radium decay, 85 Radon, 85 Raindrop, 33, 105 Rate function, 36 Ratio test, 232 Rational roots of a polynomial equation, 136 RC-series circuit, differential equation of, 30, 89 Reactance, 204 Reactions, chemical, 23, 98 Rectangular pulse, 299 Rectiﬁed sine wave, 299 Recurrence relation, 242 Recurrence relation, differential, 262–263 Reduced row-echelon form of a matrix, APP-11 Reduction of order, 129–131 Reduction to separation of variables, 73 Reﬂecting surface, 32 Regular singular point, 248 Regression line, 103 Relative error, 78 Relative growth rate, 95 Repeller, 42, 366 Resistance: air, 26, 30, 45, 92–93 electrical, 25, 88–89, 203–204 Resonance, pure, 202 Resonance curve, 209 Resonance frequency, 209 Response: impulse, 314 as a solution of a DE, 61, 125, 169, 193, 203 of a system, 28, 88 zero-input, 288 zero-state, 288 Rest solution, 170 Restocking of a ﬁshery, model of, 98 Riccati’s differential equation, 75 RK4 method, 78, 369 RKF45 method, 371 Robins, Benjamin, 226 Rocket motion, 31, 222, 225 Rodrigues’ formula, 267 Rotating ﬂuid, shape of, 32–33 Rotating rod and bead, 229–230 Rotating string, 214, 216 Round-off error, 363–364 Row-echelon form, APP-10 Row operations: elementary, APP-10 symbols for, APP-11 Runge-Kutta-Fehlberg method, 371 Runge-Kutta methods: ﬁrst-order, 368 fourth-order, 78, 369 second-order, 368 for systems, 376, 378 truncation errors for, 370 S Sawtooth function, 310 Schwartz, Laurent, 314 Second-order boundary-value problem, 380–381, 383 Second-order chemical reaction, 23, 98, 101 Second-order homogeneous linear system, 345 Second-order initial-value problem, 14, 375, 380, 383 Second-order ordinary differential equation as a system, 188, 376 Second-order Runge-Kutta method, 368 Second translation theorem: alternative form of, 295 form of, 294 inverse form of, 295 Semi-stable critical point, 42 Separation of variables, method of, 46–47 Series: power, 232 review of, 232–234 solutions of ordinary differential equations, 236, 240, 249 Series circuits, differential equations of, 25, 30, 88–89, 203 Shifting the summation index, 235 Shifting theorems for Laplace transforms, 290, 294 Shooting method, 383 Shroud of Turin, dating of, 90 Sifting property, 314 Signum function, 230 Simple harmonic electrical vibrations, 203 Simple harmonic motion of a spring/mass system, 194 Simple pendulum, 220 Simply supported ends of a beam, 211 Sine integral function, 63 Single-step numerical method: advantages of, 374–375 deﬁnition of, 373 disadvantages of, 374–375 Singular matrix, APP-7 Singular point: at , 239 irregular, 248 of a linear ﬁrst-order differential equation, 57 of a linear second-order differential equation, 239 regular, 248 Singular solution, 8 SIR model, 112 Sky diving, 30, 93 Sliding box on an inclined plane, 94–95 Sliding friction, 94–95, 230 Slope ﬁeld, 36 Slope function, 36 Snowplow problem, 33 Soft spring, 219 Solar collector, 102 Solution curve, 6 Solution of an ordinary differential equation: about an ordinary point, 238 about a singular point, 247 constant, 11, 38 deﬁned by an integral, 50 deﬁnition of, 5 equilibrium, 38 explicit, 6 general, 10, 57, 123, 125 graph of, 6 implicit, 6 integral, 7 interval of deﬁnition for, 5 n-parameter family of, 7 number of, 7 particular, 7 piecewise deﬁned, 8 singular, 8 trivial, 6 Solution of a system of ordinary differential equations: deﬁned, 9, 180 general, 330, 331 particular, 331 Solution vector, 327 Special functions, 59, 61, 257 Speciﬁc growth rate, 95 Spherical Bessel functions: of the ﬁrst kind, 264 of the second kind, 264 Spread of a communicable disease, 23, 97, 112 Spring constant, 193 Spring/mass system: dashpot damping of a, 197 Hooke’s law and, 193 linear models for, 193 nonlinear models for, 218–219 Springs, coupled, 229, 315–316, 319 Square matrix, APP-3 Square wave, 310 Stable critical point, 42 INDEX Copyright 2012 Cengage Learning. 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I-8 ● INDEX Stable numerical method, 374 Staircase function, 299 Standard form of a linear differential equation: ﬁrst order, 54, 157 second order, 130, 158, 160, 238, 239 Starting methods, 373 State of a system, 21, 28, 127 State variables, 28, 127 Stationary point, 38 Steady-state current, 89, 204 Steady-state solution, 204 Steady state term, 89, 201 Stefan’s law of radiation, 114 Step size, 76 Streamlines, 70 Subscript notation, 3 Substitutions in a differential equation, 71, 186 Sum of two matrices, APP-4 Summation index, shifting of, 235 Superposition principle: for homogeneous linear differential equations, 120 for homogeneous linear systems, 328 for nonhomogeneous linear differential equations, 126 Suspended cables, 26 Suspension bridge, 26, 53 Symmetric matrix, 339 Synthetic division, 136 Systematic elimination, 180 Systems of linear differential equations, methods for solving: by Laplace transforms, 315 by matrices, 333, 348 by systematic elimination, 180 Systems of linear ﬁrst-order differential equations: complementary function for, 331, 348 deﬁnition of, 9, 106, 326 existence of a unique solution for, 328 fundamental set of solutions for, 330 general solution of, 330, 331, 334 homogeneous, 326 initial-value problem for, 328 matrix form of, 326–327 nonhomogeneous, 326 normal form of, 326 particular solution for, 331, 348, 352 solution of, 327, 331, 33–334, 338, 342, 344, 348–352 superposition principle for, 328 undetermined coefﬁcients for, 348–349 variation of parameters for, 351–352 Wronskian for, 329–330 Systems of ordinary differential equations, 9, 106, 180, 187, 315, 325 Systems reduced to ﬁrst-order systems, 377 T Table of Laplace transforms, APP-21 Tangent lines, use of, 76–77 Taylor polynomial, 188, 369 Taylor series, use of, 187 Telephone wires, shape of, 217 Temperature: in a circular ring, 217 in a sphere, 217 Terminal velocity of a falling body, 45, 92, 93, 102 Theory of distributions, 314 Three-term recurrence relation, 244 Time of death, 91 Torricelli’s law, 24 Tractrix, 32, 113 Trajectories: orthogonal, 114 parametric equations of, 327, 335 Transfer function, 288 Transform of a derivative, 284 Transient solution, 204 Transient term, 59, 89, 201, 204 Translation property of an autonomous DE, 42 Translation theorems for Laplace transform: ﬁrst, 290 second, 294, 295 inverse forms of, 290, 295 Transpose of a matrix, APP-7 Triangular wave, 310 Trivial solution, 6 Truncation error: for Euler’s method, 364 global, 364 for Improved Euler’s method, 364–365 local, 364 for RK4 method, 370 Tsunami, model for, 102 Two-dimensional phase portrait, 335–336 U Undamped spring/mass system, 193–194 Underdamped series circuit, 203 Underdamped spring/mass system, 198 Undetermined coefﬁcients for linear DEs: annihilator approach, 149–155 superposition approach, 139–146 Undetermined coefﬁcients for linear systems, 348 Uniqueness theorems, 16, 117, 328 Unit impulse, 312 Unit step function: deﬁnition of, 293 Laplace transform of, 294 Universal law of gravitation, 31 Unstable critical point, 42 Unstable numerical method, 374 Unsymmetrical vibrations of a spring, 219 V Variable mass, 222 Variable spring constant, 197 Variables, separable, 46 Variation of parameters: for linear ﬁrst-order differential equations, 157 for linear higher-order differential equations, 158–159, 161 for systems of linear ﬁrst-order differential equations, 348, 351–352 Vectors, deﬁnition of, APP-3 Vectors, as solutions of systems of linear differential equations, 327 Velocity of a falling raindrop, 105 Verhulst, P. F., 96 Vibrations, spring/mass systems, 193, 197, 200 Virga, 33 Viscous damping, 26 Voltage drops, 25 Volterra integral equation, 305 W Water clock, 105 Weight, 26 Weight function of a linear system, 314 Weighted average, 368 Wire hanging under its own weight, 221 Wronskian determinant: for a set of functions, 122 for a set of solutions of a homogeneous linear differential equation, 122 for a set of solution vectors of a homogeneous linear system, 329–330 Y Young’s modulus of elasticity, 210 Z Zero-input response, 288 Zero matrix, APP-6 Zero-state response, 288 Zeros of Bessel functions, 262 INDEX Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. TABLE OF LAPLACE TRANSFORMS f (t) 1. 1 2. t 3. tn n a positive integer 4. t1/2 5. t1/2 6. ta 7. sin kt 8. cos kt 9. sin2 kt 10. cos2 kt 11. eat 12. sinh kt 13. cosh kt 14. sinh2kt 15. cosh2kt 16. teat 17. tn eat n a positive integer 18. eat sin kt 19. eat cos kt s a (s a)2 k2 k (s a)2 k2 n! (s a)n1, 1 (s a)2 s2 2k2 s(s2 4k2) 2k2 s(s2 4k2) s s2 k2 k s2 k2 1 s a s2 2k2 s(s2 4k2) 2k2 s(s2 4k2) s s2 k2 k s2 k2 ( 1) s1 , a 1 1 2s3/2 B s n! sn1, 1 s2 1 s { f (t)} F(s) f (t) 20. eat sinh kt 21. eat cosh kt 22. t sin kt 23. t cos kt 24. sin kt kt cos kt 25. sin kt kt cos kt 26. t sinh kt 27. t cosh kt 28. 29. 30. 1 cos kt 31. kt sin kt 32. 33. 34. sin kt sinh kt 35. sin kt cosh kt 36. cos kt sinh kt 37. cos kt cosh kt 38. J0(kt) 1 1s2 k2 s3 s4 4k4 k(s2 2k2) s4 4k4 k(s2 2k2) s4 4k4 2k2s s4 4k4 s (s2 a2)(s2 b2) cos bt cos at a2 b2 1 (s2 a2)(s2 b2) a sin bt b sin at ab (a2 b2) k3 s2(s2 k2) k2 s(s2 k2) s (s a)(s b) aeat bebt a b 1 (s a)(s b) eat ebt a b s2 k2 (s2 k2)2 2ks (s2 k2)2 2k3 (s2 k2)2 2ks2 (s2 k2)2 s2 k2 (s2 k2)2 2ks (s2 k2)2 s a (s a)2 k2 k (s a)2 k2 { f (t)} F(s) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. f(t) 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. eat f(t) F(s a) 51. 52. easF(s) 53. eas 54. f (n)(t) 55. tn f(t) 56. F(s)G(s) 57. d(t) 1 58. d(t t0) est0 t 0 f ( )g(t ) d (1)n dn dsn F(s) snF(s) s(n1) f (0) f (n1)(0) { g(t a)} g(t)(t a) f (t a)(t a) eas s (t a) erfc a 2 1t bea1s s(1s b) eabeb2terfc b 1t a 2 1t ea1s 1s(1s b) eabeb2t erfc b1t a 2 1t ea1s s1s 2 B t ea2/4t a erfc a 21t ea1s s erfc a 21t ea1s a 21t3 ea2/4t ea 1s 1s 1 1t ea2/4t 1 2 arctan a b s 1 2 arctan a b s sin at cos bt t arctan a s sin at t ln s2 k2 s2 2(1 cosh kt) t ln s2 k2 s2 2(1 cos kt) t ln s a s b ebt eat t { f (t)} F(s) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14547	https://isabelle.in.tum.de/doc/tutorial.pdf	Tobias Nipkow Lawrence C. Paulson Markus Wenzel λ → ∀ = Isabelle β α HOL A Proof Assistant for Higher-Order Logic March 13, 2025 Springer-Verlag Berlin Heidelberg NewYork London Paris Tokyo Hong Kong Barcelona Budapest Preface This volume is a self-contained introduction to interactive proof in higher-order logic (HOL), using the proof assistant Isabelle. It is written for potential users rather than for our colleagues in the research world. The book has three parts. – The first part, Elementary Techniques, shows how to model functional programs in higher-order logic. Early examples involve lists and the natural numbers. Most proofs are two steps long, consisting of induction on a chosen variable followed by the auto tactic. But even this elementary part covers such advanced topics as nested and mutual recursion. – The second part, Logic and Sets, presents a collection of lower-level tactics that you can use to apply rules selectively. It also describes Isa-belle/HOL’s treatment of sets, functions and relations and explains how to define sets inductively. One of the examples concerns the theory of model checking, and another is drawn from a classic textbook on formal languages. – The third part, Advanced Material, describes a variety of other topics. Among these are the real numbers, records and overloading. Advanced techniques for induction and recursion are described. A whole chapter is devoted to an extended example: the verification of a security protocol. The typesetting relies on Wenzel’s theory presentation tools. An anno-tated source file is run, typesetting the theory in the form of a L AT EX source file. This book is derived almost entirely from output generated in this way. The final chapter of Part I explains how users may produce their own formal documents in a similar fashion. Isabelle’s web site1 contains links to the download area and to documenta-tion and other information. The classic Isabelle user interface is Proof Gen-eral / Emacs by David Aspinall’s. This book says very little about Proof General, which has its own documentation. This tutorial owes a lot to the constant discussions with and the valuable feedback from the Isabelle group at Munich: Stefan Berghofer, Olaf Müller, Wolfgang Naraschewski, David von Oheimb, Leonor Prensa Nieto, Cornelia Pusch, Norbert Schirmer and Martin Strecker. Stephan Merz was also kind 1 iv Preface enough to read and comment on a draft version. We received comments from Stefano Bistarelli, Gergely Buday, John Matthews and Tanja Vos. The research has been funded by many sources, including the dfg grants NI 491/2, NI 491/3, NI 491/4, NI 491/6, bmbf project Verisoft, the epsrc grants GR/K57381, GR/K77051, GR/M75440, GR/R01156/01 GR/S57198/01 and by the esprit working groups 21900 and IST-1999-29001 (the Types project). Table of Contents Part I. Elementary Techniques 1. The Basics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Theories. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Types, Terms and Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Interaction and Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.6 Getting Started . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2. Functional Programming in HOL . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.1 An Introductory Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3 An Introductory Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.4 Some Helpful Commands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.5 Datatypes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.5.1 Lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.5.2 The General Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.5.3 Primitive Recursion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.5.4 Case Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.5.5 Structural Induction and Case Distinction . . . . . . . . . . . 19 2.5.6 Case Study: Boolean Expressions . . . . . . . . . . . . . . . . . . . 19 2.6 Some Basic Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.6.1 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.6.2 Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.6.3 Datatype option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.7 Definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.7.1 Type Synonyms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.7.2 Constant Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.8 The Definitional Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3. More Functional Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.1 Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.1.1 What is Simplification? . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.1.2 Simplification Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 vi Table of Contents 3.1.3 The simp Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.1.4 Adding and Deleting Simplification Rules . . . . . . . . . . . . 29 3.1.5 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.1.6 Rewriting with Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.1.7 Simplifying let-Expressions . . . . . . . . . . . . . . . . . . . . . . . 30 3.1.8 Conditional Simplification Rules . . . . . . . . . . . . . . . . . . . . 31 3.1.9 Automatic Case Splits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.1.10 Tracing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.1.11 Finding Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.2 Induction Heuristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3 Case Study: Compiling Expressions . . . . . . . . . . . . . . . . . . . . . . . 37 3.4 Advanced Datatypes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.4.1 Mutual Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.4.2 Nested Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.4.3 The Limits of Nested Recursion . . . . . . . . . . . . . . . . . . . . 43 3.4.4 Case Study: Tries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.5 Total Recursive Functions: fun . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 3.5.2 Termination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.5.3 Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.5.4 Induction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4. Presenting Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.1 Concrete Syntax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.1.1 Infix Annotations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.1.2 Mathematical Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.1.3 Prefix Annotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.1.4 Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4.2 Document Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4.2.1 Isabelle Sessions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.2.2 Structure Markup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4.2.3 Formal Comments and Antiquotations . . . . . . . . . . . . . . 60 4.2.4 Interpretation of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . 62 4.2.5 Suppressing Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Part II. Logic and Sets 5. The Rules of the Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 5.1 Natural Deduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 5.2 Introduction Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 5.3 Elimination Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 5.4 Destruction Rules: Some Examples . . . . . . . . . . . . . . . . . . . . . . . 71 5.5 Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 5.6 Negation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Table of Contents vii 5.7 Interlude: the Basic Methods for Rules . . . . . . . . . . . . . . . . . . . . 75 5.8 Unification and Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 5.8.1 Substitution and the subst Method . . . . . . . . . . . . . . . . 77 5.8.2 Unification and Its Pitfalls . . . . . . . . . . . . . . . . . . . . . . . . . 79 5.9 Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 5.9.1 The Universal Introduction Rule . . . . . . . . . . . . . . . . . . . 81 5.9.2 The Universal Elimination Rule . . . . . . . . . . . . . . . . . . . . 81 5.9.3 The Existential Quantifier . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.9.4 Renaming a Bound Variable: rename_tac . . . . . . . . . . . 83 5.9.5 Reusing an Assumption: frule . . . . . . . . . . . . . . . . . . . . . 84 5.9.6 Instantiating a Quantifier Explicitly . . . . . . . . . . . . . . . . 84 5.10 Description Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.10.1 Definite Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.10.2 Indefinite Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.11 Some Proofs That Fail . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.12 Proving Theorems Using the blast Method. . . . . . . . . . . . . . . . 90 5.13 Other Classical Reasoning Methods . . . . . . . . . . . . . . . . . . . . . . . 91 5.14 Finding More Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 5.15 Forward Proof: Transforming Theorems . . . . . . . . . . . . . . . . . . . 94 5.15.1 Modifying a Theorem using of, where and THEN . . . . . . 94 5.15.2 Modifying a Theorem using OF . . . . . . . . . . . . . . . . . . . . . 96 5.16 Forward Reasoning in a Backward Proof . . . . . . . . . . . . . . . . . . 97 5.16.1 The Method insert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 5.16.2 The Method subgoal_tac . . . . . . . . . . . . . . . . . . . . . . . . . 99 5.17 Managing Large Proofs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 5.17.1 Tacticals, or Control Structures . . . . . . . . . . . . . . . . . . . . 100 5.17.2 Subgoal Numbering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.18 Proving the Correctness of Euclid’s Algorithm . . . . . . . . . . . . . 103 6. Sets, Functions and Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 6.1.1 Finite Set Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.1.2 Set Comprehension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.1.3 Binding Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.1.4 Finiteness and Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . 111 6.2 Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 6.2.1 Function Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 6.2.2 Injections, Surjections, Bijections . . . . . . . . . . . . . . . . . . . 112 6.2.3 Function Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 6.3 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 6.3.1 Relation Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 6.3.2 The Reflexive and Transitive Closure . . . . . . . . . . . . . . . 114 6.3.3 A Sample Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.4 Well-Founded Relations and Induction . . . . . . . . . . . . . . . . . . . . 116 6.5 Fixed Point Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 viii Table of Contents 6.5.1 Propositional Dynamic Logic — PDL . . . . . . . . . . . . . . . 118 6.5.2 Computation Tree Logic — CTL . . . . . . . . . . . . . . . . . . . 121 7. Inductively Defined Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 7.1 The Set of Even Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 7.1.1 Making an Inductive Definition . . . . . . . . . . . . . . . . . . . . 127 7.1.2 Using Introduction Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 128 7.1.3 Rule Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 7.1.4 Generalization and Rule Induction . . . . . . . . . . . . . . . . . 129 7.1.5 Rule Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 7.1.6 Mutually Inductive Definitions . . . . . . . . . . . . . . . . . . . . . 132 7.1.7 Inductively Defined Predicates . . . . . . . . . . . . . . . . . . . . . 132 7.2 The Reflexive Transitive Closure . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.3 Advanced Inductive Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 7.3.1 Universal Quantifiers in Introduction Rules . . . . . . . . . 136 7.3.2 Alternative Definition Using a Monotone Function . . . . 137 7.3.3 A Proof of Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 7.3.4 Another Example of Rule Inversion . . . . . . . . . . . . . . . . . 140 7.4 Case Study: A Context Free Grammar . . . . . . . . . . . . . . . . . . . . 141 Part III. Advanced Material 8. More about Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.1 Pairs and Tuples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.1.1 Pattern Matching with Tuples . . . . . . . . . . . . . . . . . . . . . 149 8.1.2 Theorem Proving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 8.2 Records . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 8.2.1 Record Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 8.2.2 Extensible Records and Generic Operations . . . . . . . . . . 153 8.2.3 Record Equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 8.2.4 Extending and Truncating Records . . . . . . . . . . . . . . . . . 156 8.3 Type Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 8.3.1 Overloading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 8.3.2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 8.4 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 8.4.1 Numeric Literals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 8.4.2 The Type of Natural Numbers, nat . . . . . . . . . . . . . . . . . 165 8.4.3 The Type of Integers, int . . . . . . . . . . . . . . . . . . . . . . . . . 166 8.4.4 The Types of Rational, Real and Complex Numbers . . 167 8.4.5 The Numeric Type Classes . . . . . . . . . . . . . . . . . . . . . . . . 168 8.5 Introducing New Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 8.5.1 Declaring New Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 8.5.2 Defining New Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Table of Contents ix 9. Advanced Simplification and Induction . . . . . . . . . . . . . . . . . . . 175 9.1 Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 9.1.1 Advanced Features . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 9.1.2 How the Simplifier Works . . . . . . . . . . . . . . . . . . . . . . . . . 177 9.2 Advanced Induction Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . 178 9.2.1 Massaging the Proposition . . . . . . . . . . . . . . . . . . . . . . . . . 178 9.2.2 Beyond Structural and Recursion Induction . . . . . . . . . . 180 9.2.3 Derivation of New Induction Schemas . . . . . . . . . . . . . . . 182 9.2.4 CTL Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 10. Case Study: Verifying a Security Protocol . . . . . . . . . . . . . . . . 187 10.1 The Needham-Schroeder Public-Key Protocol . . . . . . . . . . . . . . 187 10.2 Agents and Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 10.3 Modelling the Adversary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 10.4 Event Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 10.5 Modelling the Protocol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 10.6 Proving Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 193 10.7 Proving Secrecy Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 A. Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 x Table of Contents Part I Elementary Techniques 1. The Basics 1.1 Introduction This book is a tutorial on how to use the theorem prover Isabelle/HOL as a specification and verification system. Isabelle is a generic system for imple-menting logical formalisms, and Isabelle/HOL is the specialization of Isabelle for HOL, which abbreviates Higher-Order Logic. We introduce HOL step by step following the equation HOL = Functional Programming + Logic. We do not assume that you are familiar with mathematical logic. However, we do assume that you are used to logical and set theoretic notation, as covered in a good discrete mathematics course , and that you are familiar with the basic concepts of functional programming [5, 14, 29, 35]. Although this tutorial initially concentrates on functional programming, do not be misled: HOL can express most mathematical concepts, and functional programming is just one particularly simple and ubiquitous instance. Isabelle is implemented in ML . This has influenced some of Isa-belle/HOL’s concrete syntax but is otherwise irrelevant for us: this tutorial is based on Isabelle/Isar , an extension of Isabelle which hides the im-plementation language almost completely. Thus the full name of the system should be Isabelle/Isar/HOL, but that is a bit of a mouthful. There are other implementations of HOL, in particular the one by Mike Gordon et al., which is usually referred to as “the HOL system” . For us, HOL refers to the logical system, and sometimes its incarnation Isa-belle/HOL. A tutorial is by definition incomplete. Currently the tutorial only intro-duces the rudiments of Isar’s proof language. To fully exploit the power of Isar, in particular the ability to write readable and structured proofs, you should start with Nipkow’s overview and consult the Isabelle/Isar Refer-ence Manual and Wenzel’s PhD thesis (which discusses many proof patterns) for further details. If you want to use Isabelle’s ML level directly (for example for writing your own proof procedures) see the Isabelle Reference Manual ; for details relating to HOL see the Isabelle/HOL manual . All manuals have a comprehensive index. 4 1. The Basics 1.2 Theories Working with Isabelle means creating theories. Roughly speaking, a theory is a named collection of types, functions, and theorems, much like a module in a programming language or a specification in a specification language. In fact, theories in HOL can be either. The general format of a theory T is theory T imports B1 . . . Bn begin declarations, definitions, and proofs end where B1 … Bn are the names of existing theories that T is based on and declarations, definitions, and proofs represents the newly introduced concepts (types, functions etc.) and proofs about them. The Bi are the direct parent theories of T. Everything defined in the parent theories (and their parents, recursively) is automatically visible. To avoid name clashes, identifiers can be qualified by theory names as in T.f and B.f. Each theory T must reside in a theory file named T.thy. This tutorial is concerned with introducing you to the different linguistic constructs that can fill the declarations, definitions, and proofs above. A com-plete grammar of the basic constructs is found in the Isabelle/Isar Reference Manual . ! ! HOL contains a theory Main, the union of all the basic predefined theories like arithmetic, lists, sets, etc. Unless you know what you are doing, always include Main as a direct or indirect parent of all your theories. HOL’s theory collection is available online at and is recommended browsing. In subdirectory Library you find a growing library of useful theories that are not part of Main but can be included among the parents of a theory and will then be loaded automatically. For the more adventurous, there is the Archive of Formal Proofs, a journal-like collection of more advanced Isabelle theories: We hope that you will contribute to it yourself one day. 1.3 Types, Terms and Formulae Embedded in a theory are the types, terms and formulae of HOL. HOL is a typed logic whose type system resembles that of functional programming languages like ML or Haskell. Thus there are 1.3 Types, Terms and Formulae 5 base types, in particular bool, the type of truth values, and nat, the type of natural numbers. type constructors, in particular list, the type of lists, and set, the type of sets. Type constructors are written postfix, e.g. (nat)list is the type of lists whose elements are natural numbers. Parentheses around single arguments can be dropped (as in nat list), multiple arguments are sep-arated by commas (as in (bool,nat)ty). function types, denoted by ⇒. In HOL ⇒represents total functions only. As is customary, τ1 ⇒τ2 ⇒τ3 means τ1 ⇒(τ2 ⇒τ3). Isabelle also sup-ports the notation [τ1, . . . , τn] ⇒τ which abbreviates τ1 ⇒· · · ⇒τn ⇒τ. type variables, denoted by 'a, 'b etc., just like in ML. They give rise to polymorphic types like 'a ⇒'a, the type of the identity function. ! ! Types are extremely important because they prevent us from writing nonsense. Isabelle insists that all terms and formulae must be well-typed and will print an error message if a type mismatch is encountered. To reduce the amount of explicit type information that needs to be provided by the user, Isabelle infers the type of all variables automatically (this is called type inference) and keeps quiet about it. Occasionally this may lead to misunderstandings between you and the system. If anything strange happens, we recommend that you ask Isabelle to display all type information via the Proof General menu item Isabelle > Settings > Show Types (see Sect. 1.5 for details). Terms are formed as in functional programming by applying functions to arguments. If f is a function of type τ1 ⇒τ2 and t is a term of type τ1 then f t is a term of type τ2. HOL also supports infix functions like + and some basic constructs from functional programming, such as conditional expressions: if b then t1 else t2 Here b is of type bool and t1 and t2 are of the same type. let x = t in u is equivalent to u where all free occurrences of x have been re-placed by t. For example, let x = 0 in x+x is equivalent to 0+0. Multiple bindings are separated by semicolons: let x1 = t1;…; xn = tn in u. case e of c1 ⇒e1 \| … \| cn ⇒en evaluates to ei if e is of the form ci. Terms may also contain λ-abstractions. For example, λx. x+1 is the func-tion that takes an argument x and returns x+1. Instead of λx.λy.λz. t we can write λx y z. t. Formulae are terms of type bool. There are the basic constants True and False and the usual logical connectives (in decreasing order of priority): ¬, ∧, ∨, and − →, all of which (except the unary ¬) associate to the right. In particular A − →B − →C means A − →(B − →C) and is thus logically equivalent to A ∧B − →C (which is (A ∧B) − →C). Equality is available in the form of the infix function = of type 'a ⇒'a ⇒bool. Thus t1 = t2 is a formula provided t1 and t2 are terms of the same 6 1. The Basics type. If t1 and t2 are of type bool then = acts as if-and-only-if. The formula t1 ̸= t2 is merely an abbreviation for ¬(t1 = t2). Quantifiers are written as ∀x. P and ∃x. P. There is even ∃! x. P, which means that there exists exactly one x that satisfies P. Nested quantifications can be abbreviated: ∀x y z. P means ∀x.∀y.∀z. P. Despite type inference, it is sometimes necessary to attach explicit type constraints to a term. The syntax is t::τ as in x < (y::nat). Note that :: binds weakly and should therefore be enclosed in parentheses. For in-stance, x < y::nat is ill-typed because it is interpreted as (x < y)::nat. Type constraints may be needed to disambiguate expressions involving overloaded functions such as +, and <. Section 8.3.1 discusses overloading, while Ta-ble A.2 presents the most important overloaded function symbols. In general, HOL’s concrete syntax tries to follow the conventions of func-tional programming and mathematics. Here are the main rules that you should be familiar with to avoid certain syntactic traps: – Remember that f t u means (f t) u and not f(t u)! – Isabelle allows infix functions like +. The prefix form of function application binds more strongly than anything else and hence f x + y means (f x) + y and not f(x+y). – Remember that in HOL if-and-only-if is expressed using equality. But equality has a high priority, as befitting a relation, while if-and-only-if typically has the lowest priority. Thus, ¬ ¬ P = P means ¬¬(P = P) and not (¬¬P) = P. When using = to mean logical equivalence, enclose both operands in parentheses, as in (A ∧B) = (B ∧A). – Constructs with an opening but without a closing delimiter bind very weakly and should therefore be enclosed in parentheses if they appear in subterms, as in (λx. x) = f. This includes if, let, case, λ, and quantifiers. – Never write λx.x or ∀x.x=x because x.x is always taken as a single qualified identifier. Write λx. x and ∀x. x=x instead. – Identifiers may contain the characters _ and ', except at the beginning. For the sake of readability, we use the usual mathematical symbols throughout the tutorial. Their ascii-equivalents are shown in table A.1 in the appendix. ! ! A particular problem for novices can be the priority of operators. If you are unsure, use additional parentheses. In those cases where Isabelle echoes your input, you can see which parentheses are dropped — they were superfluous. If you are unsure how to interpret Isabelle’s output because you don’t know where the (dropped) parentheses go, set the Proof General flag Isabelle > Settings > Show Brackets (see Sect. 1.5). 1.4 Variables 7 1.4 Variables Isabelle distinguishes free and bound variables, as is customary. Bound vari-ables are automatically renamed to avoid clashes with free variables. In ad-dition, Isabelle has a third kind of variable, called a schematic variable or unknown, which must have a ? as its first character. Logically, an unknown is a free variable. But it may be instantiated by another term during the proof process. For example, the mathematical theorem x = x is represented in Isa-belle as ?x = ?x, which means that Isabelle can instantiate it arbitrarily. This is in contrast to ordinary variables, which remain fixed. The programming language Prolog calls unknowns logical variables. Most of the time you can and should ignore unknowns and work with ordinary variables. Just don’t be surprised that after you have finished the proof of a theorem, Isabelle will turn your free variables into unknowns. It indicates that Isabelle will automatically instantiate those unknowns suitably when the theorem is used in some other proof. Note that for readability we often drop the ?s when displaying a theorem. ! ! For historical reasons, Isabelle accepts ? as an ASCII representation of the ∃ symbol. However, the ? character must then be followed by a space, as in ? x. f(x) = 0. Otherwise, ?x is interpreted as a schematic variable. The preferred ASCII representation of the ∃symbol is EX. 1.5 Interaction and Interfaces The recommended interface for Isabelle/Isar is the (X)Emacs-based Proof General [1, 2]. Interaction with Isabelle at the shell level, although possible, should be avoided. Most of the tutorial is independent of the interface and is phrased in a neutral language. For example, the phrase “to abandon a proof” corresponds to the obvious action of clicking on the Undo symbol in Proof General. Proof General specific information is often displayed in paragraphs identified by a miniature Proof General icon. Here are two examples: Proof General supports a special font with mathematical symbols known as “x-symbols”. All symbols have ascii-equivalents: for example, you can enter either & or \ to obtain ∧. For a list of the most frequent symbols see table A.1 in the appendix. Note that by default x-symbols are not enabled. You have to switch them on via the menu item Proof-General > Options > X-Symbols (and save the option via the top-level Options menu). Proof General offers the Isabelle menu for displaying information and setting flags. A particularly useful flag is Isabelle > Settings > Show Types which causes Isabelle to output the type information that is usually suppressed. This is indis-pensible in case of errors of all kinds because often the types reveal the source of the problem. Once you have diagnosed the problem you may no longer want to see the types because they clutter all output. Simply reset the flag. 8 1. The Basics 1.6 Getting Started Assuming you have installed Isabelle and Proof General, you start it by typing Isabelle in a shell window. This launches a Proof General window. By default, you are in HOL1. You can choose a different logic via the Isabelle > Logics menu. 1 This is controlled by the ISABELLE_LOGIC setting, see The Isabelle System Manual for more details. 2. Functional Programming in HOL This chapter describes how to write functional programs in HOL and how to verify them. However, most of the constructs and proof procedures in-troduced are general and recur in any specification or verification task. We really should speak of functional modelling rather than functional program-ming: our primary aim is not to write programs but to design abstract models of systems. HOL is a specification language that goes well beyond what can be expressed as a program. However, for the time being we concentrate on the computable. If you are a purist functional programmer, please note that all functions in HOL must be total: they must terminate for all inputs. Lazy data structures are not directly available. 2.1 An Introductory Theory Functional programming needs datatypes and functions. Both of them can be defined in a theory with a syntax reminiscent of languages like ML or Haskell. As an example consider the theory in figure 2.1. We will now examine it line by line. theory ToyList imports Main begin HOL already has a predefined theory of lists called List — ToyList is merely a small fragment of it chosen as an example. To avoid some ambiguities caused by defining lists twice, we manipulate the concrete syntax and name space of theory Main as follows. unbundle no list_syntax no notation append (infixr "@" 65) hide type list hide const rev datatype 'a list = Nil ("[]") \| Cons 'a "'a list" (infixr "#" 65) The datatype list introduces two constructors Nil and Cons, the empty list and the operator that adds an element to the front of a list. For example, 10 2. Functional Programming in HOL theory ToyList imports Main begin no_notation Nil ("[]") and Cons (infixr "#" 65) and append (infixr "@" 65) hide_type list hide_const rev datatype 'a list = Nil ("[]") \| Cons 'a "'a list" (infixr "#" 65) ( This is the append function: ) primrec app :: "'a list => 'a list => 'a list" (infixr "@" 65) where "[] @ ys = ys" \| "(x # xs) @ ys = x # (xs @ ys)" primrec rev :: "'a list => 'a list" where "rev [] = []" \| "rev (x # xs) = (rev xs) @ (x # [])" Fig. 2.1. A Theory of Lists the term Cons True (Cons False Nil) is a value of type bool list, namely the list with the elements True and False. Because this notation quickly becomes unwieldy, the datatype declaration is annotated with an alternative syntax: instead of Nil and Cons x xs we can write [] and x # xs. In fact, this alternative syntax is the familiar one. Thus the list Cons True (Cons False Nil) becomes True # False # []. The annotation infixr means that # associates to the right: the term x # y # z is read as x # (y # z) and not as (x # y) # z. The 65 is the priority of the infix #. ! ! Syntax annotations can be powerful, but they are difficult to master and are never necessary. You could drop them from theory ToyList and go back to the identifiers Nil and Cons. Novices should avoid using syntax annotations in their own theories. Next, two functions app and rev are defined recursively, in this order, because Isabelle insists on definition before use: primrec app :: "'a list ⇒'a list ⇒'a list" (infixr "@" 65) where "[] @ ys = ys" \| "(x # xs) @ ys = x # (xs @ ys)" primrec rev :: "'a list ⇒'a list" where "rev [] = []" \| "rev (x # xs) = (rev xs) @ (x # [])" Each function definition is of the form primrec name :: type (optional syntax) where equations 2.2 Evaluation 11 The equations must be separated by \|. Function app is annotated with con-crete syntax. Instead of the prefix syntax app xs ys the infix xs @ ys becomes the preferred form. The equations for app and rev hardly need comments: app appends two lists and rev reverses a list. The keyword primrec indicates that the recursion is of a particularly primitive kind where each recursive call peels off a datatype constructor from one of the arguments. Thus the recursion always terminates, i.e. the function is total. The termination requirement is absolutely essential in HOL, a logic of total functions. If we were to drop it, inconsistencies would quickly arise: the “definition” f (n) = f (n) + 1 immediately leads to 0 = 1 by subtracting f (n) on both sides. ! ! As we have indicated, the requirement for total functions is an essential char-acteristic of HOL. It is only because of totality that reasoning in HOL is com-paratively easy. More generally, the philosophy in HOL is to refrain from asserting arbitrary axioms (such as function definitions whose totality has not been proved) because they quickly lead to inconsistencies. Instead, fixed constructs for introduc-ing types and functions are offered (such as datatype and primrec) which are guaranteed to preserve consistency. A remark about syntax. The textual definition of a theory follows a fixed syntax with keywords like datatype and end. Embedded in this syntax are the types and formulae of HOL, whose syntax is extensible (see Sect. 4.1), e.g. by new user-defined infix operators. To distinguish the two levels, everything HOL-specific (terms and types) should be enclosed in "…". To lessen this burden, quotation marks around a single identifier can be dropped, unless the identifier happens to be a keyword, for example "end". When Isabelle prints a syntax error message, it refers to the HOL syntax as the inner syntax and the enclosing theory language as the outer syntax. Comments must be in enclosed in (and). 2.2 Evaluation Assuming you have processed the declarations and definitions of ToyList presented so far, you may want to test your functions by running them. For example, what is the value of rev (True # False # [])? Command value "rev (True # False # [])" yields the correct result False # True # []. But we can go beyond mere func-tional programming and evaluate terms with variables in them, executing functions symbolically: value "rev (a # b # c # [])" yields c # b # a # []. 12 2. Functional Programming in HOL 2.3 An Introductory Proof Having convinced ourselves (as well as one can by testing) that our definitions capture our intentions, we are ready to prove a few simple theorems. This will illustrate not just the basic proof commands but also the typical proof process. Main Goal. Our goal is to show that reversing a list twice produces the original list. theorem rev_rev [simp]: "rev(rev xs) = xs" This theorem command does several things: – It establishes a new theorem to be proved, namely rev (rev xs) = xs. – It gives that theorem the name rev_rev, for later reference. – It tells Isabelle (via the bracketed attribute simp) to take the eventual theorem as a simplification rule: future proofs involving simplification will replace occurrences of rev (rev xs) by xs. The name and the simplification attribute are optional. Isabelle’s response is to print the initial proof state consisting of some header information (like how many subgoals there are) followed by 1. rev (rev xs) = xs For compactness reasons we omit the header in this tutorial. Until we have finished a proof, the proof state proper always looks like this: 1. G1 . . . n. Gn The numbered lines contain the subgoals G1, …, Gn that we need to prove to establish the main goal. Initially there is only one subgoal, which is identical with the main goal. (If you always want to see the main goal as well, set the flag Proof.show_main_goal — this flag used to be set by default.) Let us now get back to rev (rev xs) = xs. Properties of recursively de-fined functions are best established by induction. In this case there is nothing obvious except induction on xs: apply(induct_tac xs) This tells Isabelle to perform induction on variable xs. The suffix tac stands for tactic, a synonym for “theorem proving function”. By default, induction acts on the first subgoal. The new proof state contains two subgoals, namely the base case (Nil) and the induction step (Cons): 1. rev (rev []) = [] 2. Vx1 x2. rev (rev x2) = x2 = ⇒rev (rev (x1 # x2)) = x1 # x2 The induction step is an example of the general format of a subgoal: 2.3 An Introductory Proof 13 i. Vx1 …xn. assumptions = ⇒conclusion The prefix of bound variables Vx1 … xn can be ignored most of the time, or simply treated as a list of variables local to this subgoal. Their deeper signif-icance is explained in Chapter 5. The assumptions are the local assumptions for this subgoal and conclusion is the actual proposition to be proved. Typical proof steps that add new assumptions are induction and case distinction. In our example the only assumption is the induction hypothesis rev (rev list) = list, where list is a variable name chosen by Isabelle. If there are multiple assumptions, they are enclosed in the bracket pair [ [ and ] ] and separated by semicolons. Let us try to solve both goals automatically: apply(auto) This command tells Isabelle to apply a proof strategy called auto to all sub-goals. Essentially, auto tries to simplify the subgoals. In our case, subgoal 1 is solved completely (thanks to the equation rev [] = []) and disappears; the simplified version of subgoal 2 becomes the new subgoal 1: 1. Vx1 x2. rev (rev x2) = x2 = ⇒rev (rev x2 @ x1 # []) = x1 # x2 In order to simplify this subgoal further, a lemma suggests itself. First Lemma. After abandoning the above proof attempt (at the shell level type oops) we start a new proof: lemma rev_app [simp]: "rev(xs @ ys) = (rev ys) @ (rev xs)" The keywords theorem and lemma are interchangeable and merely indi-cate the importance we attach to a proposition. Therefore we use the words theorem and lemma pretty much interchangeably, too. There are two variables that we could induct on: xs and ys. Because @ is defined by recursion on the first argument, xs is the correct one: apply(induct_tac xs) This time not even the base case is solved automatically: apply(auto) 1. rev ys = rev ys @ [] A total of 2 subgoals... Again, we need to abandon this proof attempt and prove another simple lemma first. In the future the step of abandoning an incomplete proof before embarking on the proof of a lemma usually remains implicit. 14 2. Functional Programming in HOL Second Lemma. We again try the canonical proof procedure: lemma app_Nil2 [simp]: "xs @ [] = xs" apply(induct_tac xs) apply(auto) It works, yielding the desired message No subgoals!: xs @ [] = xs No subgoals! We still need to confirm that the proof is now finished: done As a result of that final done, Isabelle associates the lemma just proved with its name. In this tutorial, we sometimes omit to show that final done if it is obvious from the context that the proof is finished. Notice that in lemma app_Nil2, as printed out after the final done, the free variable xs has been replaced by the unknown ?xs, just as explained in Sect. 1.4. Going back to the proof of the first lemma lemma rev_app [simp]: "rev(xs @ ys) = (rev ys) @ (rev xs)" apply(induct_tac xs) apply(auto) we find that this time auto solves the base case, but the induction step merely simplifies to 1. Vx1 x2. rev (x2 @ ys) = rev ys @ rev x2 = ⇒ (rev ys @ rev x2) @ x1 # [] = rev ys @ rev x2 @ x1 # [] Now we need to remember that @ associates to the right, and that # and @ have the same priority (namely the 65 in their infixr annotation). Thus the conclusion really is (rev ys @ rev list) @ (a # []) = rev ys @ (rev list @ (a # [])) and the missing lemma is associativity of @. Third Lemma. Abandoning the previous attempt, the canonical proof pro-cedure succeeds without further ado. lemma app_assoc [simp]: "(xs @ ys) @ zs = xs @ (ys @ zs)" apply(induct_tac xs) apply(auto) done Now we can prove the first lemma: lemma rev_app [simp]: "rev(xs @ ys) = (rev ys) @ (rev xs)" apply(induct_tac xs) apply(auto) done 2.3 An Introductory Proof 15 Finally, we prove our main theorem: theorem rev_rev [simp]: "rev(rev xs) = xs" apply(induct_tac xs) apply(auto) done The final end tells Isabelle to close the current theory because we are finished with its development: end The complete proof script is shown in Fig. 2.2. The concatenation of Figs. 2.1 and 2.2 constitutes the complete theory ToyList and should reside in file ToyList.thy. lemma app_Nil2 [simp]: "xs @ [] = xs" apply(induct_tac xs) apply(auto) done lemma app_assoc [simp]: "(xs @ ys) @ zs = xs @ (ys @ zs)" apply(induct_tac xs) apply(auto) done lemma rev_app [simp]: "rev(xs @ ys) = (rev ys) @ (rev xs)" apply(induct_tac xs) apply(auto) done theorem rev_rev [simp]: "rev(rev xs) = xs" apply(induct_tac xs) apply(auto) done end Fig. 2.2. Proofs about Lists Review This is the end of our toy proof. It should have familiarized you with – the standard theorem proving procedure: state a goal (lemma or theorem); proceed with proof until a separate lemma is required; prove that lemma; come back to the original goal. – a specific procedure that works well for functional programs: induction followed by all-out simplification via auto. – a basic repertoire of proof commands. 16 2. Functional Programming in HOL ! ! It is tempting to think that all lemmas should have the simp attribute just because this was the case in the example above. However, in that example all lemmas were equations, and the right-hand side was simpler than the left-hand side — an ideal situation for simplification purposes. Unless this is clearly the case, novices should refrain from awarding a lemma the simp attribute, which has a global effect. Instead, lemmas can be applied locally where they are needed, which is discussed in the following chapter. 2.4 Some Helpful Commands This section discusses a few basic commands for manipulating the proof state and can be skipped by casual readers. There are two kinds of commands used during a proof: the actual proof commands and auxiliary commands for examining the proof state and con-trolling the display. Simple proof commands are of the form apply(method), where method is typically induct_tac or auto. All such theorem proving oper-ations are referred to as methods, and further ones are introduced through-out the tutorial. Unless stated otherwise, you may assume that a method attacks merely the first subgoal. An exception is auto, which tries to solve all subgoals. The most useful auxiliary commands are as follows: Modifying the order of subgoals: defer moves the first subgoal to the end and prefer n moves subgoal n to the front. Printing theorems: thm name1 … namen prints the named theorems. Reading terms and types: term string reads, type-checks and prints the given string as a term in the current context; the inferred type is output as well. typ string reads and prints the given string as a type in the current context. Further commands are found in the Isabelle/Isar Reference Manual . Clicking on the State button redisplays the current proof state. This is helpful in case commands like thm have overwritten it. We now examine Isabelle’s functional programming constructs systemat-ically, starting with inductive datatypes. 2.5 Datatypes Inductive datatypes are part of almost every non-trivial application of HOL. First we take another look at an important example, the datatype of lists, before we turn to datatypes in general. The section closes with a case study. 2.5 Datatypes 17 2.5.1 Lists Lists are one of the essential datatypes in computing. We expect that you are already familiar with their basic operations. Theory ToyList is only a small fragment of HOL’s predefined theory List1. The latter contains many further operations. For example, the functions hd (“head”) and tl (“tail”) return the first element and the remainder of a list. (However, pattern matching is usually preferable to hd and tl.) Also available are higher-order functions like map and filter. Theory List also contains more syntactic sugar: [x1,…,xn] abbreviates x1#…#xn#[]. In the rest of the tutorial we always use HOL’s predefined lists by building on theory Main. ! ! Looking ahead to sets and quanifiers in Part II: The best way to express that some element x is in a list xs is x ∈set xs, where set is a function that turns a list into the set of its elements. By the same device you can also write bounded quantifiers like ∀x ∈set xs or embed lists in other set expressions. 2.5.2 The General Format The general HOL datatype definition is of the form datatype (α1, . . . , αn) t = C1 τ11 . . . τ1k1 \| . . . \| Cm τm1 . . . τmkm where αi are distinct type variables (the parameters), Ci are distinct con-structor names and τij are types; it is customary to capitalize the first letter in constructor names. There are a number of restrictions (such as that the type should not be empty) detailed elsewhere . Isabelle notifies you if you violate them. Laws about datatypes, such as [] ̸= x#xs and (x#xs = y#ys) = (x=y ∧xs=ys), are used automatically during proofs by simplification. The same is true for the equations in primitive recursive function definitions. Every2 datatype t comes equipped with a size function from t into the natural numbers (see Sect. 2.6.1 below). For lists, size is just the length, i.e. size [] = 0 and size(x # xs) = size xs + 1. In general, size returns – zero for all constructors that do not have an argument of type t, – one plus the sum of the sizes of all arguments of type t, for all other constructors. Note that because size is defined on every datatype, it is overloaded; on lists size is also called length, which is not overloaded. Isabelle will always show size on lists as length. 1 2 Except for advanced datatypes where the recursion involves “⇒” as in Sect. 3.4.3. 18 2. Functional Programming in HOL 2.5.3 Primitive Recursion Functions on datatypes are usually defined by recursion. In fact, most of the time they are defined by what is called primitive recursion over some datatype t. This means that the recursion equations must be of the form f x1 . . . (C y1 . . . yk) . . . xn = r such that C is a constructor of t and all recursive calls of f in r are of the form f . . . yi . . . for some i. Thus Isabelle immediately sees that f termi-nates because one (fixed!) argument becomes smaller with every recursive call. There must be at most one equation for each constructor. Their order is immaterial. A more general method for defining total recursive functions is introduced in Sect. 3.5. Exercise 2.5.1 Define the datatype of binary trees: datatype 'a tree = Tip \| Node "'a tree" 'a "'a tree" Define a function mirror that mirrors a binary tree by swapping subtrees recursively. Prove lemma mirror_mirror: "mirror(mirror t) = t" Define a function flatten that flattens a tree into a list by traversing it in infix order. Prove lemma "flatten(mirror t) = rev(flatten t)" 2.5.4 Case Expressions HOL also features case-expressions for analyzing elements of a datatype. For example, case xs of [] ⇒[] \| y # ys ⇒y evaluates to [] if xs is [] and to y if xs is y # ys. (Since the result in both branches must be of the same type, it follows that y is of type 'a list and hence that xs is of type 'a list list.) In general, case expressions are of the form case e of pattern1 ⇒e1 \| . . . \| patternm ⇒em Like in functional programming, patterns are expressions consisting of data-type constructors (e.g. [] and #) and variables, including the wildcard “”. Not all cases need to be covered and the order of cases matters. However, one is well-advised not to wallow in complex patterns because complex case distinctions tend to induce complex proofs. 2.5 Datatypes 19 ! ! Internally Isabelle only knows about exhaustive case expressions with non-nested patterns: patterni must be of the form Ci xi1 . . . xiki and C1, . . . , Cm must be exactly the constructors of the type of e. More complex case expressions are automatically translated into the simpler form upon parsing but are not translated back for printing. This may lead to surprising output. ! ! Like if, case-expressions may need to be enclosed in parentheses to indicate their scope. 2.5.5 Structural Induction and Case Distinction Induction is invoked by induct_tac, as we have seen above; it works for any datatype. In some cases, induction is overkill and a case distinction over all constructors of the datatype suffices. This is performed by case_tac. Here is a trivial example: lemma "(case xs of [] ⇒[] \| y#ys ⇒xs) = xs" apply(case_tac xs) results in the proof state 1. xs = [] = ⇒(case xs of [] ⇒[] \| y # ys ⇒xs) = xs 2. Va list. xs = a # list = ⇒(case xs of [] ⇒[] \| y # ys ⇒xs) = xs which is solved automatically: apply(auto) Note that we do not need to give a lemma a name if we do not intend to refer to it explicitly in the future. Other basic laws about a datatype are applied automatically during simplification, so no special methods are provided for them. ! ! Induction is only allowed on free (or V-bound) variables that should not occur among the assumptions of the subgoal; see Sect. 9.2.1 for details. Case distinc-tion (case_tac) works for arbitrary terms, which need to be quoted if they are non-atomic. However, apart from V-bound variables, the terms must not contain variables that are bound outside. For example, given the goal ∀xs. xs = [] ∨ (∃y ys. xs = y # ys), case_tac xs will not work as expected because Isabelle interprets the xs as a new free variable distinct from the bound xs in the goal. 2.5.6 Case Study: Boolean Expressions The aim of this case study is twofold: it shows how to model boolean expres-sions and some algorithms for manipulating them, and it demonstrates the constructs introduced above. 20 2. Functional Programming in HOL Modelling Boolean Expressions. We want to represent boolean expres-sions built up from variables and constants by negation and conjunction. The following datatype serves exactly that purpose: datatype boolex = Const bool \| Var nat \| Neg boolex \| And boolex boolex The two constants are represented by Const True and Const False. Variables are represented by terms of the form Var n, where n is a natural number (type nat). For example, the formula P0 ∧¬P1 is represented by the term And (Var 0) (Neg (Var 1)). The Value of a Boolean Expression. The value of a boolean expression depends on the value of its variables. Hence the function value takes an addi-tional parameter, an environment of type nat ⇒bool, which maps variables to their values: primrec "value" :: "boolex ⇒(nat ⇒bool) ⇒bool" where "value (Const b) env = b" \| "value (Var x) env = env x" \| "value (Neg b) env = (¬ value b env)" \| "value (And b c) env = (value b env ∧value c env)" If-Expressions. An alternative and often more efficient (because in a cer-tain sense canonical) representation are so-called If-expressions built up from constants (CIF), variables (VIF) and conditionals (IF): datatype ifex = CIF bool \| VIF nat \| IF ifex ifex ifex The evaluation of If-expressions proceeds as for boolex: primrec valif :: "ifex ⇒(nat ⇒bool) ⇒bool" where "valif (CIF b) env = b" \| "valif (VIF x) env = env x" \| "valif (IF b t e) env = (if valif b env then valif t env else valif e env)" Converting Boolean and If-Expressions. The type boolex is close to the customary representation of logical formulae, whereas ifex is designed for efficiency. It is easy to translate from boolex into ifex: primrec bool2if :: "boolex ⇒ifex" where "bool2if (Const b) = CIF b" \| "bool2if (Var x) = VIF x" \| "bool2if (Neg b) = IF (bool2if b) (CIF False) (CIF True)" \| "bool2if (And b c) = IF (bool2if b) (bool2if c) (CIF False)" At last, we have something we can verify: that bool2if preserves the value of its argument: lemma "valif (bool2if b) env = value b env" The proof is canonical: apply(induct_tac b) 2.5 Datatypes 21 apply(auto) done In fact, all proofs in this case study look exactly like this. Hence we do not show them below. More interesting is the transformation of If-expressions into a normal form where the first argument of IF cannot be another IF but must be a constant or variable. Such a normal form can be computed by repeatedly replacing a subterm of the form IF (IF b x y) z u by IF b (IF x z u) (IF y z u), which has the same value. The following primitive recursive functions perform this task: primrec normif :: "ifex ⇒ifex ⇒ifex ⇒ifex" where "normif (CIF b) t e = IF (CIF b) t e" \| "normif (VIF x) t e = IF (VIF x) t e" \| "normif (IF b t e) u f = normif b (normif t u f) (normif e u f)" primrec norm :: "ifex ⇒ifex" where "norm (CIF b) = CIF b" \| "norm (VIF x) = VIF x" \| "norm (IF b t e) = normif b (norm t) (norm e)" Their interplay is tricky; we leave it to you to develop an intuitive under-standing. Fortunately, Isabelle can help us to verify that the transformation preserves the value of the expression: theorem "valif (norm b) env = valif b env" The proof is canonical, provided we first show the following simplification lemma, which also helps to understand what normif does: lemma [simp]: "∀t e. valif (normif b t e) env = valif (IF b t e) env" Note that the lemma does not have a name, but is implicitly used in the proof of the theorem shown above because of the [simp] attribute. But how can we be sure that norm really produces a normal form in the above sense? We define a function that tests If-expressions for normality: primrec normal :: "ifex ⇒bool" where "normal(CIF b) = True" \| "normal(VIF x) = True" \| "normal(IF b t e) = (normal t ∧normal e ∧ (case b of CIF b ⇒True \| VIF x ⇒True \| IF x y z ⇒False))" Now we prove normal (norm b). Of course, this requires a lemma about nor-mality of normif: lemma [simp]: "∀t e. normal(normif b t e) = (normal t ∧normal e)" How do we come up with the required lemmas? Try to prove the main theorems without them and study carefully what auto leaves unproved. This can provide the clue. The necessity of universal quantification (∀t e) in the two lemmas is explained in Sect. 3.2 22 2. Functional Programming in HOL Exercise 2.5.2 We strengthen the definition of a normal If-expression as follows: the first argument of all IFs must be a variable. Adapt the above development to this changed requirement. (Hint: you may need to formulate some of the goals as implications (− →) rather than equalities (=).) 2.6 Some Basic Types This section introduces the types of natural numbers and ordered pairs. Also described is type option, which is useful for modelling exceptional cases. 2.6.1 Natural Numbers The type nat of natural numbers is predefined to have the constructors 0 and Suc. It behaves approximately as if it were declared like this: datatype nat = zero ("0") \| Suc nat In particular, there are case-expressions, for example case n of 0 ⇒0 \| Suc m ⇒m primitive recursion, for example primrec sum :: "nat ⇒nat" where "sum 0 = 0" \| "sum (Suc n) = Suc n + sum n" and induction, for example lemma "sum n + sum n = n(Suc n)" apply(induct_tac n) apply(auto) done The arithmetic operations +, -, , div, mod, min and max are predefined, as are the relations ≤and <. As usual, m - n = 0 if m < n. There is even a least number operation LEAST. For example, (LEAST n. 0 < n) = Suc 0. ! ! The constants 0 and 1 and the operations +, -, , min, max, ≤and < are overloaded: they are available not just for natural numbers but for other types as well. For example, given the goal x + 0 = x, there is nothing to indicate that you are talking about natural numbers. Hence Isabelle can only infer that x is of some arbitrary type where 0 and + are declared. As a consequence, you will be unable to prove the goal. To alert you to such pitfalls, Isabelle flags numerals without a fixed type in its output: x + 0 = x. (In the absence of a numeral, it may take you some time to realize what has happened if Show Types is not set). In this particular example, you need to include an explicit type constraint, for example x+0 = (x::nat). If there is enough contextual information this may not be necessary: Suc x = x automatically implies x::nat because Suc is not overloaded. For details on overloading see Sect. 8.3.1. Table A.2 in the appendix shows the most important overloaded operations. 2.6 Some Basic Types 23 ! ! The symbols > and ≥are merely syntax: x > y stands for y < x and similary for ≥and ≤. ! ! Constant 1::nat is defined to equal Suc 0. This definition (see Sect. 2.7.2) is unfolded automatically by some tactics (like auto, simp and arith) but not by others (especially the single step tactics in Chapter 5). If you need the full set of numerals, see Sect. 8.4.1. Novices are advised to stick to 0 and Suc. Both auto and simp (a method introduced below, Sect. 3.1) prove simple arithmetic goals automatically: lemma "[ [ ¬ m < n; m < n + (1::nat) ] ] = ⇒m = n" For efficiency’s sake, this built-in prover ignores quantified formulae, many logical connectives, and all arithmetic operations apart from addition. In consequence, auto and simp cannot prove this slightly more complex goal: lemma "m ̸= (n::nat) = ⇒m < n ∨n < m" The method arith is more general. It attempts to prove the first subgoal provided it is a linear arithmetic formula. Such formulas may involve the usual logical connectives (¬, ∧, ∨, − →, =, ∀, ∃), the relations =, ≤and <, and the operations +, -, min and max. For example, lemma "min i (max j (kk)) = max (min (kk) i) (min i (j::nat))" apply(arith) succeeds because k k can be treated as atomic. In contrast, lemma "nn = n+1 = ⇒n=0" is not proved by arith because the proof relies on properties of multiplication. Only multiplication by numerals (which is the same as iterated addition) is taken into account. ! ! The running time of arith is exponential in the number of occurrences of -, min and max because they are first eliminated by case distinctions. If k is a numeral, div k, mod k and k dvd are also supported, where the former two are eliminated by case distinctions, again blowing up the running time. If the formula involves quantifiers, arith may take super-exponential time and space. 2.6.2 Pairs HOL also has ordered pairs: (a1,a2) is of type τ1 × τ2 provided each ai is of type τi. The functions fst and snd extract the components of a pair: fst(x,y) = x and snd(x,y) = y. Tuples are simulated by pairs nested to the right: (a1,a2,a3) stands for (a1,(a2,a3)) and τ1 × τ2 × τ3 for τ1 × (τ2 × τ3). Therefore we have fst(snd(a1,a2,a3)) = a2. Remarks: 24 2. Functional Programming in HOL – There is also the type unit, which contains exactly one element denoted by (). This type can be viewed as a degenerate product with 0 components. – Products, like type nat, are datatypes, which means in particular that induct_tac and case_tac are applicable to terms of product type. Both split the term into a number of variables corresponding to the tuple structure (up to 7 components). – Tuples with more than two or three components become unwieldy; records are preferable. For more information on pairs and records see Chapter 8. 2.6.3 Datatype option Our final datatype is very simple but still eminently useful: datatype 'a option = None \| Some 'a Frequently one needs to add a distinguished element to some existing type. For example, type t option can model the result of a computation that may either terminate with an error (represented by None) or return some value v (represented by Some v). Similarly, nat extended with ∞can be modeled by type nat option. In both cases one could define a new datatype with customized constructors like Error and Infinity, but it is often simpler to use option. For an application see Sect. 3.4.4. 2.7 Definitions A definition is simply an abbreviation, i.e. a new name for an existing con-struction. In particular, definitions cannot be recursive. Isabelle offers defini-tions on the level of types and terms. Those on the type level are called type synonyms; those on the term level are simply called definitions. 2.7.1 Type Synonyms Type synonyms are similar to those found in ML. They are created by a type synonym command: type synonym number = nat type synonym gate = "bool ⇒bool ⇒bool" type synonym ('a, 'b) alist = "('a × 'b) list" Internally all synonyms are fully expanded. As a consequence Isabelle’s out-put never contains synonyms. Their main purpose is to improve the readabil-ity of theories. Synonyms can be used just like any other type. 2.8 The Definitional Approach 25 2.7.2 Constant Definitions Nonrecursive definitions can be made with the definition command, for example nand and xor gates (based on type gate above): definition nand :: gate where "nand A B ≡¬(A ∧B)" definition xor :: gate where "xor A B ≡A ∧¬B ∨¬A ∧B" The symbol ≡is a special form of equality that must be used in constant definitions. Pattern-matching is not allowed: each definition must be of the form f x1 . . . xn ≡t. Section 3.1.6 explains how definitions are used in proofs. The default name of each definition is f _def, where f is the name of the defined constant. ! ! A common mistake when writing definitions is to introduce extra free variables on the right-hand side. Consider the following, flawed definition (where dvd means “divides”): "prime p ≡1 < p ∧(m dvd p − →m = 1 ∨m = p)" Isabelle rejects this “definition” because of the extra m on the right-hand side, which would introduce an inconsistency (why?). The correct version is "prime p ≡1 < p ∧(∀m. m dvd p − →m = 1 ∨m = p)" 2.8 The Definitional Approach As we pointed out at the beginning of the chapter, asserting arbitrary ax-ioms such as f (n) = f (n) + 1 can easily lead to contradictions. In order to avoid this danger, we advocate the definitional rather than the axiomatic ap-proach: introduce new concepts by definitions. However, Isabelle/HOL seems to support many richer definitional constructs, such as primrec. The point is that Isabelle reduces such constructs to first principles. For example, each primrec function definition is turned into a proper (nonrecursive!) definition from which the user-supplied recursion equations are automatically proved. This process is hidden from the user, who does not have to understand the details. Other commands described later, like fun and inductive, work sim-ilarly. This strict adherence to the definitional approach reduces the risk of soundness errors. 3. More Functional Programming The purpose of this chapter is to deepen your understanding of the con-cepts encountered so far and to introduce advanced forms of datatypes and recursive functions. The first two sections give a structured presentation of theorem proving by simplification (Sect. 3.1) and discuss important heuris-tics for induction (Sect. 3.2). You can skip them if you are not planning to perform proofs yourself. We then present a case study: a compiler for ex-pressions (Sect. 3.3). Advanced datatypes, including those involving function spaces, are covered in Sect. 3.4; it closes with another case study, search trees (“tries”). Finally we introduce fun, a general form of recursive function def-inition that goes well beyond primrec (Sect. 3.5). 3.1 Simplification So far we have proved our theorems by auto, which simplifies all subgoals. In fact, auto can do much more than that. To go beyond toy examples, you need to understand the ingredients of auto. This section covers the method that auto always applies first, simplification. Simplification is one of the central theorem proving tools in Isabelle and many other systems. The tool itself is called the simplifier. This section introduces the many features of the simplifier and is required reading if you intend to perform proofs. Later on, Sect. 9.1 explains some more advanced features and a little bit of how the simplifier works. The serious student should read that section as well, in particular to understand why the simplifier did something unexpected. 3.1.1 What is Simplification? In its most basic form, simplification means repeated application of equations from left to right. For example, taking the rules for @ and applying them to the term [0,1] @ [] results in a sequence of simplification steps: (0#1#[]) @ [] ; 0#((1#[]) @ []) ; 0#(1#([] @ [])) ; 0#1#[] This is also known as term rewriting and the equations are referred to as rewrite rules. “Rewriting” is more honest than “simplification” because the terms do not necessarily become simpler in the process. 28 3. More Functional Programming The simplifier proves arithmetic goals as described in Sect. 2.6.1 above. Arithmetic expressions are simplified using built-in procedures that go be-yond mere rewrite rules. New simplification procedures can be coded and installed, but they are definitely not a matter for this tutorial. 3.1.2 Simplification Rules To facilitate simplification, the attribute [simp] declares theorems to be sim-plification rules, which the simplifier will use automatically. In addition, datatype and primrec declarations (and a few others) implicitly declare some simplification rules. Explicit definitions are not declared as simplifica-tion rules automatically! Nearly any theorem can become a simplification rule. The simplifier will try to transform it into an equation. For example, the theorem ¬ P is turned into P = False. The details are explained in Sect. 9.1.2. The simplification attribute of theorems can be turned on and off: declare theorem-name[simp] declare theorem-name[simp del] Only equations that really simplify, like rev (rev xs) = xs and xs @ [] = xs, should be declared as default simplification rules. More specific ones should only be used selectively and should not be made default. Distributivity laws, for example, alter the structure of terms and can produce an exponential blow-up instead of simplification. A default simplification rule may need to be disabled in certain proofs. Frequent changes in the simplification status of a theorem may indicate an unwise use of defaults. ! ! Simplification can run forever, for example if both f (x) = g(x) and g(x) = f (x) are simplification rules. It is the user’s responsibility not to include simplification rules that can lead to nontermination, either on their own or in combination with other simplification rules. ! ! It is inadvisable to toggle the simplification attribute of a theorem from a parent theory A in a child theory B for good. The reason is that if some theory C is based both on B and (via a different path) on A, it is not defined what the simplification attribute of that theorem will be in C: it could be either. 3.1.3 The simp Method The general format of the simplification method is simp list of modifiers where the list of modifiers fine tunes the behaviour and may be empty. Specific modifiers are discussed below. Most if not all of the proofs seen so far could 3.1 Simplification 29 have been performed with simp instead of auto, except that simp attacks only the first subgoal and may thus need to be repeated — use simp_all to simplify all subgoals. If nothing changes, simp fails. 3.1.4 Adding and Deleting Simplification Rules If a certain theorem is merely needed in a few proofs by simplification, we do not need to make it a global simplification rule. Instead we can modify the set of simplification rules used in a simplification step by adding rules to it and/or deleting rules from it. The two modifiers for this are add: list of theorem names del: list of theorem names Or you can use a specific list of theorems and omit all others: only: list of theorem names In this example, we invoke the simplifier, adding two distributive laws: apply(simp add: mod_mult_distrib add_mult_distrib) 3.1.5 Assumptions By default, assumptions are part of the simplification process: they are used as simplification rules and are simplified themselves. For example: lemma "[ [ xs @ zs = ys @ xs; [] @ xs = [] @ [] ] ] = ⇒ys = zs" apply simp done The second assumption simplifies to xs = [], which in turn simplifies the first assumption to zs = ys, thus reducing the conclusion to ys = ys and hence to True. In some cases, using the assumptions can lead to nontermination: lemma "∀x. f x = g (f (g x)) = ⇒f [] = f [] @ []" An unmodified application of simp loops. The culprit is the simplification rule f x = g (f (g x)), which is extracted from the assumption. (Isabelle notices certain simple forms of nontermination but not this one.) The problem can be circumvented by telling the simplifier to ignore the assumptions: apply(simp (no_asm)) done Three modifiers influence the treatment of assumptions: (no_asm) means that assumptions are completely ignored. (no_asm_simp) means that the assumptions are not simplified but are used in the simplification of the conclusion. 30 3. More Functional Programming (no_asm_use) means that the assumptions are simplified but are not used in the simplification of each other or the conclusion. Only one of the modifiers is allowed, and it must precede all other modifiers. 3.1.6 Rewriting with Definitions Constant definitions (Sect. 2.7.2) can be used as simplification rules, but by default they are not: the simplifier does not expand them automatically. Definitions are intended for introducing abstract concepts and not merely as abbreviations. Of course, we need to expand the definition initially, but once we have proved enough abstract properties of the new constant, we can forget its original definition. This style makes proofs more robust: if the definition has to be changed, only the proofs of the abstract properties will be affected. For example, given definition xor :: "bool ⇒bool ⇒bool" where "xor A B ≡(A ∧¬B) ∨(¬A ∧B)" we may want to prove lemma "xor A (¬A)" Typically, we begin by unfolding some definitions: apply(simp only: xor_def) In this particular case, the resulting goal 1. A ∧¬ ¬ A ∨¬ A ∧¬ A can be proved by simplification. Thus we could have proved the lemma out-right by apply(simp add: xor_def) Of course we can also unfold definitions in the middle of a proof. ! ! If you have defined f x y ≡t then you can only unfold occurrences of f with at least two arguments. This may be helpful for unfolding f selectively, but it may also get in the way. Defining f ≡λx y. t allows to unfold all occurrences of f . There is also the special method unfold which merely unfolds one or several definitions, as in apply(unfold xor_def). This is can be useful in situations where simp does too much. Warning: unfold acts on all subgoals! 3.1.7 Simplifying let-Expressions Proving a goal containing let-expressions almost invariably requires the let-constructs to be expanded at some point. Since let…=…in… is just syntac-tic sugar for the predefined constant Let, expanding let-constructs means rewriting with Let_def: 3.1 Simplification 31 lemma "(let xs = [] in xs@ys@xs) = ys" apply(simp add: Let_def) done If, in a particular context, there is no danger of a combinatorial explosion of nested lets, you could even simplify with Let_def by default: declare Let_def [simp] 3.1.8 Conditional Simplification Rules So far all examples of rewrite rules were equations. The simplifier also accepts conditional equations, for example lemma hd_Cons_tl[simp]: "xs ̸= [] = ⇒ hd xs # tl xs = xs" apply(case_tac xs, simp, simp) done Note the use of “,” to string together a sequence of methods. Assuming that the simplification rule (rev xs = []) = (xs = []) is present as well, the lemma below is proved by plain simplification: lemma "xs ̸= [] = ⇒hd(rev xs) # tl(rev xs) = rev xs" The conditional equation hd_Cons_tl above can simplify hd (rev xs) # tl (rev xs) to rev xs because the corresponding precondition rev xs ̸= [] sim-plifies to xs ̸= [], which is exactly the local assumption of the subgoal. 3.1.9 Automatic Case Splits Goals containing if-expressions are usually proved by case distinction on the boolean condition. Here is an example: lemma "∀xs. if xs = [] then rev xs = [] else rev xs ̸= []" The goal can be split by a special method, split: apply(split if_split) 1. ∀xs. (xs = [] − →rev xs = []) ∧(xs ̸= [] − →rev xs ̸= []) where if_split is a theorem that expresses splitting of ifs. Because splitting the ifs is usually the right proof strategy, the simplifier does it automatically. Try apply(simp) on the initial goal above. This splitting idea generalizes from if to case. Let us simplify a case analysis over lists: lemma "(case xs of [] ⇒zs \| y#ys ⇒y#(ys@zs)) = xs@zs" apply(split list.split) 1. (xs = [] − →zs = xs @ zs) ∧ (∀x21 x22. xs = x21 # x22 − →x21 # x22 @ zs = xs @ zs) 32 3. More Functional Programming The simplifier does not split case-expressions, as it does if-expressions, be-cause with recursive datatypes it could lead to nontermination. Instead, the simplifier has a modifier split for adding splitting rules explicitly. The lemma above can be proved in one step by apply(simp split: list.split) whereas apply(simp) alone will not succeed. Every datatype t comes with a theorem t.split which can be declared to be a split rule either locally as above, or by giving it the split attribute globally: declare list.split [split] The split attribute can be removed with the del modifier, either locally apply(simp split del: if_split) or globally: declare list.split [split del] Polished proofs typically perform splitting within simp rather than in-voking the split method. However, if a goal contains several if and case expressions, the split method can be helpful in selectively exploring the ef-fects of splitting. The split rules shown above are intended to affect only the subgoal’s conclusion. If you want to split an if or case-expression in the assumptions, you have to apply if_split_asm or t.split_asm: lemma "if xs = [] then ys ̸= [] else ys = [] = ⇒xs @ ys ̸= []" apply(split if_split_asm) Unlike splitting the conclusion, this step creates two separate subgoals, which here can be solved by simp_all: 1. [ [xs = []; ys ̸= []] ] = ⇒xs @ ys ̸= [] 2. [ [xs ̸= []; ys = []] ] = ⇒xs @ ys ̸= [] If you need to split both in the assumptions and the conclusion, use t.splits which subsumes t.split and t.split_asm. Analogously, there is if_splits. ! ! The simplifier merely simplifies the condition of an if but not the then or else parts. The latter are simplified only after the condition reduces to True or False, or after splitting. The same is true for case-expressions: only the selector is simplified at first, until either the expression reduces to one of the cases or it is split. 3.1.10 Tracing Using the simplifier effectively may take a bit of experimentation. Set the Proof General flag Isabelle > Settings > Trace Simplifier to get a better idea of what is going on: 3.1 Simplification 33 lemma "rev [a] = []" apply(simp) produces the following trace in Proof General’s Trace buffer: Applying instance of rewrite rule "List.rev.simps_2": rev (?x1 # ?xs1) ≡rev ?xs1 @ [?x1] Rewriting: rev [a] ≡rev [] @ [a] Applying instance of rewrite rule "List.rev.simps_1": rev [] ≡[] Rewriting: rev [] ≡[] Applying instance of rewrite rule "List.append.append_Nil": [] @ ?y ≡?y Rewriting: [] @ [a] ≡[a] Applying instance of rewrite rule ?x2 # ?t1 = ?t1 ≡False Rewriting: [a] = [] ≡False The trace lists each rule being applied, both in its general form and the instance being used. The [i] in front (where above i is always 1) indicates that we are inside the ith invocation of the simplifier. Each attempt to apply a conditional rule shows the rule followed by the trace of the (recursive!) simplification of the conditions, the latter prefixed by [i + 1] instead of [i]. Another source of recursive invocations of the simplifier are proofs of arithmetic formulae. By default, recursive invocations are not shown, you must increase the trace depth via Isabelle > Settings > Trace Simplifier Depth. Many other hints about the simplifier’s actions may appear. In more complicated cases, the trace can be very lengthy. Thus it is ad-visable to reset the Trace Simplifier flag after having obtained the desired trace. Since this is easily forgotten (and may have the unpleasant effect of swamping the interface with trace information), here is how you can switch the trace on locally in a proof: using apply simp Within the current proof, all simplifications in subsequent proof steps will be traced, but the text reminds you to remove the using clause after it has done its job. 34 3. More Functional Programming 3.1.11 Finding Theorems Isabelle’s large database of proved theorems offers a powerful search engine. Its chief limitation is its restriction to the theories currently loaded. The search engine is started by clicking on Proof General’s Find icon. You specify your search textually in the input buffer at the bottom of the window. The simplest form of search finds theorems containing specified patterns. A pattern can be any term (even a single identifier). It may contain “ ”, a wildcard standing for any term. Here are some examples: length " # _ = _ # " " + " " ( - (::nat))" Specifying types, as shown in the last example, constrains searches involving overloaded operators. ! ! Always use “ ” rather than variable names: searching for "x + y" will usually not find any matching theorems because they would need to contain x and y literally. When searching for infix operators, do not just type in the symbol, such as +, but a proper term such as " + ". This remark applies to more complicated syntaxes, too. If you are looking for rewrite rules (possibly conditional) that could sim-plify some term, prefix the pattern with simp:. simp: " ( + )" This finds all equations—not just those with a simp attribute—whose con-clusion has the form _ ( + ) = . . . It only finds equations that can simplify the given pattern at the root, not somewhere inside: for example, equations of the form _ + _ = . . . do not match. You may also search for theorems by name—you merely need to specify a substring. For example, you could search for all commutativity theorems like this: name: comm This retrieves all theorems whose name contains comm. Search criteria can also be negated by prefixing them with “-”. For exam-ple, -name: List finds theorems whose name does not contain List. You can use this to exclude particular theories from the search: the long name of a theorem contains the name of the theory it comes from. 3.2 Induction Heuristics 35 Finallly, different search criteria can be combined arbitrarily. The effect is conjuctive: Find returns the theorems that satisfy all of the criteria. For example, " + " -" - " -simp: " ( + )" name: assoc looks for theorems containing plus but not minus, and which do not simplify _ ( + ) at the root, and whose name contains assoc. Further search criteria are explained in Sect. 5.14. Proof General keeps a history of all your search expressions. If you click on Find, you can use the arrow keys to scroll through previous searches and just modify them. This saves you having to type in lengthy expressions again and again. 3.2 Induction Heuristics The purpose of this section is to illustrate some simple heuristics for inductive proofs. The first one we have already mentioned in our initial example: Theorems about recursive functions are proved by induction. In case the function has more than one argument Do induction on argument number i if the function is defined by recursion in argument number i. When we look at the proof of (xs@ys) @ zs = xs @ (ys@zs) in Sect. 2.3 we find – @ is recursive in the first argument – xs occurs only as the first argument of @ – both ys and zs occur at least once as the second argument of @ Hence it is natural to perform induction on xs. The key heuristic, and the main point of this section, is to generalize the goal before induction. The reason is simple: if the goal is too specific, the induction hypothesis is too weak to allow the induction step to go through. Let us illustrate the idea with an example. Function rev has quadratic worst-case running time because it calls func-tion @ for each element of the list and @ is linear in its first argument. A linear time version of rev reqires an extra argument where the result is accumulated gradually, using only #: primrec itrev :: "'a list ⇒'a list ⇒'a list" where "itrev [] ys = ys" \| "itrev (x#xs) ys = itrev xs (x#ys)" The behaviour of itrev is simple: it reverses its first argument by stacking its elements onto the second argument, and returning that second argument 36 3. More Functional Programming when the first one becomes empty. Note that itrev is tail-recursive: it can be compiled into a loop. Naturally, we would like to show that itrev does indeed reverse its first argument provided the second one is empty: lemma "itrev xs [] = rev xs" There is no choice as to the induction variable, and we immediately simplify: apply(induct_tac xs, simp_all) Unfortunately, this attempt does not prove the induction step: 1. Va list. itrev list [] = rev list = ⇒itrev list [a] = rev list @ [a] The induction hypothesis is too weak. The fixed argument, [], prevents it from rewriting the conclusion. This example suggests a heuristic: Generalize goals for induction by replacing constants by variables. Of course one cannot do this naïvely: itrev xs ys = rev xs is just not true. The correct generalization is lemma "itrev xs ys = rev xs @ ys" If ys is replaced by [], the right-hand side simplifies to rev xs, as required. In this instance it was easy to guess the right generalization. Other situ-ations can require a good deal of creativity. Although we now have two variables, only xs is suitable for induction, and we repeat our proof attempt. Unfortunately, we are still not there: 1. Va list. itrev list ys = rev list @ ys = ⇒ itrev list (a # ys) = rev list @ a # ys The induction hypothesis is still too weak, but this time it takes no intuition to generalize: the problem is that ys is fixed throughout the subgoal, but the induction hypothesis needs to be applied with a # ys instead of ys. Hence we prove the theorem for all ys instead of a fixed one: lemma "∀ys. itrev xs ys = rev xs @ ys" This time induction on xs followed by simplification succeeds. This leads to another heuristic for generalization: Generalize goals for induction by universally quantifying all free vari-ables (except the induction variable itself!). This prevents trivial failures like the one above and does not affect the validity of the goal. However, this heuristic should not be applied blindly. It is not always required, and the additional quantifiers can complicate matters in some cases. The variables that should be quantified are typically those that change in recursive calls. 3.3 Case Study: Compiling Expressions 37 A final point worth mentioning is the orientation of the equation we just proved: the more complex notion (itrev) is on the left-hand side, the sim-pler one (rev) on the right-hand side. This constitutes another, albeit weak heuristic that is not restricted to induction: The right-hand side of an equation should (in some sense) be simpler than the left-hand side. This heuristic is tricky to apply because it is not obvious that rev xs @ ys is simpler than itrev xs ys. But see what happens if you try to prove rev xs @ ys = itrev xs ys! If you have tried these heuristics and still find your induction does not go through, and no obvious lemma suggests itself, you may need to generalize your proposition even further. This requires insight into the problem at hand and is beyond simple rules of thumb. Additionally, you can read Sect. 9.2 to learn about some advanced techniques for inductive proofs. Exercise 3.2.1 Define the following addition function primrec add :: "nat ⇒nat ⇒nat" where "add m 0 = m" \| "add m (Suc n) = add (Suc m) n" and prove lemma "add m n = m+n" Exercise 3.2.2 In Exercise 2.5.1 we defined a function flatten from trees to lists. The straightforward version of flatten is based on @ and is thus, like rev, quadratic. A linear time version of flatten again reqires an extra argument, the accumulator. Define flatten2 :: "'a tree ⇒'a list ⇒'a list" and prove lemma "flatten2 t [] = flatten t" 3.3 Case Study: Compiling Expressions The task is to develop a compiler from a generic type of expressions (built from variables, constants and binary operations) to a stack machine. This generic type of expressions is a generalization of the boolean expressions in Sect. 2.5.6. This time we do not commit ourselves to a particular type of variables or values but make them type parameters. Neither is there a fixed set of binary operations: instead the expression contains the appropriate function itself. type synonym 'v binop = "'v ⇒'v ⇒'v" datatype (dead 'a, 'v) expr = Cex 'v 38 3. More Functional Programming \| Vex 'a \| Bex "'v binop" "('a,'v)expr" "('a,'v)expr" The three constructors represent constants, variables and the application of a binary operation to two subexpressions. The value of an expression with respect to an environment that maps variables to values is easily defined: primrec "value" :: "('a,'v)expr ⇒('a ⇒'v) ⇒'v" where "value (Cex v) env = v" \| "value (Vex a) env = env a" \| "value (Bex f e1 e2) env = f (value e1 env) (value e2 env)" The stack machine has three instructions: load a constant value onto the stack, load the contents of an address onto the stack, and apply a binary operation to the two topmost elements of the stack, replacing them by the result. As for expr, addresses and values are type parameters: datatype (dead 'a, 'v) instr = Const 'v \| Load 'a \| Apply "'v binop" The execution of the stack machine is modelled by a function exec that takes a list of instructions, a store (modelled as a function from addresses to values, just like the environment for evaluating expressions), and a stack (modelled as a list) of values, and returns the stack at the end of the execution — the store remains unchanged: primrec exec :: "('a,'v)instr list ⇒('a⇒'v) ⇒'v list ⇒'v list" where "exec [] s vs = vs" \| "exec (i#is) s vs = (case i of Const v ⇒exec is s (v#vs) \| Load a ⇒exec is s ((s a)#vs) \| Apply f ⇒exec is s ((f (hd vs) (hd(tl vs)))#(tl(tl vs))))" Recall that hd and tl return the first element and the remainder of a list. Because all functions are total, hd is defined even for the empty list, although we do not know what the result is. Thus our model of the machine always terminates properly, although the definition above does not tell us much about the result in situations where Apply was executed with fewer than two elements on the stack. The compiler is a function from expressions to a list of instructions. Its definition is obvious: primrec compile :: "('a,'v)expr ⇒('a,'v)instr list" where "compile (Cex v) = [Const v]" \| "compile (Vex a) = [Load a]" \| "compile (Bex f e1 e2) = (compile e2) @ (compile e1) @ [Apply f]" Now we have to prove the correctness of the compiler, i.e. that the exe-cution of a compiled expression results in the value of the expression: theorem "exec (compile e) s [] = [value e s]" 3.4 Advanced Datatypes 39 This theorem needs to be generalized: theorem "∀vs. exec (compile e) s vs = (value e s) # vs" It will be proved by induction on e followed by simplification. First, we must prove a lemma about executing the concatenation of two instruction sequences: lemma exec_app[simp]: "∀vs. exec (xs@ys) s vs = exec ys s (exec xs s vs)" This requires induction on xs and ordinary simplification for the base cases. In the induction step, simplification leaves us with a formula that contains two case-expressions over instructions. Thus we add automatic case splitting, which finishes the proof: apply(induct_tac xs, simp, simp split: instr.split) Note that because both simp_all and auto perform simplification, they can be modified in the same way as simp. Thus the proof can be rewritten as apply(induct_tac xs, simp_all split: instr.split) Although this is more compact, it is less clear for the reader of the proof. We could now go back and prove exec (compile e) s [] = [value e s] merely by simplification with the generalized version we just proved. How-ever, this is unnecessary because the generalized version fully subsumes its instance. 3.4 Advanced Datatypes This section presents advanced forms of datatypes: mutual and nested re-cursion. A series of examples will culminate in a treatment of the trie data structure. 3.4.1 Mutual Recursion Sometimes it is necessary to define two datatypes that depend on each other. This is called mutual recursion. As an example consider a language of arithmetic and boolean expressions where – arithmetic expressions contain boolean expressions because there are con-ditional expressions like “if m < n then n −m else m −n”, and – boolean expressions contain arithmetic expressions because of comparisons like “m < n”. In Isabelle this becomes datatype 'a aexp = IF "'a bexp" "'a aexp" "'a aexp" \| Sum "'a aexp" "'a aexp" 40 3. More Functional Programming \| Diff "'a aexp" "'a aexp" \| Var 'a \| Num nat and 'a bexp = Less "'a aexp" "'a aexp" \| And "'a bexp" "'a bexp" \| Neg "'a bexp" Type aexp is similar to expr in Sect. 3.3, except that we have added an IF constructor, fixed the values to be of type nat and declared the two binary operations Sum and Diff. Boolean expressions can be arithmetic comparisons, conjunctions and negations. The semantics is given by two evaluation func-tions: primrec evala :: "'a aexp ⇒('a ⇒nat) ⇒nat" and evalb :: "'a bexp ⇒('a ⇒nat) ⇒bool" where "evala (IF b a1 a2) env = (if evalb b env then evala a1 env else evala a2 env)" \| "evala (Sum a1 a2) env = evala a1 env + evala a2 env" \| "evala (Diff a1 a2) env = evala a1 env - evala a2 env" \| "evala (Var v) env = env v" \| "evala (Num n) env = n" \| "evalb (Less a1 a2) env = (evala a1 env < evala a2 env)" \| "evalb (And b1 b2) env = (evalb b1 env ∧evalb b2 env)" \| "evalb (Neg b) env = (¬ evalb b env)" Both take an expression and an environment (a mapping from variables 'a to values nat) and return its arithmetic/boolean value. Since the datatypes are mutually recursive, so are functions that operate on them. Hence they need to be defined in a single primrec section. Notice the and separating the declarations of evala and evalb. Their defining equations need not be split into two groups; the empty line is purely for readability. In the same fashion we also define two functions that perform substitution: primrec substa :: "('a ⇒'b aexp) ⇒'a aexp ⇒'b aexp" and substb :: "('a ⇒'b aexp) ⇒'a bexp ⇒'b bexp" where "substa s (IF b a1 a2) = IF (substb s b) (substa s a1) (substa s a2)" \| "substa s (Sum a1 a2) = Sum (substa s a1) (substa s a2)" \| "substa s (Diff a1 a2) = Diff (substa s a1) (substa s a2)" \| "substa s (Var v) = s v" \| "substa s (Num n) = Num n" \| "substb s (Less a1 a2) = Less (substa s a1) (substa s a2)" \| "substb s (And b1 b2) = And (substb s b1) (substb s b2)" \| "substb s (Neg b) = Neg (substb s b)" Their first argument is a function mapping variables to expressions, the sub-stitution. It is applied to all variables in the second argument. As a result, the type of variables in the expression may change from 'a to 'b. Note that there are only arithmetic and no boolean variables. Now we can prove a fundamental theorem about the interaction between evaluation and substitution: applying a substitution s to an expression a 3.4 Advanced Datatypes 41 and evaluating the result in an environment env yields the same result as evaluation a in the environment that maps every variable x to the value of s(x) under env. If you try to prove this separately for arithmetic or boolean expressions (by induction), you find that you always need the other theorem in the induction step. Therefore you need to state and prove both theorems simultaneously: lemma "evala (substa s a) env = evala a (λx. evala (s x) env) ∧ evalb (substb s b) env = evalb b (λx. evala (s x) env)" apply(induct_tac a and b) The resulting 8 goals (one for each constructor) are proved in one fell swoop: apply simp_all In general, given n mutually recursive datatypes τ1, …, τn, an inductive proof expects a goal of the form P1(x1) ∧· · · ∧Pn(xn) where each variable xi is of type τi. Induction is started by apply(induct_tac x1 and … and xn) Exercise 3.4.1 Define a function norma of type 'a aexp ⇒'a aexp that re-places IFs with complex boolean conditions by nested IFs; it should eliminate the constructors And and Neg, leaving only Less. Prove that norma preserves the value of an expression and that the result of norma is really normal, i.e. no more Ands and Negs occur in it. (Hint: proceed as in Sect. 2.5.6 and read the discussion of type annotations following lemma subst_id below). 3.4.2 Nested Recursion So far, all datatypes had the property that on the right-hand side of their definition they occurred only at the top-level: directly below a constructor. Now we consider nested recursion, where the recursive datatype occurs nested in some other datatype (but not inside itself!). Consider the following model of terms where function symbols can be applied to a list of arguments: datatype ('v,'f)"term" = Var 'v \| App 'f "('v,'f)term list" Note that we need to quote term on the left to avoid confusion with the Isabelle command term. Parameter 'v is the type of variables and 'f the type of function symbols. A mathematical term like f (x, g(y)) becomes App f [Var x, App g [Var y]], where f, g, x, y are suitable values, e.g. numbers or strings. What complicates the definition of term is the nested occurrence of term inside list on the right-hand side. In principle, nested recursion can be elim-inated in favour of mutual recursion by unfolding the offending datatypes, here list. The result for term would be something like 42 3. More Functional Programming datatype ('v,'f)"term" = Var 'v \| App 'f "('v,'f)term_list" and ('v,'f)term_list = Nil \| Cons "('v,'f)term" "('v,'f)term_list" Although we do not recommend this unfolding to the user, it shows how to simulate nested recursion by mutual recursion. Now we return to the initial definition of term using nested recursion. Let us define a substitution function on terms. Because terms involve term lists, we need to define two substitution functions simultaneously: primrec subst :: "('v⇒('v,'f)term) ⇒('v,'f)term ⇒('v,'f)term" and substs:: "('v⇒('v,'f)term) ⇒('v,'f)term list ⇒('v,'f)term list" where "subst s (Var x) = s x" \| subst_App: "subst s (App f ts) = App f (substs s ts)" \| "substs s [] = []" \| "substs s (t # ts) = subst s t # substs s ts" Individual equations in a primrec definition may be named as shown for subst_App. The significance of this device will become apparent below. Similarly, when proving a statement about terms inductively, we need to prove a related statement about term lists simultaneously. For example, the fact that the identity substitution does not change a term needs to be strengthened and proved as follows: lemma subst_id: "subst Var t = (t ::('v,'f)term) ∧ substs Var ts = (ts::('v,'f)term list)" apply(induct_tac t and ts rule: subst.induct substs.induct, simp_all) done Note that Var is the identity substitution because by definition it leaves vari-ables unchanged: subst Var (Var x) = Var x. Note also that the type anno-tations are necessary because otherwise there is nothing in the goal to enforce that both halves of the goal talk about the same type parameters ('v,'f). As a result, induction would fail because the two halves of the goal would be unrelated. Exercise 3.4.2 The fact that substitution distributes over composition can be expressed roughly as follows: subst (f ◦g) t = subst f (subst g t) Correct this statement (you will find that it does not type-check), strengthen it, and prove it. (Note: ◦is function composition; its definition is found in theorem o_def). Exercise 3.4.3 Define a function trev of type ('v, 'f) Nested.term ⇒ ('v, 'f) Nested.term that recursively reverses the order of arguments of all function symbols in a term. Prove that trev (trev t) = t. 3.4 Advanced Datatypes 43 The experienced functional programmer may feel that our definition of subst is too complicated in that substs is unnecessary. The App-case can be defined directly as subst s (App f ts) = App f (map (subst s) ts) where map is the standard list function such that map f [x1,...,xn] = [f x1,...,f xn]. This is true, but Isabelle insists on the conjunctive format. Fortunately, we can easily prove that the suggested equation holds: lemma [simp]: "subst s (App f ts) = App f (map (subst s) ts)" apply(induct_tac ts, simp_all) done What is more, we can now disable the old defining equation as a simplification rule: declare subst_App [simp del] The advantage is that now we have replaced substs by map, we can profit from the large number of pre-proved lemmas about map. Unfortunately, inductive proofs about type term are still awkward because they expect a conjunc-tion. One could derive a new induction principle as well (see Sect. 9.2.3), but simpler is to stop using primrec and to define functions with fun instead. Simple uses of fun are described in Sect. 3.5 below. Advanced applications, including functions over nested datatypes like term, are discussed in a sepa-rate tutorial . Of course, you may also combine mutual and nested recursion of data-types. For example, constructor Sum in Sect. 3.4.1 could take a list of expres-sions as its argument: Sum "'a aexp list". 3.4.3 The Limits of Nested Recursion How far can we push nested recursion? By the unfolding argument above, we can reduce nested to mutual recursion provided the nested recursion only involves previously defined datatypes. This does not include functions: datatype t = C "t ⇒bool" This declaration is a real can of worms. In HOL it must be ruled out because it requires a type t such that t and its power set t ⇒bool have the same cardinality — an impossibility. For the same reason it is not possible to allow recursion involving the type t set, which is isomorphic to t ⇒bool. Fortunately, a limited form of recursion involving function spaces is per-mitted: the recursive type may occur on the right of a function arrow, but never on the left. Hence the above can of worms is ruled out but the following example of a potentially infinitely branching tree is accepted: datatype (dead 'a,'i) bigtree = Tip \| Br 'a "'i ⇒('a,'i)bigtree" 44 3. More Functional Programming Parameter 'a is the type of values stored in the Branches of the tree, whereas 'i is the index type over which the tree branches. If 'i is instantiated to bool, the result is a binary tree; if it is instantiated to nat, we have an infinitely branching tree because each node has as many subtrees as there are natural numbers. How can we possibly write down such a tree? Using functional notation! For example, the term Br 0 (λi. Br i (λn. Tip)) of type (nat, nat) bigtree is the tree whose root is labeled with 0 and whose ith subtree is labeled with i and has merely Tips as further subtrees. Function map_bt applies a function to all labels in a bigtree: primrec map_bt :: "('a ⇒'b) ⇒('a,'i)bigtree ⇒('b,'i)bigtree" where "map_bt f Tip = Tip" \| "map_bt f (Br a F) = Br (f a) (λi. map_bt f (F i))" This is a valid primrec definition because the recursive calls of map_bt involve only subtrees of F, which is itself a subterm of the left-hand side. Thus termi-nation is assured. The seasoned functional programmer might try expressing λi. map_bt f (F i) as map_bt f ◦F, which Isabelle however will reject. Ap-plying map_bt to only one of its arguments makes the termination proof less obvious. The following lemma has a simple proof by induction: lemma "map_bt (g o f) T = map_bt g (map_bt f T)" apply(induct_tac T, simp_all) done Because of the function type, the proof state after induction looks unusual. Notice the quantified induction hypothesis: 1. map_bt (g ◦f) Tip = map_bt g (map_bt f Tip) 2. Vx1 F. (Vx2a. x2a ∈range F = ⇒ map_bt (g ◦f) x2a = map_bt g (map_bt f x2a)) = ⇒ map_bt (g ◦f) (Br x1 F) = map_bt g (map_bt f (Br x1 F)) If you need nested recursion on the left of a function arrow, there are alternatives to pure HOL. In the Logic for Computable Functions (LCF), types like datatype lam = C "lam →lam" do indeed make sense . Note the different arrow, →instead of ⇒, express-ing the type of continuous functions. There is even a version of LCF on top of HOL, called HOLCF . 3.4.4 Case Study: Tries Tries are a classic search tree data structure for fast indexing with strings. Figure 3.1 gives a graphical example of a trie containing the words “all”, “an”, 3.4 Advanced Datatypes 45 “ape”, “can”, “car” and “cat”. When searching a string in a trie, the letters of the string are examined sequentially. Each letter determines which subtrie to search next. In this case study we model tries as a datatype, define a lookup and an update function, and prove that they behave as expected. l e n r t QQ Q l n p a QQ Q a c QQQ Q Fig. 3.1. A Sample Trie Proper tries associate some value with each string. Since the information is stored only in the final node associated with the string, many nodes do not carry any value. This distinction is modeled with the help of the predefined datatype option (see Sect. 2.6.3). To minimize running time, each node of a trie should contain an array that maps letters to subtries. We have chosen a representation where the subtries are held in an association list, i.e. a list of (letter,trie) pairs. Abstracting over the alphabet 'a and the values 'v we define a trie as follows: datatype ('a,'v)trie = Trie "'v option" "('a ('a,'v)trie)list" The first component is the optional value, the second component the associa-tion list of subtries. This is an example of nested recursion involving products, which is fine because products are datatypes as well. We define two selector functions: primrec "value" :: "('a,'v)trie ⇒'v option" where "value(Trie ov al) = ov" primrec alist :: "('a,'v)trie ⇒('a ('a,'v)trie)list" where "alist(Trie ov al) = al" Association lists come with a generic lookup function. Its result involves type option because a lookup can fail: primrec assoc :: "('key 'val)list ⇒'key ⇒'val option" where "assoc [] x = None" \| "assoc (p#ps) x = (let (a,b) = p in if a=x then Some b else assoc ps x)" Now we can define the lookup function for tries. It descends into the trie examining the letters of the search string one by one. As recursion on lists is simpler than on tries, let us express this as primitive recursion on the search string argument: 46 3. More Functional Programming primrec lookup :: "('a,'v)trie ⇒'a list ⇒'v option" where "lookup t [] = value t" \| "lookup t (a#as) = (case assoc (alist t) a of None ⇒None \| Some at ⇒lookup at as)" As a first simple property we prove that looking up a string in the empty trie Trie None [] always returns None. The proof merely distinguishes the two cases whether the search string is empty or not: lemma [simp]: "lookup (Trie None []) as = None" apply(case_tac as, simp_all) done Things begin to get interesting with the definition of an update func-tion that adds a new (string, value) pair to a trie, overwriting the old value associated with that string: primrec update:: "('a,'v)trie ⇒'a list ⇒'v ⇒('a,'v)trie" where "update t [] v = Trie (Some v) (alist t)" \| "update t (a#as) v = (let tt = (case assoc (alist t) a of None ⇒Trie None [] \| Some at ⇒at) in Trie (value t) ((a,update tt as v) # alist t))" The base case is obvious. In the recursive case the subtrie tt associated with the first letter a is extracted, recursively updated, and then placed in front of the association list. The old subtrie associated with a is still in the association list but no longer accessible via assoc. Clearly, there is room here for optimizations! Before we start on any proofs about update we tell the simplifier to expand all lets and to split all case-constructs over options: declare Let_def[simp] option.split[split] The reason becomes clear when looking (probably after a failed proof at-tempt) at the body of update: it contains both let and a case distinction over type option. Our main goal is to prove the correct interaction of update and lookup: theorem "∀t v bs. lookup (update t as v) bs = (if as=bs then Some v else lookup t bs)" Our plan is to induct on as; hence the remaining variables are quantified. From the definitions it is clear that induction on either as or bs is required. The choice of as is guided by the intuition that simplification of lookup might be easier if update has already been simplified, which can only happen if as is instantiated. The start of the proof is conventional: apply(induct_tac as, auto) Unfortunately, this time we are left with three intimidating looking subgoals: 3.5 Total Recursive Functions: fun 47 1. … = ⇒lookup … bs = lookup t bs 2. … = ⇒lookup … bs = lookup t bs 3. … = ⇒lookup … bs = lookup t bs Clearly, if we want to make headway we have to instantiate bs as well now. It turns out that instead of induction, case distinction suffices: apply(case_tac[!] bs, auto) done All methods ending in tac take an optional first argument that specifies the range of subgoals they are applied to, where [!] means all subgoals, i.e. [1-3] in our case. Individual subgoal numbers, e.g. are also allowed. This proof may look surprisingly straightforward. However, note that this comes at a cost: the proof script is unreadable because the intermediate proof states are invisible, and we rely on the (possibly brittle) magic of auto (simp_all will not do — try it) to split the subgoals of the induction up in such a way that case distinction on bs makes sense and solves the proof. Exercise 3.4.4 Modify update (and its type) such that it allows both in-sertion and deletion of entries with a single function. Prove the correspond-ing version of the main theorem above. Optimize your function such that it shrinks tries after deletion if possible. Exercise 3.4.5 Write an improved version of update that does not suffer from the space leak (pointed out above) caused by not deleting overwritten entries from the association list. Prove the main theorem for your improved update. Exercise 3.4.6 Conceptually, each node contains a mapping from letters to optional subtries. Above we have implemented this by means of an association list. Replay the development replacing ('a × ('a, 'v) trie) list with 'a ⇒('a, 'v) trie option. 3.5 Total Recursive Functions: fun Although many total functions have a natural primitive recursive definition, this is not always the case. Arbitrary total recursive functions can be defined by means of fun: you can use full pattern matching, recursion need not involve datatypes, and termination is proved by showing that the arguments of all recursive calls are smaller in a suitable sense. In this section we restrict our-selves to functions where Isabelle can prove termination automatically. More advanced function definitions, including user supplied termination proofs, nested recursion and partiality, are discussed in a separate tutorial . 48 3. More Functional Programming 3.5.1 Definition Here is a simple example, the Fibonacci function: fun fib :: "nat ⇒nat" where "fib 0 = 0" \| "fib (Suc 0) = 1" \| "fib (Suc(Suc x)) = fib x + fib (Suc x)" This resembles ordinary functional programming languages. Note the oblig-atory where and \|. Command fun declares and defines the function in one go. Isabelle establishes termination automatically because fib’s argument decreases in every recursive call. Slightly more interesting is the insertion of a fixed element between any two elements of a list: fun sep :: "'a ⇒'a list ⇒'a list" where "sep a [] = []" \| "sep a [x] = [x]" \| "sep a (x#y#zs) = x # a # sep a (y#zs)" This time the length of the list decreases with the recursive call; the first argument is irrelevant for termination. Pattern matching need not be exhaustive and may employ wildcards: fun last :: "'a list ⇒'a" where "last [x] = x" \| "last (#y#zs) = last (y#zs)" Overlapping patterns are disambiguated by taking the order of equations into account, just as in functional programming: fun sep1 :: "'a ⇒'a list ⇒'a list" where "sep1 a (x#y#zs) = x # a # sep1 a (y#zs)" \| "sep1 _ xs = xs" To guarantee that the second equation can only be applied if the first one does not match, Isabelle internally replaces the second equation by the two possibilities that are left: sep1 a [] = [] and sep1 a [x] = [x]. Thus the functions sep and sep1 are identical. Because of its pattern matching syntax, fun is also useful for the definition of non-recursive functions: fun swap12 :: "'a list ⇒'a list" where "swap12 (x#y#zs) = y#x#zs" \| "swap12 zs = zs" After a function f has been defined via fun, its defining equations (or vari-ants derived from them) are available under the name f .simps as theorems. For example, look (via thm) at sep.simps and sep1.simps to see that they define the same function. What is more, those equations are automatically declared as simplification rules. 3.5 Total Recursive Functions: fun 49 3.5.2 Termination Isabelle’s automatic termination prover for fun has a fixed notion of the size (of type nat) of an argument. The size of a natural number is the number itself. The size of a list is its length. For the general case see Sect. 2.5.2. A recursive function is accepted if fun can show that the size of one fixed argument becomes smaller with each recursive call. More generally, fun allows any lexicographic combination of size measures in case there are multiple arguments. For example, the following version of Ackermann’s function is accepted: fun ack2 :: "nat ⇒nat ⇒nat" where "ack2 n 0 = Suc n" \| "ack2 0 (Suc m) = ack2 (Suc 0) m" \| "ack2 (Suc n) (Suc m) = ack2 (ack2 n (Suc m)) m" The order of arguments has no influence on whether fun can prove ter-mination of a function. For more details see elsewhere . 3.5.3 Simplification Upon a successful termination proof, the recursion equations become simpli-fication rules, just as with primrec. In most cases this works fine, but there is a subtle problem that must be mentioned: simplification may not terminate because of automatic splitting of if. Let us look at an example: fun gcd :: "nat ⇒nat ⇒nat" where "gcd m n = (if n=0 then m else gcd n (m mod n))" The second argument decreases with each recursive call. The termination condition n ̸= 0 = ⇒m mod n < n is proved automatically because it is already present as a lemma in HOL. Thus the recursion equation becomes a simplification rule. Of course the equation is nonterminating if we are allowed to unfold the recursive call inside the else branch, which is why programming languages and our simplifier don’t do that. Unfortunately the simplifier does something else that leads to the same problem: it splits each if-expression unless its condition simplifies to True or False. For example, simplification reduces fun0.gcd m n = k in one step to (if n = 0 then m else fun0.gcd n (m mod n)) = k where the condition cannot be reduced further, and splitting leads to (n = 0 − →m = k) ∧(n ̸= 0 − →fun0.gcd n (m mod n) = k) 50 3. More Functional Programming Since the recursive call fun0.gcd n (m mod n) is no longer protected by an if, it is unfolded again, which leads to an infinite chain of simplification steps. Fortunately, this problem can be avoided in many different ways. The most radical solution is to disable the offending theorem if_split, as shown in Sect. 3.1.9. However, we do not recommend this approach: you will often have to invoke the rule explicitly when if is involved. If possible, the definition should be given by pattern matching on the left rather than if on the right. In the case of fun0.gcd the following alternative definition suggests itself: fun gcd1 :: "nat ⇒nat ⇒nat" where "gcd1 m 0 = m" \| "gcd1 m n = gcd1 n (m mod n)" The order of equations is important: it hides the side condition n ̸= 0. Un-fortunately, not all conditionals can be expressed by pattern matching. A simple alternative is to replace if by case, which is also available for bool and is not split automatically: fun gcd2 :: "nat ⇒nat ⇒nat" where "gcd2 m n = (case n=0 of True ⇒m \| False ⇒gcd2 n (m mod n))" This is probably the neatest solution next to pattern matching, and it is always available. A final alternative is to replace the offending simplification rules by de-rived conditional ones. For fun0.gcd it means we have to prove these lemmas: lemma [simp]: "gcd m 0 = m" apply(simp) done lemma [simp]: "n ̸= 0 = ⇒gcd m n = gcd n (m mod n)" apply(simp) done Simplification terminates for these proofs because the condition of the if simplifies to True or False. Now we can disable the original simplification rule: declare gcd.simps [simp del] 3.5.4 Induction Having defined a function we might like to prove something about it. Since the function is recursive, the natural proof principle is again induction. But this time the structural form of induction that comes with datatypes is unlikely to work well — otherwise we could have defined the function by primrec. Therefore fun automatically proves a suitable induction rule f .induct that follows the recursion pattern of the particular function f . We call this re-cursion induction. Roughly speaking, it requires you to prove for each fun 3.5 Total Recursive Functions: fun 51 equation that the property you are trying to establish holds for the left-hand side provided it holds for all recursive calls on the right-hand side. Here is a simple example involving the predefined map functional on lists: lemma "map f (sep x xs) = sep (f x) (map f xs)" Note that map f xs is the result of applying f to all elements of xs. We prove this lemma by recursion induction over sep: apply(induct_tac x xs rule: sep.induct) The resulting proof state has three subgoals corresponding to the three clauses for sep: 1. Va. map f (sep a []) = sep (f a) (map f []) 2. Va x. map f (sep a [x]) = sep (f a) (map f [x]) 3. Va x y zs. map f (sep a (y # zs)) = sep (f a) (map f (y # zs)) = ⇒ map f (sep a (x # y # zs)) = sep (f a) (map f (x # y # zs)) The rest is pure simplification: apply simp_all done The proof goes smoothly because the induction rule follows the recursion of sep. Try proving the above lemma by structural induction, and you find that you need an additional case distinction. In general, the format of invoking recursion induction is apply(induct_tac x1 . . . xn rule: f .induct) where x1 . . . xn is a list of free variables in the subgoal and f the name of a function that takes n arguments. Usually the subgoal will contain the term fx1 . . . xn but this need not be the case. The induction rules do not mention f at all. Here is sep.induct: [ [ Va. P a []; Va x. P a [x]; Va x y zs. P a (y # zs) = ⇒P a (x # y # zs)] ] = ⇒P u v It merely says that in order to prove a property P of u and v you need to prove it for the three cases where v is the empty list, the singleton list, and the list with at least two elements. The final case has an induction hypothesis: you may assume that P holds for the tail of that list. 4. Presenting Theories By now the reader should have become sufficiently acquainted with elemen-tary theory development in Isabelle/HOL. The following interlude describes how to present theories in a typographically pleasing manner. Isabelle pro-vides a rich infrastructure for concrete syntax of the underlying λ-calculus language (see Sect. 4.1), as well as document preparation of theory texts based on existing PDF-L A T EX technology (see Sect. 4.2). As pointed out by Leibniz more than 300 years ago, notions are in princi-ple more important than notations, but suggestive textual representation of ideas is vital to reduce the mental effort to comprehend and apply them. 4.1 Concrete Syntax The core concept of Isabelle’s framework for concrete syntax is that of mixfix annotations. Associated with any kind of constant declaration, mixfixes affect both the grammar productions for the parser and output templates for the pretty printer. In full generality, parser and pretty printer configuration is a subtle af-fair . Your syntax specifications need to interact properly with the existing setup of Isabelle/Pure and Isabelle/HOL. To avoid creating ambiguities with existing elements, it is particularly important to give new syntactic constructs the right precedence. Below we introduce a few simple syntax declaration forms that already cover many common situations fairly well. 4.1.1 Infix Annotations Syntax annotations may be included wherever constants are declared, such as definition and primrec — and also datatype, which declares constructor operations. Type-constructors may be annotated as well, although this is less frequently encountered in practice (the infix type × comes to mind). Infix declarations provide a useful special case of mixfixes. The following example of the exclusive-or operation on boolean values illustrates typical infix declarations. 54 4. Presenting Theories definition xor :: "bool ⇒bool ⇒bool" (infixl "[+]" 60) where "A [+] B ≡(A ∧¬ B) ∨(¬ A ∧B)" Now xor A B and A [+] B refer to the same expression internally. Any curried function with at least two arguments may be given infix syntax. For partial applications with fewer than two operands, the operator is enclosed in paren-theses. For instance, xor without arguments is represented as ([+]); together with ordinary function application, this turns xor A into ([+]) A. The keyword infixl seen above specifies an infix operator that is nested to the left: in iterated applications the more complex expression appears on the left-hand side, and A [+] B [+] C stands for (A [+] B) [+] C. Similarly, infixr means nesting to the right, reading A [+] B [+] C as A [+] (B [+] C). A non-oriented declaration via infix would render A [+] B [+] C illegal, but demand explicit parentheses to indicate the intended grouping. The string [+] in our annotation refers to the concrete syntax to repre-sent the operator (a literal token), while the number 60 determines the prece-dence of the construct: the syntactic priorities of the arguments and result. Isabelle/HOL already uses up many popular combinations of ASCII symbols for its own use, including both + and ++. Longer character combinations are more likely to be still available for user extensions, such as our [+]. Operator precedences have a range of 0–1000. Very low or high priorities are reserved for the meta-logic. HOL syntax mainly uses the range of 10–100: the equality infix = is centered at 50; logical connectives (like ∨and ∧) are below 50; algebraic ones (like + and ) are above 50. User syntax should strive to coexist with common HOL forms, or use the mostly unused range 100–900. 4.1.2 Mathematical Symbols Concrete syntax based on ASCII characters has inherent limitations. Math-ematical notation demands a larger repertoire of glyphs. Several standards of extended character sets have been proposed over decades, but none has become universally available so far. Isabelle has its own notion of symbols as the smallest entities of source text, without referring to internal encodings. There are three kinds of such “generalized characters”: 1. 7-bit ASCII characters 2. named symbols: \ 3. named control symbols: \<^ident> Here ident is any sequence of letters. This results in an infinite store of symbols, whose interpretation is left to further front-end tools. For example, the Isabelle document processor (see Sect. 4.2) display the \ symbol as ∀. A list of standard Isabelle symbols is given in . You may introduce your own interpretation of further symbols by configuring the appropriate front-end tool accordingly, e.g. by defining certain L AT EX macros (see also 4.1 Concrete Syntax 55 Sect. 4.2.4). There are also a few predefined control symbols, such as \<^sub> and \<^sup> for sub- and superscript of the subsequent printable symbol, respectively. For example, A\<^sup>\, is output as A⋆. A number of symbols are considered letters by the Isabelle lexer and can be used as part of identifiers. These are the greek letters α (\), β (\), etc. (excluding λ), special letters like A (\) and A (\). Moreover the control symbol \<^sub> may be used to subscript a single letter or digit in the trailing part of an identifier. This means that the input \\\<^sub>1. \\<^sub>1 = \\<^sub>\ is recognized as the term ∀α1. α1 = ΠA by Isabelle. Replacing our previous definition of xor by the following specifies an Isa-belle symbol for the new operator: definition xor :: "bool ⇒bool ⇒bool" (infixl "⊕" 60) where "A ⊕B ≡(A ∧¬ B) ∨(¬ A ∧B)" It is possible to provide alternative syntax forms through the print mode concept . By convention, the mode of “xsymbols” is enabled whenever Proof General’s X-Symbol mode or L A T EX output is active. Now consider the following hybrid declaration of xor: definition xor :: "bool ⇒bool ⇒bool" (infixl "[+]" 60) where "A [+] B ≡(A ∧¬ B) ∨(¬ A ∧B)" notation (xsymbols) xor (infixl "⊕" 60) The notation command associates a mixfix annotation with a known con-stant. The print mode specification, here (xsymbols), is optional. We may now write A [+] B or A ⊕B in input, while output uses the nicer syntax of xsymbols whenever that print mode is active. Such an arrange-ment is particularly useful for interactive development, where users may type ASCII text and see mathematical symbols displayed during proofs. 4.1.3 Prefix Annotations Prefix syntax annotations are another form of mixfixes , without any template arguments or priorities — just some literal syntax. The following example associates common symbols with the constructors of a datatype. datatype currency = Euro nat ("e") \| Pounds nat ("£") \| Yen nat ("¥") \| Dollar nat ("$") Here the mixfix annotations on the rightmost column happen to consist of a single Isabelle symbol each: \, \, \, and $. Recall that a constructor like Euro actually is a function nat ⇒currency. The expression 56 4. Presenting Theories Euro 10 will be printed as e 10; only the head of the application is subject to our concrete syntax. This rather simple form already achieves conformance with notational standards of the European Commission. Prefix syntax works the same way for other commands that introduce new constants, e.g. primrec. 4.1.4 Abbreviations Mixfix syntax annotations merely decorate particular constant application forms with concrete syntax, for instance replacing xor A B by A ⊕B. Occa-sionally, the relationship between some piece of notation and its internal form is more complicated. Here we need abbreviations. Command abbreviation introduces an uninterpreted notational constant as an abbreviation for a complex term. Abbreviations are unfolded upon parsing and re-introduced upon printing. This provides a simple mechanism for syntactic macros. A typical use of abbreviations is to introduce relational notation for mem-bership in a set of pairs, replacing (x, y) ∈sim by x ≈y. We assume that a constant sim of type ('a × 'a) set has been introduced at this point. abbreviation sim2 :: "'a ⇒'a ⇒bool" (infix "≈" 50) where "x ≈y ≡ (x, y) ∈sim" The given meta-equality is used as a rewrite rule after parsing (replacing x ≈y by (x,y) ∈sim) and before printing (turning (x,y) ∈sim back into x ≈y). The name of the dummy constant sim2 does not matter, as long as it is unique. Another common application of abbreviations is to provide variant ver-sions of fundamental relational expressions, such as ̸= for negated equalities. The following declaration stems from Isabelle/HOL itself: abbreviation not_equal :: "'a ⇒'a ⇒bool" (infixl "~=" 50) where "x ~= y ≡ ¬ (x = y)" notation (xsymbols) not_equal (infix "̸=" 50) The notation ̸= is introduced separately to restrict it to the xsymbols mode. Abbreviations are appropriate when the defined concept is a simple vari-ation on an existing one. But because of the automatic folding and unfolding of abbreviations, they do not scale up well to large hierarchies of concepts. Abbreviations do not replace definitions. Abbreviations are a simplified form of the general concept of syntax trans-lations; even heavier transformations may be written in ML . 4.2 Document Preparation 57 4.2 Document Preparation Isabelle/Isar is centered around the concept of formal proof documents. The outcome of a formal development effort is meant to be a human-readable record, presented as browsable PDF file or printed on paper. The overall document structure follows traditional mathematical articles, with sections, intermediate explanations, definitions, theorems and proofs. The Isabelle document preparation system essentially acts as a front-end to L AT EX. After checking specifications and proofs formally, the theory sources are turned into typesetting instructions in a schematic manner. This lets you write authentic reports on theory developments with little effort: many technical consistency checks are handled by the system. Here is an example to illustrate the idea of Isabelle document preparation. The following datatype definition of 'a bintree models binary trees with nodes being decorated by elements of type 'a. datatype 'a bintree = Leaf \| Branch 'a "'a bintree" "'a bintree" The datatype induction rule generated here is of the form [ [P Leaf; Vx1 x2 x3. [ [P x2; P x3] ] = ⇒P (Branch x1 x2 x3)] ] = ⇒P bintree The above document output has been produced as follows: text { The following datatype definition of @{text "'a bintree"} models binary trees with nodes being decorated by elements of type @{typ 'a}. } datatype 'a bintree = Leaf \| Branch 'a "'a bintree" "'a bintree" text { \noindent The datatype induction rule generated here is of the form @{thm [display] bintree.induct [no_vars]} } Here we have augmented the theory by formal comments (using text blocks), the informal parts may again refer to formal entities by means of “antiquota-tions” (such as @{text "'a bintree"} or @{typ 'a}), see also Sect. 4.2.3. 58 4. Presenting Theories 4.2.1 Isabelle Sessions In contrast to the highly interactive mode of Isabelle/Isar theory develop-ment, the document preparation stage essentially works in batch-mode. An Isabelle session consists of a collection of source files that may contribute to an output document. Each session is derived from a single parent, usually an object-logic image like HOL. This results in an overall tree structure, which is reflected by the output location in the file system (the root directory is determined by the Isabelle settings variable ISABELLE_BROWSER_INFO). The easiest way to manage Isabelle sessions is via isabelle mkroot (to generate an initial session source setup) and isabelle build (to run sessions as specified in the corresponding ROOT file). These Isabelle tools are described in further detail in the Isabelle System Manual . For example, a new session MySession (with document preparation) may be produced as follows: isabelle mkroot MySession isabelle build -D MySession The isabelle build job also informs about the file-system location of the ultimate results. The above dry run should be able to produce some document.pdf (with dummy title, empty table of contents etc.). Any failure at this stage usually indicates technical problems of the L AT EX installation. The detailed arrangement of the session sources is as follows. – Directory MySession holds the required theory files T1.thy, …, Tn.thy. – File MySession/ROOT specifies the session options and content, with decla-rations for all wanted theories; it is sufficient to specify the terminal nodes of the theory dependency graph. – Directory MySession/document contains everything required for the L A T EX stage; only root.tex needs to be provided initially. The latter file holds appropriate L AT EX code to commence a document (\documentclass etc.), and to include the generated files Ti.tex for each theory. Isabelle will generate a file session.tex holding L A T EX commands to include all generated theory output files in topologically sorted order, so \input{session} in the body of root.tex does the job in most situations. One may now start to populate the directory MySession and its ROOT file accordingly. The file MySession/document/root.tex should also be adapted at some point; the default version is mostly self-explanatory. Note that \isabellestyle enables fine-tuning of the general appearance of characters and mathematical symbols (see also Sect. 4.2.4). Especially observe the included L AT EX packages isabelle (mandatory), isabellesym (required for mathematical symbols), and the final pdfsetup (provides sane defaults for hyperref, including URL markup). All three are 4.2 Document Preparation 59 distributed with Isabelle. Further packages may be required in particular applications, say for unusual mathematical symbols. Any additional files for the L AT EX stage go into the MySession/document directory as well. In particular, adding a file named root.bib causes an au-tomatic run of bibtex to process a bibliographic database; see also isabelle document . Any failure of the document preparation phase in an Isabelle batch ses-sion leaves the generated sources in their target location, identified by the accompanying error message. This lets you trace L AT EX problems with the generated files at hand. 4.2.2 Structure Markup The large-scale structure of Isabelle documents follows existing L AT EX conven-tions, with chapters, sections, subsubsections etc. The Isar language includes separate markup commands, which do not affect the formal meaning of a theory (or proof), but result in corresponding L A T EX elements. From the Isabelle perspective, each markup command takes a single text argument (delimited by " . . . " or { . . . }). After stripping any surrounding white space, the argument is passed to a L A T EX macro \isamarkupXYZ for command XYZ. These macros are defined in isabelle.sty according to the meaning given in the rightmost column above. The following source fragment illustrates structure markup of a theory. Note that L AT EX labels may be included inside of section headings as well. 60 4. Presenting Theories section { Some properties of Foo Bar elements } theory Foo_Bar imports Main begin subsection { Basic definitions } definition foo :: … definition bar :: … subsection { Derived rules } lemma fooI: … lemma fooE: … subsection { Main theorem \label{sec:main-theorem} } theorem main: … end 4.2.3 Formal Comments and Antiquotations Isabelle source comments, which are of the form ( . . . ), essentially act like white space and do not really contribute to the content. They mainly serve technical purposes to mark certain oddities in the raw input text. In contrast, formal comments are portions of text that are associated with formal Isabelle/Isar commands (marginal comments), or as standalone paragraphs within a theory or proof context (text blocks). Marginal comments are part of each command’s concrete syntax ; the common form is “-- text” where text is delimited by ". . . " or { . . . } as before. Multiple marginal comments may be given at the same time. Here is a simple example: lemma "A --> A" — a triviality of propositional logic — (should not really bother) by (rule impI) — implicit assumption step involved here The above output has been produced as follows: lemma "A --> A" -- "a triviality of propositional logic" -- "(should not really bother)" by (rule impI) -- "implicit assumption step involved here" 4.2 Document Preparation 61 From the L AT EX viewpoint, “--” acts like a markup command, associated with the macro \isamarkupcmt (taking a single argument). Text blocks are introduced by the commands text and txt. Each takes again a single text argument, which is interpreted as a free-form paragraph in L A T EX (surrounded by some additional vertical space). The typesetting may be changed by redefining the L AT EX environments of isamarkuptext or isamarkuptxt, respectively (via \renewenvironment). The text part of Isabelle markup commands essentially inserts quoted ma-terial into a formal text, mainly for instruction of the reader. An antiquo-tation is again a formal object embedded into such an informal portion. The interpretation of antiquotations is limited to some well-formedness checks, with the result being pretty printed to the resulting document. Quoted text blocks together with antiquotations provide an attractive means of referring to formal entities, with good confidence in getting the technical details right (especially syntax and types). The general syntax of antiquotations is as follows: @{name arguments}, or @{name [options] arguments} for a comma-separated list of options con-sisting of a name or name=value each. The syntax of arguments depends on the kind of antiquotation, it generally follows the same conventions for types, terms, or theorems as in the formal part of a theory. This sentence demonstrates quotations and antiquotations: λx y. x is a well-typed term. The output above was produced as follows: text { This sentence demonstrates quotations and antiquotations: @{term "%x y. x"} is a well-typed term. } The notational change from the ASCII character % to the symbol λ reveals that Isabelle printed this term, after parsing and type-checking. Document preparation enables symbolic output by default. The next example includes an option to show the type of all variables. The antiquotation @{term [show_types] "%x y. x"} produces the output λ(x::'a) y::'b. x. Type inference has figured out the most general typings in the present theory context. Terms may acquire different typings due to con-straints imposed by their environment; within a proof, for example, variables are given the same types as they have in the main goal statement. Several further kinds of antiquotations and options are available . Here are a few commonly used combinations: 62 4. Presenting Theories @{typ τ} print type τ @{const c} check existence of c and print it @{term t} print term t @{prop φ} print proposition φ @{prop [display] φ} print large proposition φ (with linebreaks) @{prop [source] φ} check proposition φ, print its input @{thm a} print fact a @{thm a [no_vars]} print fact a, fixing schematic variables @{thm [source] a} check availability of fact a, print its name @{text s} print uninterpreted text s Note that no_vars given above is not an antiquotation option, but an attribute of the theorem argument given here. This might be useful with a diagnostic command like thm, too. The @{text s} antiquotation is particularly interesting. Embedding un-interpreted text within an informal body might appear useless at first sight. Here the key virtue is that the string s is processed as Isabelle output, inter-preting Isabelle symbols appropriately. For example, @{text "\\"} produces ∀∃, according to the standard interpretation of these symbol (cf. Sect. 4.2.4). Thus we achieve consistent mathematical notation in both the formal and informal parts of the document very easily, independently of the term language of Isa-belle. Manual L AT EX code would leave more control over the typesetting, but is also slightly more tedious. 4.2.4 Interpretation of Symbols As has been pointed out before (Sect. 4.1.2), Isabelle symbols are the smallest syntactic entities — a straightforward generalization of ASCII characters. While Isabelle does not impose any interpretation of the infinite collection of named symbols, L A T EX documents use canonical glyphs for certain standard symbols . The L AT EX code produced from Isabelle text follows a simple scheme. You can tune the final appearance by redefining certain macros, say in root.tex of the document. 1. 7-bit ASCII characters: letters A…Z and a…z are output directly, digits are passed as an argument to the \isadigit macro, other characters are replaced by specifically named macros of the form \isacharXYZ. 2. Named symbols: \ is turned into {\isasymXYZ}; note the additional braces. 3. Named control symbols: \<^XYZ> is turned into \isactrlXYZ; subsequent symbols may act as arguments if the control macro is defined accordingly. You may occasionally wish to give new L AT EX interpretations of named symbols. This merely requires an appropriate definition of \isasymXYZ, 4.2 Document Preparation 63 for \ (see isabelle.sty for working examples). Control symbols are slightly more difficult to get right, though. The \isabellestyle macro provides a high-level interface to tune the general appearance of individual symbols. For example, \isabellestyle{it} uses the italics text style to mimic the general appearance of the L A T EX math mode; double quotes are not printed at all. The resulting quality of type-setting is quite good, so this should be the default style for work that gets distributed to a broader audience. 4.2.5 Suppressing Output By default, Isabelle’s document system generates a L AT EX file for each theory that gets loaded while running the session. The generated session.tex will include all of these in order of appearance, which in turn gets included by the standard root.tex. Certainly one may change the order or suppress unwanted theories by ignoring session.tex and load individual files directly in root.tex. On the other hand, such an arrangement requires additional maintenance whenever the collection of theories changes. Alternatively, one may tune the theory loading process in ROOT itself: some sequential order of theories sections may enforce a certain traversal of the dependency graph, although this could degrade parallel processing. The nodes of each sub-graph that is specified here are presented in some topological order of their formal dependencies. Moreover, the system build option document=false allows to disable doc-ument generation for some theories. Its usage in the session ROOT is like this: theories [document = false] T Theory output may be suppressed more selectively, either via tagged command regions or ignored material. Tagged command regions works by annotating commands with named tags, which correspond to certain L AT EX markup that tells how to treat par-ticular parts of a document when doing the actual type-setting. By default, certain Isabelle/Isar commands are implicitly marked up using the predefined tags “theory” (for theory begin and end), “proof ” (for proof commands), and “ML” (for commands involving ML code). Users may add their own tags us-ing the %tag notation right after a command name. In the subsequent example we hide a particularly irrelevant proof: lemma "x = x" The original source has been “lemma "x = x" by %invisible (simp)”. Tags observe the structure of proofs; adjacent commands with the same tag are joined into a single region. The Isabelle document preparation system allows the user to specify how to interpret a tagged region, in order to keep, 64 4. Presenting Theories drop, or fold the corresponding parts of the document. See the Isabelle System Manual for further details, especially on isabelle build and isabelle document. Ignored material is specified by delimiting the original formal source with special source comments (<) and (>). These parts are stripped before the type-setting phase, without affecting the formal checking of the theory, of course. For example, we may hide parts of a proof that seem unfit for general public inspection. The following “fully automatic” proof is actually a fake: lemma "x ̸= (0::int) = ⇒0 < x x" by (auto) The real source of the proof has been as follows: by (auto(<)simp add: zero_less_mult_iff(>)) Suppressing portions of printed text demands care. You should not mis-represent the underlying theory development. It is easy to invalidate the visible text by hiding references to questionable axioms, for example. Part II Logic and Sets 5. The Rules of the Game This chapter outlines the concepts and techniques that underlie reasoning in Isabelle. Until now, we have proved everything using only induction and simplification, but any serious verification project requires more elaborate forms of inference. The chapter also introduces the fundamentals of predicate logic. The first examples in this chapter will consist of detailed, low-level proof steps. Later, we shall see how to automate such reasoning using the methods blast, auto and others. Backward or goal-directed proof is our usual style, but the chapter also introduces forward reasoning, where one theorem is transformed to yield another. 5.1 Natural Deduction In Isabelle, proofs are constructed using inference rules. The most familiar inference rule is probably modus ponens: P →Q P Q This rule says that from P →Q and P we may infer Q. Natural deduction is an attempt to formalize logic in a way that mirrors human reasoning patterns. For each logical symbol (say, ∧), there are two kinds of rules: introduction and elimination rules. The introduction rules allow us to infer this symbol (say, to infer conjunctions). The elimination rules allow us to deduce consequences from this symbol. Ideally each rule should mention one symbol only. For predicate logic this can be done, but when users define their own concepts they typically have to refer to other symbols as well. It is best not to be dogmatic. Our system is not based on pure natural deduction, but includes elements from the sequent calculus and free-variable tableaux. Natural deduction generally deserves its name. It is easy to use. Each proof step consists of identifying the outermost symbol of a formula and applying the corresponding rule. It creates new subgoals in an obvious way from parts of the chosen formula. Expanding the definitions of constants can blow up the goal enormously. Deriving natural deduction rules for such constants 68 5. The Rules of the Game lets us reason in terms of their key properties, which might otherwise be obscured by the technicalities of its definition. Natural deduction rules also lend themselves to automation. Isabelle’s classical reasoner accepts any suitable collection of natural deduction rules and uses them to search for proofs automatically. Isabelle is designed around natural deduction and many of its tools use the terminology of introduction and elimination rules. 5.2 Introduction Rules An introduction rule tells us when we can infer a formula containing a specific logical symbol. For example, the conjunction introduction rule says that if we have P and if we have Q then we have P ∧Q. In a mathematics text, it is typically shown like this: P Q P ∧Q The rule introduces the conjunction symbol (∧) in its conclusion. In Isabelle proofs we mainly reason backwards. When we apply this rule, the subgoal already has the form of a conjunction; the proof step makes this conjunction symbol disappear. In Isabelle notation, the rule looks like this: [ [?P; ?Q] ] = ⇒?P ∧?Q (conjI) Carefully examine the syntax. The premises appear to the left of the arrow and the conclusion to the right. The premises (if more than one) are grouped using the fat brackets. The question marks indicate schematic variables (also called unknowns): they may be replaced by arbitrary formulas. If we use the rule backwards, Isabelle tries to unify the current subgoal with the conclusion of the rule, which has the form ?P ∧?Q. (Unification is discussed below, Sect. 5.8.) If successful, it yields new subgoals given by the formulas assigned to ?P and ?Q. The following trivial proof illustrates how rules work. It also introduces a style of indentation. If a command adds a new subgoal, then the next command’s indentation is increased by one space; if it proves a subgoal, then the indentation is reduced. This provides the reader with hints about the subgoal structure. lemma conj_rule: "[ [P; Q] ] = ⇒P ∧(Q ∧P)" apply (rule conjI) apply assumption apply (rule conjI) apply assumption apply assumption At the start, Isabelle presents us with the assumptions (P and Q) and with the goal to be proved, P ∧(Q ∧P). We are working backwards, so when we 5.3 Elimination Rules 69 apply conjunction introduction, the rule removes the outermost occurrence of the ∧symbol. To apply a rule to a subgoal, we apply the proof method rule — here with conjI, the conjunction introduction rule. 1. [ [P; Q] ] = ⇒P 2. [ [P; Q] ] = ⇒Q ∧P Isabelle leaves two new subgoals: the two halves of the original conjunction. The first is simply P, which is trivial, since P is among the assumptions. We can apply the assumption method, which proves a subgoal by finding a matching assumption. 1. [ [P; Q] ] = ⇒Q ∧P We are left with the subgoal of proving Q ∧P from the assumptions P and Q. We apply rule conjI again. 1. [ [P; Q] ] = ⇒Q 2. [ [P; Q] ] = ⇒P We are left with two new subgoals, Q and P, each of which can be proved using the assumption method. 5.3 Elimination Rules Elimination rules work in the opposite direction from introduction rules. In the case of conjunction, there are two such rules. From P ∧Q we infer P. also, from P ∧Q we infer Q: P ∧Q P P ∧Q Q Now consider disjunction. There are two introduction rules, which resem-ble inverted forms of the conjunction elimination rules: P P ∨Q Q P ∨Q What is the disjunction elimination rule? The situation is rather different from conjunction. From P ∨Q we cannot conclude that P is true and we cannot conclude that Q is true; there are no direct elimination rules of the sort that we have seen for conjunction. Instead, there is an elimination rule that works indirectly. If we are trying to prove something else, say R, and we know that P ∨Q holds, then we have to consider two cases. We can assume that P is true and prove R and then assume that Q is true and prove R a second time. Here we see a fundamental concept used in natural deduction: that of the assumptions. We have to prove R twice, under different assumptions. The assumptions are local to these subproofs and are visible nowhere else. 70 5. The Rules of the Game In a logic text, the disjunction elimination rule might be shown like this: P ∨Q [P] . . . . R [Q] . . . . R R The assumptions [P] and [Q] are bracketed to emphasize that they are local to their subproofs. In Isabelle notation, the already-familiar = ⇒syntax serves the same purpose: [ [?P ∨?Q; ?P = ⇒?R; ?Q = ⇒?R] ] = ⇒?R (disjE) When we use this sort of elimination rule backwards, it produces a case split. (We have seen this before, in proofs by induction.) The following proof illustrates the use of disjunction elimination. lemma disj_swap: "P ∨Q = ⇒Q ∨P" apply (erule disjE) apply (rule disjI2) apply assumption apply (rule disjI1) apply assumption We assume P ∨Q and must prove Q ∨P. Our first step uses the disjunction elimination rule, disjE. We invoke it using erule, a method designed to work with elimination rules. It looks for an assumption that matches the rule’s first premise. It deletes the matching assumption, regards the first premise as proved and returns subgoals corresponding to the remaining premises. When we apply erule to disjE, only two subgoals result. This is better than applying it using rule to get three subgoals, then proving the first by assumption: the other subgoals would have the redundant assumption P ∨Q. Most of the time, erule is the best way to use elimination rules, since it replaces an assumption by its subformulas; only rarely does the original assumption remain useful. 1. P = ⇒Q ∨P 2. Q = ⇒Q ∨P These are the two subgoals returned by erule. The first assumes P and the second assumes Q. Tackling the first subgoal, we need to show Q ∨P. The second introduction rule (disjI2) can reduce this to P, which matches the assumption. So, we apply the rule method with disjI2 … 1. P = ⇒P 2. Q = ⇒Q ∨P …and finish off with the assumption method. We are left with the other sub-goal, which assumes Q. 1. Q = ⇒Q ∨P 5.4 Destruction Rules: Some Examples 71 Its proof is similar, using the introduction rule disjI1. The result of this proof is a new inference rule disj_swap, which is neither an introduction nor an elimination rule, but which might be useful. We can use it to replace any goal of the form Q ∨P by one of the form P ∨Q. 5.4 Destruction Rules: Some Examples Now let us examine the analogous proof for conjunction. lemma conj_swap: "P ∧Q = ⇒Q ∧P" apply (rule conjI) apply (drule conjunct2) apply assumption apply (drule conjunct1) apply assumption Recall that the conjunction elimination rules — whose Isabelle names are conjunct1 and conjunct2 — simply return the first or second half of a con-junction. Rules of this sort (where the conclusion is a subformula of a premise) are called destruction rules because they take apart and destroy a premise.1 The first proof step applies conjunction introduction, leaving two sub-goals: 1. P ∧Q = ⇒Q 2. P ∧Q = ⇒P To invoke the elimination rule, we apply a new method, drule. Think of the d as standing for destruction (or direct, if you prefer). Applying the second conjunction rule using drule replaces the assumption P ∧Q by Q. 1. Q = ⇒Q 2. P ∧Q = ⇒P The resulting subgoal can be proved by applying assumption. The other sub-goal is similarly proved, using the conjunct1 rule and the assumption method. Choosing among the methods rule, erule and drule is up to you. Isabelle does not attempt to work out whether a rule is an introduction rule or an elimination rule. The method determines how the rule will be interpreted. Many rules can be used in more than one way. For example, disj_swap can be applied to assumptions as well as to goals; it replaces any assumption of the form P ∨Q by a one of the form Q ∨P. Destruction rules are simpler in form than indirect rules such as disjE, but they can be inconvenient. Each of the conjunction rules discards half of the 1 This Isabelle terminology is not used in standard logic texts, although the dis-tinction between the two forms of elimination rule is well known. Girard [9, page 74], for example, writes “The elimination rules [for ∨and ∃] are very bad. What is catastrophic about them is the parasitic presence of a formula [R] which has no structural link with the formula which is eliminated.” These Isabelle rules are inspired by the sequent calculus. 72 5. The Rules of the Game formula, when usually we want to take both parts of the conjunction as new assumptions. The easiest way to do so is by using an alternative conjunction elimination rule that resembles disjE. It is seldom, if ever, seen in logic books. In Isabelle syntax it looks like this: [ [?P ∧?Q; [ [?P; ?Q] ] = ⇒?R] ] = ⇒?R (conjE) Exercise 5.4.1 Use the rule conjE to shorten the proof above. 5.5 Implication At the start of this chapter, we saw the rule modus ponens. It is, in fact, a destruction rule. The matching introduction rule looks like this in Isabelle: (?P = ⇒?Q) = ⇒?P − →?Q (impI) And this is modus ponens: [ [?P − →?Q; ?P] ] = ⇒?Q (mp) Here is a proof using the implication rules. This lemma performs a sort of uncurrying, replacing the two antecedents of a nested implication by a conjunction. The proof illustrates how assumptions work. At each proof step, the subgoals inherit the previous assumptions, perhaps with additions or deletions. Rules such as impI and disjE add assumptions, while applying erule or drule deletes the matching assumption. lemma imp_uncurry: "P − →(Q − →R) = ⇒P ∧Q − →R" apply (rule impI) apply (erule conjE) apply (drule mp) apply assumption apply (drule mp) apply assumption apply assumption First, we state the lemma and apply implication introduction (rule impI), which moves the conjunction to the assumptions. 1. [ [P − →Q − →R; P ∧Q] ] = ⇒R Next, we apply conjunction elimination (erule conjE), which splits this con-junction into two parts. 1. [ [P − →Q − →R; P; Q] ] = ⇒R Now, we work on the assumption P − →(Q − →R), where the parentheses have been inserted for clarity. The nested implication requires two applica-tions of modus ponens: drule mp. The first use yields the implication Q − → R, but first we must prove the extra subgoal P, which we do by assumption. 5.6 Negation 73 1. [ [P; Q] ] = ⇒P 2. [ [P; Q; Q − →R] ] = ⇒R Repeating these steps for Q − →R yields the conclusion we seek, namely R. 1. [ [P; Q; Q − →R] ] = ⇒R The symbols = ⇒and − →both stand for implication, but they differ in many respects. Isabelle uses = ⇒to express inference rules; the symbol is built-in and Isabelle’s inference mechanisms treat it specially. On the other hand, − →is just one of the many connectives available in higher-order logic. We reason about it using inference rules such as impI and mp, just as we reason about the other connectives. You will have to use − →in any context that requires a formula of higher-order logic. Use = ⇒to separate a theorem’s preconditions from its conclusion. The by command is useful for proofs like these that use assumption heav-ily. It executes an apply command, then tries to prove all remaining subgoals using assumption. Since (if successful) it ends the proof, it also replaces the done symbol. For example, the proof above can be shortened: lemma imp_uncurry: "P − →(Q − →R) = ⇒P ∧Q − →R" apply (rule impI) apply (erule conjE) apply (drule mp) apply assumption by (drule mp) We could use by to replace the final apply and done in any proof, but typically we use it to eliminate calls to assumption. It is also a nice way of expressing a one-line proof. 5.6 Negation Negation causes surprising complexity in proofs. Its natural deduction rules are straightforward, but additional rules seem necessary in order to han-dle negated assumptions gracefully. This section also illustrates the intro method: a convenient way of applying introduction rules. Negation introduction deduces ¬P if assuming P leads to a contradiction. Negation elimination deduces any formula in the presence of ¬P together with P: (?P = ⇒False) = ⇒¬ ?P (notI) [ [¬ ?P; ?P] ] = ⇒?R (notE) Classical logic allows us to assume ¬P when attempting to prove P: (¬ ?P = ⇒?P) = ⇒?P (classical) 74 5. The Rules of the Game The implications P →Q and ¬Q →¬P are logically equivalent, and each is called the contrapositive of the other. Four further rules support reasoning about contrapositives. They differ in the placement of the negation symbols: [ [?Q; ¬ ?P = ⇒¬ ?Q] ] = ⇒?P (contrapos_pp) [ [?Q; ?P = ⇒¬ ?Q] ] = ⇒¬ ?P (contrapos_pn) [ [¬ ?Q; ¬ ?P = ⇒?Q] ] = ⇒?P (contrapos_np) [ [¬ ?Q; ?P = ⇒?Q] ] = ⇒¬ ?P (contrapos_nn) These rules are typically applied using the erule method, where their effect is to form a contrapositive from an assumption and the goal’s conclusion. The most important of these is contrapos_np. It is useful for applying introduction rules to negated assumptions. For instance, the assumption ¬(P →Q) is equivalent to the conclusion P →Q and we might want to use conjunction introduction on it. Before we can do so, we must move that assumption so that it becomes the conclusion. The following proof demon-strates this technique: lemma "[ [¬(P− →Q); ¬(R− →Q)] ] = ⇒R" apply (erule_tac Q = "R− →Q" in contrapos_np) apply (intro impI) by (erule notE) There are two negated assumptions and we need to exchange the conclusion with the second one. The method erule contrapos_np would select the first assumption, which we do not want. So we specify the desired assumption explicitly using a new method, erule_tac. This is the resulting subgoal: 1. [ [¬ (P − →Q); ¬ R] ] = ⇒R − →Q The former conclusion, namely R, now appears negated among the assump-tions, while the negated formula R − →Q becomes the new conclusion. We can now apply introduction rules. We use the intro method, which repeatedly applies the given introduction rules. Here its effect is equivalent to rule impI. 1. [ [¬ (P − →Q); ¬ R; R] ] = ⇒Q We can see a contradiction in the form of assumptions ¬ R and R, which suggests using negation elimination. If applied on its own, notE will select the first negated assumption, which is useless. Instead, we invoke the rule using the by command. Now when Isabelle selects the first assumption, it tries to prove P − →Q and fails; it then backtracks, finds the assumption ¬ R and finally proves R by assumption. That concludes the proof. The following example may be skipped on a first reading. It involves a peculiar but important rule, a form of disjunction introduction: (¬ ?Q = ⇒?P) = ⇒?P ∨?Q (disjCI) 5.7 Interlude: the Basic Methods for Rules 75 This rule combines the effects of disjI1 and disjI2. Its great advantage is that we can remove the disjunction symbol without deciding which disjunction to prove. This treatment of disjunction is standard in sequent and tableau calculi. lemma "(P ∨Q) ∧R = ⇒P ∨(Q ∧R)" apply (rule disjCI) apply (elim conjE disjE) apply assumption by (erule contrapos_np, rule conjI) The first proof step to applies the introduction rules disjCI. The resulting subgoal has the negative assumption ¬(Q ∧R). 1. [ [(P ∨Q) ∧R; ¬ (Q ∧R)] ] = ⇒P Next we apply the elim method, which repeatedly applies elimination rules; here, the elimination rules given in the command. One of the subgoals is trivial (apply assumption), leaving us with one other: 1. [ [¬ (Q ∧R); R; Q] ] = ⇒P Now we must move the formula Q ∧R to be the conclusion. The combination (erule contrapos_np, rule conjI) is robust: the conjI forces the erule to select a conjunction. The two subgoals are the ones we would expect from applying conjunction introduction to Q ∧R: 1. [ [R; Q; ¬ P] ] = ⇒Q 2. [ [R; Q; ¬ P] ] = ⇒R They are proved by assumption, which is implicit in the by command. 5.7 Interlude: the Basic Methods for Rules We have seen examples of many tactics that operate on individual rules. It may be helpful to review how they work given an arbitrary rule such as this: P1 . . . Pn Q Below, we refer to P1 as the major premise. This concept applies only to elimination and destruction rules. These rules act upon an instance of their major premise, typically to replace it by subformulas of itself. Suppose that the rule above is called R. Here are the basic rule methods, most of which we have already seen: – Method rule R unifies Q with the current subgoal, replacing it by n new subgoals: instances of P1, …, Pn. This is backward reasoning and is appro-priate for introduction rules. 76 5. The Rules of the Game – Method erule R unifies Q with the current subgoal and simultaneously unifies P1 with some assumption. The subgoal is replaced by the n −1 new subgoals of proving instances of P2, …, Pn, with the matching assump-tion deleted. It is appropriate for elimination rules. The method (rule R, assumption) is similar, but it does not delete an assumption. – Method drule R unifies P1 with some assumption, which it then deletes. The subgoal is replaced by the n −1 new subgoals of proving P2, …, Pn; an nth subgoal is like the original one but has an additional assumption: an instance of Q. It is appropriate for destruction rules. – Method frule R is like drule R except that the matching assumption is not deleted. (See Sect. 5.9.5 below.) Other methods apply a rule while constraining some of its variables. The typical form is rule_tac v1 = t1 and …and vk = tk in R This method behaves like rule R, while instantiating the variables v1, …, vk as specified. We similarly have erule_tac, drule_tac and frule_tac. These methods also let us specify which subgoal to operate on. By default it is the first subgoal, as with nearly all methods, but we can specify that rule R should be applied to subgoal number i: rule_tac [i] R 5.8 Unification and Substitution As we have seen, Isabelle rules involve schematic variables, which begin with a question mark and act as placeholders for terms. Unification — well known to Prolog programmers — is the act of making two terms identical, possibly replacing their schematic variables by terms. The simplest case is when the two terms are already the same. Next simplest is pattern-matching, which replaces variables in only one of the terms. The rule method typically matches the rule’s conclusion against the current subgoal. The assumption method matches the current subgoal’s conclusion against each of its assumptions. Unification can instantiate variables in both terms; the rule method can do this if the goal itself contains schematic variables. Other occurrences of the variables in the rule or proof state are updated at the same time. Schematic variables in goals represent unknown terms. Given a goal such as ∃x. P, they let us proceed with a proof. They can be filled in later, some-times in stages and often automatically. If unification fails when you think it should succeed, try setting the Proof Gen-eral flag Isabelle > Settings > Trace Unification, which makes Isabelle show the cause of unification failures (in Proof General’s Trace buffer). 5.8 Unification and Substitution 77 For example, suppose we are trying to prove this subgoal by assumption: 1. P (a, f (b, g (e, a), b), a) = ⇒P (a, f (b, g (c, a), b), a) The assumption method having failed, we try again with the flag set: apply assumption In this trivial case, the output clearly shows that e clashes with c: Clash: e =/= c Isabelle uses higher-order unification, which works in the typed λ-calculus. The procedure requires search and is potentially undecidable. For our purposes, however, the differences from ordinary unification are straight-forward. It handles bound variables correctly, avoiding capture. The two terms λx. f(x,z) and λy. f(y,z) are trivially unifiable because they differ only by a bound variable renaming. The two terms λx. ?P and λx. t x are not unifiable; replacing ?P by t x is forbidden because the free occurrence of x would become bound. Unfortunately, even if trace_unify_fail is set, Isabelle displays no information about this type of failure. ! ! Higher-order unification sometimes must invent λ-terms to replace function variables, which can lead to a combinatorial explosion. However, Isabelle proofs tend to involve easy cases where there are few possibilities for the λ-term being constructed. In the easiest case, the function variable is applied only to bound variables, as when we try to unify λx y. f(?h x y) and λx y. f(x+y+a). The only solution is to replace ?h by λx y. x+y+a. Such cases admit at most one unifier, like ordinary unification. A harder case is unifying ?h a with a+b; it admits two solutions for ?h, namely λx. a+b and λx. x+b. Unifying ?h a with a+a+b admits four solutions; their number is exponential in the number of occurrences of a in the second term. 5.8.1 Substitution and the subst Method Isabelle also uses function variables to express substitution. A typical sub-stitution rule allows us to replace one term by another if we know that two terms are equal. s = t P[s/x] P[t/x] The rule uses a notation for substitution: P[t/x] is the result of replacing x by t in P. The rule only substitutes in the positions designated by x. For example, it can derive symmetry of equality from reflexivity. Using x = s for P replaces just the first s in s = s by t: s = t s = s t = s The Isabelle version of the substitution rule looks like this: 78 5. The Rules of the Game [ [?t = ?s; ?P ?s] ] = ⇒?P ?t (ssubst) Crucially, ?P is a function variable. It can be replaced by a λ-term with one bound variable, whose occurrences identify the places in which s will be replaced by t. The proof above requires ?P to be replaced by λx. x=s; the second premise will then be s=s and the conclusion will be t=s. The simp method also replaces equals by equals, but the substitution rule gives us more control. Consider this proof: lemma "[ [x = f x; odd(f x)] ] = ⇒odd x" by (erule ssubst) The assumption x = f x, if used for rewriting, would loop, replacing x by f x and then by f(f x) and so forth. (Here simp would see the danger and would re-orient the equality, but in more complicated cases it can be fooled.) When we apply the substitution rule, Isabelle replaces every x in the subgoal by f x just once. It cannot loop. The resulting subgoal is trivial by assumption, so the by command proves it implicitly. We are using the erule method in a novel way. Hitherto, the conclusion of the rule was just a variable such as ?R, but it may be any term. The conclusion is unified with the subgoal just as it would be with the rule method. At the same time erule looks for an assumption that matches the rule’s first premise, as usual. With ssubst the effect is to find, use and delete an equality assumption. The subst method performs individual substitutions. In simple cases, it closely resembles a use of the substitution rule. Suppose a proof has reached this point: 1. [ [P x y z; Suc x < y] ] = ⇒f z = x y Now we wish to apply a commutative law: ?m ?n = ?n ?m (mult.commute) Isabelle rejects our first attempt: apply (simp add: mult.commute) The simplifier notices the danger of looping and refuses to apply the rule.2 The subst method applies mult.commute exactly once. apply (subst mult.commute) 1. [ [P x y z; Suc x < y] ] = ⇒f z = y x As we wanted, x y has become y x. This use of the subst method has the same effect as the command apply (rule mult.commute [THEN ssubst]) 2 More precisely, it only applies such a rule if the new term is smaller under a specified ordering; here, x y is already smaller than y x. 5.8 Unification and Substitution 79 The attribute THEN, which combines two rules, is described in Sect. 5.15.1 be-low. The subst method is more powerful than applying the substitution rule. It can perform substitutions in a subgoal’s assumptions. Moreover, if the sub-goal contains more than one occurrence of the left-hand side of the equality, the subst method lets us specify which occurrence should be replaced. 5.8.2 Unification and Its Pitfalls Higher-order unification can be tricky. Here is an example, which you may want to skip on your first reading: lemma "[ [x = f x; triple (f x) (f x) x] ] = ⇒triple x x x" apply (erule ssubst) back back back back apply assumption done By default, Isabelle tries to substitute for all the occurrences. Applying erule ssubst yields this subgoal: 1. triple (f x) (f x) x = ⇒triple (f x) (f x) (f x) The substitution should have been done in the first two occurrences of x only. Isabelle has gone too far. The back command allows us to reject this possibility and demand a new one: 1. triple (f x) (f x) x = ⇒triple x (f x) (f x) Now Isabelle has left the first occurrence of x alone. That is promising but it is not the desired combination. So we use back again: 1. triple (f x) (f x) x = ⇒triple (f x) x (f x) This also is wrong, so we use back again: 1. triple (f x) (f x) x = ⇒triple x x (f x) And this one is wrong too. Looking carefully at the series of alternatives, we see a binary countdown with reversed bits: 111, 011, 101, 001. Invoke back again: 1. triple (f x) (f x) x = ⇒triple (f x) (f x) x At last, we have the right combination! This goal follows by assumption. This example shows that unification can do strange things with function variables. We were forced to select the right unifier using the back command. That is all right during exploration, but back should never appear in the final version of a proof. You can eliminate the need for back by giving Isabelle less freedom when you apply a rule. 80 5. The Rules of the Game One way to constrain the inference is by joining two methods in a apply command. Isabelle applies the first method and then the second. If the second method fails then Isabelle automatically backtracks. This process continues until the first method produces an output that the second method can use. We get a one-line proof of our example: lemma "[ [x = f x; triple (f x) (f x) x] ] = ⇒triple x x x" apply (erule ssubst, assumption) done The by command works too, since it backtracks when proving subgoals by assumption: lemma "[ [x = f x; triple (f x) (f x) x] ] = ⇒triple x x x" by (erule ssubst) The most general way to constrain unification is by instantiating variables in the rule. The method rule_tac is similar to rule, but it makes some of the rule’s variables denote specified terms. Also available are drule_tac and erule_tac. Here we need erule_tac since above we used erule. lemma "[ [x = f x; triple (f x) (f x) x] ] = ⇒triple x x x" by (erule_tac P = "λu. triple u u x" in ssubst) To specify a desired substitution requires instantiating the variable ?P with a λ-expression. The bound variable occurrences in λu. P u u x indicate that the first two arguments have to be substituted, leaving the third unchanged. With this instantiation, backtracking is neither necessary nor possible. An alternative to rule_tac is to use rule with a theorem modified us-ing of, described in Sect. 5.15 below. But rule_tac, unlike of, can express instantiations that refer to V-bound variables in the current subgoal. 5.9 Quantifiers Quantifiers require formalizing syntactic substitution and the notion of arbi-trary value. Consider the universal quantifier. In a logic book, its introduction rule looks like this: P ∀x. P Typically, a proviso written in English says that x must not occur in the as-sumptions. This proviso guarantees that x can be regarded as arbitrary, since it has not been assumed to satisfy any special conditions. Isabelle’s under-lying formalism, called the meta-logic, eliminates the need for English. It provides its own universal quantifier (V) to express the notion of an arbitrary value. We have already seen another operator of the meta-logic, namely = ⇒, which expresses inference rules and the treatment of assumptions. The only other operator in the meta-logic is ≡, which can be used to define constants. 5.9 Quantifiers 81 5.9.1 The Universal Introduction Rule Returning to the universal quantifier, we find that having a similar quantifier as part of the meta-logic makes the introduction rule trivial to express: (Vx. ?P x) = ⇒∀x. ?P x (allI) The following trivial proof demonstrates how the universal introduction rule works. lemma "∀x. P x − →P x" apply (rule allI) by (rule impI) The first step invokes the rule by applying the method rule allI. 1. Vx. P x − →P x Note that the resulting proof state has a bound variable, namely x. The rule has replaced the universal quantifier of higher-order logic by Isabelle’s meta-level quantifier. Our goal is to prove P x − →P x for arbitrary x; it is an implication, so we apply the corresponding introduction rule (impI). 1. Vx. P x = ⇒P x This last subgoal is implicitly proved by assumption. 5.9.2 The Universal Elimination Rule Now consider universal elimination. In a logic text, the rule looks like this: ∀x. P P[t/x] The conclusion is P with t substituted for the variable x. Isabelle expresses substitution using a function variable: ∀x. ?P x = ⇒?P ?x (spec) This destruction rule takes a universally quantified formula and removes the quantifier, replacing the bound variable x by the schematic variable ?x. Re-call that a schematic variable starts with a question mark and acts as a placeholder: it can be replaced by any term. The universal elimination rule is also available in the standard elimination format. Like conjE, it never appears in logic books: [ [∀x. ?P x; ?P ?x = ⇒?R] ] = ⇒?R (allE) The methods drule spec and erule allE do precisely the same inference. To see how ∀-elimination works, let us derive a rule about reducing the scope of a universal quantifier. In mathematical notation we write ∀x. P →Q P →∀x. Q 82 5. The Rules of the Game with the proviso “x not free in P.” Isabelle’s treatment of substitution makes the proviso unnecessary. The conclusion is expressed as P − →(∀x. Q x). No substitution for the variable P can introduce a dependence upon x: that would be a bound variable capture. Let us walk through the proof. lemma "(∀x. P − →Q x) = ⇒P − →(∀x. Q x)" First we apply implies introduction (impI), which moves the P from the con-clusion to the assumptions. Then we apply universal introduction (allI). apply (rule impI, rule allI) 1. Vx. [ [∀x. P − →Q x; P] ] = ⇒Q x As before, it replaces the HOL quantifier by a meta-level quantifier, producing a subgoal that binds the variable x. The leading bound variables (here x) and the assumptions (here ∀x. P − →Q x and P) form the context for the conclusion, here Q x. Subgoals inherit the context, although assumptions can be added or deleted (as we saw earlier), while rules such as allI add bound variables. Now, to reason from the universally quantified assumption, we apply the elimination rule using the drule method. This rule is called spec because it specializes a universal formula to a particular term. apply (drule spec) 1. Vx. [ [P; P − →Q (?x2 x)] ] = ⇒Q x Observe how the context has changed. The quantified formula is gone, re-placed by a new assumption derived from its body. We have removed the quantifier and replaced the bound variable by the curious term ?x2 x. This term is a placeholder: it may become any term that can be built from x. (Formally, ?x2 is an unknown of function type, applied to the argument x.) This new assumption is an implication, so we can use modus ponens on it, which concludes the proof. by (drule mp) Let us take a closer look at this last step. Modus ponens yields two subgoals: one where we prove the antecedent (in this case P) and one where we may assume the consequent. Both of these subgoals are proved by the assumption method, which is implicit in the by command. Replacing the by command by apply (drule mp, assumption) would have left one last subgoal: 1. Vx. [ [P; Q (?x2 x)] ] = ⇒Q x The consequent is Q applied to that placeholder. It may be replaced by any term built from x, and here it should simply be x. The assumption need not be identical to the conclusion, provided the two formulas are unifiable. 5.9.3 The Existential Quantifier The concepts just presented also apply to the existential quantifier, whose introduction rule looks like this in Isabelle: 5.9 Quantifiers 83 ?P ?x = ⇒∃x. ?P x (exI) If we can exhibit some x such that P(x) is true, then ∃x.P(x) is also true. It is a dual of the universal elimination rule, and logic texts present it using the same notation for substitution. The existential elimination rule looks like this in a logic text: ∃x. P [P] . . . . Q Q It looks like this in Isabelle: [ [∃x. ?P x; Vx. ?P x = ⇒?Q] ] = ⇒?Q (exE) Given an existentially quantified theorem and some formula Q to prove, it creates a new assumption by removing the quantifier. As with the universal introduction rule, the textbook version imposes a proviso on the quantified variable, which Isabelle expresses using its meta-logic. It is enough to have a universal quantifier in the meta-logic; we do not need an existential quantifier to be built in as well. Exercise 5.9.1 Prove the lemma ∃x. P ∧Q(x) = ⇒P ∧(∃x. Q(x)). Hint: the proof is similar to the one just above for the universal quantifier. 5.9.4 Renaming a Bound Variable: rename_tac When you apply a rule such as allI, the quantified variable becomes a new bound variable of the new subgoal. Isabelle tries to avoid changing its name, but sometimes it has to choose a new name in order to avoid a clash. The result may not be ideal: lemma "x < y = ⇒∀x y. P x (f y)" apply (intro allI) 1. Vxa ya. x < y = ⇒P xa (f ya) The names x and y were already in use, so the new bound variables are called xa and ya. You can rename them by invoking rename_tac: apply (rename_tac v w) 1. Vv w. x < y = ⇒P v (f w) Recall that rule_tac instantiates a theorem with specified terms. These terms may involve the goal’s bound variables, but beware of referring to variables like xa. A future change to your theories could change the set of names pro-duced at top level, so that xa changes to xb or reverts to x. It is safer to rename automatically-generated variables before mentioning them. If the subgoal has more bound variables than there are names given to rename_tac, the rightmost ones are renamed. 84 5. The Rules of the Game 5.9.5 Reusing an Assumption: frule Note that drule spec removes the universal quantifier and — as usual with elimination rules — discards the original formula. Sometimes, a universal formula has to be kept so that it can be used again. Then we use a new method: frule. It acts like drule but copies rather than replaces the selected assumption. The f is for forward. In this example, going from P a to P(h(h a)) requires two uses of the quantified assumption, one for each h in h(h a). lemma "[ [∀x. P x − →P (h x); P a] ] = ⇒P(h (h a))" Examine the subgoal left by frule: apply (frule spec) 1. [ [∀x. P x − →P (h x); P a; P ?x − →P (h ?x)] ] = ⇒P (h (h a)) It is what drule would have left except that the quantified assumption is still present. Next we apply mp to the implication and the assumption P a: apply (drule mp, assumption) 1. [ [∀x. P x − →P (h x); P a; P (h a)] ] = ⇒P (h (h a)) We have created the assumption P(h a), which is progress. To continue the proof, we apply spec again. We shall not need it again, so we can use drule. apply (drule spec) 1. [ [P a; P (h a); P ?x2 − →P (h ?x2)] ] = ⇒P (h (h a)) The new assumption bridges the gap between P(h a) and P(h(h a)). by (drule mp) A final remark. Replacing this by command with apply (drule mp, assumption) would not work: it would add a second copy of P(h a) instead of the desired assumption, P(h(h a)). The by command forces Isabelle to backtrack until it finds the correct one. Alternatively, we could have used the apply command and bundled the drule mp with two calls of assumption. Or, of course, we could have given the entire proof to auto. 5.9.6 Instantiating a Quantifier Explicitly We can prove a theorem of the form ∃x. P x by exhibiting a suitable term t such that P t is true. Dually, we can use an assumption of the form ∀x. P x to generate a new assumption P t for a suitable term t. In many cases, Isabelle makes the correct choice automatically, constructing the term by unification. In other cases, the required term is not obvious and we must specify it our-selves. Suitable methods are rule_tac, drule_tac and erule_tac. We have seen (just above, Sect. 5.9.5) a proof of this lemma: 5.9 Quantifiers 85 lemma "[ [∀x. P x − →P (h x); P a] ] = ⇒P(h (h a))" We had reached this subgoal: 1. [ [∀x. P x − →P (h x); P a; P (h a)] ] = ⇒P (h (h a)) The proof requires instantiating the quantified assumption with the term h a. apply (drule_tac x = "h a" in spec) 1. [ [P a; P (h a); P (h a) − →P (h (h a))] ] = ⇒P (h (h a)) We have forced the desired instantiation. Existential formulas can be instantiated too. The next example uses the divides relation of number theory: ?m dvd ?n ≡∃k. ?n = ?m k (dvd_def) Let us prove that multiplication of natural numbers is monotone with respect to the divides relation: lemma mult_dvd_mono: "[ [i dvd m; j dvd n] ] = ⇒ij dvd (mn :: nat)" apply (simp add: dvd_def) Unfolding the definition of divides has left this subgoal: 1. [ [∃k. m = i k; ∃k. n = j k] ] = ⇒∃k. m n = i j k Next, we eliminate the two existential quantifiers in the assumptions: apply (erule exE) 1. Vk. [ [∃k. n = j k; m = i k] ] = ⇒∃k. m n = i j k apply (erule exE) 1. Vk ka. [ [m = i k; n = j ka] ] = ⇒∃k. m n = i j k The term needed to instantiate the remaining quantifier is kka. But ka is an automatically-generated name. As noted above, references to such variable names makes a proof less resilient to future changes. So, first we rename the most recent variable to l: apply (rename_tac l) 1. Vk l. [ [m = i k; n = j l] ] = ⇒∃k. m n = i j k We instantiate the quantifier with kl: apply (rule_tac x="kl" in exI) 1. Vk ka. [ [m = i k; n = j ka] ] = ⇒m n = i j (k ka) The rest is automatic, by arithmetic. apply simp done 86 5. The Rules of the Game 5.10 Description Operators HOL provides two description operators. A definite description formalizes the word “the,” as in “the greatest divisior of n.” It returns an arbitrary value unless the formula has a unique solution. An indefinite description formalizes the word “some,” as in “some member of S.” It differs from a definite description in not requiring the solution to be unique: it uses the axiom of choice to pick any solution. ! ! Description operators can be hard to reason about. Novices should try to avoid them. Fortunately, descriptions are seldom required. 5.10.1 Definite Descriptions A definite description is traditionally written ιx.P(x). It denotes the x such that P(x) is true, provided there exists a unique such x; otherwise, it returns an arbitrary value of the expected type. Isabelle uses THE for the Greek letter ι. We reason using this rule, where a is the unique solution: [ [P a; Vx. P x = ⇒x = a] ] = ⇒(THE x. P x) = a (the_equality) For instance, we can define the cardinality of a finite set A to be that n such that A is in one-to-one correspondence with {1, . . . , n}. We can then prove that the cardinality of the empty set is zero (since n = 0 satisfies the description) and proceed to prove other facts. A more challenging example illustrates how Isabelle/HOL defines the least number operator, which denotes the least x satisfying P: (LEAST x. P x) = (THE x. P x ∧(∀y. P y − →x ≤y)) Let us prove the analogue of the_equality for LEAST. theorem Least_equality: "[ [P (k::nat); ∀x. P x − →k ≤x] ] = ⇒(LEAST x. P x) = k" apply (simp add: Least_def) 1. [ [P k; ∀x. P x − →k ≤x] ] = ⇒(THE x. P x ∧(∀y. P y − →x ≤y)) = k The first step has merely unfolded the definition. apply (rule the_equality) 1. [ [P k; ∀x. P x − →k ≤x] ] = ⇒P k ∧(∀y. P y − →k ≤y) 2. Vx. [ [P k; ∀x. P x − →k ≤x; P x ∧(∀y. P y − →x ≤y)] ] = ⇒x = k As always with the_equality, we must show existence and uniqueness of the claimed solution, k. Existence, the first subgoal, is trivial. Uniqueness, the second subgoal, follows by antisymmetry: [ [x ≤y; y ≤x] ] = ⇒x = y (order_antisym) 5.10 Description Operators 87 The assumptions imply both k ≤x and x ≤k. One call to auto does it all: by (auto intro: order_antisym) 5.10.2 Indefinite Descriptions An indefinite description is traditionally written εx.P(x) and is known as Hilbert’s ε-operator. It denotes some x such that P(x) is true, provided one exists. Isabelle uses SOME for the Greek letter ε. Here is the definition of inv,3 which expresses inverses of functions: inv f ≡λy. SOME x. f x = y (inv_def) Using SOME rather than THE makes inv f behave well even if f is not injective. As it happens, most useful theorems about inv do assume the function to be injective. The inverse of f, when applied to y, returns some x such that f x = y. For example, we can prove inv Suc really is the inverse of the Suc function lemma "inv Suc (Suc n) = n" by (simp add: inv_def) The proof is a one-liner: the subgoal simplifies to a degenerate application of SOME, which is then erased. In detail, the left-hand side simplifies to SOME x. Suc x = Suc n, then to SOME x. x = n and finally to n. We know nothing about what inv Suc returns when applied to zero. The proof above still treats SOME as a definite description, since it only reasons about situations in which the value is described uniquely. Indeed, SOME satis-fies this rule: [ [P a; Vx. P x = ⇒x = a] ] = ⇒(SOME x. P x) = a (some_equality) To go further is tricky and requires rules such as these: P x = ⇒P (SOME x. P x) (someI) [ [P a; Vx. P x = ⇒Q x] ] = ⇒Q (SOME x. P x) (someI2) Rule someI is basic: if anything satisfies P then so does SOME x. P x. The repetition of P in the conclusion makes it difficult to apply in a backward proof, so the derived rule someI2 is also provided. For example, let us prove the axiom of choice: theorem axiom_of_choice: "(∀x. ∃y. P x y) = ⇒∃f. ∀x. P x (f x)" apply (rule exI, rule allI) 1. Vx. ∀x. ∃y. P x y = ⇒P x (?f x) We have applied the introduction rules; now it is time to apply the elimination rules. 3 In fact, inv is defined via a second constant inv_into, which we ignore here. 88 5. The Rules of the Game apply (drule spec, erule exE) 1. Vx y. P (?x2 x) y = ⇒P x (?f x) The rule someI automatically instantiates f to λx. SOME y. P x y, which is the choice function. It also instantiates ?x2 x to x. by (rule someI) Historical Note. The original purpose of Hilbert’s ε-operator was to ex-press an existential destruction rule: ∃x. P P[(εx.P)/ x] This rule is seldom used for that purpose — it can cause exponential blow-up — but it is occasionally used as an introduction rule for the ε-operator. Its name in HOL is someI_ex. 5.11 Some Proofs That Fail Most of the examples in this tutorial involve proving theorems. But not every conjecture is true, and it can be instructive to see how proofs fail. Here we attempt to prove a distributive law involving the existential quantifier and conjunction. lemma "(∃x. P x) ∧(∃x. Q x) = ⇒∃x. P x ∧Q x" The first steps are routine. We apply conjunction elimination to break the as-sumption into two existentially quantified assumptions. Applying existential elimination removes one of the quantifiers. apply (erule conjE) apply (erule exE) 1. Vx. [ [∃x. Q x; P x] ] = ⇒∃x. P x ∧Q x When we remove the other quantifier, we get a different bound variable in the subgoal. (The name xa is generated automatically.) apply (erule exE) 1. Vx xa. [ [P x; Q xa] ] = ⇒∃x. P x ∧Q x The proviso of the existential elimination rule has forced the variables to differ: we can hardly expect two arbitrary values to be equal! There is no way to prove this subgoal. Removing the conclusion’s existential quantifier yields two identical placeholders, which can become any term involving the variables x and xa. We need one to become x and the other to become xa, but Isabelle requires all instances of a placeholder to be identical. 5.11 Some Proofs That Fail 89 apply (rule exI) apply (rule conjI) 1. Vx xa. [ [P x; Q xa] ] = ⇒P (?x3 x xa) 2. Vx xa. [ [P x; Q xa] ] = ⇒Q (?x3 x xa) We can prove either subgoal using the assumption method. If we prove the first one, the placeholder changes into x. apply assumption 1. Vx xa. [ [P x; Q xa] ] = ⇒Q x We are left with a subgoal that cannot be proved. Applying the assumption method results in an error message: empty result sequence -- proof command failed When interacting with Isabelle via the shell interface, you can abandon a proof using the oops command. Here is another abortive proof, illustrating the interaction between bound variables and unknowns. If R is a reflexive relation, is there an x such that R x y holds for all y? Let us see what happens when we attempt to prove it. lemma "∀y. R y y = ⇒∃x. ∀y. R x y" First, we remove the existential quantifier. The new proof state has an un-known, namely ?x. apply (rule exI) 1. ∀y. R y y = ⇒∀y. R ?x y It looks like we can just apply assumption, but it fails. Isabelle refuses to substitute y, a bound variable, for ?x; that would be a bound variable capture. We can still try to finish the proof in some other way. We remove the universal quantifier from the conclusion, moving the bound variable y into the subgoal. But note that it is still bound! apply (rule allI) 1. Vy. ∀y. R y y = ⇒R ?x y Finally, we try to apply our reflexivity assumption. We obtain a new assump-tion whose identical placeholders may be replaced by any term involving y. apply (drule spec) 1. Vy. R (?z2 y) (?z2 y) = ⇒R ?x y This subgoal can only be proved by putting y for all the placeholders, making the assumption and conclusion become R y y. Isabelle can replace ?z2 y by y; this involves instantiating ?z2 to the identity function. But, just as two steps earlier, Isabelle refuses to substitute y for ?x. This example is typical of how Isabelle enforces sound quantifier reasoning. 90 5. The Rules of the Game 5.12 Proving Theorems Using the blast Method It is hard to prove many theorems using the methods described above. A proof may be hundreds of steps long. You may need to search among different ways of proving certain subgoals. Often a choice that proves one subgoal renders another impossible to prove. There are further complications that we have not discussed, concerning negation and disjunction. Isabelle’s classical reasoner is a family of tools that perform such proofs automatically. The most important of these is the blast method. In this section, we shall first see how to use the classical reasoner in its default mode and then how to insert additional rules, enabling it to work in new problem domains. We begin with examples from pure predicate logic. The following exam-ple is known as Andrew’s challenge. Peter Andrews designed it to be hard to prove by automatic means. It is particularly hard for a resolution prover, where converting the nested biconditionals to clause form produces a combi-natorial explosion . However, the blast method proves it in a fraction of a second. lemma "((∃x. ∀y. p(x)=p(y)) = ((∃x. q(x))=(∀y. p(y)))) = ((∃x. ∀y. q(x)=q(y)) = ((∃x. p(x))=(∀y. q(y))))" by blast The next example is a logic problem composed by Lewis Carroll. The blast method finds it trivial. Moreover, it turns out that not all of the assumptions are necessary. We can experiment with variations of this formula and see which ones can be proved. lemma "(∀x. honest(x) ∧industrious(x) − →healthy(x)) ∧ ¬ (∃x. grocer(x) ∧healthy(x)) ∧ (∀x. industrious(x) ∧grocer(x) − →honest(x)) ∧ (∀x. cyclist(x) − →industrious(x)) ∧ (∀x. ¬healthy(x) ∧cyclist(x) − →¬honest(x)) − →(∀x. grocer(x) − →¬cyclist(x))" by blast The blast method is also effective for set theory, which is described in the next chapter. The formula below may look horrible, but the blast method proves it in milliseconds. lemma "(S i∈I. A(i)) ∩(S j∈J. B(j)) = (S i∈I. S j∈J. A(i) ∩B(j))" by blast Few subgoals are couched purely in predicate logic and set theory. We can extend the scope of the classical reasoner by giving it new rules. Extending it effectively requires understanding the notions of introduction, elimination and destruction rules. Moreover, there is a distinction between safe and un-safe rules. A safe rule is one that can be applied backwards without losing 5.13 Other Classical Reasoning Methods 91 information; an unsafe rule loses information, perhaps transforming the sub-goal into one that cannot be proved. The safe/unsafe distinction affects the proof search: if a proof attempt fails, the classical reasoner backtracks to the most recent unsafe rule application and makes another choice. An important special case avoids all these complications. A logical equiv-alence, which in higher-order logic is an equality between formulas, can be given to the classical reasoner and simplifier by using the attribute iff. You should do so if the right hand side of the equivalence is simpler than the left-hand side. For example, here is a simple fact about list concatenation. The result of appending two lists is empty if and only if both of the lists are themselves empty. Obviously, applying this equivalence will result in a simpler goal. When stating this lemma, we include the iff attribute. Once we have proved the lemma, Isabelle will make it known to the classical reasoner (and to the simplifier). lemma [iff]: "(xs@ys = []) = (xs=[] ∧ys=[])" apply (induct_tac xs) apply (simp_all) done This fact about multiplication is also appropriate for the iff attribute: (?m ?n = 0) = (?m = 0 ∨?n = 0) A product is zero if and only if one of the factors is zero. The reasoning involves a disjunction. Proving new rules for disjunctive reasoning is hard, but translating to an actual disjunction works: the classical reasoner handles disjunction properly. In more detail, this is how the iff attribute works. It converts the equiv-alence P = Q to a pair of rules: the introduction rule Q = ⇒P and the destruction rule P = ⇒Q. It gives both to the classical reasoner as safe rules, ensuring that all occurrences of P in a subgoal are replaced by Q. The simpli-fier performs the same replacement, since iff gives P = Q to the simplifier. Classical reasoning is different from simplification. Simplification is deter-ministic. It applies rewrite rules repeatedly, as long as possible, transforming a goal into another goal. Classical reasoning uses search and backtracking in order to prove a goal outright. 5.13 Other Classical Reasoning Methods The blast method is our main workhorse for proving theorems automatically. Other components of the classical reasoner interact with the simplifier. Still others perform classical reasoning to a limited extent, giving the user fine control over the proof. Of the latter methods, the most useful is clarify. It performs all obvious reasoning steps without splitting the goal into multiple parts. It does not 92 5. The Rules of the Game apply unsafe rules that could render the goal unprovable. By performing the obvious steps, clarify lays bare the difficult parts of the problem, where human intervention is necessary. For example, the following conjecture is false: lemma "(∀x. P x) ∧(∃x. Q x) − →(∀x. P x ∧Q x)" apply clarify The blast method would simply fail, but clarify presents a subgoal that helps us see why we cannot continue the proof. 1. Vx xa. [ [∀x. P x; Q xa] ] = ⇒P x ∧Q x The proof must fail because the assumption Q xa and conclusion Q x refer to distinct bound variables. To reach this state, clarify applied the introduction rules for − →and ∀and the elimination rule for ∧. It did not apply the introduction rule for ∧because of its policy never to split goals. Also available is clarsimp, a method that interleaves clarify and simp. Also there is safe, which like clarify performs obvious steps but even applies those that split goals. The force method applies the classical reasoner and simplifier to one goal. Unless it can prove the goal, it fails. Contrast that with the auto method, which also combines classical reasoning with simplification. The latter’s pur-pose is to prove all the easy subgoals and parts of subgoals. Unfortunately, it can produce large numbers of new subgoals; also, since it proves some subgoals and splits others, it obscures the structure of the proof tree. The force method does not have these drawbacks. Another difference: force tries harder than auto to prove its goal, so it can take much longer to terminate. Older components of the classical reasoner have largely been superseded by blast, but they still have niche applications. Most important among these are fast and best. While blast searches for proofs using a built-in first-order reasoner, these earlier methods search for proofs using standard Isabelle inference. That makes them slower but enables them to work in the presence of the more unusual features of Isabelle rules, such as type classes and function unknowns. For example, recall the introduction rule for Hilbert’s ε-operator: ?P ?x = ⇒?P (SOME x. ?P x) (someI) The repeated occurrence of the variable ?P makes this rule tricky to apply. Consider this contrived example: lemma "[ [Q a; P a] ] = ⇒P (SOME x. P x ∧Q x) ∧Q (SOME x. P x ∧Q x)" apply (rule someI) We can apply rule someI explicitly. It yields the following subgoal: 1. [ [Q a; P a] ] = ⇒P ?x ∧Q ?x The proof from this point is trivial. Could we have proved the theorem with a single command? Not using blast: it cannot perform the higher-order uni-fication needed here. The fast method succeeds: 5.14 Finding More Theorems 93 apply (fast intro!: someI) The best method is similar to fast but it uses a best-first search instead of depth-first search. Accordingly, it is slower but is less susceptible to di-vergence. Transitivity rules usually cause fast to loop where best can often manage. Here is a summary of the classical reasoning methods: – blast works automatically and is the fastest – clarify and clarsimp perform obvious steps without splitting the goal; safe even splits goals – force uses classical reasoning and simplification to prove a goal; auto is similar but leaves what it cannot prove – fast and best are legacy methods that work well with rules involving un-usual features A table illustrates the relationships among four of these methods. no split split no simp clarify safe simp clarsimp auto 5.14 Finding More Theorems In Sect. 3.1.11, we introduced Proof General’s Find button for finding theo-rems in the database via pattern matching. If we are inside a proof, we can be more specific; we can search for introduction, elimination and destruction rules with respect to the current goal. For this purpose, Find provides three aditional search criteria: intro, elim and dest. For example, given the goal 1. A ∧B you can click on Find and type in the search expression intro. You will be shown a few rules ending in = ⇒?P ∧?Q, among them conjI. You may even discover that the very theorem you are trying to prove is already in the database. Given the goal 1. A − →A the search for intro finds not just impI but also imp_refl: ?P − →?P. As before, search criteria can be combined freely: for example, " @ " intro searches for all introduction rules that match the current goal and mention the @ function. Searching for elimination and destruction rules via elim and dest is anal-ogous to intro but takes the assumptions into account, too. 94 5. The Rules of the Game 5.15 Forward Proof: Transforming Theorems Forward proof means deriving new facts from old ones. It is the most funda-mental type of proof. Backward proof, by working from goals to subgoals, can help us find a difficult proof. But it is not always the best way of presenting the proof thus found. Forward proof is particularly good for reasoning from the general to the specific. For example, consider this distributive law for the greatest common divisor: k × gcd(m, n) = gcd(k × m, k × n) Putting m = 1 we get (since gcd(1, n) = 1 and k × 1 = k) k = gcd(k, k × n) We have derived a new fact; if re-oriented, it might be useful for simplification. After re-orienting it and putting n = 1, we derive another useful law: gcd(k, k) = k Substituting values for variables — instantiation — is a forward step. Re-orientation works by applying the symmetry of equality to an equation, so it too is a forward step. 5.15.1 Modifying a Theorem using of, where and THEN Let us reproduce our examples in Isabelle. Recall that in Sect. 3.5.3 we de-clared the recursive function gcd: fun gcd :: "nat ⇒nat ⇒nat" where "gcd m n = (if n=0 then m else gcd n (m mod n))" From this definition, it is possible to prove the distributive law. That takes us to the starting point for our example. ?k gcd ?m ?n = gcd (?k ?m) (?k ?n) (gcd_mult_distrib2) The first step in our derivation is to replace ?m by 1. We instantiate the theorem using of, which identifies variables in order of their appearance from left to right. In this case, the variables are ?k, ?m and ?n. So, the expression [of k 1] replaces ?k by k and ?m by 1. lemmas gcd_mult_0 = gcd_mult_distrib2 [of k 1] The keyword lemmas declares a new theorem, which can be derived from an existing one using attributes such as [of k 1]. The command thm gcd_mult_0 displays the result: k gcd 1 ?n = gcd (k 1) (k ?n) 5.15 Forward Proof: Transforming Theorems 95 Something is odd: k is an ordinary variable, while ?n is schematic. We did not specify an instantiation for ?n. In its present form, the theorem does not allow substitution for k. One solution is to avoid giving an instantiation for ?k: instead of a term we can put an underscore (). For example, gcd_mult_distrib2 [of _ 1] replaces ?m by 1 but leaves ?k unchanged. An equivalent solution is to use the attribute where. gcd mult distrib2 [where m=1] While of refers to variables by their position, where refers to variables by name. Multiple instantiations are separated by and, as in this example: gcd mult distrib2 [where m=1 and k=1] We now continue the present example with the version of gcd_mult_0 shown above, which has k instead of ?k. Once we have replaced ?m by 1, we must next simplify the theorem gcd_mult_0, performing the steps gcd(1, n) = 1 and k × 1 = k. The simplified attribute takes a theorem and returns the result of simplifying it, with respect to the default simplification rules: lemmas gcd_mult_1 = gcd_mult_0 [simplified] Again, we display the resulting theorem: k = gcd k (k ?n) To re-orient the equation requires the symmetry rule: ?s = ?t = ⇒?t = ?s (sym) The following declaration gives our equation to sym: lemmas gcd_mult = gcd_mult_1 [THEN sym] Here is the result: gcd k (k ?n) = k THEN sym gives the current theorem to the rule sym and returns the result-ing conclusion. The effect is to exchange the two operands of the equality. Typically THEN is used with destruction rules. Also useful is THEN spec, which removes the quantifier from a theorem of the form ∀x. P, and THEN mp, which converts the implication P →Q into the rule P Q. Similar to mp are the follow-ing two rules, which extract the two directions of reasoning about a boolean equivalence: [ [?Q = ?P; ?Q] ] = ⇒?P (iffD1) [ [?P = ?Q; ?Q] ] = ⇒?P (iffD2) Normally we would never name the intermediate theorems such as gcd_mult_0 and gcd_mult_1 but would combine the three forward steps: lemmas gcd_mult = gcd_mult_distrib2 [of k 1, simplified, THEN sym] 96 5. The Rules of the Game The directives, or attributes, are processed from left to right. This declaration of gcd_mult is equivalent to the previous one. Such declarations can make the proof script hard to read. Better is to state the new lemma explicitly and to prove it using a single rule method whose operand is expressed using forward reasoning: lemma gcd mult [simp]: "gcd k (kn) = k" by (rule gcd_mult_distrib2 [of k 1, simplified, THEN sym]) Compared with the previous proof of gcd_mult, this version shows the reader what has been proved. Also, the result will be processed in the normal way. In particular, Isabelle generalizes over all variables: the resulting theorem will have ?k instead of k. At the start of this section, we also saw a proof of gcd(k, k) = k. Here is the Isabelle version: lemma gcd self [simp]: "gcd k k = k" by (rule gcd_mult [of k 1, simplified]) ! ! To give of a nonatomic term, enclose it in quotation marks, as in [of "km"]. The term must not contain unknowns: an attribute such as [of "?km"] will be rejected. 5.15.2 Modifying a Theorem using OF Recall that of generates an instance of a rule by specifying values for its variables. Analogous is OF, which generates an instance of a rule by specifying facts for its premises. We again need the divides relation of number theory, which as we recall is defined by ?m dvd ?n ≡∃k. ?n = ?m k (dvd_def) Suppose, for example, that we have proved the following rule. It states that if k and n are relatively prime and if k divides m × n then k divides m. [ [gcd ?k ?n = 1; ?k dvd ?m ?n] ] = ⇒?k dvd ?m (relprime_dvd_mult) We can use OF to create an instance of this rule. First, we prove an instance of its first premise: lemma relprime 20 81: "gcd 20 81 = 1" by (simp add: gcd.simps) We have evaluated an application of the gcd function by simplification. Ex-pression evaluation involving recursive functions is not guaranteed to termi-nate, and it can be slow; Isabelle performs arithmetic by rewriting symbolic bit strings. Here, however, the simplification takes less than one second. We can give this new lemma to OF. The expression relprime_dvd_mult [OF relprime_20_81] 5.16 Forward Reasoning in a Backward Proof 97 yields the theorem 20 dvd (?m 81) = ⇒20 dvd ?m OF takes any number of operands. Consider the following facts about the divides relation: [ [?k dvd ?m; ?k dvd ?n] ] = ⇒?k dvd ?m + ?n (dvd_add) ?m dvd ?m (dvd_refl) Let us supply dvd_refl for each of the premises of dvd_add: dvd_add [OF dvd_refl dvd_refl] Here is the theorem that we have expressed: ?k dvd (?k + ?k) As with of, we can use the _ symbol to leave some positions unspecified: dvd_add [OF _ dvd_refl] The result is ?k dvd ?m = ⇒?k dvd ?m + ?k You may have noticed that THEN and OF are based on the same idea, namely to combine two rules. They differ in the order of the combination and thus in their effect. We use THEN typically with a destruction rule to extract a subformula of the current theorem. We use OF with a list of facts to generate an instance of the current theorem. Here is a summary of some primitives for forward reasoning: – of instantiates the variables of a rule to a list of terms – OF applies a rule to a list of theorems – THEN gives a theorem to a named rule and returns the conclusion – simplified applies the simplifier to a theorem – lemmas assigns a name to the theorem produced by the attributes above 5.16 Forward Reasoning in a Backward Proof We have seen that the forward proof directives work well within a backward proof. There are many ways to achieve a forward style using our existing proof methods. We shall also meet some new methods that perform forward reasoning. The methods drule, frule, drule_tac, etc., reason forward from a subgoal. We have seen them already, using rules such as mp and spec to operate on formulae. They can also operate on terms, using rules such as these: x = y = ⇒f x = f y (arg_cong) i ≤j = ⇒i k ≤j k (mult_le_mono1) For example, let us prove a fact about divisibility in the natural numbers: 98 5. The Rules of the Game lemma "2 ≤u = ⇒um ̸= Suc(un)" apply (intro notI) 1. [ [2 ≤u; u m = Suc (u n)] ] = ⇒False The key step is to apply the function …mod u to both sides of the equation um = Suc(un): apply (drule_tac f="λx. x mod u" in arg_cong) 1. [ [2 ≤u; u m mod u = Suc (u n) mod u] ] = ⇒False Simplification reduces the left side to 0 and the right side to 1, yielding the required contradiction. apply (simp add: mod_Suc) done Our proof has used a fact about remainder: Suc m mod n = (if Suc (m mod n) = n then 0 else Suc (m mod n)) (mod_Suc) 5.16.1 The Method insert The insert method inserts a given theorem as a new assumption of all sub-goals. This already is a forward step; moreover, we may (as always when using a theorem) apply of, THEN and other directives. The new assumption can then be used to help prove the subgoals. For example, consider this theorem about the divides relation. The first proof step inserts the distributive law for gcd. We specify its variables as shown. lemma relprime dvd mult: "[ [ gcd k n = 1; k dvd mn ] ] = ⇒k dvd m" apply (insert gcd_mult_distrib2 [of m k n]) In the resulting subgoal, note how the equation has been inserted: 1. [ [gcd k n = 1; k dvd m n; m gcd k n = gcd (m k) (m n)] ] = ⇒k dvd m The next proof step utilizes the assumption gcd k n = 1 (note that Suc 0 is another expression for 1): apply(simp) 1. [ [gcd k n = Suc 0; k dvd m n; m = gcd (m k) (m n)] ] = ⇒k dvd m Simplification has yielded an equation for m. The rest of the proof is omitted. Here is another demonstration of insert. Division and remainder obey a well-known law: (?m div ?n) ?n + ?m mod ?n = ?m (div_mult_mod_eq) We refer to this law explicitly in the following proof: 5.16 Forward Reasoning in a Backward Proof 99 lemma div_mult_self_is_m: "0<n = ⇒(mn) div n = (m::nat)" apply (insert div_mult_mod_eq [of "mn" n]) apply (simp) done The first step inserts the law, specifying mn and n for its variables. Notice that non-trivial expressions must be enclosed in quotation marks. Here is the resulting subgoal, with its new assumption: 1. [ [0 < n; (m n) div n n + (m n) mod n = m n] ] = ⇒(m n) div n = m Simplification reduces (m n) mod n to zero. Then it cancels the factor n on both sides of the equation (m n) div n n = m n, proving the theorem. ! ! Any unknowns in the theorem given to insert will be universally quantified in the new assumption. 5.16.2 The Method subgoal_tac A related method is subgoal_tac, but instead of inserting a theorem as an assumption, it inserts an arbitrary formula. This formula must be proved later as a separate subgoal. The idea is to claim that the formula holds on the basis of the current assumptions, to use this claim to complete the proof, and finally to justify the claim. It gives the proof some structure. If you find yourself generating a complex assumption by a long series of forward steps, consider using subgoal_tac instead: you can state the formula you are aiming for, and perhaps prove it automatically. Look at the following example. lemma "[ [(z::int) < 37; 66 < 2z; zz ̸= 1225; Q(34); Q(36)] ] = ⇒Q(z)" apply (subgoal_tac "z = 34 ∨z = 36") apply blast apply (subgoal_tac "z ̸= 35") apply arith apply force done The first assumption tells us that z is no greater than 36. The second tells us that z is at least 34. The third assumption tells us that z cannot be 35, since 35 × 35 = 1225. So z is either 34 or 36, and since Q holds for both of those values, we have the conclusion. The Isabelle proof closely follows this reasoning. The first step is to claim that z is either 34 or 36. The resulting proof state gives us two subgoals: 1. [ [z < 37; 66 < 2 z; z z ̸= 1225; Q 34; Q 36; z = 34 ∨z = 36] ] = ⇒Q z 2. [ [z < 37; 66 < 2 z; z z ̸= 1225; Q 34; Q 36] ] = ⇒z = 34 ∨z = 36 100 5. The Rules of the Game The first subgoal is trivial (blast), but for the second Isabelle needs help to eliminate the case z=35. The second invocation of subgoal_tac leaves two subgoals: 1. [ [z < 37; 66 < 2 z; z z ̸= 1225; Q 34; Q 36; z ̸= 35] ] = ⇒z = 34 ∨z = 36 2. [ [z < 37; 66 < 2 z; z z ̸= 1225; Q 34; Q 36] ] = ⇒z ̸= 35 Assuming that z is not 35, the first subgoal follows by linear arithmetic (arith). For the second subgoal we apply the method force, which proceeds by assuming that z=35 and arriving at a contradiction. Summary of these methods: – insert adds a theorem as a new assumption – subgoal_tac adds a formula as a new assumption and leaves the subgoal of proving that formula 5.17 Managing Large Proofs Naturally you should try to divide proofs into manageable parts. Look for lemmas that can be proved separately. Sometimes you will observe that they are instances of much simpler facts. On other occasions, no lemmas suggest themselves and you are forced to cope with a long proof involving many subgoals. 5.17.1 Tacticals, or Control Structures If the proof is long, perhaps it at least has some regularity. Then you can express it more concisely using tacticals, which provide control structures. Here is a proof (it would be a one-liner using blast, but forget that) that contains a series of repeated commands: lemma "[ [P− →Q; Q− →R; R− →S; P] ] = ⇒S" apply (drule mp, assumption) apply (drule mp, assumption) apply (drule mp, assumption) apply (assumption) done Each of the three identical commands finds an implication and proves its antecedent by assumption. The first one finds P− →Q and P, concluding Q; the second one concludes R and the third one concludes S. The final step matches the assumption S with the goal to be proved. Suffixing a method with a plus sign (+) expresses one or more repetitions: 5.17 Managing Large Proofs 101 lemma "[ [P− →Q; Q− →R; R− →S; P] ] = ⇒S" by (drule mp, assumption)+ Using by takes care of the final use of assumption. The new proof is more concise. It is also more general: the repetitive method works for a chain of implications having any length, not just three. Choice is another control structure. Separating two methods by a vertical bar (\|) gives the effect of applying the first method, and if that fails, trying the second. It can be combined with repetition, when the choice must be made over and over again. Here is a chain of implications in which most of the antecedents are proved by assumption, but one is proved by arithmetic: lemma "[ [Q− →R; P− →Q; x<5− →P; Suc x < 5] ] = ⇒R" by (drule mp, (assumption\|arith))+ The arith method can prove x < 5 from x+1 < 5, but it cannot duplicate the effect of assumption. Therefore, we combine these methods using the choice operator. A postfixed question mark (?) expresses zero or one repetitions of a method. It can also be viewed as the choice between executing a method and doing nothing. It is useless at top level but can be valuable within other control structures; for example, (m+)? performs zero or more repetitions of method m. 5.17.2 Subgoal Numbering Another problem in large proofs is contending with huge subgoals or many subgoals. Induction can produce a proof state that looks like this: 1. bigsubgoal1 2. bigsubgoal2 3. bigsubgoal3 4. bigsubgoal4 5. bigsubgoal5 6. bigsubgoal6 If each bigsubgoal is 15 lines or so, the proof state will be too big to scroll through. By default, Isabelle displays at most 10 subgoals. The pr command lets you change this limit: pr 2 1. bigsubgoal1 2. bigsubgoal2 A total of 6 subgoals... All methods apply to the first subgoal. Sometimes, not only in a large proof, you may want to focus on some other subgoal. Then you should try the commands defer or prefer. In the following example, the first subgoal looks hard, while the others look as if blast alone could prove them: 102 5. The Rules of the Game 1. hard 2. ¬ ¬ P = ⇒P 3. Q = ⇒Q The defer command moves the first subgoal into the last position. defer 1 1. ¬ ¬ P = ⇒P 2. Q = ⇒Q 3. hard Now we apply blast repeatedly to the easy subgoals: apply blast+ 1. hard Using defer, we have cleared away the trivial parts of the proof so that we can devote attention to the difficult part. The prefer command moves the specified subgoal into the first position. For example, if you suspect that one of your subgoals is invalid (not a theo-rem), then you should investigate that subgoal first. If it cannot be proved, then there is no point in proving the other subgoals. 1. ok1 2. ok2 3. doubtful We decide to work on the third subgoal. prefer 3 1. doubtful 2. ok1 3. ok2 If we manage to prove doubtful, then we can work on the other subgoals, confident that we are not wasting our time. Finally we revise the proof script to remove the prefer command, since we needed it only to focus our ex-ploration. The previous example is different: its use of defer stops trivial subgoals from cluttering the rest of the proof. Even there, we should consider proving hard as a preliminary lemma. Always seek ways to streamline your proofs. Summary: – the control structures +, ? and \| help express complicated proofs – the pr command can limit the number of subgoals to display – the defer and prefer commands move a subgoal to the last or first position Exercise 5.17.1 Explain the use of ? and + in this proof. lemma "[ [P∧Q− →R; P− →Q; P] ] = ⇒R" by (drule mp, (intro conjI)?, assumption+)+ 5.18 Proving the Correctness of Euclid’s Algorithm 103 5.18 Proving the Correctness of Euclid’s Algorithm A brief development will demonstrate the techniques of this chapter, includ-ing blast applied with additional rules. We shall also see case_tac used to perform a Boolean case split. Let us prove that gcd computes the greatest common divisor of its two ar-guments. We use induction: gcd.induct is the induction rule returned by fun. We simplify using rules proved in Sect. 3.5.3, since rewriting by the definition of gcd can loop. lemma gcd dvd both: "(gcd m n dvd m) ∧(gcd m n dvd n)" The induction formula must be a conjunction. In the inductive step, each conjunct establishes the other. 1. Vm n. (n ̸= 0 = ⇒ gcd n (m mod n) dvd n ∧ gcd n (m mod n) dvd m mod n) = ⇒ gcd m n dvd m ∧gcd m n dvd n The conditional induction hypothesis suggests doing a case analysis on n=0. We apply case_tac with type bool — and not with a datatype, as we have done until now. Since nat is a datatype, we could have written case_tac n instead of case_tac "n=0". However, the definition of gcd makes a Boolean decision: "gcd m n = (if n=0 then m else gcd n (m mod n))" Proofs about a function frequently follow the function’s definition, so we perform case analysis over the same formula. apply (case_tac "n=0") 1. Vm n. [ [n ̸= 0 = ⇒ gcd n (m mod n) dvd n ∧gcd n (m mod n) dvd m mod n; n = 0] ] = ⇒gcd m n dvd m ∧gcd m n dvd n 2. Vm n. [ [n ̸= 0 = ⇒ gcd n (m mod n) dvd n ∧gcd n (m mod n) dvd m mod n; n ̸= 0] ] = ⇒gcd m n dvd m ∧gcd m n dvd n Simplification leaves one subgoal: apply (simp_all) 1. Vm n. [ [gcd n (m mod n) dvd n ∧gcd n (m mod n) dvd m mod n; 0 < n] ] = ⇒gcd n (m mod n) dvd m Here, we can use blast. One of the assumptions, the induction hypothesis, is a conjunction. The two divides relationships it asserts are enough to prove the conclusion, for we have the following theorem at our disposal: [ [?k dvd (?m mod ?n); ?k dvd ?n] ] = ⇒?k dvd ?m (dvd_mod_imp_dvd) 104 5. The Rules of the Game This theorem can be applied in various ways. As an introduction rule, it would cause backward chaining from the conclusion (namely ?k dvd ?m) to the two premises, which also involve the divides relation. This process does not look promising and could easily loop. More sensible is to apply the rule in the forward direction; each step would eliminate an occurrence of the mod symbol, so the process must terminate. apply (blast dest: dvd_mod_imp_dvd) done Attaching the dest attribute to dvd_mod_imp_dvd tells blast to use it as de-struction rule; that is, in the forward direction. We have proved a conjunction. Now, let us give names to each of the two halves: lemmas gcd_dvd1 [iff] = gcd_dvd_both [THEN conjunct1] lemmas gcd_dvd2 [iff] = gcd_dvd_both [THEN conjunct2] Here we see lemmas used with the iff attribute, which supplies the new theorems to the classical reasoner and the simplifier. Recall that THEN is frequently used with destruction rules; THEN conjunct1 extracts the first half of a conjunctive theorem. Given gcd_dvd_both it yields gcd ?m1 ?n1 dvd ?m1 The variable names ?m1 and ?n1 arise because Isabelle renames schematic variables to prevent clashes. The second lemmas declaration yields gcd ?m1 ?n1 dvd ?n1 To complete the verification of the gcd function, we must prove that it returns the greatest of all the common divisors of its arguments. The proof is by induction, case analysis and simplification. lemma gcd greatest [rule format]: "k dvd m − →k dvd n − →k dvd gcd m n" The goal is expressed using HOL implication, − →, because the induction af-fects the two preconditions. The directive rule_format tells Isabelle to replace each − →by = ⇒before storing the eventual theorem. This directive can also remove outer universal quantifiers, converting the theorem into the usual for-mat for inference rules. It can replace any series of applications of THEN to the rules mp and spec. We did not have to write this: lemma gcd_greatest [THEN mp, THEN mp]: "k dvd m − →k dvd n − →k dvd gcd m n" Because we are again reasoning about gcd, we perform the same induction and case analysis as in the previous proof: 1. Vm n. [ [n ̸= 0 = ⇒ k dvd n − →k dvd m mod n − →k dvd gcd n (m mod n); n = 0] ] 5.18 Proving the Correctness of Euclid’s Algorithm 105 = ⇒k dvd m − →k dvd n − →k dvd gcd m n 2. Vm n. [ [n ̸= 0 = ⇒ k dvd n − →k dvd m mod n − →k dvd gcd n (m mod n); n ̸= 0] ] = ⇒k dvd m − →k dvd n − →k dvd gcd m n Simplification proves both subgoals. apply (simp_all add: dvd_mod) done In the first, where n=0, the implication becomes trivial: k dvd gcd m n goes to k dvd m. The second subgoal is proved by an unfolding of gcd, using this rule about divides: [ [?f dvd ?m; ?f dvd ?n] ] = ⇒?f dvd ?m mod ?n (dvd_mod) The facts proved above can be summarized as a single logical equivalence. This step gives us a chance to see another application of blast. theorem gcd greatest iff [iff]: "(k dvd gcd m n) = (k dvd m ∧k dvd n)" by (blast intro!: gcd_greatest intro: dvd_trans) This theorem concisely expresses the correctness of the gcd function. We state it with the iff attribute so that Isabelle can use it to remove some occur-rences of gcd. The theorem has a one-line proof using blast supplied with two additional introduction rules. The exclamation mark (intro!) signifies safe rules, which are applied aggressively. Rules given without the exclama-tion mark are applied reluctantly and their uses can be undone if the search backtracks. Here the unsafe rule expresses transitivity of the divides relation: [ [?m dvd ?n; ?n dvd ?p] ] = ⇒?m dvd ?p (dvd_trans) Applying dvd_trans as an introduction rule entails a risk of looping, for it multiplies occurrences of the divides symbol. However, this proof relies on transitivity reasoning. The rule gcd greatest is safe to apply aggressively because it yields simpler subgoals. The proof implicitly uses gcd_dvd1 and gcd_dvd2 as safe rules, because they were declared using iff. 6. Sets, Functions and Relations This chapter describes the formalization of typed set theory, which is the basis of much else in HOL. For example, an inductive definition yields a set, and the abstract theories of relations regard a relation as a set of pairs. The chapter introduces the well-known constants such as union and intersection, as well as the main operations on relations, such as converse, composition and transitive closure. Functions are also covered. They are not sets in HOL, but many of their properties concern sets: the range of a function is a set, and the inverse image of a function maps sets to sets. This chapter will be useful to anybody who plans to develop a substantial proof. Sets are convenient for formalizing computer science concepts such as grammars, logical calculi and state transition systems. Isabelle can prove many statements involving sets automatically. This chapter ends with a case study concerning model checking for the temporal logic CTL. Most of the other examples are simple. The chapter presents a small selection of built-in theorems in order to point out some key properties of the various constants and to introduce you to the notation. Natural deduction rules are provided for the set theory constants, but they are seldom used directly, so only a few are presented here. 6.1 Sets HOL’s set theory should not be confused with traditional, untyped set theory, in which everything is a set. Our sets are typed. In a given set, all elements have the same type, say τ, and the set itself has type τ set. We begin with intersection, union and complement. In addition to the membership relation, there is a symbol for its negation. These points can be seen below. Here are the natural deduction rules for intersection. Note the resemblance to those for conjunction. [ [c ∈A; c ∈B] ] = ⇒c ∈A ∩B (IntI) c ∈A ∩B = ⇒c ∈A (IntD1) c ∈A ∩B = ⇒c ∈B (IntD2) Here are two of the many installed theorems concerning set complement. Note that it is denoted by a minus sign. 108 6. Sets, Functions and Relations (c ∈- A) = (c / ∈A) (Compl_iff) - (A ∪B) = - A ∩- B (Compl_Un) Set difference is the intersection of a set with the complement of another set. Here we also see the syntax for the empty set and for the universal set. A ∩(B - A) = {} (Diff_disjoint) A ∪- A = UNIV (Compl_partition) The subset relation holds between two sets just if every element of one is also an element of the other. This relation is reflexive. These are its natural deduction rules: (Vx. x ∈A = ⇒x ∈B) = ⇒A ⊆B (subsetI) [ [A ⊆B; c ∈A] ] = ⇒c ∈B (subsetD) In harder proofs, you may need to apply subsetD giving a specific term for c. However, blast can instantly prove facts such as this one: (A ∪B ⊆C) = (A ⊆C ∧B ⊆C) (Un_subset_iff) Here is another example, also proved automatically: lemma "(A ⊆-B) = (B ⊆-A)" by blast This is the same example using ascii syntax, illustrating a pitfall: lemma "(A <= -B) = (B <= -A)" The proof fails. It is not a statement about sets, due to overloading; the relation symbol <= can be any relation, not just subset. In this general form, the statement is not valid. Putting in a type constraint forces the variables to denote sets, allowing the proof to succeed: lemma "((A:: 'a set) <= -B) = (B <= -A)" Section 8.3 below describes overloading. Incidentally, A ⊆-B asserts that the sets A and B are disjoint. Two sets are equal if they contain the same elements. This is the principle of extensionality for sets. (Vx. (x ∈A) = (x ∈B)) = ⇒A = B (set_ext) Extensionality can be expressed as A = B ⇐ ⇒(A ⊆B) ∧(B ⊆A). The following rules express both directions of this equivalence. Proving a set equa-tion using equalityI allows the two inclusions to be proved independently. [ [A ⊆B; B ⊆A] ] = ⇒A = B (equalityI) [ [A = B; [ [A ⊆B; B ⊆A] ] = ⇒P] ] = ⇒P (equalityE) 6.1 Sets 109 6.1.1 Finite Set Notation Finite sets are expressed using the constant insert, which is a form of union: insert a A = {a} ∪A (insert_is_Un) The finite set expression {a,b} abbreviates insert a (insert b {}). Many facts about finite sets can be proved automatically: lemma "{a,b} ∪{c,d} = {a,b,c,d}" by blast Not everything that we would like to prove is valid. Consider this attempt: lemma "{a,b} ∩{b,c} = {b}" apply auto The proof fails, leaving the subgoal b=c. To see why it fails, consider a correct version: lemma "{a,b} ∩{b,c} = (if a=c then {a,b} else {b})" apply simp by blast Our mistake was to suppose that the various items were distinct. Another remark: this proof uses two methods, namely simp and blast. Calling simp eliminates the if-then-else expression, which blast cannot break down. The combined methods (namely force and auto) can prove this fact in one step. 6.1.2 Set Comprehension The set comprehension {x. P} expresses the set of all elements that satisfy the predicate P. Two laws describe the relationship between set comprehension and the membership relation: (a ∈{x. P x}) = P a (mem_Collect_eq) {x. x ∈A} = A (Collect_mem_eq) Facts such as these have trivial proofs: lemma "{x. P x ∨x ∈A} = {x. P x} ∪A" lemma "{x. P x − →Q x} = -{x. P x} ∪{x. Q x}" Isabelle has a general syntax for comprehension, which is best described through an example: lemma "{pq \| p q. p∈prime ∧q∈prime} = {z. ∃p q. z = pq ∧p∈prime ∧q∈prime}" The left and right hand sides of this equation are identical. The syntax used in the left-hand side abbreviates the right-hand side: in this case, all numbers that are the product of two primes. The syntax provides a neat way of ex-pressing any set given by an expression built up from variables under specific 110 6. Sets, Functions and Relations constraints. The drawback is that it hides the true form of the expression, with its existential quantifiers. Remark. We do not need sets at all. They are essentially equivalent to predicate variables, which are allowed in higher-order logic. The main benefit of sets is their notation; we can write x∈A and {z. P} where predicates would require writing A(x) and λz. P. 6.1.3 Binding Operators Universal and existential quantifications may range over sets, with the obvi-ous meaning. Here are the natural deduction rules for the bounded universal quantifier. Occasionally you will need to apply bspec with an explicit instan-tiation of the variable x: (Vx. x ∈A = ⇒P x) = ⇒∀x∈A. P x (ballI) [ [∀x∈A. P x; x ∈A] ] = ⇒P x (bspec) Dually, here are the natural deduction rules for the bounded existential quan-tifier. You may need to apply bexI with an explicit instantiation: [ [P x; x ∈A] ] = ⇒∃x∈A. P x (bexI) [ [∃x∈A. P x; Vx. [ [x ∈A; P x] ] = ⇒Q] ] = ⇒Q (bexE) Unions can be formed over the values of a given set. The syntax is S x∈A. B or UN x:A. B in ascii. Indexed union satisfies this basic law: (b ∈(S x∈A. B x)) = (∃x∈A. b ∈B x) (UN_iff) It has two natural deduction rules similar to those for the existential quanti-fier. Sometimes UN_I must be applied explicitly: [ [a ∈A; b ∈B a] ] = ⇒b ∈(S x∈A. B x) (UN_I) [ [b ∈(S x∈A. B x); Vx. [ [x ∈A; b ∈B x] ] = ⇒R] ] = ⇒R (UN_E) The following built-in abbreviation (see Sect. 4.1.4) lets us express the union over a type: (S x. B x) ≡(S x∈UNIV. B x) We may also express the union of a set of sets, written Union C in ascii: (A ∈S C) = (∃X∈C. A ∈X) (Union_iff) Intersections are treated dually, although they seem to be used less often than unions. The syntax below would be INT x: A. B and Inter C in ascii. Among others, these theorems are available: (b ∈(T x∈A. B x)) = (∀x∈A. b ∈B x) (INT_iff) (A ∈T C) = (∀X∈C. A ∈X) (Inter_iff) Isabelle uses logical equivalences such as those above in automatic proof. Unions, intersections and so forth are not simply replaced by their definitions. Instead, membership tests are simplified. For example, x ∈A ∪B is replaced by x ∈A ∨x ∈B. 6.2 Functions 111 The internal form of a comprehension involves the constant Collect, which occasionally appears when a goal or theorem is displayed. For example, Collect P is the same term as {x. P x}. The same thing can happen with quantifiers: All P is ∀x. P x and Ex P is ∃x. P x; also Ball A P is ∀x∈A. P x and Bex A P is ∃x∈A. P x. For indexed unions and intersections, you may see the constants UNION and INTER. The internal constant for εx.P(x) is Eps. We have only scratched the surface of Isabelle/HOL’s set theory, which provides hundreds of theorems for your use. 6.1.4 Finiteness and Cardinality The predicate finite holds of all finite sets. Isabelle/HOL includes many familiar theorems about finiteness and cardinality (card). For example, we have theorems concerning the cardinalities of unions, intersections and the powerset: [ [finite A; finite B] ] = ⇒card A + card B = card (A ∪B) + card (A ∩B) (card_Un_Int) finite A = ⇒card (Pow A) = 2 ^ card A (card_Pow) finite A = ⇒ card {B. B ⊆A ∧card B = k} = card A choose k (n_subsets) Writing \|A\| as n, the last of these theorems says that the number of k-element subsets of A is n k . 6.2 Functions This section describes a few concepts that involve functions. Some of the more important theorems are given along with the names. A few sample proofs appear. Unlike with set theory, however, we cannot simply state lemmas and expect them to be proved using blast. 6.2.1 Function Basics Two functions are equal if they yield equal results given equal arguments. This is the principle of extensionality for functions: (Vx. f x = g x) = ⇒f = g (ext) Function update is useful for modelling machine states. It has the obvi-ous definition and many useful facts are proved about it. In particular, the following equation is installed as a simplification rule: (f(x:=y)) z = (if z = x then y else f z) (fun_upd_apply) 112 6. Sets, Functions and Relations Two syntactic points must be noted. In (f(x:=y)) z we are applying an up-dated function to an argument; the outer parentheses are essential. A series of two or more updates can be abbreviated as shown on the left-hand side of this theorem: f(x:=y, x:=z) = f(x:=z) (fun_upd_upd) Note also that we can write f(x:=z) with only one pair of parentheses when it is not being applied to an argument. The identity function and function composition are defined: id ≡λx. x (id_def) f ◦g ≡λx. f (g x) (o_def) Many familiar theorems concerning the identity and composition are proved. For example, we have the associativity of composition: f ◦(g ◦h) = f ◦g ◦h (o_assoc) 6.2.2 Injections, Surjections, Bijections A function may be injective, surjective or bijective: inj_on f A ≡∀x∈A. ∀y∈A. f x = f y − →x = y (inj_on_def) surj f ≡∀y. ∃x. y = f x (surj_def) bij f ≡inj f ∧surj f (bij_def) The second argument of inj_on lets us express that a function is injective over a given set. This refinement is useful in higher-order logic, where functions are total; in some cases, a function’s natural domain is a subset of its domain type. Writing inj f abbreviates inj_on f UNIV, for when f is injective everywhere. The operator inv expresses the inverse of a function. In general the inverse may not be well behaved. We have the usual laws, such as these: inj f = ⇒inv f (f x) = x (inv_f_f) surj f = ⇒f (inv f y) = y (surj_f_inv_f) bij f = ⇒inv (inv f) = f (inv_inv_eq) Theorems involving these concepts can be hard to prove. The following example is easy, but it cannot be proved automatically. To begin with, we need a law that relates the equality of functions to equality over all arguments: (f = g) = (∀x. f x = g x) (fun_eq_iff) This is just a restatement of extensionality. Our lemma states that an injec-tion can be cancelled from the left side of function composition: lemma "inj f = ⇒(f o g = f o h) = (g = h)" apply (simp add: fun_eq_iff inj_on_def) apply auto done The first step of the proof invokes extensionality and the definitions of injectiveness and composition. It leaves one subgoal: 6.3 Relations 113 1. ∀x y. f x = f y − →x = y = ⇒ (∀x. f (g x) = f (h x)) = (∀x. g x = h x) This can be proved using the auto method. 6.2.3 Function Image The image of a set under a function is a most useful notion. It has the obvious definition: f A ≡{y. ∃x∈A. y = f x} (image_def) Here are some of the many facts proved about image: (f ◦g) r = f g r (image_compose) f(A ∪B) = fA ∪fB (image_Un) inj f = ⇒f(A ∩B) = fA ∩fB (image_Int) Laws involving image can often be proved automatically. Here are two examples, illustrating connections with indexed union and with the general syntax for comprehension: lemma "fA ∪gA = (S x∈A. {f x, g x})" lemma "f {(x,y). P x y} = {f(x,y) \| x y. P x y}" A function’s range is the set of values that the function can take on. It is, in fact, the image of the universal set under that function. There is no constant range. Instead, range abbreviates an application of image to UNIV: range f ⇌fUNIV Few theorems are proved specifically for range; in most cases, you should look for a more general theorem concerning images. Inverse image is also useful. It is defined as follows: f -B ≡{x. f x ∈B} (vimage_def) This is one of the facts proved about it: f - (- A) = - f -` A (vimage_Compl) 6.3 Relations A relation is a set of pairs. As such, the set operations apply to them. For instance, we may form the union of two relations. Other primitives are defined specifically for relations. 114 6. Sets, Functions and Relations 6.3.1 Relation Basics The identity relation, also known as equality, has the obvious definition: Id ≡{p. ∃x. p = (x,x)} (Id_def) Composition of relations (the infix O) is also available: r O s = {(x,z). ∃y. (x,y) ∈s ∧(y,z) ∈r} (relcomp_unfold) This is one of the many lemmas proved about these concepts: R O Id = R (R_O_Id) Composition is monotonic, as are most of the primitives appearing in this chapter. We have many theorems similar to the following one: [ [r' ⊆r; s' ⊆s] ] = ⇒r' O s' ⊆r O s (relcomp_mono) The converse or inverse of a relation exchanges the roles of the two operands. We use the postfix notation r−1 or r^-1 in ASCII. ((a,b) ∈r−1) = ((b,a) ∈r) (converse_iff) Here is a typical law proved about converse and composition: (r O s)−1 = s−1 O r−1 (converse_relcomp) The image of a set under a relation is defined analogously to image under a function: (b ∈r A) = (∃x∈A. (x,b) ∈r) (Image_iff) It satisfies many similar laws. The domain and range of a relation are defined in the standard way: (a ∈Domain r) = (∃y. (a,y) ∈r) (Domain_iff) (a ∈Range r) = (∃y. (y,a) ∈r) (Range_iff) Iterated composition of a relation is available. The notation overloads that of exponentiation. Two simplification rules are installed: R ^ 0 = Id R ^ Suc n = R O R^n 6.3.2 The Reflexive and Transitive Closure The reflexive and transitive closure of the relation r is written with a postfix syntax. In ASCII we write r^ and in symbol notation r∗. It is the least solution of the equation r∗= Id ∪(r O r∗) (rtrancl_unfold) Among its basic properties are three that serve as introduction rules: (a, a) ∈r∗ (rtrancl_refl) p ∈r = ⇒p ∈r∗ (r_into_rtrancl) [ [(a,b) ∈r∗; (b,c) ∈r∗] ] = ⇒(a,c) ∈r∗ (rtrancl_trans) 6.3 Relations 115 Induction over the reflexive transitive closure is available: [ [(a, b) ∈r∗; P a; Vy z. [ [(a, y) ∈r∗; (y, z) ∈r; P y] ] = ⇒P z] ] = ⇒P b (rtrancl_induct) Idempotence is one of the laws proved about the reflexive transitive closure: (r∗)∗= r∗ (rtrancl_idemp) The transitive closure is similar. The ASCII syntax is r^+. It has two introduction rules: p ∈r = ⇒p ∈r+ (r_into_trancl) [ [(a, b) ∈r+; (b, c) ∈r+] ] = ⇒(a, c) ∈r+ (trancl_trans) The induction rule resembles the one shown above. A typical lemma states that transitive closure commutes with the converse operator: (r−1)+ = (r+)−1 (trancl_converse) 6.3.3 A Sample Proof The reflexive transitive closure also commutes with the converse operator. Let us examine the proof. Each direction of the equivalence is proved separately. The two proofs are almost identical. Here is the first one: lemma rtrancl_converseD: "(x,y) ∈(r−1)∗= ⇒(y,x) ∈r∗" apply (erule rtrancl_induct) apply (rule rtrancl_refl) apply (blast intro: rtrancl_trans) done The first step of the proof applies induction, leaving these subgoals: 1. (x, x) ∈r∗ 2. Vy z. [ [(x,y) ∈(r−1)∗; (y,z) ∈r−1; (y,x) ∈r∗] ] = ⇒(z,x) ∈r∗ The first subgoal is trivial by reflexivity. The second follows by first elimi-nating the converse operator, yielding the assumption (z,y) ∈r, and then applying the introduction rules shown above. The same proof script handles the other direction: lemma rtrancl_converseI: "(y,x) ∈r∗= ⇒(x,y) ∈(r−1)∗" apply (erule rtrancl_induct) apply (rule rtrancl_refl) apply (blast intro: rtrancl_trans) done Finally, we combine the two lemmas to prove the desired equation: lemma rtrancl_converse: "(r−1)∗= (r∗)−1" by (auto intro: rtrancl_converseI dest: rtrancl_converseD) 116 6. Sets, Functions and Relations ! ! This trivial proof requires auto rather than blast because of a subtle issue involving ordered pairs. Here is a subgoal that arises internally after the rules equalityI and subsetI have been applied: 1. Vx. x ∈(r−1)∗= ⇒x ∈(r∗)−1 We cannot apply rtrancl_converseD. It refers to ordered pairs, while x is a variable of product type. The simp and blast methods can do nothing, so let us try clarify: 1. Va b. (a,b) ∈(r−1)∗= ⇒(b,a) ∈r∗ Now that x has been replaced by the pair (a,b), we can proceed. Other methods that split variables in this way are force, auto, fast and best. Section 8.1 will discuss proof techniques for ordered pairs in more detail. 6.4 Well-Founded Relations and Induction A well-founded relation captures the notion of a terminating process. Com-plex recursive functions definitions must specify a well-founded relation that justifies their termination . Most of the forms of induction found in math-ematics are merely special cases of induction over a well-founded relation. Intuitively, the relation ≺is well-founded if it admits no infinite de-scending chains · · · ≺a2 ≺a1 ≺a0. Well-foundedness can be hard to show. The various formulations are all com-plicated. However, often a relation is well-founded by construction. HOL pro-vides theorems concerning ways of constructing a well-founded relation. The most familiar way is to specify a measure function f into the natural num-bers, when x ≺y ⇐ ⇒ f x < f y; we write this particular relation as measure f. ! ! You may want to skip the rest of this section until you need to perform a complex recursive function definition or induction. The induction rule returned by fun is good enough for most purposes. We use an explicit well-founded induction only in Sect. 9.2.4. Isabelle/HOL declares less_than as a relation object, that is, a set of pairs of natural numbers. Two theorems tell us that this relation behaves as expected and that it is well-founded: ((x,y) ∈less_than) = (x < y) (less_than_iff) wf less_than (wf_less_than) The notion of measure generalizes to the inverse image of a relation. Given a relation r and a function f, we express a new relation using f as a measure. An infinite descending chain on this new relation would give rise to an infinite descending chain on r. Isabelle/HOL defines this concept and proves a theorem stating that it preserves well-foundedness: 6.5 Fixed Point Operators 117 inv_image r f ≡{(x,y). (f x, f y) ∈r} (inv_image_def) wf r = ⇒wf (inv_image r f) (wf_inv_image) A measure function involves the natural numbers. The relation measure size justifies primitive recursion and structural induction over a datatype. Isabelle/HOL defines measure as shown: measure ≡inv_image less_than (measure_def) wf (measure f) (wf_measure) Of the other constructions, the most important is the lexicographic product of two relations. It expresses the standard dictionary ordering over pairs. We write ra <lex> rb, where ra and rb are the two operands. The lexicographic product satisfies the usual definition and it preserves well-foundedness: ra <lex> rb ≡ {((a,b),(a',b')). (a,a') ∈ra ∨ a=a' ∧(b,b') ∈rb} (lex_prod_def) [ [wf ra; wf rb] ] = ⇒wf (ra <lex> rb) (wf_lex_prod) The multiset ordering, useful for hard termination proofs, is available in the Library . Baader and Nipkow [3, Sect. 2.5] discuss it. Induction comes in many forms, including traditional mathematical in-duction, structural induction on lists and induction on size. All are instances of the following rule, for a suitable well-founded relation ≺: [∀y. y ≺x →P(y)] . . . . P(x) P(a) To show P(a) for a particular term a, it suffices to show P(x) for arbi-trary x under the assumption that P(y) holds for y ≺x. Intuitively, the well-foundedness of ≺ensures that the chains of reasoning are finite. In Isabelle, the induction rule is expressed like this: [ [wf r; Vx. ∀y. (y,x) ∈r − →P y = ⇒P x] ] = ⇒P a (wf_induct) Here wf r expresses that the relation r is well-founded. Many familiar induction principles are instances of this rule. For exam-ple, the predecessor relation on the natural numbers is well-founded; induc-tion over it is mathematical induction. The “tail of” relation on lists is well-founded; induction over it is structural induction. 6.5 Fixed Point Operators Fixed point operators define sets recursively. They are invoked implicitly when making an inductive definition, as discussed in Chap. 7 below. However, 118 6. Sets, Functions and Relations they can be used directly, too. The least or strongest fixed point yields an inductive definition; the greatest or weakest fixed point yields a coinductive definition. Mathematicians may wish to note that the existence of these fixed points is guaranteed by the Knaster-Tarski theorem. ! ! Casual readers should skip the rest of this section. We use fixed point operators only in Sect. ??. The theory applies only to monotonic functions. Isabelle’s definition of monotone is overloaded over all orderings: mono f ≡∀A B. A ≤B − →f A ≤f B (mono_def) For fixed point operators, the ordering will be the subset relation: if A ⊆B then we expect f (A) ⊆f (B). In addition to its definition, monotonicity has the obvious introduction and destruction rules: (VA B. A ≤B = ⇒f A ≤f B) = ⇒mono f (monoI) [ [mono f; A ≤B] ] = ⇒f A ≤f B (monoD) The most important properties of the least fixed point are that it is a fixed point and that it enjoys an induction rule: mono f = ⇒lfp f = f (lfp f) (lfp_unfold) [ [a ∈lfp f; mono f; Vx. x ∈f (lfp f ∩{x. P x}) = ⇒P x] ] = ⇒P a (lfp_induct) The induction rule shown above is more convenient than the basic one derived from the minimality of lfp. Observe that both theorems demand mono f as a premise. The greatest fixed point is similar, but it has a coinduction rule: mono f = ⇒gfp f = f (gfp f) (gfp_unfold) [ [mono f; a ∈X; X ⊆f (X ∪gfp f)] ] = ⇒a ∈gfp f (coinduct) A bisimulation is perhaps the best-known concept defined as a greatest fixed point. Exhibiting a bisimulation to prove the equality of two agents in a process algebra is an example of coinduction. The coinduction rule can be strengthened in various ways. 6.5.1 Propositional Dynamic Logic — PDL The formulae of PDL are built up from atomic propositions via negation and conjunction and the two temporal connectives AX and EF. Since formulae are essentially syntax trees, they are naturally modelled as a datatype:1 datatype formula = Atom "atom" \| Neg formula \| And formula formula 1 The customary definition of PDL looks quite different from ours, but the two are easily shown to be equivalent. 6.5 Fixed Point Operators 119 \| AX formula \| EF formula This resembles the boolean expression case study in Sect. 2.5.6. A validity relation between states and formulae specifies the semantics. The syntax an-notation allows us to write s \| = f instead of valid s f. The definition is by recursion over the syntax: primrec valid :: "state ⇒formula ⇒bool" ("(_ \| = _)" [80,80] 80) where "s \| = Atom a = (a ∈L s)" \| "s \| = Neg f = (¬(s \| = f))" \| "s \| = And f g = (s \| = f ∧s \| = g)" \| "s \| = AX f = (∀t. (s,t) ∈M − →t \| = f)" \| "s \| = EF f = (∃t. (s,t) ∈M ∗∧t \| = f)" The first three equations should be self-explanatory. The temporal formula AX f means that f is true in All neXt states whereas EF f means that there Exists some Future state in which f is true. The future is expressed via ∗, the reflexive transitive closure. Because of reflexivity, the future includes the present. Now we come to the model checker itself. It maps a formula into the set of states where the formula is true. It too is defined by recursion over the syntax: primrec mc :: "formula ⇒state set" where "mc(Atom a) = {s. a ∈L s}" \| "mc(Neg f) = -mc f" \| "mc(And f g) = mc f ∩mc g" \| "mc(AX f) = {s. ∀t. (s,t) ∈M − →t ∈mc f}" \| "mc(EF f) = lfp(λT. mc f ∪(M −1 T))" Only the equation for EF deserves some comments. Remember that the postfix −1 and the infix are predefined and denote the converse of a relation and the image of a set under a relation. Thus M −1 T is the set of all predecessors of T and the least fixed point (lfp) of λT. mc f ∪M −1 T is the least set T containing mc f and all predecessors of T. If you find it hard to see that mc (EF f) contains exactly those states from which there is a path to a state where f is true, do not worry — this will be proved in a moment. First we prove monotonicity of the function inside lfp in order to make sure it really has a least fixed point. lemma mono_ef: "mono(λT. A ∪(M −1 T))" apply(rule monoI) apply blast done Now we can relate model checking and semantics. For the EF case we need a separate lemma: lemma EF_lemma: "lfp(λT. A ∪(M −1 T)) = {s. ∃t. (s,t) ∈M ∗∧t ∈A}" 120 6. Sets, Functions and Relations The equality is proved in the canonical fashion by proving that each set includes the other; the inclusion is shown pointwise: apply(rule equalityI) apply(rule subsetI) apply(simp) Simplification leaves us with the following first subgoal 1. Vs. s ∈lfp (λT. A ∪M −1 T) = ⇒∃t. (s, t) ∈M ∗∧t ∈A A total of 2 subgoals... which is proved by lfp-induction: apply(erule lfp_induct_set) apply(rule mono_ef) apply(simp) Having disposed of the monotonicity subgoal, simplification leaves us with the following goal: 1. Vx. x ∈A ∨ x ∈M −1 (lfp (…) ∩{x. ∃t. (x, t) ∈M ∗∧t ∈A}) = ⇒∃t. (x, t) ∈M ∗∧t ∈A It is proved by blast, using the transitivity of M ∗. apply(blast intro: rtrancl_trans) We now return to the second set inclusion subgoal, which is again proved pointwise: apply(rule subsetI) apply(simp, clarify) After simplification and clarification we are left with 1. Vx t. [ [(x, t) ∈M ∗; t ∈A] ] = ⇒x ∈lfp (λT. A ∪M −1 T) This goal is proved by induction on (s, t) ∈M ∗. But since the model checker works backwards (from t to s), we cannot use the induction theorem rtrancl_induct: it works in the forward direction. Fortunately the converse induction theorem converse_rtrancl_induct already exists: [ [(a, b) ∈r∗; P b; Vy z. [ [(y, z) ∈r; (z, b) ∈r∗; P z] ] = ⇒P y] ] = ⇒P a It says that if (a, b) ∈r∗and we know P b then we can infer P a provided each step backwards from a predecessor z of b preserves P. apply(erule converse_rtrancl_induct) The base case 1. Vx t. t ∈A = ⇒t ∈lfp (λT. A ∪M −1 T) A total of 2 subgoals... 6.5 Fixed Point Operators 121 is solved by unrolling lfp once apply(subst lfp_unfold[OF mono_ef]) 1. Vx t. t ∈A = ⇒t ∈A ∪M −1 lfp (λT. A ∪M −1 T) A total of 2 subgoals... and disposing of the resulting trivial subgoal automatically: apply(blast) The proof of the induction step is identical to the one for the base case: apply(subst lfp_unfold[OF mono_ef]) apply(blast) done The main theorem is proved in the familiar manner: induction followed by auto augmented with the lemma as a simplification rule. theorem "mc f = {s. s \| = f}" apply(induct_tac f) apply(auto simp add: EF_lemma) done Exercise 6.5.1 AX has a dual operator EN (“there exists a next state such that”)2 with the intended semantics s \| = EN f = (∃t. (s, t) ∈M ∧t \| = f) Fortunately, EN f can already be expressed as a PDL formula. How? Show that the semantics for EF satisfies the following recursion equation: s \| = EF f = (s \| = f ∨s \| = EN (EF f)) 6.5.2 Computation Tree Logic — CTL The semantics of PDL only needs reflexive transitive closure. Let us be ad-venturous and introduce a more expressive temporal operator. We extend the datatype formula by a new constructor \| AF formula which stands for “Always in the Future”: on all infinite paths, at some point the formula holds. Formalizing the notion of an infinite path is easy in HOL: it is simply a function from nat to state. definition Paths :: "state ⇒(nat ⇒state)set" where "Paths s ≡{p. s = p 0 ∧(∀i. (p i, p(i+1)) ∈M)}" This definition allows a succinct statement of the semantics of AF: 3 2 We cannot use the customary EX: it is reserved as the ascii-equivalent of ∃. 3 Do not be misled: neither datatypes nor recursive functions can be extended by new constructors or equations. This is just a trick of the presentation (see Sect. 4.2.5). In reality one has to define a new datatype and a new function. 122 6. Sets, Functions and Relations "s \| = AF f = (∀p ∈Paths s. ∃i. p i \| = f)" Model checking AF involves a function which is just complicated enough to warrant a separate definition: definition af :: "state set ⇒state set ⇒state set" where "af A T ≡A ∪{s. ∀t. (s, t) ∈M − →t ∈T}" Now we define mc (AF f) as the least set T that includes mc f and all states all of whose direct successors are in T: "mc(AF f) = lfp(af(mc f))" Because af is monotone in its second argument (and also its first, but that is irrelevant), af A has a least fixed point: lemma mono_af: "mono(af A)" apply(simp add: mono_def af_def) apply blast done All we need to prove now is mc (AF f) = {s. s \| = AF f}, which states that mc and \| = agree for AF. This time we prove the two inclusions separately, starting with the easy one: theorem AF_lemma1: "lfp(af A) ⊆{s. ∀p ∈Paths s. ∃i. p i ∈A}" In contrast to the analogous proof for EF, and just for a change, we do not use fixed point induction. Park-induction, named after David Park, is weaker but sufficient for this proof: f S ≤S = ⇒lfp f ≤S (lfp_lowerbound) The instance of the premise f S ⊆S is proved pointwise, a decision that auto takes for us: apply(rule lfp_lowerbound) apply(auto simp add: af_def Paths_def) 1. Vp. [ [∀t. (p 0, t) ∈M − → (∀p. t = p 0 ∧(∀i. (p i, p (Suc i)) ∈M) − → (∃i. p i ∈A)); ∀i. (p i, p (Suc i)) ∈M] ] = ⇒∃i. p i ∈A In this remaining case, we set t to p 1. The rest is automatic, which is sur-prising because it involves finding the instantiation λi. p (i + 1) for ∀p. apply(erule_tac x = "p 1" in allE) apply(auto) done The opposite inclusion is proved by contradiction: if some state s is not in lfp (af A), then we can construct an infinite A-avoiding path starting from s. The reason is that by unfolding lfp we find that if s is not in lfp (af A), then s is not in A and there is a direct successor of s that is again not in 6.5 Fixed Point Operators 123 lfp (af A). Iterating this argument yields the promised infinite A-avoiding path. Let us formalize this sketch. The one-step argument in the sketch above is proved by a variant of contraposition: lemma not_in_lfp_afD: "s / ∈lfp(af A) = ⇒s / ∈A ∧(∃t. (s,t) ∈M ∧t / ∈lfp(af A))" apply(erule contrapos_np) apply(subst lfp_unfold[OF mono_af]) apply(simp add: af_def) done We assume the negation of the conclusion and prove s ∈lfp (af A). Unfold-ing lfp once and simplifying with the definition of af finishes the proof. Now we iterate this process. The following construction of the desired path is parameterized by a predicate Q that should hold along the path: primrec path :: "state ⇒(state ⇒bool) ⇒(nat ⇒state)" where "path s Q 0 = s" \| "path s Q (Suc n) = (SOME t. (path s Q n,t) ∈M ∧Q t)" Element n + 1 on this path is some arbitrary successor t of element n such that Q t holds. Remember that SOME t. R t is some arbitrary but fixed t such that R t holds (see Sect. 5.10). Of course, such a t need not exist, but that is of no concern to us since we will only use path when a suitable t does exist. Let us show that if each state s that satisfies Q has a successor that again satisfies Q, then there exists an infinite Q-path: lemma infinity_lemma: "[ [ Q s; ∀s. Q s − →(∃t. (s,t) ∈M ∧Q t) ] ] = ⇒ ∃p∈Paths s. ∀i. Q(p i)" First we rephrase the conclusion slightly because we need to prove simulta-neously both the path property and the fact that Q holds: apply(subgoal_tac "∃p. s = p 0 ∧(∀i::nat. (p i, p(i+1)) ∈M ∧Q(p i))") From this proposition the original goal follows easily: apply(simp add: Paths_def, blast) The new subgoal is proved by providing the witness path s Q for p: apply(rule_tac x = "path s Q" in exI) apply(clarsimp) After simplification and clarification, the subgoal has the following form: 1. Vi. [ [Q s; ∀s. Q s − →(∃t. (s, t) ∈M ∧Q t)] ] = ⇒(path s Q i, SOME t. (path s Q i, t) ∈M ∧Q t) ∈M ∧ Q (path s Q i) It invites a proof by induction on i: 124 6. Sets, Functions and Relations apply(induct_tac i) apply(simp) After simplification, the base case boils down to 1. [ [Q s; ∀s. Q s − →(∃t. (s, t) ∈M ∧Q t)] ] = ⇒(s, SOME t. (s, t) ∈M ∧Q t) ∈M A total of 2 subgoals... The conclusion looks exceedingly trivial: after all, t is chosen such that (s, t) ∈M holds. However, we first have to show that such a t actually exists! This reasoning is embodied in the theorem someI2_ex: [ [∃a. ?P a; Vx. ?P x = ⇒?Q x] ] = ⇒?Q (SOME x. ?P x) When we apply this theorem as an introduction rule, ?P x becomes (s, x) ∈ M ∧Q x and ?Q x becomes (s, x) ∈M and we have to prove two subgoals: ∃a. (s, a) ∈M ∧Q a, which follows from the assumptions, and (s, x) ∈M ∧Q x = ⇒(s, x) ∈M, which is trivial. Thus it is not surprising that fast can prove the base case quickly: apply(fast intro: someI2_ex) What is worth noting here is that we have used fast rather than blast. The reason is that blast would fail because it cannot cope with someI2_ex: unify-ing its conclusion with the current subgoal is non-trivial because of the nested schematic variables. For efficiency reasons blast does not even attempt such unifications. Although fast can in principle cope with complicated unifica-tion problems, in practice the number of unifiers arising is often prohibitive and the offending rule may need to be applied explicitly rather than auto-matically. This is what happens in the step case. The induction step is similar, but more involved, because now we face nested occurrences of SOME. As a result, fast is no longer able to solve the subgoal and we apply someI2_ex by hand. We merely show the proof com-mands but do not describe the details: apply(simp) apply(rule someI2_ex) apply(blast) apply(rule someI2_ex) apply(blast) apply(blast) done Function path has fulfilled its purpose now and can be forgotten. It was merely defined to provide the witness in the proof of the infinity_lemma. Aficionados of minimal proofs might like to know that we could have given the witness without having to define a new function: the term rec_nat s (λn t. SOME u. (t, u) ∈M ∧Q u) is extensionally equal to path s Q, where rec_nat is the predefined primitive recursor on nat. 6.5 Fixed Point Operators 125 At last we can prove the opposite direction of AF_lemma1: theorem AF_lemma2: "{s. ∀p ∈Paths s. ∃i. p i ∈A} ⊆lfp(af A)" The proof is again pointwise and then by contraposition: apply(rule subsetI) apply(erule contrapos_pp) apply simp 1. Vx. x / ∈lfp (af A) = ⇒∃p∈Paths x. ∀i. p i / ∈A Applying the infinity_lemma as a destruction rule leaves two subgoals, the second premise of infinity_lemma and the original subgoal: apply(drule infinity_lemma) 1. Vx. ∀s. s / ∈lfp (af A) − →(∃t. (s, t) ∈M ∧t / ∈lfp (af A)) 2. Vx. ∃p∈Paths x. ∀i. p i / ∈lfp (af A) = ⇒ ∃p∈Paths x. ∀i. p i / ∈A Both are solved automatically: apply(auto dest: not_in_lfp_afD) done If you find these proofs too complicated, we recommend that you read Sect. 9.2.4, where we show how inductive definitions lead to simpler argu-ments. The main theorem is proved as for PDL, except that we also derive the necessary equality lfp(af A) = ... by combining AF_lemma1 and AF_lemma2 on the spot: theorem "mc f = {s. s \| = f}" apply(induct_tac f) apply(auto simp add: EF_lemma equalityI[OF AF_lemma1 AF_lemma2]) done The language defined above is not quite CTL. The latter also includes an until-operator EU f g with semantics “there Exists a path where f is true Until g becomes true”. We need an auxiliary function: primrec until:: "state set ⇒state set ⇒state ⇒state list ⇒bool" where "until A B s [] = (s ∈B)" \| "until A B s (t#p) = (s ∈A ∧(s,t) ∈M ∧until A B t p)" Expressing the semantics of EU is now straightforward: s \| = EU f g = (∃p. until {t. t \| = f} {t. t \| = g} s p) Note that EU is not definable in terms of the other operators! Model checking EU is again a least fixed point construction: mc(EU f g) = lfp(λT. mc g ∪mc f ∩(M −1 T)) 126 6. Sets, Functions and Relations Exercise 6.5.2 Extend the datatype of formulae by the above until operator and prove the equivalence between semantics and model checking, i.e. that mc (EU f g) = {s. s \| = EU f g} For more CTL exercises see, for example, Huth and Ryan . Let us close this section with a few words about the executability of our model checkers. It is clear that if all sets are finite, they can be represented as lists and the usual set operations are easily implemented. Only lfp requires a little thought. Fortunately, theory While_Combinator in the Library pro-vides a theorem stating that in the case of finite sets and a monotone func-tion F, the value of lfp F can be computed by iterated application of F to {} until a fixed point is reached. It is actually possible to generate executable functional programs from HOL definitions, but that is beyond the scope of the tutorial. 7. Inductively Defined Sets This chapter is dedicated to the most important definition principle after recursive functions and datatypes: inductively defined sets. We start with a simple example: the set of even numbers. A slightly more complicated example, the reflexive transitive closure, is the subject of Sect. 7.2. In particular, some standard induction heuristics are discussed. Advanced forms of inductive definitions are discussed in Sect. 7.3. To demon-strate the versatility of inductive definitions, the chapter closes with a case study from the realm of context-free grammars. The first two sections are required reading for anybody interested in mathematical modelling. ! ! Predicates can also be defined inductively. See Sect. 7.1.7. 7.1 The Set of Even Numbers The set of even numbers can be inductively defined as the least set containing 0 and closed under the operation +2. Obviously, even can also be expressed using the divides relation (dvd). We shall prove below that the two formula-tions coincide. On the way we shall examine the primary means of reasoning about inductively defined sets: rule induction. 7.1.1 Making an Inductive Definition Using inductive set, we declare the constant even to be a set of natural numbers with the desired properties. inductive set even :: "nat set" where zero[intro!]: "0 ∈even" \| step[intro!]: "n ∈even = ⇒(Suc (Suc n)) ∈even" An inductive definition consists of introduction rules. The first one above states that 0 is even; the second states that if n is even, then so is n + 2. Given this declaration, Isabelle generates a fixed point definition for Even.even and proves theorems about it, thus following the definitional approach (see Sect. 2.8). These theorems include the introduction rules specified in the dec-laration, an elimination rule for case analysis and an induction rule. We can 128 7. Inductively Defined Sets refer to these theorems by automatically-generated names. Here are two ex-amples: 0 ∈Even.even (even.zero) n ∈Even.even = ⇒Suc (Suc n) ∈Even.even (even.step) The introduction rules can be given attributes. Here both rules are spec-ified as intro!, directing the classical reasoner to apply them aggressively. Obviously, regarding 0 as even is safe. The step rule is also safe because n +2 is even if and only if n is even. We prove this equivalence later. 7.1.2 Using Introduction Rules Our first lemma states that numbers of the form 2×k are even. Introduction rules are used to show that specific values belong to the inductive set. Such proofs typically involve induction, perhaps over some other inductive set. lemma two_times_even[intro!]: "2k ∈even" apply (induct_tac k) apply auto done The first step is induction on the natural number k, which leaves two subgoals: 1. 2 0 ∈Even.even 2. Vn. 2 n ∈Even.even = ⇒2 Suc n ∈Even.even Here auto simplifies both subgoals so that they match the introduction rules, which are then applied automatically. Our ultimate goal is to prove the equivalence between the traditional definition of even (using the divides relation) and our inductive definition. One direction of this equivalence is immediate by the lemma just proved, whose intro! attribute ensures it is applied automatically. lemma dvd_imp_even: "2 dvd n = ⇒n ∈even" by (auto simp add: dvd_def) 7.1.3 Rule Induction From the definition of the set Even.even, Isabelle has generated an induction rule: [ [x ∈Even.even; P 0; Vn. [ [n ∈Even.even; P n] ] = ⇒P (Suc (Suc n))] ] = ⇒P x (even.induct) A property P holds for every even number provided it holds for 0 and is closed under the operation Suc(Suc ·). Then P is closed under the introduction rules for Even.even, which is the least set closed under those rules. This type of inductive argument is called rule induction. 7.1 The Set of Even Numbers 129 Apart from the double application of Suc, the induction rule above resem-bles the familiar mathematical induction, which indeed is an instance of rule induction; the natural numbers can be defined inductively to be the least set containing 0 and closed under Suc. Induction is the usual way of proving a property of the elements of an inductively defined set. Let us prove that all members of the set Even.even are multiples of two. lemma even_imp_dvd: "n ∈even = ⇒2 dvd n" We begin by applying induction. Note that even.induct has the form of an elimination rule, so we use the method erule. We get two subgoals: apply (erule even.induct) 1. semiring_parity_class.even 0 2. Vn. [ [n ∈Even.even; semiring_parity_class.even n] ] = ⇒semiring_parity_class.even (Suc (Suc n)) We unfold the definition of dvd in both subgoals, proving the first one and simplifying the second: apply (simp_all add: dvd_def) 1. Vn. [ [n ∈Even.even; ∃k. n = 2 k] ] = ⇒∃k. Suc (Suc n) = 2 k The next command eliminates the existential quantifier from the assumption and replaces n by 2 k. apply clarify 1. Vn k. 2 k ∈Even.even = ⇒∃ka. Suc (Suc (2 k)) = 2 ka To conclude, we tell Isabelle that the desired value is Suc k. With this hint, the subgoal falls to simp. apply (rule_tac x = "Suc k" in exI, simp) Combining the previous two results yields our objective, the equivalence relating Even.even and dvd. theorem even_iff_dvd: "(n ∈even) = (2 dvd n)" by (blast intro: dvd_imp_even even_imp_dvd) 7.1.4 Generalization and Rule Induction Before applying induction, we typically must generalize the induction for-mula. With rule induction, the required generalization can be hard to find and sometimes requires a complete reformulation of the problem. In this ex-ample, our first attempt uses the obvious statement of the result. It fails: lemma "Suc (Suc n) ∈even = ⇒n ∈even" 130 7. Inductively Defined Sets apply (erule even.induct) oops Rule induction finds no occurrences of Suc (Suc n) in the conclusion, which it therefore leaves unchanged. (Look at even.induct to see why this happens.) We have these subgoals: 1. n ∈Even.even 2. Vna. [ [na ∈Even.even; n ∈Even.even] ] = ⇒n ∈Even.even The first one is hopeless. Rule induction on a non-variable term discards information, and usually fails. How to deal with such situations in general is described in Sect. 9.2.1 below. In the current case the solution is easy because we have the necessary inverse, subtraction: lemma even_imp_even_minus_2: "n ∈even = ⇒n - 2 ∈even" apply (erule even.induct) apply auto done This lemma is trivially inductive. Here are the subgoals: 1. 0 - 2 ∈Even.even 2. Vn. [ [n ∈Even.even; n - 2 ∈Even.even] ] = ⇒Suc (Suc n) - 2 ∈Even.even The first is trivial because 0 - 2 simplifies to 0, which is even. The second is trivial too: Suc (Suc n) - 2 simplifies to n, matching the assumption. Using our lemma, we can easily prove the result we originally wanted: lemma Suc_Suc_even_imp_even: "Suc (Suc n) ∈even = ⇒n ∈even" by (drule even_imp_even_minus_2, simp) We have just proved the converse of the introduction rule even.step. This suggests proving the following equivalence. We give it the iff attribute be-cause of its obvious value for simplification. lemma [iff]: "((Suc (Suc n)) ∈even) = (n ∈even)" by (blast dest: Suc_Suc_even_imp_even) 7.1.5 Rule Inversion Case analysis on an inductive definition is called rule inversion. It is fre-quently used in proofs about operational semantics. It can be highly effective when it is applied automatically. Let us look at how rule inversion is done in Isabelle/HOL. Recall that Even.even is the minimal set closed under these two rules: 0 ∈Even.even n ∈Even.even = ⇒Suc (Suc n) ∈Even.even Minimality means that Even.even contains only the elements that these rules force it to contain. If we are told that a belongs to Even.even then there are only two possibilities. Either a is 0 or else a has the form Suc (Suc n), for some suitable n that belongs to Even.even. That is the gist of the cases rule, which Isabelle proves for us when it accepts an inductive definition: 7.1 The Set of Even Numbers 131 [ [a ∈Even.even; a = 0 = ⇒P; Vn. [ [a = Suc (Suc n); n ∈Even.even] ] = ⇒P] ] = ⇒P (even.cases) This general rule is less useful than instances of it for specific patterns. For example, if a has the form Suc (Suc n) then the first case becomes irrele-vant, while the second case tells us that n belongs to Even.even. Isabelle will generate this instance for us: inductive cases Suc_Suc_cases [elim!]: "Suc(Suc n) ∈even" The inductive cases command generates an instance of the cases rule for the supplied pattern and gives it the supplied name: [ [Suc (Suc n) ∈Even.even; n ∈Even.even = ⇒P] ] = ⇒P (Suc_Suc_cases) Applying this as an elimination rule yields one case where even.cases would yield two. Rule inversion works well when the conclusions of the introduc-tion rules involve datatype constructors like Suc and # (list “cons”); freeness reasoning discards all but one or two cases. In the inductive cases command we supplied an attribute, elim!, in-dicating that this elimination rule can be applied aggressively. The original cases rule would loop if used in that manner because the pattern a matches everything. The rule Suc_Suc_cases is equivalent to the following implication: Suc (Suc n) ∈Even.even = ⇒n ∈Even.even Just above we devoted some effort to reaching precisely this result. Yet we could have obtained it by a one-line declaration, dispensing with the lemma even_imp_even_minus_2. This example also justifies the terminology rule in-version: the new rule inverts the introduction rule even.step. In general, a rule can be inverted when the set of elements it introduces is disjoint from those of the other introduction rules. For one-off applications of rule inversion, use the ind_cases method. Here is an example: apply (ind_cases "Suc(Suc n) ∈even") The specified instance of the cases rule is generated, then applied as an elimination rule. To summarize, every inductive definition produces a cases rule. The in-ductive cases command stores an instance of the cases rule for a given pattern. Within a proof, the ind_cases method applies an instance of the cases rule. The even numbers example has shown how inductive definitions can be used. Later examples will show that they are actually worth using. 132 7. Inductively Defined Sets 7.1.6 Mutually Inductive Definitions Just as there are datatypes defined by mutual recursion, there are sets defined by mutual induction. As a trivial example we consider the even and odd natural numbers: inductive set Even :: "nat set" and Odd :: "nat set" where zero: "0 ∈Even" \| EvenI: "n ∈Odd = ⇒Suc n ∈Even" \| OddI: "n ∈Even = ⇒Suc n ∈Odd" The mutually inductive definition of multiple sets is no different from that of a single set, except for induction: just as for mutually recursive datatypes, induction needs to involve all the simultaneously defined sets. In the above case, the induction rule is called Even_Odd.induct (simply concatenate the names of the sets involved) and has the conclusion (?x ∈Even − →?P ?x) ∧(?y ∈Odd − →?Q ?y) If we want to prove that all even numbers are divisible by two, we have to generalize the statement as follows: lemma "(m ∈Even − →2 dvd m) ∧(n ∈Odd − →2 dvd (Suc n))" The proof is by rule induction. Because of the form of the induction theorem, it is applied by rule rather than erule as for ordinary inductive definitions: apply(rule Even_Odd.induct) 1. even 0 2. Vn. [ [n ∈Odd; even (Suc n)] ] = ⇒even (Suc n) 3. Vn. [ [n ∈Even; even n] ] = ⇒even (Suc (Suc n)) The first two subgoals are proved by simplification and the final one can be proved in the same manner as in Sect. 7.1.3 where the same subgoal was encountered before. We do not show the proof script. 7.1.7 Inductively Defined Predicates Instead of a set of even numbers one can also define a predicate on nat: inductive evn :: "nat ⇒bool" where zero: "evn 0" \| step: "evn n = ⇒evn(Suc(Suc n))" Everything works as before, except that you write inductive instead of in-ductive set and evn n instead of n ∈Even. When defining an n-ary rela-tion as a predicate, it is recommended to curry the predicate: its type should 7.2 The Reflexive Transitive Closure 133 be τ 1 ⇒. . . ⇒τ n ⇒bool rather than τ 1 × . . . × τ n ⇒bool. The curried version facilitates inductions. When should you choose sets and when predicates? If you intend to com-bine your notion with set theoretic notation, define it as an inductive set. If not, define it as an inductive predicate, thus avoiding the ∈notation. But note that predicates of more than one argument cannot be combined with the usual set theoretic operators: P ∪Q is not well-typed if P, Q :: τ 1 ⇒τ 2 ⇒bool, you have to write λx y. P x y ∧Q x y instead. 7.2 The Reflexive Transitive Closure An inductive definition may accept parameters, so it can express functions that yield sets. Relations too can be defined inductively, since they are just sets of pairs. A perfect example is the function that maps a relation to its reflexive transitive closure. This concept was already introduced in Sect. 6.3, where the operator ∗was defined as a least fixed point because inductive definitions were not yet available. But now they are: inductive set rtc :: "('a × 'a)set ⇒('a × 'a)set" ("" 999) for r :: "('a × 'a)set" where rtc_refl[iff]: "(x,x) ∈r" \| rtc_step: "[ [ (x,y) ∈r; (y,z) ∈r ] ] = ⇒(x,z) ∈r" The function rtc is annotated with concrete syntax: instead of rtc r we can write r. The actual definition consists of two rules. Reflexivity is obvious and is immediately given the iff attribute to increase automation. The second rule, rtc_step, says that we can always add one more r-step to the left. Although we could make rtc_step an introduction rule, this is dangerous: the recursion in the second premise slows down and may even kill the automatic tactics. The above definition of the concept of reflexive transitive closure may be sufficiently intuitive but it is certainly not the only possible one: for a start, it does not even mention transitivity. The rest of this section is devoted to proving that it is equivalent to the standard definition. We start with a simple lemma: lemma [intro]: "(x,y) ∈r = ⇒(x,y) ∈r" by(blast intro: rtc_step) Although the lemma itself is an unremarkable consequence of the basic rules, it has the advantage that it can be declared an introduction rule without the danger of killing the automatic tactics because r occurs only in the conclusion and not in the premise. Thus some proofs that would otherwise need rtc_step can now be found automatically. The proof also shows that blast is able to handle rtc_step. But some of the other automatic tactics are 134 7. Inductively Defined Sets more sensitive, and even blast can be lead astray in the presence of large numbers of rules. To prove transitivity, we need rule induction, i.e. theorem rtc.induct: [ [(?x1.0, ?x2.0) ∈?r; Vx. ?P x x; Vx y z. [ [(x, y) ∈?r; (y, z) ∈?r; ?P y z] ] = ⇒?P x z] ] = ⇒?P ?x1.0 ?x2.0 It says that ?P holds for an arbitrary pair (?x1.0, ?x2.0) ∈?r if ?P is preserved by all rules of the inductive definition, i.e. if ?P holds for the con-clusion provided it holds for the premises. In general, rule induction for an n-ary inductive relation R expects a premise of the form (x1, . . . , xn) ∈R. Now we turn to the inductive proof of transitivity: lemma rtc_trans: "[ [ (x,y) ∈r; (y,z) ∈r ] ] = ⇒(x,z) ∈r" apply(erule rtc.induct) Unfortunately, even the base case is a problem: 1. Vx. (y, z) ∈r = ⇒(x, z) ∈r A total of 2 subgoals... We have to abandon this proof attempt. To understand what is going on, let us look again at rtc.induct. In the above application of erule, the first premise of rtc.induct is unified with the first suitable assumption, which is (x, y) ∈r rather than (y, z) ∈r. Although that is what we want, it is merely due to the order in which the assumptions occur in the subgoal, which it is not good practice to rely on. As a result, ?xb becomes x, ?xa becomes y and ?P becomes λu v. (u, z) ∈r, thus yielding the above subgoal. So what went wrong? When looking at the instantiation of ?P we see that it does not depend on its second parameter at all. The reason is that in our original goal, of the pair (x, y) only x appears also in the conclusion, but not y. Thus our induction statement is too general. Fortunately, it can easily be specialized: transfer the additional premise (y, z) ∈r into the conclusion: lemma rtc_trans[rule_format]: "(x,y) ∈r = ⇒(y,z) ∈r − →(x,z) ∈r" This is not an obscure trick but a generally applicable heuristic: When proving a statement by rule induction on (x1, . . . , xn) ∈R, pull all other premises containing any of the xi into the conclusion using − →. A similar heuristic for other kinds of inductions is formulated in Sect. 9.2.1. The rule_format directive turns − →back into = ⇒: in the end we obtain the original statement of our lemma. apply(erule rtc.induct) Now induction produces two subgoals which are both proved automatically: 7.2 The Reflexive Transitive Closure 135 1. Vx. (x, z) ∈r − →(x, z) ∈r 2. Vx y za. [ [(x, y) ∈r; (y, za) ∈r; (za, z) ∈r − →(y, z) ∈r] ] = ⇒(za, z) ∈r − →(x, z) ∈r apply(blast) apply(blast intro: rtc_step) done Let us now prove that r is really the reflexive transitive closure of r, i.e. the least reflexive and transitive relation containing r. The latter is easily formalized inductive set rtc2 :: "('a × 'a)set ⇒('a × 'a)set" for r :: "('a × 'a)set" where "(x,y) ∈r = ⇒(x,y) ∈rtc2 r" \| "(x,x) ∈rtc2 r" \| "[ [ (x,y) ∈rtc2 r; (y,z) ∈rtc2 r ] ] = ⇒(x,z) ∈rtc2 r" and the equivalence of the two definitions is easily shown by the obvious rule inductions: lemma "(x,y) ∈rtc2 r = ⇒(x,y) ∈r" apply(erule rtc2.induct) apply(blast) apply(blast) apply(blast intro: rtc_trans) done lemma "(x,y) ∈r = ⇒(x,y) ∈rtc2 r" apply(erule rtc.induct) apply(blast intro: rtc2.intros) apply(blast intro: rtc2.intros) done So why did we start with the first definition? Because it is simpler. It contains only two rules, and the single step rule is simpler than transitivity. As a consequence, rtc.induct is simpler than rtc2.induct. Since inductive proofs are hard enough anyway, we should always pick the simplest induction schema available. Hence rtc is the definition of choice. Exercise 7.2.1 Show that the converse of rtc_step also holds: [ [(x, y) ∈r; (y, z) ∈r] ] = ⇒(x, z) ∈r Exercise 7.2.2 Repeat the development of this section, but starting with a definition of rtc where rtc_step is replaced by its converse as shown in exercise 7.2.1. 136 7. Inductively Defined Sets 7.3 Advanced Inductive Definitions The premises of introduction rules may contain universal quantifiers and monotone functions. A universal quantifier lets the rule refer to any number of instances of the inductively defined set. A monotone function lets the rule refer to existing constructions (such as “list of”) over the inductively defined set. The examples below show how to use the additional expressiveness and how to reason from the resulting definitions. 7.3.1 Universal Quantifiers in Introduction Rules As a running example, this section develops the theory of ground terms: terms constructed from constant and function symbols but not variables. To simplify matters further, we regard a constant as a function applied to the null argument list. Let us declare a datatype gterm for the type of ground terms. It is a type constructor whose argument is a type of function symbols. datatype 'f gterm = Apply 'f "'f gterm list" To try it out, we declare a datatype of some integer operations: integer constants, the unary minus operator and the addition operator. datatype integer_op = Number int \| UnaryMinus \| Plus Now the type integer_op gterm denotes the ground terms built over those symbols. The type constructor gterm can be generalized to a function over sets. It returns the set of ground terms that can be formed over a set F of function symbols. For example, we could consider the set of ground terms formed from the finite set {Number 2, UnaryMinus, Plus}. This concept is inductive. If we have a list args of ground terms over F and a function symbol f in F, then we can apply f to args to obtain another ground term. The only difficulty is that the argument list may be of any length. Hitherto, each rule in an inductive definition referred to the induc-tively defined set a fixed number of times, typically once or twice. A universal quantifier in the premise of the introduction rule expresses that every element of args belongs to our inductively defined set: is a ground term over F. The function set denotes the set of elements in a given list. inductive set gterms :: "'f set ⇒'f gterm set" for F :: "'f set" where step[intro!]: "[ [∀t ∈set args. t ∈gterms F; f ∈F] ] = ⇒(Apply f args) ∈gterms F" To demonstrate a proof from this definition, let us show that the function gterms is monotone. We shall need this concept shortly. 7.3 Advanced Inductive Definitions 137 lemma gterms_mono: "F⊆G = ⇒gterms F ⊆gterms G" apply clarify apply (erule gterms.induct) apply blast done Intuitively, this theorem says that enlarging the set of function symbols enlarges the set of ground terms. The proof is a trivial rule induction. First we use the clarify method to assume the existence of an element of gterms F. (We could have used intro subsetI.) We then apply rule induction. Here is the resulting subgoal: 1. Vx args f. [ [F ⊆G; ∀t∈set args. t ∈gterms F ∧t ∈gterms G; f ∈F] ] = ⇒Apply f args ∈gterms G The assumptions state that f belongs to F, which is included in G, and that every element of the list args is a ground term over G. The blast method finds this chain of reasoning easily. ! ! Why do we call this function gterms instead of gterm? A constant may have the same name as a type. However, name clashes could arise in the theorems that Isabelle generates. Our choice of names keeps gterms.induct separate from gterm.induct. Call a term well-formed if each symbol occurring in it is applied to the correct number of arguments. (This number is called the symbol’s arity.) We can express well-formedness by generalizing the inductive definition of gterms. Suppose we are given a function called arity, specifying the arities of all symbols. In the inductive step, we have a list args of such terms and a function symbol f. If the length of the list matches the function’s arity then applying f to args yields a well-formed term. inductive set well_formed_gterm :: "('f ⇒nat) ⇒'f gterm set" for arity :: "'f ⇒nat" where step[intro!]: "[ [∀t ∈set args. t ∈well_formed_gterm arity; length args = arity f] ] = ⇒(Apply f args) ∈well_formed_gterm arity" The inductive definition neatly captures the reasoning above. The univer-sal quantification over the set of arguments expresses that all of them are well-formed. 7.3.2 Alternative Definition Using a Monotone Function An inductive definition may refer to the inductively defined set through an arbitrary monotone function. To demonstrate this powerful feature, let us change the inductive definition above, replacing the quantifier by a use of the 138 7. Inductively Defined Sets function lists. This function, from the Isabelle theory of lists, is analogous to the function gterms declared above: if A is a set then lists A is the set of lists whose elements belong to A. In the inductive definition of well-formed terms, examine the one intro-duction rule. The first premise states that args belongs to the lists of well-formed terms. This formulation is more direct, if more obscure, than using a universal quantifier. inductive set well_formed_gterm' :: "('f ⇒nat) ⇒'f gterm set" for arity :: "'f ⇒nat" where step[intro!]: "[ [args ∈lists (well_formed_gterm' arity); length args = arity f] ] = ⇒(Apply f args) ∈well_formed_gterm' arity" monos lists_mono We cite the theorem lists_mono to justify using the function lists.1 A ⊆B = ⇒lists A ⊆lists B (lists_mono) Why must the function be monotone? An inductive definition describes an iterative construction: each element of the set is constructed by a finite num-ber of introduction rule applications. For example, the elements of even are constructed by finitely many applications of the rules 0 ∈Even.even n ∈Even.even = ⇒Suc (Suc n) ∈Even.even All references to a set in its inductive definition must be positive. Applications of an introduction rule cannot invalidate previous applications, allowing the construction process to converge. The following pair of rules do not constitute an inductive definition: 0 ∈Even.even n / ∈Even.even = ⇒Suc n ∈Even.even Showing that 4 is even using these rules requires showing that 3 is not even. It is far from trivial to show that this set of rules characterizes the even numbers. Even with its use of the function lists, the premise of our introduction rule is positive: args ∈lists (well_formed_gterm' arity) To apply the rule we construct a list args of previously constructed well-formed terms. We obtain a new term, Apply f args. Because lists is mono-tone, applications of the rule remain valid as new terms are constructed. Further lists of well-formed terms become available and none are taken away. 1 This particular theorem is installed by default already, but we include the monos declaration in order to illustrate its syntax. 7.3 Advanced Inductive Definitions 139 7.3.3 A Proof of Equivalence We naturally hope that these two inductive definitions of “well-formed” co-incide. The equality can be proved by separate inclusions in each direction. Each is a trivial rule induction. lemma "well_formed_gterm arity ⊆well_formed_gterm' arity" apply clarify apply (erule well_formed_gterm.induct) apply auto done The clarify method gives us an element of well_formed_gterm arity on which to perform induction. The resulting subgoal can be proved automati-cally: 1. Vx args f. [ [∀t∈set args. t ∈well_formed_gterm arity ∧t ∈well_formed_gterm' arity; length args = arity f] ] = ⇒Apply f args ∈well_formed_gterm' arity This proof resembles the one given in Sect. 7.3.1 above, especially in the form of the induction hypothesis. Next, we consider the opposite inclusion: lemma "well_formed_gterm' arity ⊆well_formed_gterm arity" apply clarify apply (erule well_formed_gterm'.induct) apply auto done The proof script is virtually identical, but the subgoal after applying in-duction may be surprising: 1. Vx args f. [ [args ∈lists (well_formed_gterm' arity ∩ {a. a ∈well_formed_gterm arity}); length args = arity f] ] = ⇒Apply f args ∈well_formed_gterm arity The induction hypothesis contains an application of lists. Using a monotone function in the inductive definition always has this effect. The subgoal may look uninviting, but fortunately lists distributes over intersection: lists (A ∩B) = lists A ∩lists B (lists_Int_eq) Thanks to this default simplification rule, the induction hypothesis is quickly replaced by its two parts: args ∈lists (well_formed_gterm' arity) args ∈lists (well_formed_gterm arity) 140 7. Inductively Defined Sets Invoking the rule well_formed_gterm.step completes the proof. The call to auto does all this work. This example is typical of how monotone functions can be used. In par-ticular, many of them distribute over intersection. Monotonicity implies one direction of this set equality; we have this theorem: mono f = ⇒f (A ∩B) ⊆f A ∩f B (mono_Int) 7.3.4 Another Example of Rule Inversion Does gterms distribute over intersection? We have proved that this function is monotone, so mono_Int gives one of the inclusions. The opposite inclusion asserts that if t is a ground term over both of the sets F and G then it is also a ground term over their intersection, F ∩G. lemma gterms_IntI: "t ∈gterms F = ⇒t ∈gterms G − →t ∈gterms (F∩G)" Attempting this proof, we get the assumption Apply f args ∈gterms G, which cannot be broken down. It looks like a job for rule inversion: inductive cases gterm_Apply_elim [elim!]: "Apply f args ∈gterms F" Here is the result. [ [Apply f args ∈gterms F; [ [∀t∈set args. t ∈gterms F; f ∈F] ] = ⇒P] ] = ⇒P (gterm_Apply_elim) This rule replaces an assumption about Apply f args by assumptions about f and args. No cases are discarded (there was only one to begin with) but the rule applies specifically to the pattern Apply f args. It can be applied repeatedly as an elimination rule without looping, so we have given the elim! attribute. Now we can prove the other half of that distributive law. lemma gterms_IntI [rule_format, intro!]: "t ∈gterms F = ⇒t ∈gterms G − →t ∈gterms (F∩G)" apply (erule gterms.induct) apply blast done The proof begins with rule induction over the definition of gterms, which leaves a single subgoal: 1. Vargs f. [ [∀t∈set args. t ∈gterms F ∧(t ∈gterms G − →t ∈gterms (F ∩G)); f ∈F] ] = ⇒Apply f args ∈gterms G − → Apply f args ∈gterms (F ∩G) 7.4 Case Study: A Context Free Grammar 141 To prove this, we assume Apply f args ∈gterms G. Rule inversion, in the form of gterm_Apply_elim, infers that every element of args belongs to gterms G; hence (by the induction hypothesis) it belongs to gterms (F ∩G). Rule inversion also yields f ∈G and hence f ∈F ∩G. All of this reasoning is done by blast. Our distributive law is a trivial consequence of previously-proved results: lemma gterms_Int_eq [simp]: "gterms (F ∩G) = gterms F ∩gterms G" by (blast intro!: mono_Int monoI gterms_mono) Exercise 7.3.1 A function mapping function symbols to their types is called a signature. Given a type ranging over type symbols, we can represent a function’s type by a list of argument types paired with the result type. Com-plete this inductive definition: inductive set well_typed_gterm :: "('f ⇒'t list 't) ⇒('f gterm 't)set" for sig :: "'f ⇒'t list 't" 7.4 Case Study: A Context Free Grammar Grammars are nothing but shorthands for inductive definitions of nontermi-nals which represent sets of strings. For example, the production A →Bc is short for w ∈B = ⇒wc ∈A This section demonstrates this idea with an example due to Hopcroft and Ullman, a grammar for generating all words with an equal number of a’s and b’s: S → ϵ \| bA \| aB A → aS \| bAA B → bS \| aBB At the end we say a few words about the relationship between the original proof [13, p. 81] and our formal version. We start by fixing the alphabet, which consists only of a’s and b’s: datatype alfa = a \| b For convenience we include the following easy lemmas as simplification rules: lemma [simp]: "(x ̸= a) = (x = b) ∧(x ̸= b) = (x = a)" by (case_tac x, auto) 142 7. Inductively Defined Sets Words over this alphabet are of type alfa list, and the three nonterminals are declared as sets of such words. The productions above are recast as a mutual inductive definition of S, A and B: inductive set S :: "alfa list set" and A :: "alfa list set" and B :: "alfa list set" where "[] ∈S" \| "w ∈A = ⇒b#w ∈S" \| "w ∈B = ⇒a#w ∈S" \| "w ∈S = ⇒a#w ∈A" \| "[ [ v∈A; w∈A ] ] = ⇒b#v@w ∈A" \| "w ∈S = ⇒b#w ∈B" \| "[ [ v ∈B; w ∈B ] ] = ⇒a#v@w ∈B" First we show that all words in S contain the same number of a’s and b’s. Since the definition of S is by mutual induction, so is the proof: we show at the same time that all words in A contain one more a than b and all words in B contain one more b than a. lemma correctness: "(w ∈S − →size[x←w. x=a] = size[x←w. x=b]) ∧ (w ∈A − →size[x←w. x=a] = size[x←w. x=b] + 1) ∧ (w ∈B − →size[x←w. x=b] = size[x←w. x=a] + 1)" These propositions are expressed with the help of the predefined filter func-tion on lists, which has the convenient syntax [x←xs. P x], the list of all el-ements x in xs such that P x holds. Remember that on lists size and length are synonymous. The proof itself is by rule induction and afterwards automatic: by (rule S_A_B.induct, auto) This may seem surprising at first, and is indeed an indication of the power of inductive definitions. But it is also quite straightforward. For example, consider the production A →bAA: if v, w ∈A and the elements of A contain one more a than b’s, then bvw must again contain one more a than b’s. As usual, the correctness of syntactic descriptions is easy, but complete-ness is hard: does S contain all words with an equal number of a’s and b’s? It turns out that this proof requires the following lemma: every string with two more a’s than b’s can be cut somewhere such that each half has one more a than b. This is best seen by imagining counting the difference between the number of a’s and b’s starting at the left end of the word. We start with 0 and end (at the right end) with 2. Since each move to the right increases or decreases the difference by 1, we must have passed through 1 on our way from 0 to 2. Formally, we appeal to the following discrete intermediate value theorem nat0_intermed_int_val 7.4 Case Study: A Context Free Grammar 143 [ [∀i<n. \|f (i + 1) - f i\| ≤1; f 0 ≤k; k ≤f n] ] = ⇒∃i≤n. f i = k where f is of type nat ⇒int, int are the integers, \|.\| is the absolute value function2, and 1 is the integer 1 (see Sect. 8.4). First we show that our specific function, the difference between the num-bers of a’s and b’s, does indeed only change by 1 in every move to the right. At this point we also start generalizing from a’s and b’s to an arbitrary prop-erty P. Otherwise we would have to prove the desired lemma twice, once as stated above and once with the roles of a’s and b’s interchanged. lemma step1: "∀i < size w. \|(int(size[x←take (i+1) w. P x])-int(size[x←take (i+1) w. ¬P x])) - (int(size[x←take i w. P x])-int(size[x←take i w. ¬P x]))\| ≤1" The lemma is a bit hard to read because of the coercion function int :: nat ⇒int. It is required because size returns a natural number, but subtraction on type nat will do the wrong thing. Function take is predefined and take i xs is the prefix of length i of xs; below we also need drop i xs, which is what remains after that prefix has been dropped from xs. The proof is by induction on w, with a trivial base case, and a not so trivial induction step. Since it is essentially just arithmetic, we do not discuss it. apply(induct_tac w) apply(auto simp add: abs_if take_Cons split: nat.split) done Finally we come to the above-mentioned lemma about cutting in half a word with two more elements of one sort than of the other sort: lemma part1: "size[x←w. P x] = size[x←w. ¬P x]+2 = ⇒ ∃i≤size w. size[x←take i w. P x] = size[x←take i w. ¬P x]+1" This is proved by force with the help of the intermediate value theorem, instantiated appropriately and with its first premise disposed of by lemma step1: apply(insert nat0_intermed_int_val[OF step1, of "P" "w" "1"]) by force Lemma part1 tells us only about the prefix take i w. An easy lemma deals with the suffix drop i w: lemma part2: "[ [size[x←take i w @ drop i w. P x] = size[x←take i w @ drop i w. ¬P x]+2; size[x←take i w. P x] = size[x←take i w. ¬P x]+1] ] = ⇒size[x←drop i w. P x] = size[x←drop i w. ¬P x]+1" by(simp del: append_take_drop_id) In the proof we have disabled the normally useful lemma 2 See Table A.1 in the Appendix for the correct ascii syntax. 144 7. Inductively Defined Sets take n xs @ drop n xs = xs (append_take_drop_id) to allow the simplifier to apply the following lemma instead: [x∈xs@ys. P x] = [x∈xs. P x] @ [x∈ys. P x] To dispose of trivial cases automatically, the rules of the inductive defini-tion are declared simplification rules: declare S_A_B.intros[simp] This could have been done earlier but was not necessary so far. The completeness theorem tells us that if a word has the same number of a’s and b’s, then it is in S, and similarly for A and B: theorem completeness: "(size[x←w. x=a] = size[x←w. x=b] − →w ∈S) ∧ (size[x←w. x=a] = size[x←w. x=b] + 1 − →w ∈A) ∧ (size[x←w. x=b] = size[x←w. x=a] + 1 − →w ∈B)" The proof is by induction on w. Structural induction would fail here because, as we can see from the grammar, we need to make bigger steps than merely appending a single letter at the front. Hence we induct on the length of w, using the induction rule length_induct: apply(induct_tac w rule: length_induct) apply(rename_tac w) The rule parameter tells induct_tac explicitly which induction rule to use. For details see Sect. 9.2.2 below. In this case the result is that we may assume the lemma already holds for all words shorter than w. Because the induction step renames the induction variable we rename it back to w. The proof continues with a case distinction on w, on whether w is empty or not. apply(case_tac w) apply(simp_all) Simplification disposes of the base case and leaves only a conjunction of two step cases to be proved: if w = a # v and length (if x = a then [x ∈v] else []) = length (if x = b then [x ∈v] else []) + 2 then b # v ∈A, and similarly for w = b # v. We only consider the first case in detail. After breaking the conjunction up into two cases, we can apply part1 to the assumption that w contains two more a’s than b’s. apply(rule conjI) apply(clarify) apply(frule part1[of "λx. x=a", simplified]) apply(clarify) This yields an index i ≤length v such that 7.4 Case Study: A Context Free Grammar 145 length (filter (λx. x = a) (take i v)) = length (filter (λx. x = b) (take i v)) + 1 With the help of part2 it follows that length (filter (λx. x = a) (drop i v)) = length (filter (λx. x = b) (drop i v)) + 1 apply(drule part2[of "λx. x=a", simplified]) apply(assumption) Now it is time to decompose v in the conclusion b # v ∈A into take i v @ drop i v, apply(rule_tac n1=i and t=v in subst[OF append_take_drop_id]) (the variables n1 and t are the result of composing the theorems subst and append_take_drop_id) after which the appropriate rule of the grammar re-duces the goal to the two subgoals take i v ∈A and drop i v ∈A: apply(rule S_A_B.intros) Both subgoals follow from the induction hypothesis because both take i v and drop i v are shorter than w: apply(force simp add: min_less_iff_disj) apply(force split: nat_diff_split) The case w = b # v is proved analogously: apply(clarify) apply(frule part1[of "λx. x=b", simplified]) apply(clarify) apply(drule part2[of "λx. x=b", simplified]) apply(assumption) apply(rule_tac n1=i and t=v in subst[OF append_take_drop_id]) apply(rule S_A_B.intros) apply(force simp add: min_less_iff_disj) by(force simp add: min_less_iff_disj split: nat_diff_split) We conclude this section with a comparison of our proof with Hopcroft and Ullman’s [13, p. 81]. For a start, the textbook grammar, for no good reason, excludes the empty word, thus complicating matters just a little bit: they have 8 instead of our 7 productions. More importantly, the proof itself is different: rather than separating the two directions, they perform one induction on the length of a word. This deprives them of the beauty of rule induction, and in the easy direction (correctness) their reasoning is more detailed than our auto. For the hard part (completeness), they consider just one of the cases that our simp_all disposes of automatically. Then they conclude the proof by saying about the remaining cases: “We do this in a manner similar to our method of proof for part (1); this part is left to the reader”. But this is precisely the part that requires the intermediate value theorem and thus is not at all similar to the other cases (which are automatic in Isabelle). The authors are at least 146 7. Inductively Defined Sets cavalier about this point and may even have overlooked the slight difficulty lurking in the omitted cases. Such errors are found in many pen-and-paper proofs when they are scrutinized formally. Part III Advanced Material 8. More about Types So far we have learned about a few basic types (for example bool and nat), type abbreviations (types) and recursive datatypes (datatype). This chap-ter will introduce more advanced material: – Pairs (Sect. 8.1) and records (Sect. 8.2), and how to reason about them. – Type classes: how to specify and reason about axiomatic collections of types (Sect. 8.3). This section leads on to a discussion of Isabelle’s numeric types (Sect. 8.4). – Introducing your own types: how to define types that cannot be constructed with any of the basic methods (Sect. 8.5). The material in this section goes beyond the needs of most novices. Serious users should at least skim the sections as far as type classes. That material is fairly advanced; read the beginning to understand what it is about, but consult the rest only when necessary. 8.1 Pairs and Tuples Ordered pairs were already introduced in Sect. 2.6.2, but only with a minimal repertoire of operations: pairing and the two projections fst and snd. In any non-trivial application of pairs you will find that this quickly leads to unreadable nests of projections. This section introduces syntactic sugar to overcome this problem: pattern matching with tuples. 8.1.1 Pattern Matching with Tuples Tuples may be used as patterns in λ-abstractions, for example λ(x,y,z).x+y+z and λ((x,y),z).x+y+z. In fact, tuple patterns can be used in most variable binding constructs, and they can be nested. Here are some typical examples: let (x, y) = f z in (y, x) case xs of [] ⇒0 \| (x, y) # zs ⇒x + y ∀(x,y)∈A. x=y {(x,y,z). x=z} S (x, y)∈A {x + y} 150 8. More about Types The intuitive meanings of these expressions should be obvious. Unfortunately, we need to know in more detail what the notation really stands for once we have to reason about it. Abstraction over pairs and tuples is merely a convenient shorthand for a more complex internal representation. Thus the internal and external form of a term may differ, which can affect proofs. If you want to avoid this complication, stick to fst and snd and write λp. fst p + snd p instead of λ(x,y). x+y. These terms are distinct even though they denote the same function. Internally, λ(x, y). t becomes case_prod (λx y. t), where split is the uncurrying function of type ('a ⇒'b ⇒'c) ⇒'a × 'b ⇒'c defined as case_prod = (λc p. c (fst p) (snd p)) (split_def) Pattern matching in other variable binding constructs is translated similarly. Thus we need to understand how to reason about such constructs. 8.1.2 Theorem Proving The most obvious approach is the brute force expansion of split: lemma "(λ(x,y).x) p = fst p" by(simp add: split_def) This works well if rewriting with split_def finishes the proof, as it does above. But if it does not, you end up with exactly what we are trying to avoid: nests of fst and snd. Thus this approach is neither elegant nor very practical in large examples, although it can be effective in small ones. If we consider why this lemma presents a problem, we realize that we need to replace variable p by some pair (a, b). Then both sides of the equation would simplify to a by the simplification rules (case (a, b) of (c, d) ⇒f c d) = f a b and fst (x1, x2) = x1. To reason about tuple patterns requires some way of converting a variable of product type into a pair. In case of a subterm of the form case p of (x, xa) ⇒f x xa this is easy: the split rule prod.split replaces p by a pair: lemma "(λ(x,y).y) p = snd p" apply(split prod.split) 1. ∀x1 x2. p = (x1, x2) − →x2 = snd p This subgoal is easily proved by simplification. Thus we could have combined simplification and splitting in one command that proves the goal outright: by(simp split: prod.split) Let us look at a second example: lemma "let (x,y) = p in fst p = x" apply(simp only: Let_def) 8.1 Pairs and Tuples 151 1. case p of (x, y) ⇒fst p = x A paired let reduces to a paired λ-abstraction, which can be split as above. The same is true for paired set comprehension: lemma "p ∈{(x,y). x=y} − →fst p = snd p" apply simp 1. (case p of (x, xa) ⇒x = xa) − →fst p = snd p Again, simplification produces a term suitable for prod.split as above. If you are worried about the strange form of the premise: case_prod (=) is short for λ(x, y). x = y. The same proof procedure works for lemma "p ∈{(x,y). x=y} = ⇒fst p = snd p" except that we now have to use prod.split_asm, because split occurs in the assumptions. However, splitting split is not always a solution, as no split may be present in the goal. Consider the following function: primrec swap :: "'a × 'b ⇒'b × 'a" where "swap (x,y) = (y,x)" Note that the above primrec definition is admissible because × is a datatype. When we now try to prove lemma "swap(swap p) = p" simplification will do nothing, because the defining equation for swap expects a pair. Again, we need to turn p into a pair first, but this time there is no split in sight. The only thing we can do is to split the term by hand: apply(case_tac p) 1. Va b. p = (a, b) = ⇒swap (swap p) = p Again, case_tac is applicable because × is a datatype. The subgoal is easily proved by simp. Splitting by case_tac also solves the previous examples and may thus appear preferable to the more arcane methods introduced first. However, see the warning about case_tac in Sect. 2.5.5. Alternatively, you can split all V-quantified variables in a goal with the rewrite rule split_paired_all: lemma "Vp q. swap(swap p) = q − →p = q" apply(simp only: split_paired_all) 1. Va b aa ba. swap (swap (a, b)) = (aa, ba) − →(a, b) = (aa, ba) apply simp done Note that we have intentionally included only split_paired_all in the first simplification step, and then we simplify again. This time the reason was not 152 8. More about Types merely pedagogical: split_paired_all may interfere with other functions of the simplifier. The following command could fail (here it does not) where two separate simp applications succeed. apply(simp add: split_paired_all) Finally, the simplifier automatically splits all ∀and ∃-quantified variables: lemma "∀p. ∃q. swap p = swap q" by simp To turn off this automatic splitting, disable the responsible simplification rules: (∀x. P x) = (∀a b. P (a, b)) (split_paired_All) (∃x. P x) = (∃a b. P (a, b)) (split_paired_Ex) 8.2 Records Records are familiar from programming languages. A record of n fields is essentially an n-tuple, but the record’s components have names, which can make expressions easier to read and reduces the risk of confusing one field for another. A record of Isabelle/HOL covers a collection of fields, with select and update operations. Each field has a specified type, which may be polymorphic. The field names are part of the record type, and the order of the fields is significant — as it is in Pascal but not in Standard ML. If two different record types have field names in common, then the ambiguity is resolved in the usual way, by qualified names. Record types can also be defined by extending other record types. Exten-sible records make use of the reserved pseudo-field more, which is present in every record type. Generic record operations work on all possible extensions of a given type scheme; polymorphism takes care of structural sub-typing behind the scenes. There are also explicit coercion functions between fixed record types. 8.2.1 Record Basics Record types are not primitive in Isabelle and have a delicate internal rep-resentation , based on nested copies of the primitive product type. A record declaration introduces a new record type scheme by specifying its fields, which are packaged internally to hold up the perception of the record as a distinguished entity. Here is a simple example: record point = Xcoord :: int 8.2 Records 153 Ycoord :: int Records of type point have two fields named Xcoord and Ycoord, both of type int. We now define a constant of type point: definition pt1 :: point where "pt1 ≡(\| Xcoord = 999, Ycoord = 23 \|)" We see above the ASCII notation for record brackets. You can also use the symbolic brackets ( \| and \| ). Record type expressions can be also written di-rectly with individual fields. The type name above is merely an abbreviation. definition pt2 :: "( \|Xcoord :: int, Ycoord :: int\| )" where "pt2 ≡( \|Xcoord = -45, Ycoord = 97\| )" For each field, there is a selector function of the same name. For exam-ple, if p has type point then Xcoord p denotes the value of the Xcoord field of p. Expressions involving field selection of explicit records are simplified automatically: lemma "Xcoord ( \|Xcoord = a, Ycoord = b\| ) = a" by simp The update operation is functional. For example, p( \|Xcoord := 0\| ) is a record whose Xcoord value is zero and whose Ycoord value is copied from p. Updates of explicit records are also simplified automatically: lemma "( \|Xcoord = a, Ycoord = b\| )( \|Xcoord := 0\| ) = ( \|Xcoord = 0, Ycoord = b\| )" by simp ! ! Field names are declared as constants and can no longer be used as variables. It would be unwise, for example, to call the fields of type point simply x and y. 8.2.2 Extensible Records and Generic Operations Now, let us define coloured points (type cpoint) to be points extended with a field col of type colour: datatype colour = Red \| Green \| Blue record cpoint = point + col :: colour The fields of this new type are Xcoord, Ycoord and col, in that order. definition cpt1 :: cpoint where "cpt1 ≡( \|Xcoord = 999, Ycoord = 23, col = Green\| )" We can define generic operations that work on arbitrary instances of a record scheme, e.g. covering point, cpoint, and any further extensions. Every record structure has an implicit pseudo-field, more, that keeps the extension 154 8. More about Types as an explicit value. Its type is declared as completely polymorphic: 'a. When a fixed record value is expressed using just its standard fields, the value of more is implicitly set to (), the empty tuple, which has type unit. Within the record brackets, you can refer to the more field by writing “. . . ” (three dots): lemma "Xcoord ( \|Xcoord = a, Ycoord = b, . . . = p\| ) = a" by simp This lemma applies to any record whose first two fields are Xcoord and Ycoord. Note that ( \|Xcoord = a, Ycoord = b, . . . = ()\| ) is exactly the same as ( \|Xcoord = a, Ycoord = b\| ). Selectors and updates are always poly-morphic wrt. the more part of a record scheme, its value is just ignored (for select) or copied (for update). The more pseudo-field may be manipulated directly as well, but the iden-tifier needs to be qualified: lemma "point.more cpt1 = ( \|col = Green\| )" by (simp add: cpt1_def) We see that the colour part attached to this point is a rudimentary record in its own right, namely ( \|col = Green\| ). In order to select or update col, this fragment needs to be put back into the context of the parent type scheme, say as more part of another point. To define generic operations, we need to know a bit more about records. Our definition of point above has generated two type abbreviations: point = ( \|Xcoord :: int, Ycoord :: int\| ) 'a point_scheme = ( \|Xcoord :: int, Ycoord :: int, . . . :: 'a\| ) Type point is for fixed records having exactly the two fields Xcoord and Ycoord, while the polymorphic type 'a point_scheme comprises all possible extensions to those two fields. Note that unit point_scheme coincides with point, and ( \|col :: colour\| ) point_scheme with cpoint. In the following example we define two operations — methods, if we regard records as objects — to get and set any point’s Xcoord field. definition getX :: "'a point_scheme ⇒int" where "getX r ≡Xcoord r" definition setX :: "'a point_scheme ⇒int ⇒'a point_scheme" where "setX r a ≡r( \|Xcoord := a\| )" Here is a generic method that modifies a point, incrementing its Xcoord field. The Ycoord and more fields are copied across. It works for any record type scheme derived from point (including cpoint etc.): definition incX :: "'a point_scheme ⇒'a point_scheme" where "incX r ≡ ( \|Xcoord = Xcoord r + 1, Ycoord = Ycoord r, . . . = point.more r\| )" Generic theorems can be proved about generic methods. This trivial lemma relates incX to getX and setX: lemma "incX r = setX r (getX r + 1)" 8.2 Records 155 by (simp add: getX_def setX_def incX_def) ! ! If you use the symbolic record brackets ( \| and \| ), then you must also use the symbolic ellipsis, “. . . ”, rather than three consecutive periods, “...”. Mixing the ASCII and symbolic versions causes a syntax error. (The two versions are more distinct on screen than they are on paper.) 8.2.3 Record Equality Two records are equal if all pairs of corresponding fields are equal. Concrete record equalities are simplified automatically: lemma "(( \|Xcoord = a, Ycoord = b\| ) = ( \|Xcoord = a', Ycoord = b'\| )) = (a = a' ∧b = b')" by simp The following equality is similar, but generic, in that r can be any instance of 'a point_scheme: lemma "r( \|Xcoord := a, Ycoord := b\| ) = r( \|Ycoord := b, Xcoord := a\| )" by simp We see above the syntax for iterated updates. We could equivalently have written the left-hand side as r( \|Xcoord := a\| )( \|Ycoord := b\| ). Record equality is extensional: a record is determined entirely by the values of its fields. lemma "r = ( \|Xcoord = Xcoord r, Ycoord = Ycoord r\| )" by simp The generic version of this equality includes the pseudo-field more: lemma "r = ( \|Xcoord = Xcoord r, Ycoord = Ycoord r, . . . = point.more r\| )" by simp The simplifier can prove many record equalities automatically, but general equality reasoning can be tricky. Consider proving this obvious fact: lemma "r( \|Xcoord := a\| ) = r( \|Xcoord := a'\| ) = ⇒a = a'" apply simp? oops Here the simplifier can do nothing, since general record equality is not elimi-nated automatically. One way to proceed is by an explicit forward step that applies the selector Xcoord to both sides of the assumed record equality: lemma "r( \|Xcoord := a\| ) = r( \|Xcoord := a'\| ) = ⇒a = a'" apply (drule_tac f = Xcoord in arg_cong) 1. Xcoord (r( \|Xcoord := a\| )) = Xcoord (r( \|Xcoord := a'\| )) = ⇒a = a' 156 8. More about Types Now, simp will reduce the assumption to the desired conclusion. apply simp done The cases method is preferable to such a forward proof. We state the desired lemma again: lemma "r( \|Xcoord := a\| ) = r( \|Xcoord := a'\| ) = ⇒a = a'" The cases method adds an equality to replace the named record term by an explicit record expression, listing all fields. It even includes the pseudo-field more, since the record equality stated here is generic for all extensions. apply (cases r) 1. VXcoord Ycoord more. [ [r( \|Xcoord := a\| ) = r( \|Xcoord := a'\| ); r = ( \|Xcoord = Xcoord, Ycoord = Ycoord, . . . = more\| )] ] = ⇒a = a' Again, simp finishes the proof. Because r is now represented as an explicit record construction, the updates can be applied and the record equality can be replaced by equality of the corresponding fields (due to injectivity). apply simp done The generic cases method does not admit references to locally bound parameters of a goal. In longer proof scripts one might have to fall back on the primitive rule_tac used together with the internal field representation rules of records. The above use of (cases r) would become (rule_tac r = r in point.cases_scheme). 8.2.4 Extending and Truncating Records Each record declaration introduces a number of derived operations to refer collectively to a record’s fields and to convert between fixed record types. They can, for instance, convert between types point and cpoint. We can add a colour to a point or convert a cpoint to a point by forgetting its colour. – Function make takes as arguments all of the record’s fields (including those inherited from ancestors). It returns the corresponding record. – Function fields takes the record’s very own fields and returns a record fragment consisting of just those fields. This may be filled into the more part of the parent record scheme. – Function extend takes two arguments: a record to be extended and a record containing the new fields. – Function truncate takes a record (possibly an extension of the original record type) and returns a fixed record, removing any additional fields. 8.2 Records 157 These functions provide useful abbreviations for standard record expressions involving constructors and selectors. The definitions, which are not unfolded by default, are made available by the collective name of defs (point.defs, cpoint.defs, etc.). For example, here are the versions of those functions gen-erated for record point. We omit point.fields, which happens to be the same as point.make. point.make Xcoord Ycoord ≡ ( \|point.Xcoord = Xcoord, Ycoord = Ycoord\| ) point.extend r more ≡ ( \|Xcoord = Xcoord r, Ycoord = Ycoord r, . . . = more\| ) point.truncate r ≡( \|Xcoord = Xcoord r, Ycoord = Ycoord r\| ) Contrast those with the corresponding functions for record cpoint. Observe cpoint.fields in particular. cpoint.make Xcoord Ycoord col ≡ ( \|point.Xcoord = Xcoord, Ycoord = Ycoord, cpoint.col = col\| ) cpoint.fields col ≡( \|cpoint.col = col\| ) cpoint.extend r more ≡ ( \|Xcoord = Xcoord r, Ycoord = Ycoord r, col = col r, . . . = more\| ) cpoint.truncate r ≡ ( \|Xcoord = Xcoord r, Ycoord = Ycoord r, col = col r\| ) To demonstrate these functions, we declare a new coloured point by extending an ordinary point. Function point.extend augments pt1 with a colour value, which is converted into an appropriate record fragment by cpoint.fields. definition cpt2 :: cpoint where "cpt2 ≡point.extend pt1 (cpoint.fields Green)" The coloured points cpt1 and cpt2 are equal. The proof is trivial, by unfolding all the definitions. We deliberately omit the definition of pt1 in order to reveal the underlying comparison on type point. lemma "cpt1 = cpt2" apply (simp add: cpt1_def cpt2_def point.defs cpoint.defs) 1. Xcoord pt1 = 999 ∧Ycoord pt1 = 23 apply (simp add: pt1_def) done In the example below, a coloured point is truncated to leave a point. We use the truncate function of the target record. lemma "point.truncate cpt2 = pt1" by (simp add: pt1_def cpt2_def point.defs) Exercise 8.2.1 Extend record cpoint to have a further field, intensity, of type nat. Experiment with generic operations (using polymorphic selectors and updates) and explicit coercions (using extend, truncate etc.) among the three record types. 158 8. More about Types Exercise 8.2.2 (For Java programmers.) Model a small class hierarchy using records. 8.3 Type Classes The programming language Haskell has popularized the notion of type classes: a type class is a set of types with a common interface: all types in that class must provide the functions in the interface. Isabelle offers a sim-ilar type class concept: in addition, properties (class axioms) can be specified which any instance of this type class must obey. Thus we can talk about a type τ being in a class C, which is written τ :: C. This is the case if τ satisfies the axioms of C. Furthermore, type classes can be organized in a hierarchy. Thus there is the notion of a class D being a subclass of a class C, written D < C. This is the case if all axioms of C are also provable in D. In this section we introduce the most important concepts behind type classes by means of a running example from algebra. This should give you an intuition how to use type classes and to understand specifications involving type classes. Type classes are covered more deeply in a separate tutorial . 8.3.1 Overloading Type classes allow overloading; thus a constant may have multiple definitions at non-overlapping types. Overloading. We can introduce a binary infix addition operator ⊕for ar-bitrary types by means of a type class: class plus = fixes plus :: "'a ⇒'a ⇒'a" (infixl "⊕" 70) This introduces a new class plus, along with a constant plus with nice infix syntax. plus is also named class operation. The type of plus carries a class constraint "'a :: plus" on its type variable, meaning that only types of class plus can be instantiated for "'a". To breathe life into plus we need to declare a type to be an instance of plus: instantiation nat :: plus begin Command instantiation opens a local theory context. Here we can now instantiate plus on nat: primrec plus_nat :: "nat ⇒nat ⇒nat" where "(0::nat) ⊕n = n" \| "Suc m ⊕n = Suc (m ⊕n)" 8.3 Type Classes 159 Note that the name plus carries a suffix _nat; by default, the local name of a class operation f to be instantiated on type constructor κ is mangled as f_κ. In case of uncertainty, these names may be inspected using the print context command. Although class plus has no axioms, the instantiation must be formally concluded by a (trivial) instantiation proof “..”: instance .. More interesting instance proofs will arise below. The instantiation is finished by an explicit end From now on, terms like Suc (m ⊕2) are legal. instantiation prod :: (plus, plus) plus begin Here we instantiate the product type prod to class plus, given that its type arguments are of class plus: fun plus_prod :: "'a × 'b ⇒'a × 'b ⇒'a × 'b" where "(x, y) ⊕(w, z) = (x ⊕w, y ⊕z)" Obviously, overloaded specifications may include recursion over the syntactic structure of types. instance .. end This way we have encoded the canonical lifting of binary operations to prod-ucts by means of type classes. 8.3.2 Axioms Attaching axioms to our classes lets us reason on the level of classes. The results will be applicable to all types in a class, just as in axiomatic mathe-matics. ! ! Proofs in this section use structured Isar proofs, which are not covered in this tutorial; but see . Semigroups. We specify semigroups as subclass of plus: class semigroup = plus + assumes assoc: "(x ⊕y) ⊕z = x ⊕(y ⊕z)" This class specification requires that all instances of semigroup obey assoc: "Vx y z :: 'a::semigroup. (x ⊕y) ⊕z = x ⊕(y ⊕z)". 160 8. More about Types We can use this class axiom to derive further abstract theorems relative to class semigroup: lemma assoc_left: fixes x y z :: "'a::semigroup" shows "x ⊕(y ⊕z) = (x ⊕y) ⊕z" using assoc by (rule sym) The semigroup constraint on type 'a restricts instantiations of 'a to types of class semigroup and during the proof enables us to use the fact assoc whose type parameter is itself constrained to class semigroup. The main advantage of classes is that theorems can be proved in the abstract and freely reused for each instance. On instantiation, we have to give a proof that the given operations obey the class axioms: instantiation nat :: semigroup begin instance proof The proof opens with a default proof step, which for instance judgements invokes method intro_classes. fix m n q :: nat show "(m ⊕n) ⊕q = m ⊕(n ⊕q)" by (induct m) simp_all qed end Again, the interesting things enter the stage with parametric types: instantiation prod :: (semigroup, semigroup) semigroup begin instance proof fix p1 p2 p3 :: "'a::semigroup × 'b::semigroup" show "p1 ⊕p2 ⊕p3 = p1 ⊕(p2 ⊕p3)" by (cases p1, cases p2, cases p3) (simp add: assoc) Associativity of product semigroups is established using the hypothetical as-sociativity assoc of the type components, which holds due to the semigroup constraints imposed on the type components by the instance proposition. Indeed, this pattern often occurs with parametric types and type classes. qed end Monoids. We define a subclass monoidl (a semigroup with a left-hand neu-tral) by extending semigroup with one additional parameter neutral together with its property: class monoidl = semigroup + 8.3 Type Classes 161 fixes neutral :: "'a" ("0") assumes neutl: "0 ⊕x = x" Again, we prove some instances, by providing suitable parameter definitions and proofs for the additional specifications. instantiation nat :: monoidl begin definition neutral_nat_def: "0 = (0::nat)" instance proof fix n :: nat show "0 ⊕n = n" unfolding neutral_nat_def by simp qed end In contrast to the examples above, we here have both specification of class operations and a non-trivial instance proof. This covers products as well: instantiation prod :: (monoidl, monoidl) monoidl begin definition neutral_prod_def: "0 = (0, 0)" instance proof fix p :: "'a::monoidl × 'b::monoidl" show "0 ⊕p = p" by (cases p) (simp add: neutral_prod_def neutl) qed end Fully-fledged monoids are modelled by another subclass which does not add new parameters but tightens the specification: class monoid = monoidl + assumes neutr: "x ⊕0 = x" Corresponding instances for nat and products are left as an exercise to the reader. Groups. To finish our small algebra example, we add a group class: class group = monoidl + fixes inv :: "'a ⇒'a" ("÷ " 80) assumes invl: "÷ x ⊕x = 0" We continue with a further example for abstract proofs relative to type classes: 162 8. More about Types lemma left_cancel: fixes x y z :: "'a::group" shows "x ⊕y = x ⊕z ← →y = z" proof assume "x ⊕y = x ⊕z" then have "÷ x ⊕(x ⊕y) = ÷ x ⊕(x ⊕z)" by simp then have "(÷ x ⊕x) ⊕y = (÷ x ⊕x) ⊕z" by (simp add: assoc) then show "y = z" by (simp add: invl neutl) next assume "y = z" then show "x ⊕y = x ⊕z" by simp qed Any group is also a monoid; this can be made explicit by claiming an additional subclass relation, together with a proof of the logical difference: instance group ⊆monoid proof fix x from invl have "÷ x ⊕x = 0" . then have "÷ x ⊕(x ⊕0) = ÷ x ⊕x" by (simp add: neutl invl assoc [symmetric]) then show "x ⊕0 = x" by (simp add: left_cancel) qed The proof result is propagated to the type system, making group an instance of monoid by adding an additional edge to the graph of subclass relation; see also Figure 8.1. semigroup monoidl monoid group ? B B B B B B N semigroup monoidl monoid group ? PPPPP P q Fig. 8.1. Subclass relationship of monoids and groups: before and after establishing the relationship group ⊆monoid; transitive edges are left out. Inconsistencies. The reader may be wondering what happens if we attach an inconsistent set of axioms to a class. So far we have always avoided to add new axioms to HOL for fear of inconsistencies and suddenly it seems that we are throwing all caution to the wind. So why is there no problem? The point is that by construction, all type variables in the axioms of a class are automatically constrained with the class being defined (as shown for 8.4 Numbers 163 axiom refl above). These constraints are always carried around and Isabelle takes care that they are never lost, unless the type variable is instantiated with a type that has been shown to belong to that class. Thus you may be able to prove False from your axioms, but Isabelle will remind you that this theorem has the hidden hypothesis that the class is non-empty. Even if each individual class is consistent, intersections of (unrelated) classes readily become inconsistent in practice. Now we know this need not worry us. Syntactic Classes and Predefined Overloading. In our algebra exam-ple, we have started with a syntactic class plus which only specifies operations but no axioms; it would have been also possible to start immediately with class semigroup, specifying the ⊕operation and associativity at the same time. Which approach is more appropriate depends. Usually it is more conve-nient to introduce operations and axioms in the same class: then the type checker will automatically insert the corresponding class constraints when-ever the operations occur, reducing the need of manual annotations. However, when operations are decorated with popular syntax, syntactic classes can be an option to re-use the syntax in different contexts; this is indeed the way most overloaded constants in HOL are introduced, of which the most impor-tant are listed in Table A.2 in the appendix. Section 8.4.5 covers a range of corresponding classes with axioms. Further note that classes may contain axioms but no operations. An ex-ample is class finite from theory HOL.Finite_Set which specifies a type to be finite: "finite (UNIV :: 'a::finite set)". 8.4 Numbers Until now, our numerical examples have used the type of natural num-bers, nat. This is a recursive datatype generated by the constructors zero and successor, so it works well with inductive proofs and primitive recursive function definitions. HOL also provides the type int of integers, which lack induction but support true subtraction. With subtraction, arithmetic reason-ing is easier, which makes the integers preferable to the natural numbers for complicated arithmetic expressions, even if they are non-negative. There are also the types rat, real and complex: the rational, real and complex num-bers. Isabelle has no subtyping, so the numeric types are distinct and there are functions to convert between them. Most numeric operations are over-loaded: the same symbol can be used at all numeric types. Table A.2 in the appendix shows the most important operations, together with the priorities of the infix symbols. Algebraic properties are organized using type classes around algebraic concepts such as rings and fields; a property such as the 164 8. More about Types commutativity of addition is a single theorem (add.commute) that applies to all numeric types. Many theorems involving numeric types can be proved automatically by Isabelle’s arithmetic decision procedure, the method arith. Linear arithmetic comprises addition, subtraction and multiplication by constant factors; sub-terms involving other operators are regarded as variables. The procedure can be slow, especially if the subgoal to be proved involves subtraction over type nat, which causes case splits. On types nat and int, arith can deal with quantifiers—this is known as Presburger arithmetic—whereas on type real it cannot. The simplifier reduces arithmetic expressions in other ways, such as di-viding through by common factors. For problems that lie outside the scope of automation, HOL provides hundreds of theorems about multiplication, di-vision, etc., that can be brought to bear. You can locate them using Proof General’s Find button. A few lemmas are given below to show what is avail-able. 8.4.1 Numeric Literals The constants 0 and 1 are overloaded. They denote zero and one, respectively, for all numeric types. Other values are expressed by numeric literals, which consist of one or more decimal digits optionally preceeded by a minus sign (-). Examples are 2, -3 and 441223334678. Literals are available for the types of natural numbers, integers, rationals, reals, etc.; they denote integer values of arbitrary size. Literals look like constants, but they abbreviate terms representing the number in a two’s complement binary notation. Isabelle performs arithmetic on literals by rewriting rather than using the hardware arithmetic. In most cases arithmetic is fast enough, even for numbers in the millions. The arith-metic operations provided for literals include addition, subtraction, multipli-cation, integer division and remainder. Fractions of literals (expressed using division) are reduced to lowest terms. ! ! The arithmetic operators are overloaded, so you must be careful to ensure that each numeric expression refers to a specific type, if necessary by inserting type constraints. Here is an example of what can go wrong: lemma "2 m = m + m" Carefully observe how Isabelle displays the subgoal: 1. (2::'a) m = m + m The type 'a given for the literal 2 warns us that no numeric type has been specified. The problem is underspecified. Given a type constraint such as nat, int or real, it becomes trivial. 8.4 Numbers 165 ! ! Numeric literals are not constructors and therefore must not be used in pat-terns. For example, this declaration is rejected: function h where "h 3 = 2" \|"h i = i" You should use a conditional expression instead: "h i = (if i = 3 then 2 else i)" 8.4.2 The Type of Natural Numbers, nat This type requires no introduction: we have been using it from the beginning. Hundreds of theorems about the natural numbers are proved in the theories Nat and Divides. Basic properties of addition and multiplication are available through the axiomatic type class for semirings (Sect. 8.4.5). Literals. The notational options for the natural numbers are confusing. Re-call that an overloaded constant can be defined independently for each type; the definition of 1 for type nat is 1 ≡Suc 0 (One_nat_def) This is installed as a simplification rule, so the simplifier will replace every occurrence of 1::nat by Suc 0. Literals are obviously better than nested Sucs at expressing large values. But many theorems, including the rewrite rules for primitive recursive functions, can only be applied to terms of the form Suc n. The following default simplification rules replace small literals by zero and successor: 2 + n = Suc (Suc n) (add_2_eq_Suc) n + 2 = Suc (Suc n) (add_2_eq_Suc') It is less easy to transform 100 into Suc 99 (for example), and the simpli-fier will normally reverse this transformation. Novices should express natural numbers using 0 and Suc only. Division. The infix operators div and mod are overloaded. Isabelle/HOL pro-vides the basic facts about quotient and remainder on the natural numbers: m mod n = (if m < n then m else (m - n) mod n) (mod_if) m div n n + m mod n = m (div_mult_mod_eq) Many less obvious facts about quotient and remainder are also provided. Here is a selection: a b div c = a (b div c) + a (b mod c) div c (div_mult1_eq) a b mod c = a (b mod c) mod c (mod_mult_right_eq) a div (bc) = a div b div c (div_mult2_eq) a mod (bc) = b (a div b mod c) + a mod b (mod_mult2_eq) 0 < c = ⇒(c a) div (c b) = a div b (div_mult_mult1) (m mod n) k = (m k) mod (n k) (mod_mult_distrib) m ≤n = ⇒m div k ≤n div k (div_le_mono) 166 8. More about Types Surprisingly few of these results depend upon the divisors’ being nonzero. That is because division by zero yields zero: a div 0 = 0 (DIVISION_BY_ZERO_DIV) a mod 0 = a (DIVISION_BY_ZERO_MOD) In div_mult_mult1 above, one of the two divisors (namely c) must still be nonzero. The divides relation has the standard definition, which is overloaded over all numeric types: m dvd n ≡∃k. n = m k (dvd_def) Section 5.18 discusses proofs involving this relation. Here are some of the facts proved about it: [ [m dvd n; n dvd m] ] = ⇒m = n (dvd_antisym) [ [k dvd m; k dvd n] ] = ⇒k dvd (m + n) (dvd_add) Subtraction. There are no negative natural numbers, so m - n equals zero unless m exceeds n. The following is one of the few facts about m - n that is not subject to the condition n ≤m. (m - n) k = m k - n k (diff_mult_distrib) Natural number subtraction has few nice properties; often you should remove it by simplifying with this split rule. P(a-b) = ((a<b − →P 0) ∧(∀d. a = b+d − →P d)) (nat_diff_split) For example, splitting helps to prove the following fact. lemma "(n - 2) (n + 2) = n n - (4::nat)" apply (simp split: nat_diff_split, clarify) 1. Vd. [ [n < 2; n n = 4 + d] ] = ⇒d = 0 The result lies outside the scope of linear arithmetic, but it is easily found if we explicitly split n<2 as n=0 or n=1: apply (subgoal_tac "n=0 \| n=1", force, arith) done 8.4.3 The Type of Integers, int Reasoning methods for the integers resemble those for the natural numbers, but induction and the constant Suc are not available. HOL provides many lemmas for proving inequalities involving integer multiplication and division, similar to those shown above for type nat. The laws of addition, subtraction and multiplication are available through the axiomatic type class for rings (Sect. 8.4.5). The absolute value function abs is overloaded, and is defined for all types that involve negative numbers, including the integers. The arith method can prove facts about abs automatically, though as it does so by case analysis, the cost can be exponential. 8.4 Numbers 167 lemma "abs (x+y) ≤abs x + abs (y :: int)" by arith For division and remainder, the treatment of negative divisors follows mathematical practice: the sign of the remainder follows that of the divisor: 0 < b = ⇒0 ≤a mod b (pos_mod_sign) 0 < b = ⇒a mod b < b (pos_mod_bound) b < 0 = ⇒a mod b ≤0 (neg_mod_sign) b < 0 = ⇒b < a mod b (neg_mod_bound) ML treats negative divisors in the same way, but most computer hardware treats signed operands using the same rules as for multiplication. Many facts about quotients and remainders are provided: (a + b) div c = a div c + b div c + (a mod c + b mod c) div c (div_add1_eq) (a + b) mod c = (a mod c + b mod c) mod c (mod_add_eq) (a b) div c = a (b div c) + a (b mod c) div c (zdiv_zmult1_eq) (a b) mod c = a (b mod c) mod c (zmod_zmult1_eq) 0 < c = ⇒a div (bc) = a div b div c (zdiv_zmult2_eq) 0 < c = ⇒a mod (bc) = b(a div b mod c) + a mod b (zmod_zmult2_eq) The last two differ from their natural number analogues by requiring c to be positive. Since division by zero yields zero, we could allow c to be zero. However, c cannot be negative: a counterexample is a = 7, b = 2 and c = −3, when the left-hand side of zdiv_zmult2_eq is −2 while the right-hand side is −1. The prefix z in many theorem names recalls the use of Z to denote the set of integers. Induction is less important for integers than it is for the natural numbers, but it can be valuable if the range of integers has a lower or upper bound. There are four rules for integer induction, corresponding to the possible re-lations of the bound (≥, >, ≤and <): [ [k ≤i; P k; Vi. [ [k ≤i; P i] ] = ⇒P(i+1)] ] = ⇒P i (int_ge_induct) [ [k < i; P(k+1); Vi. [ [k < i; P i] ] = ⇒P(i+1)] ] = ⇒P i (int_gr_induct) [ [i ≤k; P k; Vi. [ [i ≤k; P i] ] = ⇒P(i-1)] ] = ⇒P i (int_le_induct) [ [i < k; P(k-1); Vi. [ [i < k; P i] ] = ⇒P(i-1)] ] = ⇒P i (int_less_induct) 8.4.4 The Types of Rational, Real and Complex Numbers These types provide true division, the overloaded operator /, which differs from the operator div of the natural numbers and integers. The rationals and reals are dense: between every two distinct numbers lies another. This property follows from the division laws, since if x ̸= y then (x + y)/2 lies between them: a < b = ⇒∃r. a < r ∧r < b (dense) 168 8. More about Types The real numbers are, moreover, complete: every set of reals that is bounded above has a least upper bound. Completeness distinguishes the reals from the rationals, for which the set {x \| x2 < 2} has no least upper bound. (It could only be √2, which is irrational.) The formalization of completeness, which is complicated, can be found in theory RComplete. Numeric literals for type real have the same syntax as those for type int and only express integral values. Fractions expressed using the division operator are automatically simplified to lowest terms: 1. P ((3 / 4) (8 / 15)) apply simp 1. P (2 / 5) Exponentiation can express floating-point values such as 2 10^6, which will be simplified to integers. ! ! Types rat, real and complex are provided by theory HOL-Complex, which is Main extended with a definitional development of the rational, real and complex numbers. Base your theory upon theory Complex_Main, not the usual Main. Available in the logic HOL-NSA is the theory Hyperreal, which define the type hypreal of non-standard reals. These hyperreals include infinitesimals, which represent infinitely small and infinitely large quantities; they facili-tate proofs about limits, differentiation and integration . The development defines an infinitely large number, omega and an infinitely small positive num-ber, epsilon. The relation x ≈y means “x is infinitely close to y.” Theory Hyperreal also defines transcendental functions such as sine, cosine, expo-nential and logarithm — even the versions for type real, because they are defined using nonstandard limits. 8.4.5 The Numeric Type Classes Isabelle/HOL organises its numeric theories using axiomatic type classes. Hundreds of basic properties are proved in the theory Ring_and_Field. These lemmas are available (as simprules if they were declared as such) for all numeric types satisfying the necessary axioms. The theory defines dozens of type classes, such as the following: – semiring and ordered_semiring: a semiring provides the associative oper-ators + and , with 0 and 1 as their respective identities. The operators enjoy the usual distributive law, and addition (+) is also commutative. An ordered semiring is also linearly ordered, with addition and multiplication respecting the ordering. Type nat is an ordered semiring. – ring and ordered_ring: a ring extends a semiring with unary minus (the additive inverse) and subtraction (both denoted -). An ordered ring in-cludes the absolute value function, abs. Type int is an ordered ring. 8.4 Numbers 169 – field and ordered_field: a field extends a ring with the multiplicative inverse (called simply inverse and division (/)). An ordered field is based on an ordered ring. Type complex is a field, while type real is an ordered field. – division_by_zero includes all types where inverse 0 = 0 and a / 0 = 0. These include all of Isabelle’s standard numeric types. However, the basic properties of fields are derived without assuming division by zero. Hundreds of basic lemmas are proved, each of which holds for all types in the corresponding type class. In most cases, it is obvious whether a property is valid for a particular type. No abstract properties involving subtraction hold for type nat; instead, theorems such as diff_mult_distrib are proved specifically about subtraction on type nat. All abstract properties involving division require a field. Obviously, all properties involving orderings required an ordered structure. The class ring_no_zero_divisors of rings without zero divisors satisfies a number of natural cancellation laws, the first of which characterizes this class: (a b = (0::'a)) = (a = (0::'a) ∨b = (0::'a)) (mult_eq_0_iff) (a c = b c) = (c = (0::'a) ∨a = b) (mult_cancel_right) Setting the flag Isabelle > Settings > Show Sorts will display the type classes of all type variables. Here is how the theorem mult_cancel_left appears with the flag set. ((c::'a::ring_no_zero_divisors) (a::'a::ring_no_zero_divisors) = c (b::'a::ring_no_zero_divisors)) = (c = (0::'a::ring_no_zero_divisors) ∨a = b) Simplifying with the AC-Laws. Suppose that two expressions are equal, differing only in associativity and commutativity of addition. Simplifying with the following equations sorts the terms and groups them to the right, making the two expressions identical. a + b + c = a + (b + c) (add.assoc) a + b = b + a (add.commute) a + (b + c) = b + (a + c) (add.left_commute) The name ac_simps refers to the list of all three theorems; similarly there is ac_simps. They are all proved for semirings and therefore hold for all numeric types. Here is an example of the sorting effect. Start with this goal, which in-volves type nat. 1. Suc (i + j l k + m n) = f (n m + i + k j l) Simplify using ac_simps and ac_simps. apply (simp add: ac_simps ac_simps) Here is the resulting subgoal. 1. Suc (i + (m n + j (k l))) = f (i + (m n + j (k l))) 170 8. More about Types Division Laws for Fields. Here is a selection of rules about the division operator. The following are installed as default simplification rules in order to express combinations of products and quotients as rational expressions: a (b / c) = a b / c (times_divide_eq_right) b / c a = b a / c (times_divide_eq_left) a / (b / c) = a c / b (divide_divide_eq_right) a / b / c = a / (b c) (divide_divide_eq_left) Signs are extracted from quotients in the hope that complementary terms can then be cancelled: - (a / b) = - a / b (minus_divide_left) - (a / b) = a / - b (minus_divide_right) The following distributive law is available, but it is not installed as a simplification rule. (a + b) / c = a / c + b / c (add_divide_distrib) Absolute Value. The absolute value function abs is available for all ordered rings, including types int, rat and real. It satisfies many properties, such as the following: \|x y\| = \|x\| \|y\| (abs_mult) (\|a\| ≤b) = (a ≤b ∧- a ≤b) (abs_le_iff) \|a + b\| ≤\|a\| + \|b\| (abs_triangle_ineq) ! ! The absolute value bars shown above cannot be typed on a keyboard. They can be entered using the X-symbol package. In ascii, type abs x to get \|x\|. Raising to a Power. Another type class, ordered idom, specifies rings that also have exponentation to a natural number power, defined using the obvi-ous primitive recursion. Theory Power proves various theorems, such as the following. a ^ (m + n) = a ^ m a ^ n (power_add) a ^ (m n) = (a ^ m) ^ n (power_mult) \|a ^ n\| = \|a\| ^ n (power_abs) 8.5 Introducing New Types For most applications, a combination of predefined types like bool and ⇒ with recursive datatypes and records is quite sufficient. Very occasionally you may feel the need for a more advanced type. If you are certain that your type is not definable by any of the standard means, then read on. ! ! Types in HOL must be non-empty; otherwise the quantifier rules would be unsound, because ∃x. x = x is a theorem. 8.5 Introducing New Types 171 8.5.1 Declaring New Types The most trivial way of introducing a new type is by a type declaration: typedecl my_new_type This does not define my_new_type at all but merely introduces its name. Thus we know nothing about this type, except that it is non-empty. Such declara-tions without definitions are useful if that type can be viewed as a parameter of the theory. A typical example is given in Sect. ??, where we define a tran-sition relation over an arbitrary type of states. In principle we can always get rid of such type declarations by making those types parameters of every other type, thus keeping the theory generic. In practice, however, the resulting clutter can make types hard to read. If you are looking for a quick and dirty way of introducing a new type to-gether with its properties: declare the type and state its properties as axioms. Example: axiomatization where just_one: "∃x::my_new_type. ∀y. x = y" However, we strongly discourage this approach, except at explorative stages of your development. It is extremely easy to write down contradictory sets of axioms, in which case you will be able to prove everything but it will mean nothing. In the example above, the axiomatic approach is unnecessary: a one-element type called unit is already defined in HOL. 8.5.2 Defining New Types Now we come to the most general means of safely introducing a new type, the type definition. All other means, for example datatype, are based on it. The principle is extremely simple: any non-empty subset of an existing type can be turned into a new type. More precisely, the new type is specified to be isomorphic to some non-empty subset of an existing type. Let us work a simple example, the definition of a three-element type. It is easily represented by the first three natural numbers: typedef three = "{0::nat, 1, 2}" In order to enforce that the representing set on the right-hand side is non-empty, this definition actually starts a proof to that effect: 1. ∃x. x ∈{0, 1, 2} Fortunately, this is easy enough to show, even auto could do it. In general, one has to provide a witness, in our case 0: apply(rule_tac x = 0 in exI) by simp 172 8. More about Types This type definition introduces the new type three and asserts that it is a copy of the set {0, 1, 2}. This assertion is expressed via a bijection between the type three and the set {0, 1, 2}. To this end, the command declares the following constants behind the scenes: Rep_three :: three ⇒nat Abs_three :: nat ⇒three The situation is best summarized with the help of the following diagram, where squares denote types and the irregular region denotes a set: three nat {0,1,2} Finally, typedef asserts that Rep_three is surjective on the subset and Abs_three and Rep_three are inverses of each other: Rep_three x ∈{0, 1, 2} (Rep_three) Abs_three (Rep_three x) = x (Rep_three_inverse) y ∈{0, 1, 2} = ⇒Rep_three (Abs_three y) = y (Abs_three_inverse) From this example it should be clear what typedef does in general given a name (here three) and a set (here {0, 1, 2}). Our next step is to define the basic functions expected on the new type. Although this depends on the type at hand, the following strategy works well: – define a small kernel of basic functions that can express all other functions you anticipate. – define the kernel in terms of corresponding functions on the representing type using Abs and Rep to convert between the two levels. In our example it suffices to give the three elements of type three names: definition A :: three where "A ≡Abs_three 0" definition B :: three where "B ≡Abs_three 1" definition C :: three where "C ≡Abs_three 2" So far, everything was easy. But it is clear that reasoning about three will be hell if we have to go back to nat every time. Thus our aim must be to raise our level of abstraction by deriving enough theorems about type three to characterize it completely. And those theorems should be phrased in terms of A, B and C, not Abs_three and Rep_three. Because of the simplicity of the 8.5 Introducing New Types 173 example, we merely need to prove that A, B and C are distinct and that they exhaust the type. In processing our typedef declaration, Isabelle proves several helpful lem-mas. The first two express injectivity of Rep_three and Abs_three: (Rep_three x = Rep_three y) = (x = y) (Rep_three_inject) [ [x ∈{0, 1, 2}; y ∈{0, 1, 2} ] ] = ⇒(Abs_three x = Abs_three y) = (x = y) (Abs_three_inject) The following ones allow to replace some x::three by Abs_three(y::nat), and conversely y by Rep_three x: [ [y ∈{0, 1, 2}; Vx. y = Rep_three x = ⇒P] ] = ⇒P (Rep_three_cases) (Vy. [ [x = Abs_three y; y ∈{0, 1, 2}] ] = ⇒P) = ⇒P (Abs_three_cases) [ [y ∈{0, 1, 2}; Vx. P (Rep_three x)] ] = ⇒P y (Rep_three_induct) (Vy. y ∈{0, 1, 2} = ⇒P (Abs_three y)) = ⇒P x (Abs_three_induct) These theorems are proved for any type definition, with three replaced by the name of the type in question. Distinctness of A, B and C follows immediately if we expand their defini-tions and rewrite with the injectivity of Abs_three: lemma "A ̸= B ∧B ̸= A ∧A ̸= C ∧C ̸= A ∧B ̸= C ∧C ̸= B" by(simp add: Abs_three_inject A_def B_def C_def) Of course we rely on the simplifier to solve goals like 0 ̸= 1. The fact that A, B and C exhaust type three is best phrased as a case distinction theorem: if you want to prove P x (where x is of type three) it suffices to prove P A, P B and P C: lemma three_cases: "[ [ P A; P B; P C ] ] = ⇒P x" Again this follows easily using the induction principle stemming from the type definition: apply(induct_tac x) 1. Vy. [ [P A; P B; P C; y ∈{0, 1, 2}] ] = ⇒P (Abs_three y) Simplification leads to the disjunction y = 0 ∨y = 1 ∨y = 2 which auto separates into three subgoals, each of which is easily solved by simplification: apply(auto simp add: A_def B_def C_def) done This concludes the derivation of the characteristic theorems for type three. The attentive reader has realized long ago that the above lengthy defini-tion can be collapsed into one line: datatype better_three = A \| B \| C In fact, the datatype command performs internally more or less the same derivations as we did, which gives you some idea what life would be like without datatype. 174 8. More about Types Although three could be defined in one line, we have chosen this exam-ple to demonstrate typedef because its simplicity makes the key concepts particularly easy to grasp. If you would like to see a non-trivial example that cannot be defined more directly, we recommend the definition of finite multisets in the Library . Let us conclude by summarizing the above procedure for defining a new type. Given some abstract axiomatic description P of a type ty in terms of a set of functions F, this involves three steps: 1. Find an appropriate type τ and subset A which has the desired properties P, and make a type definition based on this representation. 2. Define the required functions F on ty by lifting analogous functions on the representation via Abs ty and Rep ty. 3. Prove that P holds for ty by lifting P from the representation. You can now forget about the representation and work solely in terms of the abstract functions F and properties P. 9. Advanced Simplification and Induction Although we have already learned a lot about simplification and induction, there are some advanced proof techniques that we have not covered yet and which are worth learning. The sections of this chapter are independent of each other and can be read in any order. 9.1 Simplification This section describes features not covered until now. It also outlines the simplification process itself, which can be helpful when the simplifier does not do what you expect of it. 9.1.1 Advanced Features Congruence Rules. While simplifying the conclusion Q of P = ⇒Q, it is legal to use the assumption P. For = ⇒this policy is hardwired, but contextual information can also be made available for other operators. For example, xs = [] − →xs @ xs = xs simplifies to True because we may use xs = [] when simplifying xs @ xs = xs. The generation of contextual information during simplification is controlled by so-called congruence rules. This is the one for − →: [ [P = P'; P' = ⇒Q = Q'] ] = ⇒(P − →Q) = (P' − →Q') It should be read as follows: In order to simplify P − →Q to P' − →Q', sim-plify P to P' and assume P' when simplifying Q to Q'. Here are some more examples. The congruence rules for bounded quanti-fiers supply contextual information about the bound variable: [ [A = B; Vx. x ∈B = ⇒P x = Q x] ] = ⇒(∀x∈A. P x) = (∀x∈B. Q x) One congruence rule for conditional expressions supplies contextual informa-tion for simplifying the then and else cases: [ [b = c; c = ⇒x = u; ¬ c = ⇒y = v] ] = ⇒(if b then x else y) = (if c then u else v) 176 9. Advanced Simplification and Induction An alternative congruence rule for conditional expressions actually prevents simplification of some arguments: b = c = ⇒(if b then x else y) = (if c then x else y) Only the first argument is simplified; the others remain unchanged. This makes simplification much faster and is faithful to the evaluation strategy in programming languages, which is why this is the default congruence rule for if. Analogous rules control the evaluation of case expressions. You can declare your own congruence rules with the attribute cong, either globally, in the usual manner, declare theorem-name [cong] or locally in a simp call by adding the modifier cong: list of theorem names The effect is reversed by cong del instead of cong. ! ! The congruence rule conj_cong [ [P = P'; P' = ⇒Q = Q'] ] = ⇒(P ∧Q) = (P' ∧Q') is occasionally useful but is not a default rule; you have to declare it explicitly. Permutative Rewrite Rules. An equation is a permutative rewrite rule if the left-hand side and right-hand side are the same up to renaming of variables. The most common permutative rule is commutativity: x + y = y + x. Other examples include x - y - z = x - z - y in arithmetic and insert x (insert y A) = insert y (insert x A) for sets. Such rules are problematic because once they apply, they can be used forever. The simplifier is aware of this danger and treats permutative rules by means of a special strategy, called ordered rewriting: a permutative rewrite rule is only applied if the term becomes smaller with respect to a fixed lexicographic ordering on terms. For example, commutativity rewrites b + a to a + b, but then stops because a + b is strictly smaller than b + a. Permutative rewrite rules can be turned into simplification rules in the usual manner via the simp attribute; the simplifier recognizes their special status automatically. Permutative rewrite rules are most effective in the case of associative-commutative functions. (Associativity by itself is not permutative.) When dealing with an AC-function f , keep the following points in mind: – The associative law must always be oriented from left to right, namely f (f (x, y), z) = f (x, f (y, z)). The opposite orientation, if used with commu-tativity, can lead to nontermination. – To complete your set of rewrite rules, you must add not just associativ-ity (A) and commutativity (C) but also a derived rule, left-commutativ-ity (LC): f (x, f (y, z)) = f (y, f (x, z)). 9.1 Simplification 177 Ordered rewriting with the combination of A, C, and LC sorts a term lexi-cographically: f (f (b, c), a) A ; f (b, f (c, a)) C ; f (b, f (a, c)) LC ; f (a, f (b, c)) Note that ordered rewriting for + and on numbers is rarely necessary because the built-in arithmetic prover often succeeds without such tricks. 9.1.2 How the Simplifier Works Roughly speaking, the simplifier proceeds bottom-up: subterms are simplified first. A conditional equation is only applied if its condition can be proved, again by simplification. Below we explain some special features of the rewrit-ing process. Higher-Order Patterns. So far we have pretended the simplifier can deal with arbitrary rewrite rules. This is not quite true. For reasons of feasibility, the simplifier expects the left-hand side of each rule to be a so-called higher-order pattern . This restricts where unknowns may occur. Higher-order patterns are terms in β-normal form. (This means there are no subterms of the form (λx.M)(N).) Each occurrence of an unknown is of the form ? f x1 . . . xn, where the xi are distinct bound variables. Thus all ordinary rewrite rules, where all unknowns are of base type, for example ?a + ?b + ?c = ?a + (?b + ?c), are acceptable: if an unknown is of base type, it cannot have any arguments. Additionally, the rule (∀x. ?P x ∧?Q x) = ((∀x. ?P x) ∧(∀x. ?Q x)) is also acceptable, in both directions: all arguments of the unknowns ?P and ?Q are distinct bound variables. If the left-hand side is not a higher-order pattern, all is not lost. The simplifier will still try to apply the rule provided it matches directly: with-out much λ-calculus hocus pocus. For example, (?f ?x ∈range ?f) = True rewrites g a ∈range g to True, but will fail to match g(h b) ∈range(λx. g(h x)). However, you can eliminate the offending subterms — those that are not patterns — by adding new variables and conditions. In our example, we eliminate ?f ?x and obtain ?y = ?f ?x = ⇒(?y ∈range ?f) = True, which is fine as a conditional rewrite rule since conditions can be arbitrary terms. However, this trick is not a panacea because the newly introduced conditions may be hard to solve. There is no restriction on the form of the right-hand sides. They may not contain extraneous term or type variables, though. The Preprocessor. When a theorem is declared a simplification rule, it need not be a conditional equation already. The simplifier will turn it into a set of conditional equations automatically. For example, f x = g x ∧h x = k x becomes the two separate simplification rules f x = g x and h x = k x. In general, the input theorem is converted as follows: 178 9. Advanced Simplification and Induction ¬P 7→ P = False P − →Q 7→ P = ⇒Q P ∧Q 7→ P, Q ∀x. P x 7→ P ? x ∀x ∈A. P x 7→ ? x ∈A = ⇒P ? x if P then Q else R 7→ P = ⇒Q, ¬P = ⇒R Once this conversion process is finished, all remaining non-equations P are turned into trivial equations P = True. For example, the formula (p − →t = u ∧¬ r) ∧s is converted into the three rules p = ⇒t = u, p = ⇒r = False, s = True. 9.2 Advanced Induction Techniques Now that we have learned about rules and logic, we take another look at the finer points of induction. We consider two questions: what to do if the proposition to be proved is not directly amenable to induction (Sect. 9.2.1), and how to utilize (Sect. 9.2.2) and even derive (Sect. 9.2.3) new induction schemas. We conclude with an extended example of induction (Sect. 9.2.4). 9.2.1 Massaging the Proposition Often we have assumed that the theorem to be proved is already in a form that is amenable to induction, but sometimes it isn’t. Here is an example. Since hd and last return the first and last element of a non-empty list, this lemma looks easy to prove: lemma "xs ̸= [] = ⇒hd(rev xs) = last xs" apply(induct_tac xs) But induction produces the warning Induction variable occurs also among premises! and leads to the base case 1. xs ̸= [] = ⇒hd (rev []) = last [] A total of 2 subgoals... Simplification reduces the base case to this: 1. xs ̸= [] = ⇒hd [] = last [] 9.2 Advanced Induction Techniques 179 We cannot prove this equality because we do not know what hd and last return when applied to []. We should not have ignored the warning. Because the induction formula is only the conclusion, induction does not affect the occurrence of xs in the premises. Thus the case that should have been trivial becomes unprovable. Fortunately, the solution is easy:1 Pull all occurrences of the induction variable into the conclusion using − →. Thus we should state the lemma as an ordinary implication (− →), letting rule_format (Sect. 5.15) convert the result to the usual = ⇒form: lemma hd_rev [rule_format]: "xs ̸= [] − →hd(rev xs) = last xs" This time, induction leaves us with a trivial base case: 1. [] ̸= [] − →hd (rev []) = last [] A total of 2 subgoals... And auto completes the proof. If there are multiple premises A1, …, An containing the induction variable, you should turn the conclusion C into A1 − →· · · An − →C. Additionally, you may also have to universally quantify some other variables, which can yield a fairly complex conclusion. However, rule_format can remove any number of occurrences of ∀and − →. A second reason why your proposition may not be amenable to induction is that you want to induct on a complex term, rather than a variable. In general, induction on a term t requires rephrasing the conclusion C as ∀y1 . . . yn. x = t − →C. (9.1) where y1 . . . yn are the free variables in t and x is a new variable. Now you can perform induction on x. An example appears in Sect. 9.2.2 below. The very same problem may occur in connection with rule induction. Remember that it requires a premise of the form (x1, . . . , xk) ∈R, where R is some inductively defined set and the xi are variables. If instead we have a premise t ∈R, where t is not just an n-tuple of variables, we replace it with (x1, . . . , xk) ∈R, and rephrase the conclusion C as ∀y1 . . . yn. (x1, . . . , xk) = t − →C. For an example see Sect. 9.2.4 below. Of course, all premises that share free variables with t need to be pulled into the conclusion as well, under the ∀, again using − →as shown above. 1 A similar heuristic applies to rule inductions; see Sect. 7.2. 180 9. Advanced Simplification and Induction Readers who are puzzled by the form of statement (9.1) above should remember that the transformation is only performed to permit induction. Once induction has been applied, the statement can be transformed back into something quite intuitive. For example, applying wellfounded induction on x (w.r.t. ≺) to (9.1) and transforming the result a little leads to the goal ^ y. ∀z. t z ≺t y − →C z = ⇒C y where y stands for y1 . . . yn and the dependence of t and C on the free vari-ables of t has been made explicit. Unfortunately, this induction schema cannot be expressed as a single theorem because it depends on the number of free variables in t — the notation y is merely an informal device. 9.2.2 Beyond Structural and Recursion Induction So far, inductive proofs were by structural induction for primitive recursive functions and recursion induction for total recursive functions. But some-times structural induction is awkward and there is no recursive function that could furnish a more appropriate induction schema. In such cases a general-purpose induction schema can be helpful. We show how to apply such induc-tion schemas by an example. Structural induction on nat is usually known as mathematical induc-tion. There is also complete induction, where you prove P(n) under the assumption that P(m) holds for all m < n. In Isabelle, this is the theorem nat_less_induct: (Vn. ∀m<n. P m = ⇒P n) = ⇒P n As an application, we prove a property of the following function: axiomatization f :: "nat ⇒nat" where f_ax: "f(f(n)) < f(Suc(n))" for n :: nat ! ! We discourage the use of axioms because of the danger of inconsistencies. Ax-iom f_ax does not introduce an inconsistency because, for example, the identity function satisfies it. Axioms can be useful in exploratory developments, say when you assume some well-known theorems so that you can quickly demonstrate some point about methodology. If your example turns into a substantial proof develop-ment, you should replace axioms by theorems. The axiom for f implies n ≤f n, which can be proved by induction on f n. Following the recipe outlined above, we have to phrase the proposition as follows to allow induction: lemma f_incr_lem: "∀i. k = f i − →i ≤f i" To perform induction on k using nat_less_induct, we use the same general induction method as for recursion induction (see Sect. 3.5.4): apply(induct_tac k rule: nat_less_induct) We get the following proof state: 9.2 Advanced Induction Techniques 181 1. Vn. ∀m<n. ∀i. m = f i − →i ≤f i = ⇒∀i. n = f i − →i ≤f i After stripping the ∀i, the proof continues with a case distinction on i. The case i = 0 is trivial and we focus on the other case: apply(rule allI) apply(case_tac i) apply(simp) 1. Vn i nat. [ [∀m<n. ∀i. m = f i − →i ≤f i; i = Suc nat] ] = ⇒n = f i − →i ≤ f i by(blast intro!: f_ax Suc_leI intro: le_less_trans) If you find the last step puzzling, here are the two lemmas it employs: m < n = ⇒Suc m ≤n (Suc_leI) [ [x ≤y; y < z] ] = ⇒x < z (le_less_trans) The proof goes like this (writing j instead of nat). Since i = Suc j it suf-fices to show j < f (Suc j), by Suc_leI. This is proved as follows. From f_ax we have f (f j) < f (Suc j) (1) which implies f j ≤f (f j) by the induc-tion hypothesis. Using (1) once more we obtain f j < f (Suc j) (2) by the transitivity rule le_less_trans. Using the induction hypothesis once more we obtain j ≤f j which, together with (2) yields j < f (Suc j) (again by le_less_trans). This last step shows both the power and the danger of automatic proofs. They will usually not tell you how the proof goes, because it can be hard to translate the internal proof into a human-readable format. Automatic proofs are easy to write but hard to read and understand. The desired result, i ≤f i, follows from f_incr_lem: lemmas f_incr = f_incr_lem[rule_format, OF refl] The final refl gets rid of the premise ?k = f ?i. We could have included this derivation in the original statement of the lemma: lemma f_incr[rule_format, OF refl]: "∀i. k = f i − →i ≤f i" Exercise 9.2.1 From the axiom and lemma for f, show that f is the identity function. Method induct_tac can be applied with any rule r whose conclusion is of the form ?P ?x1 . . .?xn, in which case the format is apply(induct_tac y1 . . . yn rule: r) where y1, . . . , yn are variables in the conclusion of the first subgoal. A further useful induction rule is length_induct, induction on the length of a list (Vxs. ∀ys. length ys < length xs − →P ys = ⇒P xs) = ⇒P xs 182 9. Advanced Simplification and Induction which is a special case of measure_induct (Vx. ∀y. f y < f x − →P y = ⇒P x) = ⇒P a where f may be any function into type nat. 9.2.3 Derivation of New Induction Schemas Induction schemas are ordinary theorems and you can derive new ones whenever you wish. This section shows you how, using the example of nat_less_induct. Assume we only have structural induction available for nat and want to derive complete induction. We must generalize the statement as shown: lemma induct_lem: "(Vn::nat. ∀m<n. P m = ⇒P n) = ⇒∀m<n. P m" apply(induct_tac n) The base case is vacuously true. For the induction step (m < Suc n) we dis-tinguish two cases: case m < n is true by induction hypothesis and case m = n follows from the assumption, again using the induction hypothesis: apply(blast) by(blast elim: less_SucE) The elimination rule less_SucE expresses the case distinction: [ [m < Suc n; m < n = ⇒P; m = n = ⇒P] ] = ⇒P Now it is straightforward to derive the original version of nat_less_induct by manipulating the conclusion of the above lemma: instantiate n by Suc n and m by n and remove the trivial condition n < Suc n. Fortunately, this happens automatically when we add the lemma as a new premise to the desired goal: theorem nat_less_induct: "(Vn::nat. ∀m<n. P m = ⇒P n) = ⇒P n" by(insert induct_lem, blast) HOL already provides the mother of all inductions, well-founded in-duction (see Sect. 6.4). For example theorem nat_less_induct is a special case of wf_induct where r is < on nat. The details can be found in theory Wellfounded_Recursion. 9.2.4 CTL Revisited The purpose of this section is twofold: to demonstrate some of the induction principles and heuristics discussed above and to show how inductive defi-nitions can simplify proofs. In Sect. 6.5.2 we gave a fairly involved proof of the correctness of a model checker for CTL. In particular the proof of the infinity_lemma on the way to AF_lemma2 is not as simple as one might expect, 9.2 Advanced Induction Techniques 183 due to the SOME operator involved. Below we give a simpler proof of AF_lemma2 based on an auxiliary inductive definition. Let us call a (finite or infinite) path A-avoiding if it does not touch any node in the set A. Then AF_lemma2 says that if no infinite path from some state s is A-avoiding, then s ∈lfp (af A). We prove this by inductively defining the set Avoid s A of states reachable from s by a finite A-avoiding path: inductive set Avoid :: "state ⇒state set ⇒state set" for s :: state and A :: "state set" where "s ∈Avoid s A" \| "[ [ t ∈Avoid s A; t / ∈A; (t,u) ∈M ] ] = ⇒u ∈Avoid s A" It is easy to see that for any infinite A-avoiding path f with f 0 ∈Avoid s A there is an infinite A-avoiding path starting with s because (by definition of Avoid) there is a finite A-avoiding path from s to f 0. The proof is by induction on f 0 ∈Avoid s A. However, this requires the following reformu-lation, as explained in Sect. 9.2.1 above; the rule_format directive undoes the reformulation after the proof. lemma ex_infinite_path[rule_format]: "t ∈Avoid s A = ⇒ ∀f∈Paths t. (∀i. f i / ∈A) − →(∃p∈Paths s. ∀i. p i / ∈A)" apply(erule Avoid.induct) apply(blast) apply(clarify) apply(drule_tac x = "λi. case i of 0 ⇒t \| Suc i ⇒f i" in bspec) apply(simp_all add: Paths_def split: nat.split) done The base case (t = s) is trivial and proved by blast. In the induction step, we have an infinite A-avoiding path f starting from u, a successor of t. Now we simply instantiate the ∀f∈Paths t in the induction hypothesis by the path starting with t and continuing with f. That is what the above λ-term expresses. Simplification shows that this is a path starting with t and that the instantiated induction hypothesis implies the conclusion. Now we come to the key lemma. Assuming that no infinite A-avoiding path starts from s, we want to show s ∈lfp (af A). For the inductive proof this must be generalized to the statement that every point t “between” s and A, in other words all of Avoid s A, is contained in lfp (af A): lemma Avoid_in_lfp[rule_format(no_asm)]: "∀p∈Paths s. ∃i. p i ∈A = ⇒t ∈Avoid s A − →t ∈lfp(af A)" The proof is by induction on the “distance” between t and A. Remember that lfp (af A) = A ∪M −1 `` lfp (af A). If t is already in A, then t ∈lfp (af A) is trivial. If t is not in A but all successors are in lfp (af A) (induction hypothesis), then t ∈lfp (af A) is again trivial. The formal counterpart of this proof sketch is a well-founded induction on M restricted to Avoid s A - A, roughly speaking: 184 9. Advanced Simplification and Induction {(y, x). (x, y) ∈M ∧x ∈Avoid s A ∧x / ∈A} As we shall see presently, the absence of infinite A-avoiding paths starting from s implies well-foundedness of this relation. For the moment we assume this and proceed with the induction: apply(subgoal_tac "wf{(y,x). (x,y) ∈M ∧x ∈Avoid s A ∧x / ∈A}") apply(erule_tac a = t in wf_induct) apply(clarsimp) 1. Vt. [ [∀p∈Paths s. ∃i. p i ∈A; ∀y. (t, y) ∈M ∧t / ∈A − → y ∈Avoid s A − →y ∈lfp (af A); t ∈Avoid s A] ] = ⇒t ∈lfp (af A) 2. ∀p∈Paths s. ∃i. p i ∈A = ⇒ wf {(y, x). (x, y) ∈M ∧x ∈Avoid s A ∧x / ∈A} Now the induction hypothesis states that if t / ∈A then all successors of t that are in Avoid s A are in lfp (af A). Unfolding lfp in the conclusion of the first subgoal once, we have to prove that t is in A or all successors of t are in lfp (af A). But if t is not in A, the second Avoid-rule implies that all successors of t are in Avoid s A, because we also assume t ∈Avoid s A. Hence, by the induction hypothesis, all successors of t are indeed in lfp (af A). Mechanically: apply(subst lfp_unfold[OF mono_af]) apply(simp (no_asm) add: af_def) apply(blast intro: Avoid.intros) Having proved the main goal, we return to the proof obligation that the relation used above is indeed well-founded. This is proved by contradiction: if the relation is not well-founded then there exists an infinite A-avoiding path all in Avoid s A, by theorem wf_iff_no_infinite_down_chain: wf r = (∄f. ∀i. (f (Suc i), f i) ∈r) From lemma ex_infinite_path the existence of an infinite A-avoiding path starting in s follows, contradiction. apply(erule contrapos_pp) apply(simp add: wf_iff_no_infinite_down_chain) apply(erule exE) apply(rule ex_infinite_path) apply(auto simp add: Paths_def) done The (no_asm) modifier of the rule_format directive in the statement of the lemma means that the assumption is left unchanged; otherwise the ∀p would be turned into a Vp, which would complicate matters below. As it is, Avoid_in_lfp is now [ [∀p∈Paths s. ∃i. p i ∈A; t ∈Avoid s A] ] = ⇒t ∈lfp (af A) 9.2 Advanced Induction Techniques 185 The main theorem is simply the corollary where t = s, when the assumption t ∈Avoid s A is trivially true by the first Avoid-rule. Isabelle confirms this: theorem AF_lemma2: "{s. ∀p ∈Paths s. ∃i. p i ∈A} ⊆lfp(af A)" by(auto elim: Avoid_in_lfp intro: Avoid.intros) 10. Case Study: Verifying a Security Protocol Communications security is an ancient art. Julius Caesar is said to have encrypted his messages, shifting each letter three places along the alphabet. Mary Queen of Scots was convicted of treason after a cipher used in her letters was broken. Today’s postal system incorporates security features. The envelope provides a degree of secrecy. The signature provides authenticity (proof of origin), as do departmental stamps and letterheads. Networks are vulnerable: messages pass through many computers, any of which might be controlled by an adversary, who thus can capture or redirect messages. People who wish to communicate securely over such a network can use cryptography, but if they are to understand each other, they need to follow a protocol: a pre-arranged sequence of message formats. Protocols can be attacked in many ways, even if encryption is unbreak-able. A splicing attack involves an adversary’s sending a message composed of parts of several old messages. This fake message may have the correct for-mat, fooling an honest party. The adversary might be able to masquerade as somebody else, or he might obtain a secret key. Nonces help prevent splicing attacks. A typical nonce is a 20-byte random number. Each message that requires a reply incorporates a nonce. The reply must include a copy of that nonce, to prove that it is not a replay of a past message. The nonce in the reply must be cryptographically protected, since otherwise an adversary could easily replace it by a different one. You should be starting to see that protocol design is tricky! Researchers are developing methods for proving the correctness of security protocols. The Needham-Schroeder public-key protocol has become a standard test case. Proposed in 1978, it was found to be defective nearly two decades later . This toy protocol will be useful in demonstrating how to verify protocols using Isabelle. 10.1 The Needham-Schroeder Public-Key Protocol This protocol uses public-key cryptography. Each person has a private key, known only to himself, and a public key, known to everybody. If Alice wants to send Bob a secret message, she encrypts it using Bob’s public key (which 188 10. Case Study: Verifying a Security Protocol everybody knows), and sends it to Bob. Only Bob has the matching private key, which is needed in order to decrypt Alice’s message. The core of the Needham-Schroeder protocol consists of three messages: 1. A →B : { \|Na, A\| }Kb 2. B →A : { \|Na, Nb\| }Ka 3. A →B : { \|Nb\| }Kb First, let’s understand the notation. In the first message, Alice sends Bob a message consisting of a nonce generated by Alice (Na) paired with Alice’s name (A) and encrypted using Bob’s public key (Kb). In the second message, Bob sends Alice a message consisting of Na paired with a nonce generated by Bob (Nb), encrypted using Alice’s public key (Ka). In the last message, Alice returns Nb to Bob, encrypted using his public key. When Alice receives Message 2, she knows that Bob has acted on her mes-sage, since only he could have decrypted { \|Na, A\| }Kb and extracted Na. That is precisely what nonces are for. Similarly, message 3 assures Bob that Alice is active. But the protocol was widely believed to satisfy a further prop-erty: that Na and Nb were secrets shared by Alice and Bob. (Many protocols generate such shared secrets, which can be used to lessen the reliance on slow public-key operations.) Lowe found this claim to be false: if Alice runs the protocol with someone untrustworthy (Charlie say), then he can start a new run with another agent (Bob say). Charlie uses Alice as an oracle, masquerading as Alice to Bob . 1. A →C : { \|Na, A\| }Kc 1′. C →B : { \|Na, A\| }Kb 2. B →A : { \|Na, Nb\| }Ka 3. A →C : { \|Nb\| }Kc 3′. C →B : { \|Nb\| }Kb In messages 1 and 3, Charlie removes the encryption using his private key and re-encrypts Alice’s messages using Bob’s public key. Bob is left thinking he has run the protocol with Alice, which was not Alice’s intention, and Bob is unaware that the “secret” nonces are known to Charlie. This is a typical man-in-the-middle attack launched by an insider. Whether this counts as an attack has been disputed. In protocols of this type, we normally assume that the other party is honest. To be hon-est means to obey the protocol rules, so Alice’s running the protocol with Charlie does not make her dishonest, just careless. After Lowe’s attack, Alice has no grounds for complaint: this protocol does not have to guarantee any-thing if you run it with a bad person. Bob does have grounds for complaint, however: the protocol tells him that he is communicating with Alice (who is honest) but it does not guarantee secrecy of the nonces. Lowe also suggested a correction, namely to include Bob’s name in mes-sage 2: 10.2 Agents and Messages 189 1. A →B : { \|Na, A\| }Kb 2. B →A : { \|Na, Nb, B\| }Ka 3. A →B : { \|Nb\| }Kb If Charlie tries the same attack, Alice will receive the message { \|Na, Nb, B\| }Ka when she was expecting to receive { \|Na, Nb, C\| }Ka. She will abandon the run, and eventually so will Bob. Below, we shall look at parts of this protocol’s correctness proof. In ground-breaking work, Lowe showed how such attacks could be found automatically using a model checker. An alternative, which we shall examine below, is to prove protocols correct. Proofs can be done under more realistic assumptions because our model does not have to be finite. The strat-egy is to formalize the operational semantics of the system and to prove security properties using rule induction. 10.2 Agents and Messages All protocol specifications refer to a syntactic theory of messages. Datatype agent introduces the constant Server (a trusted central machine, needed for some protocols), an infinite population of friendly agents, and the Spy: datatype agent = Server \| Friend nat \| Spy Keys are just natural numbers. Function invKey maps a public key to the matching private key, and vice versa: type synonym key = nat consts invKey :: "key ⇒key" Datatype msg introduces the message forms, which include agent names, nonces, keys, compound messages, and encryptions. datatype msg = Agent agent \| Nonce nat \| Key key \| MPair msg msg \| Crypt key msg The notation { \|X1, . . . Xn−1, Xn\| } abbreviates MPair X1 . . . (MPair Xn−1 Xn). Since datatype constructors are injective, we have the theorem Crypt K X = Crypt K' X' = ⇒K = K' ∧X = X' A ciphertext can be decrypted using only one key and can yield only one plaintext. In the real world, decryption with the wrong key succeeds but yields garbage. Our model of encryption is realistic if encryption adds some redundancy to the plaintext, such as a checksum, so that garbage can be detected. 190 10. Case Study: Verifying a Security Protocol 10.3 Modelling the Adversary The spy is part of the system and must be built into the model. He is a mali-cious user who does not have to follow the protocol. He watches the network and uses any keys he knows to decrypt messages. Thus he accumulates ad-ditional keys and nonces. These he can use to compose new messages, which he may send to anybody. Two functions enable us to formalize this behaviour: analz and synth. Each function maps a sets of messages to another set of messages. The set analz H formalizes what the adversary can learn from the set of messages H. The closure properties of this set are defined inductively. inductive set analz :: "msg set ⇒msg set" for H :: "msg set" where Inj [intro,simp] : "X ∈H = ⇒X ∈analz H" \| Fst: "{ \|X,Y\| } ∈analz H = ⇒X ∈analz H" \| Snd: "{ \|X,Y\| } ∈analz H = ⇒Y ∈analz H" \| Decrypt [dest]: "[ [Crypt K X ∈analz H; Key(invKey K) ∈analz H] ] = ⇒X ∈analz H" Note the Decrypt rule: the spy can decrypt a message encrypted with key K if he has the matching key, K −1. Properties proved by rule induction include the following: G ⊆H = ⇒analz G ⊆analz H (analz_mono) analz (analz H) = analz H (analz_idem) The set of fake messages that an intruder could invent starting from H is synth(analz H), where synth H formalizes what the adversary can build from the set of messages H. inductive set synth :: "msg set ⇒msg set" for H :: "msg set" where Inj [intro]: "X ∈H = ⇒X ∈synth H" \| Agent [intro]: "Agent agt ∈synth H" \| MPair [intro]: "[ [X ∈synth H; Y ∈synth H] ] = ⇒{ \|X,Y\| } ∈synth H" \| Crypt [intro]: "[ [X ∈synth H; Key K ∈H] ] = ⇒Crypt K X ∈synth H" The set includes all agent names. Nonces and keys are assumed to be unguessable, so none are included beyond those already in H. Two elements of synth H can be combined, and an element can be encrypted using a key present in H. Like analz, this set operator is monotone and idempotent. It also satisfies an interesting equation involving analz: 10.4 Event Traces 191 analz (synth H) = analz H ∪synth H (analz_synth) Rule inversion plays a major role in reasoning about synth, through declara-tions such as this one: inductive cases Nonce_synth [elim!]: "Nonce n ∈synth H" The resulting elimination rule replaces every assumption of the form Nonce n ∈synth H by Nonce n ∈H, expressing that a nonce cannot be guessed. A third operator, parts, is useful for stating correctness properties. The set parts H consists of the components of elements of H. This set includes H and is closed under the projections from a compound message to its immediate parts. Its definition resembles that of analz except in the rule corresponding to the constructor Crypt: Crypt K X ∈parts H = ⇒X ∈parts H The body of an encrypted message is always regarded as part of it. We can use parts to express general well-formedness properties of a protocol, for example, that an uncompromised agent’s private key will never be included as a component of any message. 10.4 Event Traces The system’s behaviour is formalized as a set of traces of events. The most important event, Says A B X, expresses A →B : X, which is the attempt by A to send B the message X. A trace is simply a list, constructed in reverse using #. Other event types include reception of messages (when we want to make it explicit) and an agent’s storing a fact. Sometimes the protocol requires an agent to generate a new nonce. The probability that a 20-byte random number has appeared before is effectively zero. To formalize this important property, the set used evs denotes the set of all items mentioned in the trace evs. The function used has a straightforward recursive definition. Here is the case for Says event: used (Says A B X # evs) = parts {X} ∪used evs The function knows formalizes an agent’s knowledge. Mostly we only care about the spy’s knowledge, and knows Spy evs is the set of items available to the spy in the trace evs. Already in the empty trace, the spy starts with some secrets at his disposal, such as the private keys of compromised users. After each Says event, the spy learns the message that was sent: knows Spy (Says A B X # evs) = insert X (knows Spy evs) Combinations of functions express other important sets of messages derived from evs: – analz (knows Spy evs) is everything that the spy could learn by decryption 192 10. Case Study: Verifying a Security Protocol – synth (analz (knows Spy evs)) is everything that the spy could generate The function pubK maps agents to their public keys. The function priK maps agents to their private keys. It is merely an abbreviation (cf. Sect. 4.1.4) defined in terms of invKey and pubK. consts pubK :: "agent ⇒key" abbreviation priK :: "agent ⇒key" where "priK x ≡ invKey(pubK x)" The set bad consists of those agents whose private keys are known to the spy. Two axioms are asserted about the public-key cryptosystem. No two agents have the same public key, and no private key equals any public key. axiomatization where inj_pubK: "inj pubK" and priK_neq_pubK: "priK A ̸= pubK B" 10.5 Modelling the Protocol inductive set ns_public :: "event list set" where Nil: "[] ∈ns_public" \| Fake: "[ [evsf ∈ns_public; X ∈synth (analz (knows Spy evsf))] ] = ⇒Says Spy B X # evsf ∈ns_public" \| NS1: "[ [evs1 ∈ns_public; Nonce NA / ∈used evs1] ] = ⇒Says A B (Crypt (pubK B) { \|Nonce NA, Agent A\| }) # evs1 ∈ ns_public" \| NS2: "[ [evs2 ∈ns_public; Nonce NB / ∈used evs2; Says A' B (Crypt (pubK B) { \|Nonce NA, Agent A\| }) ∈set evs2] ] = ⇒Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) # evs2 ∈ ns_public" \| NS3: "[ [evs3 ∈ns_public; Says A B (Crypt (pubK B) { \|Nonce NA, Agent A\| }) ∈set evs3; Says B' A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) ∈set evs3] ] = ⇒Says A B (Crypt (pubK B) (Nonce NB)) # evs3 ∈ns_public" Fig. 10.1. An Inductive Protocol Definition 10.6 Proving Elementary Properties 193 Let us formalize the Needham-Schroeder public-key protocol, as corrected by Lowe: 1. A →B : { \|Na, A\| }Kb 2. B →A : { \|Na, Nb, B\| }Ka 3. A →B : { \|Nb\| }Kb Each protocol step is specified by a rule of an inductive definition. An event trace has type event list, so we declare the constant ns_public to be a set of such traces. Figure 10.1 presents the inductive definition. The Nil rule introduces the empty trace. The Fake rule models the adversary’s sending a message built from components taken from past traffic, expressed using the functions synth and analz. The next three rules model how honest agents would perform the three protocol steps. Here is a detailed explanation of rule NS2. A trace containing an event of the form Says A' B (Crypt (pubK B) { \|Nonce NA, Agent A\| }) may be extended by an event of the form Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) where NB is a fresh nonce: Nonce NB / ∈used evs2. Writing the sender as A' indicates that B does not know who sent the message. Calling the trace vari-able evs2 rather than simply evs helps us know where we are in a proof after many case-splits: every subgoal mentioning evs2 involves message 2 of the protocol. Benefits of this approach are simplicity and clarity. The semantic model is set theory, proofs are by induction and the translation from the informal notation to the inductive rules is straightforward. 10.6 Proving Elementary Properties Secrecy properties can be hard to prove. The conclusion of a typical secrecy theorem is X / ∈analz (knows Spy evs). The difficulty arises from having to reason about analz, or less formally, showing that the spy can never learn X. Much easier is to prove that X can never occur at all. Such regularity prop-erties are typically expressed using parts rather than analz. The following lemma states that A’s private key is potentially known to the spy if and only if A belongs to the set bad of compromised agents. The state-ment uses parts: the very presence of A’s private key in a message, whether protected by encryption or not, is enough to confirm that A is compromised. The proof, like nearly all protocol proofs, is by induction over traces. 194 10. Case Study: Verifying a Security Protocol lemma Spy_see_priK [simp]: "evs ∈ns_public = ⇒(Key (priK A) ∈parts (knows Spy evs)) = (A ∈bad)" apply (erule ns_public.induct, simp_all) The induction yields five subgoals, one for each rule in the definition of ns_public. The idea is to prove that the protocol property holds initially (rule Nil), is preserved by each of the legitimate protocol steps (rules NS1–3), and even is preserved in the face of anything the spy can do (rule Fake). The proof is trivial. No legitimate protocol rule sends any keys at all, so only Fake is relevant. Indeed, simplification leaves only the Fake case, as indicated by the variable name evsf: 1. Vevsf X. [ [evsf ∈ns_public; (Key (priK A) ∈parts (knows Spy evsf)) = (A ∈bad); X ∈synth (analz (knows Spy evsf))] ] = ⇒(Key (priK A) ∈parts (insert X (knows Spy evsf))) = (A ∈bad) by blast The Fake case is proved automatically. If priK A is in the extended trace then either (1) it was already in the original trace or (2) it was generated by the spy, who must have known this key already. Either way, the induction hypothesis applies. Unicity lemmas are regularity lemmas stating that specified items can occur only once in a trace. The following lemma states that a nonce cannot be used both as Na and as Nb unless it is known to the spy. Intuitively, it holds because honest agents always choose fresh values as nonces; only the spy might reuse a value, and he doesn’t know this particular value. The proof script is short: induction, simplification, blast. The first line uses the rule rev_mp to prepare the induction by moving two assumptions into the induction formula. lemma no_nonce_NS1_NS2: "[ [Crypt (pubK C) { \|NA', Nonce NA, Agent D\| } ∈parts (knows Spy evs); Crypt (pubK B) { \|Nonce NA, Agent A\| } ∈parts (knows Spy evs); evs ∈ns_public] ] = ⇒Nonce NA ∈analz (knows Spy evs)" apply (erule rev_mp, erule rev_mp) apply (erule ns_public.induct, simp_all) apply (blast intro: analz_insertI)+ done The following unicity lemma states that, if NA is secret, then its appearance in any instance of message 1 determines the other components. The proof is similar to the previous one. lemma unique_NA: "[ [Crypt(pubK B) { \|Nonce NA, Agent A \| } ∈parts(knows Spy evs); Crypt(pubK B') { \|Nonce NA, Agent A'\| } ∈parts(knows Spy evs); 10.7 Proving Secrecy Theorems 195 Nonce NA / ∈analz (knows Spy evs); evs ∈ns_public] ] = ⇒A=A' ∧B=B'" 10.7 Proving Secrecy Theorems The secrecy theorems for Bob (the second participant) are especially impor-tant because they fail for the original protocol. The following theorem states that if Bob sends message 2 to Alice, and both agents are uncompromised, then Bob’s nonce will never reach the spy. theorem Spy_not_see_NB [dest]: "[ [Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) ∈set evs; A / ∈bad; B / ∈bad; evs ∈ns_public] ] = ⇒Nonce NB / ∈analz (knows Spy evs)" To prove it, we must formulate the induction properly (one of the assump-tions mentions evs), apply induction, and simplify: apply (erule rev_mp, erule ns_public.induct, simp_all) The proof states are too complicated to present in full. Let’s examine the simplest subgoal, that for message 1. The following event has just occurred: 1. A′ →B′ : { \|Na′, A′\| }Kb′ The variables above have been primed because this step belongs to a different run from that referred to in the theorem statement — the theorem refers to a past instance of message 2, while this subgoal concerns message 1 being sent just now. In the Isabelle subgoal, instead of primed variables like B′ and Na′ we have Ba and NAa: 1. Vevs1 NAa Ba. [ [A / ∈bad; B / ∈bad; evs1 ∈ns_public; Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) ∈set evs1 − → Nonce NB / ∈analz (knows Spy evs1); Nonce NAa / ∈used evs1] ] = ⇒Ba ∈bad − → Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) ∈set evs1 − → NB ̸= NAa The simplifier has used a default simplification rule that does a case analysis for each encrypted message on whether or not the decryption key is compro-mised. analz (insert (Crypt K X) H) = (if Key (invKey K) ∈analz H then insert (Crypt K X) (analz (insert X H)) else insert (Crypt K X) (analz H)) (analz_Crypt_if) 196 10. Case Study: Verifying a Security Protocol The simplifier has also used Spy_see_priK, proved in Sect. 10.6 above, to yield Ba ∈bad. Recall that this subgoal concerns the case where the last message to be sent was 1. A′ →B′ : { \|Na′, A′\| }Kb′. This message can compromise Nb only if Nb = Na′ and B′ is compromised, allowing the spy to decrypt the message. The Isabelle subgoal says precisely this, if we allow for its choice of variable names. Proving NB ̸= NAa is easy: NB was sent earlier, while NAa is fresh; formally, we have the assumption Nonce NAa / ∈used evs1. Note that our reasoning concerned B’s participation in another run. Agents may engage in several runs concurrently, and some attacks work by interleaving the messages of two runs. With model checking, this possibility can cause a state-space explosion, and for us it certainly complicates proofs. The biggest subgoal concerns message 2. It splits into several cases, such as whether or not the message just sent is the very message mentioned in the theorem statement. Some of the cases are proved by unicity, others by the induction hypothesis. For all those complications, the proofs are automatic by blast with the theorem no_nonce_NS1_NS2. The remaining theorems about the protocol are not hard to prove. The following one asserts a form of authenticity: if B has sent an instance of message 2 to A and has received the expected reply, then that reply really originated with A. The proof is a simple induction. theorem B_trusts_NS3: "[ [Says B A (Crypt (pubK A) { \|Nonce NA, Nonce NB, Agent B\| }) ∈set evs; Says A' B (Crypt (pubK B) (Nonce NB)) ∈set evs; A / ∈bad; B / ∈bad; evs ∈ns_public] ] = ⇒Says A B (Crypt (pubK B) (Nonce NB)) ∈set evs" From similar assumptions, we can prove that A started the protocol run by sending an instance of message 1 involving the nonce NA. For this theorem, the conclusion is Says A B (Crypt (pubK B) { \|Nonce NA, Agent A\| }) ∈set evs Analogous theorems can be proved for A, stating that nonce NA remains se-cret and that message 2 really originates with B. Even the flawed protocol establishes these properties for A; the flaw only harms the second participant. Detailed information on this protocol verification technique can be found elsewhere , including proofs of an Internet protocol . We must stress that the protocol discussed in this chapter is trivial. There are only three messages; no keys are exchanged; we merely have to prove that encrypted data remains secret. Real world protocols are much longer and distribute many secrets to their participants. To be realistic, the model has to include the possibility of keys being lost dynamically due to carelessness. If those keys have been used to encrypt other sensitive information, there may be 197 cascading losses. We may still be able to establish a bound on the losses and to prove that other protocol runs function correctly . Proofs of real-world protocols follow the strategy illustrated above, but the subgoals can be much bigger and there are more of them. 198 199 You know my methods. Apply them! Sherlock Holmes A. Appendix [ [ [\| \ ] ] \|] \ = ⇒ ==> \ V !! \ ≡ == \ ⇌ == \ ⇀ => \ ↽ <= \ λ % \ ⇒ => \ ∧ & \ ∨ \| \ − → --> \ ¬ ~ \ ̸= ~= \ ∀ ALL, ! \ ∃ EX, ? \ ∃! EX!, ?! \! ε SOME, @ \ ◦ o \ \| \| abs \ \ ≤ <= \ × \ ∈ : \ / ∈ ~: \ ⊆ <= \ ⊂ < \ ∪ Un \ ∩ Int \ S UN, Union \ T INT, Inter \ ∗ ^ \<^sup> −1 ^-1 \ Table A.1. Mathematical Symbols, Their ascii-Equivalents and Internal Names Constant Type Syntax 0 'a::zero 1 'a::one plus 'a::plus ⇒'a::plus ⇒'a::plus (infixl + 65) minus 'a::minus ⇒'a::minus ⇒'a::minus (infixl −65) uminus 'a::uminus ⇒'a::uminus −x times 'a::times ⇒'a::times ⇒'a::times (infixl ∗70) inverse_divide 'a::inverse ⇒'a::inverse ⇒'a::inverse (infixl / 70) divide 'a::divide ⇒'a::divide ⇒'a::divide (infixl div 70) modulo 'a::modulo ⇒'a::modulo ⇒'a::modulo (infixl mod 70) abs 'a::abs ⇒'a::abs \|x\| sgn 'a::sgn ⇒'a::sgn less_eq 'a::ord ⇒'a::ord ⇒bool (infixl ≤50) less 'a::ord ⇒'a::ord ⇒bool (infixl < 50) top 'a::top bot 'a::bot Table A.2. Important Overloaded Constants in Main ALL BIT CHR EX GREATEST INT Int LEAST O OFCLASS PI PROP SIGMA SOME THE TYPE UN Un WRT case choose div dvd else funcset if in let mem mod o of op then Table A.3. Reserved Words in HOL Terms Bibliography David Aspinall. Proof General. David Aspinall. Proof General: A generic tool for proof development. In Tools and Algorithms for the Construction and Analysis of Systems (TACAS), volume 1785 of Lecture Notes in Computer Science, pages 38–42. Springer-Verlag, 2000. Franz Baader and Tobias Nipkow. Term Rewriting and All That. Cam-bridge University Press, 1998. Gertrud Bauer, Tobias Nipkow, David von Oheimb, Lawrence C Paulson, Thomas M Rasmussen, Christophe Tabacznyj, and Markus Wenzel. The supplemental Isabelle/HOL library. Part of the Isabelle distribution, 2002. Richard Bird. Introduction to Functional Programming using Haskell. Prentice-Hall, 1998. Lukas Bulwahn, Alexander Krauss, and Tobias Nipkow. Finding lexi-cographic orders for termination proofs in Isabelle/HOL. In K. Schnei-der and J. Brandt, editors, Theorem Proving in Higher Order Logics: TPHOLs 2007, volume 4732 of Lecture Notes in Computer Science, pages 38–53. Springer-Verlag, 2007. M. Burrows, M. Abadi, and R. M. Needham. A logic of authentication. Proceedings of the Royal Society of London, 426:233–271, 1989. Jacques Fleuriot and Lawrence C. Paulson. Mechanizing nonstandard real analysis. LMS Journal of Computation and Mathematics, 3:140–190, 2000. Jean-Yves Girard. Proofs and Types. Cambridge University Press, 1989. Translated by Yves Lafont and Paul Taylor. M. J. C. Gordon and T. F. Melham, editors. Introduction to HOL: A Theorem Proving Environment for Higher Order Logic. Cambridge University Press, 1993. Florian Haftmann. Haskell-style type classes with Isabelle/Isar. https: //isabelle.in.tum.de/doc/classes.pdf. David Harel, Dexter Kozen, and Jerzy Tiuryn. Dynamic Logic. MIT Press, 2000. John E. Hopcroft and Jeffrey D. Ullman. Introduction to Automata Theory, Languages, and Computation. Addison-Wesley, 1979. 204 BIBLIOGRAPHY Paul Hudak. The Haskell School of Expression. Cambridge University Press, 2000. Michael Huth and Mark Ryan. Logic in Computer Science. Modelling and reasoning about systems. Cambridge University Press, 2000. Donald E. Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching. Addison-Wesley, 1975. Alexander Krauss. Defining Recursive Functions in Isabelle/HOL. https: //isabelle.in.tum.de/doc/functions.pdf. Gavin Lowe. Breaking and fixing the Needham-Schroeder public-key protocol using CSP and FDR. In T. Margaria and B. Steffen, editors, Tools and Algorithms for the Construction and Analysis of Systems: second international workshop, TACAS ’96, LNCS 1055, pages 147–166. Springer, 1996. Robin Milner, Mads Tofte, and Robert Harper. The Definition of Stan-dard ML. MIT Press, 1990. Olaf Müller, Tobias Nipkow, David von Oheimb, and Oscar Slotosch. HOLCF = HOL + LCF. Journal of Functional Programming, 9:191–223, 1999. Wolfgang Naraschewski and Markus Wenzel. Object-oriented verifica-tion based on record subtyping in higher-order logic. In Jim Grundy and Malcom Newey, editors, Theorem Proving in Higher Order Log-ics: TPHOLs ’98, volume 1479 of Lecture Notes in Computer Science. Springer-Verlag, 1998. Roger M. Needham and Michael D. Schroeder. Using encryption for authentication in large networks of computers. Communications of the ACM, 21(12):993–999, December 1978. Tobias Nipkow. Functional unification of higher-order patterns. In M. Vardi, editor, Eighth Annual Symposium on Logic in Computer Sci-ence, pages 64–74. IEEE Computer Society Press, 1993. Tobias Nipkow. Structured Proofs in Isar/HOL. In H. Geuvers and F. Wiedijk, editors, Types for Proofs and Programs (TYPES 2002), volume 2646 of Lecture Notes in Computer Science, pages 259–278. Springer-Verlag, 2003. Tobias Nipkow, Lawrence C. Paulson, and Markus Wenzel. Isabelle’s Logics: HOL. Lawrence C. Paulson. The Old Isabelle Reference Manual. https:// isabelle.in.tum.de/doc/ref.pdf. Lawrence C. Paulson. Logic and Computation: Interactive proof with Cambridge LCF. Cambridge University Press, 1987. Lawrence C. Paulson. Isabelle: A Generic Theorem Prover. Springer, 1994. LNCS 828. Lawrence C. Paulson. ML for the Working Programmer. Cambridge University Press, 2nd edition, 1996. MLbook. BIBLIOGRAPHY 205 Lawrence C. Paulson. The inductive approach to verifying cryptographic protocols. Journal of Computer Security, 6:85–128, 1998. Lawrence C. Paulson. Inductive analysis of the Internet protocol TLS. ACM Transactions on Information and System Security, 2(3):332–351, August 1999. Lawrence C. Paulson. Relations between secrets: Two formal analyses of the Yahalom protocol. Journal of Computer Security, 9(3):197–216, 2001. F. J. Pelletier. Seventy-five problems for testing automatic theorem provers. Journal of Automated Reasoning, 2:191–216, 1986. Errata, JAR 4 (1988), 235–236 and JAR 18 (1997), 135. Kenneth H. Rosen. Discrete Mathematics and Its Applications. McGraw-Hill, 1998. Simon Thompson. Haskell: The Craft of Functional Programming. Addison-Wesley, 1999. Makarius Wenzel. The Isabelle System Manual. de/doc/system.pdf. Makarius Wenzel. The Isabelle/Isar Reference Manual. in.tum.de/doc/isar-ref.pdf. Markus Wenzel. Isabelle/Isar — a versatile environment for human-readable formal proof documents. PhD thesis, Institut für Informatik, Technische Universität München, 2002. doc/601724/601724.pdf. Index !, 201 ?, 201 ∃! , 201 ?!, 201 &, 201 ~, 201 ~=, 201 \|, 201 , 22 +, 22 -, 22 <=, 201 <, 22 >, 23 ≥, 23 ≤, 22 [], 10 #, 10 @, 11, 201 / ∈, 201 ~:, 201 T , 201 S , 201 −1, 201 ^-1, 201 ∗, 201 ^, 201 V, 13, 201 !!, 201 ⇒, 5 [\|, 201 \|], 201 %, 201 () (constant), 24 + (tactical), 100 <lex>, see lexicographic product ? (tactical), 101 \| (tactical), 101 0 (constant), 22, 164 1 (constant), 22, 164, 165 abandoning a proof, 13 abbreviation (command), 56 abs (constant), 166, 168, 170 abs, 201 absolute value, 166, 170 ac_simps (theorems), 169 Ackermann’s function, 49 add (modifier), 29 add.assoc (theorem), 169 add.commute (theorem), 169 ALL, 201 All (constant), 111 allE (theorem), 81 allI (theorem), 81 antiquotation, 61 append function, 11–15 apply (command), 16 arg_cong (theorem), 97 arith (method), 23, 164 arithmetic operations – for nat, 22 ascii symbols, 201 Aspinall, David, iii associative-commutative function, 176 assumption (method), 69 assumptions – of subgoal, 13 – renaming, 83 – reusing, 84 auto (method), 39, 93 axclass, 158 axiom of choice, 87 axiomatic type classes, 158 back (command), 79 Ball (constant), 111 ballI (theorem), 110 best (method), 93 Bex (constant), 111 bexE (theorem), 110 bexI (theorem), 110 Index 207 bij_def (theorem), 112 bijections, 112 binary trees, 18 binomial coefficients, 111 bisimulations, 118 blast (method), 90–91, 93 bool (type), 5 boolean expressions example, 19–22 bspec (theorem), 110 by (command), 73 card (constant), 111 card_Pow (theorem), 111 card_Un_Int (theorem), 111 cardinality, 111 case (symbol), 31, 32 case expressions, 5, 6, 18 case distinctions, 19 case splits, 31 case_tac (method), 19, 103, 151 cases (method), 156 clarify (method), 91, 93 clarsimp (method), 92, 93 class, 163 classical (theorem), 73 coinduction, 118 Collect (constant), 111 comment, 11 compiling expressions example, 37–39 Compl_iff (theorem), 108 complement – of a set, 107 complex (type), 167–168 complex numbers, 167–168 Complex_Main (theory), 168 composition – of functions, 112 – of relations, 114 conclusion – of subgoal, 13 conditional expressions, see if expressions conditional simplification rules, 31 cong (attribute), 176 congruence rules, 175 conjE (theorem), 72 conjI (theorem), 68 Cons (constant), 9 contrapositives, 74 converse – of a relation, 114 converse_iff (theorem), 114 CTL, 121–126, 182–185 datatype (command), 9, 39–44 datatypes, 16–22 – and nested recursion, 41, 45 – mutually recursive, 39 defer (command), 16, 102 definition (command), 25 Definitional Approach, 25 definitions, 25 – unfolding, 30 del (modifier), 29 description operators, 86–88 descriptions – definite, 86 – indefinite, 87 dest (attribute), 104 destruction rules, 71–72 diff_mult_distrib (theorem), 166 difference – of sets, 108 disjCI (theorem), 74 disjE (theorem), 70 div (symbol), 22, 23 div_mult_mod_eq (theorem), 165 divides relation, 85, 96, 103–105, 166 division – by negative numbers, 167 – by zero, 166 – for type nat, 165 division_by_zero (type class), 169 documents, 57 domain – of a relation, 114 Domain_iff (theorem), 114 done (command), 14 drule_tac (method), 76, 98 dvd (symbol), 23 dvd_add (theorem), 166 dvd_antisym (theorem), 166 dvd_def (theorem), 166 elim! (attribute), 131 elimination rules, 69–71 end (command), 15 Eps (constant), 111 equality, 5 – of functions, 111 – of records, 155 – of sets, 108 equalityE (theorem), 108 equalityI (theorem), 108 erule (method), 70 erule_tac (method), 76 Euclid’s algorithm, 103–105 208 Index evaluation, 11 even numbers – defining inductively, 127–131 EX, 201 Ex (constant), 111 exE (theorem), 83 exI (theorem), 83 ext (theorem), 111 extend (constant), 156 extensionality – for functions, 111, 112 – for records, 155 – for sets, 108 EX!, 201 False (constant), 5 fast (method), 92, 93, 124 Fibonacci function, 48 field (type class), 169 fields (constant), 156 finding theorems, 34, 93 finite (symbol), 111 fixed points, 117–118 force (method), 92, 93 formal comments, 60 formal proof documents, 57 formulae, 5–6 forward proof, 94–100 frule (method), 84 frule_tac (method), 76 fst (constant), 23 fun (command), 47–51 function (command) – and numeric literals, 165 function types, 5 functions, 111–113 – total, 11, 47–51 gcd (constant), 94–96, 103–105 generalizing for induction, 129 generalizing induction formulae, 36 Girard, Jean-Yves, 71n Gordon, Mike, 3 grammars – defining inductively, 141–146 ground terms example, 136–141 hd (constant), 17, 38 Hilbert’s ε-operator, 87 HOLCF, 44 Hopcroft, J. E., 145 hypreal (type), 168 Id_def (theorem), 114 id_def (theorem), 112 identifiers, 6 – qualified, 4 identity function, 112 identity relation, 114 if expressions, 5, 6 – simplification of, 32 – splitting of, 31, 49 if-and-only-if, 6 if_split (theorem), 31 if_split_asm (theorem), 32 iff (attribute), 91, 104, 130 iffD1 (theorem), 95 iffD2 (theorem), 95 ignored material, 63 image – under a function, 113 – under a relation, 114 image_def (theorem), 113 Image_iff (theorem), 114 impI (theorem), 72 implication, 72–73 imports (command), 4 ind_cases (method), 131 induct_tac (method), 12, 19, 51, 181 induction, 178–185 – complete, 180 – deriving new schemas, 182 – on a term, 179 – recursion, 50–51 – structural, 19 – well-founded, 117 induction heuristics, 35–37 inductive (command), 132 inductive definition – simultaneous, 142 inductive definitions, 127–146 inductive predicates, 132–133 inductive cases (command), 131, 140 inductive set (command), 127 infinitely branching trees, 43 infix annotations, 53 infixr (annotation), 10 inj_on_def (theorem), 112 injections, 112 insert (constant), 109 insert (method), 98–100 instance, 158 INT, 201 Int, 201 int (type), 166–167 INT_iff (theorem), 110 IntD1 (theorem), 107 Index 209 IntD2 (theorem), 107 integers, 166–167 INTER (constant), 111 Inter, 201 Inter_iff (theorem), 110 intersection, 107 – indexed, 110 IntI (theorem), 107 intro (method), 74 intro! (attribute), 128 introduction rules, 68–69 inv (constant), 87 inv_image_def (theorem), 117 inverse – of a function, 112 – of a relation, 114 inverse (constant), 169 inverse image – of a function, 113 – of a relation, 116 itrev (constant), 35 λ expressions, 5 LCF, 44 LEAST (symbol), 22, 86 least number operator, see LEAST Leibniz, Gottfried Wilhelm, 53 lemma (command), 13 lemmas (command), 94, 104 length (symbol), 17 length_induct, 181 less_than (constant), 116 less_than_iff (theorem), 116 let expressions, 5, 6, 30 Let_def (theorem), 30 lex_prod_def (theorem), 117 lexicographic product, 117 lfp – applications of, see CTL linear arithmetic, 22–23, 164 List (theory), 17 list (type), 5, 9, 17 list.split (theorem), 31 Lowe, Gavin, 188–189 Main (theory), 4 major premise, 75 make (constant), 156 marginal comments, 60 markup commands, 59 max (constant), 22, 23 measure functions, 116 measure_def (theorem), 117 meta-logic, 80 methods, 16 min (constant), 22, 23 mixfix annotations, 53 mod (symbol), 22, 23 mod_mult_distrib (theorem), 165 model checking example, 118–126 modus ponens, 67, 72 mono_def (theorem), 118 monotone functions, 118, 140 – and inductive definitions, 137–138 more (constant), 152, 153 mp (theorem), 72 multiset ordering, 117 nat (type), 5, 22, 165–166 nat_less_induct (theorem), 180 natural deduction, 67–68 natural numbers, 22, 165–166 Needham-Schroeder protocol, 187–189 negation, 73–75 Nil (constant), 9 no_asm (modifier), 29 no_asm_simp (modifier), 29 no_asm_use (modifier), 30 no_vars (attribute), 62 non-standard reals, 168 None (constant), 24 notation (command), 55 notE (theorem), 73 notI (theorem), 73 numbers, 163–170 numeric literals, 164–165 – for type nat, 165 – for type real, 168 O (symbol), 114 o, 201 o_def (theorem), 112 OF (attribute), 96–97 of (attribute), 94, 97 only (modifier), 29 oops (command), 13 option (type), 24 ordered rewriting, 176 ordered_field (type class), 169 ordered_ring (type class), 168 ordered_semiring (type class), 168 overloading, 22, 158–159 – and arithmetic, 164 pairs and tuples, 23, 149–152 parent theories, 4 pattern matching 210 Index – and fun, 48 patterns – higher-order, 177 PDL, 118–121 Power (theory), 170 pr (command), 101 prefer (command), 16, 102 prefix annotation, 55 primitive recursion, see recursion, primitive primrec (command), 11, 39–44 print mode, 55 product type, see pairs and tuples Proof General, 7 proof state, 12 proofs – abandoning, 13 – examples of failing, 88–89 protocols – security, 187–197 quantifiers, 6 – and inductive definitions, 136–137 – existential, 82–83 – for sets, 110 – instantiating, 84 – universal, 80–82 r_into_rtrancl (theorem), 114 r_into_trancl (theorem), 115 range – of a function, 113 – of a relation, 114 range (symbol), 113 Range_iff (theorem), 114 rat (type), 167–168 rational numbers, 167–168 real (type), 167–168 real numbers, 167–168 record (command), 152 records, 152–158 – extensible, 153–155 recursion – primitive, 18 recursion induction, 50–51 reflexive and transitive closure, 114–116 reflexive transitive closure – defining inductively, 133–135 relations, 113–116 – well-founded, 116–117 relcomp_unfold (theorem), 114 rename_tac (method), 83 rev (constant), 10–15, 35 rewrite rules, 27 – permutative, 176 rewriting, 27 ring (type class), 168 ring_no_zero_divisors (type class), 169 rtrancl_refl (theorem), 114 rtrancl_trans (theorem), 114 rule induction, 128–130 rule inversion, 130–131, 140–141 rule_format (attribute), 179 rule_tac (method), 76 – and renaming, 83 safe (method), 92, 93 safe rules, 90 searching theorems, 34, 93 selector – record, 153 semiring (type class), 168 session, 58 set (type), 5, 107 set comprehensions, 109–110 set_ext (theorem), 108 sets, 107–111 – finite, 111 – notation for finite, 109 Show Types (Proof General), 7 show_main_goal (flag), 12 simp (attribute), 12, 28 simp (method), 28 simp del (attribute), 28 simp_all (method), 29, 39 simplification, 27–35, 175–178 – of let-expressions, 30 – with definitions, 30 – with/of assumptions, 29 simplification rule, 177–178 simplification rules, 28 – adding and deleting, 29 simplified (attribute), 95, 97 size (constant), 17 snd (constant), 23 SOME (symbol), 87 SOME, 201 Some (constant), 24 some_equality (theorem), 87 someI (theorem), 87 someI2 (theorem), 87 someI_ex (theorem), 88 source comments, 60 spec (theorem), 81 split (attribute), 32 Index 211 split (constant), 150 split (method), 31, 150 split (modifier), 32 split rule, 32 ssubst (theorem), 78 structural induction, see induction, structural subclasses, 158 subgoal numbering, 47 subgoal_tac (method), 99, 100 subgoals, 12 subset relation, 108 subsetD (theorem), 108 subsetI (theorem), 108 subst (method), 78, 79 substitution, 77–80 Suc (constant), 22 surj_def (theorem), 112 surjections, 112 sym (theorem), 95 symbols, 54 syntax, 6, 11 tacticals, 100–101 tactics, 12 tagged command regions, 63 term (command), 16 term rewriting, 27 termination, see functions, total terms, 5 text, 61 text blocks, 60 THE (symbol), 86 the_equality (theorem), 86 THEN (attribute), 95, 97, 104 theorem (command), 12, 13 theories, 4 theory (command), 4 theory files, 4 thm (command), 16 tl (constant), 17 ToyList example, 9–15 tracing the simplifier, 32 trancl_trans (theorem), 115 tries, 44–47 True (constant), 5 truncate (constant), 156 tuples, see pairs and tuples txt, 61 typ (command), 16 type classes, 163 type constraints, 6 type constructors, 5 type inference, 5 type synonyms, 24 type variables, 5 type synonym (command), 24 typedecl (command), 171 typedef (command), 171–174 types, 4–5 – declaring, 171 – defining, 171–174 Ullman, J. D., 145 UN, 201 Un, 201 UN_E (theorem), 110 UN_I (theorem), 110 UN_iff (theorem), 110 Un_subset_iff (theorem), 108 unfold (method), 30 unification, 76–79 UNION (constant), 111 Union, 201 union – indexed, 110 Union_iff (theorem), 110 unit (type), 24 unknowns, 7, 68 unsafe rules, 91 update – record, 153 updating a function, 111 variables, 7 – schematic, 7 – type, 5 vimage_def (theorem), 113 wf_induct (theorem), 117 wf_inv_image (theorem), 117 wf_less_than (theorem), 116 wf_lex_prod (theorem), 117 wf_measure (theorem), 117
14548	https://svn.bmj.com/content/9/4/344	Intended for healthcare professionals Home / Archive / Volume 9, Issue 4 Email alerts Review • • Request permissions White matter disease derived from vascular and demyelinating origins • , , , , , , . Abstract Damage or microstructural alterations of the white matter can cause dysfunction of the intrinsic neural networks in a condition termed as white matter disease (WMD). Frequently detected on brain computed tomography and magnetic resonance imaging scans, WMD is commonly presented in inflammatory demyelinating diseases like multiple sclerosis (MS) and vascular diseases such as cerebral small vessel disease (CSVD). Prevention of MS and CSVD progression requires early treatments with drastically different medications and approaches, as such, early and accurate diagnosis of WMD, derived from vascular or demyelinating etiologies, is of paramount importance. However, the clinical and imaging similarities between MS, especially during the early stage, and CSVD, pose a significant dilemma in differentiating these two conditions. In this review, we attempt to summarize and contrast the distinguishing features of MS and CSVD for aiding accurate diagnosis to ensure timely corresponding management in the early stages of MS and CSVD. Introduction The white matter of the central nervous system (CNS) is primarily constituted of myelinated axons (tracts) and myelin-producing glial cells, connecting and communicating between disparate regions of grey matter,1 which is composed of bodies of neurons. In the human brain, white matter occupies approximately 4×105 mm3, which comprises 45% of the total volume.2 The medullary arteries supplying the cerebral white matter are long-end arteries absent in anastomotic connections and arise perpendicularly from pial arteries. Most of the subcortical arteries penetrate straight through the cortex, but become coiled, looped and/or spiralled on entering the white matter. This coiling architecture loses the ability to react in autoregulation and may serve as a trap for inflamed cells via the bloodstream, contributing to vascular and inflammatory diseases.3 4 Damage or microstructural alterations to the white matter frequently detected by brain CT or MRI scans are designated as white matter diseases (WMD). With the popularisation and application of imaging technology, more and more symptomatic or asymptomatic white matter lesions have been found clinically. The two common causes of WMD are vascular diseases, which are especially common in northern China, and inflammatory demyelinating causes like multiple sclerosis (MS). Differing from acute ischaemic and haemorrhagic strokes, cerebral small vessel disease (CSVD) is relatively mild symptomatically, but its progressive neurological dysfunction has received increasing attention in recent years. The differential diagnosis of MS can be challenging. Early identification of MS can help with the early application of disease-modifying therapy to reduce disease disability. The similarity of image manifestations requires careful screening and early judgement by clinicians, which is conducive to early treatment and long-term management of patients. In this review, we focus on CSVD for its high prevalence and its similar neuroimaging characteristics with MS, which obscures and hinders the differential diagnosis between the two conditions. To this end, we summarise the similarities and differences in the aetiology, clinical features, and imaging characteristics between CSVD and MS to assist accurate diagnoses and treatment. Pathological basis for white matter lesions in MS and CSVD Inflammation Multiple focal areas of myelin loss within the CNS are the main pathologic hallmarks of MS. Demyelination is accompanied by inflammation, gliosis and relative axonal preservation at the early stage of MS. Lesions are disseminated in areas throughout the CNS, with the majority affecting the white matter of the brainstem, cerebellum, juxtacortical and periventricular regions of the brain.5 Evidence of sequential inflammation in subcortical and cortical regions has been shown in patients with MS.6 Formation of plaques in MS is due to a confluence of inflammation, myelin breakdown, astrogliosis, oligodendrocyte injury, neurodegeneration, axonal loss and remyelination5 (figure 1). Acute, chronic active and chronic silent lesions are thought to occur along a continuous timeline, eventually bearing scarred and hardened areas within the CNS. Dysregulation of the blood-brain barrier (BBB) and trans-endothelial migration of activated leukocytes are among the earliest cerebrovascular abnormalities observed in the MS brain and coincide with the release of inflammatory cytokines.7 As MS progresses, inflammation wanes and is followed by neurodegeneration, mostly induced by persistent glial activation, mitochondrial injury, oxidative stress and ionic imbalance.8 Figure 1 Request permissions White matter in normal, cerebral small vessel disease or multiple sclerosis. The figure is designed and drawn by the authors. The role of inflammation in the genesis and progression of CSVD has been proposed recently. Inflammatory cells in the white matter around vessels and in the vicinity of demyelination were found in postmortem samples. Plasma inflammatory cytokines, markers of oxidative stress and vascular inflammation are associated with the white matter hyperintensity volume on MRI.9 In addition, atherosclerosis is an important cause of CSVD. Plenty of evidence supports the involvement of inflammatory responses in atherosclerosis. Inflammation during the genesis of atherosclerosis is triggered by trapping low-density lipoproteins in the intima,10 the accumulation of which exacerbates the stenosis of blood vessels and reduces blood flow leading to ischemia. BBB impairment, chronic inflammatory responses and leucocyte infiltration are classical pathological features of CSVD (figure 1, table 1). Recent studies suggest that components of the coagulation system may trigger inflammation in CSVD.11 Table 1 • Contrast the pathological processes between inflammatory demyelinating white matter disease Cerebral amyloid angiopathy (CAA), characterised by the progressive accumulation of amyloid-β fibrils in the walls of small to medium-sized arteries and arterioles, is another important cause of CSVD.12 Recurrent large lobar haemorrhages, microbleeds, white matter lesions and microinfarcts are main features of CAA.13 Different forms of CAA-associated inflammation have been described: one is a non-vasculitic form called perivascular infiltration, which is characterised by perivascular infiltration of the parenchyma by multinucleated giant cells; the other is a vasculitic form called transmural granulomatous angiitis, which is characterised by inflammation of the vessel wall, with the occasional presence of granulomas.14 Conformational transitions occurring in native soluble amyloid molecules increase their content in β-sheet structures, thereby favouring the formation and deposition of more insoluble oligomeric structures. In turn, these deposits trigger a secondary cascade of events including, among others, the release of inflammatory components, activation of the complement system, oxidative stress, alterations of BBB permeability and cell toxicity.15 Vascular damage Vascular damage of CSVD encompasses a group of pathological processes with various aetiologies affecting small arteries, arterioles, venules and capillaries of the brain. Disrupted BBB integrity may play a pivotal role in the aetiology of CSVD. A consensus was endorsed on BBB impairment appearing in the MRI of white matter hyperintensity and normal-appearing white matter.16 Pathological findings of CSVD due to arteriolosclerosis present as loss of smooth muscle cells from the tunica media, deposits of fibrohyaline material, narrowing of the lumen and thickening of the vessel wall, which initiate on a damaged endothelium; such blood vessel changes, therefore, occlude the lumen, inducing lacunar infarction.17 Brain damage arising from small-vessel endothelial leakage has been suggested to be a potential cause of CSVD. Further studies are required to test whether BBB failure and perforating artery endothelial leakage might be common pathogenetic mechanisms in lacunar stroke, white matter hyperintensities and dementia. For example, in a stroke-prone spontaneously hypertensive rat model,18 dysfunctional endothelial cells secrete heat shock protein 90α, which blocks oligodendroglia differentiation to contribute to impaired myelination as the first step in the development of SVD. Acute lesions of MS are characterised by robust inflammatory infiltration combined with demyelination and parenchymal oedema distributed throughout the lesions. Lesions were around vessels forming the specific ‘central vein sign’ on MRI are characteristics of inflammatory demyelination (figure 2). Dysregulation of the BBB and transendothelial migration of activated leukocytes are among the acute phase in MS brains and coincide with the release of inflammatory cytokines.7 Perivascular cuffing is formed by an influx of immune cells centred around vessels, including lymphocytes (predominantly T cells), monocytes and natural killer (NK) cells (figure 1). A subset of the major histocompatibility class II-expressing cells, distributed evenly throughout the lesions, are loaded with lipids (foamy macrophages) and participate in active stripping of myelin from axons. Figure 2 Request permissions Key points for differentiating cerebral small vessel disease and multiple sclerosis. FLAIR, fluid-attenuated inversion recovery. Decreased cerebral perfusion in MS in normal-appearing white matter and grey matter has been demonstrated in MR perfusion imaging. Extracranial arterial studies in MS patients via magnetic resonance angiography demonstrate a significant mid-term decrease in the number and size of the secondary neck vessels.19 Lipoprotein(a) levels are associated with the size of extracranial arteries in MS,20 while the clinical relevance of these findings are currently unknown. The abundant cytokines may be the potential cause for the accelerated narrowing of the cerebral and extracranial arterial vessels in MS (figure 1). The association of neck venous drainage abnormalities with MS is uncertain.21 Optical coherence tomography angiography can investigate the retinal microvascular structures and evaluate the vascular changes in the foveal and peripapillary region in MS. Retinal macular superficial vascular plexus density is reduced in the eyes of MS patients and correlates with disability and visual function.22 This reflects early vascular damage in patients with MS. Clinical overlap between CSVD and MS Presenting symptoms and the clinical course of CSVD are variable. Cerebral infarction includes asymptomatic infarcts, episodes of transient cerebral ischemia and neurological deficits manifested as lacunar syndrome. Asymptomatic infarctions are clinically silent and conspicuously termed as ‘silent stroke’, generally occur in hypertensive patients. Lacunar stroke is the most common type of ischaemic stroke, resulting from the occlusion of small penetrating arteries, ranging from 100 µm to 400 µm in diameter, which supplies blood to the brain’s deep structures. Transient cerebral ischaemia may be the only manifestation.23 No matter the acute attack or the asymptomatic lesions, chronic, progressive and accumulated cognitive decline, gait disorder and neuropsychiatric symptoms gradually progress and affect patients’ daily life. MS undergoes a characteristic relapse-remitting and progressive disease course. Attacks of MS, as well as CSVD, manifest as a focal loss of perception or motor ability during their acute phase. Gait abnormity is a persistent disability affecting the motor system. Non-motor MS symptoms include cognitive decline, depression, pain, fatigue, anxiety, and sleep disorder, all of which overlap CSVD symptoms.6 Therefore, CSVD and MS may both present as acute and as chronic, progressive neurological deficits. Misdiagnoses of MS as an acute stroke or vice versa will result in a vastly different evaluation and treatment. Clinically silent but MRI-positive lesions are another common feature in these two diseases. White matter hyperintensities incidentally found without neurological manifestations are called ‘silent strokes’, with vascular origins and ‘radiologically isolated syndrome’, with inflammatory origins. It is difficult to distinguish the origins of these isolated clinically silent lesions, hindering appropriate diagnosis and prevention of subsequent clinical attacks. Other common presentations include white matter lesion load progression with years, cerebral atrophy and cognition dysfunction. It is prudent for clinicians to consider MS as a differential diagnosis for older individuals or those with vascular risk factors. Furthermore, patients may present both of these risk factors, increasing the difficulty to determine the origin of peri-ventricular lesions. However, CSVD rarely attacks optic nerve and spinal cord, which are commonly damaged in MS patients. Hence, symptoms of optic neuritis or myelitis, such as transverse plane of paraesthesias, dysfunction of the bladder and dysporia, were red alerts for diagnosing CSVD. (figure 2) Similarities and differences of CSVD and MS in MRI White matter abnormalities on T2-weighted images are observed in 20%–78% of healthy and demented elderly patients. White matter changes are easily identified in proton density-weighted, T2-weighted and fluid-attenuated inversion recovery (FLAIR) imaging. It is worth noting that incidentally found lesions lead to puzzle of management, especially when these lesions are not relevant to disease presentation or symptoms.24 In patients with CSVD, white matter changes are bilateral, patchy or diffusely distributed hypodensity lesions on CT or hyperintensities on T2-weighted MRI involving the periventricular and centrum semiovale white matter (leukoaraiosis).25 These changes relate to the decline of cognitive and motor performances, and depressive symptoms, urinary disturbances and various neurological abnormalities. In 2013, STandards for ReportIng Vascular changes on nEuroimaging (STRIVE) provided as the standardised imaging features of CSVD.26 27 Correspondingly, European Magnetic Resonance Network in MS (MAGNIMS) and the Consortium of MS Centers recently proposed standardised MRI protocols for the diagnostic process, from prognosis to follow-ups.28 In the modified criteria, symptomatic lesions demonstrating dissemination in space and time include lesions of the periventricular, cortical/juxtacortical, infratentorial, spinal cord and optic nerve. Cortical lesions are considered evidence of dissemination in space, further, when cortical or juxtacortical lesions are present CNS vasculitis and occasionally haemorrhage should be included in the differential diagnosis.6 The 2018 MAGNIMS update in the imaging features differentiates the most common mimics of MS, particularly age-related cerebrovascular diseases and neuromyelitis optica spectrum disorders.29 We here summarise the key differences found in MRI to distinguish CVSD and MS (figures 2–4). Multiple, T2-hyperintense white matter lesions on conventional MRI are common features shared by CSVD and MS, mainly presenting as white matter hyperintensity with similar shapes and locations, especially the periventricular lesions. Lacunes of presumed vascular origin are typically white matter lesions seen on MRI in basal ganglia, internal capsule, thalamus and pons, which can occur silently or after an acute small subcortical infarct or haemorrhage,30 which are 3–15 mm cerebrospinal fluid (CSF)-filled cavities. White matter hyperintensities are patchy or confluent depending on the stage of development and severity. A hypointense hole surrounded by a hyperintense rim on FLAIR sequence is distinctive for lacunes. These are usually symmetrically and bilaterally distributed in the white matter, pons, brain stem and deep grey matter in patients with CSVD. But in contrast, Dawson’s finger sign, which is defined as wedge-shaped areas with a broad base to the ventricle, and extensions into adjoining tissue in the form of finger-like processes or ampullae, is specific in MS periventricular lesions. Certain lesions may appear hypointense on T1-weighted images, so-called black holes.31 Similar to a ‘silent stroke’, individuals with T2 hyperintensities on brain MRI strongly suggestive of MS but without neurological presentations or other explanations are distinguished as radiologically isolated syndrome. Radiologically isolated syndromes could provide an early indication and diagnosis of MS and also increases the rate of misdiagnosis and the possibility of adverse effects of immune therapies.32 Figure 3 Request permissions Cardinal MRI difference between cerebral small vessel disease and multiple sclerosis. T2 weighted image; #FLAIR: fluid attenuation inversion recovery; &SWI: susceptibility-weighted imaging; §T1 gadolinium enhancement. The MRI images are collected from the authors’ centres. Figure 4 Request permissions Additional distinguishing MRI features between cerebral small vessel disease and multiple sclerosis. T2 weighted image; #FLAIR: fluid attenuation inversion recovery; &SWI: susceptibility-weighted imaging; §T1 gadolinium enhancement. The MRI images are collected from the authors’ centres. Enlarged perivascular spaces and fluid-filled subarachnoid spaces that surround cerebral microvessels in the brain parenchyma (also called Virchow-Robin spaces) are commonly found as hyperintense on T2 MRI and hypointense on T1 MRI, less than 3 mm, in the basal ganglia, centrum semiovale and brainstem.33 Enlarged perivascular spaces are commonly seen in CSVD and may be associated with cognitive impairment or hypertension. In MS, perivascular spaces are associated with increased inflammatory markers, but enlarged perivascular spaces are rarely visible. Lesion is common enhanced on T1-weighted imaging after contrast injection in MS patients with acute attack. The nodular or ring-like (closed and open) enhancement correlates with altered BBB permeability in acute perivascular inflammation, thus discriminating active or chronic lesions.34 Gadolinium enhancement varies in size and shape and, usually, lasts a few days to weeks, with an average duration of 3 weeks (97% of lesions enhance in less than 2 months).35 Less than 40% of the lesions transform to black holes over 6 months, correlating pathologically with permanent demyelination and severe axonal loss.31 Black holes in MS are T1 white matter lesions with a hypointense appearance relative to normal-appearing white matter. Persistent black holes may be a marker of axonal loss and tissue destruction, different from the CSF-filled cavities of lacunae.31 Open ring sign can be used as a key point for differentiation, in particular, lesion enhancement is rare in CSVD. In addition, cerebral microbleeds are small, round, 5–10 mm areas and homogeneous foci of hypointensity found on T2-weighted (gradient echo) (T2) and susceptibility-weighted imaging, which are sensitive markers indicating BBB dysfunction in CAA or arteriosclerotic CSVD. Microbleeds are rarely found in MS patients. Central vein sign is the imaging manifestation of the perivenular nature of demyelinating plaques. The central vein sign distinguishes MS from inflammatory vasculopathy with a diagnostic accuracy of 100%.36 Ultra-high field MR and T2 sequences are helpful in finding this sign. Both MS and CSVD undergo a chronically progressed disease course. Cognitive decline and brain atrophy are common findings among MS and CSVD.37 It is often hard to distinguish them solely through clinical manifestation and brain atrophy signs, that is, reduced grey matter volume and enlarged ventricles. However, brain atrophy caused by MS is often associated with lesion load,38 which is not typically related in CSVD. Utilisation of high-field intensity MRI scanners is conducive to the detection of subtle structural changes in clinical practice and investigation, and perhaps helpful in diagnosing CSVD or MS. With regard to the lesion locations, the brain stem, periventricular and cerebral cortex are commonly attacked by MS and CSVD; lesions in juxtacortical areas, optic nerves and spinal cord are rarely found in CSVD but commonly damaged in MS. At present, 7T MRI scanning can assist to detect CSVD specific microinfarctions (50 µm to a few mm in diameter) in areas appearing normal in 3T MRI scans.39 Similarly, 7T MRI can also detect some microlesions in cortical and juxtacortical areas in MS patients, which are characteristic signs for differentiating from CSVD. Conclusions and future perspective Clinical similarities but divergent managements of CSVD and the early phase of MS highlight the need to timely differentiate WMD of these two origins. Patterns and locations of lesions as determined by multimodal MRI are instrumental to reveal the underlying aetiology. Periventricular white matter lesions are common in MS and CSVD. The most characteristic pathological lesions in MS are ‘Dawson fingers’. Age-related vascular white matter lesions mostly show a broad base along the ventricular margin, but occasionally one or more lesions may be found periventricular. Periventricular lesions can increase and fuse over time in both diseases, whereas corpus callosum lesions are usually found in MS but are rare in CSVD. Additionally, oligoclonal band presence in cerebral spinal fluid, spinal cord and optic lesions of MS offer additional clues distinguishing it from CSVD. Juxtacortical white matter is a characteristic location of MS lesions. Cortical lesions and juxtacortical lesions extending into the cortex are typical of MS in histopathological studies; conversely, cortical lesions have been found in CSVD, but juxtacortical lesions are rare. The similar features of CSVD and MS in brain pathology, MRI findings and clinical presentation cloud the differential diagnosis in real-world busy clinic. New insight into previously unknown pathogenic aspects would offer new knowledge to better understand the white matter changes of these two origins and thus provide additional tools for differential diagnosis. 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Neuroimage 2018; 168:477–89. doi:10.1016/j.neuroimage.2016.11.031• Google Scholar 40. Ziemssen T, Akgün K, Brück W, et al. Molecular biomarkers in multiple sclerosis. J Neuroinflammation 2019; 16:272. doi:10.1186/s12974-019-1674-2• Google Scholar Contributors: F-DS, YW, LY, YOL formulated the concept, L-JZ, TD-C and KS searched the literature, all authors drafted the manuscript. Funding: We thank members of the Jing-Jin Neuroimmunology Team for helpful discussions; Dr. Luc Van Kaer and Elaine Shi for editing the manuscript; Work in the Authors’ Laboratories is supported in part by the National Science Foundation of China (Grants 91642205 and 81830038), funds of Advanced Innovation Center for Human Brain Protection, Capital Medical University, Beijing, China; funded by Tianjin Key Medical Discipline (Specialty) Construction Project (TJYXZDXK-004A). Competing interests: None declared. Provenance and peer review: Not commissioned; internally peer reviewed. Ethics statements Patient consent for publication: Not applicable. Ethics approval: Not applicable. Received: 17 August 2023 Accepted: 21 August 2023 First published: 12 September 2023 Online issue publication: 27 August 2024 Article versions Earlier version - 12 September 2023 You are viewing the most recent version of this article - 27 August 2024 Altmetric Altmetric scores reflect online attention articles receive, shown as a central number with coloured threads for each source Posted by 65 X users Highlighted by 65 platforms 26 readers on Mendeley See more details Dimensions Dimensions shows where and when an article has been cited, helping readers understand its academic influence over time 10 CITATIONS 10 Total citations 10 Recent citations 7.4 Field Citation Ratio n/a Relative Citation Ratio See more details Overview Abstract Introduction Conclusions and future perspective References Footnotes Publication history Metrics Responses Overview Abstract Introduction Conclusions and future perspective References Footnotes Publication history Metrics Responses Cookies and privacy We and our 216 partners store and access personal data, like browsing data or unique identifiers, on your device. 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14549	https://www.studocu.com/my/document/universiti-teknologi-mara/financial-management/tutorial-chapter-7-time-value-of-money/62993491	Tutorial EIB: Chapter 7 - Time Value of Money Concepts and Problems - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Tutorial EIB: Chapter 7 - Time Value of Money Concepts and Problems TUTORIAL Original title: Tutorial Chapter 7 Time Value of Money Course Financial Management (FIN 420) 999+documents University Universiti Teknologi MARA Academic year:2022/2023 Uploaded by: Anonymous Student Universiti Teknologi MARA Recommended for you 9 FIN420 Final Exam Question Paper - Financial Management July 2021 Financial Management Tutorial work 100% (11) 1 Tutorial question and answer on short term financing Financial Management Tutorial work 100% (8) 49 Test Bank Principles of Managerial Finance 12th Edition Gitman Financial Management Tutorial work 93% (15) 5 Cash Budget Tutorial for Gemilang, Kojima & Terus Maju Holdings Financial Management Tutorial work 100% (6) 11 FIN420 Income Statement & Financial Ratios Analysis - April 2011 Financial Management Tutorial work 100% (5) Comments Please sign in or register to post comments. Report Document Students also viewed CASH Budget Analysis Past Year Qs for FIN420 Exam Review 6-Month Analysis of 7-Eleven Malaysia's Market Trends 21/22 Conclusion - Trading Strategy Insights (FIN 552 Tutorial) Understanding Riba & Tawarruq in Islamic Finance - FIN 546 Tuto CASE Study 1: Prospects & PE Ratio Analysis of VS Industry (ECO 745) Customer Relationship Management in Logistics: Study S162002 2020 Related documents Fm-2019-sepdec-sample-a-amended 060220 2 Dilm report Financial history - nestle IND CASE Study Chapter 10 Tutorial Manual Solution Capital A - CG Report 2021 Preview text Chapter 7 : Time Value of Money EIB CHAPTER 7 : TIME VALUE OF MONEY QUESTION 1 a) Find the future value of RM100 deposit in the bank for ten (10) years if the interest rate to be earned is 5% per annum. (5 marks) b) If you borrow RM100,000 at an annual interest rate of 10% and you must pay off the loan in ten (10) years. What is the required annual payment? (5 marks) c) If you would like to accumulate RM 10,000 over the next five (5) years, how much you must deposit each year, given 7% interest rate. (5 marks) d) A dollar received today is worth more than a dollar received one (1) year from today. Explain this statement. (5 marks) QUESTION 2 a) Syirah must have RM30 in order to buy a piece of vacant land. She plans to save RM1,500 per year in her savings account with Amanah Bank that pays 12% interest rate compounded annually. How many years will it takes if the first saving being made a year from today? Do interpolation. (6 marks) b) Hazman borrows RM24,000 in the form of education loan from Bijak Bank. The payments are to be made in installment basis at the end of year. The tenure of loan is for 5 years at an interest rate of 10% per annum. Determine the amount that Hazman will have to pay each year for 5 years? (4 marks) QUESTION 3 a) A friend borrows from you RM85,000 now and promises to pay RM201,229 in the future with an interest rate of 9 percent annually. When will your friend pay you? (3 marks) b) You decide to save to buy a computer at the end of 5 years at a cost of RM5.975. You have decided to save 5 equal amounts with the first deposit being made today. How much is your yearly savings if the savings account is offering you 6 percent annually? (3 marks) c) You expect to receive RM1,000 each at the end of the next three years. You will deposit these payments into an account, which pays 10 percent compounded annually. What is the value at the third year? (4 marks) QUESTION 4 Chapter 7 : Time Value of Money EIB a) Dora borrowed RM200,000 to buy a house. The interest charged on the loan was 12% and she promised to repay the loan in 20 equal payments. What is the annual payment of the loan? (3 marks) b) Zoom Corporation expects to receive RM5 000 per year for 10 years and RM 6,000 per year for the next 15 years. What is the present value of this cash flow? The discount rate is 10%. (6 marks) c) Qistina invests RM10,000 in a saving account paying 4% interest per year compounded annually. How much is the money in the account after four years? (3 marks) d) The expected returns from an investment are as follows; RM15,000 per year for the first 5 years, RM20,000 at the end of 6 years and RM30,000 per year for year 7 and 8. If the interest rate is at 14%, what is the future value of the cash flows at the end of year 9? (6 marks) e) Explain what is annuity? (2 marks) QUESTION 5 a) Suppose you borrow RM18,000 education loan from Standard Charted Bank. Payments are being made in installments at the end of each year. You borrow for 7 years at an interest rate of 8% per annum. Determine the amount that you will have to pay for each year for 7 years? (4 marks) b) Yusri won a grand contest organized by Malaysian Academy. He was given two choices: i) To collect RM100,000 a year for 4 years, or ii) To collect RM50,000 a year for 10 years. If the amount is discounted at 10%, which alternative is the best? (8 marks) c) Nurul must have RM20,000 in order to continue her degree. To achieve the goal, Nurul plans to save RM1,200 per year in her savings account with a bank that pays 10% annual interest. How many years will it take if the first payment being made a year from today? (Do interpolation) (6 marks) Tutorial EIB: Chapter 7 - Time Value of Money Concepts and Problems Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Tutorial EIB: Chapter 7 - Time Value of Money Concepts and Problems Course: Financial Management (FIN 420) 999+documents University: Universiti Teknologi MARA Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Chapter 7 : Time Value of Money EIB10803 CHAPTER 7 : TIME VALU E OF MONEY QUESTION 1 a) Find the future value of RM100 deposit in the bank for ten (10) years if the interest rate to be earned is 5% per annum. (5 marks) b) If you borrow RM100,000 at an annual interest rate of 10% and you must pay off the loan in ten (10) years. What is the required annual payment? (5 marks) c)If you would like to accumulate RM 10,000 over the next five(5)years,how much you must deposit each year, given 7% interest rate. (5 marks) d) A dollar received today is worth more than a dollar received one (1) year from today. Explain this statement. (5 marks) QUESTION 2 a)Syirah must have RM30.000 in order to buy a piece of vacant land.She plans to save RM1,500 per year in her savings account with Amanah Bank that pays 12%interest rate compounded annually.How many years will it takes if t he first saving being made a year from today? Do interpolation. (6 marks) b) Hazman borrows RM24,000 in the f orm of education loan from Bijak Bank. The payments are t o be made in installment basis at the end of year. The tenure of loan is for 5 years at an interest rate of 10%per annum.Determine the amount that Hazman will have t o pay each year for 5 years? (4 marks) QUESTION 3 a)A friend borrows from you RM85,000 now and promises to pay RM201,229 in t he future with an interest rate of 9 percent annually. When will your friend pay you? (3 marks) b) You decide to save to buy a com puter at the end of 5 years at a cost of RM5.975.32. You have decided to save 5 equal amounts with the first deposit being made today. How much is your yearly savings if the savings account is offering you 6 percent annually? (3 marks) c)You expect to receive RM1,000 each at t he end of the next three years.You will deposit these payments into an account,which pays 10 percent compounded annually.What is the value at the third year? (4 marks) Chapter 7 : Time Value of Money EIB10803 QUESTION 4 a)Dora borrowed RM200,000 to buy a house.The interest charged on the loan was 12% and she promised to repay the loan in 20 equal payments. What is the annual payment of the loan? (3 marks) b)Zoom Corporation expects t o r eceive RM5 000 per year for 10 years and RM 6,000 per year for the next 15 years.What is the present value of this cash flow? The discount rate is 10%. (6 marks) c)Qistina invests RM10,000 in a saving account paying 4%interest per year compounded annually. How much is the money in the account after four years? (3 marks) d) The expected returns from an investment are as follows; RM15,000 per year f or the first 5 years,RM20,000 at the end of 6 years and RM30,000 per year for year 7 and 8.If the interest rate is at 14%, what is the future value of the cash flows at the end of year 9? (6 marks) e) Explain what is annuity? (2 marks) QUESTION 5 a)Suppose you borrow RM18,000 education loan from Standard Charted Bank.Payments are being made in installments at the end of each year. You borrow for 7 years at an interest rate of 8%per annum.Determ ine the amount that you will have to pay for each year for 7 years? (4 marks) b) Yusri won a grand contest organized by Malaysian Academy. He was given two choices: i) To collect RM100,000 a year for 4 years, or ii) To collect RM50,000 a year for 10 years. If the amount is discounted at 10%, which alternative is the best? (8 marks) c)Nurul must have RM20,000 in order to continue her degree.To achieve the goal,Nurul plans to save RM1,200 per year in her savings account with a bank that pays 10%annual interest. How many years will it take if the f irst payment being made a year from today? (Do interpolation) (6 marks) 1 out of 2 Share Download Download More from:Financial Management(FIN 420) More from: Financial ManagementFIN 420Universiti Teknologi MARA 999+documents Go to course 48 FIN420- Company Analysis MCD & Domino'S Financial Management Mandatory assignments 100% (35) 12 FIN420 - Financial Management (Question & Answer)Financial Management Practical 97% (35) 13 FINANCIAL Ratio Analysis (SIME Darby Property)Financial Management Mandatory assignments 100% (16) 9 FIN420 Final Exam Question Paper - Financial Management July 2021 Financial Management Tutorial work 100% (11) More from: Financial ManagementFIN 420Universiti Teknologi MARA999+documents Go to course 48 FIN420- Company Analysis MCD & Domino'S Financial Management 100% (35) 12 FIN420 - Financial Management (Question & Answer) Financial Management 97% (35) 13 FINANCIAL Ratio Analysis (SIME Darby Property) Financial Management 100% (16) 9 FIN420 Final Exam Question Paper - Financial Management July 2021 Financial Management 100% (11) 21 FIN420 Group Assignment (Group 1) Financial Management 100% (9) 68 FIN420 CASE Study ANALYSIS OF CORPORATE FINANCIAL PERFORMANCE : A TREND AND COMPARATIVE STUDY OF ASTRO MALAYSIA HOLDINGS BERHAD AND MEDIA PRIMA BERHAD Financial Management 100% (9) Recommended for you 9 FIN420 Final Exam Question Paper - Financial Management July 2021 Financial Management Tutorial work 100% (11) 1 Tutorial question and answer on short term financing Financial Management Tutorial work 100% (8) 49 Test Bank Principles of Managerial Finance 12th Edition Gitman Financial Management Tutorial work 93% (15) 5 Cash Budget Tutorial for Gemilang, Kojima & Terus Maju Holdings Financial Management Tutorial work 100% (6) 9 FIN420 Final Exam Question Paper - Financial Management July 2021 Financial Management 100% (11) 1 Tutorial question and answer on short term financing Financial Management 100% (8) 49 Test Bank Principles of Managerial Finance 12th Edition Gitman Financial Management 93% (15) 5 Cash Budget Tutorial for Gemilang, Kojima & Terus Maju Holdings Financial Management 100% (6) 11 FIN420 Income Statement & Financial Ratios Analysis - April 2011 Financial Management 100% (5) 2 Pro Form Balance Sheet Tutorial - Analysis & Exercises Financial Management 100% (5) Students also viewed CASH Budget Analysis Past Year Qs for FIN420 Exam Review 6-Month Analysis of 7-Eleven Malaysia's Market Trends 21/22 Conclusion - Trading Strategy Insights (FIN 552 Tutorial) Understanding Riba & Tawarruq in Islamic Finance - FIN 546 Tuto CASE Study 1: Prospects & PE Ratio Analysis of VS Industry (ECO 745) Customer Relationship Management in Logistics: Study S162002 2020 Related documents Fm-2019-sepdec-sample-a-amended 060220 2 Dilm report Financial history - nestle IND CASE Study Chapter 10 Tutorial Manual Solution Capital A - CG Report 2021 Get homework AI help with the Studocu App Open the App English Malaysia Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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14550	https://link.aps.org/accepted/10.1103/PhysRevD.109.104066	This is the accepted manuscript made available via CHORUS. The article has been published as: Extremal Kerr black hole dark matter from Hawking evaporation Quinn Taylor, Glenn D. Starkman, Michael Hinczewski, Deyan P. Mihaylov, Joseph Silk, and Jose de Freitas Pacheco Phys. Rev. D 109 , 104066 — Published 21 May 2024 DOI: 10.1103/PhysRevD.109.104066 Extremal Kerr Black Hole Dark Matter from Hawking Evaporation Quinn Taylor, Glenn D. Starkman, Michael Hinczewski, and Deyan P. Mihaylov Department of Physics/CERCA/Institute for the Science of Origins, Case Western Reserve University, Cleveland, OH 44106-7079 – USA Joseph Silk Institut d’Astrophysique de Paris (UMR7095: CNRS & UPMC- Sorbonne Universities), F-75014, Paris, France William H. Miller III Department of Physics and Astronomy, The Johns Hopkins University, Baltimore MD 21218, USA and Beecroft Institute of Particle Astrophysics and Cosmology, Department of Physics, University of Oxford, Oxford OX1 3RH, UK Jose de Freitas Pacheco Universit´ e de la Cˆ ote d’Azur - Observatoire de la Cˆ ote d’Azur, 06304 Nice Cedex - France (Dated: April 10, 2024) The Hawking process results in a monotonic decrease of the black hole mass, but a biased random walk of the black hole angular momentum. We demonstrate that this stochastic process leads to a significant fraction of primordial black holes becoming extremal Kerr black holes ( ekbh s) of one to a few Planck masses regardless of their initial mass. For these ekbh s, the probability of ever absorbing a photon or other particle from the cosmic environment is small, even in the cores of galaxies. Assuming that ekbh s are stable, they behave as cold dark matter, and can comprise all of the dark matter if they are formed with the correct initial abundance. Keywords: dark matter, black hole, extremal, primordial, mass, angular momentum, stochastic, Kerr I. INTRODUCTION Black holes ( bh s) have long been considered a promising class of dark matter candidates [1–11]. To constitute all the dark matter, they must be formed before Big Bang Nucleosynthesis ( bbn ), i.e. before the temperature of the Universe has fallen below ∼ 1 MeV, and endure from their epoch of formation to the present [6, 11, 12]. They must be stable, or at least have lifetimes much longer than the current age of the Universe. Given evidence for the role of dark matter in the Uni-verse back to the epoch of bbn , plausible candidates include bh s formed in the early Universe, with masses therefore unrelated to those of the stellar mass black holes that emerge from core-collapse supernovae. Most recent attention has focused on bh s of large mass, M ≫ 10 15 gbecause they are expected to be stable, with decay life-times much longer than the current age of the Universe . As first pointed out in , if bh evaporation leaves behind Planck mass stable relics, then they could be the dark matter. Barrow et al. discussed this possibility of Planck relics from bh evaporation and concluded that a substantial relic density could remain if the initial mass function of primordial bh s was relatively narrow, and this was followed up by others (e.g. ). Others argued more recently [15–17] that bh decay inevitably resulted in a Planck relic. In this paper we argue that the Hawking process itself is likely to “strand” a small fraction of all primordial bh sas extremal Kerr black holes ( ekbh ), which are expected to be stable. Recently, Dai and Stojkovic have ar-gued that ekbh s undergo super-radiance and therefore are not stable, even though they do not undergo the Hawking process. We comment on this possibility below. One way or another, the scenario we present should be instructive for how bh decay even in semiclassical Gen-eral Relativity may result in naturally Planck relic dark matter if Planck relics are stable. bh s are expected to decrease in mass via the Hawking evaporation process [19–28]. For a Schwarzschild (i.e. uncharged, non-rotating) bh of mass Mbh , the Hawking temperature is TSchw = M 2Pl 8πM ≃ 1.1 × 10 13 GeV g M (1) (in units where ℏ = c = k = 1) and the horizon radius is RSchw = 2MM 2Pl . (2) Assuming that the spectrum of radiation is that of a black body, this results in a Hawking luminosity (for one mass-less degree of freedom) LSchw = 4 πR 2Schw σ T 4Schw = M 4Pl 15360 πM 2 , (3) and a Hawking lifetime of tSchw = 5120 π3M 3 M 4Pl = 4 M 10 9g 3 s . (4) The Hawking temperature characterizes the effective temperature as inversely proportional to PBH mass and 2is constrained by soft gamma ray constraints at the ∼ 10 17 gm PBH mass scale, incorporating appropriate spec-tral corrections due to nonthermal aspects of the com-plex emission processes . The direct gamma ray sig-nals associated with Hawking evaporation include the 511 keV gamma-ray line, produced by electron-positron pair-annihilation. The INTEGRAL detection of the Large Magellanic Cloud provides one of the strongest bounds attainable with present observations . Similar limits come from Voyager constraints as well as the diffuse soft gamma x-ray background . The temperature of a black hole is not, however, deter-mined exclusively by its mass [33–36]. For a rotating (but uncharged) Kerr black hole with angular momentum J,the temperature is TKerr = M 2Pl 4πM p1 − a2 ∗ 1 + p1 − a2 ∗ , (5) where a∗ ≡ JM 2Pl M 2 . (6) If the Hawking process involved only photons (they pre-dominate), was entirely in the orbital angular momentum zero channel (it predominates), and was unbiased as to whether each photon was emitted spin-aligned or anti-aligned with the black-hole spin (it is not, but we will return to this point), then the evolution of J would be an unbiased random walk with step-size 1 (i.e., ℏ). We can investigate the fate of a black hole under this simplified assumption – does it remain Schwarzschild like if it begins uncharged and with J = 0. Since TSchw ≪ M for M ≫ MPl ≃ 2.17 × 10 −5 g, a radiating Schwarzschild black hole will emit N particles N (M ) ≃ 1 γ Z MMi −dMTSchw = 4πγM 2Pl M 2 i − M 2 , (7) where Mi is the black hole’s initial mass, M is the black hole’s current mass, and γ = π4 30 ζ(3) ≈ 2.70 . (8) In the meantime, the black hole’s spin will undergo an un-biased random walk, and the expected root mean square angular momentum will grow to J2 1/2 = pN (ϵ) = 2 MPl r πγ M 2 i − M 21/2 (9) A black hole becomes an ekbh once a⋆ = 1 or when JM 2Pl = M 2. Using Eq. (9), and for Mi ≫ MPl , this oc-curs when M ≃ 1.5 √MiMPl . In other words, the typical Schwarzschild black hole would be expected to become an ekbh by the time its mass falls from some initial value Mi to M ≃ √MiMPl !In order for these ekbh s to be stable against Hawking radiation before bbn , we would require tSchw ≲ 1 s, im-plying Mi ≲ 10 7 g. Essentially every single Schwarzschild black hole ever formed with initial mass between MPl and 10 7 g would have become an ekbh by 1 s under this as-sumption of unbiased emission of photons. Hawking particle emission is not however expected to be unbiased with respect to the alignment between the spin of the photon and the spin of the black hole. Rather, the emission is calculated to be biased to prefer the emis-sion of photons with spins parallel to that of the black hole [19, 24]. This will reduce the probability of a black hole evolving to extremality. According to , for a∗ ≪ 1, the probabilities of emit-ting photons (in the s-state) with spin (and thus angular momentum) aligned/anti-aligned with J are P↑↓ (a∗) = 12 ∓ a∗ + O(a2 ∗ ). (10) To evaluate the efficacy of this bias in suppressing the formation of ekbh , we must extend these approximate formulae to \|a∗\| = 1. Consider P↑↓ (a∗) = 12 ∓ a∗ ± a∗\|a∗\| 2 . (11) This correctly reduces to Eq. (10) for \|a∗\| ≪ 1, is mono-tonic on a∗ ∈ [−1, 1], and gives P↑↓ (∓a∗) = 0 as required. Of course, a Kerr black hole doesn’t just emit photons but will emit all particles with masses m < T Kerr . They will similarly preferentially be emitted in the s-wave, and with spins preferentially aligned with the angular mo-mentum of the black hole [19, 24]. For simplicity we focus on the s-wave photons with the expectation that we will correctly capture the qualitative behaviour. The authors of [28, 35, 38] have studied the effect of this spin-down bias on the expected evolution of a∗. They conclude that even black holes with initial \|a∗\| very near 1 would be expected to evolve to very near a∗ = 0 once Hawking evaporation becomes significant. However, the question we address here is not the expected evolution of the black hole, but the evolution of a large number of black holes governed by the stochastic nature of the Hawking process. Unsurprisingly, spin-down bias sub-stantially reduces the probability of reaching extremality, but by how much? Do enough black holes become ekbh sfor these to constitute the dark matter, and what would we expect their mass distribution to be? II. THE DISTRIBUTION OF EKBH To answer these questions, we simulated the evolution of a large number of initially Schwarzchild black holes as they evaporated through the Hawking process. Starting from an initial mass of M (t = 0) ≫ MPl , and J(t = 0) = 0, each black hole was followed until either: a. the mass of the bh reaches some cut-off value Mcut [16, 33, 39]; 3b. the black hole becomes extremal i.e. \|J\| ≥ (M/M Pl )2 (at which point we set \|J\| = ( M/M Pl )2). We assume that the terminal M = Mcut black holes will decay away, and focus on the properties of the ekbh 1.As expected the black holes remained nearly Schwarzchild (i.e. a2 ∗ ≪ 1) for the majority of their evolution until M fell to just a few times MPl , i.e. un-til the the last few Hawking particle emissions. Using 2 × 10 9 simulations of black holes, we found that approx-imately 21.70% of them became extremal, independent of the value of M0, so long as M0 ≫ MPl . P (Mext ), the probability density of a black hole be-coming extremal at mass Mext , is well-approximated by P (Mext ) = KM ext √ν 1 − ⌊M 2ext ⌋ ∼ M 2 !2 exp − ν⌊M 2ext ⌋2 M 4Pl (12) where K = 8√πγ , (13a) ν = ln " 1 + 2M 2Pl ∼ M 2 − M 4Pl ∼ M 4 , (13b) ∼ M = 12 Mext + r M 2ext + γ 2π M 2Pl ! (13c) and ⌊x⌋ is the greatest integer ≤ x. Black holes with √n − 1 ≤ Mext /M Pl < √n have J = n, for n ∈ Z>1.The final distribution of ekbh masses Pc(Mext ), where Pc(Mext ) = P (Mext ) R ∞ 1 P (M ′ ext ) d M ′ ext (14) is shown in Fig. 1 for Mcut = MPl . The jagtooth shape of the distribution seems extraordinary, however in Ap-pendix A we show that this can be understood as emerg-ing from the doubly stochastic process of black hole mass loss and black hole angular momentum-biased random walk. We see that the distribution is heavily weighted toward the lightest possible ekbh , and so the fraction of initially Schwarzschild black holes that reach Kerr extremality is a steeply decreasing function of Mcut . The average mass of the ekbh is M ≃ 1.3 MPl . 1The code can be found at BH_Extremal . The outline of the code is as follows: At each time step t→t+δt , the mass M(t) is updated via M(t+δt ) = M(t)−δM (t), where δM (t)=x(t)TKerr (M(t), a ∗(t)). Here x(t) is a random value sampled from the Planck distribution p(x) = (1 /2ζ(3)) x2/(ex−1), and TKerr is given by Eq. (5). Meanwhile, the angular momentum Jof the bh is changed by ±1 with probabilities given by Eq. (11). We have checked that our results don’t depend materially on whether P↑↑/↑↓(a∗) is cal-culated using M(t) or M(t+δt ). 1 √2 √3 2 √5 √6 √7 √8 3 10 −8 10 −5 10 −2 10 1 Mext /M Pl Pc simulated data integrated weight analytical prediction Fig. 1: The mass distribution function Pc(Mext ) (scale on left vertical axis) of extremal Kerr black holes emerg-ing from the Hawking-process evolution of an ensemble of initially Schwarzshild black holes of mass M0 ≫ MPl (al-lowing only for s-wave photon emission). Numerical re-sults from 2 × 10 9 ekbh s (shaded histogram) are well de-scribed by Eq. (12) (red curve), which is the result of an approximate analytic treatment of the evolution of the en-semble (see Appendix A)). Black dashed vertical lines mark Mext /M Pl = √n, for n ∈ Z>0, and the black holes with √n − 1 ≤ Mext /M Pl < √n have J = n. Since consequently a2 ∗ 1 except at Mext /M Pl = √n, which is unphysical, we also plot the integrated weight between √n − 1 and √n and display the values as black dots. Of note is the structure of the distribution of final masses. The distibution is a sawtooth that monotically increases toward a local maximum as M increases from √n − 1 MPl to √n M Pl (for integers n ≥ 2), then falls vertically to a lower value for the next sawtooth maxi-mum. This distribution can be explained by considering a black hole in the last time step before it becomes ex-tremal. With each Hawking photon emitted, the value of J changes by ±1 while the mass changes by an amount drawn randomly from the Planck distribution of the ap-propriate temperature. With the final photon, the black hole spin changes from J = n−1 to J = n (in the spin up case), and the black hole becomes extremal with a mass M ≤ √n M Pl . The evolution of a⋆ as a function of M is shown in Fig. 2. Because we have quantized J but not Mbh , the black holes with √n − 1 ≤ Mbh /M Pl < √n and J = n have a2 ∗ This is unphysical, and reflects the need to prop-erly deal with the quantum mechanical nature of the de-cay. One way to resolve this issue is to take the final mass of the black hole to be √n M Pl so that a∗ = ±1exactly. For illustrative purposes we therefore also plot in Fig. 1 the integrated weight under each “tooth” of the mass distribution function, which corresponds to this res-olution (ignoring the low-likelihood possibility that the final photon jumps the black hole across an entire tooth). While the final black holes are extremal and no longer radiating, if they were to absorb any energy they would 412345 −1 01 Mbh /M Pl a∗ = J/M 2 non-extremal bh ekbh Fig. 2: The evolution of a⋆ for 25 black holes, each with an initial mass of 5 MPl , as they evaporate and either become a ekbh (blue) or reach M = MPl (orange). be kicked out of extremality, begin to radiate again, and would once again be more likely to decay away than re-turn to extremality. This would be a problem if these black holes were of a much larger mass, but with a geo-metric cross section σ = 16 πM 2 bh /M 4Pl these black holes are extremely weakly interacting, and are unlikely to ever be knocked out of extremality once they reach it . For example, the last time that a typical primordial 10 9 gSchwarzschild black hole would have even encountered a cosmic microwave background ( cmb ) photon was well before recombination ; see Fig. 3. III. EKBHS AS DARK MATTER The formation of black holes during the early Universe has been a rich area of study with many proposed mech-anisms [42–56]. One common mechanism for the forma-tion of these primordial black holes ( pbh s) comes from density perturbations in the early Universe [2, 51, 57–69]. A black hole will form if the density of the perturbed re-gion reaches some critical density δc. The resulting bh mass will be proportional to the horizon mass MH ∼ 5 × 10 20 MPl t 10 −23 s (15) In order for ekbh s to be the dark matter, and in order not to disturb bbn , they need to have been in place be-fore the temperature of the Universe has fallen to 1 MeV, when MH ∼ 5 × 10 43 MPl . However, since a bh will only evaporate through Hawking emission when its Hawking temperature is greater than the ambient temperature [21, 70], the initial black holes must be light enough that TH (Mi) ≫ 1 MeV, i.e. Mi ≪ 5 × 10 20 MPl . The upper limit on Mi is lowered even more due to the decay life-times τ of ekbh s being constrained to less than 1 second. Thus Mi ≤ 10 13 MPl .1 MPl 10 15 g 1 M hubble time 10 2 10 5 10 8 10 11 10 14 10 17 10 −62 10 −32 10 −2 10 28 10 58 t [s] δt [s] Fig. 3: The time δt between cmb photons hitting different mass ekbh s as a function of the age of the Universe (solid curves), compared to the Hubble time (dotted curve). Naively one might then simply solve for the initial number density of black holes ni bh required to constitute all of the dark matter using ni bh = Ωdm ρc Pext M (1 + zi)3 (16) where the current average density of dark matter is ρdm = Ω dm ρc ≃ 2.5 × 10 −30 g/cm 3, zi is the redshift of bh formation, Pext ≈ 0.11 is the probability of a primor-dial black hole becoming extremal, and M ≈ 10 −5 g is the average mass of the ekbh that ultimately remains after bh decay. For ekbh s made at t = 10 −22 s after the Big Bang, the initial number density of black holes is ni bh = 2 .27 × 10 12 cm −3. Before these black holes be-come extremal, however, they radiate away almost all their mass and contribute to the overall radiation energy density of the Universe 2.dρrad dt = − dMbh dt ni bh aai −3 − 4ρrad ˙aa (17) where − dMbh dt = σA bh T 4 bh (18) is the energy radiated away from the black hole. Because decaying bh s remain nearly-Schwarzschild (i.e. a2 ∗ ≪ 1) throughout their lifetime until nearly 2For simplicity, we will take the number of effective relativistic de-grees of freedom of the Universe to always be equal to the number for the Standard Model of particle physics at high temperature, 106 .75. 5their final moments, and because each Hawking parti-cle typically reduces the mass by only a small fraction (i.e. Tbh ≪ Mbh ), we can approximate the time evolu-tion of the bh mass as nearly the expected evolution of the mass. For a Schwarzschild bh , this gives MSchw (t) = M 3Schw (ti) − M 4Pl 5120 π (t − ti) 1/3 (19) To find ρ(t), we utilize the Friedmann equation − ˙TT !2 = 8π 3M 2Pl (ρrad + ρm) , (20) where we used a/a 0 = T0/T . Taking the time derivative of both sides ¨TT − 3 ˙TT !2 = − 4π 3M 2Pl Mbh (t) ni bh TTi 3 . (21) Solving Eq. (21) numerically gives T (t) for t < τ , en-abling us to find ρrad (t = τ ), as a function of the initial mass and number density of the black holes. (For sim-plicity we assume that all the black holes were formed at the same time with a unique mass.) 1 = ρm(teq ) ρrad (teq ) ≃ ρbh (t = τ ) ρrad (t = τ )1 + zτ 1 + zeq (22) where we have substituted in ρbh for ρm since the matter energy density is dominated by the energy density from dm . Eq. (22) is then used to find the values for ni that satisfy the equality. For a single set of black holes made at time t = 10 −15 s, the behavior of ρbh /ρ rad as the black holes evaporate is shown in Fig. 4. 10 −14 10 −9 10 −4 10 1 10 6 10 11 10 −14 10 −7 10 0 10 7 age of the universe, [s] ρbh (t)/ρ rad (t) 10 8MPl 10 10 MPl 10 12 MPl Fig. 4: The evolution of ρbh (t)/ρ rad (t) as black holes formed at ti= 10 −15 s with an initial mass Mievaporate. Here ρrad is taken to be the radiation energy density in a radiation-dominated Universe (with the 106.75 effective relativistic de-grees of freedom of the Standard Model at high temperature). IV. CONCLUSIONS AND CONSIDERATIONS We have shown that the biased random walk of a black hole’s angular momentum can result in extremal Kerr black holes when the mass of these black holes is ini-tially of order MPl . If the minimum mass of a stable black hole is Mi < 10 13 MPl , then of the initial number of black holes simulated around 21.70% become extremal Kerr black holes with average mass ¯M = 1 .3 MPl . Us-ing this percentage and average mass, we have calculated the initial density, as a function of initial mass, of ekbh sneeded to be the dark matter in the Universe today. For our analysis, however, we have considered that the contribution of the black hole entropy is only due to the outer horizon of the Kerr black holes. References [71, 72] show that the inner horizon of black holes ef-fects the entropy of the black hole, ultimately resulting in the temperature of a black hole being only propor-tional to mass. This will make all black holes with inner horizons radiate like a Schwarzchild black hole, making them unstable. In , it is argued that waves incident on black hole horizons undergo super-radiance and take away energy from the black hole. This would also cause the ekbh to radiate and make them unsatisfactory as dark matter candidates. However, the calculation of assumed a single isolated ekbh , whereas dark matter of mass μ M Pl would have a cosmological average abun-dance of 2 /μ ×10 −19 cm −3, and thus a mean separation of just ∼ 15 km. Their individual Kerr metrics would dom-inate the instantaneous local geometry at most over that distance. The prospect that the dark matter is stabilized by its own many-body effects is intriguing. While ekbh s are a good candidate for the dark matter in the Universe, they are a troubling candidate. Their small geometric cross-section is a double-edged sword: while this ensures the stability of ekbh ’s due to their low interaction rate with other objects in the Universe, it also means that detecting such dark matter would also be extremely difficult, though perhaps not impossible[9, 40, 73–75]. Acknowledgments G. D. S. and Q. T. acknowledge support from DOE grant DESC0009946; Q.T. from the GEM Fellowship. D. P. M. acknowledges support from NASA ATP grant RES240737. 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Koutsangelas, New Mass Window for Primordial Black Holes as Dark Matter from Memory Burden Effect (2024), arXiv:2402.14069 [astro-ph, physics:gr-qc, physics:hep-ph, physics:hep-th], URL . Appendices Appendix A: Approximating the Extremal Mass Distribution We want to approximate analytically the distribution of extremal black hole masses that emerges from the nu-merical evolution of an ensemble of ekbh ’s (blue dots in Fig. 1). The evolution of a black hole is driven by two stochastic processes: the random decrease in the black hole’s mass, and the biased random walk of the black hole’s angular momentum. 3 3In principle, we should add a third random process – the random walk of the black hole’s electromagnetic charge. Indeed, to really understand the formation of extremal black holes near the Planck mass we might rightly consider the random walk of all the black hole’s SU (3) ×SU (2) ×U(1) gauge charges, however we defer those complications to future considerations. Note that for convenience, in this appendix we set MPl = 1 in addition to ℏ = c = k = 1, and take the minimum mass of a black hole to be Mcut = 1. Approximating the mass trajectory We start by defining an affine parameter λ that increases monotonically as the black hole decays. The black hole starts out at λ = 0 with mass M (0) = M0 and angular momentum J(0) = 0. Our first approximation is related to the evolution of the mass of the black hole M (λ). Though this evo-lution is stochastic in the simulations, M (λ) stays in the vicinity of a deterministic curve M (λ), with only small discrepancies until near the end of the evolution. For the majority of the black hole’s Hawking-process evolution, M (λ) ≫ 1, hence T (λ) ≪ M (λ). Also \|a⋆(λ)\| = \|J(λ)/M 2(λ)\| ≪ 1, so the black hole is nearly Schwarzschild. The individual decrements in M (λ), δM (λ), are each stochastic draws from a black body distribution of tem-perature T (λ). However, since a large number of these decrements are needed to substantially decrease M (λ), the stochasticity of δM (λ) gets averaged out. We can thus replace the stochastic mass decrement by a deter-ministic Schwarzschild approximation: δM (λ) ≈ γT (λ) , (A1) where γ = π4 30 ζ(3) and T (t) = 18πM (t) . This deterministic approximation Eq. (A1) for the black hole mass decrement allows us to define λ in such a way that M (λ) obeys a simple differential equation: dM (λ)dλ = γ 8πM (λ) , (A2) which has the solution M (λ) = M0 s 1 − γλ 4πM 20 . (A3) We will always restrict ourselves to λ < 4πM 20 /γ , so that M (λ) as given by Eq. (A2) is real and positive. We can easily relate the affine parameter λ to the time t, because for a Schwarzschild black hole Eq. (3) gives dM dt = − 115360 π 1 M 2 . (A4) Comparing to Eq. (A1) dλ dt = 1920 π2γM 0 s 1 − γλ 4πM 20 , (A5) 91 2 3 4 5 black hole mass data fit \|M fit − M data \| /M fit −100 −80 −60 −40 −20 0 10 −3 10 −2 10 −1 time steps fractional difference Fig. 5: The average mass values of 488 ekbh trajectories plotted against Eq. (A3). so that t(λ) = M0 240 πγ 2 1 − s 1 − γλ 4πM 20  (A6) Near the end of the black hole’s evolutionary trajec-tory, the mass decrements are no longer small compared to the black hole mass, and it is no longer necessarily the case that \|a⋆(λ)\| ≪ 1, indeed for an ekbh we have \|a⋆(λ)\| → 1. There are therefore noticeable differences between the actual M (λ) and M (λ). However, in order to enable an analytical approach, we will ignore these differences and assume M (λ) ≈ M (λ). Trajectory end time distribution Unlike the evolution of the black holes mass, the evolu-tion of the angular momentum cannot be approximated as deterministic. With each photon emission, the spin J changes by ±1 with probability given by (11), which we can rewrite in terms of the probability to emit a photon with spin up (+1) or down ( −1): ρ↑(J, M ) =  (J − M 2)2 2M 4 J ≥ 0(J − M 2)2 − 2J2 2M 4 J < 0(A7) ρ↓(J, M ) = 1 − ρ↑(J, M ) (A8) In the simulations, the black hole evolutionary trajectory starts with J = 0 at λ = 0 and ends at some λf when \|J\| ≥ M 2 (or M = 1). We can study this in the ap-proximation that M changes deterministically as M (λ). We can model the terminal-λ distribution with a survival probability, Σ( λ), or the cumulative probability that the spin trajectory has not reached λf . This is related to the probability density P(λ) of trajectory termination λ, i.e. λf , by Σ( λ) = 1 − Z λ 0 dλ′ P(λ′) (A9) Note that Σ(0) = 1. If we find Σ( λ), then P(λ) = −dΣ( λ)/dλ. We can then change variables to get the probability density at end times of final masses, P(M ). The survival probability satisfies dΣ( λ)dλ = −k(λ) Σ( λ), (A10) where k(λ) is the rate of loss, or the rate at which a tra-jectory reaches the end conditions at λ. In other words, the fraction of surviving trajectories will be diminished by k(λ)Σ( λ) in times between λ and λ + d λ. With the boundary conditions specified above, Σ( λ) = exp − Z λ 0 dλ′ k(λ′) ! , (A11) and hence P(λ) = − dΣ( λ)dλ = k(λ) exp − Z λ 0 dλ′k(λ′) ! . (A12) We are only interested in λ < λ max = (4 π/γ ) ( M 20 −1), so that M (λ) > 1. Because only about 21% of trajecto-ries end up as extremal, we can approximate the expo-nential term in (A12) with 1. In the code, the condition of extremality is \|J\| ≥ M 2, if the final spin jump occurs between increment λ and λ+δλ the mass that determines ρ↑ is the mass M (λ) ≈ M (λ)and not M (λ + δλ ). To distinguish these two slightly dif-ferent masses, we will denote the ultimate mass as Mext ,and the penultimate mass as ∼ M (Mext ). The expected difference between them is dM ( ∼ M ) = γT ( ∼ M ) = γ 8π ∼ M (A13) Thus ∼ M (Mext ) = Mext + γ 8π ∼ M . (A14) so that ∼ M (Mext ) = 12 Mext + r M 2ext + γ 2π (A15) We can determine λ(Mext ) from (A3), which we write as λ(Mext ) = 4πγ M 20 − M 2ext − 1. (A16) Using the change of variable theorem for probability densities, the distribution of final masses P (Mext ) is: P (Mext ) = dλ dMext P(λ(Mext )) ≈ 8πM ext γ k(λ(Mext )) . (A17) 10 M = 1 .1MPl −1 −0.5 0 0.5 1 0 5 10 15 energy E(J) M = 2 .1MPl −4 −2 0 2 4 angular momentum J M = 3 .1MPl −8 −4 0 4 8 Fig. 6: Results of spin dynamics simulations at three different fixed mass values, with reflecting boundary conditions at the smallest and largest possible values of the spin, averaged over both even and odd time steps. The points show E(J) = −log p(J), where p(J) is the distribution of spin values. The curves show parabolic fits to the simulation data. Approximating the trajectory loss rate There are two major contributions to the loss function k(λ) at late λ: i) the spin J = ⌊M 2ext ⌋, and the an-gular momentum increases J → J + 1; ii) the spin J = −⌊ M 2ext ⌋, and the angular momentum decreases J → J − 1. By symmetry, both of these contributions have the same probability, so we can focus on the positive spin case and multiply by a factor of 2. Since Mext is slightly smaller than ∼ M , we know that ⌊M 2ext ⌋ ≤ ⌊ ∼ M 2 ⌋, with the two floor values usually equal. However there are narrow ranges of λ when ⌊ ∼ M 2 ⌋ = ⌊M 2 f ⌋ + 1, and in these ranges J can go as high as ⌊ ∼ M 2 ⌋. During these times, we also have possible tran-sitions from J = ⌊ ∼ M 2 ⌋ → ⌊ ∼ M 2 ⌋ + 1 that would end the trajectory (and their negative spin counterparts). How-ever, as we will see below, the probability of being at J = ⌊ ∼ M 2 ⌋ = ⌊M 2ext ⌋ + 1 will be much smaller than J = ⌊M 2ext ⌋, so we can safely ignore the contribution of these transitions. Thus the loss rate k(λ) can be expressed as: k(λ) = 2 pλ(J = ⌊M 2ext ⌋)ρ↑(J = ⌊M 2ext ⌋, ∼ M )= pλ(J = ⌊M 2ext ⌋) (⌊M 2ext ⌋− ∼ M 2 )2 ∼ M 4 . (A18) ρup is given by (11), and pλ(J) is the probability to be at spin J at affine parameter λ. The spin dynamics can be seen as a discrete random walk under a force ρup − ρdown that is approximately given by −2J/M 2(λ) for small J.This acts like a Hookean restoring force, biasing the walk toward J = 0. Since M 2(λ) decreases with λ, the ef-fective spring constant gradually increases. Because the change in M (λ) per λ increment is small, and the frac-tion of trajectories lost per λ step is also small, we will assume the system is roughly in quasi-equilibrium. In other words, we will approximate pλ(J) by the equilib-rium distribution of J that would have occurred if the mass was fixed at a value M = M (λ), and the boundary conditions at the smallest and largest allowed spin values were reflecting rather than absorbing. If the spin dynam-ics were a continuous Wiener process, this equilibrium distribution under a Hookean force would be a Gaussian. In the actual system the discreteness of the dynamics complicates matters and prevents a simple closed form solution. However inspired by the continuum case, we will assume a Gaussian ansatz, pλ(J) ∝ exp( −Eλ(J)), where Eλ(J) = Eλ(0) + ν(M (λ)) J2 for some function ν(M ). In Fig. 6, we show simulation results for three differ-ent fixed values of the mass, where the points represent E(J) = − log p(J) from the numerically determined equi-librium distribution p(J) (under reflecting boundary con-ditions). In each case a parabola (red curve) provides an excellent fit, validating the Gaussian ansatz. Note that the spin dynamics has the property that if J = 0 at λ = 0 then only even values of J are possible at even steps in λ, and odd values of J at odd steps in λ. Thus, strictly speaking, the distribution forever oscillates between even and odd values from λ-step to λ-step. We simplify the situation by averaging over both even and odd λ-steps, which gives an approximately Gaussian stationary distri-bution. To find an expression for ν(M ), we note that the sta-tionary distribution should satisfy local detailed balance: in equilibrium the probability of observing a transition from J to J + 1 should be the same as the probability of observing a transition from J + 1 to J. Namely, pλ(J)ρ↑(J, M ) = pλ(J + 1) ρ↓(J + 1 , M ). (A19) 11 where M = M (λ). This in turn implies that pλ(J + 1) pλ(J) = (M 2 − J)2 M 4 + 2( J + 1) M 2 − (J + 1) 2 (A20) for 0 ≤ J < ⌊M 2⌋. We can thus solve for the successive differences Eλ(J + 1) − Eλ(J) = ln M 4 + 2( J + 1) M 2 − (J + 1) 2 (M 2 − J)2 (A21) for 0 ≤ J < ⌊M 2⌋. To estimate ν(M ) we only need to know Eλ(1) − Eλ(0) = ν(M ), and hence plug in J = 0 into Eq. (A21) to find ν(M ) = ln M 4 + 2 M 2 − 1 M 4 . (A22) Since Eλ(J) is not exactly parabolic, different choices of J would give somewhat different estimates of ν(M ), but the J = 0 expression is the simplest, and works well compared to the numerics. In Fig. 7 we show ν(M ) from Eq. (A22) versus numerical estimates (based on the same type of simulations as in Fig. 6). 1 2 3 401234567M ν(M) Fig. 7: Estimates of ν(M ) based on spin dynamics simula-tions (points) versus the analytical expression in Eq. (A22) (blue curve). With proper normalization, the expression for pλ(J) is given by pλ(J) = e−ν(M )J2 P⌊M 2⌋ J′=−⌊ M2⌋ e−ν(M )J′2 ≈ r ν(M ) π e−ν(M )J2 , (A23) where we have approximated the sum in the denominator by an integral over J′, letting the bounds go to ±∞ since the contribution from the Gaussian tails is negligible. Analytical expression for final mass distribution Putting together Eqs. (A17), (A18), and (A23), we get an expression for the final mass distribution P (Mext ) = 8√πγ Mext q ν( ∼ M ) 1 − ⌊M 2ext ⌋ ∼ M 2 !2 e−ν( ∼ M)⌊M2ext ⌋2 , (A24) where ν( ∼ M ) is given by Eq. (A22) and ∼ M (Mext ) is given by Eq. (A15). Note that since P (Mext ) d Mext is the prob-ability of a black hole having a final mass between Mext and Mext + d Mext , the integral R ∞ 1 P (Mext ) d Mext gives the overall probability of a black hole becoming extremal. The integral can be evaluated numerically to give a value of 21.1% for this probability.
14551	https://plutuseducation.com/blog/consumer-surplus-formula/	Consumer Surplus Formula: Concept, Examples, Theories & Factors Skip to content Home Programs ACCA ACCA Registration ACCA Exam Dates ACCA Eligibility ACCA Exemptions ACCA Subjects ACCA Syllabus ACCA Exam Pattern ACCA Books ACCA Study Material US CMA CMA Foundation Registration CMA Final Registration CMA Exam Dates CMA Eligibility CMA Syllabus CMA Exam Pattern CMA Subjects CMA Notes CFA CFA Eligibility CFA Syllabus CFA Exam Pattern CFA Salary CFA Course Fees US CPA CPA Eligibility CPA Exam Dates CPA Course Duration CPA Course Fees CPA Subjects CPA Syllabus CPA Books CPA Results Webinar Blog Study Material ACCA Exemption Calculator Posted inStudy Material Consumer Surplus Formula: Concept, Examples, Theories & Factors Posted bySukhpreet Monga The consumer surplus formula helps understand the benefit a buyer receives when they pay less than they are willing to pay. Simply put, consumer surplus shows the difference between what a consumer is ready to pay and what they pay. It is a key part of economic studies and helps to know how markets work. The formula is helpful in various areas like policymaking, pricing strategy, and understanding market changes. It is also a significant part of understanding consumer behavior and demand. The economic consumer surplus formula uses graphs, calculus, or simple pricing to find the value. What is the Consumer Surplus? Consumer surplus is the extra value or benefit from buying a product at a price lower than the maximum they were willing to pay. It shows how happy or satisfied a consumer feels after a purchase. This gap becomes the consumer surplus when the market price exceeds what the buyer is ready to pay. This surplus explains why buyers think they got a good deal. Download Consumer Surplus Formula PDF Understanding the Basic Meaning of Consumer Surplus Consumer surplus means savings for the buyer. If a student wants to buy a book for ₹500 but finds it for ₹300, then the consumer surplus is ₹200. The student paid less and gained value. Many such buyers in the market create a large consumer surplus. This shows how much benefit the market gives to buyers. Consumer surplus is a big part of how the economy works. When prices go down, consumer surplus goes up. This makes people feel they saved money. When prices go up, surplus goes down, and buyers may feel less happy. So, understanding this surplus helps in seeing how pricing affects consumers. Consumer surplus also shows how well the market works. If the market gives a high consumer surplus, prices are reasonable for buyers. If the surplus is low, it may mean the market has problems like high prices or less competition. So, the consumer surplus formula helps study all these changes. Economic Theory Behind Consumer Surplus Consumer surplus comes from demand and utility theory in economics. The idea is that people buy goods to gain utility or satisfaction. They are ready to pay a high price for the first unit, but their need decreases with each extra unit. This shows how demand drops as quantity goes up. Marginal Utility and Demand Curve The economic idea behind consumer surplus is marginal utility. Marginal utility means the extra happiness or satisfaction from buying one more unit. As people buy more, marginal utility becomes less. This is called the law of diminishing marginal utility. Because of this, people are ready to pay more for the first unit and less for the next. This creates a downward sloping demand curve. The consumer surplus is the top part of the demand curve, above the market price. This area between the demand curve and the market price shows the total gain to all buyers. Demand and Market Price Interaction In the real market, many buyers pay one price, but each buyer values the product differently. Some people may be ready to pay more, and some may pay only the market price. So, the consumer surplus is he extra amount saved by buyers who would pay more. The demand curve tells us how much quantity people buy at different prices. Using this curve, economists find the area that shows consumer surplus. This area shows how many buyers gain from the current price. The theory also helps compare this gain with what producers get, called producer surplus. People learn how consumer behaviour, pricing, and utility are linked using the consumer surplus formula in economics. When policy changes or taxes happen, the theory helps predict how consumer surplus will change. Calculating the Consumer Surplus A proper method or formula must be used to find the consumer surplus. The consumer surplus formula uses simple math or advanced tools like calculus. It depends on how the demand is shown: with a straight line, graph, or equation. Simple Consumer Surplus Formula When the demand curve is straight, use this basic formula: Consumer Surplus = ½ × (Maximum Price – Market Price) × Quantity This is like finding the area of a triangle. It fits well when demand is linear and the market price is clear. Here’s a small table to explain: Max Price (₹)Market Price (₹)QuantityConsumer Surplus (₹) 100 60 10 ½ × (100-60) × 10 = 200 This table shows how buyers gain ₹200 by paying less. Using Calculus in the Consumer Surplus Formula For more exact values, use the consumer surplus formula in calculus. In this case: Consumer Surplus = ∫ from 0 to Q (D(q) dq) – (P × Q) D(q) is the demand function, P is the price, and Q is the quantity. This finds the area under the demand curve and subtracts the total amount paid. Consumer Surplus Formula from Demand Function If demand is a function, say D(q) = 100 – 2q, and market price is ₹40, find where D(q) = 40: 100 – 2q = 40 → q = 30 Then apply: From 0 to 30 (100 – 2q dq) – (40 × 30) = [100q – q²] from 0 to 30 – 1200 = (3000 – 900) – 1200 = 900 So, consumer surplus = ₹900. This consumer surplus formula from the demand function helps in exact answers. Using the change in consumer surplus formula, small price or quantity changes show how surplus increases or decreases. This helps in seeing what policies or taxes do to buyers. Factors Affecting Consumer Surplus Many reasons change the value of consumer surplus. Some are in buyer behaviour, others in market price or product type. Understanding these helps firms and the government make intelligent decisions. Price Changes The biggest reason for change is price. If the price goes down, buyers gain more. The gap between what they would pay and what they pay becomes larger. This increases consumer surplus. If the price goes up, the surplus becomes small or even zero. Income and Preferences When people earn more, they are ready to pay more. Their demand shifts, and consumer surplus may change. Also, changes in likes, trends, or habits shift the demand curve, which changes surplus. New tech or popular items often raise buyer value. Product Substitutes If there are many options in the market, people will choose the cheaper ones. This keeps prices low and surplus high. If no substitutes exist, firms raise prices, and buyers lose surplus. Monopoly markets give low consumer surplus. Government Policies Tax, subsidy, or price control affects the market price. A tax may raise prices and lower surplus. A subsidy reduces price and raises surplus. So, changing the consumer surplus formula helps measure these policy results. Elasticity of Demand When demand is elastic, minor price cuts bring significant quantity changes. This creates more consumer surplus. If demand is inelastic, surplus does not grow much with a lower price. Elastic demand helps buyers more. Download Consumer Surplus Formula PDF Consumer Surplus vs Producer Surplus Consumer surplus and producer surplus are both parts of the total market benefit. They show how gains are split between buyers and sellers. Understanding both gives a complete market picture. What Is Producer Surplus? Producer surplus is the extra money a seller gets over the minimum amount they need. If a seller sells for ₹50 but gets ₹80, their surplus is ₹30. It shows how much gain sellers get from the market price. Consumer Surplus vs Producer Surplus The consumer and producer surplus formula helps measure both values. Together, they form total economic surplus. Policymakers use this to check if a market is doing well. When the government changes laws or taxes, it studies the consumer and producer surplus formula to check who gains or loses. If both fail, the policy is not good. FeatureConsumer SurplusProducer Surplus Who Gains?Buyers Sellers Based on Max Willing Price Min Acceptable Price Shape on Graph Below Demand Curve Above Supply Curve Increases When Price Decreases Price Increases Consumer Surplus Formula FAQs 1. What is the consumer surplus formula? The consumer surplus formula is ½ × (Willing Price – Actual Price) × Quantity. This formula finds how much value a buyer gets when the price is lower than expected. It works for simple cases. 2. How to calculate consumer surplus using calculus? Use the formula: ∫ from 0 to Q (Demand Function) dq – (P × Q). This helps find the area between the demand curve and the price line. It gives more accurate answers. 3. What is the change in the consumer surplus formula? This shows how much consumer surplus changes due to price shifts. It uses the difference in old and new surplus values to show how policies or market changes affect buyers. 4. How is the consumer surplus formula from the demand function used? Use the demand function to find the quantity at the market price, then calculate the area under the curve up to that point. Subtract the total price paid to get the surplus. 5. What is the difference between consumer surplus and producer surplus? Consumer surplus benefits buyers, and producer surplus benefits sellers. Both depend on the price. Together, they make the total surplus, which shows how well the market works. 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14552	https://chem.libretexts.org/Courses/Oregon_Tech_PortlandMetro_Campus/OT_-_PDX_-_Metro%3A_General_Chemistry_I/07%3A_Ideal_Gas_Behavior/7.03%3A_Applications_of_the_Ideal_Gas_Law_and_Partial_Pressures	7.3: Applications of the Ideal Gas Law and Partial Pressures - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: Ideal Gas Behavior OT - PDX - Metro: General Chemistry I { "7.3.01:_Practice_Problems-_Applications_of_the_Ideal_Gas_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "7.01:_Temperature_and_Pressure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_The_Gas_Laws" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Applications_of_the_Ideal_Gas_Law_and_Partial_Pressures" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Kinetic_Molecular_Theory" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Matter_and_Measurement" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Atoms_and_Elements" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Nuclei_Ions_and_the_Periodic_Table" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Molecules_Compounds_and_Quantifying_Chemicals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Transformations_of_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Aqueous_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Ideal_Gas_Behavior" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Thermochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Thu, 28 May 2020 20:53:42 GMT 7.3: Applications of the Ideal Gas Law and Partial Pressures 217292 217292 Chelsea LeBrun Gustafson { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "transcluded:yes", "source-chem-98789" ] [ "article:topic", "showtoc:no", "transcluded:yes", "source-chem-98789" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Oregon Tech Portland-Metro Campus 4. OT - PDX - Metro: General Chemistry I 5. 7: Ideal Gas Behavior 6. 7.3: Applications of the Ideal Gas Law and Partial Pressures Expand/collapse global location 7.3: Applications of the Ideal Gas Law and Partial Pressures Last updated May 28, 2020 Save as PDF 7.2.1: Practice Problems- The Gas Laws 7.3.1: Practice Problems- Applications of the Ideal Gas Law Page ID 217292 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Density of a Gas 2. 3. Molar Mass of a Gas 4. 5. The Pressure of a Mixture of Gases: Dalton’s Law 6. 7. Collection of Gases over Water 8. 9. Chemical Stoichiometry and Gases 10. Avogadro’s Law Revisited 1. Summary 2. Key Equations 3. Footnotes 4. Glossary 5. Contributors 6. Feedback Skills to Develop Use the ideal gas law to compute gas densities and molar masses Perform stoichiometric calculations involving gaseous substances State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it." As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed. Density of a Gas Recall that the density of a gas is its mass to volume ratio, ρ=m V. Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example 7.3.1. Example 7.3.1: Derivation of a Density Formula from the Ideal Gas Law Use PV = nRT to derive a formula for the density of gas in g/L S olution (7.3.1)P⁢V=n⁢R⁢T Rearrange to get (mol/L): (7.3.2)n v=P R⁢T Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained: (7.3.3)(ℳ)⁢(n V)=(P R⁢T)⁢(ℳ) (7.3.4)ℳ/V=ρ=P⁢ℳ R⁢T Exercise 7.3.1 A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas? Answer ρ=P⁢ℳ R⁢T (7.3.5)0.0847⁢g⁡/L=760⁢torr×1⁢atm 760⁢torr×ℳ 0.0821 L⁢atm/mol K×290 K ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H 2, 2.02 g/mol) We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP. Example 7.3.2: Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane? Solution Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula: (7.3.6)85.7⁢g⁡C×1 mol C 12.01⁢g⁡C=7.136 mol C 7.136 7.136=1.00 mol C (7.3.7)14.3⁢g⁡H×1 mol H 1.01⁢g⁡H=14.158 mol⁢H⁡14.158 7.136=1.98 mol H Empirical formula is CH 2 [empirical mass (EM) of 14.03 g/empirical unit]. Next, use the density equation related to the ideal gas law to determine the molar mass: (7.3.8)d=P⁢ℳ R⁢T⁢1.56 g 1.00 L=0.984 atm×ℳ 0.0821 L atm/mol K×323 K ℳ = 42.0 g/mol, ℳ E⁢ℳ=42.0 14.03=2.99, so (3)(CH 2) = C 3 H 6 (molecular formula) Exercise 7.3.2 Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene? Answer Empirical formula, CH; Molecular formula, C 2 H 2 Molar Mass of a Gas Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n: (7.3.9)ℳ=grams of substance moles of substance=m n The ideal gas equation can be rearranged to isolate n: (7.3.10)n=P⁢V R⁢T and then combined with the molar mass equation to yield: (7.3.11)ℳ=m⁢R⁢T P⁢V This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass. Example 7.3.3: Determining the Molar Mass of a Volatile Liquid The approximate molar mass of a volatile liquid can be determined by: Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (Figure 7.3.1) Figure 7.3.1: When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At t l⟶g, the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott) Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm 3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform? Solution Since ℳ=m n and n=P⁢V R⁢T substituting and rearranging gives (7.3.12)ℳ=m⁢R⁢T P⁢V then (7.3.13)ℳ=m⁢R⁢T P⁢V=(0.494 g)×0.08206 L⋅atm/mol K×372.8 K 0.976 atm×0.129 L=120⁢g⁡/mol Exercise 7.3.3 A sample of phosphorus that weighs 3.243 × 10−2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor? Answer 124 g/mol P 4 The Pressure of a Mixture of Gases: Dalton’s Law Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ( Figure 7.3.2). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases: (7.3.14)P T⁢o⁢t⁢a⁢l=P A+P B+P C+...=∑i P i In the equation P Total is the total pressure of a mixture of gases, P A is the partial pressure of gas A; P B is the partial pressure of gas B; P C is the partial pressure of gas C; and so on. Figure 7.3.2: If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the mixture is 1350 kPa. The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components: (7.3.15)P A=X A×P T⁢o⁢t⁢a⁢l where⁢X A=n A n T⁢o⁢t⁢a⁢l where P A, X A, and n A are the partial pressure, mole fraction, and number of moles of gas A, respectively, and n Total is the number of moles of all components in the mixture. Example 7.3.4: The Pressure of a Mixture of Gases A 10.0-L vessel contains 2.50 × 10−3 mol of H 2, 1.00 × 10−3 mol of He, and 3.00 × 10−4 mol of Ne at 35 °C. What are the partial pressures of each of the gases? What is the total pressure in atmospheres? Solution The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using P=n⁢R⁢T V: (7.3.16)P H 2=(2.50×10−3 mol)⁢(0.08206⁢L⁢atm⁢mol−1 K−1)⁢(308⁢K)10.0⁢L=6.32×10−3 atm (7.3.17)P He=(1.00×10−3⁢mol)⁢(0.08206⁢L⁢atm⁢mol−1 K−1)⁢(308⁢K)10.0⁢L=2.53×10−3 atm (7.3.18)P Ne=(3.00×10−4⁢mol)⁢(0.08206⁢L⁢atm⁢mol−1 K−1)⁢(308⁢K)10.0⁢L=7.58×10−4 atm The total pressure is given by the sum of the partial pressures: (7.3.19)P T=P H 2+P He+P Ne=(0.00632+0.00253+0.00076)⁢atm=9.61×10−3 atm Exercise 7.3.4 A 5.73-L flask at 25 °C contains 0.0388 mol of N 2, 0.147 mol of CO, and 0.0803 mol of H 2. What is the total pressure in the flask in atmospheres? Answer 1.137 atm Here is another example of this concept, but dealing with mole fraction calculations. Example 7.3.5: The Pressure of a Mixture of Gases A gas mixture used for anesthesia contains 2.83 mol oxygen, O 2, and 8.41 mol nitrous oxide, N 2 O. The total pressure of the mixture is 192 kPa. What are the mole fractions of O 2 and N 2 O? What are the partial pressures of O 2 and N 2 O? Solution The mole fraction is given by X A=n A n T⁢o⁢t⁢a⁢l and the partial pressure is P A=X A×P T⁢o⁢t⁢a⁢l For O 2, X O 2=n O 2 n T⁢o⁢t⁢a⁢l=2.83⁢mol(2.83+8.41)⁢mol=0.252 and P O 2=X O 2×P T⁢o⁢t⁢a⁢l=0.252×192 kPa=48.4 kPa For N 2 O, X N 2⁢O=n N 2⁢O n T⁢o⁢t⁢a⁢l=8.41 mol(2.83+8.41)⁢mol=0.748 and P N 2⁢O=X N 2⁢O×P T⁢o⁢t⁢a⁢l=(0.748)×192 kPa=143.6 kPa Exercise 7.3.5 What is the pressure of a mixture of 0.200 g of H 2, 1.00 g of N 2, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C? Answer 1.87 atm Collection of Gases over Water A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 7.3.3), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer. Figure 7.3.3:When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers). However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. Figure 7.3.4: This graph shows the vapor pressure of water at sea level as a function of temperature. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 7.3.4); more detailed information on the temperature dependence of water vapor can be found in Table 7.3.1, and vapor pressure will be discussed in more detail in the next chapter on liquids. Table 7.3.1 : Vapor Pressure of Ice and Water in Various Temperatures at Sea Level\| Temperature (°C) \| Pressure (torr) \| \| Temperature (°C) \| Pressure (torr) \| \| Temperature (°C) \| Pressure (torr) \| --- --- --- --- \| \| –10 \| 1.95 \| \| 18 \| 15.5 \| \| 30 \| 31.8 \| \| –5 \| 3.0 \| 19 \| 16.5 \| 35 \| 42.2 \| \| –2 \| 3.9 \| 20 \| 17.5 \| 40 \| 55.3 \| \| 0 \| 4.6 \| 21 \| 18.7 \| 50 \| 92.5 \| \| 2 \| 5.3 \| 22 \| 19.8 \| 60 \| 149.4 \| \| 4 \| 6.1 \| 23 \| 21.1 \| 70 \| 233.7 \| \| 6 \| 7.0 \| 24 \| 22.4 \| 80 \| 355.1 \| \| 8 \| 8.0 \| 25 \| 23.8 \| 90 \| 525.8 \| \| 10 \| 9.2 \| 26 \| 25.2 \| 95 \| 633.9 \| \| 12 \| 10.5 \| 27 \| 26.7 \| 99 \| 733.2 \| \| 14 \| 12.0 \| 28 \| 28.3 \| 100.0 \| 760.0 \| \| 16 \| 13.6 \| 29 \| 30.0 \| 101.0 \| 787.6 \| Example 7.3.6: Pressure of a Gas Collected Over Water If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 7.3.3, what is the partial pressure of argon? Solution According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water: (7.3.20)P T=P Ar+P H 2⁡O Rearranging this equation to solve for the pressure of argon gives: (7.3.21)P Ar=P T−P H 2⁡O The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so: (7.3.22)P Ar=750 torr−25.2 torr=725 torr Exercise 7.3.6 A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure? Answer 0.583 L Chemical Stoichiometry and Gases Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions. We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure. Avogadro’s Law Revisited Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure. We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to (7.3.23)N⁢A 2⁢(g)+3⁢H⁡A 2⁢(g)⟶2 NH⁢A 3⁢(g) a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant. The explanation for this is illustrated in Figure 7.3.4. According to Avogadro’s law, equal volumes of gaseous N 2, H 2, and NH 3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N 2 reacts with three molecules of H 2 to produce two molecules of NH 3, the volume of H 2 required is three times the volume of N 2, and the volume of NH 3 produced is two times the volume of N 2. Figure 7.3.5:One volume of N 2 combines with three volumes of H 2 to form two volumes of NH 3. Example 7.3.7: Reaction of Gases Propane, C 3 H 8(g), is used in gas grills to provide the heat for cooking. What volume of O 2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion. Solution The ratio of the volumes of C 3 H 8 and O 2 will be equal to the ratio of their coefficients in the balanced equation for the reaction: (7.3.24)C⁢A 3⁢H⁡A 8⁢(g)+5 O⁢A 2⁢(g)⟶3 CO⁢A 2⁢(g)+4⁢H⁡A 2⁢O⁢(l)(7.3.25)1 volume+5 volumes 3 volumes+4 volumes From the equation, we see that one volume of C 3 H 8 will react with five volumes of O 2: (7.3.26)2.7⁢L C 3⁢H 8×5 L⁢O⁢A 2 1⁢L C 3⁢H 8=13.5 L⁢O⁢A 2 A volume of 13.5 L of O 2 will be required to react with 2.7 L of C 3 H 8. Exercise 7.3.7 An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C 2 H 2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 10 3 L of O 2 at 0 °C and 1 atm, will be required to burn the acetylene? 2 C⁢A 2⁢H⁡A 2+5 O⁢A 24 CO⁢A 2+2⁢H⁡A 2⁢O Answer 3.34 tanks (2.34 × 10 4 L) Example 7.3.8: Volumes of Reacting Gases Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H 2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N 2? N⁢A 2⁢(g)+3⁢H⁡A 2⁢(g)⟶2 NH⁢A 3⁢(g) Solution Because equal volumes of H 2 and NH 3 contain equal numbers of molecules and each three molecules of H 2 that react produce two molecules of NH 3, the ratio of the volumes of H 2 and NH 3 will be equal to 3:2. Two volumes of NH 3, in this case in units of billion ft 3, will be formed from three volumes of H 2: (7.3.27)683⁢billion ft 3 NH 3×3 billion ft 3 H 2 2⁢billion ft 3 NH 3=1.02×10 3 billion ft 3 H 2 The manufacture of 683 billion ft 3 of NH 3 required 1020 billion ft 3 of H 2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.) Exercise 7.3.8 What volume of O 2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C 2 H 4(g), measured under the same conditions of temperature and pressure? The products are CO 2 and water vapor. Answer 51.0 L Example 7.3.9: Volume of Gaseous Product What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid? (7.3.28)2 Ga⁢(s)+6 HCl⁢(a⁢q)⟶2 GaCl⁢A 3⁢(a⁢q)+3⁢H⁡A 2⁢(g) Sol ution To convert from the mass of gallium to the volume of H 2(g), we need to do something like this: The first two conversions are: (7.3.29)8.88⁢g⁡Ga×1⁢mol Ga 69.723⁢g⁡Ga×3 mol H 2 2⁢mol Ga=0.191 mol H 2 Finally, we can use the ideal gas law: (7.3.30)V H 2=(n⁢R⁢T P)H 2=0.191⁢mol×0.08206 L⁢atm mol−1 K−1×300 K 0.951 atm=4.94 L Exercise 7.3.9 Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO 2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen? Answer 1.30 × 10 3 L Greenhouse Gases and Climate Change The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost 1 3 is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 7.3.6). Figure 7.3.6: Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events. There is strong evidence from multiple sources that higher atmospheric levels of CO 2 are caused by human activity, with fossil fuel burning accounting for about 3 4 of the recent increase in CO 2. Reliable data from ice cores reveals that CO 2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO 2 concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure 7.3.7). Figure 7.3.7: CO 2 levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years. Portrait of a Chemist: Susan Solomon Atmospheric and climate scientist Susan Solomon (Figure 7.3.8) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA. Figure 7.3.8: Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration) Summary Video 7.3.1: An overview of the topics of partial pressure, vapor pressure, and collecting gases over water. The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products. Key Equations P Total = P A + P B + P C + … = Ʃ i P i P A = X A P Total X A=n A n T⁢o⁢t⁢a⁢l Footnotes “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, Glossary Dalton’s law of partial pressures total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.mole fraction (X)concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components partial pressure pressure exerted by an individual gas in a mixture vapor pressure of water pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature Contributors Paul Flowers (University of North Carolina - Pembroke),Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at Adelaide Clark, Oregon Institute of Technology Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes. Feedback Have feedback to give about this text? Click here. Found a typo and want extra credit? Click here. 7.3: Applications of the Ideal Gas Law and Partial Pressures is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 7.2.1: Practice Problems- The Gas Laws 7.3.1: Practice Problems- Applications of the Ideal Gas Law Was this article helpful? Yes No Recommended articles 2.6.4: Applications of the Ideal Gas Law and Partial Pressures 7.3.1: Practice Problems- Applications of the Ideal Gas Law 7.1: Temperature and Pressure 7.2: The Gas Laws 7.4: Kinetic Molecular Theory Article typeSection or PageShow Page TOCno on pageTranscludedyes Tags source-chem-98789 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status×
14553	https://www.rush.edu/conditions/hyperparathyroidism	Hyperparathyroidism Signs & Symptoms \| Rush Skip to main content Call (888) 352-RUSH (7874) Schedule Appointment Call (312) 942-5555 MyChart Pay Bill Refer a Patient Giving Rush University I am a… Close Select from the list below to customize your experience: Select a new category I am a: Medical Professional Referring Patients Nursing Career Opportunities Residencies & Fellowships Research Conflict of Interest Disclosures Patient Plan Your Stay Patient Resources Services & Treatments MyChart Access Clinical Trials & Studies Visitor Locations and Directions Resources For Your Visit Visiting Hours & Policies Job Candidate Careers Home Page RUSH University Medical Center Careers RUSH Copley Medical Center Careers RUSH Oak Park Hospital Careers Provider and Faculty Positions I am a Close Select from the list below to customize your experience: Select a new category I am a: Medical Professional Referring Patients Nursing Career Opportunities Residencies & Fellowships Research Conflict of Interest Disclosures Patient Plan Your Stay Patient Resources Services & Treatments MyChart Access Clinical Trials & Studies Visitor Locations and Directions Resources For Your Visit Visiting Hours & Policies Job Candidate Careers Home Page RUSH University Medical Center Careers RUSH Copley Medical Center Careers RUSH Oak Park Hospital Careers Provider and Faculty Positions Doctors Services Services Adults Primary Care Cancer Care Digestive Diseases Heart & Vascular Care Neurology & Neurosurgery Care See All Children Pediatric Primary Care Children's Hospital Pediatric Endocrinology Pediatric Gastroenterology See All Quick Links Clinical Trials & Studies Rush Connect+ Virtual Health Membership Locations Locations Adults Primary Care Emergency Centers Walk-In Care Lab Facilities See All Children Pediatric Primary Care Children's Hospital Pediatric Specialty Care Family Medicine See All Quick Links Rush University Medical Center Rush Copley Medical Center Rush Oak Park Hospital Rush Oak Brook Rush North & Harlem Rush Munster Rush Oak Lawn Rush South Loop Patients & Visitors Schedule Appointment Search Rush Find a Doctor Popular Services Primary Care Psychiatry and Psychology Services General Neurology Services Obstetrics and Gynecology (OB-GYN) Services Cancer Care View All Services Popular Locations Rush Copley Medical Center Rush Oak Park Hospital Rush University Medical Center Rush Oak Brook Rush South Loop View All Locations Your Saved Locations (0 Locations) Located Near: Close Schedule Appointment Call (312) 942-5555 MyChart Doctors Services Adults Children Locations Adults Children Patients & Visitors Pay Bill About Refer a Patient For Medical Professionals News & More Classes & Events Clinical Trials Career Opportunities Contact Us Volunteer Giving to Rush Rush University COVID-19 Schedule Appointment Call (312) 942-5555 MyChart Close Doctors Services Adults Children Locations Adults Children Patients & Visitors Pay Bill About Refer a Patient For Medical Professionals News & More Classes & Events Clinical Trials Career Opportunities Contact Us Volunteer Giving to Rush Rush University COVID-19 Home Services & Treatments Hyperparathyroidism Hyperparathyroidism Our nationally recognized experts in parathyroid disease offer minimally-invasive techniques, which means smaller scars, less pain and faster healing for you. Primary hyperparathyroidism occurs when one or more of the four parathyroid glands next to the thyroid gland secrete too much parathyroid hormone (PTH), which regulates the amount of calcium, phosphorus and vitamin D in your blood and bones. Hyperparathyroidism can cause osteoporosis, kidney stones, chronic fatigue and difficulty with memory and concentration. It has also been linked to heart disease, high blood pressure, depression, anxiety and other chronic conditions. Symptoms of Hyperparathyroidism Symptoms are caused by too much calcium in the blood and include the following: Fatigue Bone and joint pain Bone fractures Kidney stones Muscle weakness Depression Confusion or memory issues Increased thirst and urination A routine calcium blood test with results in the above normal range is often the first indicator of primary hyperparathyroidism – before any symptoms appear. Hyperparathyroidism Treatment at Rush Your primary care provider can diagnose primary hyperparathyroidism if your blood tests show high levels of calcium and PTH at the same time, although additional testing may be needed to assess possible complications such as osteoporosis and kidney stones. Your provider may refer you to an endocrinologist for further evaluation and treatment. In nearly all cases, the cause of primary hyperparathyroidism is a benign (noncancerous) tumor in a parathyroid gland. Treatment options include the following: Monitoring:If you have no symptoms or kidney stones, normal bone density and only mildly elevated levels of PTH and calcium, your provider may decide to monitor your health until something changes. Medications:Your provider may prescribe calcimimetics. Other medications are also being studied. Surgery:If you have primary hyperparathyroidism, you may benefit from surgery to remove overactive parathyroid glands, even if you have no symptoms. Two types of surgery for hyperparathyroidism exist, including: Minimally invasive parathyroidectomy:Performed through a very small neck incision which allows for rapid recovery, this surgery is used in more than 95% of cases. Endocrine surgeons use this technique to examine all four parathyroid glands to determine which needs to be removed. Standard neck exploration:This traditional surgery is used in the most complicated cases. Hyperparathyroidism Providers at Rush Learn more about hyperparathyroidism providers at Rush. Meet our hyperparathyroidism providers Meet our hyperparathyroidism providers Rush Excellence in Hyperparathyroidism Care Team-based approach:Working collaboratively to provide the best care possible, your team is led by an endocrinologist and may include an endocrine surgeon, a nephrologist and other specialists. Pulling together all the right experts means your treatment is comprehensive, personalized and highly effective. Nationally recognized experts:The Rush endocrinology and endocrine surgery teams include nationally recognized experts in the care of parathyroid conditions. These leaders bring you ground-breaking discoveries and treatments in the field as soon as they're available. U.S. News & World Report ranked Rush University Medical Center among the best in the nation for endocrinology. Minimally invasive surgery:Rush offers minimally invasive techniques for parathyroid surgery. This means you'll have smaller scars, faster healing and less pain. 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14554	https://askfilo.com/user-question-answers-smart-solutions/if-then-prove-that-3132343730363932	Question asked by Filo student If tan−1x+tan−1y+tan−1z=π then prove that x+y+z=xyz. Views: 5,058 students Updated on: Sep 23, 2024 Text SolutionText solutionverified iconVerified Concepts: Trigonometry, Inverse trigonometric functions, Properties of arctangent Explanation: To prove that x+y+z=xyz given that tan−1x+tan−1y+tan−1z=π, we will use the properties of the arctangent function and the tangent addition formula. Step by Step Solution: Step 1 Given that tan−1x+tan−1y+tan−1z=π, we know that tan(π)=0. Step 2 Using the tangent addition formula, we have: tan(tan−1x+tan−1y+tan−1z)=tan(π)=0. Step 3 Let A=tan−1x, B=tan−1y, and C=tan−1z. Then, A+B+C=π. Using the tangent addition formula for three angles, we get: tan(A+B+C)=1−(tanAtanB+tanBtanC+tanCtanA)tanA+tanB+tanC−tanAtanBtanC. Step 4 Since A+B+C=π, we have: tan(π)=0. Therefore, the numerator of the tangent addition formula must be zero: x+y+z−xyz=0. Step 5 Rearranging the equation, we get: x+y+z=xyz. Final Answer: x+y+z=xyz Students who ask this question also asked Views: 5,356 Topic: Smart Solutions View solution Views: 5,777 Topic: Smart Solutions View solution Views: 5,245 Topic: Smart Solutions View solution Views: 5,096 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| If tan−1x+tan−1y+tan−1z=π then prove that x+y+z=xyz. \| \| Updated On \| Sep 23, 2024 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 11 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
14555	https://www.thoughtco.com/definition-of-independent-variable-605238	Independent Variable Definition and Examples Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Chemistry› Chemical Laws› Independent Variable Definition and Examples Understand the Independent Variable in an Experiment Print In a science experiment, the independent variable is the one you intentionally change or control.Hero Images / Getty Images Science Chemistry Chemical Laws Basics Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. Learn about ourEditorial Process Updated on August 16, 2024 Close The independent variable and the dependent variable are the two main variables in a science experiment. Below is the definition of an independent variable and a look at how you might use it. Key Takeaways: Independent Variable The independent variable is the factor that you purposely change or control to see what effect it has. The variable that responds to the change in the independent variable is called the dependent variable. The dependent variable depends on the independent variable. The independent variable is graphed on the x-axis. Independent Variable Definition An independent variable is defined as a variable that is changed or controlled in a scientific experiment. The independent variable represents the cause or reason for an outcome. Independent variables are the variables that the experimenter changes to test his or her dependent variable. A change in the independent variable directly causes a change in the dependent variable. The effect on the dependent variable is measured and recorded. Common misspellings:independant variable Independent Variable Examples Here are some examples of an independent variable. A scientist is testing the effect of light and dark on the behavior of moths by turning a light on and off. The independent variable is the amount of light (cause) and the moth's reaction is the dependent variable (the effect). In a study to determine the effect of temperature on plant pigmentation, the independent variable is the temperature, while the amount of pigment or color is the dependent variable. Graphing the Independent Variable When graphing data for an experiment, the independent variable is plotted on the x-axis, while the dependent variable is recorded on the y-axis. An easy way to keep the two variables straight is to use the acronym DRY MIX, which stands for: Dependent variable that Responds to change goes on the Y axis Manipulated or Independent variable goes on the X axis Practice Identifying the Independent Variable Students are often asked to identify the independent and dependent variable in an experiment. The difficulty is that the value of both of these variables can change. It is even possible for the dependent variable to remain unchanged in response to controlling the independent variable. Example: You are asked to identify the independent and dependent variable in an experiment to see if there is a relationship between hours of sleep and student test scores. There are two ways to identify the independent variable. The first is to write the hypothesis and see if it makes sense. For example: Student test scores do not affect the number of hours the students sleep. The number of hours students sleep do not affect their test scores. Only one of these statements makes sense. This type of hypothesis is constructed to state the independent variable followed by the predicted impact on the dependent variable. So, the number of hours of sleep is the independent variable. The other way to identify the independent variable is more intuitive. Remember, the independent variable is the one the experimenter controls to measure its effect on the dependent variable. A researcher can control the number of hours a student sleeps. On the other hand, the scientist has no control over the students' test scores. The independent variable always changes in an experiment, even if there is just a control and an experimental group. The dependent variable may or may not change in response to the independent variable. In the example regarding sleep and student test scores, the data might show no change in test scores, no matter how much sleep students get (although this outcome seems unlikely). The point is that a researcher knows the values of the independent variable. The value of the dependent variable is measured. Sources Babbie, Earl R. (2009). The Practice of Social Research (12th ed.). Wadsworth Publishing. ISBN 0-495-59841-0. Dodge, Y. (2003). The Oxford Dictionary of Statistical Terms. OUP. ISBN 0-19-920613-9. Everitt, B. S. (2002). The Cambridge Dictionary of Statistics (2nd ed.). Cambridge UP. ISBN 0-521-81099-X. Gujarati, Damodar N.; Porter, Dawn C. (2009). "Terminology and Notation". Basic Econometrics (5th international ed.). New York: McGraw-Hill. p. 21. ISBN 978-007-127625-2. Shadish, William R.; Cook, Thomas D.; Campbell, Donald T. (2002). Experimental and quasi-experimental designs for generalized causal inference. (Nachdr. ed.). Boston: Houghton Mifflin. ISBN 0-395-61556-9. Cite this Article Format mlaapachicago Your Citation Helmenstine, Anne Marie, Ph.D. "Independent Variable Definition and Examples." ThoughtCo, Aug. 16, 2024, thoughtco.com/definition-of-independent-variable-605238.Helmenstine, Anne Marie, Ph.D. (2024, August 16). Independent Variable Definition and Examples. Retrieved from Helmenstine, Anne Marie, Ph.D. "Independent Variable Definition and Examples." ThoughtCo. (accessed September 29, 2025). copy citation Sponsored Stories Top Cardiologist Begs: Quit Eating Blueberries Before This Happens thehealthyfat.com 56-Year-Old Replaced $21,500 Facelift With This $38 Drugstore Item The Skincare Magazine Warren Buffett, Elon Musk and other moguls all use the 5-hour rule. Here's what I learned from it Blinkist Magazine Heart Surgeon Begs: Throw Out Your Olive Oil Now (Here’s Why)enrichyourfood.com Oxidation Definition and Example in Chemistry Dependent Variable vs. Independent Variable: What Is the Difference? Heavy Metal Definition and List Melting Definition in Chemistry Stoichiometry Definition in Chemistry The Combined Gas Law in Chemistry pKa Definition in Chemistry Avogadro's Law Example Problem Sponsored Stories Aging Seniors: Quit Eating Cottage Cheese Before This Happens thehealthyfat.com This 3-Ingredient Pink Salt Mix Is Being Called the “Homemade Bariatric Drink"Health Guide Thinning Skin? Drink This Every Morning NativePath This One Habit Breaks Vets’ Hearts (Most Dog Parents Don’t Realize)Dr. Marty Nature's Blend Element Symbol Definition in Chemistry What Is an Element in Chemistry? Definition and Examples Absolute Temperature Definition Beer's Law Definition and Equation The Chemical Composition of Air Why Is the Carbon Cycle Important? 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14556	https://www.quora.com/An-arithmetic-progression-has-10-terms-The-sum-of-the-odd-terms-is-245-whereas-the-sum-of-the-even-terms-is-305-what-is-the-common-difference	An arithmetic progression has 10 terms. The sum of the odd terms is 245 whereas the sum of the even terms is 305. what is the common difference? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Patterns and Sequence Application of Arithmetic... Sum of Series Odd Numbers Mathematics Word Problems Arithmetic Arithmetic and Geometric ... Sums and Series (mathemat... 5 An arithmetic progression has 10 terms. The sum of the odd terms is 245 whereas the sum of the even terms is 305. what is the common difference? All related (32) Sort Recommended Rafiq Teli Maths Teacher (1984–present) · Author has 690 answers and 500.7K answer views ·3y a + (a + 2d) + (a + 4d) + (a + 6d) + (a + 8d) = 245 So, 5a + 20d = 245…. (1) (a + d) + (a + 3d) + (a + 5d) + (a + 7d) + (a + 9d) = 305 So, 5a + 25 d = 305…(2) Subtracting (1) from (2), 5d = 60 So, d = 12 So, the common difference is 12. Upvote · 9 3 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 623 Related questions More answers below What is the common difference of 1st term = -5 and 17th term = 800 in arithmetic sequence? The eighth term of an arithmetic progression is four times the fifth term and the sum of the first eight terms is 20. What is the sum for the first ten terms? Given the sum of the first 14 terms of arimethic progression is 224 and the sum of even term is 245/2. What is the first term and c mmon difference? The first term of an arithmetic progression (AP) is 5 and the common difference is 3. What is the sum of the first 50 terms? An arithmetical progression is such that the fifth term is 22 and the sum of the first eight terms is 166. What is the first term? Bruce Stevens Author has 2.8K answers and 1.5M answer views ·1y Related In an arithmetic progression, the 13th term is 27 and the 7th terms is three times the second term. What is the first, the common difference and the sum of the first ten terms? Set the question into AS equations: t(13) = 27 = ( a + 12d ), t(7) = ( a + 6d ) = ( 3 ( a + d ) ). Solve for the value of the constant difference, d: t(7) = ( a + 6d ) = ( 3a + 3d ), a = ( 3d / 2 ). Substitute a = ( 3d / 2 ) into t(13): 27 = ( (3d / 2) + 12d ), 54 = ( 3d + 24d ), 54 = 27d, d = 2. Solve for the value of the 1st term, a: a = ( 3d / 2 ), a = ( (3 2) / 2 ), a = 3. Solve for the sum value of the first 10 terms: Number of differences = ((n (n — 1)) / 2), Number of differences = ((10 (10 — 1)) / 2), Number of differences = 45. Number of 1st terms: 10. Sum(10) = ( 10a + 45d ), Sum(10) = ( (10 Continue Reading Set the question into AS equations: t(13) = 27 = ( a + 12d ), t(7) = ( a + 6d ) = ( 3 ( a + d ) ). Solve for the value of the constant difference, d: t(7) = ( a + 6d ) = ( 3a + 3d ), a = ( 3d / 2 ). Substitute a = ( 3d / 2 ) into t(13): 27 = ( (3d / 2) + 12d ), 54 = ( 3d + 24d ), 54 = 27d, d = 2. Solve for the value of the 1st term, a: a = ( 3d / 2 ), a = ( (3 2) / 2 ), a = 3. Solve for the sum value of the first 10 terms: Number of differences = ((n (n — 1)) / 2), Number of differences = ((10 (10 — 1)) / 2), Number of differences = 45. Number of 1st terms: 10. Sum(10) = ( 10a + 45d ), Sum(10) = ( (10 3) + (45 2) ), Sum(10) = ( 30 + 90 ), Sum(10) = 120. P.S. The arithmetic sequence is: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, . . . The general term rule, of this arithmetic sequence, when n = 1, 2, 3, . . ., is: t(n) = ( 2n + 1 ). Upvote · 9 4 9 1 Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·Aug 4 Related The sum of the first 20 terms of an arithmetic sequence is 760. What is the first term if the last term is 95? What is its common difference? The sum of the first 20 terms of an arithmetic sequence is 760. What is the first term if the last term is 95? What is its common difference? As few formulas as possible. Call [math]a[/math] your first term. Then you know that the terms are [math]a,\quad a+d,\quad \dots,\quad a+19d[/math] where [math]d[/math] is the common difference. Then we see that [math]a+19d=95[/math] [math]20a+(0+1+2+\dots+19)d=760[/math] You surely know how to compute the sum in parentheses, don’t you? It’s [math]\dfrac{19\cdot20}{2}=190[/math] Now the second equation simplifies to [math]2a+19d=76[/math] and subtracting yields [math]a=76-95=-19[/math] and therefore [math]d=\dfrac{95-a}{19}=6[/math] Upvote · 9 3 Assistant Bot · Sep 6 Let the arithmetic progression be a, a + d, a + 2d, …, a + 9d (10 terms). Odd-position terms: positions 1,3,5,7,9 — five terms: their sequence is a, a+2d, a+4d, a+6d, a+8d. Sum_odd = 5a + (0+2+4+6+8)d = 5a + 20d = 245. (1) Even-position terms: positions 2,4,6,8,10 — five terms: their sequence is a+d, a+3d, a+5d, a+7d, a+9d. Sum_even = 5a + (1+3+5+7+9)d = 5a + 25d = 305. (2) Subtract (1) from (2): (5a+25d) - (5a+20d) = 305 - 245 → 5d = 60 → d = 12. Common difference = 12. Upvote · Related questions What is the common difference of 1st term = -5 and 17th term = 800 in arithmetic sequence? The eighth term of an arithmetic progression is four times the fifth term and the sum of the first eight terms is 20. What is the sum for the first ten terms? Given the sum of the first 14 terms of arimethic progression is 224 and the sum of even term is 245/2. What is the first term and c mmon difference? The first term of an arithmetic progression (AP) is 5 and the common difference is 3. What is the sum of the first 50 terms? An arithmetical progression is such that the fifth term is 22 and the sum of the first eight terms is 166. What is the first term? An arithmetic progression consists of 10 terms. The first term is 4. The sum of the 10 terms is 130. What is the last term? What is the 7th term of the arithmetic progression whose 1st term is 24 and the common difference is 4 and the sum of the first ten terms? An arithmetic progression has 20 terms, the sum of all even numbers is 350. Given that the first term is 5. What is the common difference and the eleventh term? In an arithmetic progression, the sixth term is one more than twice the third term. The sum of the fourth and fifth terms is five times the second term. What is the tenth term of the arithmetic progression? The sum of all terms of the arithmetic progression having ten terms except for the first term, is 99, and except for the sixth term, is 89. what is the 8th term of the progression if the sum of the first and the fifth term is equal to 10? What is the first term if the sum for the first terms of an Arithmetic progression is 100. The 21st term of the Arithmetic progression is 41? The sum of the 4th and 6th terms of an arithmetical progression is 42. The sum of the 3rd and 9th terms of the progression is 52. How do I find the first term, common difference, and sum of the ten terms of the progression? The sum of the first 10 terms of an arithmetic sequence is 530. What is the first term if the last term is 80? What is the common difference? What is the formula for summing an arithmetic progression given its initial term and common difference? The 3rd term and 7th term of an arithmetic progression are 6 and 30 respectively. What is the common difference between the first term and 10th term? Related questions What is the common difference of 1st term = -5 and 17th term = 800 in arithmetic sequence? The eighth term of an arithmetic progression is four times the fifth term and the sum of the first eight terms is 20. What is the sum for the first ten terms? Given the sum of the first 14 terms of arimethic progression is 224 and the sum of even term is 245/2. What is the first term and c mmon difference? The first term of an arithmetic progression (AP) is 5 and the common difference is 3. What is the sum of the first 50 terms? An arithmetical progression is such that the fifth term is 22 and the sum of the first eight terms is 166. What is the first term? An arithmetic progression consists of 10 terms. The first term is 4. The sum of the 10 terms is 130. What is the last term? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
14557	https://www.youtube.com/watch?v=PJ4nYMmRJZ0	Cone Shape: Problem Solving Using Trigonometry Yes, I Can Do Math 1110 subscribers 10 likes Description 1891 views Posted: 23 Jan 2023 Learn how to solve cone problems using trigonometry, including the Pythagorean Theorem and sine, cosine, tangent ratios. The last of four cone shape examples is on finding the cone volume. In this step-by-step video we'll go over four problems. If you're looking to solve problems about things that are cone shape, you're in the right place. This is part of a series in answering questions on geometric solids using trigonometric properties. Look at the time stamps if you just want to skip to the cone volume calculation problem. Once you learn what the volume of a cone is, remember to go back and watch the other videos on this series on geometric solids. • The longest rod problem (solving rectangular solid problems) • Square right pyramid problems part 1 • Click here to watch the second video using SOH-CAH-TOA to find angle measures in a square right pyramid: • Click here for part 3 which uses cosine rule to solve for the missing angle: • Click here to review how to use SOH-CAH-TOA: • Click here to review using the Pythagorean Theorem: • Right cylinders • Right cylinder volume • Cone shape 🎒 My favorite school supplies: 📱 TI-84 Color Graphing Calculator 👉 ✏️ Graphing Notebook 👉 🖊️ Pilot G2 Pens (0.7mm, pack of 4) 👉 h These links are affiliate links which help support the channel—thank you! 2 comments Transcript: let's go ahead and talk about Combs on your screen you're going to see an example of a problem where you're going to have to use trigonometric principles in order to be able to solve the problem it says if a b equals 11.3 centimeters and angle ABC is 53 degrees calculate the radius of the cone well clearly we have an image we have a drawing we have a diagram that we can use to visualize we can see that DC represents the radius right here of the problem we also know that a b is 11.3 and because this is a cone that is our slant height and that means that BC is also going to be 11.3 centimeters so what we have to work with right now is angle ABC angle ABC is the entire top angle of that cone it is 53 degrees one of the things that we have been working with during the series of 3D figures is trying to identify right triangles or any type of triangle within the images within the drawings so that we can use trigonometric properties in order to be able to apply them to those triangles and then get the solutions for these problems so I'm going to go ahead and create a right triangle because I keep telling you it's always a great idea to redraw these so that you get a clear picture of what you are doing at the very top where angle B is right here remember that we took angle ABC and we cut it in half so now I'm going to work with 26.5 degrees then I have D which is my 90 degree angle and c b c is 11.3 so my job is to figure out the radius which is basically DC because this is a right triangle one of the things that I can use are the trigonometric ratios sohcahtoa sine cosine tangent and that is exactly what we're going to work with here we're going to start by identifying the angle which is 26.5 according to that angle that makes the radius the opposite side the hypotenuse is going to be 11.3 remember the hypotenuse is always across from the 90 degree angle it's the longest side of the right triangle so now I have to determine what is going to connect opposite with hypotenuse and opposite hypotenuse that is going to be sine so I'm going to set up the problem as the sine of 26.5 is equal to opposite which is R over hypotenuse which is 11.3 take your calculators let's multiply by 11.3 on both sides so we get 11.3 times the sine of 26.5 make sure your calculator is in degree mode and you should get an answer of approximately 5.04 centimeters so this means that the radius is going to be approximately 5.04 centimeters long okay let's take a look at this second problem it says vaob is a right circular cone with radius 20 centimeters and a slant height of 60 centimeters calculate the vertical height vo calculate angle a v b so this problem is very similar to the one that we just finished doing right now in this case I'm going to once again work with this right triangle on the right hand side of the cone I'm going to take the time to redraw it on the side of the screen so that we can very clearly see exactly what it is that we're working with the slant height is 60 centimeters and the radius is 20 centimeters notice that vo is the vertical height of this cone therefore it is a 90 degree angle and they have also gone ahead and shown us the 90 degree angle by that little square that's right there on your screen so our job is to find vo we're going to do this by applying the Pythagorean theorem we've seen over and over again how in a right triangle whenever you have two side lengths you can go ahead and use the Pythagorean theorem to figure out the third one so in this case we're going to set up the problem 20 squared plus vo squared is equal to 60 squared because 60 is the hypotenuse and you have to be very careful with that when filling in the Pythagorean theorem so you don't make a mistake so 20 squared is 400 plus vo squared is equal to 60 squared which is going to be 3600 let's go ahead and subtract 400 from both sides we get that vo squared is equal to 3200 take the square root of 3200 and we get that vo is approximately 56.6 centimeters which makes sense because the slant height which is the longest side of that right triangle is 60. so we know that our answer for vo has to be less than that and so the vertical height vo is 56.6 centimeters now the second part of the question Part B says to calculate angle a v b that is the entire angle up here in the previous problem they had given it to us and then we decided to take half of it to work with the right triangle in this case we're kind of going backwards so what I'm going to do in the right triangle that you see on the right on the screen is that I'm going to label angle ovb as Theta because this is a right triangle once again I can use trigonometric ratio sohcahtoa to be able to determine a relationship that will connect Theta with these sides opposite of theta is 20 and the hypotenuse of course is 60. which one of these trig ratios connects opposite with hypotenuse sine so the sine of theta is equal to opposite over hypotenuse only when we're looking for the angle are we going to press second or shift in the calculator to figure out what's called the arc sine so in your calculators right now go ahead and press second sine of 20 divided by 60 and you're going to get that Theta is approximately 19.47 degrees now of course that is not our final answer remember that our job is to find angle a v b so we have to take Theta and we have to multiply that answer by 2. keep the entire decimal in your calculator so that you don't have round off error and so the final answer if I multiply this by 2 it is going to be approximately 38.9 degrees so that means that angle AVB is approximately 38.9 degrees okay so now we get to this third problem a rescue helicopter is hovering 156 meters above the sea it is dark and the helicopter is shining a light down onto the surface of the sea if the diameter of the cone formed by the beam of the light is 140 meters what is the angle at the apex of the cone all right so once again we have a cone but our job is going to be to work with one of these triangles either the right one or the left one so that we are able to figure out the angle at the apex of the cone one of the things that we know in this problem is that the diameter of the cone formed by the beam of the light is 140 meters the diameter is twice the radius that tells me the radius is going to be half of 140 which is 70 meters so that's information that we're going to need when we redraw this triangle notice that I have placed Theta at the top because we're interested in the angle at the apex of the cone we know the height is 156 and we also know the radius is 70 which is half of the diameter which is 140. let's go ahead and work with our trigonometric ratios sohcahtoa sine cosine tangent once again according to Theta 70 is our opposite side and 156 is the adjacent side in this case we do not know the hypotenuse and we don't necessarily need to find it in order to get our answer so we're going to go ahead and set up our work as the tangent of theta is equal to opposite which is 70 over adjacent so here Theta is the arc tangent of 70 over 156 take your calculator second tan of 70 divided by 156 and we're going to get the Theta is about 24.16629 and it continues once again because we want the entire angle then that means that we need to make we need to take this angle and multiply that by two when I multiply by 2 I get approximately 48.3 degrees So my answer is 48.3 degrees and now for our final problem each of the following three containers has a volume of a thousand cubic centimeters shape a is a cube of side 20. shape B is a cylinder of diameter 20 centimeters shape c is a right cone of diameter 20 centimeters work out the heights of shapes B and C all right so the information that we know is that we know the volume is 8 000 cubic centimeters shape B and shape c the information that we know for both of them is the diameter which is also measured in centimeters which is great because we don't have to worry about any conversions our job is to figure out the height of these shapes so let's go ahead and start with the cylinder find the formula for the volume of a cylinder which is pi radius squared times the height we're going to fill in the information that we know we know volume is eight thousand Pi we're going to write as Pi because remember there is a pi button in the calculator that we can use the radius in this case has to be half of the diameter the diameter is 20. so half of the diameter is going to be 10. so in this case the radius is 10 squared and then the height is exactly what we are looking for this means that 8 000 is equal to 100 pi times the height let's go ahead and divide by a hundred Pi on both sides take your calculators do 8 000 divided by 100 times pi and you're going to get that the height is approximately 25.5 if you round because you're going to get 25.46 so about 25.5 centimeters is going to be the height of the cylinder so now moving on to find the height of the cone for the cone let's go ahead and look up the volume formula the volume of a cone is 1 3 Pi radius squared times the height notice how similar it is to the cylinder formula except for the fact that it is one third so now we're going to fill in the information we have 8 000 is equal to 1 3 Pi here once again the diameter is 20 so the radius is 10. squared times the height before I divide by 100 Pi I'm going to multiply both sides by three I get 24 000 it's equal to 100 pi times H that is so that I could get rid of the one-third and now I'm going to go ahead and divide by 100 Pi 24 000 divided by 100 Pi is approximately equal to 76.4 centimeters and so we have figured out that the height of the cylinder shape B is about 25.5 centimeters and then the height of shape C is about 76.4 centimeters hope you found this video helpful remember to subscribe to my channel for more help with math so that you can say yes I can do math with confidence until next time thanks for watching
14558	https://math.stackexchange.com/questions/2905526/midpoint-of-segment-in-trilinear-coordinates	geometry - Midpoint of segment in trilinear coordinates - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Midpoint of segment in trilinear coordinates Ask Question Asked 7 years ago Modified7 years ago Viewed 124 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Suppose you have two points p 1=x 1:y 1:z 1 p 1=x 1:y 1:z 1 and p 2=x 2:y 2:z 2 p 2=x 2:y 2:z 2 in trilinear coordinates. If p 1=(a 1,b 1,c 1)p 1=(a 1,b 1,c 1) and p 2=(a 2,b 2,c 2)p 2=(a 2,b 2,c 2) in actual directed distances, then the midpoint of the segment p 1 p 2 p 1 p 2 is obviously (a 1+a 2 2,b 1+b 2 2,c 1+c 2 2)(a 1+a 2 2,b 1+b 2 2,c 1+c 2 2), by considering the right trapezoids formed when dropping the perpendiculars. However, this fails when considering the proportional coordinates. How can you calculate the midpoint of a segment if you are not given the absolute trilinear coordinates of the endpoints? geometry analytic-geometry coordinate-systems Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Sep 4, 2018 at 20:39 Cois MonisCois Monis 123 5 5 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. No problem at all. If you´re given only the proportional coordinates of a point, nothing prevents you from calculating the absolute trilinear coordinates. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 15, 2018 at 1:17 MrDudulexMrDudulex 738 4 4 silver badges 12 12 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry analytic-geometry coordinate-systems See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1What is the maximum point for which number of way to reach is given 1Proof for points of intersecting circles. 0Let A(a 1,a 2),B(b 1,b 2)A(a 1,a 2),B(b 1,b 2). 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14559	https://www2.math.uconn.edu/~stein/math1070/Slides/math1070-140notes.pdf	Linear Programming Problems Linear programming problems come up in many applications. In a linear programming problem, we have a function, called the objective function, which depends linearly on a number of independent variables, and which we want to optimize in the sense of either ﬁnding its mini-mum value or maximum value. Depending linearly means that the objective function is in the form of a linear polynomial, a polynomial in which each of the variables occurs to the ﬁrst power and none of the variables are multiplied together. Constraints The variables are also subject to conditions, in the form of linear inequalities. These are called constraints. The variables must also satisfy the non-negativity condition: they can’t be negative. The set of points, or values of the variables, which satisfy the con-straints and the non-negativity condition is called the feasible set. Fundamental Theorem of Linear Programming Theorem 1 (Fundamental Theorem of Linear Programming). The optimal value of the objective functions must occur at one of the vertices of the feasible set. We can understand this fairly easily for a case where there are two independent variables. The notation gets more complicated when there are more variables, but the basic ideas carry through. Consider the values of an objective function p = αx + βy along a line y = mx + b. If we plug mx + b in for y in the formula for p, we get p = αx + β(mx + b) = αx + βmx + βb or p = (α + βm)x + b. If α +βm > 0, then p increases when x increases and p decreases when x decreases. If α +βm < 0, then p decreases when x increases and p increases when x decreases. In either case, p increases when we go along the line in one direction and decreases when we go along the line in the opposite direction. A third possibility is that α + βm = 0, in which case p never changes. In any of those cases, along any segment of the line, any extreme value for p must occur at an endpoint. 1 2 If an extreme value occurs in a closed polygon, we easily see the ex-treme value must occur along the boundary, and then if it occurs along the boundary it must occur at one of the vertices, which is what the Fundamental Theorem of Linear Programming says. In a linear programming problem with just two variables and a hand-ful of constraints, it’s easy to sketch the feasible set and ﬁnd its vertices. This is the essence of solving linear programming problems geometri-cally. • Find the feasible set. • Find the vertices. • Evaluate the objective function at each vertex. • Pick out the optimal value for the objective function. We’ll do some examples to help understand linear programming problems, but most linear programming problems that come up in real life involve numerous variables and constraints and eﬀectively require a more eﬃcient approach. The most common approach is called the Simplex Method. The Elephant in the Room Linear programming problems don’t come out of thin air; there are real problems that, when translated, turn into linear programming problems. The process of turning a real problem into a linear programming prob-lem is the same involved in any other word problem: Read the prob-lem!!! In particular, • Look for variables and unknowns. • Find connections among the variables and unknowns. In this case, the connections translate into constraints along with the objective function. The Simplex Method In most word problems studied before, the connections translated to equations, which were then solved. In linear programming problems, we generally use the Simplex Method. An Example Consider the following maximum problem: 3 2x + y ≤3 3x + y ≤4 x ≥0, y ≥0 p = 17x + 5y Vertices The vertices of the feasible set and the values of the objective func-tion p = 17x + 5y at those points are the following: (0, 0) 0 (4/3, 0) 222 3 (1, 1) 22 (0, 3) 15 . Thus, the maximum value for p is 222 3, which occurs at (4/3, 0). The Simplex Method With the Simplex Method, we do not need to evaluate the objective function at each vertex. We will illustrate the Simplex Method using this example and then generalize the method so we can perform it mechanically, as an algo-rithm. The ﬁrst step is to change the contraints from inequalities into equa-tions by introducing what are called slack variables. The slack variables essentially take up the slack in the inequalities. Looking at the inequality 2x + y ≤3, we introduce the slack variable u = 3 −(2x + y), so u ≥0 while 2x + y + u = 3. Similarly, looking at the inequality 3x + y ≤4, we introduce the slack variable v = 4 −(3x + y), so v ≥0 while 3x + y + v = 4. Restating the Problem We can thus restate the problem as 2x + y + u = 3 3x + y + v = 4 x, y, u, v ≥0 p = 17x + 5y where we ﬁnd the maximum value for the objective function p which satisﬁes the contraints, now given as equations, and the non-negativity condition. A New System of Equations 4 If we rewrite the formula for the objective function in the form −17x −5y + p = 0, we get a system of equations: 2x + y + u = 3 3x + y + v = 4 −17x −5y + p = 0. We will write down the augmented matrix for this system, use it to easily pick out one of the vertices and the value of the objective function at that vertex, and then pivot in a way that will enable us to ﬁnd another vertex at which the value for the objective function is larger. The Initial Simplex Tableau The augmented matrix for the system 2x + y + u = 3 3x + y + v = 4 −17x −5y + p = 0. is     x y u v p 2 1 1 0 0 3 3 1 0 1 0 4 −17 −5 0 0 1 0     The ﬁrst row isn’t actually part of the augmented matrix, but we’ll write it down as a reminder of which variable each column is associated with. The Initial Tableau     x y u v p 2 1 1 0 0 3 3 1 0 1 0 4 −17 −5 0 0 1 0     Looking at the matrix, it’s fairly obvious that the columns corre-sponding to x and y are the most complicated. We can easily read oﬀ a solution if we set x = 0 and y = 0. If we do that, the ﬁrst row, which corresponds to the equation 2x + y + u = 3, gives a solution u = 3. Similarly, the second row, which corresponds to the equation 3x+y+v = 4, gives a solution v = 4. The third row, which corresponds to the equation −17x −5y + p = 0, gives a solution p = 0. A Solution 5 So we have a solution of the equations giving x = 0, y = 0, u = 3, v = 4 and p = 0. This corresponds to the vertex (0, 0), which gives a value of 0 for the objective function. We will call x and y Group I variables and we will call u and v Group II variables. Each time we pivot, we will exchange one pair of variables, one from Group I and one from Group II, to get another set of Group I variables which will be set to 0. This will give another fairly obvious solution, which will correspond to another vertex and another value of the ob-jective function. The Objective Function The last row gives a formula −17x−5y+p = 0 involving the objective function. We can solve this for the objective function, getting p = 17x + 5y. It may seem as if we’re going around in circles, but right now we’re trying to understand a process; later we will synthesize the ideas into an eﬃcient algorithm. Since at the solution we found, x = y = 0, p takes on the value 0. On the other hand, it’s clear that if we found another solution where either x or y was positive, we’d get a larger value for p. It should also be clear that increasing x will increase p faster than increasing y would, so we’ll pivot in a way that will make x a Group II variable. We’ll do this by pivoting about the column corresponding to x, in other words, we’ll pivot about the ﬁrst column. Finding the Pivot Row We now have to decide which entry in the ﬁrst column to pivot around. Since the last row corresponds to the objective function, which is diﬀerent in character from the others, we won’t even consider pivoting about the last row. If we pivoted about the ﬁrst row, we’d start by dividing the ﬁrst row by 2, getting 3/2 in the last column. We’d next subtract 3 times the ﬁrst row from the second, which would give −1 2 in the last column. Since negative numbers are evil, we’ll avoid that by pivoting about the second row. So we’ll pivot about the entry in the second row, ﬁrst column. The Pivot   2 1 1 0 0 3 3 1 0 1 0 4 −17 −5 0 0 1 0   First, we divide the second row by 3 to get a 1 at the pivot point: 6   2 1 1 0 0 3 1 1/3 0 1/3 0 4/3 −17 −5 0 0 1 0   To get 0’s elsewhere in the ﬁrst column, we’ll subtract 2× the pivot row from the ﬁrst row and add 17× the pivot row to the third row: R1 ←R1 −2R2, R3 = R3 + 17R2:   0 1/3 1 −2/3 0 1/3 1 1/3 0 1/3 0 4/3 0 2/3 0 17/3 1 68/3   Interpreting the New Matrix We’ll rewrite the matrix with the columns again labeled:     x y u v p 0 1/3 1 −2/3 0 1/3 1 1/3 0 1/3 0 4/3 0 2/3 0 17/3 1 68/3     This time, the Group II variables are y and v and we’ll set them to 0. We can then interpret the equations. The ﬁrst row corresponds to the equation 1 3y + u −2 3v = 1 3. Since y = v = 0, we get u = 1 3. The second row corresponds to the equation x + 1 3y + 1 3v = 4 3. Again, since y = v = 0, we get x = 4 3. The third row corresponds to the equation 2 3y + 17 3 v + p = 68 3 . Once again, since y = v = 0, we easily get p = 68 3 . This corresponds to the vertex (4/3, 0), giving the value 68 3 = 222 3 for the objective function. Another Look at the Last Row Looking at the last row, corresponding to the equation 2 3y+ 17 3 v+p = 68 3 , more closely, we can solve for p to get: p = 68 3 −2 3y −17 3 v. We just found a vertex corresponding to values of 0 for y and v. Any other solution would either involve still having 0 values for y and v, which would give the same value for p, or would involve positive values for one or both of y and v, which would give a smaller value for p. Obviously, we have found the largest possible value for the objective function. We’ll analyze what we just did, synthesize it, and come up with the Simplex Method. 7 The Simplex Method (1) The ﬁrst step is to take each inequality and introduce a slack variable to get an equation. (2) We also rewrite the formula for the objective function, bringing all the variables over to the left side. (3) We now write down the augmented matrix for the system of equations, with the folowing optional adjustment: Since we never pivot about the last row, the next to last column never changes. Eﬀectively, the next to last column is unnecessary, so we don’t have to write it down. At this point, we have the Initial Simplex Tableau. Pivoting We then pivot, choosing where to pivot as follows: (1) We look at the bottom row, look at the negative entries, and choose the column for which the negative entry in the bottom row has the largest magnitude. That will be the pivot column. If there are no negative entries, then we are ﬁnished. (2) Look at the rows with positive entries in both the pivot column and the last column. In each such row, compute the quotient of the entry in the last column divided by the entry in the pivot column. The row for which this quotient is smallest is the pivot row. Once we have chosen the pivot row and pivot column, we pivot. The End Game We continue pivoting until we can’t pivot any more, which occurs when there are no negative entries in the bottom row. At this point, we can set the Group I variables to 0, easily ﬁnd the values for the Group II variables, and the entry in the bottom right of the matrix will be the value of the objective function. Complications This works well as long as all the contraints are in the form (Linear Polynomial) ≤(Positive Constant) and we are trying to maximize the objective function. Things get more complicated if the inequality goes the other way, which generally occurs when we are trying to minimize the objective function and can also occur occasionally in maximum problems. Dealing with minimum problems is another complication. Dealing with ≥ 8 If the inequality goes the other way, we can deal with it by multi-plying both sides by −1. In other words, if we have an equality of the form P ≥k, where P is a linear polynomial and k is a constant, we can instead use the equivalent inequality −P ≤−k. The complication is that this is likely to lead to a negative number on the right hand side. We need to see how to deal with that. Dealing with Negative Constants The problem with having a negative constant is that when we pick out what we hope will be a feasible solution, one of the variables will fail to satisfy the non-negativity condition. We deal with that via some preliminary pivoting on the Initial Simplex Tableau. Preliminary Pivoting If there is a negative entry in the last column of a tableau (excluding the bottom row), we look in that row for a negative entry. The column containing that entry becomes our pivot column. Note: • If there is more than one row with a negative entry in the last column, we can look at any of those rows. • If there is more than one negative entry in that row, we may choose any of the columns with a negative entry as the pivot column. Once we have the pivot column, we choose the pivot row in the usual way, with the following adjustment: We consider all rows where the quotient of the entry in the last column divided by the entry in the pivot column is positive. We then pivot and repeat this preliminary process until there are no more negative entries in the last column. At that point, we proceed with the Simplex Method in the usual way. Minimizing the Objective Function We minimize the objective function by maximizing its additive in-verse, that is, its negative. In other words, if we need to minimize p, we simply maximize M = −p. When we are ﬁnished pivoting, we then merely recognize that the en-try in the lower right hand corner of the ﬁnal simplex tableau is the negative of minimum value for the objective function.
14560	https://ro.scribd.com/document/486930060/ciclo-de-refrigeracion-por-compresion-de-vapor	Ciclo de Refrigeración Por Compresión de Vapor \| PDF \| Refrigeración \| Ramas de la termodinámica Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 56 views 6 pages Ciclo de Refrigeración Por Compresión de Vapor El documento describe el ciclo de refrigeración por compresión de vapor, que consta de un compresor, condensador, válvula de expansión y evaporador. Se analizaron los datos del ciclo utiliza… Full description Uploaded by Liza Bejarano AI-enhanced title and description Go to previous items Go to next items Download Save Save ciclo de refrigeración por compresión de vapor For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save ciclo de refrigeración por compresión de vapor For Later You are on page 1/ 6 Search Fullscreen FDFBA KM YM@YDNMYGFDÖO XAY FAJXYMVDÖO KM PGXAY Bdzg Barmog Imlgrgoa Yakrênumz 970=456 Kafmotm< Odkdg Artdz D.A I L M[D P A V D.7 Ai lm td v a n mo mr g b Gogbdzgr y rmfaoafmr mb fajpartgjdmota km bas fdfbas tmrjakdoçjdfas km sdstmjgs km rm`rdnmrgfdöo par fajprmsdöo km vgpar. 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14561	https://thirdspacelearning.com/us/blog/what-is-rounding-numbers-math/	What Is Rounding Numbers In Math? Explained For Elementary Math Tutoring for Schools AI Math Tutor – Elementary Programs – Middle School Programs – High School Programs How it Works Impact Use cases Scheduled Tutoring On-Demand Sessions Homework Test Prep Substitute Coverage Virtual Paraprofessional Pricing Resources Math Resources (Free) Math Topic Guides (Free) – Algebra – Measurement and Data – Geometry – Number and Quantity – Ratio and Proportion – Statistics and Probability Math Standards by State Blog Try for free Log in 🇺🇸 US 🇬🇧 UK 🇺🇸 US 🇬🇧 UK Math Tutoring for Schools AI Math Tutor – Elementary Programs – Middle School Programs – High School Programs How it Works Impact Use cases Scheduled Tutoring On-Demand Sessions Homework Test Prep Substitute Coverage Virtual Paraprofessional Pricing Resources Math Resources (Free) Math Topic Guides (Free) – Algebra – Measurement and Data – Geometry – Number and Quantity – Ratio and Proportion – Statistics and Probability Math Standards by State Blog Try for free Log in Interventions School Interventions Tutoring Administration Leadership Funding Instruction Strategies Professional Development Elementary School Operations & Algebraic Thinking Number & Operations (Base Ten) Number & Operations (Fractions) Geometry Elementary Math (Other) Elementary Worksheets & Questions Middle & High School Middle School Math High School Math Middle & High Worksheets & Questions Assessment Types of Assessments Math Tests More Math Support Home Learning Dictionary & Wiki Games & Activities Worksheets Problems & Questions Word Problems TSL Updates x NEW LOWER-COST TUTORING Introducing Skye, your students’ AI voice tutor Adaptive, dialogue-driven one-on-one tutoring built by math teachers Unlimited sessions for as many grade 3-8 students as need it One fixed low yearly cost no matter how many sessions you schedule Meet Skye A personal math tutor for every student that needs it "This innovative one-on-one math tutoring solution offers a cost-effective alternative to traditional one-on-one tutoring." Meet Skye Free ready-to-use math resources Hundreds of free math resources created by experienced math teachers to save time, build engagement and accelerate growth Explore all resources Contents Number & Operations (Base Ten) What Is Rounding in Math: Rounding Numbers Explained for Teachers, Parents and Kids March 31, 2025 \| 4 min read Sophie Bartlett Here you can find out about rounding numbers, why it is used, and how you can help children understand it. Rounding digits requires a solid understanding of place value. Students will need to be aware of the number of decimal places or significant figures a value has to go from that original number to a rounded number. What is rounding in math? Rounding to the nearest 10 Rounding to the nearest 100 Rounding to the nearest integer or whole number Rounding to the nearest tenth When do children learn about rounding numbers in elementary school? How does rounding numbers relate to other areas of math? Rounding in math in real life 3 worked examples for rounding in math 5 rounding in math practice questions and answers FAQs Rounding Numbers Check for Understanding Use this quiz to check grades 3-5 students' understanding of rounding numbers Download Free Now! What is rounding in math? Rounding numbers makes them ‘easier’ to use or understand while also keeping the number close to its original value. Instead of using exact numbers, simpler values can be used. For example, 189.2 could be rounded to 189, 190 or 200, depending on the degree of accuracy required. In context, $19.99 could be rounded to $20; 56 minutes could be rounded to an hour. Rounding to the nearest 10 Students first officially encounter this in 3rd grade. In order to round to the nearest 10, children must first understand what multiples of ten are (numbers in the ten times table, or numbers divisible by 10, e.g. 30, 250, 3430 etc.). They must then be able to identify which multiples of 10 a given number is between, e.g. 218 is between 210 and 220 (a number line is often used for this, as seen below). Finally, children should then identify the nearest multiple of 10 – in this case, 220. Children should round numbers up If they lie halfway between two multiples (and therefore end in a 5). In this case, the number will always round up to the multiple of 10 above it. Meet Skye, the voice-based AI tutor making math success possible for every student. Built by teachers and math experts, Skye uses the same pedagogy, curriculum and lesson structure as our traditional tutoring. But, with more flexibility and a low cost, schools can scale online math tutoring to support every student who needs it. Find out more Rounding to the nearest 100 Students first officially encounter this in 3rd grade. In order to round to the nearest 100, children must first understand what multiples of 100 are (numbers in the 100 times table, or numbers divisible by 100, e.g. 500, 2500, 34300 etc.). They must then be able to identify which multiples of 100 a given number is between, e.g. 218 is between 200 and 300 (a number line is often used for this). Finally, children should then identify the nearest multiple of 100 – in this case, 200. If a number lies halfway between (and therefore ends in 50), the number will always round up to the multiple of 100 above it. Rounding to the nearest integer or whole number Students following Common Core State Standards first officially encounter this in 5th grade. Children must be able to identify which whole numbers, also termed ‘integers’, are either side of a decimal number. For example, 6.2 is between 6 and 7; 14.92 is between 14 and 15 (again, a number line is a useful tool in helping to visualize this). They should then work out which of these integers is closer to the decimal number using the numbers to the right of the decimal point. As before, if the decimal number is exactly halfway between two integers (such as 3.5), it will always round up to the integer above it (in this case, 4). Rounding to the nearest tenth Students first officially encounter this in 5th grade. In a similar fashion to the methods previously explained, children should be able to identify the tenths either side of a decimal number. For example, 4.62 is between 4.6 and 4.7; 5.213 is between 5.2 and 5.3. Sometimes these decimal numbers could be written with a 0 in the hundredths place and/or thousandths place value to make it easier to work out the surrounding tenths: in the first example above, 4.62 has 62 hundredths and is therefore between 60 hundredths (4.60, or 4.6) and 70 hundredths (4.70, or 4.7). Finally, children should then work out which of these tenths is closer to the decimal number. As the nearest hundredth of 4.62 is 60, the nearest tenth is will be 6. Once again, if the decimal number is exactly halfway between two tenths (such as 2.75), it will always round up to the tenth above it (in this case, 2.8). When do children learn about rounding numbers in elementary school? Schools following Common Core: In3rd grade children are taught to: Round any number to the nearest 10 or 100. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.3 This progresses to 4th grade, where students are expected to: Use place value understanding to round whole numbers to any place. Assess the reasonableness of answers using mental computation and estimation strategies including rounding And finally to 5th grade: Use place value understanding to round decimals to any place. Other schools: Depending on the state curriculum or standards a school is using, rounding may be taught in different grade levels. If your child is in a district that follows the Texas Essential Knowledge and Skills, rounding to the 10s and 100s begins in 3rd grade, rounding to the nearest hundred thousand is taught in 4th grade, and rounding decimals to the nearest tenth and hundredth is introduced in 5th grade. Teaching 3rd grade students rounding numbers to the nearest 100 in a Third Space Learning online math lesson. How does rounding numbers relate to other areas of math? Rounding is used to estimate and check answers to problems. For example, when multiplying decimals (such as 0.7 x 0.8), students may misinterpret the magnitude of the answer (in this case, perhaps 5.6). Rounding the decimals and then multiplying (0.7 and 0.8 both round to 1, and 1 x 1 = 1) helps to check the accuracy of the answer (5.6 is much larger than 1, so the multiplication could be recalculated to achieve 0.56, which is much closer to 1). Rounding in math in real life Rounding is useful in real life. For example, rounding remainders appropriate to the context: Rounding up: A mini bus holds 12 people and you need to transport 161 people. 161 ÷ 12 = 13.4, but you can’t have 0.4 of a mini bus, so you would need to round up to work out that 14 mini buses would be needed. Rounding down: An egg carton holds 12 eggs and you have 161 eggs to package. 161 ÷ 12 = 13.4, but you can’t have 0.4 of an egg carton and they all must be fully filled, so you would need to round down to work out that 13 egg cartons would be needed. In the first example, 13.4 wouldn’t conventionally be rounded up, but in the context of the problem, it was required to be. Here’s a fun real-life activity parents could get their children to do during their next trip to the grocery store: tell your child you have $15 to spend on fruit and vegetables and they need to determine how many items you could purchase by rounding up their prices to the nearest whole dollar. Find out how accurate their estimates were when you check out! 3 worked examples for rounding in math 1) Round 218 to the nearest multiple of 10. Answer: 220 2) Round 42.4 to the nearest whole number. Answer: 42 3) Round 1.15 to the nearest tenth. Answer: 1.2 (any number that is halfway between always rounds up) Once children are confident with the concept through using a number line, they may be able to round by just looking at two digits: When rounding to the nearest 10, you only need to look at the tens place (to identify if it will change or not) and the ones place (to identify if it will round the tens digit up or down) – e.g. in 3,268, the 8 rounds the 6 ‘up’ to a 7, becoming 3,270 When rounding to the nearest 100, you only need to look at the hundreds digit (to identify if it will change or not) and the tens digit (to identify if it will round the hundreds digit up or down) – e.g. in 742,125, the 2 rounds the 1 ‘down’ to stay as 1, becoming 742,100 When rounding to the nearest 1000, you only need to look at the digits in the thousands and hundreds place – e.g. in 42,517, the 5 rounds the 2 ‘up’ to a 3, becoming 43,000 … and so on. 5 rounding in math practice questions and answers 4th grade Q: Circle the number which is closer to 1000 and explain how you know: 996, 1006 Answer: 996 is closer to 1000 as it is only 4 away, whereas 1006 is 6 away. Q: Round 73,482 to the nearest: a) 10,000 b) 1,000 c) 100 Answer: a) 70,000 b) 73,000 c) 73,500 Q: Round 3,593,174 to the nearest million. Answer: 4,000,000 5th grade Q: John thinks of a whole number. He multiplies it by 4. He rounds his answer to the nearest 10. The result is 50. Write all the possible numbers that John could have started with. Answer: 12 or 13 (Any number from 45 – 54 would be rounded to 50 when being rounded to the nearest 10. There are only two multiples of 4 within that range: 48 and 52. Dividing those by 4 gives you 12 and 13) Q: Round each of these numbers to the nearest 100: a) 20,906 b) 2,090.6 c) 209.06 Answer: a) 20,900 b) 2,100 c) 200 FAQs How do you explain rounding numbers? See relevant section above: rounding to the nearest 10, 100, whole number, tenth or worked examples for rounding in math Should 0.5 be rounded up or down? 0.5 should be rounded up to the next whole number How do you round numbers to the nearest thousand? In order to round to the nearest 1000, children must first understand what multiples of 1000 are (numbers in the 1000 times table, or numbers divisible by 1000, e.g. 5000, 25000, 343000 etc.). They must then be able to identify which multiples of 1000 a given number is between, e.g. 2185 is between 2000 and 3000 (a number line is often used for this). Finally, children should then identify the nearest multiple of 1000 – in this case, 2000. If a number lies halfway between (and therefore ends in 500), the number will always round up to the multiple of 1000 above it. Wondering about how to explain other key math vocabulary to children? Check out our Math Dictionary For Kids and our What is Math Mastery? blog. Do you have students who need extra support in math? Skye—our AI math tutor built by experienced teachers—provides students with personalized one-on-one, spoken instruction that helps them master concepts, close skill gaps, and gain confidence. Since 2013, we’ve delivered over 2 million hours of math lessons to more than 170,000 students, guiding them toward higher math achievement. Discover how our AI math tutoring can boost student success, or see how our math programs can support your school’s goals: – 3rd grade tutoring – 4th grade tutoring – 5th grade tutoring – 6th grade tutoring – 7th grade tutoring – 8th grade tutoring The content in this article was originally written by primary school teacher Sophie Bartlett and has since been revised and adapted for US schools by elementary math teacher Christi Kulesza. Share: Related articles The Best Place Value Chart Ever: Your Free Printable Place Value Accordion 3 min read What Is Place Value? Explained For Elementary School 4 min read x [FREE] Ultimate Math Vocabulary Lists (K-5) An essential guide for your Kindergarten to Grade 5 students to develop their knowledge of important terminology in math. Use as a prompt to get students started with new concepts, or hand it out in full and encourage use throughout the year. Download free Third Space Learning Inc, 3 Germay Dr, Unit 4 #2810, Wilmington 19804 email protected 298-4593Contact usPress enquiries Math Tutoring Math tutoring for schools and districts How it works Funding Impact Texas approved provider New York approved vendor Ohio approved tutoring program Arkansas approved vendor Tutoring FAQs Request a demo for your school Policies Data Protection & Privacy Policy Cookies Policy Terms of Service Popular Blogs What is a prime number? What is mean median mode? Algebra questions Math questions for 5th graders What are vertices faces edges? 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14562	https://www.houseofmath.com/encyclopedia/numbers-and-quantities/vectors/two-dimensions/products-and-vector-calculation/find-the-medians-of-a-triangle-using-vector-calculation	Log inSign up Menu Numbers & QuantitiesStatistics & ProbabilityFunctionsProofs EncyclopediaStudy TipsPen & Paper exercisesExcel / GeoGebra recipes AlgebraGeometryNumbers & QuantitiesStatistics & ProbabilityFunctionsProofs Junior MathMultiplication MasterTreasure TrailWind SurferStack n´load Book a Tutor ... Products and Vector Calculation>How to Find the Medians of a Triangle Using Vector Calculation How to Find the Medians of a Triangle Using Vector Calculation The median theorem shows you that if you draw the lines from the corners of a triangle to the midpoint of their opposite sides, the three lines will intersect at a ratio of . That means the intersection is two thirds of the way from a corner to the midpoint on its opposite side. Example 1 The points , and form a triangle. Find the point where the medians intersect. You can do this in two ways: By using what you know about the median theorem, or by doing all the dirty work yourselves. Either way, you need to find the midpoints , and and the vectors \| \| \| \| \| \| \| \| They give you This in turn gives you \| \| \| \| Because the intersection has two coordinates, and , you need two equations to find the set of coordinates. You can do this by creating two expressions, using two different sums of vectors that both become . Then you can solve the system of equations you end up with to find the two coordinates of .Solve the system of equations:That only confirms what the median theorem told us: The medians intersect of the way from each of the corners to the midpoints of their opposite sides. Finally, you can put the values you got for or back into one of the expressions you created for to find its coordinates: Previous entry What Are Basis Vectors Used For? Next entry How to Parametrize a Line Go back
14563	https://math.stackexchange.com/questions/4134106/dot-product-maximization-with-norm-and-coefficient-constraints	linear algebra - Dot Product Maximization with Norm and Coefficient Constraints - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Dot Product Maximization with Norm and Coefficient Constraints Ask Question Asked 4 years, 4 months ago Modified4 years, 4 months ago Viewed 227 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Let a,b,x∈R n a,b,x∈R n where a a and b b are constant. We wish to find a value of x x so that, a T⋅x=0 a T⋅x=0 ∥x∥=1‖x‖=1 b T⋅x b T⋅x is maximized given the previous constraints Solutions which solve a relaxed version of this optimization problem are also of interest. linear-algebra optimization Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked May 10, 2021 at 15:29 Joseph ZambranoJoseph Zambrano 1,267 11 11 silver badges 20 20 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Let y y denote the component of b b orthogonal to a a, i.e. the vector y=b−b T a a T a a.y=b−b T a a T a a. We will find that the x x that maximizes the objective function is x=y/∥y∥x=y/‖y‖. To see that this holds, let u 1 u 1 and u 2 u 2 denote the unit-vectors in the direction of a a and y y respectively. Notably, u 2=y/∥y∥u 2=y/‖y‖. Because u 1,u 2 u 1,u 2 are orthogonal with length 1 1, we note that ∥x∥2=(u T 1 x)2+(u T 2 x)2+∥x−(u T 1 x)u 1−(u T 2 x)u 2∥2.‖x‖2=(u 1 T x)2+(u 2 T x)2+‖x−(u 1 T x)u 1−(u 2 T x)u 2‖2. Now, the constraint that x T a=0 x T a=0 means that u T 1 x=0 u 1 T x=0. Thus, the above becomes ∥x∥2=(u T 2 x)2+∥x−(u T 2 x)u 2∥2,‖x‖2=(u 2 T x)2+‖x−(u 2 T x)u 2‖2, which means that we must have \|u T 2 x\|≤1\|u 2 T x\|≤1. On the other hand, for any x x that is orthogonal to a a, we have b T x=∥y∥⋅u T 2 x=∥b∥2−(b T a)2∥a∥2−−−−−−−−−−−−√⋅(u T 2 x).b T x=‖y‖⋅u 2 T x=‖b‖2−(b T a)2‖a‖2⋅(u 2 T x). Now, we can see that since we are given that \|u T 2 x\|≤1\|u 2 T x\|≤1, this expression is maximized when u T 2 x=1 u 2 T x=1. The only unit-vector x x for which u T 2 x=1 u 2 T x=1 is x=u 2 x=u 2, which is what we wanted. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited May 10, 2021 at 16:47 answered May 10, 2021 at 16:33 Ben GrossmannBen Grossmann 235k 12 12 gold badges 184 184 silver badges 358 358 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra optimization See similar questions with these tags. 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14565	https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10?srsltid=AfmBOor0bjP6x5bTkLXgYZMDgcxnQib42XA8XvDxNf_sTx4aMaQxY6zR	Art of Problem Solving 2011 AIME I Problems/Problem 10 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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Find the sum of all possible values of . Solution 1 Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater). Break up the problem into two cases: an even number of sides , or an odd number of sides . For polygons with sides, the circumdiameter has endpoints on vertices. There are points on one side of a diameter, plus of the endpoints of the diameter for a total of points. For polygons with points, the circumdiameter has endpoint on a vertex and endpoint on the midpoint of the opposite side. There are also points on one side of the diameter, plus the vertex for a total of points on one side of the diameter. Case 1: -sided polygon. There are clearly different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously choices for this point. From there, the other two points must be within the points remaining on the same side of the diameter. So our desired probability is so . and so the polygon has sides. Case 2: -sided polygon. Similarly, total triangles. Again choose the leftmost point, with choices. For the other two points, there are again possibilities. The probability is so and our polygon has sides. Adding, Solution 2 We use casework on the locations of the vertices, if we choose the locations of vertices on the n-gon (where the vertices of the n-gon are in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that . By symmetry, we can assume WLOG that the location of vertex A is vertex . Now, vertex B can be any of . We start in on casework. Case 1: vertex B is at one of the locations . (The ceiling function is necessary for the cases in which n is odd.) Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle. There are choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at , there are possible places for vertex C. Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is . Case 2: vertex B is at one of the locations not covered in the first case. Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in , then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at , then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at , and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case. Therefore, there are total obtuse triangles obtainable. The total number of triangles obtainable is . The ratio of obtuse triangles obtainable to all triangles obtainable is therefore . So, . Now, we have that is divisible by . It is now much easier to perform trial-and-error on possible values of n, because we see that . We find that and both work, so the final answer is . Video Solution 2011 AIME I #10 MathProblemSolvingSkills.com Video Solution ~IceMatrix See also 2011 AIME I (Problems • Answer Key • Resources) Preceded by Problem 9Followed by Problem 11 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
14566	https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-addnformulae-2009-1.pdf	The addition formulae mc-TY-addnformulae-2009-1 There are six so-called addition formulae often needed in the solution of trigonometric problems. In this unit we start with one and derive a second from that. Then we take another one as given, and derive a second one from that. Finally we use these four to help us derive the ﬁnal two. This exercise will improve your familiarity and conﬁdence in working with the addition formulae. The proofs of the formulae are left as structured exercises for you to complete. In order to master the techniques explained here it is vital that you undertake the practice exercises provided. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: • work with the six addition formulae Contents 1. Introduction 2 2. The ﬁrst two addition formulae: sin(A ± B) 2 3. The second two addition formulae: cos(A ± B) 3 4. Deriving the two formulae for tan(A ± B) 5 5. Examples of the use of the formulae 6 www.mathcentre.ac.uk 1 c ⃝mathcentre 2009 1. Introduction There are six so-called addition formulae often needed in the solution of trigonometric problems. In this unit we start with one and derive a second from that. Then we take another one as given, and derive a second one from that. And then we are going to use these four to help us derive the ﬁnal two. This exercise will improve your familiarity and conﬁdence in working with the addition formulae. 2. The ﬁrst two addition formulae: sin(A ± B) The formula we are going to start with is sin(A + B) = sin A cos B + cos A sin B This is called an addition formula because of the sum A + B appearing the formula. Note that it enables us to express the sine of the sum of two angles in terms of the sines and cosines of the individual angles. Key Point sin(A + B) = sin A cos B + cos A sin B We now want to look at sin(A −B). We can obtain a formula for sin(A −B) by replacing the B in the formula for sin(A + B) by −B. Then sin(A −B) = sin A cos(−B) + cos A sin(−B) We now use the following important facts: cos(−B) = cos B, but sin(−B) = −sin B. Then sin(A −B) = sin A cos B −cos A sin B This is the second of our addition formulae. Key Point sin(A −B) = sin A cos B −cos A sin B www.mathcentre.ac.uk 2 c ⃝mathcentre 2009 Exercise 1 O B A P Q R S T 1 1. By using right angled triangle OSR, in which the length of OS equals 1, determine the length of OR in terms of angle B. 2. By using the answer of part 1 and right angled triangle ORQ determine the length of QR in terms of angles A and B. 3. By using the answer of part 2 determine the length of PT. 4. What is ∠TRO? 5. What is ∠TRS? 6. What is ∠RST? 7. By using right angled triangle OSR determine the length of RS. 8. By using the answer of part 7 and right angled triangle RST determine the length of TS. 9. By using the answers of parts 3 and 8 determine the length of PS. 10. By using the answer of part 9 and right angled triangle OSP determine sin(A + B). 3. The second two addition formulae: cos(A ± B) This time, the addition formula we are going to start with is cos(A + B) = cos A cos B −sin A sin B Key Point cos(A + B) = cos A cos B −sin A sin B www.mathcentre.ac.uk 3 c ⃝mathcentre 2009 We want to use this to derive another formula for cos(A −B). To do this, as before, we replace B with −B. This gives cos(A −B) = cos A cos(−B) −sin A sin(−B) But cos(−B) = cos B and sin(−B) = −sin B, and so cos(A −B) = cos A cos B + sin A sin B Key Point cos(A −B) = cos A cos B + sin A sin B So we’ve now got four addition formulae. We will summarise them all here: Key Point sin(A + B) = sin A cos B + cos A sin B sin(A −B) = sin A cos B −cos A sin B cos(A + B) = cos A cos B −sin A sin B cos(A −B) = cos A cos B + sin A sin B Exercise 2 Refer back to the ﬁgure in Exercise 1. Use a similar strategy to that of exercise 1 to determine lengths PQ (=TR), OQ and hence OP. From this determine cos(A + B). www.mathcentre.ac.uk 4 c ⃝mathcentre 2009 4. Deriving the two formulae for tan(A ± B) From the four formulae we have seen already, it is possible to derive two more formulae. We can derive a formula for tan(A + B) from the earlier formulae by noting that tan(A + B) = sin(A + B) cos(A + B) Then, tan(A + B) = sin(A + B) cos(A + B) = sin A cos B + cos A sin B cos A cos B −sin A sin B This result gives tan(A + B) in terms of sines and cosines. We now look at how we can write it directly in terms of tan A and tan B. We do this by dividing every term, both top and bottom, on the right-hand side by cos A cos B. This produces tan(A + B) = sin A cos B cos A cos B + cos A sin B cos A cos B cos Acos B cos Acos B −sin A sin B cos A cos B Cancelling common factors where possible produces tan(A + B) = sin A cos B cos A cos B + cos A sin B cos A cos B cos A cos B cos A cos B −sin A sin B cos A cos B so that tan(A + B) = tan A + tan B 1 −tan A tan B We can do the same with tan(A −B) which would produce tan(A −B) = tan A −tan B 1 + tan A tan B Key Point tan(A + B) = tan A + tan B 1 −tan A tan B tan(A −B) = tan A −tan B 1 + tan A tan B www.mathcentre.ac.uk 5 c ⃝mathcentre 2009 5. Examples of the use of the formulae Let’s have a look at some fairly typical examples of when we need to use the addition formulae. Example Suppose we know that sin A = 3 5 and that cos B = 5 13 where A and B are acute angles. Suppose we want to use this information to ﬁnd sin(A+B) and cos(A−B). Before we can use the addition formulae we need to know expressions for cos A and sin B. We can ﬁnd these by referring to the right-angled triangle in Figure 1. 3 5 A Figure 1. A right-angled triangle constructed from the given information: sin A = 3 5 Using Pythagoras’ theorem we can deduce that the length of the third side is 4 as shown in Figure 2. Hence cos A = 4 5. 3 4 A 5 Figure 2. From the right-angled triangle, cos A = 4 5 Similarly, given that cos B = 5 13, then by reference to the triangle in Figure 3 and by using Pythagoras’ theorem we can deduce that sin B = 12 13. 12 13 5 B Figure 3. From the triangle sin B = 12 13. We are now in a position to use the addition formulae: sin(A + B) = sin A cos B + cos A sin B = 3 5 × 5 13 + 4 5 × 12 13 = 15 65 + 48 65 = 63 65 cos(A −B) = cos A cos B + sin A sin B = 4 5 × 5 13 + 3 5 × 12 13 = 20 65 + 36 65 = 56 65 This is one way in which the formulae can be used. www.mathcentre.ac.uk 6 c ⃝mathcentre 2009 Example Suppose we are asked to ﬁnd an expression for sin 75◦, not by using a calculator but by using a combination of other known quantities. Note that we can rewrite sin 75◦as sin(45◦+ 30◦) and then use an addition formula. We have speciﬁcally chosen the values 45◦and 30◦because of the standard results that sin 45◦= cos 45◦= 1 √ 2, sin 30◦= 1 2 and cos 30◦= √ 3 2 . Then sin(A + B) = sin A cos B + cos A sin B sin(45◦+ 30◦) = sin 45◦cos 30◦+ cos 45◦sin 30◦ = 1 √ 2 × √ 3 2 + 1 √ 2 × 1 2 = √ 3 2 √ 2 + 1 2 √ 2 = √ 3 + 1 2 √ 2 Example Suppose we wish to ﬁnd an expression for tan 15◦using known results. Note that 15◦= 60◦−45◦ and also that tan 60◦= √ 3 and tan 45◦= 1. tan 15◦ = tan(60◦−45◦) = tan 60◦−tan 45◦ 1 + tan 60◦tan 45◦ = √ 3 −1 1 + √ 3 × 1 = √ 3 −1 √ 3 + 1 It would be more usual to tidy this result up to avoid leaving a root in the denominator. This can be done by multiplying top and bottom by the same quantity, as follows: √ 3 −1 √ 3 + 1 = ( √ 3 −1) √ 3 + 1 × ( √ 3 −1) √ 3 −1 = 3 − √ 3 − √ 3 + 1 3 −1 = 4 −2 √ 3 2 = 2 − √ 3 www.mathcentre.ac.uk 7 c ⃝mathcentre 2009 Example In this Example we use an addition formula to simplify an expression. Suppose we have sin(90◦+ A) and we want to write it in a diﬀerent form. We can use the ﬁrst addition formula as follows: sin(90◦+ A) = sin 90◦cos A + cos 90◦sin A = cos A since sin 90◦= 1 and cos 90◦= 0. So sin(90◦+ A) can be written in the simpler form cos A. Example Suppose we wish to simplify cos(180◦−A). cos(180◦−A) = cos 180◦cos A + sin 180◦sin A = −cos A since cos 180◦= −1 and sin 180◦= 0. So we can see that these addition formulae help us to simplify quite complicated expressions. Exercise 3 1. Verify each of the three addition formulae (i.e. for sin(A + B), cos(A + B), tan(A + B)) for the cases: a) A = 60◦, B = 30◦ and b) A = 45◦, B = 45◦. 2. Verify each of the three subtraction formulae (i.e. for sin(A−B), cos(A−B), tan(A−B)) for the cases: a) A = 90◦, B = 60◦ and b) A = 90◦, B = 45◦. 3. Angles A, B and C are acute angles such that sin A = 0.1, cos B = 0.4, sin C = 0.7. Without ﬁnding angles A, B and C, use the addition formulae to calculate, to 2 decimal places, a) sin(A + B) b) cos(B −C) c) sin(C −A) d) cos(A + C) e) tan(B −A) f) tan(C + B) [Hint: Work to 4 decimal places when ﬁnding cos A, tan A, etc.] 4. By ﬁnding the angles A, B and C in question 3 verify your answers. Answers Exercise 1 1. cos B 2. sin A cos B 3. sin A cos B 4. A 5. 90o −A 6. A 7. sin B 8. cos A sin B 9. sin A cos B + cos A sin B 10. sin A cos B + cos A sin B Exercise 2 PQ = TR = RS sin A = sin A sin B OQ = OR cos A = cos A cos B OP = OQ - PQ = cos A cos B −sin A sin B cos(A + B) = OP = cos A cos B −sin A sin B Exercise 3 3. a) 0.95 b) 0.93 c) 0.63 d) 0.64 e) 1.78 f) -2.63 www.mathcentre.ac.uk 8 c ⃝mathcentre 2009
14567	https://www.collegevine.com/faq/18214/struggling-with-grid-in-questions-on-math-tests	Struggling with Grid-In Questions on Math Tests \| CollegeVine Skip to main content SageChancing Schools expand_more Explore CollegesRankings Resources expand_more Homework HelpExpert FAQBlog ArticlesEssay GuidesLivestreamsScholarshipsCommunityPeer essay review Calculate your chances My feedCollegeVine Expert FAQ What are your chances of acceptance? ### Calculate for all schools Your chance of acceptance Duke University Loading… UCLA Loading… + add school Your chancing factors Unweighted GPA:3.7 1.0 4.0 SAT: 720 math 200 800 \| 800 verbal 200 800 Gender not specified Extracurriculars + add Low accuracy (4 of 18 factors) Add more factors › Struggling with Grid-In Questions on Math Tests Hey guys, I've been having trouble with grid-in questions on math tests. Does anyone have any tips or resources to help me practice and improve my accuracy on these types of questions? Any help would be appreciated! 2 years ago Hey there! I understand that grid-in questions can be a bit challenging, but with practice, you'll definitely improve your accuracy. Here are a few tips and resources to help you out: Read the question carefully: Make sure you understand what is being asked before trying to solve the problem. Break it down into smaller steps if necessary. Show your work: Grid-in questions typically require more complex calculations, so it's essential to write down your steps as you solve the problem. This will make it easier for you to trace any mistakes if you get the wrong answer and need to backtrack. Manage your time efficiently: Time management is crucial for any math test, but grid-in questions can be particularly time-consuming. Make sure you allocate enough time to work through them but don't spend all your time on them, as you'll need to complete other question types as well. Double-check your answers: Sometimes, errors can come from simple math mistakes or misreading the question. Take a moment to go back and double-check your work before moving on to the next question. As for resources, there are several online test prep sites that offer practice grid-in questions: Khan Academy: They have SAT Math practice problems that cover a variety of topics. Look for the questions labeled "Grid-Ins" for targeted practice. ( CollegeVine: CollegeVine has free SAT and ACT prep courses that cover the math sections. Although not every question is a grid-in, there are still ample math practice questions to work through. ( College Board: While not explicitly labeled as grid-in questions, you can find some free-response math questions in their official SAT practice tests. Look through the Math section for questions where a numerical answer is required rather than selecting from multiple-choice options. ( 2 years ago About CollegeVine’s Expert FAQ CollegeVine’s Q&A seeks to offer informed perspectives on commonly asked admissions questions. Every answer is refined and validated by our team of admissions experts to ensure it resonates with trusted knowledge in the field.
14568	https://www.cuemath.com/geometry/difference-between-square-and-rhombus/	Difference Between Square and Rhombus The main difference between a square and a rhombus is that all the angles of a square are equal to 90°, whereas, the angles of a rhombus are not equal to 90°. It is often said that a square is a rhombus, but a rhombus isn’t always a square because a square fulfills all the properties of a rhombus. When a rhombus satisfies all the properties of a square, it becomes a square. Let us learn more about the difference between square and rhombus. \| \| \| --- \| \| 1. \| Difference Between Square and Rhombus \| \| 2. \| Square and Rhombus Formulas \| \| 3. \| Similarities Between Square and Rhombus \| \| 4. \| FAQs on Difference Between a Square and Rhombus \| Difference Between Square and Rhombus A square is a two-dimensional figure that has four equal sides and four equal angles. The sides of a square are perpendicular to each other and its diagonals are of equal length. A rhombus is a quadrilateral in which the opposite sides are parallel and the opposite angles are equal. The sides of a rhombus are not perpendicular to each other and the diagonals of a rhombus are not equal in measure. The following figure helps to identify and differentiate between a square and a rhombus. Observe the following table that shows the difference between a square and a rhombus. \| SQUARE \| RHOMBUS \| --- \| \| \| \| --- \| \| All the angles of a square are equal and measure 90° each. \| Only the opposite angles of a rhombus are equal to each other. \| \| The diagonals of a square are of equal length. \| The diagonals of a rhombus are of unequal length. \| \| A square can be inscribed in a circle. \| A rhombus cannot be inscribed in a circle. \| \| A square has four lines of symmetry. \| A rhombus has two lines of symmetry. \| \| The sides of a square are perpendicular to each other. \| The sides of a rhombus are not perpendicular to each other. \| A square can be inscribed in a circle. A rhombus cannot be inscribed in a circle. A square has four lines of symmetry. A rhombus has two lines of symmetry. Square and Rhombus Formulas Once we know the length of the sides and the length of the diagonals of a given square and a rhombus, we can use the following formulas to calculate their area and perimeter. Similarities between Square and Rhombus Squares and rhombuses have many differences but they are also similar in many ways. The following table shows the similarities between a square and a rhombus. \| SQUARE \| RHOMBUS \| --- \| \| \| \| --- \| \| A square is a quadrilateral. \| A rhombus is also a quadrilateral. \| \| The diagonals of a square bisect each other at right angles. \| The diagonals of a rhombus also bisect each other at right angles. \| \| The sum of the four interior angles of a square is 360°. \| The sum of the four interior angles of a rhombus is 360°. \| \| The opposite sides of a square are parallel. \| The opposite sides of a rhombus are also parallel. \| \| All the sides of a square are of equal length. \| All the sides of a rhombus are also of equal length. \| Related Articles Check out the following articles related to the difference between square and rhombus. Solved Examples Example 1: Identify if the given parallelogram ABCD is a square or a rhombus if the sides AB, BC, CD, and DA are equal to each other and the interior angles ∠A, ∠B, ∠C, ∠B are all equal to 90°. Solution: The question mentions that ABCD is a parallelogram. It is given that the sides AB = BC = CD = DA; and ∠A = 90° , ∠B = 90°, ∠C = 90°, ∠D = 90° We know the difference between square and rhombus and the fact that all the sides of a square and a rhombus are of equal length. The interior angles of a square are 90°, but the angles of a rhombus are not 90°. Therefore, we can conclude that the given parallelogram is a square. Example 2: Using the properties of a square and rhombus, write true or false for the following statements: a.) All the angles of a square are equal and measure 90° each. b.) All the sides of a rhombus are of equal length. Solution: go to slidego to slide Book a Free Trial Class Practice Questions go to slidego to slide FAQs on Difference Between Square and Rhombus What is the Difference Between Square and Rhombus? The main difference between a square and a rhombus is that all the angles of a square are equal to 90°, and hence are equal in measure, but in the case of a rhombus, only the opposite angles are equal. Apart from this, the diagonals of a square are equal, whereas, the diagonals of a rhombus are not equal in length. Can a Square be a Rhombus? Yes, a square comes under the category of a rhombus since it fulfills the properties of a rhombus in which all the sides are equal in length, the diagonals are perpendicular to each other, and the opposite angles are of equal measure. Can a Rhombus be a Square? No, a rhombus cannot be a square unless its diagonals are equal and the interior angles are 90° each. Are Squares and Rhombuses Parallelograms? Yes, a square and a rhombus satisfy all the properties of a parallelogram. Therefore, both squares and rhombuses are parallelograms whose opposite sides are parallel and equal to each other, and the opposite angles are of equal measure. What are Some of the Real-life Examples of a Square and a Rhombus? A few real-life examples of a square are a chessboard, bread slice, and so on. Some real-life examples of a rhombus are, a kite, car windows, and so on. What is Common Between a Square and a Rhombus? There are many similarities between a square and a rhombus. What is the Difference Between the Diagonals of Square and Rhombus? The diagonals of a square are equal, whereas, the diagonals of a rhombus are not equal in length.
14569	https://www.youtube.com/watch?v=LPGVm3x5vEM	Number Theory Basics: Factors and Divisibility Ms. Hearn 9930 subscribers 12 likes Description 592 views Posted: 19 Oct 2019 The goal of this video is to define the terms number theory, divisible by, divides, factor, multiple, and factorization. We also discuss some sample homework questions that involve these terms. 13 comments Transcript: hi I'm miss Hearne let's get started in this video our goal is just to familiarize ourselves with the terms number theory divisible by divides factor multiple and factorization this comes from an area of mathematics known as number theory number theory is a branch of mathematics devoted to the study of the properties of the natural numbers which start with the number one two three and so on continuing indefinitely in that pattern natural numbers are also known as counting numbers or positive integers we'd like to discuss the divisibility of counting numbers a counting number is divisible by another counting number if dividing the first by the second leaves are a manor of zero so for example we would say that ten is divisible by 5 because 5 goes into 10 twice with a remainder of 0 the formal definition of what it means to be divisible is that the natural number a is divisible by the natural number B if there exists a natural number K such that a equals B times K so that sounds a little complicated but all we're saying is that for example 10 is divisible by 5 because 10 can be written as 5 times some number K in this case the number K would be 2 but it doesn't really matter what K is this is the same reason that 5 divides into 10 twice so as long as you can find that number K the number of times at 5 goes into 10 10 is divisible by 5 we don't like having to write 10 is divisible by 5 over and over again so the way that we write it is 5 divides 10 so that's this notation here B divides a there are lots of different terms that we can use to explain the same relationship since five divides ten meaning that ten can be written as a product of five in another number we can explain that same relationship by saying that five is a factor of ten we could also say five is a divisor of ten or we can say ten is a multiple of five all of these really imply the same relationship which is to say the five goes evenly into the number 10 the number 30 equals 6 times 5 this product is called a factorization of 30 because 30 is equal to the product of its two factors 6 is a factor and 5 is a factor and any time you write a number as a product we call that a factorization an example of what you might need to do in your class is decide whether the first number is divisible by the second so we just want to make sure you know what divisible by means and all it means is that it goes evenly into the other number so you would check what's 56 divided by 8 and you can write it in either format and you would see that it goes in 7 times or another way of thinking about it is 56 can be written as the product of eight times seven so since 56 is evenly divided by eight with no remainder we would say yes 56 is divisible by 8 we would write this as 8 divides 56 here's another example is 70 divisible by 6 so if you try dividing 6 into 70 you're going to see that it does not in fact go in evenly six does not go evenly into seventy six goes into 7 once 1 times 6 is 6 subtract bring down 6 goes into 10 once 1 times 6 is 6 subtract and we have a remainder of 4 so 6 does not go evenly into 70 another way of saying that is because the quotient 70 divided by 6 is not a natural number in other words it results in a fraction 70 is not divisible by 6 and we can write that by crossing off the divisibility symbol so 6 does not divide 70 could be written like this you might also be asked to find all the natural number factors of each number in other words all the numbers that divide evenly into the number so for example for Part A we have the number 24 I find it easiest to find all the factors starting with the number 1 and thinking of what it multiplies by to give me 24 so for example 1 times 24 is 24 now usually we like to list all the factors in numerical order so what I do is I have one at the beginning and then I put the number 24 a distance away to make it at the end of my list and then I go up to the number 2 does 2 go into 24 yes 2 times 12 is 24 so both 2 and 12 are factors of 24 and then does 3 go into 24 yes it goes in 8 times and then does 4 go into 24 yes it goes in 6 times does 5 go into 24 no and then once we reach 6 there's no need to check anymore we know we have the whole list once you work your way up to the larger factor the smallest of the larger factors now for Part B you're gonna find that only the number 1 and 13 go in evenly 2 3 4 5 6 and so on do not divide into 13 so the only factors of 13 are 1 and 13 so let's say we were asked to find all the factors of 75 so don't let the fact that it's a larger number intimidate you just be methodical about it so we have one always goes into every natural number evenly 1 incent 1 times 75 is 75 so I'm gonna start and end my list with one in 75 now does 2 go into 75 no 75 is not an even number so it's definitely not gonna go in how about 3 well you can confirm this on your calculator but 3 goes into 75 25 times how about 4 no because no even numbers are gonna go in if 2 doesn't go in right so we're going to skip the even numbers and just check the odds how about 5 you can confirm on your calculator 5 goes into 75 15 times we're gonna skip 6 because it's even how about the number 7 well 7 is not going to go into 75 evenly 7 is going to go into 75 10 times with a remainder of 5 so 7 doesn't work how about 9 9 goes into 75 8 times 8 times 9 is 72 with remainder of 3 so 9 doesn't go in evenly either 10 doesn't go in evenly because 10 goes into 75 seven times with a remainder of 5 let's check 11 11 goes into 75 let's see it looks like it would go in about 6 times 6 times 11 would be 66 and subtracting we would get a remainder of 9 so 11 doesn't work we could try 13 13 goes into 75 we could try maybe 5 times 5 times 13 though is going to give us 65 which has a remainder of 10 so in fact the only factors are 1 3 5 15 25 and 75 how about if we find all the factors of 45 why don't you try pausing the video and see if you can come up with that list of factors yourself okay so to find the factors of 45 we're gonna have one and 45 it's not an even number so 2 doesn't go in so that means none of the even numbers will 3 goes in 15 times 5 goes in 9 times does 7 go in 7 goes into 45 6 times with a remainder of 3 does and we've already checked 9 8 is even so we skip over it we've already checked 9 so here are all the factors of 45 sometimes you're gonna be asked what numbers which factors two numbers have in common in this case they have the factors 1 3 5 and 15 in common so these are known as common factors you're also in our homework for my classes that I teach sometimes you'll have true or false statements about the divisibility of numbers for example here it says if N is a natural number and 9 divides n then 3 divides N and we're supposed to determine if that's true or false well let's think about some numbers that 9 goes into evenly 9 goes into 9 evenly does 3 go into 9 yes it does 3 times 9 times 2 is 18 so 9 divides 18 does 3 divide 18 yeah 3 times 6 is 18 9 goes into 27 evenly 3 times does 3 go into 27 evenly yes it does now why is that happening well the real reason that it's happening is because to say that it's divisible by like the number 27 is divisible by 9 because it can be written as 9 times some number but if 27 can be written as nine times some number 9 is 3 times 3 so 27 can be written as 3 times some number that number being 3 and instead of n so it's always gonna be the case because the 3 goes evenly into the 9 that 3 will also go evenly into any number that 9 goes evenly into another way of saying that is if 9 divides n then n equals 9 a for some natural number a so N equals 3 times 3 times a and that means 3 is a factor of n so this statement is true again that worked because 3 divided 9 now this next one you might confuse with the previous one so you have to be careful it says if N is a natural number and 5 divides n then 10 divides n ok now since 5 divides 10 you might think this is true but let's check what our numbers that 5 goes into evenly that 5 divides 5 divides 5 5 divides 10 5 divides 25 and 5 divides 30 and so on 5 divides all the multiples of 5 but 10 only goes evenly into multiples of 10 10 20 30 oh I skipped they skipped 15 and 20 on my was there hold on I wanna add them in there just to be complete so 10 goes into the 10 evenly it goes into the 20 evenly it goes into 30 evenly but it doesn't go into the 5 the 15 and the 25 so what has happened here well in the previous problem we had the order reversed the question was more similar to if n is divisible by 10 then it's divisible by five that would work because 10 is the larger number to say that ten goes in guarantees at five does also but not the other way around as we can see there are definitely numbers that are divisible by five that are not divisible by 10 so in this case this would actually be false the fact that 10 divides n means n is equal to 10 a which is 2 times 5 times a we would need 5 and 2 both to be factors of n when we say that 5 divides n it only guarantees that N equals 5 a for some natural number a so 5 is a factor but not necessarily the 2 I hope you found this video helpful if you did please remember to give it a thumbs up that helps other students to find the video
14570	http://moniquelowesib.weebly.com/topic-5-energeticsthermochemistry.html	Topic 5: Energetics/Thermochemistry - Monique Lowes' IB Blog Monique Lowes' IB Blog Home About HL subjects Biology > Topic 1: Cell Biology Topic 2: Molecular Biology Topic 3: Genetics Topic 4: Ecology Topic 5: Evolution and Biodiversity Topic 6: Human Physiology (AHL) Topic 7: Nucleic Acids (AHL) Topic 8: Metabolism, Cell Respiration and Photosynthesis (AHL) Topic 9: Plant Biology (AHL) Topic 10: Genetics and Evolution (AHL) Topic 11: Animal Physiology Option D: Human physiology English Language and Literature Business and Management > Unit 1: Business Organization and Environment > Introduction to Business Why Start a Business and What is a Business Plan Types of Organisations Internal and External Stakeholders Business Sectors and Activity Growth of Businesses PEST Analysis Organisational Objectives Threats to Businesses SWOT Analysis Decision Trees Economies and Diseconomies of Scale Globalisation and Multinational Companies The Ansoff Matrix Fishbone Analysis Gantt Charts Force Field Analysis Globalisation CUEGIS Concepts Unit 2: Human Resource Management > Change Unit 3: Finance and Accounts > Sources of Finance Classification of Costs Break-even Analysis Profit and Loss Accounts + Balance Sheets Tangible and Intangible Assets Depreciation Ratios Cash Flow Forecasts Investment Appraisal Budgeting and Variances Cost and Profit Centres Unit 4: Marketing Unit 5: Operations Management > Methods of Production Just-in-time and Just-in-case Capacity Utilisation Stock Control Total Quality Culture SL subjects Chemistry > Topic 1: Stoichiometric Relationships Topic 2: Atomic Structure Topic 3: Periodicity Topic 4: Chemical Bonding and Structure Topic 5: Energetics/Thermochemistry Topic 6: Chemical Kinetics Topic 7: Equilibrium Topic 8: Acids and Bases Topic 9: Redox Processes Topic 10: Organic Chemistry Topic 11: Measurement and Data Processing Mathematics French > Tenses Pronouns Prepositions CAS Feedback and comments energetics/thermochemistry ESSENTIAL IDEA: The enthalpy changes from chemical reactions can be calculated from their effect on the temperature of their surroundings. 5.1 Measuring energy changes UNDERSTANDINGS: 5.1.1 Heat is a form of energy. - 5.1.2 Temperature is a measure of the average kinetic energy of the particles. - 5.1.3 Total energy is conserved in chemical reactions. The Law of Conservation of Mass-Energy: This law was later amended by Einstein in the law of conservation of mass-energy, which describes the fact that the total mass and energy in a system remains constant. This amendment incorporates the fact that mass and energy can be converted from one to another. However, the law of conservation of mass remains a useful concept in chemistry, since the energy produced or consumed in a typical chemical reaction accounts for a minute (small) amount of mass. 5.1.4 Chemical reactions that involve transfer of heat between the system and the surroundings are described as endothermic or exothermic. 5.1.5 The enthalpy change (∆H) for chemical reactions is indicated in kJ mol-1 (kJ per mol). 5.1.6 ∆H values are usually expressed under standard conditions, given by ∆H°, including standard states. Endothermic and exothermic reactions: Chemical reactions involve breaking chemical bonds in the reactants and forming chemical bonds in the products. Chemical energy is absorbed by bond breaking, for example, and as the energy is absorbed, the temperature of the surroundings increase, an endothermic process. Chemical energy is released by bond making, for example,and as the energy is released, the temperature of the surroundings increase, an exothermic process. Enthalpy is the chemical energy stored in bonds that can be converted to heat. It is not possible to measure the absolute enthalpy (H) in a system but the 'change in enthalpy' (∆H), or 'heat of reaction' can be meaured (unit of enthalpy = kJ). Superscript theta (Ɵ) = standard reaction conditions = 1 mole of reactants at standard temperature (298K) and standard pressure (100kPa, 1 atm pressure). A positive value for∆H means that the enthalpy of the system has increased i.e. the reaction is endothermic. A negative value for∆H means that the enthalpy of the system has decreased i.e. the reaction is exothermic. APPLICATION AND SKILLS: Calculation of the heat change when the temperature of a pure substance is changed using q=mc∆T q= energy change for the quanitites used in kJ m = mass of the substance being heated in kg i.e. the water, as the reaction is taking place in dilute solution. The density of pure water is 1.00g cm-3 so the mass of water being heated can be assumed to be the same as the volume of the solution used. c = specific heat capacity of water - 4.18 kJ/kg/K i.e. the energy needed to raise the temperature of one kg of water by 1 Kelvin (1 degree = 1 Kelvin) ∆T = change in temperature of the water NOTE: specific heat capacity is a limitation to calulating enthalpy changes - it is always assumed to be 4.18 kJ/kg/K. If the reaction happens at a slow rate, heat is lost to the environment, making temperature readings unreliable. To compensate for this heat loss,∆T is determined by plotting temperature against time and extrapolating: Sources of error in experimental enthalpy invesitgations: Heat loss - to reduce it, a closed beaker should be used, by insulating around the apparatus or by utilising a polystyrene cup with a lid. Specific heat capacity Essential idea: In chemical transformations energy can neither be created nor destroyed (the first law of thermodynamics) 5.2 Hess's Law UNDERSTANDINGS: 5.2.1 The enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. Hess's law is an application of the law of conservation of energy and states that 'the total enthalpy change for a reaction depends only on the enthalpy difference between the products and the reactants, and is independent of the pathway taken.' ∆Hc° : standard enthalpy of conbustion (enthalpy change for the complete combustion of one mole of a reactant) ∆Hf° : standard enthalpy of formation (enthalpy change for the formation of one mole of a substance from its elements in their standard states).∆H f° values can be used to calculate∆H for a reaction. ∆H =Σ(∆Hf° products ) -Σ(∆Hf° reactants ) Hess's Law is saying that if you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps. If you look at the change on an enthalpy diagram, that is actually fairly obvious. This shows the enthalpy changes for an exothermic reaction using two different ways of getting from reactants A to products B. In one case, you do a direct conversion; in the other, you use a two-step process involving some intermediates. In either case, the overall enthalpy change must be the same, because it is governed by the relative positions of the reactants and products on the enthalpy diagram. If you go via the intermediates, you do have to put in some extra heat energy to start with, but you get it back again in the second stage of the reaction sequence. However many stages the reaction is done in, ultimately the overall enthalpy change will be the same, because the positions of the reactants and products on an enthalpy diagram will always be the same. With these reactions,calculate standard change in enthalpy∆HforC2H6 (g) --> C2H4 (g) + H2 (g) 2C2H6 (g) + 7O2 (g) --> 4CO2 (g) + 6H2O (l) ∆H = -3120 2H2 (g) + O2 (g) --> 2H2O (l) ∆H = -572 C2H4 (g) + 3O2 (g) --> 2CO2 (g) + 2H2O (l) ∆H = -1411 state symbols are essential. Step 1 - divide 1. by 2 to get C2H6 which means the enthalpymust also be divided by 2. Step 2 - reverse reaction 3. which means the enthalpy valuemust change signs i.e. from negative to positive Step 3 - reverse and divide 2. by 2, meaning the enthalpy valuemust change signs AND be halved. Once the steps above are completes, all of the products and reactants OTHER than the ones required for the equation in the opening question should CANCEL OUT. This means the calculations have been done correctly. Essential idea: Energy is absorbed when bonds are broken and is released when bonds are formed. 5.3 Bond enthalpies UNDERSTANDINGS: 5.3.1 Bond-forming releases energy and bond-breaking requires energy. 5.3.2Average bond enthalpy is the energy needed to break one mol of a bond in a gaseous molecule averaged over similar compounds. Definition of bond enthalpy: the energy required to break 1 mole of bonds in gaseous covalent molecules under standard conditions. Because the precise enthalpy of a bond is influenced by other bonds in a molecule, average bond enthalpiesare used in thermochemical calculations. Average bond enthalpies:t he energy required to break 1 mole of a bond in a gaseous molecule averaged over similar compounds. As chemical reactions involve both bond-breaking and bond-making/formation, both the endothermic and exothermic processes occur. ∆H (reaction) =ΣBB -ΣBM i.e. sum of bond breaking in reactant molecules - sum of bond making in the product molecules. A reaction is exothermic over if the bond formation enthalpy is greater than the bond breaking enthalpy i.e. if MORE bonds are made than broken. Enthalpy diagrams (reaction profiles): In an endothermic reaction the enthalpy of the products is greater than that of the reactants i.e. the products are energetically less stable than the reactants. In an exothermic reaction the enthalpy of the products is lower than that of the reactants i.e. the products are energetically more stable than the reactants. Examples of exothermic reactions: combustion respiration 3.neutralisation of acids with alkalis reactions of metals with acids the Thermite Process (s hows aluminium and iron oxide reacting violently in a thermit to produce metallic iron, using magnesium as a fuse. The aluminium changes to aluminium oxide. This is called a redox reaction as it combines reduction with oxidation). bond making Examples of endothermic reactions: thermal decomposition (blast furnace - i ron is extracted from iron ore in a huge container called a blast furnace. Iron ores such ashaematitecontain iron oxide. The oxygen must be removed from the iron oxide to leave the iron behind. Reactions in which oxygen is removed are calledreduction reactions). photosynthesis some types of electrolysis sherbet! bond breaking Proudly powered by Weebly Home About HL subjects Biology > Topic 1: Cell Biology Topic 2: Molecular Biology Topic 3: Genetics Topic 4: Ecology Topic 5: Evolution and Biodiversity Topic 6: Human Physiology (AHL) Topic 7: Nucleic Acids (AHL) Topic 8: Metabolism, Cell Respiration and Photosynthesis (AHL) Topic 9: Plant Biology (AHL) Topic 10: Genetics and Evolution (AHL) Topic 11: Animal Physiology Option D: Human physiology English Language and Literature Business and Management > Unit 1: Business Organization and Environment > Introduction to Business Why Start a Business and What is a Business Plan Types of Organisations Internal and External Stakeholders Business Sectors and Activity Growth of Businesses PEST Analysis Organisational Objectives Threats to Businesses SWOT Analysis Decision Trees Economies and Diseconomies of Scale Globalisation and Multinational Companies The Ansoff Matrix Fishbone Analysis Gantt Charts Force Field Analysis Globalisation CUEGIS Concepts Unit 2: Human Resource Management > Change Unit 3: Finance and Accounts > Sources of Finance Classification of Costs Break-even Analysis Profit and Loss Accounts + Balance Sheets Tangible and Intangible Assets Depreciation Ratios Cash Flow Forecasts Investment Appraisal Budgeting and Variances Cost and Profit Centres Unit 4: Marketing Unit 5: Operations Management > Methods of Production Just-in-time and Just-in-case Capacity Utilisation Stock Control Total Quality Culture SL subjects Chemistry > Topic 1: Stoichiometric Relationships Topic 2: Atomic Structure Topic 3: Periodicity Topic 4: Chemical Bonding and Structure Topic 5: Energetics/Thermochemistry Topic 6: Chemical Kinetics Topic 7: Equilibrium Topic 8: Acids and Bases Topic 9: Redox Processes Topic 10: Organic Chemistry Topic 11: Measurement and Data Processing Mathematics French > Tenses Pronouns Prepositions CAS Feedback and comments
14571	https://www.youtube.com/watch?v=w7VONBEE0g0	Find the Vertices, Foci, Endpoints of the Minor Axis, and Graph the Ellipse The Math Sorcerer 1220000 subscribers 96 likes Description 14751 views Posted: 20 Oct 2020 Find the Vertices, Foci, Endpoints of the Minor Axis, and Graph the Ellipse If you enjoyed this video please consider liking, sharing, and subscribing. Udemy Courses Via My Website: My FaceBook Page: There are several ways that you can help support my channel:) Consider becoming a member of the channel: My GoFundMe Page: My Patreon Page: Donate via PayPal: Udemy Courses(Please Use These Links If You Sign Up!) Abstract Algebra Course Advanced Calculus Course Calculus 1 Course Calculus 2 Course Calculus 3 Course Calculus Integration Insanity Differential Equations Course College Algebra Course How to Write Proofs with Sets Course How to Write Proofs with Functions Course Statistics with StatCrunch Course Math Graduate Programs, Applying, Advice, Motivation Daily Devotionals for Motivation with The Math Sorcerer Thank you:) mathsorcerer #onlinemathhelp 2 comments Transcript: Intro in this problem we're given the equation of an ellipse and we're asked to find the vertices the endpoints of the minor axis and the foci and graph it so in other words we pretty much have to do everything um so let's start Finding the Center by finding the center so in an ellipse typically when you're looking for the center the general form would be something like this x minus h squared over a squared plus y minus k squared over b squared equals 1. that's the case where a is bigger than b which is what we have here the 4 is bigger than the 1. so in this case you can see that the center of this ellipse is 0 0 because the h and k are not here it's x minus 0 squared and also y minus 0 squared in an ellipse a is always bigger than b and so a is always the square root of the bigger number so in this case the bigger number is 4 so a is the square root of 4 which is 2. and then so b is the square root of the other number so b is the square root of 1 which is 1. because the bigger number is under the x the major axis is said to be horizontal if it was the case that the bigger number was under the y then the major axis is vertical the vertices are the endpoints of the major axis okay again in an ellipse a is the square root of the bigger number because the bigger number is under the x the major axis is horizontal that means it looks like this okay if the bigger number is another y it's vertical so it would look different it would look like an egg standing up Finding the Graph all right so now i guess we can you know do a rough sketch of our ellipse so there's the y-axis and there's the x-axis so x and y and the center is 0 0. and this is an ellipse that has a major axis that's horizontal so we go left and right by a from from the center so we'll go left to put a dot go right to put a dot so this is the major axis here this horizontal axis and then we go up and down by one so up one down one so this would be the graph of our ellipse so the vertices let's go ahead and specify them completely those are the endpoints of the major axis so it would be the ones on this one here and this one here so that would be negative two comma zero and two comma zero those would be the vertices and the endpoints of the minor axis so of minor axis that would be this number here this point there in this point here so it'd be 0 1 and 0 negative 1. that that those would be the endpoints of the major axis so we've got the sketch we've got the vertices the endpoints of the major axis the only thing we need is the foci so there's a formula that will give us the foci the formula is c squared equals a squared and then here's the trick it's the other symbol okay there's a plus here so the formula has a minus you just switch it so a squared is four you see it matches the the form of an ellipse up here so four minus and then b squared is one so you just get 3. so we have c squared equal to 3. take the square root we get c equals plus or minus the square root of 3. and the key is that the foci lie on the major axis and they're a distance of c in each direction from the center so one of them will be over here and one of them will be because these yellow dots would be the foci so those yellow dots are a distance of a square root of three from the set from the center so the foci in this case would be negative root 3 comma 0 and then root 3 comma 0. this would be the foci right plural focus and there's two of them right there's one here and there's one here and you know it's like this you know what's the x you know the x has the c because it's on the major axis right so that's why you know to put the c as the x coordinate if it was like this uh then the focus would be here and here and then you would put the c in in the y coordinate so that's why it's important to know like where they are okay they're on the major axis so you're going to go left and right by c that's it i hope this video has been helpful
14572	https://math.stackexchange.com/questions/4002698/find-the-least-multiplier-of-a-fraction-that-gives-a-remainder-of-1	modular arithmetic - Find the least multiplier of a fraction that gives a remainder of 1 - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find the least multiplier of a fraction that gives a remainder of 1 Ask Question Asked 4 years, 8 months ago Modified4 years, 8 months ago Viewed 43 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I have two natural numbers 2≤b<a 2≤b<a. I want to know the least natural number x 1 x 1 such that x a b x a b gives a remainder of 1 1, and the least natural number x 2 x 2 such that the remainder is b−1 b−1 (I'm looking for a formula, not an algorithmic procedure). In other words, that x 1 a b−⌊x 1 a b⌋=1 b x 2 a b−⌊x 2 a b⌋=b−1 b x 1 a b−⌊x 1 a b⌋=1 b x 2 a b−⌊x 2 a b⌋=b−1 b Assume that a a is not divisible by b b, so that both x x exist. Under that assumption, I know that x 1,x 2∈[1,…,b−1]x 1,x 2∈[1,…,b−1] because all possible remainders from 0 0 up to b−1 b−1 happens exactly once when the multipliers goes from 1 1 to b b. For example, for 19 7 19 7 iterating a multiplier x x over 1,2,…,7 1,2,…,7 gives the following list of remainders: 5 3 1 6 4 2 0 5 3 1 6 4 2 0 where the position ocuppied by the remainder 1 1 (x 1=3 x 1=3 in this case), is the "step size" of the remainder list, because it is the number of forward positions, moving in a circular fashion, to reach the next remainder (2 2 appears three positions later, 3, other three positions later, 4, other three positions later, and so on). About x 2 x 2, I know that x 2=m o d(x 1(b−1),b)x 2=m o d(x 1(b−1),b), but that's not a very useful expression that I could use comfortably in other equations that I'm working on, so I wonder if there's a way to express x 2 x 2 that doesn't involve the explicit use of modular arithmetic. modular-arithmetic Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 28, 2021 at 3:00 ABuABu asked Jan 28, 2021 at 2:40 ABuABu 471 2 2 silver badges 10 10 bronze badges 6 2 Seems you're looking for a multiplicative inverse of a a modulo b b (as x 1 x 1). Is there any particular reason you aren't comfortable with modular arithmetics?Alexey Burdin –Alexey Burdin 2021-01-28 02:46:26 +00:00 Commented Jan 28, 2021 at 2:46 "I'm looking for a formula, not an algorithmic procedure" Why should there be one.fleablood –fleablood 2021-01-28 02:54:50 +00:00 Commented Jan 28, 2021 at 2:54 " but that's not a very useful expression that I could use comfortably in other equations that I'm working on" Why not?fleablood –fleablood 2021-01-28 02:56:27 +00:00 Commented Jan 28, 2021 at 2:56 @AlexeyBurdin Because I'm not proficient with it, and I'm working with some inequations involving a set of floor functions and fractional parts trying to find/proof properties of a specific problem of mine. Modular arithmetic will just make the problem a bit more harder for me, so I'm trying to avoid it if I can.ABu –ABu 2021-01-28 03:07:12 +00:00 Commented Jan 28, 2021 at 3:07 1 modular arithmetic makes everything easier and avoiding modular arithmetic makes everything harder. And you can learn everything you need to know about modular arithmetic in two sentences: If you divide an integer by an integer n n there will be a remainder. You can do all arithmetic by only paying attention to the remainders and ignoring everything else. Now you know everything you need to know and everything will be easier. (I serious. It will.)fleablood –fleablood 2021-01-28 07:08:44 +00:00 Commented Jan 28, 2021 at 7:08 \|Show 1 more comment 0 Sorted by: Reset to default You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions modular-arithmetic See similar questions with these tags. 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14573	https://www.hinkali.com/Education/CrossingNumber.pdf	The Crossing Number of Graphs: Theory and Computation Petra Mutzel Technische Universität Dortmund, Informatik LS11, Algorithm Engineering Otto-Hahn-Str. 14, 44227 Dortmund, Germany petra.mutzel@tu-dortmund.de Abstract. This survey concentrates on selected theoretical and com-putational aspects of the crossing number of graphs. Starting with its introduction by Turán, we will discuss known results for complete and complete bipartite graphs. Then we will focus on some historical con-fusion on the crossing number that has been brought up by Pach and Tóth as well as Székely. A connection to computational geometry is made in the section on the geometric version, namely the rectilinear crossing number. We will also mention some applications of the crossing num-ber to geometrical problems. This review ends with recent results on approximation and exact computations. 1 Introduction The crossing number cr(G) of a graph is the smallest number of edge crossings achievable when laying out G in the 2-dimensional plane. The problem originated from Turán in 1944 when he worked in a labor camp : “There were some kilns where the bricks were made and some open storage yards where the bricks were stored. All the kilns were connected by rail with all the storage yards. The bricks were carried on small wheeled trucks to the storage yards. All we had to do was to put the bricks on the trucks at the kilns, push the trucks to the storage yards, and unload them there . . . the trouble was only at the crossings. The trucks generally jumped the rails there, and the bricks fell out of them; in short, this caused a lot of trouble and loss of time . . . the idea occured to me that this loss of time could have been minimized if the number of crossings of the rails had been minimized.” In 1952, Turán mentioned this problem to Zarankiewicz, who presented a solution for the crossing number of complete bipartite graphs in 1954 . Un-fortunately, Ringel found a gap in the published proof that has not been closed yet. S. Albers, H. Alt, and S. N¨ aher (Eds.): Festschrift Mehlhorn, LNCS 5760, pp. 305–317, 2009. © Springer-Verlag Berlin Heidelberg 2009 306 P. Mutzel 1.1 The Crossing Number for Complete Bipartite Graphs However, the formula presented by Zarankiewicz is conjectured to be correct. His construction provides drawings with exactly Z(m, n) := m 2 m −1 2 n 2 n −1 2 crossings. Figure 1 shows the construction for K6,6. Fig. 1. Zarankiewicz’s construction for K6,6 The conjecture is known to be true for Km,n with min(m, n) ≤6 and for m, n ≤7. The smallest unsolved cases are K7,11 and K9,9 with conjectured values 225 and 256, respectively. New bounding techniques using semi-deﬁnite programming have shown that 0.8594Z(m, n) ≤cr(Km,n) ≤Z(m, n). 1.2 The Crossing Number for Complete Graphs The crossing number for the complete graph Kn is not known either. It is gen-erally believed to be given by the formula provided by Guy : Z(n) := 1 4 n 2 n −1 2 n −2 2 n −3 2 . For odd n the formula can be written as Z(n) = 1 64(n −1)2(n −3)2. Also Guy presented a general drawing scheme for Kn that produces drawings with exactly Z(n) crossings. Guy’s construction for K8 is shown in Figure 2. The Crossing Number of Graphs: Theory and Computation 307 Fig. 2. Guy’s construction for K8 Combinatorial arguments show the following fact: If Guy’s conjecture is true for K2k−1, then it is also true for K2k. This is the reason why the proofs con-centrate on Kn for odd n. Guy was able to prove his conjecture for Kn with n ≤10. For 35 years, this result could not be improved. Recently, Pan and Richter showed that cr(K11) = 100 thus getting cr(K12) = 150 for free. The smallest unsolved case is K13 with conjectured crossing number 225. For many graph classes the situation is similar: cr(G) is known for small instances only, while for general n not much is known (e.g., hypercube graphs, toroidal graphs, or generalized Peterson graphs). For a bibliography on the cross-ing number, see . 2 Confusion on the Crossing Number In their interesting article “What crossing number is it anyway?” Pach and Toth stated that “. . . some authors might have thought of . . . ” diﬀerent crossing numbers. They pointed out that the deﬁnitions for the crossing number provided in the literature were not always the same. In order to investigate this further, we will use the formal deﬁnitions of Szégely . A drawing D of a graph G into the plane is an injection from the vertex set V (G) into the plane, and a mapping Φ of the edge set E(G) into the set of simple planar curves, such that the curve corresponding to the edge has end points Φ(u) and Φ(v), and contains no other vertices. The number of crossings cr(D) in D is the number of intersection points of all unordered pairs of interiors of edges. The crossing number cr(G) of a graph G is the minimum cr(D) over all drawings D of G. A drawing D is optimal if it realizes cr(D) = cr(G). A drawing D is called normal if it satisﬁes i any two of the curves have ﬁnitely many points in common ii no two curves have a point in common in a tangential way iii no three curves cross each other in the same point 308 P. Mutzel A drawing is nice, if it is normal, and satisﬁes iv no two adjacent edges cross v any two edges cross at most once It can be observed that an optimal drawing must satisfy i, ii, iv, and v, and can be transformed to satisfy iii. Therefore, we can restrict ourselves to consider normal or even nice drawings when interested in cr(G). Pach and Tóth introduced two variants of crossing numbers: The pairwise crossing number cr-pair(G) is the minimum number of edge pairs that cross each other at least once, over all normal drawings of G. The odd crossing number cr-odd(G) is equal to the minimum number of edge pairs that cross each other an odd number of times, over all normal drawings of G. In , Tutte introduces yet another version that Szégely calls the independent odd crossing number cr-iodd(G). It is equal to the minimum number of non-adjacent edge pairs that cross each other odd times, over all normal drawings of G. The reason why Tutte introduced this crossing number was his “. . . view that crossings of adjacent edges are trivial, and easily get rid of.” But so far nobody has shown that this can be done in this setting. The following relation between these variants is obvious: cr-iodd(G) ≤cr-odd(G) ≤cr-pair(G) ≤cr(G) Pach and Tóth mention that “. . . perhaps the most exciting open problem in the area . . . ” is the question: “Are they all equal?” One of the reasons why researchers thought that these numbers might be equal is an old theorem by Hanani , rediscovered by Tutte . It states that every graph that can be drawn such that every pair of non-adjacent edges intersects an even number of times, is planar. Pach and Tóth have generalized this result by showing that one can redraw even edges without crossings even in the presence of odd edges. Pelsmajer, Schaefer, and Štefankovič have shown that this redrawing can be performed without adding pairs of edges that intersect an odd number of times; in particular, the odd crossing number does not increase by the redrawing. It is known that cr(G) ≤2(cr-odd(G))2 and that for graphs with cr-odd(G) ≤3 indeed cr-odd(G) = cr(G) . Some authors have stated the conjecture that cr-odd(G) = cr(G). A surprising result by Pelsmajer, Schaefer, and Štefankovič showed that equality of both crossing number variants does not hold. The authors have presented a quite simple inﬁnite family of graphs with cr-odd(G) < cr-pair(G) = cr(G). Figure 3 shows an example of such a graph G. The four distinguished edges a, b, c and d have weights wa = 1, wb = wc = 3, and wd = 4. We assume that the weights of the edges e along the two main cycles are heavy so that they are not crossed in an optimal drawing (e.g., we = 15). We can think of replacing an edge with weight w by w parallel edges. It is also possible to get rid of parallel edges by subdividing these edges. The left drawing shows an optimal drawing for cr(G) and cr-pair(G). The crossing number and the pairwise crossing number is cr(G) = cr-pair(G) = 15. In the right drawing, the edges a and c cross The Crossing Number of Graphs: Theory and Computation 309 a c d b a c d b Fig. 3. A graph with cr-odd(G) < cr(G). The left drawing shows an optimal drawing for cr(G) and cr-pair(g) with value 15. The right drawing shows an optimal drawing for cr-odd(G) with value 13. each other exactly twice providing the amount of 0 to cr-odd(G) thus giving cr-odd(G) = 13 < cr(G). However, the question if cr(G) = cr-pair(G) is still open. The bound of cr(G) = O(cr-pair(G)2/ log(cr-pair(G))) has been provided by Valtr (mentioned in ). Now that we know that the four crossing numbers are not equal, a ques-tion already stated by Szégely is catching our interest : “How is it possible that decades in research of crossing numbers passed by and no major confusion resulted from these foundational problems?” — Perhaps the graph theory com-munity was just lucky that the bounds they provided in all these years apply for all kinds of crossing numbers. 3 The Rectilinear Crossing Number The geometric version of the crossing number is called the rectilinear crossing number, denoted by cr-lin, and requires a drawing in the plane with straight line segments. It is well known that a planar graph always has a planar drawing with straight line segments. Therefore, one may think that cr-lin(G) and cr(G) are equal or close together. However, already Guy has shown that cr(K8) = 18 while cr-lin(K8) = 19. Later, Bienstock and Dean have shown that the two numbers are equal (cr-lin(G) = cr(G)) for graphs with small crossing number cr(G) ≤3. On the other hand, the authors have also shown that there are graphs with crossing number 4 and arbitrarily large rectilinear crossing number. How-ever, for graphs with bounded degree, the crossing number and the rectilinear crossing number are bounded as functions of one another . In detail, if a graph has maximum degree d and crossing number k, its rectilinear crossing number is at most O(dk2). It is conjectured that the construction by Zarankiewicz for the crossing num-ber of complete bipartite graphs provides the correct numbers for cr-lin(Km,n). Due to the nature of the construction, a proof for cr(Km,n) = Z(m, n) would directly lead to cr-lin(Km,n) = Z(m, n). 310 P. Mutzel Until 2001, the rectilinear crossing number for the class of complete graphs Kn was known only for n ≤9. Then, two groups of researchers independently showed that cr-lin(K10) = 62. Aichholzer, Aurenhammer and Krasser exhaustively enumerated all combinatorial inequivalent point sets (so-called order types) of size 10. Similar methods have been successful for showing cr-lin(K11) = 102 and cr-lin(K12) = 153. The authors initiated the Rectilinear Crossing Number Project in which users provide their own computing power to the project. The main goal of the current project is to use sophisticated mathematical meth-ods (abstract extension of order types) to determine the rectilinear crossing number for small values of n, and to compute all existing combinatorial non-isomorphic minimal drawings. Currently, the rectilinear crossing number is known for all Kn with n ≤21 (cr-lin(K21) = 2055). In contrast to cr(Kn) there is no conjecture following some formula for arbitrarily large n for cr-lin(Kn). However, the gaps between the lower and upper bounds for n up to 100 are quite small, e.g., 1.459.912 ≤cr-lin(K100) ≤1.463.970. 4 Applications of the Crossing Number Székely has used bounds for the crossing number cr(G) for providing a simple proof of the Szemerédi-Trotter theorem, that is an important result in combi-natorial geometry. It asks for the maximum number of incidences of n points and m curves in the plane such that each pair of curves intersects at maximal O(1) points, and there are no more than O(1) curves passing through each pair of points. The answer in this case is O(m + n + (mn)2/3). The idea of Szégely’s proof was to build a graph in which the vertex set is associated with the point set and the edges with the curve segments. Using bounds for the crossing number for complete graphs essentially provided the solution. Szemerédi-Trotter like theorems can be used for proving hard Erdős problems in combinatorics, in number theory, in analysis or geometric measure theory. E.g., Elekes used it to show that any n distinct real numbers have Ω(n1.25) distinct sums or products. And a famous problem of Erdős in geometry asks for the maximum number of unit distances that are possible among n points in the plane. Application of the Szemerédi-Trotter theorem provides O(n4/3), and this is the best known estimate so far . Dey used the bounds known for the rectilinear crossing number for proving upper bounds on geometric k-sets. This led to considerable improvement on this bound after its early solution about 27 years ago. The rectilinear crossing number of a complete graph is essentially the same as the minimum number of convex quadrilaterals determined by a set of n points in general position. It is known that the number of quadrilaterals is proportional to the fourth power of n, but the precise constant is unknown . An important application of crossing numbers is in graph drawing and VLSI-design. This is why the author started research in this area in 1995. Figure 4 shows a graph with 120 objects and 161 edges that originated at an insurance company. The original drawing had 122 crossings, while the crossing minimal drawing has only 6 crossings. The Crossing Number of Graphs: Theory and Computation 311 Fig. 4. A drawing of the insurance graph with 6 crossings 5 Approximation Garey and Johnson have shown NP-completeness of the decision variant of the crossing number problem . Pach and Tóth have shown the same for the crossing number variants cr-odd and cr-pair. Bienstock has shown that also the computation of cr-lin(G) is NP-hard . No polynomial time algorithm is known for approximating cr(G) for general graphs within some non-trivial factor. Bhatt and Leighton suggested the ﬁrst algorithm which approximates \|V \|+ cr(G) for a bounded degree graph G = (V, E) with a factor of O(log4 \|V \|). This approximation factor has been improved by Even, Guha and Schieber in 2002 to O(log3 \|V \|). For sparse graphs, when cr(G) = o(\|V \|), this approximation does not guarantee good results. Until recently, polynomial time algorithm approximating cr(G) was known, not even for special graph classes. Recently, the ﬁrst approximation results have been achieved that do depend on the maximum degree and cr(G) only. The approximation results concern the graph classes of almost planar graphs and apex graphs. A graph G = (V, E) is called almost planar if G is non-planar, but there does exist an edge e ∈E so that G −{e} is planar. Given a planar embedding Π of the remaining graph G −{e}, the edge e can be re-inserted with the minimal number of crossings via a shortest path in the extended geometric dual graph of Π. Gutwenger, Mutzel, and Weiskircher have presented a linear time algorithm (based on the data structure of SPQR-trees) which is able to ﬁnd the optimal embedding Π0 of G −{e}, so that inserting e into Π0 leads to a crossing minimum drawing over the set of all possible planar embeddings Π. The natural question arises, if this approach does approximate the crossing number cr(G) by some small factor. 312 P. Mutzel ... m dense planar graphs v e Fig. 5. Inserting edge e back into the planar graph G −{e} yields m crossings if deg(v) ≥2m. Note that the shaded blocks are dense triconnected subgraphs. The crossing minimal drawing of G with 2 crossings can be found when ﬂipping the two blocks adacent to v. Figure 5 shows an example for which the edge e will be inserted into the pla-nar graph G −{e} with m crossings. However, the minimum crossing number of G is 2. This can be achieved by ﬂipping the two lower components adjacent to v. Hliněný and Salazar observed that the approximation factor in this case de-pends on the degree of vertex v. If the degree is bounded, then this example does not hurt the approximation anymore. Hliněný and Salazar have shown that the above algorithm provides crossing numbers of at most Δ(G −{e})cr(G), where Δ(G) denotes the maximum degree of G. This number has later been improved to (Δ(G −{e})/2)cr(G) by Cabello and Mohar . This provides the ﬁrst constant approximation algorithm for almost planar graphs with bounded degree graphs. Very recently, these results could be generalized to apex-graphs. A graph G = (V, E) is called an apex graph if G is non-planar, but there does exist a vertex v ∈V so that G −{v} is planar. Chimani, Gutwenger, Mutzel, and Wolf have shown that v and all its incident edges can be re-inserted into an optimal embedding Π0 of G −{v} (which the algorithm will identify) with the minimum number of crossings in polynomial time. Chimani, Hliněný, and Mutzel have shown that this algorithm will ﬁnd solutions which are at most a factor of deg(v)Δ(G −{v})/2 away from the optimum solution cr(G). Both approximation results are (almost) tight: for almost planar graphs, there is an example showing that the approximation factor can be reached, while for apex graphs the example is still a factor of 2 away. Some open questions arise: – Is there a polynomial time algorithm for computing the crossing number cr(G) for almost planar graphs? Cabello and Mohar have shown that the weighted version is NP-hard. – Can the above results be generalized, e.g., for graphs that are planar after deleting a ﬁxed number of edges? The Crossing Number of Graphs: Theory and Computation 313 6 Exact Computation Very few publications exist for computing the crossing number of general graphs exactly. Grohe has shown in 2001 that the crossing number problem is ﬁxed-parameter tractable. However, the used concepts are based on the theoretical results of Robertson and Seymour. Recently, Kawarabayashi and Reed 2007 have improved the quadratic running time to a linear time algorithm for ﬁxed k. Both approaches reduce the graph to one with bounded tree-width and the same crossing number and then test if the graph has crossing number at most k. There is common agreement that this approach is purely theoretical. Related to this, the following open problem is among the most important ones in the area: Can cr(G) be computed in polynomial time for graphs with bounded tree-width? Until 2005, no practically eﬃcient algorithm for computing the crossing num-ber was known. Today, there exist two approaches for computing the exact cross-ing number for general graphs. The approaches are based on two integer linear programming (ILP) formulations of the crossing number problem that can be solved by branch-and-cut algorithms. The ILP formulation by Buchheim et al. is called the subdivision cross-ing minimization approach (SOCM) and optimizes over the set of all simple drawings. A drawing is called simple if every edge is only crossed at most once. In order to provide an optimal solution for cr(G), we need to subdivide all edges in G into a path of length \|E\|. The variables xe,f are associated with all non-adjacent edge pairs (e, f) ∈E2. The constraints come essentially from Ku-ratowski’s theorem stating that a graph G is planar if and only if it does not contain a subdivision of K3,3 or K5. Besides the Kuratowski-constraints and the 0/1-constraints, the ILP also contains constraints that guarantee to get simple drawings of the subdivided graph. The second approach by Bomze, Chimani and Mutzel is called the ordering-based ILP model (OOCM). This is not restricted to simple drawings. In-stead, additional linear ordering variables yefg are introduced for each edge e ∈E that may be crossed more than once. The variables yefg for edges e, f, g ∈E pro-vide the information in which order an edge e is crossed by f and g. In this ILP we need Kuratowski constraints on the x variables, linear ordering constraints on the y variables, 0/1-constraints for all variables, and additional linking con-straints between the x and y variables. We solve both ILP models with branch-and-cut algorithms. In order to get these algorithms to work in practice, we needed to come up with new prepro-cessing techniques as well as new combinatorial column generation methods. Our computational experiments on a benchmark set of about 11,000 graphs show that we can compute the exact crossing numbers for general sparse graphs with up to 100 vertices and crossing number up to 37 within 30 minutes. Figure 6 shows the percentage of instances that have been solved within 30 minutes of computation time for about 11.000 graphs of the Rome library. The x-asis shows the number of vertices \|V \|. The number of edges of the Rome graphs is below 1.5\|V \| on average. The crossing number of almost all graphs with up to 60 vertices could be solved to provable optimality within 30 minutes CPU-time. 314 P. Mutzel 40% 50% 60% 70% 80% 90% 100% 10 20 30 40 50 60 70 80 90 100 % solved nodes best published results improved SOCM OOCM Fig. 6. The average percentage of instances solved within 30 minutes of computation time of the ILP models SOCM and OOCM 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% -500 -250 0 250 500 750 1000 1250 1500 1750 2000 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 generated variables (%) # start variables nodes # start variables improved-SOCM increase (%) OOCM increase (%) Fig. 7. The dark line shows the average number of start variables in our branch-and-cut approach (left axis). The right axis shows the percentage of additional variables generated by the two approaches. The Crossing Number of Graphs: Theory and Computation 315 The graphs with 100 vertices are much harder to solve. But still, more than 50% of the instances with 100 vertices could be solved to optimality. It seems that the second formulation based on linear ordering dominates our ﬁrst ILP model. For most instances, we need far more variables in our SOCM model than in the OOCM model. The dark line in Figure 7 shows the average number of start variables (left axis) in our branch-and-cut approach. For graphs with 100 vertices the average number was about 1800 variables. During the run of the algorithm, column generation adds in additional variables. These numbers are much higher for the SOCM approach. While SOCM added about 50% new variables with respect to the start variables, OOCM only had to add 18% additional variables for the 100-vertex graphs. We ﬁnd it surprising that in the OOCM model only very few y-variables are needed in order to ﬁnd the optimum solution. Detailed experiments and results can be found in [8, 11]. 7 Solved Open Problems We close our survey with selected experimental results for special graph classes. While we could verify the crossing number of the complete graph on 11 (and 12) vertices (with an alternative optimum solution), one more vertex is still a challenge. On the other hand, we are able to compute the crossing number of generalized Petersen graphs Pn,4 up to n = 44 which was unknown before. Based on our computed results, we came up with a conjecture of the crossing number of this graph class. Moreover, we are conﬁdent that we are about to be able to compute the crossing number of the smallest toroidal grid graph T8,8 whose crossing number is still unknown (and conjectured to be 48). Fig. 8. Knuth’s musical graph 316 P. Mutzel In Exercise 133 of The Art of Computer Programming, Volume 4, Draft of Section 7, Donald E. Knuth uses the “musical graph” on page 73 of Graphs by R. J. Wilson and J. J. Watkins (1990) (see Fig. 8). It represents simple modulations between key signatures. While all kinds of properties of this graph are easily analyzed, the question “Can it be drawn with fewer than 12 crossings?” remained open. After 9.71 seconds of computation time, our program proved that the crossing number is indeed 12 and produced an alternative embedding that is not as nice as the original, though. Acknowledgments. I am grateful to Michael Jünger who pointed out the open problem of Knuth’s forthcoming book. Many thanks to Markus Chimani who was co-author of most cited papers on the crossing number, and who did all computations with his code. References 1. Aichholzer, O., Aurenhammer, F., Krasser, H.: On the crossing number of complete graphs. In: Symposium on Computational Geometry, pp. 19–24 (2002) 2. Bhatt, S.N., Leighton, F.T.: A framework for solving vlsi layout problems. Journal of Computer and System Sciences 28, 300–343 (1984) 3. Bienstock, D.: Some provably hard crossing number problems. Discrete & Compu-tational Geometry 6, 443–459 (1991) 4. Bienstock, D., Dean, N.: New results on rectilinear crossing numbers and plane embeddings. J. Graph Theory 16(5), 389–398 (1992) 5. Bienstock, D., Dean, N.: Bounds for rectilinear crossing numbers. J. Graph The-ory 17(3), 333–348 (1993) 6. Buchheim, C., Chimani, M., Ebner, D., Gutwenger, C., Jünger, M., Klau, G.W., Mutzel, P., Weiskircher, R.: A branch-and-cut approach to the crossing num-ber problem. Discrete Optimization, Special Issue in memory of George B. Dantzig 5(2), 373–388 (2008) 7. Cabello, S., Mohar, B.: Crossing and weighted crossing number of near-planar graphs. In: Tollis, I.G., Patrignani, M. (eds.) GD 2008. LNCS, vol. 5417, pp. 38– 49. Springer, Heidelberg (2009) 8. Chimani, M.: Exact Crossing Minimization. PhD thesis, Fakultät für Informatik, Technische Universität Dortmund (2008) 9. Chimani, M., Gutwenger, C., Mutzel, P., Wolf, C.: Inserting a vertex into a planar graph. In: SODA, pp. 375–383. SIAM, Philadelphia (2009) 10. Chimani, M., Hliněný, P., Mutzel, P.: Vertex insertion approximates the crossing number for apex graphs (submitted, 2009) 11. Chimani, M., Mutzel, P., Bomze, I.: A new approach to exact crossing minimiza-tion. In: Halperin, D., Mehlhorn, K. (eds.) ESA 2008. LNCS, vol. 5193, pp. 284–296. Springer, Heidelberg (2008) 12. de Klerk, E., Pasechnik, D.V., Schrijver, A.: Reduction of symmetric semideﬁnite programs using the regular -representation. Math. Program. 109(2-3), 613–624 (2007) 13. Dey, T.K.: Improved bounds for planar k -sets and related problems. Discrete & Computational Geometry 19(3), 373–382 (1998) The Crossing Number of Graphs: Theory and Computation 317 14. Elekes, G.: On the number of sums and products. Acta Arithm. 81(4), 365–367 (1997) 15. Even, G., Guha, S., Schieber, B.: Improved approximations of crossings in graph drawings and VLSI layout areas. SIAM Journal on Computing 32(1), 231–252 (2002) 16. Gutwenger, C., Mutzel, P., Weiskircher, R.: Inserting an edge into a planar graph. Algorithmica 41(4), 289–308 (2005) 17. Guy, R.K.: The decline and fall of Zarankiewicz’s theorem. In: Proof techniques in Graph Theory, pp. 63–69. Academic Press, London (1969) 18. Guy, R.K.: Crossing numbers of graphs. In: Proc. Graph Theory and Applications, pp. 111–124. LNM (1972) 19. Chojnacki, C., Hanani, H.: Über wesentlich unplättbare Kurven im drei-dimensionalen Raume. Fundam. Math. (1934) 20. Hliněný, P., Salazar, G.: Crossing and weighted crossing number of near-planar graphs. In: Kaufmann, M., Wagner, D. (eds.) GD 2006. LNCS, vol. 4372, pp. 162– 173. Springer, Heidelberg (2007) 21. Johnson, M.R., Johnson, D.S.: Crossing number is NP-complete. SIAM J. Algebraic Discrete Methods 4(3), 312–316 (1983) 22. Kolman, P., Matousek, J.: Crossing number, pair-crossing number, and expansion. J. Combin. Theory Ser. B 92, 99–113 (2003) 23. Pach, J., Tóth, G.: Which crossing number is it anyway? Journal of Combinatorial Theory, Series B 80, 225–246 (2000) 24. Pan, S., Richter, R.B.: The crossing number of K11 is 100. Journal of Graph Theory 56(2), 128–134 (2007) 25. Pelsmajer, M.J., Schaefer, M., Štefankovič, D.: Removing even crossings on sur-faces. Electronic Notes in Discrete Mathematics 29, 85–90 (2007) 26. Pelsmajer, M.J., Schaefer, M., Štefankovič, D.: Odd crossing number and crossing number are not the same. Discrete & Computational Geometry 39(1-3), 442–454 (2008) 27. The rectilinear crossing number project, 28. Scheinerman, E.R., Wilf, H.: The rectilinear crossing number of a complete graph and sylvester’s “four point problem" of geometric probability. American Mathe-matical Monthly 101, 939–943 (1994) 29. Spencer, J., Szemerédi, E., Trotter, W.T.: Unit distances in the euclidean plane. 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14574	https://en.wikipedia.org/wiki/Factor_theorem	Factor theorem - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Factorization of polynomialsToggle Factorization of polynomials subsection 1.1 Example 2 ProofsToggle Proofs subsection 2.1 Proof 1 2.2 Proof 2 2.3 Proof 3 2.4 Corollary of other theorems 3 References Factor theorem [x] 10 languages العربية Español 한국어 Magyar 日本語 ਪੰਜਾਬੀ ភាសាខ្មែរ Svenska தமிழ் 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Polynomial zeros related to linear factors In algebra, the factor theorem connects polynomial factors with polynomial roots. Specifically, if f(x){\displaystyle f(x)} is a (univariate) polynomial, then x−a{\displaystyle x-a} is a factor of f(x){\displaystyle f(x)} if and only if f(a)=0{\displaystyle f(a)=0} (that is, a{\displaystyle a} is a root of the polynomial). The theorem is a special case of the polynomial remainder theorem. The theorem results from basic properties of addition and multiplication. It follows that the theorem holds also when the coefficients and the element a{\displaystyle a} belong to any commutative ring, and not just a field. In particular, since multivariate polynomials can be viewed as univariate in one of their variables, the following generalization holds: If f(X 1,…,X n){\displaystyle f(X_{1},\ldots ,X_{n})} and g(X 2,…,X n){\displaystyle g(X_{2},\ldots ,X_{n})} are multivariate polynomials and g{\displaystyle g} is independent of X 1{\displaystyle X_{1}}, then X 1−g(X 2,…,X n){\displaystyle X_{1}-g(X_{2},\ldots ,X_{n})} is a factor of f(X 1,…,X n){\displaystyle f(X_{1},\ldots ,X_{n})} if and only if f(g(X 2,…,X n),X 2,…,X n){\displaystyle f(g(X_{2},\ldots ,X_{n}),X_{2},\ldots ,X_{n})} is the zero polynomial. Factorization of polynomials [edit] Main article: Factorization of polynomials Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent. The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows: Deduce the candidate of zero a{\displaystyle a} of the polynomial f{\displaystyle f} from its leading coefficient a n{\displaystyle a_{n}} and constant term a 0{\displaystyle a_{0}}. (See Rational Root Theorem.) Use the factor theorem to conclude that (x−a){\displaystyle (x-a)} is a factor of f(x){\displaystyle f(x)}. Compute the polynomial g(x)=f(x)(x−a){\textstyle g(x)={\dfrac {f(x)}{(x-a)}}}, for example using polynomial long division or synthetic division. Conclude that any root x≠a{\displaystyle x\neq a} of f(x)=0{\displaystyle f(x)=0} is a root of g(x)=0{\displaystyle g(x)=0}. Since the polynomial degree of g{\displaystyle g} is one less than that of f{\displaystyle f}, it is "simpler" to find the remaining zeros by studying g{\displaystyle g}. Continuing the process until the polynomial f{\displaystyle f} is factored completely, which all its factors is irreducible on R[x]{\displaystyle \mathbb {R} [x]} or C[x]{\displaystyle \mathbb {C} [x]}. Example [edit] Find the factors of x 3+7 x 2+8 x+2.{\displaystyle x^{3}+7x^{2}+8x+2.} Solution: Let p(x){\displaystyle p(x)} be the above polynomial Constant term = 2 Coefficient of x 3=1{\displaystyle x^{3}=1} All possible factors of 2 are ±1{\displaystyle \pm 1} and ±2{\displaystyle \pm 2}. Substituting x=−1{\displaystyle x=-1}, we get: (−1)3+7(−1)2+8(−1)+2=0{\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0} So, (x−(−1)){\displaystyle (x-(-1))}, i.e, (x+1){\displaystyle (x+1)} is a factor of p(x){\displaystyle p(x)}. On dividing p(x){\displaystyle p(x)} by (x+1){\displaystyle (x+1)}, we get Quotient = x 2+6 x+2{\displaystyle x^{2}+6x+2} Hence, p(x)=(x 2+6 x+2)(x+1){\displaystyle p(x)=(x^{2}+6x+2)(x+1)} Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic −3±7.{\displaystyle -3\pm {\sqrt {7}}.} Thus the three irreducible factors of the original polynomial are x+1,{\displaystyle x+1,}x−(−3+7),{\displaystyle x-(-3+{\sqrt {7}}),} and x−(−3−7).{\displaystyle x-(-3-{\sqrt {7}}).} Proofs [edit] Several proofs of the theorem are presented here. If x−a{\displaystyle x-a} is a factor of f(x),{\displaystyle f(x),} it is immediate that f(a)=0.{\displaystyle f(a)=0.} So, only the converse will be proved in the following. Proof 1 [edit] This proof begins by verifying the statement for a=0{\displaystyle a=0}. That is, it will show that for any polynomial f(x){\displaystyle f(x)} for which f(0)=0{\displaystyle f(0)=0}, there exists a polynomial g(x){\displaystyle g(x)} such that f(x)=x⋅g(x){\displaystyle f(x)=x\cdot g(x)}. To that end, write f(x){\displaystyle f(x)} explicitly as c 0+c 1 x 1+…+c n x n{\displaystyle c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}}. Now observe that 0=f(0)=c 0{\displaystyle 0=f(0)=c_{0}}, so c 0=0{\displaystyle c_{0}=0}. Thus, f(x)=x(c 1+c 2 x 1+…+c n x n−1)=x⋅g(x){\displaystyle f(x)=x(c_{1}+c_{2}x^{1}+\dotsc +c_{n}x^{n-1})=x\cdot g(x)}. This case is now proven. What remains is to prove the theorem for general a{\displaystyle a} by reducing to the a=0{\displaystyle a=0} case. To that end, observe that f(x+a){\displaystyle f(x+a)} is a polynomial with a root at x=0{\displaystyle x=0}. By what has been shown above, it follows that f(x+a)=x⋅g(x){\displaystyle f(x+a)=x\cdot g(x)} for some polynomial g(x){\displaystyle g(x)}. Finally, f(x)=f((x−a)+a)=(x−a)⋅g(x−a){\displaystyle f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)}. Proof 2 [edit] First, observe that whenever x{\displaystyle x} and y{\displaystyle y} belong to any commutative ring (the same one) then the identity x n−y n=(x−y)(y n−1+x 1 y n−2+…+x n−2 y 1+x n−1){\displaystyle x^{n}-y^{n}=(x-y)(y^{n-1}+x^{1}y^{n-2}+\dotsc +x^{n-2}y^{1}+x^{n-1})} is true. This is shown by multiplying out the brackets. Let f(X)∈R[X]{\displaystyle f(X)\in R\left[X\right]} where R{\displaystyle R} is any commutative ring. Write f(X)=∑i c i X i{\displaystyle f(X)=\sum {i}c{i}X^{i}} for a sequence of coefficients (c i)i{\displaystyle (c_{i}){i}}. Assume f(a)=0{\displaystyle f(a)=0} for some a∈R{\displaystyle a\in R}. Observe then that f(X)=f(X)−f(a)=∑i c i(X i−a i){\displaystyle f(X)=f(X)-f(a)=\sum {i}c_{i}(X^{i}-a^{i})}. Observe that each summand has X−a{\displaystyle X-a} as a factor by the factorisation of expressions of the form x n−y n{\displaystyle x^{n}-y^{n}} that was discussed above. Thus, conclude that X−a{\displaystyle X-a} is a factor of f(X){\displaystyle f(X)}. Proof 3 [edit] The theorem may be proved using Euclidean division of polynomials: Perform a Euclidean division of f(x){\displaystyle f(x)} by (x−a){\displaystyle (x-a)} to obtain f(x)=(x−a)Q(x)+R(x){\displaystyle f(x)=(x-a)Q(x)+R(x)} where deg⁡(R)<deg⁡(x−a){\displaystyle \deg(R)<\deg(x-a)}. Since deg⁡(R)<deg⁡(x−a){\displaystyle \deg(R)<\deg(x-a)}, it follows that R{\displaystyle R} is constant. Finally, observe that 0=f(a)=R{\displaystyle 0=f(a)=R}. So f(x)=(x−a)Q(x){\displaystyle f(x)=(x-a)Q(x)}. The Euclidean division above is possible in every commutative ring since (x−a){\displaystyle (x-a)} is a monic polynomial, and, therefore, the polynomial long division algorithm does not involve any division of coefficients. Corollary of other theorems [edit] It is also a corollary of the polynomial remainder theorem, but conversely can be used to show it. When the polynomials are multivariate but the coefficients form an algebraically closed field, the Nullstellensatz is a significant and deep generalisation. References [edit] ^Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p.381, ISBN0-13-370149-2 ^Sehgal, V K; Gupta, Sonal (September 2009), Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p.119, ISBN978-81-317-2816-1. ^Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p.142, ISBN81-7008-629-9. Retrieved from " Category: Theorems about polynomials Hidden categories: Articles with short description Short description is different from Wikidata This page was last edited on 25 July 2025, at 14:53(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Factor theorem 10 languagesAdd topic
14575	https://www.youtube.com/watch?v=bltnuzbs2JA	Chemical Reactions (9 of 11) Stoichiometry: Grams to Grams Step by Step Science 218000 subscribers 3287 likes Description 296058 views Posted: 15 Feb 2013 Shows how to use stoichiometry to determine the grams of the other substances in the chemical equation if you are given the grams of one of the substances. A chemical reaction is a process that leads to the chemical change of one set of chemical substances to another. Chemical reactions encompass changes that only involve the positions of electrons in the forming and breaking of chemical bonds between atoms. In a chemical reaction there is no change to the nuclei of the atoms. They can often be described by a chemical equation. Chemical equations are used to graphically illustrate chemical reactions. They consist of the chemical formulas of the of the reactants on the left and those of the products on the right. They are separated by an arrow (→) which indicates the direction and type of the reaction. The most common types of chemical reactions are: double replacement, single replacement, combustion, decomposition and synthesis. You can also see listing of all my videos from my website, Also, please don't forget to do all of the following; (1) Subscribe to my channel, Step-By-Step Science. (2) Give me a thumbs up for this video. (3) Leave me a nice positive comment. (4) Sharing is Caring! Share this video with others in need. Thank you! I greatly appreciate all of your support. 373 comments Transcript: okay in today's video we're going to go every over everybody's favorite chemistry topic stochiometry and we're going to call I call this grams to grams because we're going to start with the grams of one of the compounds in our chemical equation and we want to find the number of grams that we're going to need to react that mass and the mass of each of the products okay so here's our chemical equation 4 NH3 plus 702 is going to yield 4 NO2 plus 6 uh molecules of water all right you have to remember that the chemical equation shows you a particle or I like to think of it just as a mole relationship it's not a mass relationship okay but we do want to find the masses of the other compounds or elements in our chemical equation so you can see in this equation we've been given that we have 9.75 G of NH3 and we want to know how many gram of oxygen do we need to react all of that and therefore how many grams of NO2 are we going to produce and how many gram of H2O are we going to produce okay now there are three simple steps that you need to follow and if you can follow these steps you can do this stochiometry problem and just about every other one the first step is this we have been given grams therefore we have to convert from grams to moles because the chemical equation is a molar relationship not a gram relationship if you've been given the moles which sometimes you will be given the moles you don't need to do this step because it's already done but if you've been given grams which is the typical way to start then you have to convert from grams to moles then step two use the molar ratio to determine the moles of the other substance or the other compound or the other element that you want to figure out the mass for and then the third one is to convert back from moles of that substance into grams of that substance okay now it doesn't matter which Mass you've been given this one or if we were given the mass of oxygen the mass of the na2 and the mass of the H2O and we want to find all the other masses it doesn't matter which one you use the same three steps okay so let's go through and let's get started okay first step we're going to convert from grams to moles using the molar mass we have been given 9.75 G of NH3 and you'll see I wrote that down 9.75 G and NH3 write all three of those things down the number that unit and the chemical formula now we're going to convert so I make this like to make this little t- chart I call it a railroad track it's not really a t- chart I don't think I make a little railroad track you can put a multiplication sign and a division line it's the same thing we want to convert from grams to moles and we're going to use the marar mass and we have NH3 on the top or grams on the top of that one so we need to put the grams on the bottom we put the marar mass one nitrogen three Hy hydrogens is 177 G and put the mole on the top that allows us to C uh cancel the grams we're left with moles and that is how many moles of NH3 we have 0574 now I'm going to write that down right here so that reminds me that 9.75 g is 574 moles okay you always have to convert from grams to moles first because we're going to use this number of moles to solve for the mass the mass and the mass we always start with the moles of the substance that we have been given next slide you see I have it right there the moles I'm going to write down the moles because in this case we're going to figure out first the mass of the O2 so I write that down now I'm going to convert I done step one in step two I'm going to use my molar mass to figure out the moles of oxygen all right so not the M Mass I'm going to use the molar ratio to figure out the moles of oxygen excuse me step two is you use the m m molar ratio put down my little conversion chart my t- chart my railroad track whatever you want to call it now I have moles of NH3 on the top over here that means I have to have moles of NH3 on the bottom so I'm going to write it down like that I want to get to moles of oxygen so I put the moles of oxygen the thing I want on the top and then I look at my chemical equation what is the mol ratio well it's four to 7 so I put the four in front of the NH3 that coefficient goes there the seven goes there and now I can do my math because you'll notice my moles of NH3 cancel and I'm moles of o2 5.74 7 / 4 gives me 1 mole of o2 but I don't want the number of moles I want the number of grams so now I'm going to use that to convert using the molar mass of oxygen which is 32 my moles of o2 cancel and now I'm left with 32 G of oxygen okay so that tells me that now now I'm going to need 32 G of oxygen to fully react 9.75 G of NH3 all right now we're going to do the same thing for the other two products not for the other two products but for the two products okay you can see we've got our step one done we're going to go through through steps two and step three again but I'm going to start with the moles of the NH3 that was the thing I was originally given yes I could convert this to moles and use that but I've already done this so why do I want to do that let's just go on we're going to convert we want to get out of moles of NH3 so NH3 goes in the bottom we want to get into moles of NO2 so the NO2 goes in the top the molar ratio is 4 to4 just like that basically one: one so you will notice if they have the same coefficient that the number of moles is the same and we just proved it right there the number of moles is the same now that was step two we use the mol ratio to get the moles of the other material the other substance and now we can convert that to grams we use the molar ratio molar mass to convert to grams we want to get out of the moles of NO2 so that goes in the bottom 1 mole of NO2 has a molar mass of 46 G cancel cancel and you're left with 26.4 G of NO2 so that tells us that if we react 9.75 G of NH3 and which we would need 32 G of oxygen we would produce 26.4 G of NO2 all right now let's do the last one now this one I'm going to string out in one string and I'm going to kind of skip the intermediate steps so to speak hopefully you'll see what I mean here but we are going to once again start with the moles of the original thing we were given which was the NH3 we're we're going to now do step two again because we already did step one when we figured out it was 74574 moles we're going to get rid of the NH3 again we put that on the bottom we're want to figure out how many moles of oxygen we need so that goes on the top we look at the molar ratio we can see we have a four here so four goes on the bottom we have a six here so six goes on the top now I'm I could multiply this out and get the answer you can see my units cancel and I'm in currently in moles of H2O I'm just going to continue along because this gives me makes it a little less writing now I want to get out of moles of H2O I'm currently in moles of H2O I want to get out of moles of H2O so that goes on the bottom I want it to be in grams the grams goes on the top 18 is the gram the molar mass 18 gam is the marar mass of water and I multiply I can that's right I cancel these first you can see now I'm in grams of H2O and I can see I need 15 I'm going to produce 15.5 gr of H2O okay so that tells me that that is going to be the masses of each of the two reactants and this is how much of each product I'm going to produce okay if you add these two up if you add the mass of the reactants up and the mass of the products up they should be equal to each other because that is conservation of mass whatever Mass you put into the equation you have to get out and I didn't add the up but I think I checked it before and they're pretty close close enough okay so if you can check it by adding the two masses on each side and they should be very close now of course I could have figured out the mass of the Water by just adding up the mass of the NH3 the mass of the oxygen and subtracting the mass of the NO2 okay you could do it that way but I thought hey it's so much fun to do sto geometry let's go through and calculate the whole thing okay so that is STO gometry I don't think it's that hard follow those steps be consistent with your units cross your units out make sure you got your Moler ratios in the right way not upside down and I think you will be successful okay thanks for watching I hope that was helpful if you thought that was also helpful give me a thumbs up down there in the comment section okay thank you very much have a nice day
14576	https://www.solubilityofthings.com/common-mistakes-balancing-chemical-equations	Common Mistakes in Balancing Chemical Equations Introduction to the importance of balancing chemical equations Balancing chemical equations is a fundamental skill in chemistry that speaks directly to the core principles of the discipline. Understanding the importance of this process extends beyond mere academic success; it is essential for accurately predicting the outcomes of chemical reactions. When a chemical equation is balanced, it reflects the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a closed system. This concept underscores the critical relationship between reactants and products, as the total number of atoms of each element must remain constant throughout a chemical reaction. The significance of balancing chemical equations can be appreciated in various contexts: Accuracy in Predictions: A balanced equation facilitates the prediction of the amounts of reactants and products involved in reactions, allowing chemists to calculate yields and efficiency. Safety in Laboratory Practices: Accurate balancing aids in ensuring that the correct proportions of chemicals are combined, minimizing the risk of hazardous reactions. Foundation for Advanced Concepts: Mastery of balancing equations is a prerequisite for delving into more complex topics such as stoichiometry, thermodynamics, and kinetics. Real-World Applications: From industrial processes to biochemical pathways, balanced equations are crucial for understanding and optimizing reactions in various fields, including pharmaceuticals, environmental science, and material engineering. An effective approach to balancing equations relies on the systematic identification of reactants and products, as well as the utilization of coefficients to represent the number of molecules or moles involved. As the renowned chemist "Chemistry is the study of matter, but I prefer to see it as the study of change." emphasized, balancing equations embodies this change, translating chemical interactions into a coherent framework that can be analyzed and understood. In summary, the importance of balancing chemical equations is not merely an academic exercise; it is a critical skill that serves a range of practical applications and principles. As we progress through the exploration of stoichiometry, we will delve deeper into the intricacies of balancing equations and the common pitfalls that can hinder this essential skill. Overview of stoichiometry and its relation to balancing equations Straightforward as it may seem, the art of balancing chemical equations is deeply intertwined with the broader concept of stoichiometry. Stoichiometry is defined as the branch of chemistry that deals with the quantitative relationships between the amounts of reactants consumed and products formed in a chemical reaction. It serves as the bridge that connects the microscopic world of atoms and molecules with the macroscopic quantities scientists observe in the laboratory and the environment. At its core, stoichiometry is built on two essential principles: The Law of Conservation of Mass: This foundational law dictates that the mass of the reactants must equal the mass of the products in a closed system. Therefore, through balanced equations, stoichiometry ensures that the numbers of atoms for each element remain constant before and after the reaction. Mole Ratios: These ratios arise from the coefficients of a balanced equation and provide the relationship between moles of different substances involved in a reaction. For example, in the combustion of methane (( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} )), the mole ratio of reactants to products is essential for calculating the quantities needed or produced. Understanding stoichiometry thus directly influences one's ability to balance equations effectively. As such, it encompasses a variety of calculations and concepts, including: Determining Empirical and Molecular Formulas: Stoichiometry allows chemists to deduce the simplest ratio of elements in a compound, crucial for understanding its composition. Calculating Yields: By knowing the balanced equation and the starting quantities, one can predict the expected amount of product, leading to effective planning and resource management in both laboratory and industrial settings. Limiting Reactants: In any reaction, the reactant that gets completely consumed limits the production of the product. Stoichiometry helps identify this limiting reagent through balanced equations. The relationship between stoichiometry and balancing equations is so critical that "Stoichiometry is the language of chemistry; it interprets the relationships of elements and compounds in a reaction." Without a strong grasp of these principles, students and chemists may struggle to navigate chemical processes efficiently. As we proceed through the intricate layers of stoichiometry, remember that mastering the techniques of balancing chemical equations lays the groundwork not only for success in chemistry but for a deeper appreciation of the relationships that govern matter and its transformations. By embracing the principles of stoichiometry, one can truly unlock the potential of chemistry, making it both an art and a science. A chemical equation is a symbolic representation of a chemical reaction, articulating how reactants transform into products. This representation is not merely a sequence of letters and numbers; it embodies the essence of a chemical reaction by illustrating both the identities and quantities of the substances involved. A standard chemical equation features the reactants on the left side, followed by an arrow indicating the direction of the reaction, which leads to the products on the right side. For example, the reaction between hydrogen and oxygen to form water can be expressed as: This equation reveals that two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water. The components of a chemical equation include the following: Reactants: The starting substances in a chemical reaction, found on the left side of the equation. In the previous example, H2 and O2 are the reactants. Products: The substances formed as a result of the reaction, located on the right side of the equation. In our example, water (H2O) is the product. Coefficients: Numbers placed before the chemical formulas to indicate the relative amounts of each substance involved in the reaction. They are used to balance the equation according to the Law of Conservation of Mass. Subscripts: Small numbers indicating the number of atoms of each element in a molecule. For example, the formula for water, H2O, contains two hydrogen atoms and one oxygen atom. A chemical equation succinctly represents a complex interaction, providing valuable information about the stoichiometry of the reaction. As stated by the renowned chemist Antoine Lavoisier: "Nothing is lost, nothing is created, everything is transformed." This quote encapsulates the fundamental principle that drives the creation of balanced chemical equations, emphasizing the role of matter in reactions. It is important to note that a chemical equation must be balanced to accurately reflect the conservation of mass, making each component of a chemical equation critical to understanding the overall process. In summary, the definition of a chemical equation extends beyond simple notation; it represents the very framework through which chemists communicate the detailed aspects of chemical transformations. By mastering the components and interpretation of chemical equations, one can better appreciate the nature of reactions and the beauty of chemistry as a science. As we move forward, we will explore the common mistakes encountered in balancing these equations, ensuring a solid foundation in this essential skill. Explanation of the law of conservation of mass The Law of Conservation of Mass is a pivotal principle in chemistry, stating that in a closed system, the total mass of reactants must equal the total mass of products. This law was first formulated by the French chemist Antoine Lavoisier in the late 18th century, who famously remarked: "In chemical reactions, matter is neither created nor destroyed." This fundamental concept holds immense significance when it comes to balancing chemical equations, as it emphasizes that the number of atoms of each element involved in a reaction remains constant throughout the process. To better understand the implications of this law, consider the following key points: Implication in Reactions: Every chemical reaction involves the transformation of substances. Even though the substances change form, the total mass remains unchanged. Relationship to Balancing Equations: Balancing equations ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the conservation of mass. Real-life Applications: This law is not only theoretical; it has practical applications in various fields, such as engineering, environmental science, and pharmacology. For instance, in a closed container during combustion, the mass of the fuel and oxygen utilized equals the mass of carbon dioxide and water produced. To illustrate the law's application in a chemical reaction, consider the combustion of propane (( \text{C}_3\text{H}_8 )). When propane reacts with oxygen, it forms carbon dioxide and water. The balanced equation for this reaction is: By analyzing the coefficients, we can see that the total number of each type of atom on the reactant side equals the total on the product side, confirming the conservation of mass. This balance is crucial because it enables chemists to make accurate predictions regarding the quantities of reactants needed and products formed. Moreover, the law serves as a guiding principle in conducting experiments and in industrial processes. Chemists and engineers must account for the mass during the planning and execution of reactions to ensure efficiency and safety. Failure to consider this law can lead to dangerous outcomes, improper formulations, or wasted resources. In summary, the Law of Conservation of Mass forms the cornerstone of chemical reactions and equations. Understanding and applying this law not only aids in the accurate representation of reactions but also enhances our capability to manipulate and predict chemical behavior in various scientific and industrial contexts. As we delve deeper into the specifics of balancing equations, this foundational concept will serve as an ever-present guide. Common types of chemical equations (e.g., synthesis, decomposition, single-replacement, double-replacement, combustion) A comprehensive understanding of the types of chemical equations is essential for mastering the art of balancing them. Chemical reactions can be classified into several fundamental types, each with distinct characteristics and implications. These classifications not only facilitate communication among chemists but also provide a framework for predicting the outcomes of reactions. The major types of chemical equations include: Synthesis Reactions: In synthesis reactions, two or more reactants combine to form a single product. The general form can be represented as: Decomposition Reactions: Decomposition reactions involve a single compound breaking down into two or more simpler substances. The standard representation is: Single-Replacement Reactions: In this type of reaction, one element replaces another in a compound. It can be represented as: Double-Replacement Reactions: Here, the ions of two compounds exchange places in an aqueous solution, forming two new compounds. The general form is: Combustion Reactions: This type involves the reaction of a substance with oxygen, producing energy, carbon dioxide, and water. A common example is the combustion of hydrocarbons: Understanding these types of reactions allows chemists to predict the products given certain reactants and conditions. For instance, in synthesis reactions, recognizing the reactants' identities helps in determining the expected product. Similarly, recognizing the components of decomposition equations enables chemists to deduce the possible simpler substances formed. As the famous chemist "To the right chemist, every question is a reaction, waiting to be balanced." suggests, mastering the common types of reactions becomes a stepping stone towards proficiently balancing chemical equations. Each reaction type not only exhibits unique patterns but also adheres to the same principles governing mass conservation. Therefore, an understanding of how these reactions operate can significantly enhance one's ability to approach and solve balancing problems effectively. In conclusion, familiarizing oneself with the common types of chemical equations cultivates a deeper understanding of the underlying principles of chemistry. This knowledge, combined with effective practice, forms the foundation for developing meticulous balancing skills, ensuring that the intricacies of chemical reactions are represented accurately and comprehensively. Recognizing the components of a chemical equation (reactants, products, coefficients, subscripts) To effectively balance chemical equations, it is crucial to recognize and understand the various components that make up these symbolic representations of chemical reactions. Each part of the equation serves a specific role in conveying the identities and quantities of substances involved in the reaction. The primary components of a chemical equation include: Reactants: These are the starting materials that undergo a chemical change; they are positioned on the left side of a chemical equation. For instance, in the combustion of methane: CH4 and O2 are the reactants that react to form products. Products: These are the substances formed as a result of the chemical reaction, located on the right side of the equation. Continuing with the previous example, the products of the combustion of methane are: CO2 and H2O, which signify carbon dioxide and water respectively. Coefficients: Numbers that precede the chemical formulas in an equation, coefficients indicate the relative amounts of reactants and products involved in the reaction. For example, in the balanced equation for methane combustion: Here, the coefficients indicate that one molecule of methane reacts with two molecules of oxygen to yield one molecule of carbon dioxide and two molecules of water. Subscripts: Small numbers located to the bottom right of chemical symbols indicate the number of atoms of each element within a molecule. A prime example is in water: H2O, where the subscript "2" denotes that each molecule of water contains two hydrogen atoms and one oxygen atom. Understanding the distinction between coefficients and subscripts is vital as it ensures accuracy in balancing equations. While coefficients tell us how many molecules participate, subscripts reveal the composition of each molecule. The wise words of Lavoisier remind us: "To know is to know that you know nothing. That is the meaning of true knowledge." In mastering the components of chemical equations, one lays the groundwork for effective balancing and understanding of chemical reactions. As students and chemists delve deeper into the complexities of chemical equations, recognizing these fundamental elements will prove to be an invaluable skill that enhances both their theoretical and practical grasp of chemistry. Mistake 1: Confusing coefficients and subscripts One of the most common mistakes in balancing chemical equations stems from the confusion between coefficients and subscripts. While both play vital roles in chemical notation, they convey different information that is crucial for accurately representing a chemical reaction. Coefficients are the numbers placed before a chemical formula in an equation, signifying how many molecules or moles of a substance participate in the reaction. For example, in the balanced equation for the combustion of methane: Here, the coefficients indicate that 1 molecule of methane reacts with 2 molecules of oxygen, producing 1 molecule of carbon dioxide and 2 molecules of water. In this case, the coefficients provide a clear picture of the quantities involved in the reaction. On the other hand, subscripts are small numbers written to the bottom right of a chemical symbol, indicating the number of atoms of each element in a molecule. For instance, in the chemical formula for water (H2O), the subscript "2" reveals that each molecule contains two hydrogen atoms and one oxygen atom. This distinction is critical because: Subscript Misinterpretation: A common error occurs when students mistakenly interpret subscripts as coefficients. This can lead to incorrect assumptions about the quantities of molecules present in a balanced equation. Error in Balancing: Misreading subscripts can result in an unbalanced equation, fundamentally undermining the principle of the Law of Conservation of Mass. For example, if one were to treat H2O as just H and O rather than recognizing its composition as two hydrogen atoms and one oxygen atom, the equation would certainly fail to balance. As the esteemed chemist Linus Pauling once stated: "If you want to have good chemistry, you must be able to express it quantitatively." This quote highlights the necessity of precise language in chemistry. By conflating coefficients and subscripts, one cannot only misrepresent the reactants and products, but also lose grip on the quantitative relationships that govern chemical reactions. To avoid these pitfalls, here are some strategies: Practice Differentiation: Regularly practice writing and balancing equations, paying special attention to identifying coefficients and subscripts correctly. Use Visual Aids: Diagrams and graphical representations can assist in understanding how coefficients and subscripts function in chemical equations. Work with Peers: Collaborating with classmates can provide new perspectives and insights, helping to clarify misunderstandings. In summary, recognizing and understanding the distinction between coefficients and subscripts is essential for mastering the balancing of chemical equations. Being able to accurately interpret these components not only enhances one’s ability to balance equations effectively but also reinforces foundational concepts in stoichiometry and chemical interactions. Mistake 2: Not considering the law of conservation of mass One of the most significant mistakes students make when balancing chemical equations is failing to consider the Law of Conservation of Mass. This foundational principle asserts that in a closed system, matter cannot be created or destroyed, meaning the total mass of reactants must equal the total mass of products. Ignoring this law not only leads to inaccurate equations but also impedes a deeper understanding of chemical processes. When students overlook the conservation of mass during the balancing process, several issues can arise: Unbalanced Equations: An equation that does not adhere to the conservation of mass results in a disparity between reactants and products. This not only leads to incorrect stoichiometric calculations but also misrepresents the actual chemical reaction. Inaccurate Reaction Predictions: If the law is disregarded, it becomes impossible to predict the amounts of reactants needed or products formed accurately. This is crucial in practical applications such as pharmaceuticals, where precise measurements are vital to ensure safety and efficacy. Misleading Conclusions: Failing to consider this fundamental law can lead to misconceptions about the chemical nature of substances, potentially fostering errors in experimental design or industrial applications. As chemist Robert Berg(R): “To understand the nature of matter, one must appreciate its conservation.” To ensure that students consider the Law of Conservation of Mass when balancing chemical equations, various strategies can be employed: Visualizing the Reaction: Creating a visual representation of the reaction can help students clearly see the reactants and products involved. Practice with Real-World Examples: Engaging students with real-world chemical equations facilitates understanding how the conservation of mass applies. For instance, when burning propane, the balanced equation is: Encourage Inquiry: Prompt students to question whether the number of atoms for each element corresponds on both sides of the equation. This reflective practice reinforces the necessity of the conservation law in understanding chemical reactions. Ultimately, a firm grasp of the Law of Conservation of Mass is vital for effective balancing and will enhance students' overall chemistry competence. As the acclaimed chemist Marie Curie noted: "Nothing in life is to be feared, it is only to be understood." By reinforcing the role of this law, educators can empower students to appreciate the complexities of chemical equations, transforming mistakes into valuable learning opportunities. Mistake 3: Incorrectly balancing polyatomic ions One frequent error encountered by students when balancing chemical equations is the incorrect treatment of polyatomic ions. A polyatomic ion is a charged species composed of two or more atoms bonded together. These ions often act as a single unit during chemical reactions, and failing to recognize this can lead to significant mistakes in balancing equations. When faced with reactions containing polyatomic ions, students sometimes attempt to balance the individual atoms of the components instead of treating the polyatomic ion as a whole. This can lead to inconsistency and confusion in the overall balance of the equation. To effectively balance equations involving polyatomic ions, consider the following strategies: Identify the Polyatomic Ion: Begin by identifying any polyatomic ions present in the reactants or products. Common examples include sulfate (( \text{SO}_4^{2-} )), nitrate (( \text{NO}_3^{-} )), and ammonium (( \text{NH}_4^{+} )). Familiarizing yourself with these ions and their charges aids in the balancing process. Treat as a Unit: When balancing, treat each polyatomic ion as a single entity. For instance, in the reaction between sodium sulfate and barium chloride: Check for Consistency: After determining the coefficients for compounds containing polyatomic ions, always double-check the quantities of atoms. Ensure that the sum of each element on the reactant side matches the sum on the product side, including each component of the polyatomic ions. As chemists often remind us: "The beauty of chemistry lies in its structure, where balance must be delicately maintained." This balance applies to both individual atoms and larger units such as polyatomic ions. Failing to recognize the importance of these ions can lead to incorrect conclusions and hinder one's understanding of the chemical processes involved. For example, if one were to treat the sulfate ion (( \text{SO}_4^{2-} )) as just sulfur and oxygen, the balancing would become exceedingly difficult, if not impossible. To enhance understanding and proficiency, consider the following tips: Practice Regularly: Engaging with a variety of chemical equations involving polyatomic ions will solidify one's ability to balance them correctly. Use Visual Aids: Diagrams illustrating the structure of polyatomic ions can assist students in recognizing them as whole entities during balancing. Collaborate with Peers: Working with fellow students can enhance learning as different perspectives may illuminate common misunderstandings about polyatomic ions. In summary, accurately balancing chemical equations often hinges on the correct treatment of polyatomic ions. By adopting effective strategies, students can safeguard against this common mistake and promote a deeper understanding of the intricacies involved in chemical reactions. As one grows more adept at recognizing and balancing these ions, confidence in handling chemical equations will invariably increase. Mistake 4: Forgetting to balance all elements in the equation One of the critical oversights students encounter when balancing chemical equations is forgetting to balance all elements in the equation. This mistake not only leads to an incorrect representation of the chemical reaction but can also obscure the underlying principles of stoichiometry. Balancing an equation correctly requires that each element involved in the reaction be considered; neglecting even one can completely skew the outcome of the balancing process. This error often arises due to: Overlooking Elements: In complex reactions, it is easy to focus on a few familiar elements while unintentionally ignoring others. For instance, in the combustion of butane (( \text{C}_4\text{H}_{10} )), it's essential to ensure that both carbon and hydrogen are balanced along with oxygen. Rushing the Process: Students may become impatient during balancing tasks, leading them to incorrectly assume that they have accounted for all elements. This sense of urgency can result in a lack of thoroughness that undermines the goal of accurately reflecting the chemical equation. Misinterpretation of Products: Sometimes, students incorrectly identify the products of a reaction, which can lead to missing crucial elements that must be balanced. For example, when ammonium dichromate decomposes, it produces chromium(III) oxide, nitrogen gas, and water—the failure to recognize all these components can lead to imbalance. The importance of balancing all elements becomes clear through the lens of the Law of Conservation of Mass. As stated by the prominent chemist Lavoisier: "In all chemical reactions, no matter can be created or destroyed; it can only change form." This principle emphasizes that if any element is left unbalanced, the equation violates this fundamental law. As a consequence, chemists might draw misleading conclusions regarding the behavior and interactions of reactants and products, leading to further discrepancies in calculations. To prevent this common mistake, consider the following strategies: Systematic Approach: Adopt a methodical strategy to ensure every element is accounted for. Writing a list of all elements present in both reactants and products prior to balancing can serve as a helpful reminder to check each one. Double-Check Your Work: After completing the balancing, revisit the equation with a fresh perspective to confirm that all elements are in balance. This thorough review can help catch any oversights before finalizing the equation. Utilize Balanced Reaction Templates: Familiarizing oneself with balanced templates for common reactions can provide a reference point. Acknowledging the products and their compositions may help highlight what needs balancing in a new equation. In conclusion, adequately balancing all elements in a chemical equation is essential for a faithful representation of the reaction. By ensuring that no elements are omitted, chemists can uphold the principles of conservation of mass and maintain the integrity of chemical equations. Mastery of this skill not only strengthens one's ability to balance equations but also deepens understanding of stoichiometric relationships and the nuanced interactions that govern chemical reactions. Mistake 5: Balancing oxygen or hydrogen last One common mistake that emerges during the process of balancing chemical equations is the tendency to prioritize balancing oxygen or hydrogen last. While it might seem convenient to leave these elements until the end, this approach can often lead to confusion and inaccuracies in the final balanced equation. Here’s why this practice is generally discouraged and how it can be addressed for improved accuracy in balancing. First, it’s essential to understand the reason for this mistake and its implications: Complexity in Balancing: Oxygen and hydrogen are frequently present in many reactions, particularly in combustion and redox reactions. By saving these elements for last, students may inadvertently disrupt the balance achieved with other elements, making it challenging to reach a final balanced equation. Interdependence of Elements: The quantities of oxygen and hydrogen often depend on the quantities of other elements in the reaction. Considering them last may complicate corrections needed for earlier balanced elements. Increased Risk of Misbalancing: Leaving the most abundant elements till the end can lead to miscalculations and oversight since students might believe they are “almost done” without ensuring all elements are accurately represented. As the renowned chemist Antoine Lavoisier famously remarked: "The only thing that is constant is change." This quote speaks to the necessity of reevaluating and adjusting throughout the balancing process, especially when dealing with crucial elements like oxygen and hydrogen. To mitigate these common pitfalls, consider adopting the following strategies: Balance All Elements Simultaneously: Instead of focusing on balancing oxygen and hydrogen last, aim for a systematic approach where all elements are balanced together throughout the process. This ensures a more comprehensive understanding of the reaction dynamics. Use a Stepwise Method: For reactions that include complex compounds, it can be helpful to first balance the elements that appear in multiple reactants and products before focusing on hydrogen and oxygen. For instance, if you’re balancing the combustion of a hydrocarbon, approach the balancing as follows: Review the Entire Equation: After balancing, take a step back to ensure that every atom is accounted for. This holistic view may reveal inconsistencies or errors that need adjustment. Ultimately, avoiding the practice of balancing oxygen or hydrogen last will help streamline the balancing process and increase the overall accuracy of reactions. As noted by the esteemed chemist Robert H. Grubbs: "Chemistry is the study of transformations, and reactions are filled with interdependencies." Embracing this interconnectedness will not only improve your ability to balance chemical equations effectively but will also give you a deeper appreciation for the complexity of chemical interactions. Remember, the key to successful balancing is not just about plugging numbers into the equation, but rather understanding the nature and relationships of the substances involved. Mistake 6: Assuming elemental states without checking conditions One of the crucial mistakes students often make when balancing chemical equations is assuming the elemental states of reactants and products without checking the specific conditions under which the reactions occur. Chemical substances can exist in different physical states—such as solid, liquid, gas, or aqueous—depending on temperature, pressure, and other environmental factors. Misjudging these states can lead to significant errors in the balancing process. This oversight typically arises from: Textbook Assumptions: Many textbooks may present reactions under standard conditions (usually 25°C and 1 atm), which may not always reflect real-world scenarios in laboratories or industrial applications. Neglecting Experimental Conditions: Students might fail to consider specific temperature or pressure changes that can affect the physical state of the substances involved, such as transitioning from solid to gas during a reaction. Lack of Familiarity with Phase Changes: The transition of substances between states—like melting or evaporation—requires a solid understanding of how these changes influence chemical reactions. For instance, consider the combustion of methane (CH4). Under standard conditions, it is a gas, and its reaction with oxygen can be expressed as: However, if this reaction occurs at a high temperature, the water produced may exist in a vapor state, necessitating adjustments to the balanced equation. If one assumes that water is always a liquid at all times without checking conditions, the resulting calculations could lead to imbalance and incorrect predictions about product yields. As the physicist Albert Einstein wisely noted: "A person who never made a mistake never tried anything new." This sentiment resonates deeply in chemistry, highlighting the importance of vigilance and inquiry. To avoid this mistake, consider these strategies: Always Verify State Conditions: Before balancing an equation, check the physical state of each compound involved. Has the temperature affected its state? Is there any pressure affecting the gaseous products? Taking these variables into account will assist in accurately reflecting the reaction. Consult Reliable Sources: Use reliable references or databases that specify standard states at various temperatures and pressures to ensure you understand how the elements behave under those conditions. Practice with Diverse Scenarios: Engage in exercises where different conditions are imposed on reactions to see the impact on the state of reactants and products. This practice can reinforce knowledge about how physical states influence balancing. Considering the physical states of substances is paramount for consistently achieving accurate results in chemical reactions. By overcoming this tendency to assume states, chemists can avoid common balancing errors and deepen their overall understanding of chemical relationships and behaviors. As one becomes more adept at recognizing and incorporating these aspects into their equations, balancing will become a more intuitive and precise skill. Tips for verifying balanced equations Verifying balanced equations is as crucial as the initial balancing process itself. A chemical equation can only be deemed correct when it accurately reflects the conservation of mass, ensuring that the number of each type of atom on both sides of the equation remains equal. To aid in this verification process, here are some helpful tips: Tally All Atoms: After balancing, systematically count the number of atoms for each element present in the reactants and products. For instance, in the combustion of methane: In this reaction, confirm that both sides contain: 1 carbon (C), 4 hydrogen (H), and 4 oxygen (O) atoms. Check by Substitution: Temporarily replace coefficients with a simple number (often 1) to assess the ratio of each component. This can assist in visualizing how the reactants relate to the products without overcomplicating the situation at first. Use Visual Aids: Diagrams or charts can simplify the balance-checking process. A drawing illustrating the reactants and products can visually illustrate the conservation of mass. Apply the “What If” Method: Consider altering one of the coefficients in the balanced equation and analyze its effect on the overall balance. This helps reinforce the relationships between reactants and products. For instance, if you change the coefficient of water produced in the methane combustion from 2 to 3, note how it impacts the oxygen and carbon atoms. Review the Stoichiometric Coefficients: Ensure that the coefficients are in their simplest form. Sometimes, students might accidentally multiply all coefficients to yield whole numbers. Always simplify, as complex coefficients can obscure real relationships between substances. In the words of the chemist Marie Curie, "I am among those who think that science has great beauty." Balancing and verifying equations exemplify this beauty, where the elegance of mathematical relationships reveals order in the chemical chaos. Engaging with these verification techniques not only reinforces accuracy but also deepens one’s appreciation of chemical reactions and their underlying principles. As you practice these methods, you'll develop a more intuitive understanding of how to balance and verify equations effectively, transforming your approach to chemistry into an artful practice. Step-by-step approach to balancing equations Balancing chemical equations may initially appear daunting, but using a step-by-step approach can greatly simplify the process. By breaking down the balancing task into manageable stages, students can better visualize the components involved and ensure accuracy throughout. The following systematic strategy outlines how to effectively balance equations: Write the Unbalanced Equation: Begin by clearly writing the unbalanced chemical equation. Ensure that you accurately represent the reactants and products along with their chemical formulas. For instance, consider the combustion of propane: Count the Number of Atoms: For both reactants and products, carefully count the number of atoms of each element. Create a table to help visualize the counts, such as: Reactants: C: 3, H: 8, O: 2 Products: C: 1, H: 2, O: 3 Balance Elements One at a Time: Start with the most complex molecule and balance its elements first. In this example, balance carbon (C) by placing a coefficient of 3 in front of carbon dioxide: Adjust Coefficients: After balancing carbon, proceed to balance hydrogen next. Adjust coefficients as necessary. In this case, place a coefficient of 4 in front of water to balance the hydrogen atoms: Reassess the Oxygen Atoms: Now, sum the total oxygen atoms in the products—there are 6 (3 from CO2 and 4 from H2O). Address this by placing a coefficient of 5 in front of O2 in the reactants: Final Verification: Cross-check the balance of all elements once again. Ensure that the number of atoms of each element matches on both the reactant and product sides. Final Count: Reactants: C: 3, H: 8, O: 6 Products: C: 3, H: 8, O: 6 As you refine your balancing skills, it is essential to remember that practice leads to mastery. The wise words of renowned chemist Albert Einstein resonate here: "If you can't explain it simply, you don't understand it well enough." This sentiment underscores the importance of clarity in the balancing process. By practicing this systematic approach regularly, students will not only enhance their ability to balance equations but also foster a deeper understanding of chemical relationships as a whole. Through diligence and perseverance, balancing chemical equations can transform from a challenging task into an empowering skill. Utilizing different methods for balancing equations (e.g., inspection method, algebraic method, redox method) Balancing chemical equations can be approached in various ways, each suited for different types of equations and personal preferences. Utilizing a variety of methods not only enhances understanding but also allows for more efficient problem-solving. Here are several effective techniques for balancing chemical equations: Inspection Method: This intuitive method involves adjusting the coefficients of the reactants and products by inspection, relying on trial and error. Chemists often use this straightforward approach due to its simplicity. For example, when balancing the combustion of glucose ((C_6H_{12}O_6)), one would begin by examining the number of carbon (C), hydrogen (H), and oxygen (O) atoms on both sides. The equation can be balanced as follows: Algebraic Method: This systematic approach employs algebraic equations to represent the coefficients of the reactants and products. By assigning variables to unknown coefficients, one can create a set of linear equations based on the conservation of atoms. For instance, in a reaction like the combustion of ethylene ((C_2H_4)), one might set coefficients as variables and solve the resulting equations to find their values. Redox Method: This technique is particularly useful for balancing redox reactions, where changes in oxidation states occur. The redox method involves balancing the oxidation and reduction half-reactions separately before combining them back into a complete equation. This process typically includes identifying the substance being oxidized and the one being reduced, determining the changes in oxidation states, and ensuring the electrons lost equal the electrons gained. For example, in the reaction of hydrogen peroxide decomposition: Oxidation Number Method: Similar to the redox method, this technique focuses on the changes in oxidation numbers for each element involved in a redox reaction. By analyzing the changes for all elements, one can deduce the coefficients needed to balance the overall reaction. As the chemist Linus Pauling once remarked, “The greatest advances in science occur when a person makes up his or her mind to be a little less stupid.” This quote underscores the importance of experimentation in finding the most effective balancing method. Ultimately, the effectiveness of each method may vary depending on the individual and the specific equation in question. It is advisable for students to practice multiple techniques to determine which resonates with them the most. Engaging with these diverse methods can transform the balancing of chemical equations from a daunting challenge into a confident and calculated endeavor. By mastering these strategies, chemists can not only solve equations with greater ease but also deepen their understanding of the intricate relationships between reactants and products in chemical reactions. Common misconceptions about balancing chemical equations As students and novices in chemistry tackle the task of balancing chemical equations, several common misconceptions can hinder their understanding and ability to master this essential skill. Recognizing these misconceptions is crucial for cultivating a more accurate and holistic approach to balancing chemical reactions. Here are some of the prevalent misunderstandings: Believing that coefficients represent the number of atoms: One widespread misconception is that coefficients indicate the number of atoms present in a molecule. In reality, coefficients signify the number of molecules or moles involved in the reaction. For example, in the equation: 2H2 indicates there are two molecules of diatomic hydrogen gas, each consisting of two hydrogen atoms, not four atoms total. Understanding this difference is essential for accurate balancing. Assuming all reactions can be balanced by inspection: Many students believe that they can balance every chemical reaction simply by examining it. However, while simpler equations can often be balanced through this inspection method, more complex reactions may require systematic strategies like the algebraic or redox methods. Overlooking the role of physical states: Another common error is neglecting the significance of physical states (solid, liquid, gas, aqueous) when balancing equations. The state of the reactants and products can influence the reaction's behavior. As the physicist Albert Einstein wisely noted, "Reality is merely an illusion, albeit a very persistent one." This metaphor resonates with how a student's perception can misrepresent the reaction's real dynamics. Understanding the states ensures accurate representations and yields insightful predictions. Rushing through the process: In an effort to complete assignments quickly, students may skip essential steps, leading to erroneous conclusions. Balancing is an iterative process that demands attention and care. Chemistry requires precise thinking, and a rushed approach can obscure critical aspects of the reaction. Believing that balancing is only about matching numbers: Some individuals view balancing equations as a mere game of matching quantities. However, the essence of balancing lies in the relationship between reactants and products. Fostering a deeper understanding of chemical interactions transcends mere calculations, as these relationships govern how substances interact in reactions. As emphasized by chemist Daniel Klein, "The more you know, the more you realize you don’t know." This statement encapsulates the journey of learning chemistry, where each misconception offers an opportunity for growth and understanding. By debunking these common misconceptions, students can approach the balancing of chemical equations with clarity and confidence, paving the way for success in their chemistry endeavors. Practical examples of chemical equations and common errors Practical examples of chemical equations provide an invaluable opportunity to identify and understand common errors encountered during the balancing process. Examining real-world reactions can illuminate how missteps can unfold, enhancing our grasp of the balancing principles. Here are some illustrative examples that encapsulate frequent mistakes: 1. Combustion of Ethylene Consider the combustion of ethylene (C2H4), which can be represented by the unbalanced equation: Newcomers may falsely prioritize balancing oxygen last, subsequently leading to miscounting hydrogen and oxygen atoms due to their interconnected nature. The correct balanced equation, obtained through proper sequential balancing, is: 2. Decomposition of Ammonium Dichromate The decomposition of ammonium dichromate, represented as: Students often mistakenly attempt to balance the nitrogen or oxygen without recognizing the entirety of the products formed. Failing to account for chromium can lead to errors, resulting in: Common Mistake: Neglecting to include solid chromium at all. The balanced equation should be: Through these practical examples, we can see how vital it is to remain attentive and systematic while balancing chemical equations. As chemist Richard Feynman once opined: "The opposite of a good idea is not a bad idea; it’s a different good idea." This sentiment urges us to embrace different perspectives and approaches while verifying our understanding of chemical processes. By learning from these common errors and examples, aspiring chemists can gain confidence in their ability to balance equations accurately. Importance of practice and resources for improving balancing skills Mastering the art of balancing chemical equations is an essential skill that requires consistent practice and access to relevant resources. Just as athletes refine their techniques through repetition, aspiring chemists must engage with balancing exercises to enhance their proficiency. The importance of practice can be encapsulated in the words of the renowned chemist "The only way to learn chemistry is to do chemistry." This principle underscores the value of hands-on experience, especially when it comes to balancing equations, which is intrinsic to understanding stoichiometric relationships. Regular practice allows students to: Develop Intuition: The more one interacts with equations, the more familiar they become with recognizing patterns and common coefficients, leading to quicker and more accurate balancing. Avoid Common Mistakes: Through repeated exposure, students can identify the typical pitfalls discussed earlier, such as confusing coefficients and subscripts or neglecting to balance all elements. Build Confidence: Success in practice problems enhances self-efficacy, enabling students to tackle more complex balancing scenarios with assurance. Enhance Critical Thinking: Regularly balancing equations fosters analytical skills, as students learn to strategize which elements to balance first and how to adjust coefficients appropriately. In addition to practice, utilizing a variety of resources can significantly bolster balancing skills. Here are some valuable resources to consider: Online Tutorials and Videos: Websites like Khan Academy and YouTube feature a wealth of instructional videos where experienced educators break down the balancing process. Practice Workbooks: Many textbooks and online platforms offer workbooks specifically designed for chemical equations. These often include step-by-step examples, practice problems, and answer keys. Interactive Simulations: Platforms like PhET provide simulations where students can visually engage with chemical reactions, allowing them to balance equations dynamically while observing the laws of conservation of mass in real time. Mobile Apps: There are various chemistry apps available that allow for on-the-go practice. These apps often feature quizzes, flashcards, and game-like interfaces to make learning more interactive and enjoyable. Furthermore, collaboration in study groups can enhance learning as peers offer different perspectives and approaches to balancing equations. Discussing problems aloud can sharpen understanding and reinforce concepts. As the American chemist George Washington Carver said: "It is not the style of clothes one wears, neither the kind of automobile one drives, nor the amount of money one has in the bank, that counts. These mean nothing. It is simply service that measures success." This sentiment highlights that success in chemistry, particularly in mastering balancing equations, comes from the dedication to practice and the willingness to utilize available resources for improvement. In conclusion, the journey to mastering the balancing of chemical equations is one that demands persistent practice and the effective use of resources. By embracing the myriad of tools and opportunities available, students not only enhance their balancing skills but also cultivate a deeper appreciation and understanding of the chemical processes that govern our world. Remember, as with any skill in science, the path to proficiency is best traveled with both diligence and curiosity. Conclusion summarizing key points and encouraging practice In conclusion, mastering the art of balancing chemical equations is a critical skill that underpins many concepts in chemistry. The processes involved here extend beyond rote memorization; they require understanding the relationships between reactants and products, as well as a commitment to practice and refinement. To reinforce the key takeaways from this exploration, consider the following essential points: Understanding the Law of Conservation of Mass: This law is fundamental to balancing equations, emphasizing that matter cannot be created or destroyed. Effective balancing must mirror this principle. Recognizing Components: Distinguishing between coefficients and subscripts is crucial. Coefficients indicate the number of molecules, while subscripts denote the number of atoms in a molecule. Common Mistakes: Awareness of common errors—such as confusing coefficients with subscripts, neglecting to balance all elements, and assuming states without context—can significantly improve balancing accuracy. Practical Strategies: Employing systematic approaches such as the inspection method, algebraic method, or recognizing the structure of polyatomic ions can facilitate more efficient balancing. Regular Practice: Consistent engagement with balancing exercises fosters confidence and intuition, helping to strengthen foundational knowledge and critical-thinking skills. Leverage Resources: Utilizing diverse resources—be it online tutorials, practice workbooks, or collaborative study groups—can enhance comprehension and offer varied perspectives on complex topics. Ultimately, patience and perseverance are vital to mastering this essential chemistry skill. As noted by the esteemed chemist Albert Einstein, "A person who never made a mistake never tried anything new." Each attempt at balancing equations—whether successful or fraught with errors—provides invaluable learning opportunities. Embrace the challenges you encounter as stepping stones to greater understanding and success. Therefore, it is crucial to engage actively with the content and seek out practice exercises that enhance your skills. The journey to mastering chemical equations does not end with reading; it thrives on hands-on experience. Remember, the deeper your understanding, the more intuitive the balancing process will become. By committing to these practices, you will not only improve your ability to balance equations but also cultivate a richer appreciation of the dynamic world of chemistry. Now, with each equation you tackle, approach it with curiosity and determination, embodying the spirit of exploration that defines scientific inquiry. Transform mistakes into lessons, and with consistent practice, you will emerge as a confident and capable chemist.
14577	https://www.vaia.com/en-us/explanations/math/pure-maths/integrating-polynomials/	Integrating Polynomials: Definition & Examples We use essential cookies to make our site work. With your consent, we may also use non-essential cookies to improve user experience and analyze website traffic. By clicking “Accept,” you agree to our website's cookie use as described in our Cookie Policy. You can change your cookie settings at any time by clicking “Preferences.” Accept Find study content Learning Materials Discover learning materials by subject, university or textbook. 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You'll start by understanding and defining the fundamental elements of integrating polynomials. Then, master different techniques of polynomial integration, including a step-by-step integration method, whilst learning to avoid common mistakes. Real-life examples and complex scenarios of integrating a polynomial provide practical insight. This guide also unlocks the nuances of the Fundamental Theorem of Calculus in relation to polynomials, and thoroughly breaks down the key polynomial integration rules for easy understanding and application. Get started Millions of flashcards designed to help you ace your studies Sign up for free + Add tag Immunology Cell Biology Mo What is the role of integration in mathematics? Show Answer + Add tag Immunology Cell Biology Mo What is an example of how polynomial integration is applied in economics? Show Answer + Add tag Immunology Cell Biology Mo How are Polynomial Integration Rules applied in real scenarios? Show Answer + Add tag Immunology Cell Biology Mo What is the process of Integrating Polynomials? Show Answer + Add tag Immunology Cell Biology Mo What are key Polynomial Integration rules? Show Answer + Add tag Immunology Cell Biology Mo What are the steps to integrate a polynomial? Show Answer + Add tag Immunology Cell Biology Mo What are the common mistakes to avoid in polynomial integration? Show Answer + Add tag Immunology Cell Biology Mo What role does polynomial integration play in the field of physics? Show Answer + Add tag Immunology Cell Biology Mo How do you integrate a polynomial such as 3 x 2+2 x+1 ? Show Answer + Add tag Immunology Cell Biology Mo What is the power rule for integration? Show Answer + Add tag Immunology Cell Biology Mo What is the definition of a complex antiderivative in the context of polynomial integration? Show Answer + Add tag Immunology Cell Biology Mo What role does polynomial integration play in the field of physics? Show Answer + Add tag Immunology Cell Biology Mo How do you integrate a polynomial such as 3 x 2+2 x+1 ? Show Answer + Add tag Immunology Cell Biology Mo What is the power rule for integration? Show Answer + Add tag Immunology Cell Biology Mo What is the definition of a complex antiderivative in the context of polynomial integration? Show Answer + Add tag Immunology Cell Biology Mo What is the role of integration in mathematics? Show Answer + Add tag Immunology Cell Biology Mo What is an example of how polynomial integration is applied in economics? Show Answer + Add tag Immunology Cell Biology Mo How are Polynomial Integration Rules applied in real scenarios? Show Answer + Add tag Immunology Cell Biology Mo What is the process of Integrating Polynomials? Show Answer + Add tag Immunology Cell Biology Mo What are key Polynomial Integration rules? Show Answer + Add tag Immunology Cell Biology Mo What are the steps to integrate a polynomial? Show Answer + Add tag Immunology Cell Biology Mo What are the common mistakes to avoid in polynomial integration? Show Answer Achieve better grades quicker with Premium PREMIUM Karteikarten Spaced Repetition Lernsets AI-Tools Probeklausuren Lernplan Erklärungen Karteikarten Spaced Repetition Lernsets AI-Tools Probeklausuren Lernplan Erklärungen Kostenlos testen Geld-zurück-Garantie, wenn du durch die Prüfung fällst Did you know that StudySmarter supports you beyond learning? Find your perfect university Get started for free Find your dream job Get started for free Claim big discounts on brands Get started for free Finance your studies Get started for free Sign up for free and improve your grades Scan and solve every subject with AI Try our homework helper for free Recommended Create a study plan Get a personalized plan and set yourself up for success. Generate flashcards Upload or scan documents to turn them into flashcards. 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Save ArticleSave Article Fact Checked Content Last Updated: 09.02.2024 Published at: 10.01.2022 15 min reading time Applied Mathematics Calculus Decision Maths Discrete Mathematics Geometry Logic and Functions Mechanics Maths Probability and Statistics Pure Maths ASA Theorem Absolute Convergence Absolute Value Equations and Inequalities Absolute Values Abstract algebra Addition and Multiplication of series Addition and Subtraction of Rational Expressions Addition, Subtraction, Multiplication and Division Algebra Algebra of limits Algebra over a field Algebraic Expressions Algebraic Fractions Algebraic K-theory Algebraic Notation Algebraic Representation Algebraic curves Algebraic geometry Algebraic number theory Algebraic topology Analyzing Graphs of Polynomials Angle Conversion Angle Measure Angle Of Depression Angle Of Elevation Angles Angles in Polygons Angular geometry Approximation and Estimation Area and Perimeter of Quadrilaterals Area of Triangles Argand Diagram Arithmetic Sequences Associative algebra Average Rate of Change Banach algebras Basis Bearing In Trigonometry Bijective Functions Bilinear forms Binomial Expansion Binomial Theorem Bounded Sequence C-algebras Category theory Cauchy Sequence Cayley Hamilton Theorem Central Angles Chain Rule Circle Theorems Circles Circles Maths Clifford algebras Cofunction Identities Cohomology theory Combinatorics Common Factors Common Multiples Commutative algebra Compact Set Completing the Square Complex Numbers Composite Figures Composite Functions Composition of Functions Compound Interest Compound Units Congruence Equations Conic Sections Connected Set Construction and Loci Continuity and Uniform convergence Continuity of derivative Continuity of real valued functions Continuous Function Convergent Sequence Converting Metrics Convexity and Concavity Coordinate Geometry Coordinates in Four Quadrants Cosecant Ratios Cosine Ratios Cotangent Ratios Coupled First-order Differential Equations Cubic Function Graph Cyclic Quadrilaterals Data Transformations De Moivre's Theorem Deductive Reasoning Definite Integrals Derivative of a real function Deriving Equations Determinant Of Inverse Matrix Determinant of Matrix Determinants Diagonalising Matrix Differentiability of real valued functions Differential Equations Differential algebra Differentiation Differentiation Rules Differentiation from First Principles Differentiation of Hyperbolic Functions Dimension Direct and Inverse proportions Direct variation Discontinuity Disjoint and Overlapping Events Disproof By Counterexample Distance formula Distance from a Point to a Line Divergent Sequence Divisibility Tests Division algebras Double Angle and Half Angle Formulas Drawing Conclusions from Examples Eigenvalues and Eigenvectors Ellipse Ellipses Elliptic curves Equation of Line in 3D Equation of a Perpendicular Bisector Equation of a circle Equations Equations and Identities Equations and Inequalities Equicontinuous families of functions Estimation in Real Life Euclidean Algorithm Evaluating and Graphing Polynomials Even Functions Exponential Form of Complex Numbers Exponential Rules Exponentials and Logarithms Expression Math Expressions and Formulas Faces Edges and Vertices Factorials Factoring Polynomials Factoring Quadratic Equations Factorising expressions Factors Fermat's Little Theorem Field theory Finding Maxima and Minima Using Derivatives Finding Rational Zeros Finding The Area First Fundamental Theorem First-order Differential Equations Forms of Quadratic Functions Fourier analysis Fractional Powers Fractional Ratio Fractions Fractions and Decimals Fractions and Factors Fractions in Expressions and Equations Fractions, Decimals and Percentages Function Basics Functional Analysis Functions Fundamental Counting Principle Fundamental Theorem of Algebra Generating Terms of a Sequence Geometric Sequence Gradient and Intercept Gram-Schmidt Process Graphical Representation Graphing Rational Functions Graphing Trigonometric Functions Graphs Graphs And Differentiation Graphs Of Exponents And Logarithms Graphs of Common Functions Graphs of Trigonometric Functions Greatest Common Divisor Grothendieck topologies Group Mathematics Group representations Growth and Decay Growth of Functions Gröbner bases Harmonic Motion Hermitian algebra Higher Derivatives Highest Common Factor Homogeneous System of Equations Homological algebra Homotopy theory Hopf algebras Hyperbolas Ideal theory Imaginary Unit And Polar Bijection Implicit differentiation Indirect variation Inductive Reasoning Inequalities Maths Infinite geometric series Injective functions Injective linear transformation Instantaneous Rate of Change Integers Integrating Ex And 1x Integrating Polynomials Integrating Trigonometric Functions Integration Integration By Parts Integration By Substitution Integration Using Partial Fractions Integration of Hyperbolic Functions Interest Interest calculations Invariant Points Inverse Hyperbolic Functions Inverse Matrices Inverse and Joint Variation Inverse functions Inverse of a Matrix and System of Linear equation Invertible linear transformation Irrational numbers Isosceles Triangles Trigonometry Iterative Methods Jordan algebras Knot theory L'hopitals Rule Lattice theory Law Of Cosines In Algebra Law Of Sines In Algebra Laws of Logs Leibnitz's Theorem Lie algebras Lie groups Limits of Accuracy Linear Algebra Linear Combination Linear Expressions Linear Independence Linear Systems Linear Transformation Linear Transformations of Matrices Linear graphs Location of Roots Logarithm Base Logic Lower and Upper Bounds Lowest Common Denominator Lowest Common Multiple Math formula Matrices Matrix Addition And Subtraction Matrix Calculations Matrix Determinant Matrix Multiplication Matrix inverses Matrix operations Mean value theorem Metric and Imperial Units Midpoint formula Misleading Graphs Mixed Expressions Mode Modelling with First-order Differential Equations Modular Arithmetic Module theory Modulus Functions Modulus and Phase Monoidal categories Monotonic Function Multiples of Pi Multiplication and Division of Fractions Multiplicative Relationship Multiplicative ideal theory Multiplying And Dividing Rational Expressions Natural Logarithm Natural Numbers Non-associative algebra Normed spaces Notation Number Number Line Number Systems Number Theory Number e Numerical Methods Odd functions Open Sentences and Identities Operation with Complex Numbers Operations With Matrices Operations with Decimals Operations with Polynomials Operator algebras Order of Operations Orthogonal groups Orthogonality Parabola Parallel Lines Parametric Differentiation Parametric Equations Parametric Hyperbolas Parametric Integration Parametric Parabolas Partial Fractions Pascal's Triangle Percentage Percentage Increase and Decrease Perimeter of a Triangle Permutations and Combinations Perpendicular Lines Phase Shift Points Lines and Planes Pointwise convergence Poisson algebras Polynomial Graphs Polynomial division Polynomial rings Polynomials Powers Roots And Radicals Powers and Exponents Powers and Roots Prime Factorization Prime Numbers Problem-solving Models and Strategies Product Rule Product of polynomials Product-to-sum Formulas Proof Proof and Mathematical Induction Proof by Contradiction Proof by Deduction Proof by Exhaustion Proof by Induction Properties of Determinants Properties of Exponents Properties of Riemann Integral Properties of dimension Properties of eigenvalues and eigenvectors Proportion Proving an Identity Pythagorean Identities Quadratic Equations Quadratic Function Graphs Quadratic Graphs Quadratic forms Quadratic functions Quadrilaterals Quantum groups Quotient Rule Radian Measure Radians Radical Functions Rates of Change Ratio Ratio Fractions Ratio and Root test Rational Exponents Rational Expressions Rational Functions Rational Numbers and Fractions Rationalizing denominators Ratios as Fractions Real Numbers Rearrangement Reciprocal Graphs Recurrence Relation Recursion and Special Sequences Reduced Row Echelon Form Reducible Differential Equations Remainder and Factor Theorems Representation Of Complex Numbers Representation theory Rewriting Formulas and Equations Riemann integral for step function Riemann surfaces Riemannian geometry Right Triangle Trigonometry Ring theory Roots Of Unity Roots of Complex Numbers Roots of Polynomials Rounding SAS Theorem SSS Theorem Scalar Products Scalar Triple Product Scale Drawings and Maps Scale Factors Scalene Triangles Trigonometry Scientific Notation Secant Ratios Second Fundamental Theorem Second Order Recurrence Relation Second-order Differential Equations Sector of a Circle Segment of a Circle Sequence and series of real valued functions Sequence of Real Numbers Sequences Sequences and Series Series Maths Series of non negative terms Series of real numbers Sets Math Sigma notation Similar Triangles Similar and Congruent Shapes Similarity and diagonalisation Simple Interest Simple algebras Simplifying Fractions Simplifying Radicals Simultaneous Equations Sine Ratios Sine and Cosine Rules Sinusoidal Graphs Slope formulas Slope-intercept form Small Angle Approximation Solving Linear Equations Solving Linear Systems Solving Oblique Triangles Solving Quadratic Equations Solving Radical Inequalities Solving Rational Equations Solving Right Triangles Solving Simultaneous Equations Using Matrices Solving Systems of Inequalities Solving Trigonometric Equations Solving and Graphing Quadratic Equations Solving and Graphing Quadratic Inequalities Spanning Set Special Products Special Sequences Spherical Trigonometry Standard Form Standard Integrals Standard Unit Standard form equations Stone Weierstrass theorem Straight Line Graphs Subgroup Subsequence Subspace Substraction and addition of fractions Sum and Difference of Angles Formulas Sum of Natural Numbers Sum-to-product Formulas Summation by Parts Supremum and Infimum Surds Surjective functions Surjective linear transformation System of Linear Equations Tables and Graphs Tangent Ratios Tangent of a Circle Tangent rule Taylor theorem The Quadratic Formula and the Discriminant Topological groups Torsion theories Transformations Transformations geometry Transformations of Graphs Transformations of Roots Translations of Trigonometric Functions Transversals Triangle Rules Triangle trigonometry Trigonometric Functions Trigonometric Functions of General Angles Trigonometric Identities Trigonometric Identities Sum Angles Trigonometric Proofs Trigonometric Ratios Trigonometric Ratios Of Complementary Angles Trigonometry Trigonometry Height And Distance Trigonometry Word Problems Turning Points Types of Functions Types of Numbers Types of Triangles Uniform convergence Unit Circle Units Universal algebra Upper and Lower Bounds Valuation theory Variable expressions Variables in Algebra Vector Notation Vector Space Vector spaces Vectors Verifying Trigonometric Identities Vertex form Volume calculations Volumes of Revolution Von Neumann algebras Writing Equations Writing Linear Equations Zariski topology secant lines slope formula Statistics Theoretical and Mathematical Physics Contents Applied Mathematics Calculus Decision Maths Discrete Mathematics Geometry Logic and Functions Mechanics Maths Probability and Statistics Pure Maths ASA Theorem Absolute Convergence Absolute Value Equations and Inequalities Absolute Values Abstract algebra Addition and Multiplication of series Addition and Subtraction of Rational Expressions Addition, Subtraction, Multiplication and Division Algebra Algebra of limits Algebra over a field Algebraic Expressions Algebraic Fractions Algebraic K-theory Algebraic Notation Algebraic Representation Algebraic curves Algebraic geometry Algebraic number theory Algebraic topology Analyzing Graphs of Polynomials Angle Conversion Angle Measure Angle Of Depression Angle Of Elevation Angles Angles in Polygons Angular geometry Approximation and Estimation Area and Perimeter of Quadrilaterals Area of Triangles Argand Diagram Arithmetic Sequences Associative algebra Average Rate of Change Banach algebras Basis Bearing In Trigonometry Bijective Functions Bilinear forms Binomial Expansion Binomial Theorem Bounded Sequence C-algebras Category theory Cauchy Sequence Cayley Hamilton Theorem Central Angles Chain Rule Circle Theorems Circles Circles Maths Clifford algebras Cofunction Identities Cohomology theory Combinatorics Common Factors Common Multiples Commutative algebra Compact Set Completing the Square Complex Numbers Composite Figures Composite Functions Composition of Functions Compound Interest Compound Units Congruence Equations Conic Sections Connected Set Construction and Loci Continuity and Uniform convergence Continuity of derivative Continuity of real valued functions Continuous Function Convergent Sequence Converting Metrics Convexity and Concavity Coordinate Geometry Coordinates in Four Quadrants Cosecant Ratios Cosine Ratios Cotangent Ratios Coupled First-order Differential Equations Cubic Function Graph Cyclic Quadrilaterals Data Transformations De Moivre's Theorem Deductive Reasoning Definite Integrals Derivative of a real function Deriving Equations Determinant Of Inverse Matrix Determinant of Matrix Determinants Diagonalising Matrix Differentiability of real valued functions Differential Equations Differential algebra Differentiation Differentiation Rules Differentiation from First Principles Differentiation of Hyperbolic Functions Dimension Direct and Inverse proportions Direct variation Discontinuity Disjoint and Overlapping Events Disproof By Counterexample Distance formula Distance from a Point to a Line Divergent Sequence Divisibility Tests Division algebras Double Angle and Half Angle Formulas Drawing Conclusions from Examples Eigenvalues and Eigenvectors Ellipse Ellipses Elliptic curves Equation of Line in 3D Equation of a Perpendicular Bisector Equation of a circle Equations Equations and Identities Equations and Inequalities Equicontinuous families of functions Estimation in Real Life Euclidean Algorithm Evaluating and Graphing Polynomials Even Functions Exponential Form of Complex Numbers Exponential Rules Exponentials and Logarithms Expression Math Expressions and Formulas Faces Edges and Vertices Factorials Factoring Polynomials Factoring Quadratic Equations Factorising expressions Factors Fermat's Little Theorem Field theory Finding Maxima and Minima Using Derivatives Finding Rational Zeros Finding The Area First Fundamental Theorem First-order Differential Equations Forms of Quadratic Functions Fourier analysis Fractional Powers Fractional Ratio Fractions Fractions and Decimals Fractions and Factors Fractions in Expressions and Equations Fractions, Decimals and Percentages Function Basics Functional Analysis Functions Fundamental Counting Principle Fundamental Theorem of Algebra Generating Terms of a Sequence Geometric Sequence Gradient and Intercept Gram-Schmidt Process Graphical Representation Graphing Rational Functions Graphing Trigonometric Functions Graphs Graphs And Differentiation Graphs Of Exponents And Logarithms Graphs of Common Functions Graphs of Trigonometric Functions Greatest Common Divisor Grothendieck topologies Group Mathematics Group representations Growth and Decay Growth of Functions Gröbner bases Harmonic Motion Hermitian algebra Higher Derivatives Highest Common Factor Homogeneous System of Equations Homological algebra Homotopy theory Hopf algebras Hyperbolas Ideal theory Imaginary Unit And Polar Bijection Implicit differentiation Indirect variation Inductive Reasoning Inequalities Maths Infinite geometric series Injective functions Injective linear transformation Instantaneous Rate of Change Integers Integrating Ex And 1x Integrating Polynomials Integrating Trigonometric Functions Integration Integration By Parts Integration By Substitution Integration Using Partial Fractions Integration of Hyperbolic Functions Interest Interest calculations Invariant Points Inverse Hyperbolic Functions Inverse Matrices Inverse and Joint Variation Inverse functions Inverse of a Matrix and System of Linear equation Invertible linear transformation Irrational numbers Isosceles Triangles Trigonometry Iterative Methods Jordan algebras Knot theory L'hopitals Rule Lattice theory Law Of Cosines In Algebra Law Of Sines In Algebra Laws of Logs Leibnitz's Theorem Lie algebras Lie groups Limits of Accuracy Linear Algebra Linear Combination Linear Expressions Linear Independence Linear Systems Linear Transformation Linear Transformations of Matrices Linear graphs Location of Roots Logarithm Base Logic Lower and Upper Bounds Lowest Common Denominator Lowest Common Multiple Math formula Matrices Matrix Addition And Subtraction Matrix Calculations Matrix Determinant Matrix Multiplication Matrix inverses Matrix operations Mean value theorem Metric and Imperial Units Midpoint formula Misleading Graphs Mixed Expressions Mode Modelling with First-order Differential Equations Modular Arithmetic Module theory Modulus Functions Modulus and Phase Monoidal categories Monotonic Function Multiples of Pi Multiplication and Division of Fractions Multiplicative Relationship Multiplicative ideal theory Multiplying And Dividing Rational Expressions Natural Logarithm Natural Numbers Non-associative algebra Normed spaces Notation Number Number Line Number Systems Number Theory Number e Numerical Methods Odd functions Open Sentences and Identities Operation with Complex Numbers Operations With Matrices Operations with Decimals Operations with Polynomials Operator algebras Order of Operations Orthogonal groups Orthogonality Parabola Parallel Lines Parametric Differentiation Parametric Equations Parametric Hyperbolas Parametric Integration Parametric Parabolas Partial Fractions Pascal's Triangle Percentage Percentage Increase and Decrease Perimeter of a Triangle Permutations and Combinations Perpendicular Lines Phase Shift Points Lines and Planes Pointwise convergence Poisson algebras Polynomial Graphs Polynomial division Polynomial rings Polynomials Powers Roots And Radicals Powers and Exponents Powers and Roots Prime Factorization Prime Numbers Problem-solving Models and Strategies Product Rule Product of polynomials Product-to-sum Formulas Proof Proof and Mathematical Induction Proof by Contradiction Proof by Deduction Proof by Exhaustion Proof by Induction Properties of Determinants Properties of Exponents Properties of Riemann Integral Properties of dimension Properties of eigenvalues and eigenvectors Proportion Proving an Identity Pythagorean Identities Quadratic Equations Quadratic Function Graphs Quadratic Graphs Quadratic forms Quadratic functions Quadrilaterals Quantum groups Quotient Rule Radian Measure Radians Radical Functions Rates of Change Ratio Ratio Fractions Ratio and Root test Rational Exponents Rational Expressions Rational Functions Rational Numbers and Fractions Rationalizing denominators Ratios as Fractions Real Numbers Rearrangement Reciprocal Graphs Recurrence Relation Recursion and Special Sequences Reduced Row Echelon Form Reducible Differential Equations Remainder and Factor Theorems Representation Of Complex Numbers Representation theory Rewriting Formulas and Equations Riemann integral for step function Riemann surfaces Riemannian geometry Right Triangle Trigonometry Ring theory Roots Of Unity Roots of Complex Numbers Roots of Polynomials Rounding SAS Theorem SSS Theorem Scalar Products Scalar Triple Product Scale Drawings and Maps Scale Factors Scalene Triangles Trigonometry Scientific Notation Secant Ratios Second Fundamental Theorem Second Order Recurrence Relation Second-order Differential Equations Sector of a Circle Segment of a Circle Sequence and series of real valued functions Sequence of Real Numbers Sequences Sequences and Series Series Maths Series of non negative terms Series of real numbers Sets Math Sigma notation Similar Triangles Similar and Congruent Shapes Similarity and diagonalisation Simple Interest Simple algebras Simplifying Fractions Simplifying Radicals Simultaneous Equations Sine Ratios Sine and Cosine Rules Sinusoidal Graphs Slope formulas Slope-intercept form Small Angle Approximation Solving Linear Equations Solving Linear Systems Solving Oblique Triangles Solving Quadratic Equations Solving Radical Inequalities Solving Rational Equations Solving Right Triangles Solving Simultaneous Equations Using Matrices Solving Systems of Inequalities Solving Trigonometric Equations Solving and Graphing Quadratic Equations Solving and Graphing Quadratic Inequalities Spanning Set Special Products Special Sequences Spherical Trigonometry Standard Form Standard Integrals Standard Unit Standard form equations Stone Weierstrass theorem Straight Line Graphs Subgroup Subsequence Subspace Substraction and addition of fractions Sum and Difference of Angles Formulas Sum of Natural Numbers Sum-to-product Formulas Summation by Parts Supremum and Infimum Surds Surjective functions Surjective linear transformation System of Linear Equations Tables and Graphs Tangent Ratios Tangent of a Circle Tangent rule Taylor theorem The Quadratic Formula and the Discriminant Topological groups Torsion theories Transformations Transformations geometry Transformations of Graphs Transformations of Roots Translations of Trigonometric Functions Transversals Triangle Rules Triangle trigonometry Trigonometric Functions Trigonometric Functions of General Angles Trigonometric Identities Trigonometric Identities Sum Angles Trigonometric Proofs Trigonometric Ratios Trigonometric Ratios Of Complementary Angles Trigonometry Trigonometry Height And Distance Trigonometry Word Problems Turning Points Types of Functions Types of Numbers Types of Triangles Uniform convergence Unit Circle Units Universal algebra Upper and Lower Bounds Valuation theory Variable expressions Variables in Algebra Vector Notation Vector Space Vector spaces Vectors Verifying Trigonometric Identities Vertex form Volume calculations Volumes of Revolution Von Neumann algebras Writing Equations Writing Linear Equations Zariski topology secant lines slope formula Statistics Theoretical and Mathematical Physics Contents Fact Checked Content Last Updated: 09.02.2024 15 min reading time Content creation process designed by Lily Hulatt Content sources cross-checked by Gabriel Freitas Content quality checked by Gabriel Freitas Sign up for free to save, edit & create flashcards. Save ArticleSave Article Jump to a key chapter Understanding Integrating Polynomials Techniques for Integrating a Polynomial Practical Examples of Integrating Polynomials Decoding the Fundamental Theorem of Calculus Polynomials Breaking Down Polynomial Integration Rules Integrating Polynomials - Key takeaways Similar topics in Math Related topics to Pure Maths Flashcards in Integrating Polynomials Learn faster with the 15 flashcards about Integrating Polynomials Frequently Asked Questions about Integrating Polynomials About Vaia Test your knowledge with multiple choice flashcards 1/3 What is the role of integration in mathematics? A. Integration builds up functions by taking infinitesimally small pieces and assembling them to form a whole. It significantly contributes to diverse fields such as physics, engineering, and economics. B. Integration's role is to subtract 1 from the exponent of a term and divide the whole term by the new power. It is specifically designed for solving polynomial equations. C. Integration breaks down functions into smaller parts. It primarily focuses on subtracting functions systematically to simplify complex mathematical problems. D. Integration is used to find the derivative of a function, providing the slope of the tangent line at any point, which plays a crucial role in prediction and rate of change problems. 1/3 What is an example of how polynomial integration is applied in economics? A. In economics, polynomial integration is primarily used to evaluate the performance of stock market indices. B. In economics, polynomial integration is primarily used to calculate and predict the rate of inflation. C. Polynomial integration in economics is mainly employed to calculate the depreciation value of assets over time. D. In economics and business studies, polynomial functions and their integrals are used to optimize profit functions, model cost estimation, and forecast future revenue projections. 1/3 How are Polynomial Integration Rules applied in real scenarios? A. Polynomial Integration Rules are used in computer science for developing algorithms and in chemistry for balancing chemical reactions. B. Polynomial Integration Rules are primarily used for solving complex equations in abstract mathematics and have no real-world applications. C. Polynomial Integration Rules are used in geology for mineral identification and in astronomy for calculating distances between celestial bodies. D. Polynomial Integration Rules are used in physics for tasks like calculating displacement or work done, in engineering for calculating areas or centroids, and in economics for solving problems involving accumulation of quantities over time. Score That was a fantastic start! You can do better! Sign up to create your own flashcards Access over 700 million learning materials Study more efficiently with flashcards Get better grades with AI Sign up for free Already have an account? Log in Good job! Keep learning, you are doing great. Don't give up! Next Open in our app Understanding Integrating Polynomials It's fantastic that you're venturing into the realm of Integrating Polynomials! This subject pertains to a key aspect of calculus and constitutes an integral part of higher-level mathematics. The concept might seem complex at first glance. However, once you get started, you'll find it incredibly fascinating and logical. So, what exactly are Integrating Polynomials? Well, to put it simply, integrating a polynomial function results in the addition of one degree to each term of the polynomial, along with appropriate division to balance the multiplication that occurs during differentiation. Integration and differentiation are two sides of the same coin in the realm of calculus. While differentiation is about breaking things down, integration is about putting things together. Defining Integrating Polynomials Before we delve deeper, let's first understand the basic definitions related to Integrating Polynomials. Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. When you "integrate" a polynomial, you're essentially finding the "antiderivative" or the original function. Let's summarize essential terms: Coefficients: These are the numerical values by which the variables in polynomial equations are multiplied. Variables: In mathematics, variables are symbols used to represent unspecified numbers or values. Antiderivative: This is the function whose derivative gives the original function. Fundamental Elements of Integrating Polynomials Now, it's time to focus on the actual process of integrating polynomials, which is a step-by-step operation. Consider a simple polynomial: 3 x 2+2 x+1. To integrate this polynomial, we can work with each term separately: For 3 x 2, we add 1 to the power to get 3 x 3, then divide by the new power, resulting in 3 3 x 3=x 3. For 2 x, we follow the same procedure to get x 2. Finally, for the constant term 1, we treat it as 1 x 0 and integrate to get x. Hence, the integral of the given polynomial is x 3+x 2+x+C, where C is the constant of integration, representing an unknown constant. Let's apply these steps to another example, say 4 x 3−5 x 2+2 x−7. On integration, we find it to be equal to x 4−5 x 3 3+x 2−7 x+C. In general, the power rule for integration states that the integral of x n is x n+1 n+1 plus a constant C, assuming n is not equal to -1. This rule is a direct consequence of the Fundamental Theorem of Calculus. Techniques for Integrating a Polynomial Integration of polynomial functions is a systematic process that can be mastered with a bit of practice. The key is to understand the technique and apply it methodically. We're now moving forward to explore various techniques you can use to simplify the process of integrating a polynomial. Step-by-step Polynomial Integration Technique Polynomial integration often seems daunting to many learners breaking into calculus. However, it's just about following a consistent, step-by-step approach. Here, we'll walk you through a tried-and-true technique for integrating a polynomial. Step 1: Start by identifying the polynomial that needs to be integrated. Step 2: Break down the polynomial into individual terms, if it isn't already in this form. Step 3: Begin integration by focusing on individual terms. Keep in mind, the integration of a sum of terms is equal to the sum of the integrations. Step 4: For each term, apply the power rule for integration. This entails adding 1 to the exponent of the term and dividing the term by the new power. Step 5: After you've done this for all terms, don't forget to add the constant of integration (usually denoted as C) at the end. To illustrate these steps, let's take the example of the polynomial 2 x 3+x−1. Breaking it down, we separately integrate 2 x 3 to become x 4 2, x to become x 2 2, and −1 to become −x. Bringing it all together, the integral of the given polynomial is x 4 2+x 2 2−x+C. Common Mistakes to Avoid in Polynomial Integration When setting out on your journey of polynomial integration, it is easy to stumble upon stumbling blocks and commit errors. But you can navigate your way around these common mistakes with the right knowledge and a careful approach. Among these, three errors stand out as frequent culprits: MistakeDescription Forgetting the Constant of Integration This is perhaps the most common error. When integrating, always remember to include the constant of integration C. Mishandling of Exponents Adding 1 to the power and dividing the term by the new power is the crux of integration which can easily be muddled up. Ignoring the Power Rule The power rule ∫x n d x=x n+1 n+1+C for n≠−1 applies universally for all polynomials, and forgetting it can cause major errors. Steering clear of these common integration mistakes can make your polynomial integration journey much smoother. Keep practising, and soon you'll find integrating polynomials comes naturally to you! Contrary to differentiation, which breaks down functions, integration serves to build up functions. It takes infinitesimally small pieces and puts them together to form a whole. The result of this mathematical operation contributes significantly to many diverse fields such as physics, engineering, and economics. Stay organized and focused with your smart to do list Sign up for free Practical Examples of Integrating Polynomials After delving into the theory and methodology of integrating polynomials, it's time to see it in action with practical examples. These will not only help solidify your understanding but also showcase how polynomial integration transcends formulas and textbooks to extend to interesting real-life applications. Real-life Examples of Integrating a Polynomial Mathematics is not confined to the classroom; it permeates every aspect of life. Polynomial integration, in particular, finds wide application in various fields, from physics to economics. The following are several real-life examples where Integrating Polynomials plays an essential role. Physics: Polynomial integration is used in acceleration or velocity models to predict distance travelled. For instance, if you have an acceleration function given in polynomial form, integrating it once provides velocity, and integrating velocity gives the distance covered. Economics: In studies of economics and business, polynomial functions and their integrals are used to optimize profit functions, model cost estimation, and forecast future revenue projections. Engineering: Engineers use polynomial integration to calculate the area and volume of complex shapes, perturbations in system dynamics, or even designing gear systems. Take the example of a car accelerating from a stationary position. Suppose the acceleration a(t) at a given time t is represented by the polynomial 3 t 2. Integrating this polynomial gives the velocity function v(t)=t 3+C, where C can be determined based on initial conditions. Once more integration gives the displacement or distance travelled s(t)=1 4 t 4+C t+D. Exploring Complex Antiderivative of Polynomials Examples Moving into higher degrees of polynomial integration, let's explore more complex antiderivative examples. For more complex polynomials, the process remains the same, albeit a bit more intricate, but that's all part of the fun, isn't it? A complex antiderivative involves polynomial functions of higher degree where coefficients or powers can be any real or complex numbers. Fundamentally, irrespective of the complexity, every term of the polynomial equation is treated in the same way: add one to the exponent and divide the term by the new exponent. Consider the polynomial 3 x 4−2 x 3+x 2−5 x+1. By applying the method of polynomial integration, the integral can be computed as 3 5 x 5−2 4 x 4+1 3 x 3−5 2 x 2+x+C. Such examples allow you to encounter many different applications and variations of polynomial integrals. It’s important to note that although the calculations might be a bit lengthy for higher degree polynomials, the core process remains the same as for simpler polynomials. In advanced scientific research, polynomial integration plays a vital role in solving complex mathematical models, often providing the backbone for many groundbreaking scientific discoveries and technological advancements. Access millions of flashcards designed to help you ace your studies Sign up for free Decoding the Fundamental Theorem of Calculus Polynomials Fundamental Theorem of Calculus (FTC) serves as the cornerstone of integral calculus. It essentially bridges the gap between integration and differentiation, two primary operations in calculus. When it comes to polynomials, this theorem sheds light on a unique relationship between the integral and differential forms of polynomials. So, let's get started and decode what it exactly means. Unpacking the Fundamental Theorem of Calculus in relation to Polynomials The Fundamental Theorem of Calculus is a powerful theorem that establishes the link between the seemingly unrelated fields of differentiation and integration, more specifically, between antiderivatives and definite integrals. It essentially states that integrating a function over an interval yields the net change in the original function throughout that interval. Applied to polynomials, the FTC allows us to switch seamlessly between rates of change (derivatives) and total changes (integrals). This theorem makes the process of polynomial integration quicker and more efficient, whether you're integrating a linear polynomial or a high degree polynomial. There are two parts to the FTC: FTC Part 1: If f is a function that is continuous over an interval I and F is an antiderivative of f on I, then F(b)−F(a) is equal to the definite integral of f from a to b. FTC Part 2: This part tells us that if f is continuous over an interval I and a is any number in I, then the function F defined by the integral of f from a to x is continuous on I, differentiable on the interior of I, and F′(x)=f(x) for all x in I. To bring this principle to life, let's consider the polynomial 4 x 3. The derivative of this function is 12 x 2. So, say we want to find the definite integral of 12 x 2 from 1 to 2. Using FTC Part 1, instead of integrating, we can use the antiderivative 4 x 3 and subtract its value at 1 from its value at 2, i.e., 4(2 3)−4(1 3)=24. Impact of Fundamental Theorem on Polynomial Calculus The Fundamental Theorem of Calculus is a game-changer for polynomial calculus. It provides a faster and more direct method for integrating polynomials by using antiderivatives, thereby reducing the need for tedious computations. This makes tackling complex polynomial integrals much more manageable. Additionally, the FTC's second part can help validate whether a given polynomial function is an antiderivative of another function. An antiderivative is effectively a solution to a differential equation. If F′(x)=f(x), for all x in the function's domain, then F is the antiderivative of f. Let's summarise the crucial concepts: Continuity: A function is continuous at a point if the left-hand limit, right-hand limit, and the value of the function at that point all exist and are all equal. Definite Integral: It represents the signed area underneath the function and between two points along the function's domain. Antiderivative: If the derivative of a function F(x) gives another function f(x), F(x) is said to be an antiderivative of f(x). The importance of the Fundamental Theorem of Calculus extends beyond the arithmetic realm. It lays the foundation for many areas of mathematics and forms the cornerstone for disciplines that require a strong understanding of change and accumulation, such as physics, engineering, and economics. Suppose you have a polynomial function f(x)=2 x 2. Utilizing the FTC, you can easily find any definite integral of f(x) between a and b. Begin by determining the antiderivative of f(x), which is F(x)=2 3 x 3. According to FTC, the definite integral of f(x) from a to b is F(b)−F(a) or 2 3 b 3−2 3 a 3. Find relevant study materials and get ready for exam day Sign up for free Breaking Down Polynomial Integration Rules Let's roll up our sleeves and dive into the heart of polynomial integration: the rules! These invaluable tools can help you streamline the process of integration and operate within the mathematical framework efficiently. Whether you're dealing with a simple first-degree polynomial or facing a complex, high-degree polynomial, these rules are going to be your best companion. Understanding Crucial Polynomial Integration Rules Polynomial integration might appear complex at times, but certain essential rules help make this process simpler. These fundamental rules apply to all polynomials and can significantly reduce the time and effort required to solve integral calculus problems. A Polynomial Integration Rule is a mathematical guideline used to compute the integral or antiderivative of a polynomial. The rules essentially extrapolate the principle of the power rule for integration, which states that the integral of x n is x n+1 n+1+C, where n ≠ -1. Key Polynomial Integration rules include: Power Rule: As stated above, the integral of x n is x n+1 n+1+C, where n ≠ -1. Sum Rule: The integral of a sum of two functions is the sum of the integrals of each function. Constant Multiple Rule: The integral of a constant multiplied by a function is the constant times the integral of the function. Let's demonstrate the third rule with an example. Suppose we wish to integrate a polynomial function 5 x 3. With the constant multiple rule, we can take the constant 5 out and integrate x 3 normally. Thus, the integral becomes 5∫x 3 d x=5⋅x 3+1 3+1=5 4 x 4+C. Application of Polynomial Integration Rules in Real Scenarios Polynomial integration and its rules aren't solely confined to academia. You'll face myriad situations in real-life where these rules come into play. Below are a few scenarios where understanding and applying polynomial integration rules can prove quite fruitful: Physics: Calculating displacement given velocity, or computing work done given force. Both tasks require integration of polynomial functions. Engineering: Engineers often need to calculate the area of irregular figures, or find the centroid of a planar lamina; both scenarios require polynomial integration. Economics: Economists use polynomial integration to solve problems involving the accumulation of quantities over time, such as total growth or total cost over certain periods. Consider an engineering problem where the moment of inertia, I, of a rectangular beam about its centroidal axis needs to be calculated. Here, I=∫y 2 d A, where y is the distance from the axis and d A is the differential area element. Knowing integration rules, you substitute d A with y d y, and the integral becomes I=∫y 3 d y. You apply the power rule and easily solve for I. Precise calculations, such as those required in the fields of physics, economics, and engineering, would be nearly impossible without the polynomial integration rules we have discussed. These allow for a systematic procedure to follow, regardless of the complexity of the problem. Hence, polynomial integration rules not only form a cornerstone of calculus but also pave the way for advancements in numerous scientific and economic disciplines. Integrating Polynomials - Key takeaways Integrating polynomials is a step-by-step operation where each term of the polynomial is treated separately. The power rule for integration states that the integral of x n is x n+1 n+1 plus a constant C, where n is not equal to -1. Polynomial integration techniques involve identifying the polynomial, breaking it into individual terms, and applying the power rule for integration to each term. The Fundamental Theorem of Calculus establishes a link between integration and differentiation and is essential for the process of polynomial integration. Polynomial integration is used extensively in various fields such as physics, economics, and engineering for various applications, from predicting distances in acceleration models to optimizing profit functions. 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Expressions Right Triangle Trigonometry Composite Figures Solving Right Triangles Trigonometric Identities Sum Angles Flashcards in Integrating Polynomials ------------------------------------- 15 Start learning What is the role of integration in mathematics? Integration builds up functions by taking infinitesimally small pieces and assembling them to form a whole. It significantly contributes to diverse fields such as physics, engineering, and economics. What is an example of how polynomial integration is applied in economics? In economics and business studies, polynomial functions and their integrals are used to optimize profit functions, model cost estimation, and forecast future revenue projections. How are Polynomial Integration Rules applied in real scenarios? Polynomial Integration Rules are used in physics for tasks like calculating displacement or work done, in engineering for calculating areas or centroids, and in economics for solving problems involving accumulation of quantities over time. What is the process of Integrating Polynomials? Integrating polynomials involves adding one degree to each term of the polynomial, accompanied by suitable division to balance the multiplication that happens during differentiation. What are key Polynomial Integration rules? Key rules include the Power Rule (integral of x n is x n+1 n+1+C), the Sum Rule (integral of a sum of two functions is the sum of the integrals of each function), and the Constant Multiple Rule (integral of a constant multiplied by a function is the constant times the integral of the function). What are the steps to integrate a polynomial? Start by identifying the polynomial, break it down into individual terms, begin integration on each term, apply the power rule for integration (adding 1 to the exponent and dividing by the new power), and add the constant of integration. Learn faster with the 15 flashcards about Integrating Polynomials Sign up for free to gain access to all our flashcards. Sign up with Email Already have an account? Log in Frequently Asked Questions about Integrating Polynomials What are the steps to follow when integrating polynomials? The steps are as follows: first, identify each term of the polynomial. Then, increase the exponent of each term by one. Next, divide each term by its new exponent. Finally, add a constant of integration, represented as '+ C'. What's the significance of the constant of integration in polynomial integration? The constant of integration in polynomial integration signifies any constant value that could have been the derivative of the original function. It is necessary as indefinite integrals only produce families of solutions rather than unique solutions, due to the absence of initial conditions. How can one identify the degree of a polynomial before starting the integration process? The degree of a polynomial is identified by the highest power of the variable present in the polynomial. For example, in the polynomial x^4 + x^2 - 3, the degree is 4 because the highest power of x is 4. Why is it necessary to add a constant when integrating polynomials? Adding a constant when integrating polynomials is necessary because the process of integration only determines the antiderivative up to an arbitrary constant. This is because when you differentiate any constant, the result is zero. Thus, multiple functions differing by a constant can have the same derivative. What is the role of the Power Rule in integrating polynomials? The Power Rule, in the context of integrating polynomials, is used to find the antiderivative or integral of a function raised to a power. It states that the integral of x^n dx is (x^(n+1))/(n+1), where n is not equal to -1. Save Article How we ensure our content is accurate and trustworthy? At StudySmarter, we have created a learning platform that serves millions of students. Meet the people who work hard to deliver fact based content as well as making sure it is verified. 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14578	https://gis.stackexchange.com/questions/60302/creating-a-systematic-rectangular-grid	qgis - Creating a systematic rectangular grid - Geographic Information Systems Stack Exchange Join Geographic Information Systems By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Geographic Information Systems helpchat Geographic Information Systems Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Creating a systematic rectangular grid Ask Question Asked 12 years, 4 months ago Modified12 years, 4 months ago Viewed 497 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. How can I produce a point shapefile (or simply point coordinates) that represent a systematic rectangular grid? At least three input variables would be required: distance between points, distance between rows of points and orientation (azimuth of rows). The purpose is creation of sample locations for natural resource surveys. I guess a fourth variable would be required as well: either a starting location or some definition of the area to be encompassed by the grid. Typically an area of interest can be defined by a polygon. qgis Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications asked May 9, 2013 at 12:45 L FullerL Fuller 59 3 3 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. It doesn't have all the criteria that you have listed, but the Regular Points tool (Vector > Research Tools > Regular points) would get you close. This does allow you to define the area of interest with a polygon, but it uses the minimum bounding box rather than the actual boundary. The picture below shows the implications of bounding box method. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered May 9, 2013 at 13:15 katahdinkatahdin 2,554 1 1 gold badge 23 23 silver badges 26 26 bronze badges 1 Thanks. I have discovered several useful tools for regular spacing in square and hexagonal patterns, but none that allow for distributing points in a rectangular pattern. Though not a GIS professional or programmer I am not averse to solutions that involve scripting, but I don't know how to begin. I currently use primarily QGIS and SAGA, but don't mind tackling a more robust platform if necessary. I also have R Spatial but currently have zero proficiency in its use. Thank you for the response.L Fuller –L Fuller 2013-05-10 16:50:28 +00:00 Commented May 10, 2013 at 16:50 Add a comment\| Your Answer Thanks for contributing an answer to Geographic Information Systems Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions qgis See similar questions with these tags. 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14579	https://afoqtguide.com/wp-content/uploads/2018/03/Comp_AFOQTPracTest01.pdf	Subtest 1: Verbal Analogies Directions: The Verbal Analogies subtest measures your ability to reason and to see relationships be-tween words. You are to choose the answer that best completes the analogy developed at the beginning of each question. The best way to approach this type of test is to look for patterns or comparisons between the first phrase and the choices available to you. You have eight (8) minutes to complete this subtest. Questions: 25 Time: 8 minutes For sample Verbal Analogies questions, see page 195. 1. School is to principal as office is to a. manager b. secretary c. computer d. teacher e. building 2. Paint is to brush as clay is to a. dough b. mold c. sculpture d. artist e. rock 3. Knife is to slice as spoon is to a. fork b. stab c. stir d. cereal e. handle 4. Proud is to brag as despairing is to a. sing b. exult c. sad d. depress e. weep 5. Excited is to thrilled as angered is to a. amused b. enraged c. disinterested d. sleepy e. delayed 6. String is to guitar as key is to a. score b. organ c. orchestra d. song e. drum 7. Thesaurus is to synonyms as dictionary is to a. words b. reference c. book d. volume e. definitions 8. Pedestal is to bust as tripod is to a. film b. figure c. leg d. adjust e. camera 9. Beautician is to salon as a. follicle is to hair b. musician is to concert hall c. scissors is to blow dryer d. cinematographer is to movie e. cosmetologist is to makeup 10. Stall is to engine as a. drive is to motor b. race is to heart c. block is to progress d. freeze is to computer e. mechanical is to functional –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 57 11. Leave is to bolt as a. dance is to shuffle b. toss is to hurl c. drive is to race d. play is to frolic e. shine is to glisten 12. Disseminate is to gather as a. puncture is to prod b. resist is to tempt c. distinct is to dissimilar d. rip is to mend e. adjoined is to conjoined 13. Enjoy is to adore as a. love is to desire b. spend is to waste c. burn is to sear d. care is to concern e. dislike is to loathe 14. Gaggle is to goose as a. leap is to kangaroo b. stripe is to tiger c. barracuda is to fish d. mane is to lion e. flock is to sheep 15. Science fiction is to novel as a. tale is to story b. horror is to frightening c. country is to western d. rhythm and blues is to album e. style is to book 16. Carpenter is to constructive as a. farmer is to crops b. conservationist is to environment c. employee is to duty d. critic is to judgmental e. instructor is to lesson 17. Carelessness is to mistake as a. destruction is to damage b. meticulous is to perfection c. process is to result d. watchfulness is to sight e. reflective is to intelligence 18. Cautious is to heedless as a. oblivious is to negligent b. wary is to trusting c. alert is to cagey d. shameful is to abhorrent e. scorn is to react 19. Surprised is to jump as a. relaxed is to seated b. amused is to chuckle c. stunned is to disturbed d. energy is to sprint e. perplexed is to wonder 20. Brace is to bolster as a. revert is to rectitude b. covert is to manifest c. accelerate is to hasten d. fulcrum is to level e. halt is to increase 21. Spider is to arachnid as a. lobster is to crustacean b. arthropod is to insect c. ant is to cricket d. octopus is to squid e. porpoise is to fish 22. Psychology is to mind as a. think is to cerebral cortex b. physiology is to science c. methodology is to study d. anatomy is to body e. cognizant is to perceptive –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 58 23. Cardiologist is to heart as a. surgeon is to general practitioner b. neurologist is to brain c. epidermis is to endocrine system d. orthodontist is to braces e. dermatologist is to podiatrist 24. Bushel is to peas as a. core is to apple b. bunch is to bananas c. skin is to grape d. juice is to orange e. stem is to carrot 25. Apathy is to disinterest as a. tenacity is to resolve b. insincerity is to candor c. sobriety is to indulgence d. affection is to relationship e. believability is to reality Subtest 2: Arithmetic Reasoning Directions: The Arithmetic Reasoning subtest mea-sures mathematical reasoning and problem solving. Each problem is followed by five possible answers. Decide which one of the five answers is most nearly correct. A method for attacking each of these ques-tions is given in the answer block at the end of this chapter. You have twenty-nine (29) minutes to com-plete this subtest. Questions: 25 Time: 29 minutes For sample Arithmetic Reasoning questions, see page 196. 1. It costs $0.85 to make a single color copy at a copy center. At this price, how many copies can be purchased with $68.00? a. 9 b. 45 c. 68 d. 72 e. 80 2. An aquarium has a base length of 12 inches and a width of 5 inches. If the aquarium is 10 inches tall, what is the total volume? a. 480 cubic inches b. 540 cubic inches c. 600 cubic inches d. 720 cubic inches e. 920 cubic inches 3. A man turns a woman’s handbag in to the Lost and Found Department of a large downtown store. The man informs the clerk in charge that he found the handbag on the floor beside an entranceway. The clerk estimates that the handbag is worth approximately $150. Inside, the clerk finds the following items: one leather makeup case valued at $65, one vial of perfume, unopened, valued at $75, one pair of earrings valued at $150, and $178 in cash. The clerk is writing a report to be submitted along with the found property. What should he write as the total value of the found cash and property? a. $468 b. $608 c. $618 d. $658 e. $718 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 59 Use the following information to answer questions 4 through 6. The cost of movie theater tickets is $7.50 for adults and $5 for children ages 12 and under. On Saturday and Sunday afternoons until 4:00 P.M., there is a matinee price: $5.50 for adults and $3 for children ages 12 and under. Special group discounts are available for groups of 30 or more people. 4. Which of these can be determined from the information given in the above passage? a. how much it will cost a family of four to buy movie theater tickets on Saturday afternoon b. the difference between the cost of two movie theater tickets on Tuesday night and the cost of one ticket on Sunday at 3:00 P.M. c. how much movie theater tickets will cost each person if he or she is part of a group of 40 people d. the difference between the cost of a movie theater ticket for an adult on Friday night and a movie theater ticket for a 13-year-old on Saturday afternoon at 1:00 P.M. e. none of the above 5. Based on the passage, how much will movie theater tickets cost for two adults, one 15-year-old child, and one 10-year-old child at 7:00 P.M. on a Sunday night? a. $17.00 b. $19.50 c. $25.00 d. $27.50 e. $37.50 6. Using the passage, how can you find the difference in price between a movie theater ticket for an adult and a movie theater ticket for a child under the age of 12 if the tickets are for a show at 3:00 P.M. on a Saturday afternoon? a. Subtract $3 from $5.50. b. Subtract $5 from $7.50. c. Subtract $7.50 from $5.50. d. Add $5.50 and $3 and divide by 2. e. Add $7.50 and $5.50 and divide by 2. 7. It takes a typist 0.50 seconds to type one word. At this rate, how many words can be typed in 60 seconds? a. 2.25 b. 50 c. 90 d. 120 e. 220 8. If the average cadet burns 8.2 calories per minute while riding a bicycle, how many calories will the cadet burn if he or she rides for 35 minutes? a. 286 b. 287 c. 387 d. 980 e. 1,080 9. Dr. Drake charges $36 for an office visit, which is 3 4 of what Dr. Jean charges. How much does Dr. Jean charge? a. $27 b. $38 c. $48 d. $57 e. $68 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 60 10. Thirty percent of the cadets at the Air Force Academy are involved in athletics. If 15% of the athletes play lacrosse, what percentage of the whole academy plays lacrosse? a. 4.5% b. 9.0% c. 15% d. 30% e. 40% Use the following information to answer questions 11 and 12. Basic cable television service, which includes 16 channels, costs $15 a month. The initial labor fee to install the service is $25. A $65 deposit is required but will be refunded within two years if the customer’s bills are paid in full. Other cable services may be added to the basic service: the movie channel service is $9.40 a month; the news channels are $7.50 a month; the arts channels are $5 a month; the sports channels are $4.80 a month. 11. A customer’s cable television bill totaled $20 a month. Using the preceding passage, what portion of the bill was for basic cable service? a. 25% b. 33% c. 50% d. 75% e. 85% 12. A customer’s first bill after having cable television installed totaled $112.50. This customer chose basic cable and one additional cable service. Which additional service was chosen? a. the news channels b. the movie channels c. the arts channels d. the sports channels e. none of the above 13. Out of every 200 shoppers polled, 60 said they buy fresh vegetables every week. How many shoppers out of 40,000 could be expected to buy fresh vegetables every week? a. 3,600 b. 9,000 c. 12,000 d. 24,000 e. 36,000 Use the following pie chart to answer questions 14 and 15. Songs Downloaded 14. If 400 total songs were downloaded, how many downloads were country music? a. 11 b. 28 c. 55 d. 110 e. 270 Country 27.5% Rock 45.5% Rap 15% Jazz 7.5% 4.5% Classical –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 61 15. Based on the pie chart, which types of music represent exactly half of the songs downloaded? a. rock and jazz b. classical and rock c. rap, classical, and country d. jazz, classical, and rap e. jazz and rap 16. Last year, 220 people bought cars from a certain dealer. Of those, 60 percent reported that they were completely satisfied with their new cars. How many people reported being unsatisfied with their new car? a. 36 b. 55 c. 88 d. 132 e. 155 17. Of 1,125 OTS candidates, 135 speak fluent Spanish. What percentage of the candidates speaks fluent Spanish? a. 7.3% b. 8.3% c. 12% d. 14% e. 16% 18. The perimeter of a rectangle is 268 feet. Its two longest sides add up to 156 feet. What is the length of each of its two shortest sides? a. 43 feet b. 56 feet c. 72 feet d. 80 feet e. 112 feet 19. A piece of wire 3 feet 4 inches long was divided into 5 equal parts. How long was each part? a. 6 inches b. 7.5 inches c. 8 inches d. 10 inches e. 1 foot 2 inches 20. A middle school cafeteria has three different options for lunch. For $2, a student can get either a sandwich or two cookies. For $3, a student can get a sandwich and one cookie. For $4, a student can get either two sandwiches or a sandwich and two cookies. If Jimae has $6 to pay for lunch for her and her brother, which of the following is NOT a possible combination? a. three sandwiches and one cookie b. two sandwiches and two cookies c. one sandwich and four cookies d. three sandwiches and no cookies e. three sandwiches and two cookies 21. A bed is 4 feet wide and 6 feet long. What is the area of the bed? a. 10 square feet b. 20 square feet c. 24 square feet d. 30 square feet e. 36 square feet 22. Airman Beard’s temperature is 98 degrees Fahrenheit. Using the formula C = 5 9(F – 32), what is his temperature in degrees Celsius? a. 35.8 b. 36.7 c. 37.6 d. 41.1 e. 59.6 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 62 23. All of the rooms on the main floor of a barracks are rectangular, with 8-foot high ceilings. Captain Keira’s office is 9 feet wide by 11 feet long. What is the combined surface area of the four walls of her office, including any windows and doors? a. 99 square feet b. 160 square feet c. 320 square feet d. 792 square feet e. 640 square feet 24. A recipe serves four people and calls for 11 2 cups of broth. If you want to serve six people, how much broth do you need? a. 2 cups b. 21 4 cups c. 21 3 cups d. 21 2 cups e. 23 4 cups 25. Fort Greenville is 120 miles west and 90 miles north of Fort Johnson. How long is a direct straight line route from Fort Greenville to Fort Johnson City? a. 100 miles b. 125 miles c. 150 miles d. 180 miles e. 195 miles Subtest 3: Word Knowledge Directions: The Word Knowledge subtest measures your vocabulary comprehension. For each question you are to choose the answer that most closely means the same as the italicized word. If you are somewhat familiar with the italicized word, you can quickly eliminate the options that you know are incorrect. You have five (5) minutes to complete this subtest. Questions: 25 Time: 5 minutes For sample Word Knowledge questions, see page 200. 1. Preside a. challenge b. alter c. confuse d. preview e. lead 2. Liability a. burden b. support c. effort d. ability e. link 3. Diligent a. angry b. hardworking c. hearty d. surprised e. dainty 4. Agility a. leadership b. difficulty c. slim d. fierce e. quickness 5. Elicit a. impair b. extract c. illegal d. sketch e. crave 6. Proclamation a. performance b. study c. contract d. announcement e. fad –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 63 7. Pragmatic a. forward b. tremendous c. sensible d. clumsy e. fearless 8. Quandary a. question b. stone c. problem d. jumble e. discovery 9. Opulent a. wealthy b. overweight c. sickly d. inexpensive e. fair 10. Eloquent a. elegant b. beautiful c. well spoken d. tidy e. sharp 11. Fidelity a. falsity b. love c. clarity d. loudness e. loyalty 12. Limpid a. bright b. long c. true d. slight e. strange 13. Refute a. agree b. contest c. respond d. fuse e. shock 14. Vulnerable a. creative b. old c. upset d. weak e. feisty 15. Systematic a. orderly b. institution c. mechanism d. computerized e. symbolically 16. Sagacity a. wisdom b. age c. size d. darkness e. humor 17. Motley a. ugly b. tough c. multicolor d. dangerous e. inspire 18. Jaunt a. joke b. trip c. bend d. exercise e. story –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 64 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 65 19. Myriad a. watery b. strong c. ethical d. absurd e. many 20. Unison a. single b. under c. mystery d. harmony e. truth 21. Valor a. smoothness b. swiftness c. popularity d. smarts e. courage 22. Blatant a. secretive b. terrible c. obvious d. humble e. dignified 23. Construe a. interpret b. make c. deceive d. suspect e. order 24. Heinous a. delightful b. monstrous c. hairy d. tiny e. depressing 25. Gusto a. hunger b. clarity c. pride d. enthusiasm e. sleepiness Subtest 4: Math Knowledge Directions: The Math Knowledge subtest measures your ability to use learned mathematical relationships. Each problem is followed by five possible answers. You must decide which one of the five answers is correct. The best method for attacking each of these questions is given in the answer block at the end of this chapter. When you take the actual test, scratch paper will be provided for working out the problems. You have twenty-two (22) minutes to finish this subtest. Questions: 25 Time: 22 minutes For sample Math Knowledge questions, see page 201. 1. In this figure, angle POS measures 90°. What is the measure of angle ROQ? a. 45° b. 90° c. 180° d. 270° e. 360° 2. 41 5 + 12 5 + 3 3 10 a. 8 9 10 b. 9 1 10 c. 84 5 d. 8 6 15 e. 9 5 10 P R O S Q –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 66 3. 3 4 is equal to a. 0.25 b. 0.30 c. 0.34 d. 0.50 e. 0.75 4. 761 2 + 115 6 = a. 871 2 b. 881 3 c. 882 3 d. 885 6 e. 89 5. What is the decimal equivalent of 1 6 rounded to the nearest thousandth? a. 0.165 b. 0.666 c. 0.123 d. 0.167 e. 0.176 6. 1 6 + 7 12 + 2 3 = a. 10 24 b. 21 6 c. 15 6 d. 1 5 12 e. 25 6 7. Which of the following is equivalent to 202,436? a. 20,000 + 2,000 + 400 + 30 + 6 b. 2,000 + 40 + 300 + 6 c. 200,000 + 2,000 + 400 + 30 + 6 d. 200,000 + 2,000 + 4,000 + 300 + 6 e. 200,000 + 12,000 + 4,000 + 300 + 6 8. What are the missing integers on this number line? a. –4 and 1 b. –6 and 1 c. –6 and –1 d. 4 and 9 e. 5 and 8 9. 11 2 is equal to a. 0.50 b. 1.25 c. 2.50 d. 1.50 e. .150 10. If x 54 = 2 9, then x is a. 12 b. 14 c. 18 d. 108 e. 118 11. Which of the following is divisible by both 7 and by 8? a. 63 b. 106 c. 114 d. 112 e. 78 12. What is 3 8 equal to? a. 0.25 b. 0.333 c. 0.60 d. 0.375 e. 3.80 -5 0 13. What is another way to write 4 × 4 × 4? a. 3 × 4 b. 8 × 4 c. 43 d. 34 e. 83 14. Which of the following choices completes this number sentence? 5_ = (10 × 2) + (5 × 3) a. × (5 + 2) b. + (5 + 2) c. × (5 × 2) d. + (5 × 2) e. + (5 × 3) 15. Which of these is equivalent to 20°C? (F = 9 5C + 32) a. 68°F b. 95°F c. 45°F d. 19°F e. 70°F 16. What is the volume of a pyramid that has a rectangular base 5 feet by 3 feet and a height of 8 feet? (V = 1 3lwh) a. 16 feet3 b. 30 feet3 c. 40 feet3 d. 120 feet3 e. 220 feet3 17. What is another way to write 7.25 × 103? a. 72.5 b. 725 c. 7,250 d. 72,500 e. 720,500 18. How many inches are there in 31 3 yards? a. 126 b. 120 c. 160 d. 168 e. 313 19. 3 5 = a. 0.60 b. 0.20 c. 0.50 d. 0.80 e. 0.90 20. 0.97 is equal to a. 97% b. 9.7% c. 0.97% d. 0.097% e. 0.0097% 21. In a triangle, angle A is 70 degrees and angle B is 30 degrees. What is the measure of angle C? a. 90 degrees b. 70 degrees c. 80 degrees d. 100 degrees e. 120 degrees 22. Which value of x will make the number sentence x + 32 ≤14 true? a. –16 b. –21 c. 12 d. 38 e. none of the above –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 67 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 68 23. What is the length of a rectangle if its width is 6 feet and its area is 108 square feet? a. 1.8 feet b. 10.5 feet c. 18 feet d. 16 feet e. 68 feet 24. 37.5 percent is equal to a. 3 8 b. 5 8 c. 43 4 d. 63 4 e. 37 5 25. 0.15 is equal to a. 2 5 b. 3 20 c. 2 10 d. 1 20 e. 1.5 20 Subtest 5: Instrument Comprehension Directions: The Instrument Comprehension subtest measures your ability to determine the position of an aircraft in flight by reading instruments showing its compass heading, its amount of climb or dive, and its degree of bank to right or left. In each test item, the left-hand dial is labeled artificial horizon. The small aircraft silhouette remains stationary in the center of this dial, while the positions of the heavy black line and the black pointer vary with the changes in the position of the aircraft in which the instru-ment is located. The heavy black line represents the horizon line and the black pointer shows the degree of bank to right or left. If the aircraft is neither climbing nor div-ing, the horizon line is directly on the silhouette’s fuselage. If the aircraft has no bank, the black pointer will point to zero (Dial 1). If the aircraft is climbing, the fuselage silhouette is seen between the horizon line and the pointer. The greater the amount of climb, the greater the distance between the horizon line and the fuselage silhouette. If the aircraft is banked to the pilot’s right, the pointer will point to the left of zero (Dial 2). If the aircraft is diving, the horizon line is be-tween the fuselage silhouette and the pointer. The greater the amount of dive, the greater the distance between the horizon line and the fuselage silhouette. If the aircraft is banked to the pilot’s left, the pointer will point to the right of zero (Dial 3). The horizon line tilts as the aircraft is banked. It is always at a right angle to the pointer. In each test item, the right-hand dial is the com-pass. This dial shows the direction in which the air-craft is headed. Dial 4 shows north, Dial 5 is west, and Dial 6 is northwest. Each item in this test consists of two dials and four silhouettes of aircraft in flight. Your task is to de-termine which of the four aircraft is closest to the po-sition indicated by the two dials. Remember, you are always looking north at the same altitude as each plane. East is always to the right as you look at the N S W COMPASS Dial 4 N S W COMPASS Dial 5 N S W COMPASS Dial 6 ARTIFICIAL HORIZON Dial 1 ARTIFICIAL HORIZON Dial 2 ARTIFICIAL HORIZON Dial 3 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 69 page. (Note: B in Question 2 is the rear view of the aircraft, and B in Question 4 is the front view.) You have nine (9) minutes to complete this subtest. Questions: 20 Time: 9 minutes For sample Instrument Comprehension questions, see page 205. 1. 2. 3. 4. 5. 6. 7. N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A 8. 9. 10. 11. 12. 13. 14. 15. N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A N S W E ARTIFICIAL HORIZON COMPASS D C B A –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 70 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 71 16. 17. 18. 19. 20. Subtest 6: Block Counting Directions: The Block Counting subtest measures your ability to see into a three-dimensional stack of blocks to determine how many pieces are touched by the numbered blocks. It is also a test of your abilities to observe and deduce what you cannot specifically see. Closely study the way in which the blocks are stacked. You may find it helpful to remember that all of the blocks in a pile are the same size and shape. Each stack of blocks is followed by five questions per-taining only to that stack. You have three (3) minutes to complete this subtest. Questions: 20 Time: 3 minutes For sample Block Counting questions, see page 210. Use the following figure to answer questions 1 through 5. 1. Block 1 is touched by _ other blocks. a. 1 b. 2 c. 3 d. 4 e. 5 2. Block 2 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 3. Block 3 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 4. Block 4 is touched by other blocks. a. 1 b. 2 c. 3 d. 4 e. 5 5. Block 5 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 Use the following figure to answer questions 6 through 10. 6. Block 1 is touched by other blocks. a. 3 b. 4 c. 5 d. 6 e. 7 7. Block 2 is touched by other blocks. a. 1 b. 2 c. 3 d. 4 e. 5 8. Block 3 is touched by other blocks. a. 1 b. 2 c. 3 d. 4 e. 5 1 2 3 4 5 1 2 3 4 5 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 72 9. Block 4 is touched by other blocks. a. 3 b. 4 c. 5 d. 6 e. 7 10. Block 5 is touched by other blocks. a. 4 b. 5 c. 6 d. 7 e. 8 Use the following figure to answer questions 11 through 15. 11. Block 1 is touched by other blocks. a. 4 b. 5 c. 6 d. 7 e. 8 12. Block 2 is touched by other blocks. a. 1 b. 2 c. 3 d. 4 e. 5 13. Block 3 is touched by other blocks. a. 4 b. 5 c. 6 d. 7 e. 8 14. Block 4 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 15. Block 5 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 Use the following figure to answer questions 16 through 20. 16. Block 1 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 1 2 3 4 5 1 2 3 4 5 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 73 17. Block 2 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 18. Block 3 is touched by other blocks. a. 3 b. 4 c. 5 d. 6 e. 7 19. Block 4 is touched by other blocks. a. 3 b. 4 c. 5 d. 6 e. 7 20. Block 5 is touched by other blocks. a. 2 b. 3 c. 4 d. 5 e. 6 Subtest 7: Table Reading Directions: The Table Reading subtest measures your ability to read tables quickly and accurately. Notice that the X values in each table are shown horizontally across the top of the table and the Y values are shown vertically along the left edge of the table. In this test, you are to find the entry that occurs at the intersec-tion of the row and the column corresponding to the values given. On your answer sheet, fill in the letter that corresponds with the number at the intersection of the X and Y values. You have seven (7) minutes to complete this subtest. Questions: 40 Time: 7 minutes For sample Table Reading questions, see page 212. Use the following table to determine the correct value for the X and Y values given in questions 1 through 5. X VALUE –3 –2 –1 0 1 2 3 Y –3 76 47 65 88 93 20 22 V –2 57 43 74 77 37 57 46 A –1 43 85 94 22 13 11 35 L 0 48 30 68 67 48 58 56 U 1 98 42 76 84 10 84 57 E 2 76 35 10 62 93 37 43 3 56 86 40 41 95 99 68 1. –3,0 a. 85 b. 48 c. 76 d. 40 e. 56 2. –1,1 a. 42 b. 99 c. 35 d. 88 e. 76 3. 2,0 a. 58 b. 35 c. 20 d. 37 e. 10 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 74 Y V A L U E 4. 1,2 a. 10 b. 62 c. 37 d. 93 e. 95 5. –2,2 a. 98 b. 42 c. 35 d. 86 e. 40 Use the following table to determine the correct value for the X and Y values given in questions 6 through 10. X VALUE –3 –2 –1 0 1 2 3 Y –3 23 34 35 36 49 74 58 V –2 91 45 47 24 20 34 43 A –1 82 36 56 35 29 55 23 L 0 73 27 65 36 39 67 54 U 1 74 18 64 47 48 88 65 E 2 65 20 52 58 50 79 74 3 12 39 13 69 68 60 37 6. –3,0 a. 82 b. 73 c. 74 d. 27 e. 36 7. –2,1 a. 27 b. 74 c. 64 d. 18 e. 13 8. 1,1 a. 48 b. 49 c. 20 d. 29 e. 68 9. 0,3 a. 58 b. 47 c. 69 d. 13 e. 60 10. 2, 0 a. 36 b. 67 c. 39 d. 54 e. 48 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 75 Y V A L U E Use the following table to determine the correct value for the X and Y values given in questions 11 through 15. X VALUE –3 –2 –1 0 1 2 3 Y –3 43 35 19 21 24 90 83 V –2 46 46 59 10 35 98 49 A –1 57 54 48 99 43 78 30 L 0 68 76 37 88 45 67 36 U 1 79 58 24 77 65 87 38 E 2 10 49 11 66 67 56 48 3 34 22 50 55 87 76 74 11. –1,2 a. 37 b. 24 c. 87 d. 50 e. 11 12. 3,–1 a. 83 b. 49 c. 30 d. 57 e. 50 13. –2,3 a. 22 b. 10 c. 29 d. 11 e. 49 14. 0,0 a. 88 b. 99 c. 77 d. 66 e. 55 15. 0,3 a. 50 b. 87 c. 76 d. 55 e. 74 Use the following table to determine the correct value for the X and Y values given in questions 16 through 20. X VALUE –3 –2 –1 0 1 2 3 Y –3 84 21 22 79 24 70 99 V –2 58 25 34 86 35 60 87 A –1 95 34 35 58 46 56 67 L 0 60 64 36 46 57 77 57 U 1 70 56 47 58 54 43 46 E 2 74 45 54 69 32 55 43 3 83 37 38 70 12 78 22 16. 1,–3 a. 12 b. 78 c. 32 d. 24 e. 70 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 76 Y V A L U E Y V A L U E 17. 2,3 a. 32 b. 78 c. 22 d. 77 e. 56 18. 2,–2 a. 55 b. 60 c. 45 d. 25 e. 38 19. 1,0 a. 79 b. 95 c. 57 d. 87 e. 36 20. 0,–2 a. 25 b. 69 c. 77 d. 86 e. 54 Use the following table to determine the correct value for the X and Y values given in questions 21 through 25. X VALUE –3 –2 –1 0 1 2 3 Y –3 38 33 87 34 13 54 54 V –2 94 21 68 35 14 75 36 A –1 58 31 69 64 25 86 47 L 0 66 41 80 58 36 93 85 U 1 77 54 88 61 47 20 77 E 2 58 36 65 91 58 41 65 3 45 47 43 92 66 55 45 21. 0,2 a. 93 b. 91 c. 61 d. 92 e. 66 22. 1,1 a. 47 b. 69 c. 33 d. 77 e. 58 23. –1,–1 a. 36 b. 65 c. 54 d. 68 e. 69 24. 3,2 a. 54 b. 55 c. 20 d. 47 e. 65 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 77 Y V A L U E 25. –3,–3 a. 45 b. 54 c. 36 d. 38 e. 93 Use the following table to determine the correct value for the X and Y values given in questions 26 through 30. X VALUE –3 –2 –1 0 1 2 3 Y –3 92 57 37 83 33 48 50 V –2 84 65 55 94 20 59 33 A –1 84 68 68 85 82 60 42 L 0 10 69 77 86 92 17 35 U 1 29 40 99 62 74 63 46 E 2 39 42 80 73 85 48 47 3 44 36 75 32 55 94 58 26. –3,3 a. 58 b. 50 c. 92 d. 84 e. 44 27. 0,1 a. 86 b. 62 c. 77 d. 92 e. 74 28. 2,0 a. 17 b. 80 c. 94 d. 58 e. 75 29. 3,0 a. 46 b. 85 c. 35 d. 32 e. 75 30. 3,1 a. 39 b. 37 c. 46 d. 29 e. 69 Use the following table to determine the correct value for the X and Y values given in questions 31 through 35. X VALUE –3 –2 –1 0 1 2 3 Y –3 66 24 97 11 12 91 21 V –2 77 43 67 22 31 82 23 A –1 88 44 68 33 23 73 34 L 0 99 56 56 55 41 64 53 U 1 10 67 47 65 62 55 64 E 2 12 69 33 76 63 45 75 3 13 87 42 89 47 36 83 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 78 Y V A L U E Y V A L U E 31. –2,–2 a. 82 b. 45 c. 12 d. 43 e. 24 32. –2,3 a. 91 b. 36 c. 87 d. 24 e. 69 33. 0,–3 a. 22 b. 67 c. 76 d. 65 e. 11 34. 1,–3 a. 12 b. 47 c. 97 d. 13 e. 21 35. 1,1 a. 55 b. 52 c. 33 d. 99 e. 23 Use the following table to determine the correct value for the X and Y values given in questions 36 through 40. X VALUE –3 –2 –1 0 1 2 3 Y –3 44 13 83 35 45 99 33 V –2 65 24 25 46 65 89 44 A –1 76 32 47 57 43 70 56 L 0 83 53 65 68 27 67 57 U 1 46 28 43 97 56 60 69 E 2 57 91 78 66 33 61 80 3 45 28 97 54 43 23 43 36. –3,0 a. 13 b. 35 c. 83 d. 25 e. 24 37. 0,2 a. 67 b. 66 c. 53 d. 91 e. 78 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 79 38. –2,–1 a. 70 b. 32 c. 60 d. 28 e. 53 39. 0,–1 a. 57 b. 46 c. 27 d. 66 e. 43 40. 1,3 a. 43 b. 69 c. 56 d. 76 e. 46 Subtest 8: Aviation Information Directions: The Aviation Information subtest measures your knowledge of aviation. This subtest is common to all three service selection tests, although the number of questions varies from one service to another. Each of the questions or incomplete state-ments is followed by five choices. You must decide which one of the choices best completes the state-ment or answers the question. Eliminating any obvi-ously incorrect choices first will increase your chances of selecting the correct answer. You have eight (8) minutes to complete this subtest. Questions: 20 Time: 8 minutes For sample Aviation Information questions, see page 212. 1. An example of a high lift device would be a. flaps. b. slats. c. leading edge extensions. d. all the above. e. none of the above. 2. Aircraft performance increases in cold weather because a. cold air improves fuel flow. b. cold air is more dense. c. cold air is less dense. d. the engine exhaust is hotter than the surrounding air. e. none of the above. 3. If you wanted to roll the aircraft about its longitudinal axis, what flight control move-ment would accomplish this? a. aft stick movement b. trim tabs c. aileron deflection d. vertical stabilizer movement e. forward stick movement 4. The standard altimeter setting used in Class A airspace would be a. obtained from the nearest airfield. b. 30.00 in Hg. c. 29.92 in Hg. d. calculated from an air computer. e. none of the above. –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 80 5. During the execution of an aerobatic loop, what sort of energy will an aircraft have at the top of a loop? a. kinetic b. potential c. forced d. assumed e. none of the above 6. Airfield runways are numbered in accordance with a. their compass headings. b. FAA directives. c. airfield altitude. d. average relative wind. e. runway length. 7. In which of the following takeoff or landing situations would you be most concerned about aircraft performance? a. low altitude airport on a hot day b. low altitude airport on a cold day c. high altitude airport on a hot day d. high altitude airport on a cold day e. low altitude airport located next to water 8. Lights that outline an airport taxiway are _ in color. a. red b. blue c. green d. white e. alternating red and white 9. Extending wing flaps produces an increase in both lift and a. thrust. b. angle of attack. c. pressure. d. drag. e. energy. 10. The very thin layer of air flowing over the surface of an aircraft wing, an airfoil, or over the entire fuselage is called a. free-stream velocity air. b. the boundary layer. c. the slipstream. d. wake turbulence. e. wing drop. 11. If your cockpit turn and bank indicator shows a perfectly centered ball, a. the aircraft is in aerodynamically balanced flight. b. the aircraft weight and balance criteria have been met. c. the aircraft thrust-to-weight ratio is equal. d. the aircraft is pointed directly ahead. e. none of the above. 12. What effect do wing spoilers have when they are employed? a. true air speed is increased b. they aid in the production of lift by increasing effective wing area c. they disrupt the boundary layer airflow, increasing lift and reducing drag d. they disrupt the boundary layer airflow, reducing lift and increasing drag e. they augment boundary layer air to increase lift 13. The aircraft instrument that transmits signals that identify various aircraft parameters to air traffic control and flight monitoring organiza-tions is the a. altimeter. b. attitude indicator. c. transponder. d. tachometer. e. UHF radio. –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 81 14. How would taking off into a headwind affect your aircraft? a. Your takeoff distance would be unchanged. b. Your takeoff distance would be longer with little to no increase in climb angle. c. Your takeoff distance would be shorter with an increased climb angle. d. Your takeoff distance would be shorter and your available engine power would be unchanged. e. None of the above. 15. The acronym VSI stands for a. variable situational index. b. velocity and speed instrument. c. vertical speed indicator. d. vertical stability index. e. variable speed instrument. 16. The VSI instrument indicates a. rate of climb. b. rate of descent. c. level flight. d. all of the above. e. none of the above. 17. If one end of a runway is numbered 33, what would the other end be numbered? a. 15 b. 33 c. 24 d. 66 e. There is not enough information provided. 18. Ground speed can be affected by which of the following? a. pressure b. altitude c. wind d. heat e. rain 19. The two main types of drag an aircraft experiences in flight are a. parasite and induced. b. controlled and unlimited. c. supersonic and transonic. d. kinetic and potential. e. uncontrolled and limited. 20. Bernoulli’s Principle states that a. a body that is at rest will stay at rest unless acted upon by an outside force. b. when there is an increase in pressure, there must be a decrease in temperature. c. when there is an increase in velocity there must be a decrease in pressure. d. force times mass equals acceleration. e. air will circulate counterclockwise in the northern hemisphere. Subtest 9: General Science Directions: The General Science subtest measures your knowledge in the area of science. Each of the questions or incomplete statements is followed by five choices. You must decide which one of the choices best answers the question or completes the statement. Again, if you are unsure of an answer, use the process of elimination. Remember, there are no penalties for guessing. You have ten (10) minutes to complete this subtest. Questions: 20 Time: 10 minutes For sample General Science questions, see page 215. 1. An element’s location on the periodic table is determined by its number of a. electrons. b. neutrons. c. protons. d. nuclei. e. radons. –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 82 2. What are atoms of the same element that have different numbers of neutrons called? a. alloys b. isotopes c. alkali metals d. ions e. neurons 3. Marine biology is most closely associated with which field of science? a. geography b. botany c. oceanography d. geology e. seismology 4. Which of the following symbols represents a molecule of carbon dioxide? a. C b. O c. CO d. CO2 e. C2O 5. The driver of a car you are riding in loses control in a snowstorm. The car spins 360 degrees and you are thrown against the car door. Which of the following is the best description of what you are experiencing? a. fundamental forces b. center of mass c. centrifugal force d. Coriolis effect e. center of gravity 6. If particles of food coloring are dropped into a glass of hot water, they will spread rapidly. This is an example of a. osmosis. b. respiration c. evaporation. d. active transport. e. diffusion. 7. Which of the following represents a chemical change? a. tearing a piece of paper b. melting an ice cube c. cooking a hamburger d. dissolving sugar in water e. the wind blowing 8. Carbohydrates are much better foods for quick energy than fats because they a. are digested more easily and absorbed more quickly. b. supply essential amino acids, which provide energy. c. are high in both protein and iron. d. carry oxygen to the blood. e. all of the above. 9. Which of the following atmospheric levels is closest to the Earth’s surface? a. mesosphere b. stratosphere c. thermosphere d. troposphere e. necrosphere 10. All of the following are characteristics of reptiles except a. cold blood. b. lungs. c. land-dwelling adults. d. scaly skin. e. internal development of eggs. 11. Which of the following is a vertebrate? a. a sponge b. a starfish c. an octopus d. a snake e. an oyster –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 83 12. The process by which an organism adapts physiologically to the rigors of a new environment is known as a. natural selection. b. acclimatization. c. evolution. d. mutation. e. gestation. 13. Which of the following is the best description of what an omnivore eats? a. animal matter only b. vegetable matter only c. detritus only d. decomposing matter only e. both animal and vegetable matter 14. Which of the following has the shortest wavelength? a. ultraviolet b. x-rays c. microwave d. infrared e. visible 15. The fundamental force that is the natural force of attraction acting between objects with mass is which of the following? a. electromagnetism b. strong nuclear force c. weak nuclear force d. gravity e. radiomagnetism 16. Where is most of the mass of our solar system? a. Sun b. Earth c. Venus d. Jupiter e. Mercury 17. Absolute zero is equal to which of the following? a. 0° Fahrenheit b. 30° Fahrenheit c. 30° Kelvin d. –30° Kelvin e. –273° Celsius 18. Our solar system is made up of the Sun and how many planets? a. eight b. nine c. ten d. eleven e. twelve 19. A cell containing chloroplasts would most likely belong to which organism? a. rabbit b. fern c. roach d. lizard e. shark 20. In animal cells, what organelle contains the DNA? a. nucleus b. cytoplasm c. Golgi apparatus d. ribosomes e. endoplasmic reticulum –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 84 Subtest 10: Rotated Blocks Directions: The Rotated Blocks subtest measures your ability to visualize and manipulate objects in space. For each question in this test, you will be shown a picture of a block. You must find a second block that is identical to the first. You have thirteen (13) minutes to complete this subtest. Questions: 15 Time: 13 minutes For sample Rotated Block questions, see page 202. 1. a. b. c. d. e. 2. a. b. c. d. e. 3. a. b. c. d. e. 4. a. b. c. d. e. 5. a. b. c. d. e. 6. a. b. c. d. e. 7. a. b. c. d. e. –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 85 –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 86 8. a. b. c. d. e. 9. a. b. c. d. e. 10. a. b. c. d. e. 11. a. b. c. d. e. 12. a. b. c. d. e. 13. a. b. c. d. e. 14. a. b. c. d. e. 15. a. b. c. d. e. –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 87 Subtest 11: Hidden Figures Directions: The Hidden Figures subtest measures your ability to see a simple figure in a complex draw-ing. Above each group of questions are five figures, lettered A, B, C, D, and E. Below this set of figures are several numbered drawings. You are to determine which lettered figure is contained in each of the numbered drawings. Each numbered drawing con-tains only one of the lettered figures. The correct fig-ure in each drawing will always be of the same size and in the same position as it appears in the top set of figures. Look at each numbered drawing and de-cide which one of the five lettered figures is con-tained in it. You have eight (8) minutes to complete this subtest. Questions: 15 Time: 8 minutes For sample Hidden Figures questions, see page 217. Use the following figure to answer questions 1 through 5. 1. The hidden figure in block 1 is . a. A b. B c. C d. D e. E 2. The hidden figure in block 2 is . a. A b. B c. C d. D e. E 3. The hidden figure in block 3 is . a. A b. B c. C d. D e. E 4. The hidden figure in block 4 is . a. A b. B c. C d. D e. E 5. The hidden figure in block 5 is . a. A b. B c. C d. D e. E Use the following figure to answer questions 6 through 10. 6. The hidden figure in block 6 is . a. A b. B c. C d. D e. E 7 8 9 10 6 A B D E C 1 2 3 4 5 A B C D E –DIAGNOSTIC TEST FOR THE AIR FORCE OFFICER QUALIFYING TEST– 88 7. The hidden figure in block 7 is . a. A b. B c. C d. D e. E 8. The hidden figure in block 8 is . a. A b. B c. C d. D e. E 9. The hidden figure in block 9 is . a. A b. B c. C d. D e. E 10. The hidden figure in block 10 is . a. A b. B c. C d. D e. E Use the following figure to answer questions 11 through 15. 11. The hidden figure in block 11 is . a. A b. B c. C d. D e. E 12. The hidden figure in block 12 is . a. A b. B c. C d. D e. E 13. The hidden figure in block 13 is . a. A b. B c. C d. D e. E 14. The hidden figure in block 14 is . a. A b. B c. C d. D e. E 15. The hidden figure in block 15 is ___. a. A b. B c. C d. D e. E 12 13 14 15 11 A B D E C
14580	https://arxiv.org/pdf/2309.17424	arXiv:2309.17424v1 [math.NT] 29 Sep 2023 Representing the inverse map as a composition of quadratics in a finite field of characteristic 2 Florian Luca 1,2, Santanu Sarkar 3, Pantelimon St˘ anic˘ a4 1 School of Mathematics, University of the Witwatersrand, Private Bag X3, Wits 2050, Johannesburg, South Africa; and 2 Centro de Ciencias Matem´ aticas, UNAM, Morelia, Mexico; Florian.Luca@wits.ac.za , 3 Department of Mathematics, Indian Institute of Technology Madras, Sardar Patel Road, Chennai TN 600036, INDIA; santanu@iitm.ac.in , 4 Applied Mathematics Department, Naval Postgraduate School, Monterey 93943, USA; pstanica@nps.edu October 2, 2023 Abstract In 1953, Carlitz showed that all permutation polynomials over Fq , where q > 2is a power of a prime, are generated by the special permutation polynomials xq−2 (the inversion) and ax + b (affine functions, where 0 6 = a, b ∈ Fq). Recently, Nikova, Nikov and Rijmen proposed an algorithm (NNR) to find a decomposition of the inverse function in quadratics, and computationally covered all dimensions n ≤ 16. Petrides found a class of integers for which it is easy to decompose the inverse into quadratics, and improved the NNR algorithm, thereby extending the computation up to n ≤ 32. Here, we extend Petrides’ result, as well as we propose a number theoretical approach, which allows us to cover easily all (surely, odd) exponents up to 250, at least. Keywords: Permutations, Decompositions, Quadratics, Algorithm, Primes, Sieves MSC 2020 : 11A41, 11N13, 11N36, 20B99, 94A60, 94D10 1 Introduction This is an expanded and vastly improved version of the Extended Abstract, which was presented at the 8th International Workshop on Boolean Functions and their Applications (BFA) in Voss in September 2023. 1In , Carlitz showed that all permutation polynomials over Fq, where q > 2 is a power of a prime, are generated by the special permutation polynomials xq−2 (the inversion) and ax + b (affine functions, where 0 6 = a, b ∈ Fq). The smallest number of inversions in such a decomposition is called the Carlitz rank .Here, we ask whether the inverse in F2n (the finite field of dimension n over the two-element prime field F2) can be written as a composition of quadratics, or quadratics and cubics. That is, we ask if there are integers r ≥ 1 and a1 ≥ 0, . . . , a r ≥ 0 such that −1 ≡ r ∏ i=1 (2 ai + 1) (mod 2 n − 1) . Nikova, Nikov and Rijmen proposed an algorithm to find such a decomposition. Via Carlitz , they were able to use the algorithm and show that for n ≤ 16 any permutation can be decomposed in quadratic permutations, when n is not multiple of 4 and in cubic permutations, when n is multiple of 4. Petrides , in addition to a theoretical result, which we will discuss below, improved the complexity of the algorithm and presented a computational table of shortest decompositions for n ≤ 32, allowing also cubic permuta-tions in addition to quadratics. Here, we find a number theoretical approach which allows us to cover all (surely, odd) exponents up to 250. 2 Our results Let ν2 be the 2-valuation, that is, the largest power of 2 dividing the argument. We start with a proposition, extending one of Petrides’ results , which states that if n is an odd integer and n−1 2ν2(n−1) ≡ 2k (mod 2 n − 1), for some k, then 2n − 2 = 2 ( 2 (n−1 2ν2(n−1) ) 2ν2(n−1) − 1 ) = 2 ( 2 n−1 2ν2(n−1) − 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) ≡ 2 ( 22k − 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) = 2 k−1 ∏ j=0 ( 22j 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) . This implies, via Carlitz , that for all odd integers (coined good integers , with the coun-terparts named bad integers in ) satisfying the congruence n−1 2ν2(n−1) ≡ 2k (mod 2 n − 1), one can decompose any permutation polynomial in F2n into affine and quadratic power permutations. 2The smallest odd positive integer that is not good is n = 7. We note however that in that case 27 − 2 = 2(2 6 − 1) = 2(2 2 − 1)(2 4 + 2 2 + 1) = 2(2 + 1)(2 4 + 2 2 + 1) , and so, any permutation in F27 can be decomposed into affine, quadratic and cubic per-mutations. We are ready to generalize this observation. Theorem 1. Let n be an odd integer satisfying n − 1 2ν2(n−1) ≡ 2k3s (mod 2 n − 1) , for some non-negative integers r, s . Then, the inverse power permutation in F2n has a decomposition into affine, quadratic and cubic power permutations of length k+s+ν2(n−1) .Proof. As we have already alluded to above, using the difference of cubes factorization, a3 − b3 = ( a − b)( a2 + ab + b2), we have 2n − 2 = 2 ( 2 n−1 2ν2(n−1) − 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) ≡ 2 ( 22k 3s − 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) = 2 ( 22k 3s−1 − 1 ) ( 22k+1 3s−1 2 2k 3s−1 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) · · · · · · · · · · · · = 2 ( 22k − 1 ) s−1∏ j=0 ( 22k+1 3j 2 2k 3j 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) ≡ 2 k−1 ∏ j=0 ( 22j 1 ) s−1∏ j=0 ( 22k+1 3j 2 2k 3j 1 ) ν2(n−1) ∏ j=1 ( 2 n−1 2j 1 ) . The claim is shown. Example 1. It is natural to investigate the counting function B(x) of superbad inte-gers (that is, integers n such that n−1 2ν2(n−1) 6 ≡ 2k3s (mod 2 n − 1) ), with B(x) = {n ≤ x : n is superbad }, or its complement A(x) = {n ≤ x : n − 1 2ν2(n−1) ≡ 2k3s (mod 2 n − 1) }. 3As an example, \|B (50) \| = 16 , more precisely, B(50) = {1, 2, 3, 4, 5, 7, 9, 10 , 13 , 17 , 19 , 25 , 28 , 33 , 37 , 49 } (Petrides noted that 25 integers up to 50 are bad, so our extension surely prunes the integers better). In a recent paper , it was shown that A(x) ≪ x (log log x)1+ o(1) . For most of the remaining work we restrict our attention to n = p, a prime. Let p ≥ 3 be prime, N := Np = 2 p − 1. It is known that if q \| Np, then q ≡ 1 (mod p). We ask if we can say anything about the number of distinct prime factors ω(Np) of Np.We propose the following conjecture. Conjecture 1. There exists p0 such that for p > p 0, ω(Np) < 1.36 log p. Similar heuristics regarding lower bounds for Ω(2 n − 1) and ω(2 n − 1) can be found in and . Our Conjecture 1, is based on statistical arguments originating from sieve methods. The Tur´ an-Kubilius inequality asserts that ∑ n≤x (ω(n) − log log x)2 = O(x log log x). So, if δ > 0 is fixed, the set of n ≤ x such that ω(n) ≥ (1 + δ) log log x is of counting function Oδ (x/ log log x). One can do better using sieves. Indeed, Exercise 04 in shows that for fixed δ > 0 we in fact have that #{n ≤ x : ω(n) ≥ (1 + δ) log log x} ≪ δ x (log x)Q(δ) , where Q(δ) := (1+ δ) log((1+ δ)/e )+1. We would like to apply such heuristics to Np = 2 p−1. But note that if q \| Np, then 2 p ≡ 1 (mod q). In particular, ( 2 q ) = 1, so q ≡ ± 1 (mod 8). But then the same proof as Exercise 04 in shows that # {n ≤ x : q \| n ⇒ q ≡ ± 1 (mod 8) and ω(n) ≥ (1 + δ) log log x}≤ x (log x)Q1(δ)+ o(1) as x → ∞ , where Q1(δ) := (1 + δ) log((1 + δ)/(0 .5e)) + 1. Taking δ = 0 .36, we get Q1(δ) = 1.00086 . . . . Thus, the probability that a number having only prime factors congruent to ±1 (mod 8) to have more than 1 .36 log log n distinct prime factors is O ( 1 (log n)1.00008 ) . 4Applying this to Np, we get O ( 1 (log(2 p − 1)) 1.0008 ) ≪ 1 p1.0008 , and since the series ∑ p≥3 1 p1.0008 is convergent, we are led to believe that maybe there are at most finitely many prime numbers p such that ω(Np) ≥ 1.36 log p. It has been noted that perhaps infinitely often ω(Np) ≥ 2. For example, this is the case if p ≡ 3 (mod 4) is such that q = 2 p + 1 is prime. Indeed, then 2 is a quadratic residue modulo q so 2 (q−1) /2 ≡ 1 (mod q), showing that q \| Np. Since Np is never a perfect power, in particular it cannot be a power of q, we get the desired conclusion that ω(Np) ≥ 2. Conjecture 2. There exists p0 such that if p > p 0, then Np is squarefree. We offer some heuristic evidence for Conjecture 2. Knowing that the prime q divides Np, the conditional probability that Np is divisible by q2 is 1 /q . Thus, the probability that Np is not squarefree is bounded above by ∑ q:q\|Npfor some prime p 1 q , and it was shown by Murata and Pomerance in that the above sum is finite under GRH. So, assuming Conjecture 1 and 2, let Np := q1 · · · qk for some distinct primes q1, . . . , q k with k ≤ 1.36 log p. We take numbers of the form 2 a + 1 with an odd a ∈ [5 , p − 2]. We want to compute ( 2a + 1 2p − 1 ) . This was done by Rotkiewicz in . Namely, write the Euclidean algorithm with even quotients and signed remainders: p = (2 k1)a + ε1r1, ε1 ∈ {± 1}, 1 ≤ r1 ≤ a − 1 a = (2 k2)r1 + ε2r2, ε2 ∈ {± 1}, 1 ≤ r2 ≤ r1 − 1,. . . = . . . rℓ−2 = (2 kℓ)rℓ−1 + εℓrℓ, εℓ ∈ {± 1}, rℓ = 1 , where ℓ := ℓ(a, p ) is minimal with rℓ = 1. Note that ri are all odd for i = 1 , . . . , ℓ . In particular, rj ≥ 3 for j = 1 , . . . , ℓ − 1. Then ( 2a + 1 2p − 1 ) = ( 2p − 1 2a + 1 ) = ( (2 a)2k1 · 2ε1r1 − 1 2a + 1 ) = ( 2ε1r1 − 1 2a + 1 ) = ( 2r1 − 1 2a + 1 ) . 5The right–most equality is clear if ε1 = 1, and if ε1 = −1, then ( 2−r1 − 1 2a + 1 ) = ( 2 2a + 1 )r1 ( 2−r1 − 1 2a + 1 ) = ( 2r1 (2 −r1 − 1) 2a + 1 ) = ( 1 − 2r1 2a + 1 ) = ( −1 2a + 1 ) ( 2r1 − 1 2a + 1 ) = ( 2r1 − 1 2a + 1 ) . The above calculation shows how to transit from the first step to the second step, or more generally from a step with 2 rj − 1 in the bottom to a step with 2 rj − 1 in the top. For the next step, we write ( 2r1 − 1 2a + 1 ) = ( 2a + 1 2r1 − 1 ) = ( (2 r1 )2k2 · 2ε2r2 + 1 2r1 − 1 ) = ( 2ε2r2 + 1 2r1 − 1 ) = (2r2 + 1 2r1 − 1 ) . The last inequality above is clear if ε2 = 1. If ε2 = −1, then since r1 ≥ 3, we have ( 2−r2 + 1 2r1 − 1 ) = ( 2 2r1 − 1 )r2 ( 2−r2 + 1 2r1 − 1 ) = (2r2 (2 −r2 + 1) 2r1 − 1 ) = ( 2r2 + 1 2r1 − 1 ) . At step ℓ we end up with ( 2rℓ + ( −1) ℓ 2rℓ−1 + ( −1) ℓ−1 ) . If ℓ(a, p ) is odd, we get ( 21 − 1 2rℓ−1 + 1 ) = 1 , and if ℓ(a, p ) is even we get ( 21 + 1 2rℓ−1 − 1 ) = − ( 2rℓ−1 − 1 3 ) = − ( 1 3 ) = −1. We thus get that ( 2a + 1 2p − 1 ) = ( −1) ℓ+1 . We select the subset A(p) of odd a in the interval [5 , p − 2] such that ℓ ≡ 0 (mod 2). We assume that there are a positive proportion of such, namely that there is a constant c1 > 0such that for large p, there are > c 1p odd numbers a ∈ [5 , p − 2] such that ℓ(a, p ) ≡ 0(mod 2). So, we have k∏ i=1 ( 2a + 1 qi ) = −1 for a ∈ A (p). We next conjecture that for such a, the values are (( 2a + 1 qi ) , 1 ≤ i ≤ k ) (1) 6are uniformly distributed among the 2 k vectors ( ±1, ±1, · · · , ±1) ︸︷︷︸ ktimes . Indeed, if not, then somehow for some a the value of 2 a + 1 of the Legendre symbol ( 2a + 1 pi ) should be determined in terms of the values of the same symbol for b ≤ a − 1. This can happen for example if: (i) 2 a + 1 is a square. This never happens for a ≥ 4. (ii) 2 a + 1 is multiplicatively dependent over {2b + 1 : 0 ≤ b ≤ a − 1}. This does not happen for a ≥ 4 because of the Carmichael’s Primitive Divisor Theorem: 2 a + 1 has a prime factor pa which is primitive in the sense that pa does not divide 2 b + 1 for any b ≤ a − 1. Well, so we fix i ∈ { 1, . . . , k } and search for ai such that ( 2ai + 1 qi ) = ( −1) δij , (2) where δij is the Kronecker symbol. That is, 2 ai + 1 is a quadratic residue modulo pj for all j 6 = i but it is not a quadratic residue modulo qi. Do we expect to find it? Well, let us see. Fix i in {1, . . . , k }. The probability that 2 ai + 1 verifies the Legendre conditions given by (2) is 1 /2k so it is (1 − 1/2k ) that they are not satisfied. Note that since ( 2ai + 1 Np ) = −1we know that an odd number of the p = pj ’s satisfy that ( 2ai + 1 pj ) = 1. So, if we assume that this is so for all possible ai’s, and assuming that there events are independent, we get that the probability that this be so is ≪ ( 1 − 1 2k )c1p < ( 1 − 1 p1.36 log 2 )c1p < ( 1 − 1 p0.95 )c1p ≪ 1 ec1p0.05 . In the above, we used that k < 1.36 log p and 1 .36 · log 2 < 0.95. Of course, this is for i fixed and now we sum up over i from 1 to k introducing another logarithmic factor in the above count. Since the series ∑ p log p ec1p0.05 converges, so we expect that the above event does not occur when p > p 0. So, we have the following conjecture. Conjecture 3. Assume Conjectures 1 and 2. Write 2p − 1 = q1 . . . q k for p > p 0 with prime factors q1 < · · · < q k and k < 1.36 log p. Then for each i = 1 , . . . , k , there exists an odd ai ∈ [5 , p − 2] such that equalities (2) hold. 7The rest of the proof is unconditional. We will show that there exist integers xi such that (−1) = k ∏ i=1 (2 ai + 1) xi (mod 2 p − 1) . (3) Write qi − 1 =: 2 αi Ri for i = 1 , . . . , k , where Ri is odd. Let R := lcm[ Ri : 1 ≤ i ≤ k]and write xi = yiR for i = 1 , . . . , k . Let ρi be a primitive root modulo qi. Write 2ai + 1 = ρbij j (mod qj ). (4) Conditions (2) show that bij ≡ δij (mod 2). Equation (3) holds if and only if it holds one prime qj at a time. Thus, we want ρ(qj −1) /2 j ≡ ρR ∑ki=1 yibij j (mod qj ), which holds provided that (qj − 1) 2 ≡ R k ∑ i=1 yibij (mod qj − 1) . This in turn is equivalent to 2αj −1 ≡ (R/R j ) k ∑ i=1 yibij (mod 2 αj ). Since R/R j is odd, it follows that it is invertible modulo 2 αj . Writing ( R/R j )∗ for the inverse of ( R/R j ) modulo 2 αj , we get that 2αj −1(R/R j )∗ ≡ k ∑ i=1 yibij (mod 2 αj ). Since ( R/R j )∗ is odd the left–hand side is just 2 αj −1 (mod 2 αj ). Thus, 2αj −1 ≡ k ∑ i=1 yibij (mod 2 αj ). This is a linear system of modular equations for i = 1 , . . . , k . To see that it is nondegenerate note that the coefficient matrix B = ( bij )1≤i,j ≤k modulo 2 is in fact the identity matrix. Hence, its determinant is odd integer, so invertible modulo powers of 2, which shows that 8there exist an integer solution y1, . . . , y k. To solve it, we can generate for each i, j the number bi,j (mod 2 αj ) appearing in (4) as an integer in the interval [0 , 2αj − 1]. Having done that, we solve the linear system k ∑ i=1 yibij = 2 αj −1 for j = 1 , 2, . . . , k. This is non-degenerate since the determinant of the coefficient matrix is odd. Thus, (y1, . . . , y k) are some rational numbers. Now we treat them as residue classes modulo 2α, where α := max {αi : 1 ≤ i ≤ k} (by inverting the odd determinant modulo 2 α). These ones are the yi’s that we are looking for. We implemented this and checked it for all primes p ≤ 250. We present our approach in Algorithm 1. Algorithm 1: 1 for Prime p ≤ 250 do 2 Factor 2 p − 1 = q1 · · · qk, where qi is prime for 1 ≤ i ≤ k; 3 for j = 1 to k do 4 Find odd aj ∈ [5 , p − 2] such that the Legendre symbol ( 2aj +1 qi ) = ( −1) δij where δij is the Kronecker symbol. 5 end 6 Take a primitive root ρi modulo qi for 1 ≤ i ≤ k; 7 Find bij such that 2 ai + 1 = ρbij j (mod qj ) for 1 ≤ i, j ≤ k; 8 Find largest αi such that 2 αi is a divisior of qi − 1 for 1 ≤ i ≤ k; 9 Calculate α = max {αi : 1 ≤ i ≤ k}; 10 Solve the system of linear equations ∑ki=1 yibij = 2 αj −1 for j = 1 , 2, . . . , k. in Zα 11 end The factorization of 2 p − 1 is known for all primes p < 1000. Surely, we can use the same algorithm modulo 2 n − 1 for n ≤ 250 and odd. Note that if 2n − 1 = k ∏ j=1 qαj j , then we only want to find a relation of the form (−1) ≡ k ∏ i=1 (2 ai + 1) xi (mod q1 . . . q k). (5) 9Indeed, if we have found the above relation, then q1 . . . q k \| (2 a1 + 1) x1 · · · (2 ak + 1) xk + 1 . Writing Q := (2 n − 1) /(q1 . . . q k), we then get easily that (2 n − 1) \| (2 a1 + 1) x1Q(2 a2 + 1) x2Q · · · (2 ak + 1) xk Q + 1 . Thus, (−1) ≡ k∏ i=1 (2 ai + 1) xiQ (mod 2 n − 1) . Thus, we factor 2 n − 1, take q1, . . . , q k to be all its distinct prime factors and attempt to find some numbers a1, . . . , a k in [5 , n − 2] such that the congruences (2) are satisfied. If we are successful, then the argument based on the matrix with odd determinant will work to find a solution of (5), which in turn can be easily lifted to a solution modulo 2 n − 1. The factorizations of 2 n − 2 with weight 2 factors for odd 33 ≤ n ≤ 249 are given in Table 1. Remark 1. We have checked that Algorithm 1 works for most primes up to 250 . But there are a few primes like 47 for which there is no aj ∈ [5 , p − 2] such that ( 2aj +1 qi ) = ( −1) δij .In these cases, we use the following trick. We first find ai and calculate ( 2aj +1 qi ) = ( −1) di,j .Ideally, di,j should be Kronecker symbols, but if they are not, we can just record what they are. Because di,j are no longer Kronecker symbols, we cannot be certain that the system is solvable because it may have an even determinant and we cannot invert the matrix modulo powers of 2. However, we checked primes (and odd integers) up to 250 . We observed that in the case of failure, we can use this trick and always get suitable ai’s such that the corresponding matrix has odd determinant, and is therefore invertible. Table 1: Factorization of 2 n − 2 (mod 2 n − 1) for odd 33 ≤ n ≤ 249. n= 33 (2 5+ 1) 599478 ·(2 13 + 1) 299739 ·(2 29 + 1) 1798434 n= 35 ( (2 + 1)(2 17 + 1) )967995 ·(2 29 + 1) 276570 n= 37 (2 5+ 1) 77039772 ·(2 13 + 1) 19259943 n= 39 ( (2 11 + 1)(2 21 + 1) )1592955 n= 41 (2 9+ 1) 20111512782 ·(2 13 + 1) 3351918797 n= 43 ( (2 5+ 1)(2 17 + 1)(2 23 + 1) )593211015 n= 45 (2 + 1) 407925 ·(2 13 + 1) 349650 · ( (2 25 + 1)(2 33 + 1)(2 41 + 1) )116550 n= 47 (2 11 + 1) 1927501725 ·(2 37 + 1) 435242325 ·(2 41 + 1) 1616614350 n= 49 (2 9+ 1) 34630287489 ·(2 11 + 1) 3393768173922 n= 51 (1 + 2 29 )150009615 n= 53 (1 + 2 5)6512186850 ·(1 + 2 15 )3506562150 ·(1 + 2 21 )250468725 n= 55 (1 + 2) 6588945 ·(1 + 2 11 )5856840 ·(1 + 2 17 )732105 · (1 + 2 25 )1464210 ·(1 + 2 33 )10249470 ·(1 + 2 47 )732105 10 n = 57 (1 + 2 5 )396029391534 · (1 + 2 17 )1188088174602 · (1 + 2 21 )594044087301 · (1 + 2 47 )198014695767 n = 59 (1 + 2 7 )3663925098759300 · (1 + 2 13 )305327091563275 n = 61 (1 + 2 9 )1152921504606846975 n = 63 (1 + 2) 42958503 · (1 + 2 5 )3735522 · (1 + 2 39 )56032830 · (1 + 2 43 )44826264 · (1 + 2 47 )29884176 n = 65 (1 + 2 17 )72647571779055 · (1 + 2 23 )72647571779055 · (1 + 2 29 )72647571779055 n = 67 (1 + 2 5 )15295807610659665 n = 69 (1 + 2 11 )36566619637113225 · (1 + 2 17 )2437774642474215 · (1 + 2 53 )19502197139793720 · (1 + 2 67 )21939971782267935 n = 71 (1 + 2 11 )3659326099961865 · (1 + 2 13 )14637304399847460 n = 73 (1 + 2 31 )1726845200475585 · (1 + 2 45 )107064402429486270 n = 75 (1 + 2) 36654975 · (1 + 2 39 )17832150 · (1 + 2 41 )9906750 · (1 + 2 43 )7925400 · (1 + 2 53 )57459150 · (1 + 2 55 )15850800 · (1 + 2 63 )43589700 n = 77 (1 + 2 25 )290641821624556479 · (1 + 2 31 )290641821624556479 · (1 + 2 41 )290641821624556479 · (1 + 2 67 )581283643249112958 n = 79 (1 + 2 9 )12102186118644337359 · (1 + 2 15 )12102186118644337359 · (1 + 2 41 )12102186118644337359 n = 81 (1 + 2) 106331083505919 · (1 + 2 25 )155626336778778 · (1 + 2 37 )105108887143782 · (1 + 2 39 )155626336778778 · (1 + 2 43 )4073987873790 n = 83 (1 + 2 11 )7239076764159456135965 n = 85 (1 + 2 9 )4760486403166879215 · (1 + 2 13 )4760486403166879215 · (1 + 2 23 )4760486403166879215 n = 87 (1 + 2 39 )3371346107168004 · (1 + 2 41 )280945508930667 · (1 + 2 53 )2809455089306670 · (1 + 2 61 )4214182633960005 · (1 + 2 71 )1685673053584002 · (1 + 2 83 )280945508930667 n = 89 (1 + 2 13 )309485009821345068724781055 n = 91 (1 + 2 59 )280368506850705 · (1 + 2 67 )1682211041104230 · (1 + 2 71 )280368506850705 · (1 + 2 73 )280368506850705 · (1 + 2 81 )3364422082208460 n = 93 (1 + 2 17 )2305843010287435773 n = 95 (1 + 2 43 )7354378117756963125 · (1 + 2 51 )7354378117756963125 n = 97 (1 + 2 5 )612535370185410489825162846 · (1 + 2 9 )102089228364235081637527141 n = 99 (1 + 2) 160190876329840719 · (1 + 2 23 )160190876329840719 · (1 + 2 35 )58251227756305716 · (1 + 2 57 )29125613878152858 · (1 + 2 59 )101939648573535003 · (1 + 2 75 )58251227756305716 n = 101 (1 + 2 7 )261479084205457681314981849 · (1 + 2 9 )1045916336821830725259927396 n = 103 (1 + 2 5 )8204858250687037849538541156 · (1 + 2 9 )2051214562671759462384635289 n = 105 (1 + 2 7 )736412106675 · (1 + 2 29 )6627708960075 · (1 + 2 37 )1472824213350 · (1 + 2 55 )6627708960075 · (1 + 2 69 )15464654240175 · (1 + 2 79 )736412106675 · (1 + 2 83 )4418472640050 · (1 + 2 85 )441847264005 · (1 + 2 87 )13255417920150 n = 107 (1 + 2 5 )27043212804868893898596335048021 n = 109 (1 + 2 7 )744308608310570490215126499806 · (1 + 2 15 )372154304155285245107563249903 n = 111 (1 + 2 17 )2078233794395472907116 · (1 + 2 31 )742226355141240323970 · (1 + 2 39 )890671626169488388764 (1 + 2 71 )180254971962872650107 · (1 + 2 87 )519558448598868226779 n = 113 (1 + 2 15 )82901226266607482846190 · (1 + 2 25 )13816871044434580474365 · (1 + 2 29 )37854441217628987601 · (1 + 2 75 )13816871044434580474365 · (1 + 2 97 )82901226266607482846190 n = 115 (1 + 2 17 )23588654041464621525 · (1 + 2 23 )165120578290252350675 · (1 + 2 39 )23588654041464621525 · (1 + 2 45 )23588654041464621525 · (1 + 2 75 )188709232331716972200 n = 117 (1 + 2 5 )350280341971560 · (1 + 2 11 )481635470210895 · (1 + 2 31 )1225981196900460 · (1 + 2 55 )1269766239646905 · (1 + 2 71 )1225981196900460 · (1 + 2 87 )744345726689565 · (1 + 2 93 )1903697510715 · (1 + 2 111 )1094626068661125 · (1 + 2 115 )1182196154154015 n = 119 (1 + 2 21 )121807344007626864485535 · (1 + 2 25 )28109387078683122573585 · (1 + 2 51 )6635419517925198843570 · (1 + 2 81 )5968559856373716359791215 · (1 + 2 97 )852651408053388051398745 · (1 + 2 109 )6635419517925198843570 11 n = 121 (1 + 2 9 )99244104353509123769903900571 · (1 + 2 19 )893196939181582113929135105139 · (1 + 2 25 )893196939181582113929135105139 · (1 + 2 43 )1786393878363164227858270210278 n = 123 (1 + 2 5 )38263506571610465341512132024 · (1 + 2 9 )19131753285805232670756066012 · (1 + 2 27 )4782938321451308167689016503 · (1 + 2 53 )28697629928707849006134099018 · (1 + 2 113 )7726284980805959347805334351 n = 125 (1 + 2 23 )2898591397871459238374625 · (1 + 2 29 )644131421749213164083250 · (1 + 2 95 )1610328554373032910208125 (1 + 2 109 )4186854241369885566541125 · (1 + 2 121 )1932394265247639492249750 n = 127 (1 + 2 5 )28356863910078205288614550619314017621 n = 129 (1 + 2 9 )8471295533565243108183419405055 · (1 + 2 79 )9529016348217371325290685495 n = 131 (1 + 2 9 )1293849303881895298339827404683529321 · (1 + 2 15 )2587698607763790596679654809367058642 n = 133 (1 + 2 5 )27256203475454233141905720953493 · (1 + 2 17 )81768610426362699425717162860479 (1 + 2 45 )81768610426362699425717162860479 n = 135 (1 + 2 9 )1390256215369200900 · (1 + 2 25 )9826330930229511961200 · (1 + 2 47 )38551429107264868200 (1 + 2 55 )2813183451799578021150 · (1 + 2 89 )7686726614776311776100 (1 + 2 101 )1406591725899789010575 · (1 + 2 109 )4576375896941567062575 (1 + 2 117 )1042692161526900675 · (1 + 2 121 )2119793164384189072275 n = 137 (1 + 2 5 )39741006355730039527321333167397040041 · (1 + 2 7 )79482012711460079054642666334794080082 n = 139 (1 + 2 9 )17408530362059304982034022473992637175 · (1 + 2 17 )457116770994992690994328817697837300 n = 141 (1 + 2) 1216799735702178355508978464575 · (1 + 2 51 )110618157791107123228088951325 · (1 + 2 65 )63210375880632641844622257900 · (1 + 2 121 )56889338292569377660160032110 n = 145 (1 + 2 41 )534639083977880631530660485081925505 · (1 + 2 49 )76377011996840090218665783583132215 · (1 + 2 135 )305508047987360360874663134332528860 · (1 + 2 137 )4144489023084345980857833217689345 · (1 + 2 139 )534639083977880631530660485081925505 n = 147 (1 + 2 15 )17249119260282613026137951811234 · (1 + 2 59 )51747357780847839078413855433702 · (1 + 2 67 )43122798150706532565344879528085 · (1 + 2 71 )96786724738252439757774062940813 · (1 + 2 73 )8624559630141306513068975905617 n = 149 (1 + 2 9 )29933886172524326364132038117944134026225 n = 151 (1 + 2 35 )47657859344287051433215338407025 · (1 + 2 53 )47657859344287051433215338407025 · (1 + 2 55 )95315718688574102866430676814050 · (1 + 2 81 )95315718688574102866430676814050 · (1 + 2 119 )47657859344287051433215338407025 n = 153 (1 + 2 67 )74105228687928761744692516074690 · (1 + 2 85 )566868663181345103125787385 · (1 + 2 101 )185263071719821904361731290186725 · (1 + 2 115 )24701742895976253914897505358230 · (1 + 2 125 )66694705819135885570223264467221 · (1 + 2 147 )22231568606378628523407754822407 n = 155 (1 + 2 17 )62671642336461616797239779725 · (1 + 2 43 )35812367049406638169851302700 · (1 + 2 51 )17906183524703319084925651350 · (1 + 2 59 )2984363920783886514154275225 · (1 + 2 73 )26859275287054978627388477025 · (1 + 2 101 )17906183524703319084925651350 · (1 + 2 123 )865014224607504542831738415625 n = 157 (1 + 2 15 )1707444675887681902216221662393643900 · (1 + 2 17 )341488935177536380443244332478728780 · (1 + 2 29 )3841750520747284279986498740385698775 · (1 + 2 45 )30734004165978274239891989923085590200 n = 159 (1 + 2 57 )9362516203257056384802075 · (1 + 2 65 )44348760962796582875378250 · (1 + 2 69 )4434876096279658287537825 · (1 + 2 89 )38435592834423705158661150 · (1 + 2 101 )79907677410444293469150 · (1 + 2 123 )5913168128372877716717100 · (1 + 2 127 )2002242486366485713350 · (1 + 2 137 )48783637059076241162916075 n = 161 (1 + 2 29 )93343471924356402246389002034385 · (1 + 2 43 )62228981282904268164259334689590 · (1 + 2 47 )82971975043872357552345779586120 · (1 + 2 87 )217801434490164938574907671413565 · (1 + 2 157 )248915925131617072657037338758360 12 n = 163 (1 + 2 13 )168486137937535997136381884224759350 · (1 + 2 87 )5616204597917866571212729474158645 · (1 + 2 89 )1925555862143268538701507248282964 · (1 + 2 119 )78626864370850131996978212638221030 n = 165 (1 + 2 61 )39948352158132627380823842541450 · (1 + 2 105 )9321282170230946388858896593005 · (1 + 2 109 )13316117386044209126941280847150 · (1 + 2 113 )15535470283718243981431494321675 · (1 + 2 119 )19974176079066313690411921270725 · (1 + 2 123 )79896704316265254761647685082900 · (1 + 2 127 )3698921496123391424150355790875 · (1 + 2 135 )5326446954417683650776512338860 · (1 + 2 147 )4438705795348069708980426949050 · (1 + 2 157 )13316117386044209126941280847150 n = 167 (1 + 2 5 )350060123390813635242448130256489390771914057740 · (1 + 2 33 )17503006169540681762122406512824469538595702887 n = 169 (1 + 2 15 )7089726406583596958466287242575266870940 · (1 + 2 25 )7286663251210919096201461888202357617355 · (1 + 2 55 )29146653004843676384805847552809430469420 n = 171 (1 + 2 49 )49149123355767553400835304429946193135 · (1 + 2 55 )9829824671153510680167060885989238627 · (1 + 2 77 )9829824671153510680167060885989238627 · (1 + 2 99 )9829824671153510680167060885989238627 · (1 + 2 109 )9829824671153510680167060885989238627 · (1 + 2 159 )9829824671153510680167060885989238627 · (1 + 2 163 )9829824671153510680167060885989238627 n = 173 (1 + 2 21 )752712011377013221558430642567508556008861 · (1 + 2 61 )752712011377013221558430642567508556008861 · (1 + 2 69 )752712011377013221558430642567508556008861 · (1 + 2 77 )31613904477834555305454086987835359352372162 n = 175 (1 + 2 19 )281198684623467689204913686779425 · (1 + 2 43 )540766701198976325394064782268125 · (1 + 2 61 )12329480787336660218984677035713250 · (1 + 2 97 )129784008287754318094575547744350 · (1 + 2 105 )6651799969190141587990774087125 · (1 + 2 123 )11139794044698912303117734514723375 · (1 + 2 141 )4975053651030582193625395996866750 · (1 + 2 163 )1838606784076519506339820259711625 · (1 + 2 165 )1114982889070054279163020169625 n = 177 (1 + 2 19 )102104448867391604403906908898445293975 · (1 + 2 57 )4288386852430447384964090173734702346950 · (1 + 2 93 )81683559093913283523125527118756235180 · (1 + 2 103 )1286516055729134215489227052120410704085 · (1 + 2 133 )1837880079613048879270324360172015291550 · (1 + 2 163 )568867643689753224536052778148480923575 n = 179 (1 + 2 21 )1713334865061395551905989490523888483510297545 · (1 + 2 29 )17731656063809998410201669013040877638868476180 n = 181 (1 + 2 11 )16825262628616094460214312446168738419261880 · (1 + 2 21 )21031578285770118075267890557710923024077350 · (1 + 2 31 )2103157828577011807526789055771092302407735 · n = 183 (1 + 2 17 )5301342361191869603932356617428842175355175 · (1 + 2 75 )171011043909415148513946987658994908882425 · (1 + 2 177 )1279634363046313352673327459379375697499525 n = 185 (1 + 2 21 )24289606148175875174394125504156277451293856980 · (1 + 2 25 )56675747679077042073586292843031314053018999620 · (1 + 2 61 )6072401537043968793598531376039069362823464245 · (1 + 2 73 )28337873839538521036793146421515657026509499810 · (1 + 2 121 )3333867510533943651387428990766547885471705860 n = 187 (1 + 2 9 )52332852605905914814562661586433935229041537035 · (1 + 2 19 )52332852605905914814562661586433935229041537035 · (1 + 2 41 )203629776676676711340710745472505584548799755 · (1 + 2 89 )104665705211811829629125323172867870458083074070 n = 189 (1 + 2 7 )954808575327093401067436576289941581 · (1 + 2 11 )76741427691674298191288254054776183774 · (1 + 2 21 )862835224381042468504831900414614 · (1 + 2 33 )8687708795283882814108104232616171748 · (1 + 2 39 )6515781596462912110581078174462128811 · (1 + 2 73 )60813961566987179698756729628313202236 · (1 + 2 89 )53091553748957061641771748088209938460 · (1 + 2 125 )62503238277181268023722194340210791187 · (1 + 2 157 )121386597889660918208232678583498177479 13 n = 191 (1 + 2 55 )1025444869877616060103489665965652402208725 · (1 + 2 77 )1860992541629747664632259023419146952156575 · (1 + 2 179 )1063424309502712951218433727668083972660900 · (1 + 2 183 )2126848619005425902436867455336167945321800 · (1 + 2 189 )138245160235352683658396384596850916445917 n = 193 (1 + 2 5 )690644713229389686815238348164730529976649425988651 · (1 + 2 9 )76738301469932187423915372018303392219627713998739 · (1 + 2 13 )98663530461341383830748335452104361425235632284093 n = 195 (1 + 2 31 )397218618589975176651322156679935564968150 · (1 + 2 129 )42129247426209488432715986314538620526925 · (1 + 2 163 )66203103098329196108553692779989260828025 n = 197 (1 + 2 5 )47788807121282329843547481918370106279929779662675299482 · (1 + 2 13 )7964801186880388307257913653061684379988296610445883247 n = 199 (1 + 2 7 )9012349943070385113930109307619809450853768536749084363 · (1 + 2 27 )234321098519830012962182841998115045722197981955476193438 n = 201 (1 + 2 67 )201 · (1 + 2 107 )915031412652462933978517192307754031960084170 · (1 + 2 119 )14182986896113175476667016480770187495381304635 · (1 + 2 145 )3507620415167774580250982570513057122513655985 · (1 + 2 177 )83184673877496630361683381118886730178189470 n = 203 (1 + 2 57 )718674251279934430341052428990271148449812 · (1 + 2 81 )51205540403695328161799985565556819327049105 · (1 + 2 111 )6288399698699426265484208753664872548935855 · (1 + 2 127 )18865199096098278796452626260994617646807565 · (1 + 2 153 )2695028442299754113778946608713516806686795 · (1 + 2 175 )64680682615194098730694718609124403360483080 · (1 + 2 193 )8085085326899262341336839826140550420060385 n = 205 (1 + 2 43 )9467961424350347777980448419013907428323430550 · (1 + 2 67 )29756450190815378730795695031186566203302210300 · (1 + 2 131 )676282958882167698427174887072421959165959325 · (1 + 2 145 )7213684894743122116556532128772500897770232800 · (1 + 2 157 )772894810151048798202485585225625096189667800 · (1 + 2 187 )16907073972054192460679372176810548979148983125 n = 207 (1 + 2 5 )33192619261066535128289058132930982761836775 · (1 + 2 121 )46469666965493149179604681386103375866571485 · (1 + 2 127 )102131136187897031163966332716710716190267 · (1 + 2 179 )19915571556639921076973434879758589657102065 · (1 + 2 187 )39831143113279842153946869759517179314204130 · (1 + 2 199 )3983114311327984215394686975951717931420413 n = 209 (1 + 2 59 )270250337042154559392793016292662090505143401152585 · (1 + 2 61 )360333782722872745857057355056882787340191201536780 · (1 + 2 87 )180166891361436372928528677528441393670095600768390 · (1 + 2 97 )25738127337348053275504096789777341952870800109770 n = 211 (1 + 2 9 )1154929871973565275224108161553051512738585560465529713690 · (1 + 2 13 )64162770665198070845783786752947306263254753359196095205 · (1 + 2 55 )2309859743947130550448216323106103025477171120931059427380 n = 213 (1 + 2 11 )5497736589318700986099894638477660690688115321335 · (1 + 2 41 )140967604854325666310253708678914376684310649265 · (1 + 2 85 )1832578863106233662033298212825886896896038440445 · (1 + 2 119 )2356172823993728994042811987918997438866335137715 · (1 + 2 161 )1099547317863740197219978927695532138137623064267 · (1 + 2 203 )1570781882662485996028541325279331625910890091810 14 n = 215 (1 + 2 41 )66443114885376278757699109185773253174500402025 · (1 + 2 97 )31006786946508930086926250953360851481433520945 · (1 + 2 119 )186040721679053580521557505720165108888601125670 · (1 + 2 151 )186040721679053580521557505720165108888601125670 · (1 + 2 203 )279061082518580370782336258580247663332901688505 n = 217 (1 + 2 13 )1126940817087752014462533930787563178937611179075 · (1 + 2 57 )125215646343083557162503770087507019881956797675 · (1 + 2 89 )50086258537233422865001508035002807952782719070 · (1 + 2 185 )751293878058501342975022620525042119291740786050 · (1 + 2 215 )9537309672323317621714046370301517484866488275 n = 219 (1 + 2 19 )2066997275827785054568851212103878502402165134820 · (1 + 2 63 )15659070271422614049764024334120291684864887385 · (1 + 2 107 )8267989103311140218275404848415514009608660539280 · (1 + 2 113 )861248864928243772737021338376616042667568806175 · (1 + 2 115 )2239247048813433809116255479779201710935678896055 · (1 + 2 139 )9802832608939360014892925802660670404346311615 · (1 + 2 189 )574165909952162515158014225584410695111712537450 n = 221 (1 + 2 9 )3989486683832532366484265709728099870565523986917322701595 · (1 + 2 19 )3989486683832532366484265709728099870565523986917322701595 · (1 + 2 29 )3989486683832532366484265709728099870565523986917322701595 n = 223 (1 + 2 17 )1014848095919801109402176209373936431577327592183 · (1 + 2 91 )1503478660621927569484705495368794713447892729160 · (1 + 2 103 )1691413493199668515670293682289894052628879320305 · (1 + 2 113 )563804497733222838556764560763298017542959773435 · (1 + 2 145 )2631087656088373246598234616895390748533812276030 n = 225 (1 + 2 29 )64638048753257449056214060125 · (1 + 2 31 )788456198653595814230254674000 · (1 + 2 49 )5797472048923498634045990250 · (1 + 2 51 )84063344709390730193666858625 · (1 + 2 85 )4479864765077248944490083375 · (1 + 2 97 )1809865365091208573573993683500 · (1 + 2 99 )365703246128755015876741500 · (1 + 2 105 )5302697068866947730212751750 · (1 + 2 115 )68066717742077353015043250 · (1 + 2 131 )5270429135384998758223627500 · (1 + 2 203 )152315402012626464112662834750 · (1 + 2 207 )828774981539291054730665424375 · (1 + 2 217 )438130774024554946771130154075 · (1 + 2 219 )94077160066622227834291750875 n = 227 (1 + 2 5 )59383142438657794704063431302361624145812280978139757521155099815 · (1 + 2 7 )118766284877315589408126862604723248291624561956279515042310199630 n = 229 (1 + 2 9 )147385697387219867509609565535361987370219507869771466051 · (1 + 2 21 )1768628368646638410115314786424343848442634094437257592612 · (1 + 2 121 )589542789548879470038438262141447949480878031479085864204 n = 231 (1 + 2 41 )459189704451443127647362594916155732882807466830218 · (1 + 2 55 )1567200356489566988557551518485173149770673948226 · (1 + 2 69 )2755138226708658765884175569496934397296844800981308 (1 + 2 91 )1147974261128607819118406487290389332207018667075545 · (1 + 2 131 )535721321860016982255256360735515021696608711301921 · (1 + 2 157 )918379408902886255294725189832311465765614933660436 (1 + 2 159 )1147974261128607819118406487290389332207018667075545 · (1 + 2 209 )2350800534734350482836327277727759724656010922339 n = 233 (1 + 2 15 )17038321523200602421013846948443997824061934891243478185488130 · (1 + 2 23 )2839720253866767070168974491407332970676989148540579697581355 (1 + 2 49 )2839720253866767070168974491407332970676989148540579697581355 · (1 + 2 113 )2839720253866767070168974491407332970676989148540579697581355 15 n = 235 (1 + 2 73 )23191222403616672243206263595493280420250247335633250 · (1 + 2 105 )26891949382917205047973220552220931551141244250893875 (1 + 2 119 )8141599354461172170487305304800832487960193213573375 · (1 + 2 141 )363659699007176721748930895080669461243705878250 (1 + 2 167 )14062762521342024658114436435565074297385788277990375 · (1 + 2 171 )23684652667523409950508524523056967237702380257668000 (1 + 2 197 )2556865912971277210566261170102740781342870595998250 n = 237 (1 + 2 9 )312620627852416761979153117722703908013332701360090433 · (1 + 2 169 )3647240658278195556423453040098212260155548182534388385 (1 + 2 173 )1042068759508055873263843725742346360044442337866968110 · (1 + 2 175 )173678126584675978877307287623724393340740389644494685 (1 + 2 199 )429087136268023006638053298835083795312417433239339810 · (1 + 2 225 )416827503803222349305537490296938544017776935146787244 (1 + 2 233 )67961006054873209125902851678848675655072326382628355 n = 239 (1 + 2 59 )137903930258717580167839711391793163262983138814261779866 · (1 + 2 73 )25172939650400828125875502873105101230544541212127150293 · (1 + 2 87 )206855895388076370251759567087689744894474708221392669799 · (1 + 2 197 )298791848893888090363652708015551853736463467430900523043 · (1 + 2 199 )91935953505811720111893140927862108841988759209507853244 · (1 + 2 237 )6566853821843694293706652923418722060142054229250560946 n = 241 (1 + 2 7 )916414411031455987244373856003197176202834172185057465573737668369009 · (1 + 2 9 )5498486466188735923466243136019183057217005033110344793442426010214054 n = 243 (1 + 2 23 )785013025560884300793728032156196770104389659275 · (1 + 2 31 )22335098635205076340574688781013129860710668130 · (1 + 2 35 )847419918806310249392392603750204044715198879050 · (1 + 2 61 )132039847814006480719279777793636444176554243945 · (1 + 2 151 )41135370735706828274307725135899531886322869175 · (1 + 2 155 )725890705644164981068677385382926720473096714225 · (1 + 2 163 )160944093106624814807082316216124023996297461525 · (1 + 2 177 )387579652787382207086443128846992547582920417550 · (1 + 2 181 )91968053203785608461189894980642299426455692300 n = 245 (1 + 2 69 )404534281273826986829987345146663806009193260698421162909645 · (1 + 2 117 )652474647215849978758044105075264203240634291449066391789750 · (1 + 2 125 )1096157407322627964313514096526443861444265609634431538206780 · (1 + 2 141 )1057008928489676965588031450221928009249827552147487554699395 · (1 + 2 151 )404534281273826986829987345146663806009193260698421162909645 · (1 + 2 165 )1578988646262356948594466734282139371842334985306740668131195 · (1 + 2 167 )404534281273826986829987345146663806009193260698421162909645 n = 247 (1 + 2) 134057357388441380704540286280333486890775035828909707815645 · (1 + 2 9 )130787665744820859223941742712520475015390278857472885673800 · (1 + 2 35 )16348458218102607402992717839065059376923784857184110709225 · (1 + 2 71 )85011982734133558495562132763138308760003681257357375687970 · (1 + 2 147 )81742291090513037014963589195325296884618924285920553546125 · (1 + 2 195 )15040581560654398810753300411939854626769882068609381852487 (1 + 2 97 )292527702190729434230102491312771097283901482612310325863937852070 · (1 + 2 119 )204769391533510603961071743918939768098731037828617228104756496449 · n = 249 (1 + 2 137 )585055404381458868460204982625542194567802965224620651727875704140 · (1 + 2 173 )633810021413247107498555397844337377448453212326672372705198679485 · (1 + 2 199 )536300787349670629421854567406747011687152718122568930750552728795 16 3 Further comments One can go further than our choice of bound, namely 250, by using our method. We have yet to encounter an exponent for which we cannot apply Algorithm 1. If there is such an exponent n for which we cannot find aj , di,j as above, then we can involve cubics in the factorization of −1 (mod 2 n − 1). More precisely, we do something similar as above using 2ai + 2 bi + 1 and check if we can find such powers ai, b i such that the number above is a quadratic nonresidue modulo qi and quadratic residue modulo qj for all j 6 = i. The rest of Algorithm 1 runs unchanged. While, via Carlitz’ result, we know that any permutation can be decomposed as a composition of inverses and affine functions, it would also be interesting to check whether one can modify our method in this paper to other exponents, other than the inverse and surely, the Gold 2 k + 1 exponents, to directly find their decomposition in quadratics, or quadratics and cubics, and we leave that to future work and the interested reader. Acknowledgements. The first and the third-named authors worked on this paper during visits to the Max Planck Institute for Software Systems in Saarbr¨ ucken, Germany in Spring of 2022 and 2023. They thank Professor J. Ouaknine for the invitation and the Institute for hospitality and support. During the final stages of the preparation of this paper, the first-named author was a fellow at the Stellenbosch Institute for Advanced Study. He thanks this Institution for hospitality and support. References L. Carlitz, “Permutations in a finite field”, Proc. Amer. Math. Soc. 4 (1953), 538. R. T. Hall and G. Tenenbaum, Divisors , Cambridge Tracts in Mathematics, 90 .Cambridge University Press, Cambridge, 1988. 17 A. Kontorovich and J. Lagarias, “On toric orbits in the affine sieve”, Exp. Math . 30 (2021), 575–587. F. Luca and P. St˘ anic˘ a, “Asymptotics on a class of S-unit integers”, to appear in Periodica Math. Hungarica . F. Luca and P. St˘ anic˘ a, “Prime divisors of Lucas sequences and a conjecture of Skalba”, Int. J. Number Theory 1 (2005), no. 4, 583–591. P. Moree, “On the divisors of ak + bk”, Acta Arith. LXXX.3 (1997), 197–212. L. Murata and C. Pomerance, “On the largest prime factor of a Mersenne number”, in Number Theory , 209–218, CRM Proc. Lecture Notes, 36, Amer. Math. Soc., Providence, RI, 2004. S. Nicoka, V. Nikov, V. Rijmen, “Decomposition of permutations in a finite field”, Cryptogr. Commun. 11 (2019), 379–384. G. Petrides, “On decompositions of permutation polynomials into quadratic and cubic power permutations”, Cryptogr. Commun. 15 (2023), 199–207. A. Rotkiewicz, “Applications of Jacobi’s symbol to Lehmer’s numbers”, Acta Arith. 42 (1983), 163–187. 18
14581	https://physics.stackexchange.com/questions/640876/qualitatively-brewster-angle-explanation-with-dipoles	optics - Qualitatively Brewster angle explanation with dipoles - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Qualitatively Brewster angle explanation with dipoles Ask Question Asked 4 years, 3 months ago Modified4 years, 3 months ago Viewed 828 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. The qualitatively explanation of the Brewster angle (e.g., see is that incident light produces small electric dipoles on the surface of the dieletric medium (glas), and these dipoles produce the reflected light. However, a dipole doesn't emit in the direction of the oscillation axis. If the incident light is p p-polarized and hits the glass under the Brewster angle α B r e w s t e r=a r c t a n(n 2 n 1)α B r e w s t e r=a r c t a n(n 2 n 1) the dipole axis is in the direction of reflection and thus no light is reflected in this direction. When testing it with glass it works perfectly. However, when testing it with a mirror, it doesn't work at all. My initial expectation was that p p-polarized incident light under 45∘∘ angle should be reflected in −45∘−45∘, which is also the direction of the dipole axis. So there shouldn't be reflected light. But there is... (measured it). This was also asked here Why do mirrors not follow brewster's angle? and the answer suggested that the refractive index of a metal is almost infinity (n 2≈∞n 2≈∞), meaning that the Brewster angle is 90∘90∘, which is of course no practical incidence angle. Unfortunately, this doesn't explain physically what is happening... My explanation at the moment is this: Because the refractive index of metal is so large, the angle of refraction according to Snell's law os 0∘0∘ (so straight into the mirror...), and thus the axes of the induced dipoles on the mirror surface are all within the mirror plane (independent on the polarization or incidence angle of the incident light). And thus there is no polarization effect of the reflected (re-emitted) light. Is that explanation correct, resp. is that the mechanism how a mirror works at all? optics reflection refraction Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Jun 2, 2021 at 16:21 Charles Tucker 3Charles Tucker 3 1,868 7 7 silver badges 21 21 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. First, you confuse permittivity with refractive index. The refractive index of silver for the visible wavelengths is almost 0. Secondly, all laws still apply (ie Maxwell eqs). However, I think its easier to think of it this way: metals have conducting electrons, and they oscillate mostly freely. They are not bound in a potential which affects their mobility and thus, also their optical properties. Reflection of metals is more akin to some plasma currents than to the thoughts of dipoles like in dielectrics, electrons respond immediately to "any" applied field (of course, metals still have an atomic structure which affects different frequencies differently) and generate electrical currents, while dielectrics do not. If you go to this website, it immediately shows you Silver, for the green wavelength. If you scroll down, you can find the curves for p- and s- reflection for different incident angles. You can see the typical brewster's angle dip is around 75°, except the difference between s- and p- reflection percentages are 99% and 97% respectively. Hope this can help. (But of course, just like antennas, they generate small dipoles, except they are on the surface and along the E-field lines) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 3, 2021 at 9:49 answered Jun 3, 2021 at 9:43 José AndradeJosé Andrade 1,183 6 6 silver badges 16 16 bronze badges 5 Many thanks, that helps! To conclude: 1) Yes, I confused refractive index and permittivity... However, the answer referenced in my question <physics.stackexchange.com/questions/453820/…> is apparently incorrect, although the argumentation there is indeed with dielectric constant. Could you confirm? 2) So the dipole-explanation is in general not good for mirrors, only for dielectric materials?Charles Tucker 3 –Charles Tucker 3 2021-06-03 10:37:51 +00:00 Commented Jun 3, 2021 at 10:37 Ok, maybe I wasn't clear. Dipoles still apply, that's their definition, oscillating electrons. Just that in dielectrics they are bound in a potential, which in turn has a restitution force to the movement, effectively shaping their movement direction, range, etc. In a metal, the electrons are free, they oscillate directly with the field with almost no constraints. And all electromagnetic constants and the way they shape electromagnetic propagation are well defined and do not care if its a dielectric or metal, thats why refractive index of metals <<1 and in dielectrics >1, their formalism (1/2)José Andrade –José Andrade 2021-06-03 11:08:43 +00:00 Commented Jun 3, 2021 at 11:08 is exactly the same in both cases. Just the reason and how these constants arise is different. In dielectrics J J will be 0, in metals, that is not the case. I might also be glossing over things a bit too generally, and not in a very precise manner. I would instead challenge you to take a book on classic electrodynamics and find how maxwell eqs are used to derived metal and dielectric properties.José Andrade –José Andrade 2021-06-03 11:13:27 +00:00 Commented Jun 3, 2021 at 11:13 Yes, I will do so. It's a bit embarrassing but after all this years I realize that I don't know how a mirror actually (physically) works... And I don't want to be pushy with the dipoles, but is the equation for the Brewster angle above applicable to mirrors as well? Because with n=0.05 n=0.05 for silver (from the link you provided) the angle would be 2.8∘∘, so basically the frontal reflection should be polarized (hard to believe). Gold (same webpage) has n=0.27 n=0.27 (15∘∘) and Aluminium n=1.19 n=1.19. So this should be easy to observe (which I did not in my measurements...).Charles Tucker 3 –Charles Tucker 3 2021-06-03 15:46:04 +00:00 Commented Jun 3, 2021 at 15:46 No need to apologize. Well, although I am an optics expert, my fundamental theory is long long gone. My feeling is that the brewster formula applies because of how the refractive index arises in dielectrics, however (again, my feeling) it should not work for metals. Refractive index shows the phase shift of light, and its underlying physics are different in a conductor and in a dielectric, but of course both materials display measurable phase shifts. I am sorry, but my time is limited, but I would look into Fresnel eqs and Maxwell eqs and try to understand how EM fields and electrons interact.José Andrade –José Andrade 2021-06-03 16:59:11 +00:00 Commented Jun 3, 2021 at 16:59 Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 1Why do mirrors not follow brewster's angle? Related 0Microscopic understanding of Brewster angle 1Brewster's Angle 3What is the difference between horizontally/vertically polarized light and s/p polarized light? 1Why do mirrors not follow brewster's angle? 1Collective dipole oscillation and specular reflection 2Polarization of light and Brewster's angle in internal reflection 2Does a retroreflector depolarize light or preserve polarisation? Hot Network Questions Is it safe to route top layer traces under header pins, SMD IC? What "real mistakes" exist in the Messier catalog? Why include unadjusted estimates in a study when reporting adjusted estimates? Why do universities push for high impact journal publications? 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14582	https://ophysics.com/f3.html	\| \| \| oPhysics: Interactive Physics Simulations Atwood's Machine & Inclined Plane with Friction \| \| Description This is a simulation of two objects attached to each other with a massless string. The string passes over a massless, frictionless pulley. Use the "Run" button to start the simulation, the "Pause" button to pause it, and the "Reset" button to reset the time back to zero. Use the sliders to adjust the masses of the two objects, the angle of the incline, and the coefficient of friction between mass m2 and the incline (in the simulation it is assumed that the static and kinetic friction coefficients have the same value). Use the checkboxes to show or hide the numerical values and the free body diagrams for the two objects. \| Click here to donate to oPhysics.com to help keep the site going
14583	https://people.richland.edu/james/lecture/m170/ch07-bin.html	Stats: Normal Approximation to Binomial Stats: Normal Approximation to Binomial Recall that according to the Central Limit Theorem, the sample mean of any distribution will become approximately normal if the sample size is sufficiently large. It turns out that the binomial distribution can be approximated using the normal distribution if np and nq are both at least 5. Furthermore, recall that the mean of a binomial distribution is np and the variance of the binomial distribution is npq. Continuity Correction Factor There is a problem with approximating the binomial with the normal. That problem arises because the binomial distribution is a discrete distribution while the normal distribution is a continuous distribution. The basic difference here is that with discrete values, we are talking about heights but no widths, and with the continuous distribution we are talking about both heights and widths. The correction is to either add or subtract 0.5 of a unit from each discrete x-value. This fills in the gaps to make it continuous. This is very similar to expanding of limits to form boundaries that we did with group frequency distributions. Examples Discrete Continuous x = 6 5.5 < x < 6.5 x > 6 x > 6.5 x >= 6 x > 5.5 x < 6 x < 5.5 x <= 6 x < 6.5 As you can see, whether or not the equal to is included makes a big difference in the discrete distribution and the way the conversion is performed. However, for a continuous distribution, equality makes no difference. Steps to working a normal approximation to the binomial distribution Identify success, the probability of success, the number of trials, and the desired number of successes. Since this is a binomial problem, these are the same things which were identified when working a binomial problem. Convert the discrete x to a continuous x. Some people would argue that step 3 should be done before this step, but go ahead and convert the x before you forget about it and miss the problem. Find the smaller of np or nq. If the smaller one is at least five, then the larger must also be, so the approximation will be considered good. When you find np, you're actually finding the mean, mu, so denote it as such. Find the standard deviation, sigma = sqrt (npq). It might be easier to find the variance and just stick the square root in the final calculation - that way you don't have to work with all of the decimal places. Compute the z-score using the standard formula for an individual score (not the one for a sample mean). Calculate the probability desired. Table of Contents
14584	https://www.youtube.com/watch?v=wOrxe1VCrUk	Proof p^2==1 (modulo 24) for all prime numbers greater than or equal to 5 My One Fiftieth Of A Dollar 1480 subscribers 5 likes Description 501 views Posted: 22 Sep 2022 Credit to the vast intellect of hill runner mathematician Seamus O'Boyle for pointing out this "nice little number theoretic" result at the non country music version of Michael Penn's math channel. The talented Edinburgh Mathematics brain(Thank you Francesca Iezzi for not excluding parents in the educational teacher/student learning engagement loop) and coding guru could have chosen any career he wanted including service to his community in the classroom. I believe professor Penn did video which tangentially noted the strong correlation between the affinity for math and chess simultaneously(Susan Polgar comes to mind). Modular arithmetic solved two congruences with relatively prime moduli of 3 and 8 to show p squared is congruent to 1 (mod 24) Transcript: splitting the congruence to be proven into equivalent system of two congruences where the moduli are relatively prime was super useful here! The primes 2 and 3 are excluded since 2^2==4(mod 8) and 3^2==0(mod 3)
14585	https://math.stackexchange.com/questions/647732/how-to-teach-original-price-percentage-discount-problems-to-a-12-year-old-no-al	How to teach original price percentage discount problems to a 12 year old? No algebra please! - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to teach original price percentage discount problems to a 12 year old? No algebra please! Ask Question Asked 11 years, 8 months ago Modified11 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This is for a 12 year old who is not good with finding percentages as it is, but who needs to be able to work out the original price of some sold goods if they know the new price and the discount applied. e.g. A hat now costs €6.30 after a 10% reduction, how much was it originally (X). i.e. X=6.3 1−10 100=7 X=6.3 1−10 100=7 How can I explain that without any algebra or mention of such? algebra-precalculus Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jan 22, 2014 at 17:18 HCAIHCAI 359 1 1 gold badge 2 2 silver badges 14 14 bronze badges 2 1 The formula you have there could be good way to explain it... (at least, in my opinion)apnorton –apnorton 2014-01-22 17:24:06 +00:00 Commented Jan 22, 2014 at 17:24 2 Consider that your proscription against algebra might actually be making it harder to explain.user14972 –user14972 2014-01-22 17:47:14 +00:00 Commented Jan 22, 2014 at 17:47 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. You can try to explain your formula for X X in terms of it being the ratio of the current price (after discount) with respect to the fraction of the original price that the hat is now selling for (1−0.1)=0.9(1−0.1)=0.9, where 1 1 is taken to be 100%100% of the original price. (Of course, this may need some "reminder" prompts about navigating between how to represent a fraction/decimal as a percentage, and vice versa.) The ratio's value will not change: 6.3 0.9=7 1 6.3 0.9=7 1 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 22, 2014 at 17:33 amWhyamWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges 1 with the word "ratio", one opens another can of worms entirely. ;)HCAI –HCAI 2014-01-22 20:19:04 +00:00 Commented Jan 22, 2014 at 20:19 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. A method I found useful for students starting out: (thinking in terms of "units") Using the example, After the 10% reduction we have 90%. 90% of the hat costs 6.3. 10% of the hat costs 6.3 9 6.3 9. Hence, 100% of the hat costs 6.3 9⋅10=7 6.3 9⋅10=7. Eventually we want to proceed to the more general formula (like what you have) to handle "weirder" numbers (3.142%?) but this method could be a good start. When I guide students towards the more general formula one common mistake is confusing between multiplying or dividing by a certain percentage. I found explaining in terms of whether the costs should be larger or smaller helped. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 22, 2014 at 17:32 Kelvin SohKelvin Soh 1,825 12 12 silver badges 15 15 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I can only report on how my sister does it. They defined 3 different variables. One for the original price, one for the new price and one for the percentage. They then learned different formulas, by heart, without understanding the meaning, to compute the missing value, given the other two values. Personally I find this whole approach very unsatisfying because they don't understand the underlying principle but that's how their teacher explained it. Maybe you could try giving her/him at least a way to approximate the result. Imagine two people buying the product, one after the discount, one before. The difference between both spendings has to be 10 percent of the original price. Now let us try some value: If the original price was 8Euro then she should have saved 80 cents but this means the product would have costed 7.20, so 8 was too high, try again with 7.50, still to high, go lower, and so on, until you get to 7 where it magically works. Or start teaching her/him how to solve terms for x by always doing the same on each side of the equation. I guess this would be a good approach in the long run but right now she/he might only get confused. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 22, 2014 at 17:39 user667804user667804 231 1 1 silver badge 7 7 bronze badges 1 1 I think I'm going to do try both first and second option and see which they prefer. They do a lot of guestimation and correction these days so the second option would probably fit in well.HCAI –HCAI 2014-01-22 20:24:50 +00:00 Commented Jan 22, 2014 at 20:24 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I suggest that you offer the 12-year-old the regula falsi method. In this method, you guess, check if the guess is correct, and if it isn't, you adjust the guess. Let's try a slightly different problem: A hat now costs €6.39 after a 10% reduction. How much was it originally? The original price must have been a little more than €6.39, so we begin by guessing that it was around €7 before. But 10% of €7 is €0.70, so had it been €7 before, the price after reduction would be €7−€0.70=€6.30€7−€0.70=€6.30. This is too small by €0.09. €0.09 is small, so we must be very close. What increase in the original price would cause the discounted price to increase by €0.09? It must be a bit more than €0.09, so try increasing the supposed original price by €0.10, to €7.10. On checking, this is in fact the answer. This method is self-correcting. Suppose the 12-year-old tried increasing the guess by €0.20, to €7.20. This would give a discounted price of €7.20−€0.72=€6.48€7.20−€0.72=€6.48, which is too high. So the correct answer must be somewhere between €7.00 and €7.20. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 22, 2014 at 17:36 MJDMJD 67.8k 44 44 gold badges 309 309 silver badges 621 621 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 4Algebra equation for percentage increase needed to get the current value as a 20% discount 5How do I start from a 10% discount and find the original price? 23How to explain to a 14-year-old that √(−3)2(−3)2−−−−−√ isn't −3−3? 0How to find percentage change between original and final BPM 1how to tackle such word problems of algebra? 0A merchant gives a discount of 10% on the cost price of the tea, but uses a weight of 900 gm per kg. Find his net profit/loss percentage 1The ratio of cost price and marked price is 2:3 and ratio of percentage profit and percentage discount is 3:2. 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14586	https://artofproblemsolving.com/wiki/index.php/Talk:Rearrangement_inequality?srsltid=AfmBOoqi0Ph1Y4hKjnReYSW2ZySp8uYW9j0H8q1tP2rXBbyMh2UbXJb9	Art of Problem Solving Talk:Rearrangement inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Talk:Rearrangement inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Talk:Rearrangement inequality AoPSWiki Article of the Day Rearrangement inequality was the AoPSWiki Article of the Day for December 13th, 2007 The uses section should be partitioned into information for the greedy algorith, intermediate, and olympiad sections. A few examples will add worlds of clarity for readers.--MCrawford 01:24, 20 June 2006 (EDT) I think that "similarly sorted" has not been defined very well here. For example, the sequences {1,1,1,1,1,1}; {1,2,3,4,5,6} are similarly sorted, but the fifth 1 in the first sequence is greater than or equal to five other terms, whereas the 5 in the second sequence is only greater than or equal to four other terms. At the moment, I can't think of a good definition for "similarly sorted" that doesn't take up a great deal of space. Perhaps we should create a page for similarly vs. opposite sorting, and link to that? After all, it's kind of easy to tell what it means, and an external link would not, I think, be a severe inconvenience. Or could someone else write a more precise (but still concise) definition? Retrieved from " Category: Article of the Day Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
14587	https://www.ohiocriminaldefenselawfirm.com/blog/what-is-the-role-of-the-defense-in-a-criminal-case/	What Is the Role of the Defense in a Criminal Case? - DiCaudo, Pitchford & Yoder About Areas Served Attorneys Thomas M. DiCaudo Marshal Pitchford J. Reid Yoder Brittany Fitzgibbon Seth Marcum Practice Areas Criminal Defense Domestic Violence Drug Crimes Drug Possession Drug Trafficking 5th Degree Drug Crimes OVI Drug OVI License Suspension Multiple OVI Offenses Underage DUI/OVI Vehicular Manslaughter Sex Crimes Child Pornography Rape Defense Sexual Abuse Federal Crimes Juvenile Defense Probation Violations Theft Crimes Burglary Robbery Violent Crimes Assault and Battery Child Abuse Murder & Manslaughter White Collar Crimes Fraud Reviews Case Results Blog Contact Us 330-762-7477 Blog Home / Blog / What Is the Role of the Defense in a Criminal Case? Recent Posts What Is Considered Sexual Assault in Ohio? Can a Felony Drug Charge Be Reduced to a Misdemeanor in Ohio? How Much Does an OVI Cost in Ohio? Breath & Blood Tests Field Sobriety Tests Categories Assault Checkpoint Criminal Defense Criminal Mischief Domestic Violence Domestic Violence Archives Domestic Violence Cases Drug Charges Drug Crimes Drugged Driving DUI Featured Stories Federal Criminal Defense Felony Firm News In The News Juvenile Crimes Archives Juvenile Law Marijuana Legalization News OVI Privacy Probation Recreational Marijuana Search and Seizure Search warrants Sex Crimes Sex Crimes Archive Sexual Assault The Fourth Amendment What Is the Role of the Defense in a Criminal Case? Criminal Defense January 23, 2025 Are you facing criminal charges in Akron, Ohio? If so, you might have questions about the role acriminal defense attorney could play in your case. Here’s an overview of the defense’s responsibilities during each phase of a criminal case. Table of Contents Toggle Building a Defense Strategy Pre-Trial Responsibilities of the Defense The Role of the Defense During Trial The Role of Defense Investigators and Experts Post-Trial and Appeals Contact a Trusted Defense Attorney in Akron, Ohio Building a Defense Strategy Defense lawyers start by reviewing every detail of a criminal case to build solid defense strategies. They gather evidence, such as documents, photos, and video recordings, to back up their client’s side of the story. They speak with witnesses and ask detailed questions to uncover facts that strengthen their defense. They also review the prosecution’s evidence to identify weaknesses in the government’s case. The defense gathers all these facts and prepares arguments to create a defense plan that protects the client’s rights and addresses the prosecution’s claims. Pre-Trial Responsibilities of the Defense Next, before the trial begins, the defense lawyer handles several key tasks. They represent their client at bail hearings and present arguments that could lower the defendant’s bail or allow their release. They might also discuss plea deals with the prosecutor, which can sometimes result in reduced charges or lighter sentences. Additionally, defense lawyers can file motions with the court to exclude improper evidence or limit the prosecutor’s case. These pre-trial steps can significantly affect the trial’s outcome, and a strong pre-trial strategy can strengthen a client’s position in court. The Role of the Defense During Trial During a criminal trial, the defense lawyer actively challenges the prosecutor’s evidence and arguments. They cross-examine the prosecutor’s witnesses to identify inconsistencies or highlight facts that support the defense. They also present evidence and call witnesses who strengthen their client’s case. The defense lawyer makes both opening and closing statements to the jury to summarize their arguments and reiterate key points. Defense attorneys question evidence, present strong arguments, and respond to prosecutor’s cases to achieve the most favorable results possible for their clients. The Role of Defense Investigators and Experts Defense lawyers often rely on investigators and experts to gather present information that strengthens their cases. Defense investigators might interview new witnesses or locate evidence that supports a client’s story. They might also uncover details that challenge the prosecution’s claims. Experts can provide specialized knowledge on complex topics like DNA analysis, accident reconstruction, or mental health. Defense attorneys hire these professionals to add detailed insights and facts that might otherwise remain unknown. This added information can play a major role in building a strong defense. Post-Trial and Appeals If a trial ends with a conviction, a defense lawyer can work on their client’s appeal. In an appeal, the lawyer identifies possible mistakes that occurred during the trial and files documents to challenge the verdict or sentence. For instance, they might argue that the judge allowed improper evidence or that their client’s rights were violated. If the appeal succeeds, the court might reduce the sentence or even order a new trial. If an appeal isn’t possible, defense attorneys can still explore other options, such as requesting a lighter sentence or seeking new evidence that could impact a trial’s outcome. Contact a Trusted Defense Attorney in Akron, Ohio If you or a loved one faces criminal charges, callDiCaudo, Pitchford & Yoder today. Our team offers a free initial consultation to discuss your case and explain how we can help. Don’t face this process alone—contact us now to get experienced legal support on your side. 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14588	https://chemistry.stackexchange.com/questions/40024/calculating-the-charges-of-anions-and-cations	ions - Calculating the charges of anions and cations - Chemistry Stack Exchange Join Chemistry By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Calculating the charges of anions and cations Ask Question Asked 9 years, 11 months ago Modified9 years, 11 months ago Viewed 16k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I am new to chemistry. Could someone explain me how to calculate the charges on anions and cations. I've heard about some crossing method (the left subscript becomes the right superscript and the right subscript becomes the left super script.) Example: R b X 2 S⟶2 R b X++S X 2−R b X 2 S⟶2 R b X++S X 2− But what about: K X 2 M n O X 4 K X 2 M n O X 4 and Z n(C l O)X 2 Z n(C l O)X 2 This trick doesn't seem to apply here, so how can I find out the charges? ions Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 2, 2015 at 1:11 Melanie Shebel 6,606 11 11 gold badges 50 50 silver badges 91 91 bronze badges asked Nov 1, 2015 at 10:16 privetDruziaprivetDruzia 302 1 1 gold badge 5 5 silver badges 12 12 bronze badges 1 2 @Close flagger, I don't think this is a homework question. It looks more like questioning a [stupid] rule resulting from oversimplification to me.M.A.R. –M.A.R. 2015-11-01 16:07:29 +00:00 Commented Nov 1, 2015 at 16:07 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Some basic knowledge of the atoms in the periodic table is necessary to extract the ionic charges. In the case of K X 2 M n O X 4 K X 2 M n O X 4, for example, we know that the potassium ion is positive, and that the manganate ion is negative. We see that the compound is neutral: has zero net charge. So how can we place charges to get the molecule to be neutral, given the subscripts? Potassium is a Group 1 element, meaning it only has 1 valence electron, placed in the 4s orbital. Potassium is eager to lose this electron to fulfil the octet rule. Therefore potassium ions have one less electron than protons, and the charge is +1. Since we have two K X+K X+ ions, the manganate ion must have a charge of -2 to make sure the molecule has zero net charge. K X 2 M n O X 4↽−−⇀H X 2 O 2 K X++M n O X 4 X 2−K X 2 M n O X 4↽−−⇀H X 2 O 2 K X++M n O X 4 X 2− Zinc is one of two transition metals that only have one (natural) oxidation state: +2. The electron configuration is [Ar]3d 10 10 4s 2 2. When zinc becomes a cation, it loses the two 4s electrons. Knowing this, we see that the hypochlorite must have a charge of -1: the two hypochlorite -1 charges counter the 2+ charge of zinc. Z n(C l O)X 2↽−−⇀H X 2 O Z n X 2++2 C l O X−Z n(C l O)X 2↽−−⇀H X 2 O Z n X 2++2 C l O X− I recommend not relying on simple rules, such as the one you mentioned. Rather, try to understand why the rule came about, and understand when the rule fails, and why. It really helps to be able to "read" the periodic table, as a lot of basic information is incorporated into it. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 1, 2015 at 12:04 YodaYoda 4,833 6 6 gold badges 33 33 silver badges 79 79 bronze badges 4 Potassium problem: Is there a way to know that Mn is 2- except by deduction like you did?privetDruzia –privetDruzia 2015-11-01 13:43:20 +00:00 Commented Nov 1, 2015 at 13:43 @privetDruzia Each potassium ion has to have a 1+ charge, so x x in [M n O X 4]X x−[M n O X 4]X x− has to be 2.orthocresol –orthocresol 2015-11-01 13:59:01 +00:00 Commented Nov 1, 2015 at 13:59 Thx. This semi-deductive method seems to work indeed. But not for more "complex" molecules like: KnaCO3 or Ba(BrO2)NO2, ... Or am I wrong?privetDruzia –privetDruzia 2015-11-01 14:05:51 +00:00 Commented Nov 1, 2015 at 14:05 This may not be the best reply, but you should just memorize the charge of common ions. There are ways of estimating the charge in molecular ions, such as "Formal charge". The formal charge assumes, however, that all bond electrons are shared equally, regardless of electronegativity differences, so such an approach may not always be appropriate. In quantum mechanics, the electrons don't necessarily belong to a single atom, but rather the whole molecule. Thus, fractional charges are more common than integer charges. See this pdf: chem.ucla.edu/harding/tutorials/formalcharge.pdfYoda –Yoda 2015-11-01 15:37:24 +00:00 Commented Nov 1, 2015 at 15:37 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Elements like K, Li, Cs and Rb can only have +1 charge on them (Due to very high second ionization energy). Also, they can't have negative charge due to their low electron gain enthalpy (they are metals and are electropositive in nature). Also, Oxygen can only have -2 (in oxides, eg: H2O), -1 (in peroxides, eg: H2O2) and -0.5 (in superoxides, eg: KO2). Also, H can only have +1 or -1(if H is more electronegative of the two elements, as in the case of hydrides, eg: LiH) and Fluorine can have only -1 (since it is the most electronegative element). Similarly, Ca, Mg.. have +2 charge. You need to remember these basic rules and a few others (refer to any inorganic chemistry text book and you can find them), to calculate the charges. Take the case of K2MnO4 : K has +1. O has -2 (in is an oxide). Now, let the charge on Mn be x. So, x + 2(+1) + 4(-2) = 0 (Since the net charge on the species is zero). Hence, x = +6. I hope this answer helped. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 1, 2015 at 12:13 ShankRamShankRam 1,173 1 1 gold badge 7 7 silver badges 18 18 bronze badges 3 Note that alkali metals have been reported to have a negative oxidation state though.M.A.R. –M.A.R. 2015-11-01 12:24:47 +00:00 Commented Nov 1, 2015 at 12:24 1 Also, oxygen has an oxidation state of +2 in oxygen difluoride.M.A.R. –M.A.R. 2015-11-01 12:28:45 +00:00 Commented Nov 1, 2015 at 12:28 I clearly mentioned these are the few basic rules and the person who asked this question has to refer a book to know all the exceptions and other such rules.ShankRam –ShankRam 2015-11-02 13:51:52 +00:00 Commented Nov 2, 2015 at 13:51 Add a comment\| Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! 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14589	https://fnoschese.wordpress.com/2011/12/06/a-graph-to-visualize-average-velocity/	A Graph to Visualize Average Velocity Note: This is an expansion of the today’s Noschese 180 post. I thought it was too good not to share here. We started Constant Acceleration in college-prep today. Rather than dive right in with carts and motion detectors, I propped up one end of a lab table with textbooks (best use ever) and let a C-battery roll down. (Batteries accelerate more slowly than marbles and hot wheels cars. They also roll much straighter.) “What do you observe?” I asked “It rolls down and gets faster.” they said. “Prove it. You have 10 minutes.” I challenged them. I hate prescribing directions for activites like this. I want to see how my students approach these tasks. They wanted stopwatches and metersticks. Some wanted tape. One group wisely rolled the battery down a whiteboard and left marks at one second intervals. They were done in 2 minutes. The other groups marked out equal intervals of distance to time with a stopwatch. Most groups made data tables to show that it takes less time to travel each successive distance interval, thereby showing it continously increases in speed. Many groups added a velocity column and calculted the “velocity” for each interval to show it changes. (But velocity when? where? average? I didn’t want to go down that road just yet. I just let it be.) Some groups went further and also made distance-time graphs of their data to show the slope increases. Two groups went even further and added an average-velocity step graph like this one: It was beautiful. And something I had never considered doing. You see, over the years, I’ve tried a variety of acceleration labs. Kids would collect position-time data and make position-time and velocity-time graphs. And getting the velocity-time graph was always laborious. Here are some methods I’ve tried in years past… Method 1: Manually draw tangent lines on the position-time graphs. Calculate and graph the slopes of the tangent lines. (Tedious) Method 2: Use the slope tool in Logger Pro to get the slope of the tangent at each data point. Graph the slopes of the tangent lines. (Computer issues) Method 3: Kids calculate the average velocity for each distance/time interval. Tell them to graph it at the midpoint in time. This typically involved a lot of hand waving b/c kids didn’t quite understand why at the midtime rather than the end time. And I’d still have groups that would incorrectly graph the average velocity at the end time. One time I made a data table worksheet to avoid this issue — but it was scary table with rows in between rows for midtime data. Method 4: Method 3, but using Excel (OMG, what was I thinking?) The average velocity step-graph method is perfect. It doesn’t matter how the students took the data. They calculate the average velocity for each interval, then graph each average velocity as a step that is as long as the interval. No need to handwave about midtimes. No need to assume the acceleration is constant. The board pictured above inspired me, so I had all groups make their own average velocity step graph as well, just to see if it would work. “Is this how the velocity-time graph really looks?” I asked. “No. There wouldn’t be any steps. It would be a line. Or a curve.” they said. They made the leap on their own to draw a line through the steps. And, lo and behold, the “best fit line” cuts through the middle of each step — the midtime. You can’t miss it. A great visualization. Kids who took data at equal time intervals had equal sized step-widths and step-heights. Kids who took data at equal distance intervals had unqual step-widths and step-heights (the steps got narrower and shorter over time — which in a data table looks like non-constant acceleration). But the line still cut through the midtime of each step. Now we can talk about why that happened and what that means AFTER, rather than all the handwaving and number crunching first. Several graphs also got a y-intercept, which we chalked up to reaction time error. I love it when I learn from kids! UPDATE: There’s a mistake in the step-graphs here. Read my follow-up post “A Mistake Made in Haste.” Sorry! Share this: Related 12 responses to “A Graph to Visualize Average Velocity” “Prove it. You have 10 minutes.” – I love it. I’d never thought of looking at average velocity as a step graph before either. Thanks for sharing. Love that line about the textbooks! This is AWESOME! Already thinking of connections with chemical Kinetics. Thanks! I absolutely love the “You have 10 minutes to prove it” idea. My students get a similar idea to your step graph when we use the very old-fashioned ticker tape timers. We make them cut the tape every 5 or 10 dots and stick their strips of tape side by side. They get a bar chart which is basically the same as your average velocity graph. Totally agree, Frank. My students have been doing the step graph for a few years now — it arises naturally from the CVPM unit — and it is very effective for them. Great post. The line about the use of textbooks reminded me of this even better description: “Returning to teaching after raising a family, [a teacher] said, she had her husband and son bringing in some of her favorite equipment to the new classroom, including a sand table and a water table. Seeing this, her principal spoke to her nervously: “You know, Mrs. ______, our school has a strong academic emphasis.” “Oh yes,” she replied, “I couldn’t teach physics without this equipment!” She later found piles of fill-in-the-blanks workbooks in her closet. Asking if she would be required to use them, she was told indeed she must. “So,” she said, “we used them. We weighed them singly and by twos and threes. We weighed them dry and we weighed them wet.”” (Hawkins, 1990) Hawkins, D. (1990). Defining and Bridging the Gap. In SCIENCE EDUCATION: A Minds-On Approach for the Elementary Years (pp. 97-139). Hillsdale, New Jersey: Lawrence Erlbaum Associates. Awesome. We used our AP books last spring as the counterwieght for our trebuchet. It very helpfully saved on materials costs. Pingback: A Mistake Made in Haste \| Action-Reaction That’s also all I used the textbook for in my class last year. :oP Is there ever the issue of science literacy in your class or school? As much as I’d like to be proud of not using the textbook, it’s mostly because my students’ literacy levels aren’t quite at that level, and ideally, I would like to help them get to a level where they can learn something from the book. I know physics is not learned well from a book, even if you can comprehend the paragraphs (best case, “clear and easy to understand” like a KA video, rite? :oP heh), so it’s not like I’d try to teach them Newton’s 2nd law with that, but maybe just read and understand a passage on how Newton’s laws apply in the engineering industry, or one of those other tidbits in the book. I know you do a lot of the “sexy” stuff in your class (which is ironic, because to outsiders, modeling is mundane and KA is sexy), but is there also “mundane” stuff that goes on in your class that helps build literacy around science? I’m not saying whether you should or not, but just wondering. Hi Frank, I’ll admit that I haven’t done much to promote literacy directly in my classes. But now that NY adopted the Common Core Standards, we have been taking science department meeting time to explore the reading and writing standards and to see how well our courses align with them. Some ideas I’ve had: Current events reading and responses in writing (LHC, Astronomy, Medical Physics, etc.) Socratic seminars around pieces from these 4 great books about experiments and model building in science: – The 10 Most Beautiful Experiments (Johnson) – The Prism and the Pendulum (Crease) – Great Experiments in Physics (Shamos) – What’s the Matter (Whitfield) Pieces from popular science writing: – A Brief History of Everything (Bryson) – The Physics of Superheroes (Kakalios) – The Physics of the Impossible (Kaku) – Why Does E=mc^2 (Cox) – How to Teach Physics to Your Dog (Orzel) – Six Easy Pieces (Feynman) – Demon Haunted World (Sagan) – The Physics of ____________ (there are so many of these books now). But now how do I assess those assignments? Feedback only, no grade? Do I add reading and writing standards to my SBG system? (I can’t just assign point values, you know :)) And what’s the purpose of reading/writing? To get information we’ll use later? To broaden our horizons? To deepen our backgroud? To write better? Who’s going to read their responses? Me? Other peers? Public on the web? So overwhelming…. Ooh, that’s quite the list. I’m also considering having them do shorter readings like current events or articles, or “how stuff works” type of articles. The only attempt at writing I did so far this semester was have them write their own experiment procedures for finding out if heavy things fall faster. If I remember, I’ll post my rubric and some student work samples, but it’s definitely far from perfect, as it’s my debut to any sort of writing and one of few examples of creative work. My attempt at making it more authentic was to give their procedures to another period to follow. If it’s unclear, it may be misinterpreted by another group, giving real consequences to their miserable attempt at communication. The irony and difficulty was asking my students to create a clear cut cookbook lab when I’ve never modeled such a thing for them because I half ask them to come up with their own procedures all the time. Oh, as for grading, I tried to take from your experimentation standards, and make one of them “I can create an experiment procedure that can be repeated consistently by other parties” or some similar (or way different heh) wording. I only gave them credit for perfect papers. However, they can revise it as many times as they want. Pingback: The future of textbooks « The Official Edufy Blog Leave a comment Cancel reply Δ Search this blog Subscribe TEDx NYED RSS 180 Photo Project Essential Resources Currently Popular Recent Posts SBG Blogs Physics Teaching Blogs Modeling Instruction Blogs Science Teaching Blogs Math Teaching Blogs Archives Blog Stats Visitors Twitter Updates
14590	https://math.stackexchange.com/questions/330269/the-distance-from-a-point-to-a-line-segment	Skip to main content The distance from a point to a line segment [duplicate] Ask Question Asked Modified 3 years, 11 months ago Viewed 32k times This question shows research effort; it is useful and clear 19 Save this question. Show activity on this post. I'm pretty sure this may be a duplicate post somewhere, but I've searched all through the internet looking for a definite formula to calculate the distance between a point and a line segment. There are so many different variations of the formula that people have posted that its hard to determine which is correct. I actually have several points that I will be looping through to get their distances. Each point will be within the boundary of the segment(if that makes any sense) so there will be no need to perform this theoretical check of the point being 'within' the segment that I've somewhat heard about. So can anyone please post the correct formula to calculate the distance of a point C(x,y) from a line segment AB?? geometry trigonometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Mar 14, 2013 at 12:57 Cameron Buie 105k1010 gold badges106106 silver badges245245 bronze badges asked Mar 14, 2013 at 12:51 user1898629user1898629 29111 gold badge22 silver badges33 bronze badges 5 What do you mean by "within the segment" ? If C in on AB, distance is zero. If C is not, distance will be found by projection of C on AB, if this projection falls inside AB, or will be AC or BC in other cases. – Jean-Claude Arbaut Commented Mar 14, 2013 at 13:06 I don't think this has a general answer as it may depend, in euclidean plane or space, on the relative positions of the point and the segment. There is a general, pretty nice, formula for the distance from a point to a straight line, though... – DonAntonio Commented Mar 14, 2013 at 13:06 1 I think "within the segment" means exactly that the projection on (the line supporting) the segment falls within the segment. If you are assume this is always the case, then just use that line instead of the segment of it. And for the distance of a point to a line, see the answer by julien. – Marc van Leeuwen Commented Mar 14, 2013 at 13:34 I had first overlooked the "segment" thing. So I did the distance to a line. But note that the distance to a line segment is done here. And note that is valid in any Hilbert space, not only R3. – Julien Commented Mar 14, 2013 at 13:46 Aarbautjc - I had mis typed. WHen I meant C is within the segment AB, i had just meant that if A(-1,1) and B(4,4) then our point C is of course not on the segment but at the same time, Cx can only be greater than Ax and less than Bx. Make sense? Geometry lingo is a tad bit off. But I agree that the distance does need to be found by projection of C on AB so that a line from point to the segment would form a right angle, i.e. orthological I think. Some big word that starts with an O. lol – user1898629 Commented Mar 14, 2013 at 13:53 Add a comment \| 2 Answers 2 Reset to default This answer is useful 27 Save this answer. Show activity on this post. Let the line segment be described by two points s1,s2, and you wish to find the nearest point on the segment to the point p. We find the nearest point to the line through s1,s2, then 'project' back to the segment, then compute the distance. A point on the line can be parameterized by s(t)=s1+t(s2−s1), note that s(t) is on the line segment iff t∈[0,1]. The distance from p to the point s(t) given by the function ϕ(t)=∥s(t)−p∥. It is easier to deal with ϕ2, which is a convex quadratic in t. To find the minimizing 't', we set the derivative of ϕ2 to zero giving t^=⟨p−s1,s2−s1⟩∥s2−s1∥2. To find the t that minimizes the distance on the segment, we 'project' back to [0,1] using t∗=min(max(t^,0),1). Then the minimum distance is given by ∥s(t∗)−p∥. Addendum: Note that ϕ(t)2=ϕ(t^)2+(t−t^)2∥s2−s1∥2. Hence the minimum distance will correspond to the value of t∈[0,1] that results in the smallest (t−t^)2. It is straightforward to see that this is given by t∗ above. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Sep 5, 2021 at 21:55 answered Mar 14, 2013 at 15:02 copper.hatcopper.hat 178k1010 gold badges127127 silver badges270270 bronze badges 3 2 This is a pretty beautiful way to think about it, neat and clean. And this method is actually quite easy to convert to code, just implement basic vector operations. – Manish Chandra Joshi Commented Apr 22, 2020 at 11:01 Great answer. I am wondering why if t^ is negative or larger than 1 why is the chosen t∗ the minimal value? – IntegrateThis Commented Sep 5, 2021 at 0:50 1 @IntegrateThis I added a note elaborating why. – copper.hat Commented Sep 5, 2021 at 21:55 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. Edit: this is indeed a duplicate, I had not read the question carefully enough. Below is how you compute the distance from a point to a line, which is the major bulk when computing the distance from a point to a line segment. Let us assume we are in Rn (n≥2) equipped with its usual Euclidean inner product (x,y)=∑nk=1xkyk. Let L be a line parameterized by t⟼P+tu⃗ where P is a point belonging to this line and u⃗ is a vector giving the direction of L. If you know two points P,P′ on the line, it suffices to take P and u⃗ =PP′→. Now let Q be any point. The distance Q to L is the distance between Q and QL its orthogonal projection on L. Now QL is characterized by the vector projection formula: PQL→=(PQL→,u⃗ )∥u⃗ ∥2u⃗ . So QQL→=QP→+PQL→=QP→+(PQL→,u⃗ )∥u⃗ ∥2u⃗ . It only remains to compute the norm of the latter to get the distance from Q to L. Note: when n=2 and L is given by a cartesian equation ax+by+c=0, this yields the formula d(P,L)=\|ax+by+c\|a2+b2−−−−−−√ for every P=(x,y). Algorithm to compute the distance from Q to the line segment [P,P′]: Take an arbitrary point Q. Compute the coordinates of the projection QL on the line (which does not necessarily belong to the segment). Compute d(Q,QL)=∥QQL→∥ the distance between Q and QL. Also compute the distances d(Q,P) and d(Q,P′) to the endpoints. Then the number you are looking for (the distance from Q to [P,P′]) is the minimum of these three numbers: d(Q,QL), d(Q,P) and d(Q,P′). Share CC BY-SA 3.0 Follow this answer to receive notifications edited Apr 13, 2017 at 12:21 CommunityBot 1 answered Mar 14, 2013 at 13:24 JulienJulien 45.7k33 gold badges9595 silver badges172172 bronze badges 15 Julien -- I appreciate the mathematical breakdown. I will attempt to decipher all this so it fits into a python script. W/out full understanding, I have no script. My math/geometry skill is shamefully rusty. At this point, I'd need step by step instructions. Lol. Thanks for your feedback though. – user1898629 Commented Mar 14, 2013 at 13:57 @Gis_Guy Note that my answer in the link I provided above explains the three steps of a possible algorithm. I think it is not to hard to program it then, in any language. Good luck! – Julien Commented Mar 14, 2013 at 14:01 Julien -- Just so I can better my understanding, because I am dealing with strictly a segment, my projection Q would belong to [p1, p2] in terms of calculating the distance, correct? Also, in the link explaining the algorithm, I am having trouble understanding "3.2". Again, my math is lacking, and all the symbols and arrows aren't making things to clear. Thanks for your patience/understanding. – user1898629 Commented Mar 14, 2013 at 14:54 6 Your algorithm is wrong: If the projection is outside the segment it should be ignored. The projection could be outside the segment and still be closer than the distance to the endpoints. e.g. the line between (0,0) and (1,1) for the point (3,2). – keyser Commented May 1, 2014 at 17:20 2 As keyser said, the algorithm is wrong. When the projection is outside of the segment, its length will be always shorter than the distance to any of the endpoints. – Saul Berardo Commented Jul 6, 2015 at 16:54 \| Show 10 more comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry trigonometry See similar questions with these tags. 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14591	https://www.vdh.virginia.gov/environmental-health/environmental-health-services/shellfish-safety/regulations-and-control-plans/	Regulations and Control Plans – Environmental Health A Commonwealth of Virginia WebsiteAn official website of the Commonwealth of Virginia Here's how you knowAn official websiteHere's how you know English ▼ ኣማርኛ العربية الفصحى 简化字正體字 Français हिंदी 한국어 Español Tagalog اردو Tiếng Việt Find a Commonwealth Resource Skip to content Environmental Health Search for: × About How Do I A-Z Index Locations Data Clinicians Newsroom Contact Virginia Department of Health>Environmental Health>Shellfish Safety>Regulations and Control Plans Bedding and Upholstered Furniture Program Childhood Lead Poisoning Prevention Food Safety in Virginia Marina Program Migrant Labor Camps Public Health Toxicology Shellfish Safety Tourist Establishment Regulation Campgrounds Tourist Lodging Pool & Spa Safety Waterborne Hazards Control Onsite Water and Wastewater Services Email this page Regulations and Control Plans Note: Please be aware that these links will take you away from our site, and to some fairly dense reading. You may return by using your browser’s Back feature. If you need help interpreting these regulations, please contact your local DSS office Virginia Administrative Code: (if need be, click on the links below, then click on “Next” to see the details in each Code section) “Regulations for the Sanitary Control of Storing, Processing, Packing or Repacking of Oysters, Clams and Other Shellfish” 12VAC5-150. “Regulations for the Sanitary Control of the Picking, Packing and Marketing of Crab Meat for Human Consumption” 12VAC5-160. “Regulations for the Sanitary Control of the Repacking of Crab Meat” 12VAC5-165. Regulations “Pertaining to Restrictions on Shellfish” – these restrictions govern Public Health Restrictions on Harvest Times and Handling Procedures including those referring to warm weather harvest 4VAC20-1230. Vibrio Control: Vibrio Control Plan – The National Shellfish Sanitation Program (NSSP) requires every state from which oysters are harvested to conduct Vibrio parahaemolyticus and Vibrio vulnificus risk evaluations annually. This report is intended to present data collected and evaluated by the Division of Shellfish Safety, in cooperation with the Interstate Shellfish Sanitation Conference (ISSC), to identify and control Vibrio parahaemolyticus and Vibrio vulnificus risks associated with the consumption of oysters. Biotoxin Control: Biotoxin Control Plan – The National Shellfish Sanitation Program (NSSP) requires every state from which oysters are harvested to develop and adopt a marine biotoxin contingency plan for all marine and estuarine shellfish growing areas. This plan describes the monitoring program of potential toxic algal species, toxins in shellfish meats, and the protocols in place regarding closures and reopening of growing areas if necessary. National Shellfish Sanitation Program: Manuals of Operations FDA: HACCP Last Updated: September 15, 2022 Contact Us Health Department Locator Phone & Directions Feedback Form Job Opportunities Agency Information Accessibility Code of Ethics FOIA Policy Internships Organizational Charts Privacy Policy Non-Discrimination Policy Newsletters Health Districts Healthcare Professionals All Topics View VDH Expenditures Original text Rate this translation Your feedback will be used to help improve Google Translate
14592	https://www.youtube.com/watch?v=Td8C_cTEGkA	Differential Equations: Population Growth Math Easy Solutions 56300 subscribers 56 likes Description 8161 views Posted: 30 Jun 2016 In this video I go over an introduction to differential equations and explain a bit about how they are the most important applications of calculus to the real-world. Differential equations are simply equations that consist of a function and some of its derivatives. These types of equations model real-world applications very well because often times physical concepts change at a rate that is proportional to its current state. One such example is in population growth in which the rate of growth depends on what the current population is. In this video I use differential equations to model population growth. The first differential equation for population growth that I go over is for ideal conditions and is simply stated as the rate of growth is proportional to the current population. But a more accurate model assumes that there is a maximum carrying capacity in which the population levels off. This latter model is known as the Logistic Differential Equation and was first proposed by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s. I also go over a brief history lesson on Verhulst. This is an extremely important video as it lays the foundation for my later videos on one of the most powerful mathematical concepts, differential equations, so make sure to watch this video! Download the notes in my video: View Video Notes on Steemit: Related Videos: Natural Exponential Function: y = e^x: . Become a MES Super Fan! DONATE! ʕ •ᴥ•ʔ SUBSCRIBE via EMAIL: MES Links: MES Truth: Official Website: Hive: Email me: contact@mes.fm Free Calculators: BMI Calculator: Grade Calculator: Mortgage Calculator: Percentage Calculator: Free Online Tools: iPhone and Android Apps: 2 comments Transcript: Introduction hi it's time for another math easy solution ter SC say a very interesting video on modeling with differential equations and then uh basically look at uh population growth to illustrate more on differential equations and also go introduction basically on this whole concept of modeling and as well as uh go over what differential equations are uh basically perhaps the most important of all applications of uh calculus is to differential equations so yeah this video is going to be a lot of reading out here basically when physical scientists or social scientist use calculus more often than not it is to analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying yes and although it is often impossible to find an explicit formula for the solution of a differential equation yes so it is very very a lot of these uh equations are hard to find explicitly and uh we will see that graphical and numerical approaches provide the needed information to uh solve these uh or estimate them using approximate analysis and basically the process of modeling um involves formulating basically a mathematical model of a real world problem either through intuitive reasoning about the phenomenon or from a physical law based on evidence from experiments so again a lot of this is this is why it's really applicable to uh the physics and science because yeah because we are looking at those experimental data and then from there trying to develop mathematical models Bally the mathematical model often takes the form of a differential equation and a different and a differential equation is an equation that contains an unknown function and some of its derivatives yes so basically that function includes a different its derivative and that's why there's it's called differential equation so just write Circle that better uh so this is not surprising because uh well in a real world problem we often notice that changes occur and and we want to predict future behavior on the basis of the current values and how they change basically uh the the rate of change or what however things change depends on what the value is currently that's all often is the case in the real world and one example of how differential equations arise when we model physical phenomenon is in population growth and here we're going to look at uh some models of population growth and in fact we'll go over uh two models one simple and one a bit more uh complicated basically one model of for the growth of a population is based on an assumption that the population grows at a rate proportional to the size of the population and this is a reasonable assumption for a population of bacteria or animals under ideal conditions and condition then those conditions are basically such as unlimited environment adequate nutrition absence of predators immunity from disease Yeah so basically if you just assume let's say yeah let's say that yeah there's no way of them dying off and they just keep uh replicating on and on and Immunity from disease could include uh death as well if they just don't die but even if you include death yeah when you ignore everything else when you assume there's unlimited food Etc you'll notice that well again it it's proportional to the size of this population because when you have a large population you're and each one is reproducing so you're going to obviously get a bigger rate of increase as opposed to very small and often the case when a small population they just don't reproduce as fast because they're just not in that uh Constant Contact as when you have more condensed uh large population anyway anyway let's identify and name the variables in this model so we'll let tals time this will be the independent variable so whatever time T we would look at population which is we'll write p and that's the dependent variable so this depends on whatever time is so at a time we'll look at the population and not the other way around uh and that basically that's the Y AIS so the rate of growth of the population is the derivative DP / DT so our assumption that the rate of growth is proportional to the population size is written by the equation DP over DT so the rate of change rate of growth population is uh let's just write equals to is proportional so K P which is a proportionality constant and again I'll just write that down here so basically where K is a proportionality constant so this is basically again again uh modeling this assumption we made which is the growth growth rate is proportional to the population size at whatever time it is so this equation is in fact our first model of population growth it is a differential equation because it contains an unknown P so that's the function and it's derivative DP over DT so this is a a a differential equation so now that we've formulated this model let's look at add its consequences if we rule out a population of zero cuz we have nothing it's not going to replicate then uh then basically then basically the population is going to be greater than zero for all time T so we're not going to have a negative population that doesn't make physical sense so if uh yeah if K is greater than zero so that we're always growing it's supposed to decreasing and again it's a constant uh then this equation shows that the derivative p uh p p Prime of t or DP over DT this is always positive for all T So this means that it is uh increasing because if this is positive p is positive K is positive this is positive positive this is also positive meaning that the growth rate is increasing so this means the population is always increasing in fact as P of T increases the equation shows that DP over d T becomes larger and as you can see here if this is growing and then p is growing as well that means and it's since the derivative is related to the population as this gets bigger this number K is constant so this just gets bigger the rate gets bigger and basically it just keeps increasing so in other words yeah so DP over DT becomes larger as PFT increases in other words the growth rate increases as the population increas increases and Solution now let's try to think of a solution of the equation this equation asks us to find a function a function whose derivative is a constant multiple of itself so if you had this function P the derivative here is just P but then there's a constant there so again it's a constant multiple of itself and we know that exponential functions have that property and you can see that my earlier video in fact if we let P of T equals to C e^ of KT yeah then what we end up having is the derivative P of T so yeah the derivative P of T that just uh is equal to itself right here so this equals to C and then derivative of e of um e^ of KT where K is also constant we'll have e to the KT and then using chain rule add the K in front I'll just put that in bracket like this k e k K T which this equals 2 I'll put the C inside so what we'll have is K and then c e k t in other words what we have right here is DP over DT that's the same thing as writing P Prime of T equals to K and then C KT that's equal to P of T so there you go that's a solution to that um to this differential equation yeah so thus what we have is any exponential function of the form P of T equals to C KT is a solution to our differential equation and in later videos I will show how there is in fact no other solution but this exponential function which is pretty remarkable on all so yeah stay tuned for those videos in the future in the near future anyway so allowing C to vary through all real numbers including negative numbers we get the family of solutions of P of t e no equals to C e^ of KT whose graphs are shown in the figure below and I'll draw that like this so if we have t should put this more in the center CU we have negative values like this this is p and this is time T that's population so it's just an exponential function so if you have a c value like that goes up if you have a high one low one Etc so something like this and then in the negative side you get you're going to obviously get something like this if if we have a negative um negative C right there so we have so these are functions of the population yeah and these are for all T values all time and all uh C values when we vary through all real numbers yeah but uh but populations yeah they Physical Media don't have negative values so they only have positive values so we are interested only in the solutions of C where it's greater than uh zero so we don't have any uh negative populations and and we are probably concerned only with the values of T greater than the initial time T equals z cuz yeah I'm not sure how you would work with a negative time so now the figure below shows the physical mean physically meaningful Solutions so if we were just to look at the positives what we end up having is T P and we basically cut off all the population so we have something that looks like this for that was a past to zero so something like this let's go with slower I think this would be a bit wider out it would be something like that so yeah again these are more meaningful Solutions of the population growth differenti equation these are just looking at physically meaningful so we only have positive uh time and positive population growth or po population and also what's interesting is if we put tal 0 we end up having is p of 0 equals to well c e k and t is z now and then this goes to one e to the 0 is 1 so we have equals to C so all these uh at time is zero would we end up having a C1 C2 these are just different constants C4 over there so uh thus the constant C turns out to be the initial population P of zero so now the yeah the basically Carrying Capacity list right further the the the differential equation we use is appropriate for modeling population growth under ideal conditions but we have to recognize that a more realistic model must reflect that a given environment has limited resources yes if for on Earth you can't just keep keep growing population unlimited there's not earth isn't big enough so many populations in fact start increasing in an exponential man but the population levels off when it approaches What's called the carrying capacity we we'll call it capital K or decreases to K if it ever exceeds K like that so this is a interesting model and yeah an interesting observation about populations basically there yeah you have limited resources so you so so a environment can only uh withstand or sustain a limited number of people or animals or whatever population you're looking at bacteria for example or anything so for a model to take into account both trends that it does increase in fact exponentially initially but then it uh it uh tailor off up to this um carrying capacity you make two assumptions initially the growth rate is proportional to P or in other words what we could write is DP over DT well it's roughly proportional to K Over p and we're putting roughly proportional because it's a model when we have again a couple assumptions so we're just this is roughly what we want and now what we want is p decreases if it ever Inc if it ever exceeds K in other words what we'll have is DP over DT the growth rate has to decree increase so it's less than zero meaning population is decreasing so it's less than Z the derivative is less than Z if the population exceeds K and now what's interesting is a simple sorry that's not capital a simple expression that incorporates both these assumptions given is given by both assumptions is as given by the equation I'll draw that right here so the differential equation will look at something that's pretty simple but still matches both and in fact if we if we were to write DP over DT the derivative equals to let's say it equals to KP so it's proportional to this but we'll add a a factor of 1 - p over K capital K so this is yeah I don't know why my uh pen was doing that so that's capital K like this small K capital K just to make sure you're not uh confused I'll TR this even smaller so small k and then we have capital K here so if we look at this right here uh when p is greater than K we have this greater than one so then we'll have a 1 minus a number greater than one which makes it negative so whenever it's greater than K you'll have a decreasing and when p is really really small this goes to roughly zero so this this approach is zero when it's small so what what we'll end up having is this roughly is equal to 1 which makes it uh the proportional yes basically makes the derivative roughly proportional to the population size or just KP and here I've written that down Bic notice that if p is small compared with the capital K or the carrying capacity then P / K is close to zero so if it's close to zero this is close to 0 we have 1 - 0 where it's just one so one to end up having is so then again DP over DT is roughly proportional to KP or equals to KP like that which is our first assumption and now the second one if p is greater than k then 1 minus p over K is negative because this p over K will be well greater than one 1 minus a number greater than one is negative so this derivative is going to be well negative DP over DT is negative so the population is Logistic Equation decreasing and now some interesting notes on this equation this equation is called the logistic differential equation and was first proposed by by the Dutch mathematical biologist Pierre franois verol not sure how you pronounced it in the 1840s as a model for world population growth just a brief uh history lesson on Pierre he was born in Brussels Belgium and lived from 1804 to 1849 so only live about 45 years and yeah he published this logistic equation in 1838 and its solution in 1845 a long time to find out solution but it was not until 1920 when this model was rediscovered by American biological researchers Raymond Pearl and LEL uh Reed yeah who promoted its uh use also I think I just Overlook the step basically he was a mathematician and a doctor in number Theory from the University of G in 1825 and yeah my calcul book says biologist I researched up here I didn't see that on Wikipedia But but I guess he might have been a biologist as well anyways anyways yeah I thought it was pretty interesting so he developed this equation but again it wasn't used widely until uh these these American uh researchers promoted its use and this often is the case in mathematics and uh classical mathematics and Sciences it usually takes a while for its popularity of whatever theorem to go widespread yeah so now um basically yeah continuing further uh I will go over techniques in later videos that enable us to find explicit Solutions of this logistic equation again in my later videoos so stay tuned for that or for those better grammar but for now we could deduce qualitative characteristic characteristics of the solutions directly from the logistic differential equation so again the explicit Solutions I'll do that in later videos but for now we could try to uh roughly estimate how the graphs would look like of the solutions qualitatively by looking at that logistic equation so we first observe that the constant functions P of tal 0 and P of tal K are solutions because in either case one of the factors of the on the right side of the listic equation is equal to zero so let's just scroll back up here so if the if P of T is equal to Z so if this equals to 0 this is just Z so then the derivative is equal to zero which is true yeah so if population is zero when you plug in P that just goes zero and if K is equal to uh P I mean I mean if p is equal to K for all values of of time you plug in K we have 1 - 1 - K Over K which is 1 - 1 goes to zero yeah which equals z and the derivative of p p of tal K is also zero so I'll just draw that out actually just to be more more to clarify it further so if P of T is equal to 0 the derivative of 0 is equal to 0 and then if we plug in let's just plug in our value so DP now we look at the differential equation D of P of DT is equal to K 0 what p and then this is 1 - 0 over K this just goes to zero so this is true this is true so the the solution uh so basically this is a solution of this differential equation now if we look at part the second one if P of T is equal to K capital K this is again small K capital K what we end up having is the derivative P of T is of K as is constant equals zero as well and then similarly DP over DT is equal to k p 1 - K Over K which is just goes to zero yes this is 1 - 1 that equals to 0o so 1 - 1 is 0 so we have 0 so 0 0 and it uh fits so basically the derivative of this is equal to the derivative of this differential equation so yeah anyways let's just scroll back up here in fact this does make um yeah this does make physical sense because if the population is ever zero or at the carrying capacity it stays that way because again because again if the PO if the environment can withstand a certain amount of people then it will with withstand that in case unless there's some sort of epidemic or anything but outside of that it will just stay that way and if it's zero well nothing's uh reproducing so you just stick at zero and now these two constant Solutions are called equilibrium Solutions again because they're an equilibrium and nothing is changing and now if the inial population P of T lies between 0 and K then the right side of the logistic equation is positive and again just look at that if um if you look at this if if a population is between 0 and K this is less than K sub 1 minus this is going to be well less than one just the ratio of less than one where K is greater than P so then this is positive so we are increasing so then D of P DP over DT is greater than zero and the population increases but if the population exceeds a carrying capacity p is greater than k then 1 - p k is negative because well this is p is greater 1us the number that's going to be larger than one and it's negative so D of P DP / DT or the derivative is less than zero in the population decreases yeah but notice that in either either case if the population approaches the carrying capacity as p is approaching k then DP over DT approaches zero yeah which means the population levels off so initially it's exponential and then it levels off because again if we just look at how this function is as you get closer and closer to K we get closer and closer to one if this is closer to 1 1 - 1 is approaching zero so uh yeah so we would expect fact that the solutions of of the logistic differential equations have graphs that looks look like the ones in the figure below and I'll draw these out like this let's say we had this is T and that's P yeah something like that and now what what I'll do first is I'll draw the equilibrium equation so this one is already at the zero so p = 0 that's this right here and now we will draw let's just include the negative of t values to be more General so this is our P I'll write that here actually P equals de K uh line draw that a bit straighter so it's just a horizontal line and this and this I'll write I'll drag this up here so these two curves are our equilibrium Solutions yeah now our non equilibrium Solutions are just going to be a draw that in Red so again you'll have it if you look at the uh top of that let's just draw the first ones actually the more ones we're used to so we get a small population will increase like this exponentially and then it will approach it but never reach it as we are um yeah as we are approaching uh the the K the carrying capacity so again let's look at a more smaller one so it looks something like this again I'll draw this like that so different solution when we vary yeah when we vary whatever constant it is again we don't have the solution for but we are just guessing uh using we're not guessing approximating using these qualitative observations that we made so again it's exponential then it goes uh to leveling off near it and then if we look at if we had it higher than uh the carrying capacity we would again decrease because it's always like that and it's decreasing as we approach it it's getting closer and closer to zero the derivative so when we ever have the derivative of zero it's going to be horizontal so another one goes like that Etc so this is how the graphs we expect them to be and again yeah basically notice that the graphs move away from the equilibrium solution P equals z and move towards the solution p = k so yeah you have this uh bottom equilibrium solution and at all these graphs approach the top one of the carrying capacity P equal K but then if you look in Reverse time it's going to go from here approaching the other equilibrium so anyways these are just how these families of uh of the solutions should look like and in later videos when I find an explicit solution and formula for them you will see that this is actually the case anyways that is all for today if you learn all this pretty extensive um introduction to differential equations and how we go about modeling real world problems using them and also going about finding uh solutions to these differential equations or models any was all for today hope you learned and like always you can download these exact notes in the link below thanks for watching and stay tuned for another math easy solution
14593	https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series?srsltid=AfmBOoofigStjzFowQIfkjeBidfXBEbfYVCfFAG-RWAbB1uorFGC2ExC	Art of Problem Solving Arithmetico-geometric series - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Arithmetico-geometric series Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arithmetico-geometric series An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: , where and are the th terms of arithmetic and geometric sequences, respectively. Contents [hide] 1 Finite Sum 2 Infinite Sum 3 Example Problems 4 See Also Finite Sum The sum of the first terms of an is , where is the common difference of and is the common ratio of . Or, , where is the sum of the first terms of . Proof: Let represent the sum of the first terms. Infinite Sum The sum of an infinite arithmetico-geometric sequence is , where is the common difference of and is the common ratio of (). Or, , where is the infinite sum of the . Example Problems Mock AIME 2 2006-2007 Problem 5 1994 AIME Problem 4 See Also Sequence Arithmetic sequence Geometric sequence Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
14594	https://www.slideshare.net/slideshow/varian-microeconomic-analysis-solution-book/6822502	Varian, microeconomic analysis, solution book \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Cookie Policy Download free for 30 days Sign in UploadLanguage (EN)Support BusinessMobileSocial MediaMarketingTechnologyArt & PhotosCareerDesignEducationPresentations & Public SpeakingGovernment & NonprofitHealthcareInternetLawLeadership & ManagementAutomotiveEngineeringSoftwareRecruiting & HRRetailSalesServicesScienceSmall Business & EntrepreneurshipFoodEnvironmentEconomy & FinanceData & AnalyticsInvestor RelationsSportsSpiritualNews & PoliticsTravelSelf ImprovementReal EstateEntertainment & HumorHealth & MedicineDevices & HardwareLifestyle Change Language Language English Español Português Français Deutsche Cancel Save Submit search EN Upload Download free for 30 days Sign in Uploaded byJosé Antonio PAYANO YALE 116,749 views Varian, microeconomic analysis, solution book AI-enhanced description This document contains the answers to exercises for the third edition of the textbook "Microeconomic Analysis" by Hal R. Varian. The answers are organized by chapter and include solutions to mathematical problems as well as explanations and justifications. Key information provided in the answers includes derivations of production functions, profit functions, cost functions, and factor demand functions for various technologies. Convexity and monotonicity properties of technologies are also analyzed. In this document Powered by AI Slides 1 - 2 Answers to Exercises - Microeconomic Analysis These slides introduce the textbook "Microeconomic Analysis" by Hal R. Varian, providing copyright details. View Slides 3 - 4 Chapter 1: Technology Answers Answers to technology-related exercises include topics like production functions, convexity, homothetic functions, and convexity proof. View Slides 5 - 8 Chapter 2: Profit Maximization Answers This section discusses profit maximization, Kuhn-Tucker theorem, elasticity of scale, and various related exercises. View Slides 9 - 10 Chapter 3: Profit Function Answers Addresses the profit function, first-order conditions, and factor demand functions derived from various technologies. View Slides 11 - 12 Chapter 4: Cost Minimization Answers Discusses cost minimization conditions, including profit maximization, relevant cost functions, and efficiency constraints. View Slides 13 - 18 Chapter 5: Cost Function Answers Details the cost function, factors affecting production choices, and properties of functions such as homogeneity and concavity. View Slides 19 - 32 Chapters 6-11: Various Economic Concepts Answers regarding utility maximization, consumer surplus, uncertainty in economic decision-making, and asset markets. View Slides 33 - 50 Chapters 12-24: Market Structures and Efficiency Discussion around competitive markets, monopolies, game theory, and public goods, focusing on welfare and efficiency. View Slides 51 - 54 Chapter 25: Information and Incentives Explores the implications of information asymmetry, incentive constraints, and solution strategies in labor economics. View Read more 61 Save Share Embed Download Downloaded 4,034 times 1 / 53 2 / 53 3 / 53 4 / 53 5 / 53 6 / 53 7 / 53 8 / 53 9 / 53 10 / 53 11 / 53 12 / 53 13 / 53 14 / 53 15 / 53 16 / 53 17 / 53 18 / 53 19 / 53 Most read 20 / 53 21 / 53 22 / 53 Most read 23 / 53 24 / 53 25 / 53 26 / 53 27 / 53 28 / 53 29 / 53 30 / 53 31 / 53 32 / 53 33 / 53 34 / 53 35 / 53 36 / 53 37 / 53 38 / 53 Most read 39 / 53 40 / 53 41 / 53 42 / 53 43 / 53 44 / 53 45 / 53 46 / 53 47 / 53 48 / 53 49 / 53 50 / 53 51 / 53 52 / 53 53 / 53 Image 2: Answers to Exercises Microeconomic Analysis Third Edition Hal R. Varian University of California at Berkeley W. W. Norton & Company • New York • London Image 3: Copyright c 1992, 1984, 1978 by W. W. Norton & Company, Inc. All rights reserved Printed in the United States of America THIRD EDITION 0-393-96282-2 W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, N.Y. 10110 W. W. Norton Ltd., 10 Coptic Street, London WC1A 1PU 234567890 ![Image 4: ANSWERS Chapter 1. Technology 1.1 False. There are many counterexamples. Consider the technology generated by a production function f(x) = x2 . The production set is Y = {(y, −x) : y ≤ x2 } which is certainly not convex, but the input re- √ quirement set is V (y) = {x : x ≥ y} which is a convex set. 1.2 It doesn’t change. 1.3 1 = a and 2 = b. 1.4 Let y(t) = f(tx). Then n dy ∂f(x) = xi , dt ∂xi i=1 so that n 1 dy 1 ∂f(x) = xi . y dt f(x) ∂xi i=1 1.5 Substitute txi for i = 1, 2 to get 1 1 f(tx1 , tx2 ) = [(tx1 )ρ + (tx2 )ρ ] ρ = t[xρ + xρ ] ρ = tf(x1 , x2 ). 1 2 This implies that the CES function exhibits constant returns to scale and hence has an elasticity of scale of 1. 1.6 This is half true: if g (x) > 0, then the function must be strictly increasing, but the converse is not true. Consider, for example, the function g(x) = x3 . This is strictly increasing, but g (0) = 0. 1.7 Let f(x) = g(h(x)) and suppose that g(h(x)) = g(h(x )). Since g is monotonic, it follows that h(x) = h(x ). Now g(h(tx)) = g(th(x)) and g(h(tx )) = g(th(x )) which gives us the required result. 1.8 A homothetic function can be written as g(h(x)) where h(x) is ho- mogeneous of degree 1. Hence the TRS of a homothetic function has the ]( Image 5: 2 ANSWERS form g (h(x)) ∂h ∂h ∂x1 ∂x1 = . g (h(x)) ∂h ∂h ∂x2 ∂x2 That is, the TRS of a homothetic function is just the TRS of the un- derlying homogeneous function. But we already know that the TRS of a homogeneous function has the required property. 1.9 Note that we can write 1 1 a1 a2 ρ (a1 + a2 ) ρ xρ + xρ . a1 + a2 1 a1 + a2 2 1 Now simply deﬁne b = a1 /(a1 + a2 ) and A = (a1 + a2 ) ρ . 1.10 To prove convexity, we must show that for all y and y in Y and 0 ≤ t ≤ 1, we must have ty + (1 − t)y in Y . But divisibility implies that ty and (1 − t)y are in Y , and additivity implies that their sum is in Y . To show constant returns to scale, we must show that if y is in Y , and s > 0, we must have sy in Y . Given any s > 0, let n be a nonnegative integer such that n ≥ s ≥ n − 1. By additivity, ny is in Y ; since s/n ≤ 1, divisibility implies (s/n)ny = sy is in Y . 1.11.a This is closed and nonempty for all y > 0 (if we allow inputs to be negative). The isoquants look just like the Leontief technology except we are measuring output in units of log y rather than y. Hence, the shape of the isoquants will be the same. It follows that the technology is monotonic and convex. 1.11.b This is nonempty but not closed. It is monotonic and convex. 1.11.c This is regular. The derivatives of f(x1 , x2) are both positive so the technology is monotonic. For the isoquant to be convex to the origin, it is suﬃcient (but not necessary) that the production function is concave. To check this, form a matrix using the second derivatives of the production function, and see if it is negative semideﬁnite. The ﬁrst principal minor of the Hessian must have a negative determinant, and the second principal minor must have a nonnegative determinant. ∂ 2 f(x) 1 −3 1 ∂ 2 f(x) 1 −1 − 1 2 = − x1 2 x2 2 = x12 x2 2 ∂x1 4 ∂x1 ∂x2 4 ∂ 2 f(x) 1 1 −3 2 = − x1 x22 2 ∂x2 4 Image 6: Ch. 2 PROFIT MAXIMIZATION 3 −3/2 1/2 1 −1/2 −1/2 − 1 x1 x2 4 4 x1 x2 Hessian = 1 −1/2 −1/2 1 1/2 −3/2 4 x1 x2 − 4 x1 x2 1 −3/2 1/2 D1 = − x1 x2 < 0 4 1 −1 −1 1 D2 = x x − x−1 x−1 = 0. 16 1 2 16 1 2 So the input requirement set is convex. 1.11.d This is regular, monotonic, and convex. 1.11.e This is nonempty, but there is no way to produce any y > 1. It is monotonic and weakly convex. 1.11.f This is regular. To check monotonicity, write down the production √ function f(x) = ax1 − x1 x2 + bx2 and compute ∂f(x) 1 −1/2 1/2 = a − x1 x2 . ∂x1 2 1 x2 This is positive only if a > 2 x1 , thus the input requirement set is not always monotonic. Looking at the Hessian of f, its determinant is zero, and the determinant of the ﬁrst principal minor is positive. Therefore f is not concave. This alone is not suﬃcient to show that the input requirement sets are not convex. But we can say even more: f is convex; therefore, all sets of the form √ {x1 , x2 : ax1 − x1 x2 + bx2 ≤ y} for all choices of y are convex. Except for the border points this is just the complement of the input requirement sets we are interested in (the inequality sign goes in the wrong direction). As complements of convex sets (such that the border line is not a straight line) our input requirement sets can therefore not be themselves convex. 1.11.g This function is the successive application of a linear and a Leontief function, so it has all of the properties possessed by these two types of functions, including being regular, monotonic, and convex. Chapter 2. Profit Maximization Image 7: 4 ANSWERS 2.1 For proﬁt maximization, the Kuhn-Tucker theorem requires the follow- ing three inequalities to hold ∂f(x∗ ) p − wj x∗ = 0, j ∂xj ∂f(x∗ ) p − wj ≤ 0, ∂xj x∗ ≥ 0. j Note that if x∗ > 0, then we must have wj /p = ∂f(x∗ )/∂xj . j 2.2 Suppose that x is a proﬁt-maximizing bundle with positive proﬁts π(x ) > 0. Since f(tx ) > tf(x ), for t > 1, we have π(tx ) = pf(tx ) − twx > t(pf(x ) − wx ) > tπ(x ) > π(x ). Therefore, x could not possibly be a proﬁt-maximizing bundle. 2.3 In the text the supply function and the factor demands were computed for this technology. Using those results, the proﬁt function is given by a 1 w a−1 w a−1 π(p, w) = p −w . ap ap To prove homogeneity, note that a 1 w a−1 w a−1 π(tp, tw) = tp − tw = tπ(p, w), ap ap which implies that π(p, w) is a homogeneous function of degree 1. Before computing the Hessian matrix, factor the proﬁt function in the following way: 1 a a 1 1 a π(p, w) = p 1−a w a−1 a 1−a − a 1−a = p 1−a w a−1 φ(a), where φ(a) is strictly positive for 0 < a < 1. The Hessian matrix can now be written as ∂ 2 π(p,w) ∂ 2π(p,w) 2 ∂p2 ∂p∂w D π(p, ω) = ∂ 2 π(p,w) ∂ 2π(p,w) ∂w∂p ∂w 2  2a−1 a a 1  a (1−a)2 p 1−a w a−1 − (1−a)2 p 1−a w a−1 a     =  φ(a).   a 1 1 2−a − (1−a)2 p 1−a w a−1 a a (1−a)2 p 1−a w a−1 Image 8: Ch. 2 PROFIT MAXIMIZATION 5 The principal minors of this matrix are a 2a−1 a p 1−a w a−1 φ(a) > 0 (1 − a)2 and 0. Therefore, the Hessian is a positive semideﬁnite matrix, which implies that π(p, w) is convex in (p, w). 2.4 By proﬁt maximization, we have ∂f ∂x1 w1 \|T RS\| = = . ∂f w2 ∂x2 Now, note that ln(w2 x2 /w1 x1 ) = −(ln(w1 /w2 ) + ln(x1 /x2 )). Therefore, d ln(w2 x2 /w1 x1 ) d ln(w1 /w2 ) d ln \|T RS\| = −1= − 1 = 1/σ − 1. d ln(x1 /x2 ) d ln(x2 /x1 ) d ln(x2 /x1 ) 2.5 From the previous exercise, we know that ln(w2 x2 /w1 x1 ) = ln(w2 /w1 ) + ln(x2 /x1 ), Diﬀerentiating, we get d ln(w2 x2 /w1 x1 ) d ln(x2 /x1 ) =1− = 1 − σ. d ln(w2 /w1 ) d ln \|T RS\| 2.6 We know from the text that Y O ⊃ Y ⊃ Y I. Hence for any p, the maximum of py over Y O must be larger than the maximum over Y , and this in turn must be larger than the maximum over Y I. 2.7.a We want to maximize 20x − x2 − wx. The ﬁrst-order condition is 20 − 2x − w = 0. 2.7.b For the optimal x to be zero, the derivative of proﬁt with respect to x must be nonpositive at x = 0: 20 − 2x − w < 0 when x = 0, or w ≥ 20. 2.7.c The optimal x will be 10 when w = 0. 2.7.d The factor demand function is x = 10 − w/2, or, to be more precise, x = max{10 − w/2, 0}. ![Image 9: 6 ANSWERS 2.7.e Proﬁts as a function of output are 20x − x2 − wx = [20 − w − x]x. Substitute x = 10 − w/2 to ﬁnd w 2 π(w) = 10 − . 2 2.7.f The derivative of proﬁt with respect to w is −(10 − w/2), which is, of course, the negative of the factor demand. Chapter 3. Profit Function 3.1.a Since the proﬁt function is convex and a decreasing function of the factor prices, we know that φi(wi ) ≤ 0 and φi (wi) ≥ 0. 3.1.b It is zero. 3.1.c The demand for factor i is only a function of the ith price. Therefore the marginal product of factor i can only depend on the amount of factor i. It follows that f(x1 , x2 ) = g1 (x1 ) + g2 (x2 ). 3.2 The ﬁrst-order conditions are p/x = w, which gives us the demand function x = p/w and the supply function y = ln(p/w). The proﬁts from operating at this point are p ln(p/w) − p. Since the ﬁrm can al- ways choose x = 0 and make zero proﬁts, the proﬁt function becomes π(p, w) = max{p ln(p/w) − p, 0}. 3.3 The ﬁrst-order conditions are p a1 − w1 = 0 x1 p a2 − w2 = 0, x2 which can easily be solved for the factor demand functions. Substituting into the objective function gives us the proﬁt function. 3.4 The ﬁrst-order conditions are pa1 x1 1 −1 xa2 − w1 = 0 a 2 pa2 x2 2 −1 xa1 − w2 = 0, a 1 which can easily be solved for the factor demands. Substituting into the objective function gives us the proﬁt function for this technology. In order ]( Image 10: Ch. 4 COST MINIMIZATION 7 for this to be meaningful, the technology must exhibit decreasing returns to scale, so a1 + a2 < 1. 3.5 If wi is strictly positive, the ﬁrm will never use more of factor i than it needs to, which implies x1 = x2 . Hence the proﬁt maximization problem can be written as max pxa − w1 x1 − w2 x2 . 1 The ﬁrst-order condition is pax1 − (w1 + w2 ) = 0. a−1 The factor demand function and the proﬁt function are the same as if the production function were f(x) = xa , but the factor price is w1 + w2 rather than w. In order for a maximum to exist, a < 1. Chapter 4. Cost Minimization 4.1 Let x∗ be a proﬁt-maximizing input vector for prices (p, w). This means that x∗ must satisfy pf(x∗ ) − wx∗ ≥ pf(x) − wx for all permissible x. Assume that x∗ does not minimize cost for the output f(x∗ ); i.e., there exists a vector x∗∗ such that f(x∗∗ ) ≥ f(x∗ ) and w(x∗∗ − x∗ ) < 0. But then the proﬁts achieved with x∗∗ must be greater than those achieved with x∗ : pf(x∗∗ ) − wx∗∗ ≥ pf(x∗ ) − wx∗∗ > pf(x∗ ) − wx∗ , which contradicts the assumption that x∗ was proﬁt-maximizing. 4.2 The complete set of conditions turns out to be ∂f(x∗ ) t − wj x∗ = 0, j ∂xj ∂f(x∗ ) t − wj ≤ 0, ∂xj x∗ ≥ 0, j (y − f(x∗ )) t = 0, y − f(x∗ ) ≤ 0, t ≥ 0. If, for instance, we have x∗ > 0 and x∗ = 0, the above conditions imply i j ∂f(x∗ ) ∂xi wi ∗ ≥ . ∂f(x ) wj ∂xj Image 11: 8 ANSWERS This means that it would decrease cost to substitute xi for xj , but since there is no xj used, this is not possible. If we have interior solutions for both xi and xj , equality must hold. 4.3 Following the logic of the previous exercise, we equate marginal costs to ﬁnd y1 = 1. We also know y1 + y2 = y, so we can combine these two equations to get y2 = y −1. It appears that the cost function is c(y) = 1/2+y −1 = y −1/2. However, on reﬂection this can’t be right: it is obviously better to produce everything in plant 1 if y1 < 1. As it happens, we have ignored the implicit constraint that y2 ≥ 0. The actual cost function is y2 /2 if y < 1 c(y) = y − 1/2 if y > 1. 4.4 According to the text, we can write the cost function for the ﬁrst plant as c1 (y) = Ay and for the second plant as c2 (y) = By, where A and B depend on a, b, w1 , and w2 . It follows from the form of the cost functions that c(y) = min{A, B}y. 4.5 The cost of using activity a is a1 w1 +a2 w2 , and the cost of using activity b is b1 w1 + b2 w2 . The ﬁrm will use whichever is cheaper, so c(w1 , w2 , y) = y min{a1 w1 + a2 w2 , b1 w1 + b2 w2 }. The demand function for factor 1, for example, is given by   a1 y  if a1 w1 + a2 w2 < b1 w1 + b2 w2 b1 y if a1 w1 + a2 w2 > b1 w1 + b2 w2 x1 =   any amount between a1 y and b1 y otherwise. The cost function will not be diﬀerentiable when a1 w 1 + a2 w 2 = b 1 w 1 + b 2 w 2 . 4.6 By the now standard argument, √ √ c(y) = min{4 y1 + 2 y2 : y1 + y2 ≥ y}. It is tempting to set M C1 (y1 ) = M C2 (y2 ) to ﬁnd that y1 = y/5 and y2 = 4y/5. However, if you think about it a minute you will see that this Image 12: Ch. 5 COST FUNCTION 9 doesn’t make sense—you are producing more output in the plant with the higher costs! It turns out that this corresponds to a constrained maximum and not to the desired minimum. Check the second-order conditions to verify this. Since the cost function is concave, rather than convex, the optimal solu- tion will always occur at a boundary. That is, you will produce all output √ at the cheaper plant so c(y) = 2 y. 4.7 No, the data violate WACM. It costs 40 to produce 100 units of output, but at the same prices it would only cost 38 to produce 110 units of output. 4.8 Set up the minimization problem min x1 + x2 x1 x2 = y. Substitute to get the unconstrained minimization problem min x1 + y/x1 . The ﬁrst-order condition is 1 − y/x2 , 1 √ √ which implies x1 = y. By symmetry, x2 = y. We are given that √ √ 2 y = 4, so y = 2, from which it follows that y = 4. Chapter 5. Cost Function 5.1 The ﬁrm wants to minimize the cost of producing a given level of output: 2 2 c(y) = min y1 + y2 y1 ,y2 such that y1 + y2 = y. The solution has y1 = y2 = y/2. Substituting into the objective function yields c(y) = (y/2)2 + (y/2)2 = y2 /2. 5.2 The ﬁrst-order conditions are 6y1 = 2y2 , or y2 = 3y1 . We also require y1 + y2 = y. Solving these two equations in two unknowns yields y1 = y/4 and y2 = 3y/4. The cost function is 2 y 2 3y 3y2 c(y) = 3 + = . 4 4 4 ![Image 13: 10 ANSWERS 5.3 Consider the ﬁrst technique. If this is used, then we need to have 2x1 + x2 = y. Since this is linear, the ﬁrm will typically specialize and set x2 = y or x1 = y/2 depending on which is cheaper. Hence the cost function for this technique is y min{w1 /2, w2 }. Similarly, the cost function for the other technique is y min{w3 , w4/2}. Since both techniques must be used to produce y units of output, c(w1 , w2, y) = y [min{w1 /2, w2} + min{w3 , w4 /2}] . 5.4 The easiest way to answer this question is to sketch an isoquant. First draw the line 2x1 + x2 = y and then the line x1 + 2x2 = y. The isoquant is the upper northeast boundary of this “cross.” The slope is −2 to the left of the diagonal and −1/2 to the right of the diagonal. This means that when w1 /w2 < 1/2, we have x1 = 0 and x2 = y. When w1 /w2 < 1/2, we have x1 = y and x2 = 0. Finally, when 2 > w1 /w2 > 1/2, we have x1 = x2 = y/3. The cost function is then c(w1 , w2 , y) = min{w1 , w2 , (w1 + w2 )/3}y. 5.5 The input requirement set is not convex. Since y = max{x1 , x2 }, the ﬁrm will use whichever factor is cheaper; hence the cost function is c(w1 , w2 , y) = min{w1 , w2}y. The factor demand function for factor 1 has the form y if w1 < w2 x1 = either 0 or y if w1 = w2 . 0 if w1 > w2 5.6 We have a = 1/2 and c = −1/2 by homogeneity, and b = 3 since ∂x1 /∂w2 = ∂x2 /∂w1 . 5.7 Set up the minimization problem min x1 + x2 x1 x2 = y. Substitute to get the unconstrained minimization problem min x1 + y/x1 . The ﬁrst-order condition is 1 − y/x2 , 1 √ √ which implies x1 = y. By symmetry, x2 = y. We are given that √ √ 2 y = 4, so y = 2, from which it follows that y = 4. ]( ![Image 14: Ch. 5 COST FUNCTION 11 5.8 If p = 2, the ﬁrm will produce 1 unit of output. If p = 1, the ﬁrst- order condition suggests y = 1/2, but this yields negative proﬁts. The ﬁrm can get zero proﬁts by choosing y = 0. The proﬁt function is π(p) = max{p2 /4 − 1, 0}. 5.9.a dπ/dα = py > 0. 5.9.b dy/dα = p/c (y) > 0. 5.9.c p (α) = n[y + αp/c ]/[D (p) − nα/c ] < 0. 5.10 Let y(p, w) be the supply function. Totally diﬀerentiating, we have n n n ∂y(p, w) ∂xi (p, w) ∂xi (w, y) ∂y(p, w) dy = dwi = − dwi = − dwi. ∂wi ∂p ∂y ∂p i=1 i=1 i=1 The ﬁrst equality is a deﬁnition; the second uses the symmetry of the substitution matrix; the third uses the chain rule and the fact that the unconditional factor demand, xi (p, w), and the conditional factor demand, xi (w, y), satisfy the identity xi(w, y(p, w)) = xi (p, w). The last expression on the right shows that if there are no inferior factors then the output of the ﬁrm must increase. 5.11.a x = (1, 1, 0, 0). 5.11.b min{w1 + w2 , w3 + w4 }y. 5.11.c Constant returns to scale. 5.11.d x = (1, 0, 1, 0). 5.11.e c(w, y) = [min{w1 , w2 } + min{w3 , w4 }]y. 5.11.f Constant. 5.12.a The diagram is the same as the diagram for an inferior good in consumer theory. 5.12.b If the technology is CRS, then conditional factor demands take the form xi(w, 1)y. Hence the derivative of a factor demand function with respect to output is xi(w) ≥ 0. 5.12.c The hypothesis can be written as ∂c(w, y)2 /∂y∂wi < 0. But ∂c(w, y)2 /∂y∂wi = ∂c(w, y)2 /∂wi ∂y = ∂xi (w, y)/∂y. ]( Image 15: 12 ANSWERS 5.13.a Factor demand curves slope downward, so the demand for unskilled workers must decrease when their wage increases. 5.13.b We are given that ∂l/∂p < 0. But by duality, ∂l/∂p = −∂ 2 π/∂p∂w = −∂ 2 π/∂w∂p = −∂y/∂w. It follows that ∂y/∂w > 0. 5.14 Take a total derivative of the cost function to get: n ∂c ∂c dc = dwi + dy. ∂wi ∂y i=1 It follows that dc − n ∂c dw ∂c i=1 ∂wi i = . ∂y dy Now substitute the ﬁrst diﬀerences for the dy, dc, dwi terms and you’re done. 5.15 By the linearity of the function, we know we will use either x1 , or a combination of x2 and x3 to produce y. By the properties of the Leontief function, we know that if we use x2 and x3 to produce y, we must use 3 units of both x2 and x3 to produce one unit of y. Thus, if the cost of using one unit of x1 is less than the cost of using one unit of both x2 and x3 , then we will use only x1 , and conversely. The conditional factor demands can be written as: 3y if w1 < w2 + w3 x1 = 0 if w1 > w2 + w3 0 if w1 < w2 + w3 x2 = 3y if w1 > w2 + w3 0 if w1 < w2 + w3 x3 = 3y if w1 > w2 + w3 if w1 = w2 +w3 , then any bundle (x1 , x2 , x3 ) with x2 = x3 and x1 +x2 = 3y (or x1 + x3 = 3y) minimizes cost. The cost function is c(w, y) = 3y min(w1 , w2 + w3 ). 5.16.a Homogeneous: c(tw, y) = y1/2 (tw1 tw2 )3/4 = t3/2 (y1/2 (w1 w2 )3/4 ) = t3/2 c(w, y) No. Image 16: Ch. 5 COST FUNCTION 13 Monotone: ∂c 3 −1/4 3/4 ∂c 3 3/4 −1/4 = y1/2 w1 w2 > 0 = y1/2 w1 w2 >0 Yes. ∂w1 4 ∂w2 4 Concave: −5/4 3/4 9 1/2 −1/4 −1/4 − 16 y1/2 w1 w2 3 16 y w1 w2 Hessian = 9 1/2 −1/4 −1/4 3 1/2 3/4 −5/4 16 y w1 w2 − 16 y w1 w2 \|H1 \| < 0 9 −1/2 −1/2 81 −1/2 −1/2 \|H2 \| = yw w2 − yw w2 256 1 256 1 72 y =− √ < 0 No 256 w1 w2 Continuous: Yes 5.16.b Homogeneous: √ c(tw, y) = y(tw1 + tw1 tw2 + tw2 ) √ = ty(w1 + w1 w2 + w2 ) = tc(y, w) Yes Monotone: ∂c 1 w2 ∂c 1 w1 =y 1+ >0 =y 1+ >0 Yes ∂w1 2 w1 ∂w2 2 w2 Concave: 1/2 −3/2 −1/2 −1/2 − 1 yw2 w1 4 1 4 yw2 w1 H= −1/2 −1/2 −3/2 1/2 1 4 yw2 w1 − 4 yw2 w1 1 \|H1 \| < 0 1 1 \|H2 \| = yw−1 w−1 − yw2 w1 = 0 Yes −1 −1 16 2 1 16 Continuous: Yes Image 17: 14 ANSWERS Production Function: 1 w2 x1 (w, y) = y 1 + (1) 2 w1 1 w1 x2 (w, y) = y 1 + (2) 2 w2 Rearranging these equations: y w2 x1 − y = (1 ) 2 w1 y w1 x2 − y = (2 ) 2 w2 2 Multiply (1 ) and (2 ): (x1 − y)(x2 − y) = y . This is a quadratic equation 4 which gives y = 2 (x2 + x1 ) ± 2 x2 + x2 + 2 − x1 x2 . 3 3 1 2 5.16.c Homogeneous: c(tw, y) = y(tw1 e−tw1 + tw2 ) = ty(w1 e−tw1 + w2 ) = tc(w, y) No Monotone: ∂c = y(−w1 e−w1 + e−w1 ) = ye−w1 (1 − w1 ) ∂w1 This is positive only if w1 < 1. ∂c = y > 0 No ∂w2 Concave: y(w1 − 2)e−w1 0 H= 0 0 \|H1 \| = y(w1 − 2)e−w1 This is less than zero only if w1 < 2. \|H2 \| = 0 No Continuous: Yes Image 18: Ch. 5 COST FUNCTION 15 5.16.d Homogeneous: √ c(tw, y) = y(tw1 − tw1 tw2 + tw2 √ = ty(w1 − w1 w2 + w2 ) = tc(w, y) Yes Monotone: ∂c 1 w2 = y(1 − ) ∂w1 2 w1 1 w2 This is greater than 0 only if 1 > 2 w1 ∂c 1 w1 = y(1 − ) ∂w2 2 w2 w2 This is greater than 0 only if 2 > w1 1 √ √ w2 > w1 (by symmetry) 2 w1 > w2 4 or 1 w1 < 4w2 w1 > w2 4 Monotone only if 1 w2 < w1 < 4w2 . No. 4 Concave: −3/2 1/2 −1/2 −1/2 1 4 yw1 w2 − 1 yw1 w2 4 H= −1/2 −1/2 1/2 −1/2 − 4 yw1 w2 1 1 4 yw1 w2 1 −3/2 1/2 \|H1 \| = yw1 w2 > 0 4 \|H2 \| = 0 No (it is convex) Continuous: Yes 5.16.e Homogeneous: 1√ c(tw, y) = (y + tw1 tw2 ) y = tc(y, w) Yes Image 19: 16 ANSWERS Monotone in w: ∂c 1 1 w2 ∂c 1 1 w1 = (y + ) >0 = (y + ) >0 Yes ∂w1 2 y w1 ∂w2 2 2 w2 Concave: −3/2 1/2 −1/2 −1/2 − 1 (y + y w1 4 1 w2 1 1 4 (y + y )w1 w2 H= −1/2 −1/2 1/2 −3/2 But not in y! 1 1 4 (y, y )w1 w2 − 4 (y + y )w1 w2 1 1 \|H1\| < 0 Yes \|H2\| = 0 Continuous: Not for y = 0. √ 5.17.a y = ax1 + bx2 5.17.b Note that this function is exactly like a linear function, except that the linear combination of x1 and x2 will produce y2 , rather than just y. So, we know that if x1 is relatively cheaper, we will use all x1 and no x2 , and conversely. 5.17.c The cost function is c(w, y) = y2 min( w1 , w2 ). a b Chapter 6. Duality 6.1 The production function is f(x1 , x2 ) = x1 + x2 . The conditional factor demands have the form  y if wi < wj xi = 0 if wi > wj  any amount between 0 and y if wi = wj . 6.2 The conditional factor demands can be found by diﬀerentiating. They are x1 (w1 , w2 , y) = x2 (w1 , w2, y) = y. The production function is f(x1 , x2 ) = min{x1 , x2 }. 6.3 The cost function must be increasing in both prices, so a and b are both nonnegative. The cost function must be concave in both prices, so a and b are both less than 1. Finally, the cost function must be homogeneous of degree 1, so a = 1 − b. Image 20: Ch. 8 CHOICE 17 Chapter 7. Utility Maximization 7.1 The preferences exhibit local nonsatiation, except at (0, 0). The con- sumer will choose this consumption point when faced with positive prices. 7.2 The demand function is   m/p1 if p1 < p2 x1 = any x1 and x2 such that p1 x2 + p2 x2 = m if p1 = p2  0 if p1 > p2 The indirect utility function is v(p1 , p2 , m) = max{m/p1 , m/p2 }, and the expenditure function is e(p1 , p2 , u) = u min{p1 , p2}. 7.3 The expenditure function is e(p1 , p2 , u) = u min{p1 , p2}. The utility function is u(x1 , x2 ) = x1 + x2 (or any monotonic transformation), and the demand function is   m/p1 if p1 < p2 x1 = any x1 and x2 such that p1 x1 + p2 x2 = m if p1 = p2  0 if p1 > p2 7.4.a Demand functions are x1 = m/(p1 + p2 ), x2 = m/(p1 + p2 ). 7.4.b e(p1 , p2 , u) = (p1 + p + 2)u 7.4.c u(x1 , x2 ) = min{x1 , x2} 7.5.a Quasilinear preferences. 7.5.b Less than u(1). 7.5.c v(p1 , p2 , m) = max{u(1) − p1 + m, m} 7.6.a Homothetic. 7.6.b e(p, u) = u/A(p) 7.6.c µ(p; q, m) = mA(q)/A(p) 7.6.d It will be the same, since this is just a monotonic transformation. Chapter 8. Choice 8.1 We know that xj (p, m) ≡ hj (p, v(p, m)) ≡ ∂e(p, v(p, m))/∂pj . (0.1) ![Image 21: 18 ANSWERS (Note that the partial derivative is taken with respect to the ﬁrst occurrence of pj .) Diﬀerentiating equation (0.1) with respect to m gives us ∂xj ∂ 2 e(p, v(p, m)) ∂v(p, m) = . ∂m ∂pj ∂u ∂m Since the marginal utility of income, ∂v/∂m, must be positive, the result follows. 8.2 The Cobb-Douglas demand system with two goods has the form a1 m x1 = p1 a2 m x2 = p2 where a1 + a2 = 1. The substitution matrix is −a1 mp−2 − a2 mp−2 1 1 1 −a1 a2 mp−1 p−1 1 2 . −a1 a2 mp−1 p−1 2 1 2 −a2 mp−2 − a2 mp−2 2 2 2 This is clearly symmetric and negative deﬁnite. 8.3 The equation is dµ/dt = at + bµ + c. The indirect money metric utility function is c a c c a aq µ(q, p, m) = eb(q−p) m + + 2+ p − − 2− . b b b b b b 8.4 The demand function can be written as x = ec+ap+bm . The integrability equation is dµ = eat+bµ+c . dt Write this as dµ e−bµ = ec eat . dt Integrating both sides of this equation between p and q, we have e−bµ q ec eat q − = . b p a p Evaluating the integrals, we have bec ap ebµ(q;p,m) = e−bm − [e − eaq ] . a ]( Image 22: Ch. 8 CHOICE 19 8.5 Write the Lagrangian 3 L(x, λ) = ln x1 + ln x2 − λ(3x1 + 4x2 − 100). 2 (Be sure you understand why we can transform u this way.) Now, equating the derivatives with respect to x1 , x2 , and λ to zero, we get three equations in three unknowns 3 = 3λ, 2x1 1 = 4λ, x2 3x1 + 4x2 = 100. Solving, we get x1 (3, 4, 100) = 20, and x2 (3, 4, 100) = 10. Note that if you are going to interpret the Lagrange multiplier as the marginal utility of income, you must be explicit as to which utility function you are referring to. Thus, the marginal utility of income can be measured in original ‘utils’ or in ‘ln utils’. Let u∗ = ln u and, correspondingly, v∗ = ln v; then ∂v(p, m) ∂v∗ (p, m) ∂m µ λ= = = , ∂m v(p, m) v(p, m) where µ denotes the Lagrange multiplier in the Lagrangian 3 L(x, µ) = x1 x2 − µ(3x1 + 4x2 − 100). 2 3 20 2 1 Check that in this problem we’d get µ = 4 ,λ= 40 , and v(3, 4, 100) = 3 20 2 10. 8.6 The Lagrangian for the utility maximization problem is 1 1 L(x, λ) = x1 x2 − λ(p1 x1 + p2 x2 − m), 2 3 taking derivatives, 1 −1 3 1 x1 2 x2 = λp1 , 2 1 2 −2 1 x1 x2 3 = λp2 , 3 p1 x1 + p2 x2 = m. Solving, we get 3m 2m x1 (p, m) = , x2 (p, m) = . 5 p1 5 p2 Image 23: 20 ANSWERS Plugging these demands into the utility function, we get the indirect utility function 1 1 1 1 5 3m 2 2m 3 m 6 3 2 2 3 v(p, m) = U (x(p, m)) = = . 5 p1 5 p2 5 p1 p2 Rewrite the above expression replacing v(p, m) by u and m by e(p, u). Then solve it for e(·) to get 3 2 p1 5 p2 5 6 e(p, u) = 5 u5 . 3 2 Finally, since hi = ∂e/∂pi , the Hicksian demands are p1 −2 5 p2 2 5 6 h1 (p, u) = u5 , 3 2 and p1 3 5 p2 −3 5 6 h2 (p, u) = u5 . 3 2 8.7 Instead of starting from the utility maximization problem, let’s now start from the expenditure minimization problem. The Lagrangian is L(x, µ) = p1 x1 + p2 x2 − µ((x1 − α1 )β1 (x2 − α2 )β2 − u); the ﬁrst-order conditions are p1 = µβ1 (x1 − α1 )β1 −1 (x2 − α2 )β2 , p2 = µβ2 (x1 − α1 )β1 (x2 − α2 )β2 −1 , (x1 − α1 )β1 (x2 − α2 )β2 = u. Divide the ﬁrst equation by the second p1 β2 x2 − α 2 = , p2 β1 x1 − α 1 using the last equation 1 x2 − α2 = (x1 − α1 )−β1 u β2 ; substituting and solving, β2 p2 β1 β1 β1 +β2 h1 (p, u) = α1 + u 2 , p1 β2 Image 24: Ch. 8 CHOICE 21 and β1 p1 β2 β1 β1 +β2 h2 (p, u) = α2 + u 1 . p2 β1 Verify that 1 β2 β1 β1 +β2 ∂h1 (p, m) u β1 β2 ∂h2 (p, m) = = . ∂p2 β1 + β2 p1 p2 ∂p1 The expenditure function is  β2   β1  p2 β1 β1 β1 +β2 p1 β2 β1 β1 +β2 e(p, u) = p1 α1 + u 2  + p2 α2 + u 1 . p1 β2 p2 β1 Solving for u, we get the indirect utility function v(p, m) = β1 β2 β1 m − α 2 p2 β2 m − α 1 p1 − α1 − α2 . β1 + β2 p1 β1 + β2 p2 By Roy’s law we get the Marshallian demands 1 m − α 1 p1 x1 (p, m) = β1 α2 + β2 , β1 + β2 p2 and 1 m − α 2 p2 x2 (p, m) = β2 α1 + β1 . β1 + β2 p1 8.8 Easy—a monotonic transformation of utility doesn’t change anything about observed behavior. 8.9 By deﬁnition, the Marshallian demands x(p, m) maximize φ(x) subject to px = m. We claim that they also maximize ψ(φ(x)) subject to the same budget constraint. Suppose not. Then, there would exist some other choice x such that ψ(φ(x )) > ψ(φ(x(p, m))) and px = m. But since applying the transformation ψ−1 (·) to both sides of the inequality will preserve it, we would have φ(x ) > φ(x(p, m)) and px = m, which contradicts our initial assumption that x(p, m) maximized φ(x) subject to px = m. Therefore x(p, m) = x∗ (p, m). (Check that the reverse proposition also holds—i.e., the choice that maximizes u∗ also maximizes u when the the same budget constraint has to be veriﬁed in both cases.) v∗ (p, m) = ψ(φ(x∗ (p, m))) = ψ(φ(x(p, m)) = ψ(v(p, m)), Image 25: 22 ANSWERS the ﬁrst and last equalities hold by deﬁnition and the middle one by our previous result; now e∗ (p, u∗) = min{px : ψ(φ(x)) = u∗ } = min{px : φ(x) = ψ−1 (u∗ )} = e(p, ψ−1 (u∗ )); again, we’re using deﬁnitions at both ends and the properties of ψ(·) — namely that the inverse is well deﬁned since ψ(·) is monotonic— to get the middle equality; ﬁnally using deﬁnitions and substitutions as often as needed we get h∗(p, u∗) = x∗ (p, e∗ (p, u∗)) = x(p, e∗ (p, u∗)) = x(p, e(p, ψ−1 (u∗ ))) = h(p, ψ−1(u∗ )). 8.10.a Diﬀerentiate the identity hj (p, u) ≡ xj (p, e(p, u)) with respect to pi to get ∂hj (p, u) ∂xj (p, m) ∂xj (p, e(p, u)) ∂e(p, u) = + . ∂pi ∂pi ∂m ∂pi We must be careful with this last term. Look at the expenditure minimiza- tion problem e(p, u) = min{p(x − x) : u(x) = u}. By the envelope theorem, we have ∂e(p, u) = hi (p, u) − xi = xi (p, e(p, u)) − xi . ∂pi Therefore, we have ∂hj (p, u) ∂xj (p, m) ∂xj (p, e(p, u)) = + (xi (p, m) − xi ), ∂pi ∂pi ∂m and reorganizing we get the Slutsky equation ∂xj (p, m) ∂hj (p, u) ∂xj (p, e(p, u)) = + (xi − xi(p, m)). ∂pi ∂pi ∂m 8.10.b Draw a diagram, play with it and verify that Dave is better oﬀ when p2 goes down and worse oﬀ when p1 goes down. Just look at the sets of allocations that are strictly better or worse than the original choice—i.e., the sets SB(x) = {z : z x} and SW (x) = {z : z x}. When p1 goes down the new budget set is contained in SW (x), while when p2 goes down there’s a region of the new budget set that lies in SB(x). Image 26: Ch. 8 CHOICE 23 8.10.c The rate of return—also known as “own rate of interest”—on good x is (p1 /p2 ) − 1 8.11 No, because his demand behavior violates GARP. When prices are (2, 4) he spends 10. At these prices he could aﬀord the bundle (2, 1), but rejects it; therefore, (1, 2) (2, 1). When prices are (6, 3) he spends 15. At these prices he could aﬀord the bundle (1, 2) but rejects it; therefore, (2, 1) (1, 2). 8.12 Inverting, we have e(p, u) = u/f(p). Substituting, we have µ(p; q, y) = v(q, y)/f(p) = f(q)y/f(p). 8.13.a Draw the lines x2 + 2x1 = 20 and x1 + 2x2 = 20. The indiﬀerence curve is the northeast boundary of this X. 8.13.b The slope of a budget line is −p1 /p2 . If the budget line is steeper than 2, x1 = 0. Hence the condition is p1 /p2 > 2. 8.13.c Similarly, if the budget line is ﬂatter than 1/2, x2 will equal 0, so the condition is p1 /p2 < 1/2. 8.13.d If the optimum is unique, it must occur where x2 − 2x1 = x1 − 2x2 . This implies that x1 = x2 , so that x1 /x2 = 1. α 8.14.a This is an ordinary Cobb-Douglas demand: S1 = α+β+γ Y and β S2 = α+β+γ Y . 8.14.b In this case the utility function becomes U (C, S1 , L) = S1 Lβ C γ . α The L term is just a constant, so applying the standard Cobb-Douglas α formula S1 = α+γ Y . s 8.15 Use Slutsky’s equation to write: ∂L = ∂L + (L − L) ∂L . Note that ∂w ∂w ∂m the substitution eﬀect is always negative, (L − L) is always positive, and hence if leisure is inferior, ∂L is necessarily negative. Thus the slope of ∂w the labor supply curve is positive. 8.16.a True. With the grant, the consumer will maximize u(x1 , x2 ) subject to x1 + x2 ≤ m + g1 and x1 ≥ g1 . We know that when he maximizes his utility subject to x1 + x2 ≤ m, he chooses x∗ ≥ g1 . Since x1 is a normal 1 good, the amount of good 1 that he will choose if given an unconstrained grant of g1 is some number x1 > x∗ ≥ g1 . Since this choice satisﬁes the 1 constraint x1 ≥ g1 , it is also the choice he would make when forced to spend g1 on good 1. Image 27: 24 ANSWERS 8.16.b False. Suppose for example that g1 = x∗ . Then if he gets an 1 unconstrained grant of g1 , since good 1 is inferior, he will choose to reduce his consumption to less than x∗ = g1 . But with the constrained grant, he 1 must consume at least g1 units of good 1. Incidentally, he will accept the grant, since with the grant he can always consume at least as much of both goods as without the grant. 8.16.c If he got an unconstrained grant of g1 , he would spend (48 + g1 )/4 on good 1. This is exactly what he will spend if g1 ≤ (48 + g1 )4. But if g1 > (48 + g1 )/4, he will spend g1 on good 1. The curve therefore has slope 1/4 if g1 < 16 and slope 1 if g1 > 16. Kink is at g1 = 16. Chapter 9. Demand 9.1 If preferences are homothetic, demand functions are linear in income, so we can write xi (p)m and xj (p)m. Applying Slutsky symmetry, we have ∂xi (p) ∂xj + xi(p)xj (p)m = + xj (p)xi (p)m. ∂pj ∂pi Subtracting xi (p)xj (p)m from each side of the equation establishes the result. 9.2 Note that p is the relative price of good x with respect to the other good which we’ll call z. Also, let m be income measured in units of z. Thus the consumer’s budget constraint is px + z = m. How do you know that there must be another good around? From dµ(p; q, m) = a + bp, dp we ﬁnd bp2 µ(p; q, m) = ap + + C. 2 Here C is a constant of integration. Since µ(q; q, m) = m, note that bq 2 C = m − aq − ; 2 therefore, bp2 bq 2 µ(p; q, m) = ap + + m − aq − . 2 2 A money metric utility function behaves like an indirect utility function with respect to q and m when holding p ﬁxed. Therefore, an indirect utility function consistent with the demand function given above is bq 2 v(q, m) = m − aq − . 2 Image 28: Ch. 9 DEMAND 25 We can drop the terms in p since they are just constants. (Use Roy’s identity to check that this indirect utility is indeed consistent with our original demand function.) To get the direct utility function∗ we must solve bq 2 u(x, z) = min{v(q, m) : qx + z = m} = min{m − aq − : qx + z = m}; q q 2 use the budget constraint to eliminate m from the objective function and get the optimal value of q x−a q∗ = . b Thus (x − a)2 u(x, z) = z + . 2b This is, of course, a quasilinear utility function. On the back of an envelope, solve maxx,z {u(x, z) : px + z = m} and check that you get precisely the original demand function for x. What’s the demand for z? 9.3 We have to solve dµ(p; q, m) = a + bp + cµ(p; q, m). dp The homogeneous part has a solution of the form Aecp . A particular solu- tion to the nonhomogeneous equation is given by (a + bp)c + b µ=− . c2 Therefore the general solution to the diﬀerential equation is given by (a + bp)c + b µ(p; q, m) = Aecp − . c2 Since µ(q; q, m) = m we get (a + bq)c + b (a + bp)c + b µ(p; q, m) = m + ec(p−q) − . c2 c2 Hence, the indirect utility function is (a + bq)c + b v(q, m) = m + e−cq . c2 ∗ Strictly speaking, we should be saying “a utility function consistent with the given demand,” but we’ll just say “the utility function” with the understanding that any monotonic transformation of it would also generate the same demand function. Image 29: 26 ANSWERS (Verify that using Roy’s identity we get the original demand function.) To get the direct utility function, we must solve (a + bq)c + b min m+ e−cq q c2 such that qx + z = m. The optimal value is given by x − cz − a q∗ = , b + cx which implies that b + cx ac − cx + c2 z u(x, z) = exp . c2 b + cx (Again, substitute z by m−px above, equate the derivative of the resulting expression with respect to x to zero, solve for x and recover the original demand function.) 9.4 Now the budget constraint is given by z + p1 x1 + p2 x2 = m. The symmetry of the substitution eﬀects implies ∂x1 ∂x2 = =⇒ b12 = b21 . ∂p2 ∂p1 The negative semideﬁniteness of the substitution matrix implies b1 < 0 and b1 b2 − b2 > 0. (Prove that these two conditions together imply that b2 < 0 must also hold.) We have to solve the following system of partial diﬀerential equations ∂µ(p; q, m) = a1 + b1 p1 + bp2 , ∂p1 and ∂µ(p; q, m) = a2 + bp1 + b2 p2 . ∂p2 The ﬁrst equation implies b1 2 µ(p; q, m) = a1 p1 + p + bp1 p2 + C1 , 2 1 where C1 is a constant of integration. The second implies b2 2 µ(p; q, m) = a2 p2 + p + bp1 p2 + C2 . 2 2 ![Image 30: Ch. 9 DEMAND 27 Therefore, we must have b1 2 b2 µ(p; q, m) = a1 p1 + p + bp1 p2 + a2 p2 + p2 + C 2 1 2 2 a 1 b b p1 = [p1 , p2 ] 1 + [p1 , p2 ] 1 + C. a2 2 b b2 p2 Using µ(q; q, m) = m, we have b1 2 µ(p; q, m) = m + a1 (p1 − q1 ) + (p − q1 ) + b(p1 p2 − q1 q2 ) 2 2 1 b2 + a2 (p2 − q2 ) + (p2 − q2 ). 2 2 2 The indirect utility function is given by b1 2 b2 2 v(q, m) = m − a1 q1 − q1 − bq1 q2 − a2 q2 − q2 2 2 a1 1 b1 b q1 = m − [q1 , q2 ] − [q1 , q2] . a2 2 b b2 q2 9.5 To get the direct utility function we must solve u(x, z) = min{v(q, m) : z + q1 x1 + q2 x2 = m}. q After a few minutes of algebraic fun, we get ∗ b2 (x1 − a1 ) − b(x2 − a2 ) q1 = , b1 b2 − b2 and ∗ b1 (x2 − a2 ) − b(x1 − a1 ) q2 = . b1 b2 − b2 Substituting these values back into v(·), we get u(x, z) b2 (x1 − a1 )2 + b1 (x2 − a2 )2 b(a1 x2 + a2 x1 − x1 x2 − a1 a2 ) =z+ + 2(b1 b2 − b2 ) b1 b2 − b2 1 b −b x 1 − a1 =z+ [x1 − a1 , x2 − a2 ] 2 . 2(b1 b2 − b2 ) −b b1 x 2 − a2 9.6 Write the indirect utility function as v(p) = v(q/m) and diﬀerentiate with respect to qi and m: ∂v ∂v 1 = ∂qi ∂pi m k ∂v ∂v qi =− ∂m i=1 ∂pi m2 k ∂v 1 =− pi . i=1 ∂pi m ]( Image 31: 28 ANSWERS Dividing ∂v by ∂v yields the result. ∂qi ∂m 9.7 The function is weakly separable, and the subutility for the z-good b c consumption is z2 z3 . The conditional demands for the z-goods are Cobb- Douglas demands: b mz z1 = b + c p2 c mz z2 = . b + c p3 9.8.a dµ dp = a − bp + cµ 9.8.b µ(q, q, y) ≡ y 9.9.a The function V (x, y) = min{x, y}, and U (V, z) = V + z. 9.9.b The demand function for the z-good is z = m/pz if pz < px + py . If pz > px + py , then the demand for the x-good and the y-good is given by x = y = m/(px + py ). If pz = px + py , then take any convex combination of these demands. 9.9.c The indirect utility function is m m v(px , py , pz , m) = max , . px + py pz 9.11.a There are a variety of ways to solve this problem. The easiest is to solve for the indirect utility function to get v1 (p1 , p2 , m1 ) = m1 (p1 p2 )−1/2 . Now use Roy’s identity to calculate: 1 m1 x1 = 2 p1 1 m1 x2 = . 2 p2 Note that these are Cobb-Douglas demands. Recognizing that person 2 has Cobb-Douglas utility, we can write down the demands immediately: 3 m2 x1 = 3 + a p1 a m2 x2 = . 3 + a p2 9.11.b We must have the marginal propensity to consume each good the same for each consumer. This means that 1 3 = , 2 3+a which implies that a = 3. Image 32: Ch. 11 UNCERTAINTY 29 Chapter 10. Consumers’ Surplus 10.1 We saw that in this case the indirect utility function takes the form v(p)+m. Hence the expenditure function takes the form e(p, u) = u−v(p). The expenditure function is necessarily a concave function of prices, which implies that v(p) is a convex function. 10.2 Ellsworth’s demand functions for the x-good and the y-good take the form 150 x=y= px + py . Plugging this into the utility function, we ﬁnd that the indirect utility function takes the form 150 v(px , py , 150) = . px + py Hence A is the solution to 150 − A 150 = 1+1 1+2 and B is the solution to 150 150 + B = . 1+1 1+2 Solving, we have A = 50 and B = 75. Chapter 11. Uncertainty 11.1 The proof of Pratt’s theorem established that 1 π(t) ≈ r(w)σ2 t2 . 2 But the σ2 t2 is simply the variance of the gamble t˜. 11.2 If risk aversion is constant, we must solve the diﬀerential equation u (x)/u (x) = −r. The answer is u(x) = −e−rx , or any aﬃne transforma- tion of this. If relative risk aversion is constant, the diﬀerential equation is u (x)x/u (x) = −r. The solution to this is u(x) = x1−r /(1 − r) for r = 1 and u(x) = ln x for r = 1. 11.3 We have seen that investment in a risky asset will be independent of wealth if risk aversion is constant. In an earlier problem, we’ve seen that ![Image 33: 30 ANSWERS constant absolute risk aversion implies that the utility function takes the form u(w) = −e−rw . 11.4 Marginal utility is u (w) = 1 − 2bw; when w is large enough this is a negative number. Absolute risk aversion is 2b/(1 − 2bw). This is an increasing function of wealth. 11.5.a The probability of heads occurring for the ﬁrst time on the j th toss ∞ is (1 − p)j−1p. Hence the expected value of the bet is j=1 (1 − p)j−1p2j = ∞ −j j ∞ j=1 2 2 = j=1 1 = ∞. 11.5.b The expected utility is ∞ ∞ (1 − p)j−1 p ln(2j ) = p ln(2) j(1 − p)j−1 . j=1 j=1 11.5.c By standard summation formulas: ∞ 1 (1 − p)j = . p j=0 Diﬀerentiate both sides of this expression with respect to p to obtain ∞ 1 j(1 − p)j−1 = . p2 j=1 Therefore, ∞ ln(2) p ln(2) j(1 − p)j−1 = . p j=1 11.5.d In order to solve for the amount of money required, we equate the utility of participating in the gamble with the utility of not participating. This gives us: ln(2) ln(w0 ) = , p Now simply solve this equation for w0 to ﬁnd w0 = eln(2)/p. 11.6.a Note that ∞ 2 1 1 s−µ E[u(R)] = u(s) √ exp − ds = φ(µ, σ2 ). −∞ σ 2π 2 σ ]( ![Image 34: Ch. 11 UNCERTAINTY 31 11.6.b Normalize u(·) such that u(µ) = 0. Diﬀerentiating, we have ∞ ∂E[u(R)] 1 = 2 u(s)(s − µ)f(s)ds > 0, ∂µ σ −∞ since the terms [u(s)(s − µ)] and f(s) are positive for all s. 11.6.c Now we have ∞ ∂E[u(R)] 1 = 3 u(s)((s − µ)2 − σ2 )f(s) ds ∂σ2 σ −∞ ∞ 1 < 3 u (µ)(s − µ)((s − µ)2 − σ2 )f(s)ds σ −∞ ∞ ∞ u (µ) = (s − µ)3 f(s)ds − σ2 (s − µ)f(s)ds σ3 −∞ −∞ = 0. The ﬁrst inequality follows from the concavity of u(·) and the normaliza- tion imposed; the last equality follows from the fact that R is normally distributed and, hence, E[(R − E[R])k ] = 0 for k odd. 11.7 Risk aversion implies a concave utility function. Denote by α ∈ [0, 1] the proportion of the initial wealth invested in asset 1. We have E[u(αw0 (1 + R1 ) + (1 − α)w0 (1 + R2 ))] = u(αw0 (1 + r1 ) + (1 − α)w0 (1 + r2 ))f(r1 )f(r2 )dr1 dr2 > [αu(w0 (1 + r1 )) + (1 − α)u(w0 (1 + r2 ))]f(r1 )f(r2 )dr1 dr2 = u(w0 (1 + r1 ))f(r1 )dr1 = u(w0 (1 + r2 ))f(r2 )dr2 = = E[u(w0 (1 + R1 ))] = E[u(w0 (1 + R2 ))]. The inequality follows from the concavity of u(·). For part b, proceed as before reversing the inequality since now u(·) is convex. 11.8.a Start by expanding both sides of E[u(w − πu )] = E[u(w + ˜)]: ˜ ˜ E[u(w − πu )] = pu(w1 − πu ) + (1 − p)u(w2 − πu ) ˜ ≈ p(u(w1 ) − u (w1 )πu ) + (1 − p)(u(w2 ) − u (w2 )πu ); p E[u(w + ˜)] = ˜ (u(w1 − ) + u(w1 + )) + (1 − p)u(w2 ) 2 u (w1 ) 2 ≈ p u(w1 ) + + (1 − p)u(w2 ). 2 ]( Image 35: 32 ANSWERS Combining, we obtain 1 −(pu (w1 ) + (1 − p)u (w2 ))πu ≈ pu (w1 ) 2 , 2 or − 1 pu (w1 ) 2 πu ≈ 2 . pu (w1 ) + (1 − p)u (w2 ) 11.8.b For these utility functions, the Arrow-Pratt measures are −u /u = a, and −v /v = b. 11.8.c We are given a > b and we want to show that a value of (w1 − w2 ) large enough will eventually imply πv > πu , thus we want to get ae−aw1 be−bw1 < −bw1 ; pe−aw1 + (1 − p)e−aw2 pe + (1 − p)e−bw2 cross-multiplying we get ape−w1 (a+b) + a(1 − p)e−(aw1 +bw2 ) < bpe−w1 (a+b) + b(1 − p)e−(aw1 +bw2 ) , which implies p (a − b) < bea(w1 −w2 ) − aeb(w1 −w2 ) . 1−p The derivative of the RHS of this last inequality with respect to w1 − w2 is ab ea(w1 −w2 ) − eb(w1 −w2 ) > 0 whenever w1 > w2 ; the LHS does not depend on w1 or w2 . Therefore, this inequality will eventually hold for (w1 − w2 ) large enough. According to the Arrow-Pratt measure, u exhibits a higher degree of risk aversion than v. We’ve shown that v could imply a higher risk premium than u to avoid a fair lottery provided there’s an additional risk “big” enough. In this case, higher risk premium would no longer be synonymous with higher absolute risk aversion. 11.9 Initially the person has expected utility of 1√ 1√ 4 + 12 + 4 + 0 = 3. 2 2 If he sells his ticket for price p, he needs to get at least this utility. To ﬁnd the breakeven price we write the equation 4 + p = 3. Image 36: Ch. 13 COMPETITIVE MARKETS 33 Solving, we have p = 5. 11.10 The utility maximization problem is max π ln(w+x)+(1−π) ln(w−x). The ﬁrst-order condition is π 1−π = , w+x w−x which gives us x = w(2π − 1). If π = 1/2, x = 0. 11.11 We want to solve the equation p 1−p 1 + = . w1 w2 w After some manipulation we have w1 w2 w= . pw2 + (1 − p)w1 11.12.a max αpx x such that x is in X. 11.12.b v(p, α) has the same form as the proﬁt function. 11.12.c Just mimic the proof used for the proﬁt function. 11.12.d It must be monotonic and convex, just as in the case of the proﬁt function. Chapter 13. Competitive Markets 13.1 The ﬁrst derivative of welfare is v (p) + π (p) = 0. Applying Roy’s law and Hotelling’s lemma, we have −x(p) + y(p) = 0, which is simply the condition that demand equals supply. The second derivative of this welfare measure is −x (p)+y (p) which is clearly positive; hence, we have a welfare minimum rather than a welfare maximum. The intuition behind this is that at any price other than the equilibrium price, the ﬁrm wants to supply a diﬀerent amount than the consumer wants to demand; hence, the “welfare” associated with all prices other than the equilibrium price is not attainable. 13.2 By Hotelling’s law we know that ∂π(p, w)/∂p = y(p); therefore, p1 y(p)dp = π(p1 , w) − π(p0 , w). p0 Image 37: 34 ANSWERS 13.3.a The average cost curve is just c(w, y) y2 + 1 y2 + 2 = w1 + w2 . y y y You should verify that it is a convex function that has a unique minimum at w1 /w2 + 2 ym = . w1 /w2 + 1 The derivative of ym with respect to w1 /w2 is negative, so the minimum of the average cost shifts to the left (right) as w1 /w2 increases (decreases). √ In fact it converges to 1 as the ratio approaches ∞ and to 2 as it goes down to 0. 13.3.b The marginal cost is ∂c(w, y) = 2y(w1 + w2 ), ∂y so short-run supply schedule is given by p y(p) = . 2(w1 + w2 ) 13.3.c The long-run supply curve is Y (p) = arbitrarily large amount if p > 2ym (w1 + w2 ) 0 otherwise. 13.3.d From the cost function we have that x1 = y2 + 1 and x2 = y2 + 2. Also, we see that x1 and x2 are not substitutes at any degree. Therefore, the input requirement set for an individual ﬁrm is √ √ V (y) = (x1 , x2) ∈ [1, ∞) × [2, ∞): y ≤ min x1 − 1, x2 − 2 . 13.4.a y(p) = p/2 13.4.b Y (p) = 50p 13.4.c Equating D(p) = Y (p), we get p∗ = 2 and y∗ = 1 (Y ∗ = 100). 13.4.d The equilibrium rent on land r must equal the diﬀerence between each ﬁrm revenues and labor costs at the competitive equilibrium. There- fore, r = 2 − 1 = 1. ![Image 38: Ch. 13 COMPETITIVE MARKETS 35 13.5.a In order to oﬀset the output subsidy, the U.S. should choose a tax of the same size as the subsidy; that is, choose t(s) = s. 13.5.b In the case of the capital subsidy, the producers receive p − t(s). If y∗ is to remain optimal, we must have p − t(s) = ∂c(w, r − s, y∗ )/∂y. 13.5.c Diﬀerentiate the above expression to get ∂ 2 c(w, r − s, y∗ ) ∂K(w, r − s, y∗ ) t (s) = = . ∂y∂r ∂y 13.5.d Since K(w, r − s, y) = K(w, r − s, 1)y, the formula reduces to t (s) = K(w, r − s, 1). 13.5.e In this case ∂K/∂y < 0 so that an increase in the subsidy rate implies a decrease in the tariﬀ. 13.6 Each ﬁrm that has marginal cost less than 25 will produce to capacity. What about the ﬁrm that has marginal cost equal to 25? If it produces a positive amount, it will just cover its variable cost, but lose the quasiﬁxed cost. Hence it prefers to stay out of business. This means that there will be 24 ﬁrms in the market, each producing 12 units of output, giving a total supply of 288. 13.7.a ym = 500 13.7.b p = 5 13.7.c yc = 50 × 5 = 250 13.8.a Price equals marginal cost gives us p = y, so Y = p + p = 2p. 13.8.b Set demand equal to supply 90−p = 2p to ﬁnd p∗ = 30 and Y ∗ = 60. 13.8.c Let p be the price paid by consumers. Then the domestic ﬁrms receive a price of p and the foreign ﬁrms receive a price of p − 3. Demand equals supply gives us 90 − p = p + [p − 3]. Solving we have p∗ = 31. 13.8.d The supply of umbrellas by domestic ﬁrms is 31 and by foreign ﬁrms is 28. ]( Image 39: 36 ANSWERS Chapter 14. Monopoly 14.1 The proﬁt-maximizing level of output is 5 units. If the monopolist only has 4 to sell, then it would ﬁnd it most proﬁtable to charge a price of 6. This is the same as the competitive solution. If, however, the monopolist had 6 units to sell, it would be most proﬁtable to dispose of one unit and only sell 5 units at a price of 5. 14.2 The monopolist has zero marginal costs up until 7 units of output and inﬁnite marginal costs for any output greater than 7 units. The proﬁt- maximizing price is 5 and the proﬁts are 25. 14.3 For this constant elasticity demand function revenue is constant at 10, regardless of the level of output. Hence output should be as small as possible—that is, a proﬁt-maximizing level of output doesn’t exist. 14.4 According to the formula given in the text, we must have 1 = 1, 2 + yp (y)/p (y) or yp (y) = −p (y). The required inverse demand function is the solution to this diﬀerential equation. It turns out that it is given by p(y) = a − b ln x. The direct demand function then takes the form ln x = a/b − p/b, which is sometimes called a semilog demand function. 14.5 The monopolist’s proﬁt maximization problem is max p(y, t)y − cy. y The ﬁrst-order condition for this problem is ∂p(y, t) p(y, t) + y − c = 0. ∂y According to the standard comparative statics calculations, the sign of dy/dt is the same as the sign of the derivative of the ﬁrst-order expression with respect to t. That is, dy ∂p ∂p2 sign = sign + y. dt ∂t ∂y∂t For the special case p(y, t) = a(p) + b(t), the second term on the right-hand side is zero. Image 40: Ch. 14 MONOPOLY 37 14.6 There is no proﬁt-maximizing level of output since the elasticity of demand is constant at −1. This means that revenue is independent of output, so reductions in output will lower cost but have no eﬀect on revenue. 14.7 Since the elasticity of demand is −1, revenues are constant at any price less than or equal to 20. Marginal costs are constant at c so the monopolist will want to produce the smallest possible output. This will happen when p = 20, which implies y = 1/2. 14.8 For this to occur, the derivative of consumer’s surplus with respect to quality must be zero. Hence ∂u/∂q − ∂p/∂qx ≡ 0. Substituting for the deﬁnition of the inverse demand function, this means that we must have ∂u/∂q ≡ x∂ 2 u/∂x∂q. It is easy to verify that this implies that u(x, q) = f(q)x. 14.9 The integral to evaluate is x x ∂ 2 u(z, q) ∂p(x, q) dz < dz. 0 ∂z∂q 0 ∂q Carrying out the integration gives ∂u(x, q) ∂p(x, q) < x, ∂q ∂q which is what is required. 14.10 If the ﬁrm produces x units of output which it sells at price p(x), then the most that it can charge for entry is the consumer’s surplus, u(x)−p(x)x. Once the consumer has chosen to enter, the ﬁrm makes a proﬁt of p(x) − c on each unit of output purchased. Thus the proﬁt maximization problem of the ﬁrm is max u(x) − p(x)x + (p(x) − c(x))x = u(x) − c(x). x It follows that the monopolist will choose the eﬃcient level of output where u (x) = c (x). The entry fee is set equal to the consumer’s surplus. 14.11 The ﬁgure depicts the situation where the monopolist has reduced the price to the point where the marginal beneﬁt from further reductions just balance the marginal cost. This is the point where p2 = 2p1 . If the high-demand consumer’s inverse demand curve is always greater than twice the low-demand consumer’s inverse demand curve, this condition cannot be satisﬁed and the low-demand consumer will be pushed to a zero level of consumption. ![Image 41: 38 ANSWERS 14.12 Area B is what the monopolist would gain by selling only to the high-demand consumer. Area A is what the monopolist would lose by doing this. 14.13 This is equivalent to the price discrimination problem with x = q and wt = rt . All of the results derived there translate on a one-to-one basis; e.g., the consumer who values quality the more highly ends up consuming the socially optimal amount, etc. 14.14 The maximization problem is maxp py(p) − c(y(p)). Diﬀerentiating, we have py (p) + y(p) − c (y)y (p) = 0. This can also be written as p + y(p)/y (p) − c (y) = 0, or p[1 + 1/ ] = c (y). 14.15 Under the ad valorem tax we have 1 (1 − τ )PD = 1+ c. Under the output tax we have 1 PD − t = 1 + . Solve each equation for PD , set the results equal to each other, and solve for t to ﬁnd τ kc 1 t= k= 1−τ 1+ 1 14.16.a The monopolist’s proﬁt maximization problem is max p(y, t)y − cy. y The ﬁrst-order condition for this problem is ∂p(y, t) p(y, t) + y − c = 0. ∂y According to the standard comparative statics calculations, the sign of dy/dt is the same as the sign of the derivative of the ﬁrst-order expression with respect to t. That is, dy ∂p ∂p2 sign = sign + y. dt ∂t ∂y∂t ]( Image 42: Ch. 14 MONOPOLY 39 14.16.b For the special case p(y, t) = a(y) + b(t), the second term on the right-hand side is zero, so that ∂p/∂t = ∂b/∂t. 14.17.a Diﬀerentiating the ﬁrst-order conditions in the usual way gives ∂x1 1 = <0 ∂t1 p1 − c1 ∂x2 1 = < 0. ∂t2 2p2 + p2 x2 − c2 14.17.b The appropriate welfare function is W = u1 (x1 )+u2 (x2 )−c1 (x1 )− c2 (x2 ). The total diﬀerential is dW = (u1 − c1 )dx1 + (u2 − c2 )dx2 . 14.17.c Somewhat surprisingly, we should tax the competitive industry and subsidize the monopoly! To see this, combine the answers to the ﬁrst two questions to get the change in welfare from a tax policy (t1 , t2 ). dx1 dx2 dW = (p1 − c1 ) dt1 + (p2 − c2 ) dt2 . dt1 dt2 The change in welfare from a small tax or subsidy on the competitive in- dustry is zero, since price equals marginal cost. But for the monopolized industry, price exceeds marginal cost, so we want the last term to be posi- tive. But this can only happen if dt2 is negative—i.e., we subsidize industry 2. 14.18.a The proﬁt maximization problem is max r1 + r2 such that a1 x1 − r1 ≥ 0 a2 x 2 − r 2 ≥0 a1 x 1 − r 1 ≥ a1 x 2 − r 2 a2 x2 − r2 ≥ a2 x 1 − r 1 x1 + x2 ≤ 10. 14.18.b The binding constraints will be a1 x1 = r1 and a2 x2 −r2 = a2 x1 −r1 , and x1 + x2 = 10. 14.18.c The expression is a2 x2 + (2a1 − a2 )x1 . ![Image 43: 40 ANSWERS 14.18.d Formally, our problem is to solve max a2 x2 + (2a1 − a2 )x1 subject to the constraint that x1 + x2 = 10. Solve the constraint for x2 = 10 − x1 and substitute into the objective function to get the problem max 10a2 + 2(a1 − a2 )x1 . x1 Since a2 > a1 the coeﬃcient on the second term is negative, which means that x∗ = 0 and, therefore, x∗ = 10. Since x∗ = 10, we must have r2 = 1 2 2 ∗ ∗ ∗ 10a2 . Since x1 = 0, we must have r1 = 0. 14.19.a The proﬁt-maximizing choices of p1 and p2 are p1 = a1 /2b1 p2 = a2 /2b2 . These will be equal when a1 /b1 = a2 /b2 . 14.19.b We must have p1 (1 − 1/b1 ) = c = p2 (1 − 1/b2 ). Hence p1 = p2 if and only if b1 = b2 . 14.20.a The ﬁrst-order condition is (1 − t)[p(x) + p (x)x] = c (x), or p(x) + p (x)x = c (x)/(1 − t). This expression shows that the revenue tax is equivalent to an increase in the cost function, which can easily be shown to reduce output. 14.20.b The consumer’s maximization problem is maxx u(x) − m − px + tpx = maxx u(x) − m − (1 − t)px. Hence the inverse demand function satisﬁes u (x) − (1 − t)p(x), or p(x) = u (x)/(1 − t). 14.20.c Substituting the inverse demand function into the monopolist’s ob- jective function, we have (1 − t)p(x)x − c(x) = (1 − t)u (x)x/(1 − t) − c(x) = u (x)x − c(x). Since this is independent of the tax rate, the monopolist’s behavior is the same with or without the tax. 14.21 Under the ad valorem tax we have 1 (1 − τ )PD = 1+ c. Under the output tax we have 1 PD − t = 1 + . ]( Image 44: Ch. 15 GAME THEORY 41 Solve each equation for PD , set the results equal to each other, and solve for t to ﬁnd τ kc 1 t= k= 1−τ 1+ 1 14.22.a Note that his revenue is equal to 100 for any price less than or equal to 20. Hence the monopolist will want to produce as little output as possible in order to keep its costs down. Setting p = 20 and solving for demand, we ﬁnd that D(20) = 5. 14.22.b They should set price equal to marginal cost, so p = 1. 14.22.c D(1) = 100. 14.23.a If c < 1, then proﬁts are maximized at p = 3/2 + c/2 and the monopolist sells to both types of consumers. The best he can do if he sells only to Type A consumers is to sell at a price of 2 + c/2. He will do this if c ≥ 1. 14.23.b If a consumer has utility ax1 −x2 /2+x2, then she will choose to pay 1 k if (a−p)2 /2 > k. If she buys, she will buy a−p units. So if k < (2−p)2 /2, then demand is N (4 − p) + N (2 − p). If (2 − p)2 < k < (4 − p)2 /2, then demand is N (4 − p). If k > (4 − p)2 /2, then demand is zero. 14.23.c Set p = c and k = (4 − c)2 /2. The proﬁt will be N (4 − c)2 /2. 14.23.d In this case, if both types of consumers buy the good, then the proﬁt-maximizing prices will have the Type B consumers just indiﬀerent between buying and not buying. Therefore k = (2 − p)2 /2. Total proﬁts will then be N ((6 − 2p)(p − c) + (2 − p)2 /2). This is maximized when p = 2(c + 2)/3. Chapter 15. Game Theory 15.1 There are no pure strategy equilibria and the unique mixed strategy equilibrium is for each player to choose Head or Tails with probability 1/2. 15.2 Simply note that the dominant strategy on the last move is to defect. Given that this is so, the dominant strategy on the next to the last move is to defect, and so on. 15.3 The unique equilibrium that remains after eliminating weakly domi- nant strategies is (Bottom, Right). 15.4 Since each player bids v/2, he has probability v of getting the item, giving him an expected payoﬀ of v2 /2. Image 45: 42 ANSWERS 15.5.a a ≥ e, c ≥ g, b ≥ d, f ≥ h 15.5.b Only a ≥ e, b ≥ d. 15.5.c Yes. 15.6.a There are two pure strategy equilibria, (Swerve, Stay) and (Stay, Swerve). 15.6.b There is one mixed strategy equilibrium in which each player chooses Stay with probability .25. 15.6.c This is 1 − .252 = .9375.. 15.7 If one player defects, he receives a payoﬀ of πd this period and πc forever after. In order for the punishment strategy to be an equilibrium the payoﬀs must satisfy πc πj πd + ≤ πj + . r r Rearranging, we ﬁnd πj − πc r≤ πd − πj . 15.8.a Bottom. 15.8.b Middle. 15.8.c Right. 15.8.d If we eliminate Right, then Row is indiﬀerent between his two re- maining strategies. 15.9.a (Top, Left) and (Bottom, Right) are both equilibria. 15.9.b Yes. (Top, Left) dominates (Bottom, Right). 15.9.c Yes. 15.9.d (Top, Left). Chapter 16. Oligopoly 16.1 The Bertrand equilibrium has price equal to the lowest marginal cost, c1 , as does the competitive equilibrium. Image 46: Ch. 16 OLIGOPOLY 43 16.2 ∂F (p, u)/∂u = 1 − r/p. Since r is the largest possible price, this expression will be nonpositive. Hence, increasing the ratio of uninformed consumers decreases the probability that low prices will be charged, and increases the probability that high prices will be charged. 16.3 Let δ = β1 β2 − γ 2 . Then by direct calculation: ai = (αi βj − αj γ)/δ, bi = βj /δ, and c = γ/δ. 16.4 The calculations are straightforward and may be found in Singh & Vives (1984). Let ∆ = 4β1 β2 − γ 2 , and D = 4b1 b2 − c2 . Then it turns out that pc − pb = αiγ 2 /∆ and qi − qi = aic2 /D, where superscripts refer to i i b c Bertrand and Cournot. 16.5 The argument is analogous to the argument given on page 297. 16.6 The problem is that the thought experiment is phrased wrong. Firms in a competitive market would like to reduce joint output, not increase it. A conjectural variation of −1 means that when one ﬁrm reduces its output by one unit, it believes that the other ﬁrm will increase its output by one unit, thereby keeping joint output—and the market price—unchanged. 16.7 In a cartel the ﬁrms must equate the marginal costs. Due to the assumption about marginal costs, such an equality can only be established when y1 > y2 . 16.8 Constant market share means that y1 /(y1 + y2 ) = 1/2, or y1 = y2 . Hence the conjectural variation is 1. We have seen that the conjectural variation that supports the cartel solution is y2 /y1 . In the case of identical ﬁrms, this is equal to 1. Hence, if each ﬁrm believes that the other will attempt to maintain a constant market share, the collusive outcome is “stable.” 16.9 In the Prisoner’s Dilemma, (Defect, Defect) is a dominant strategy equilibrium. In the Cournot game, the Cournot equilibrium is only a Nash equilibrium. 16.10.a Y = 100 16.10.b y1 = (100 − y2 )/2 16.10.c y = 100/3) 16.10.d Y = 50 16.10.e y1 = 25, y2 = 50 16.11.a P (Y ) + P (Y )yi = c + ti ![Image 47: 44 ANSWERS 16.11.b Sum the ﬁrst order conditions to get nP (Y ) + P (Y )Y = nc + n i=1 ti , and note that industry output Y can only depend on the sum of the taxes. 16.11.c Since total output doesn’t change, ∆yi must satisfy P (Y ) + P (Y )[yi + ∆yi] = c + ti + ∆ti . Using the original ﬁrst order condition, this becomes P (Y )∆yi = ∆ti, or ∆yi = ∆ti /P (Y ). 16.12.a y = p 16.12.b y = 50p 16.12.c Dm (p) = 1000 − 100p 16.12.d ym = 500 16.12.e p = 5 16.12.f yc = 50 × 5 = 250 16.12.g Y = ym + yc = 750. Chapter 17. Exchange 17.1 In the proof of the theorem, we established that x∗ ∼i xi . If x∗ and xi i i were distinct, a convex combination of the two bundles would be feasible and strictly preferred by every agent. This contradicts the assumption that x∗ is Pareto eﬃcient. 17.2 The easiest example is to use Leontief indiﬀerence curves so that there are an inﬁnite number of prices that support a given optimum. 17.3 Agent 2 holds zero of good 2. 17.4 x1 = ay/p1 = ap2 /p1 , x1 = x2 so from budget constraint, (p1 + A B B p2 )x1 = p1 , so x1 = p1 /(p1 + p2 ). Choose p1 = 1 an numeraire and solve B B ap2 + 1/(1 + p2 ) = 1. 17.5 There is no way to make one person better oﬀ without hurting someone else. 17.6 x1 = ay1 /p1 , x2 = by2 /p1 1 y1 = y2 = p1 + p2 . Solve x1 + x1 = 2. 1 2 ]( Image 48: Ch. 18 PRODUCTION 45 17.7 The Slutsky equation for consumer i is ∂xi ∂hi = . ∂pj ∂pj 17.8 The strong Pareto set consists of 2 allocations: in one person A gets all of good 1 and person B gets all of good 2. The other Pareto eﬃcient allocation is exactly the reverse of this. The weak Pareto set consists of all allocations where one of the consumers has 1 unit of good 1 and the other consumer has at least 1 unit of good units of good 2. 17.9 In equilibrium we must have p2 /p1 = x2 /x1 = 5/10 = 1/2. 3 3 17.10 Note that the application of Walras’ law in the proof still works. 17.11.a The diagram is omitted. 17.11.b We must have p1 = p2 . 17.11.c The equilibrium allocation must give one agent all of one good and the other agent all of the other good. Chapter 18. Production 18.1.a Consider the following two possibilities. (i) Land is in excess supply. (ii) All land is used. If land is in excess supply, then the price of land is zero. Constant returns requires zero proﬁts in both the apple and the bandanna industry. This means that pA = pB = 1 in equilibrium. Every consumer will have income of 15. Each will choose to consume 15c units of apples and 15(1 − c) units of bandannas. Total demand for land will be 15cN . Total demand for labor will be 15N . There will be excess supply of land if c < 2/3. So if c < 2/3, this is a competitive equilibrium. If all land is used, then the total outputs must be 10 units of apples and 5 units of bandannas. The price of bandannas must equal the wage which is 1. The price of apples will be 1 + r where r is the price of land. Since preferences are homothetic and identical, it will have to be that each person consumes twice as many apples as bandannas. People will want to consume c twice as much apples as bandannas if pA /pB = (1−c) (1/2). Then it also must be that in equilibrium, r = (pA /pB ) − 1 ≥ 0. This last inequality will hold if and only if c ≥ 2/3. This characterizes equilibrium for c ≥ 2/3. 18.1.b For c < 2/3. 18.1.c For c < 2/3. Image 49: 46 ANSWERS 18.2.a Let the price of oil be 1. Then the zero-proﬁt condition implies that pg 2x − x = 0. This means that pg = 1/2. A similar argument shows that pb = 1/3. 18.2.b Both utility functions are Cobb-Douglas, and each consumer has an endowment worth 10. From this we can easily calculate that xg = 8, 1 xb = 18, xg = 10, xb = 15. 1 2 2 18.2.c To make 18 guns, ﬁrm 1 needs 9 barrels of oil. To make 33 units of butter, ﬁrm 2 needs 11 barrels of oil. Chapter 19. Time 19.1 See Ingersoll (1987), page 238. 19.2.a Apartments will be proﬁtable to construct as long as the present value of the stream of rents is at least as large as the cost of construction. In equations: (1 + π)p p+ ≥ c. 1+r In equilibrium, this condition must be satisﬁed as an equality, so that 1+r p= c. 2+r+π 19.2.b Now the condition becomes 1+r p= c. 2 + r + 3π 4 19.2.c Draw the ﬁrst period demand curve and subtract oﬀ the K rent controlled apartments to get the residual demand for new apartments. Look for the intersection of this curve with the two ﬂat marginal cost curves derived above. 19.2.d Fewer. 19.2.e The equilibrium price of new apartments will be higher. Image 50: Ch. 22 WELFARE 47 Chapter 20. Asset Markets 20.1 The easiest way to show this is to write the ﬁrst-order conditions as ˜ ˜ ˜ Eu (C)Ra = Eu (C)R0 ˜ ˜ ˜ Eu (C)Rb = Eu (C)R0 and subtract. 20.2 Dividing both sides of the equation by pa and using the deﬁnition ˜ ˜ Ra = Va /pa , we have ˜ ˜ Ra = R0 − R0 cov(F (C), Ra ). Chapter 21. Equilibrium Analysis 21.1 The core is simply the initial endowment. 21.2 Since the income eﬀects are zero, the matrix of derivatives of the Marshallian demand function is equal to the matrix of derivatives of the Hicksian demand function. It follows from the discussion in the text that the index of every equilibrium must be +1, which means there can be only one equilibrium. 21.3 Diﬀerentiating V (p), we have dV (p) = −2z(p)Dz(p)p˙ dt = −2z(p)Dz(p)Dz(p)−1z(p) = −2z(p)z(p) < 0. Chapter 22. Welfare 22.1 We have the equation k ∂hj θxi = tj . ∂pi j=1 Multiply both sides of this equation by ti and sum to get k k ∂hj θR = θ tixi = ti tj . i j=1 i=1 ∂pi Image 51: 48 ANSWERS The right-hand side of expression is nonpositive (and typically negative) since the Slutsky matrix is negative semideﬁnite. Hence θ has the same sign as R. 22.2 The problem is max v(p, m) k such that (pi − ci )xi(pi ) = F. i=1 This is almost the same as the optimal tax problem, where pi − ci plays the role of ti. Applying the inverse elasticity rule gives us the result. Chapter 23. Public goods 23.1 Suppose that it is eﬃcient to provide the public good together, but neither agent wants to provide it alone. Then any set of bids such that b1 + b2 = c and bi ≤ ri is an equilibrium to the game. However, there are also many ineﬃcient equilibria, such as b1 = b2 = 0. 23.2 If utility is homothetic, the the consumption of each good will be proportional to wealth. Let the demand function for the public good be given by ai fi (w) = w. 1 + ai Then the equilibrium amount of the public good is the same as in the Cobb-Douglas example given in the text. 23.3 Agent 1 will contribute g1 = αw1 . Agent 2’s reaction function is f2 (w2 + g1 ) = max{α(w2 + g1 ) − g1 , 0}. Solving f2 (w2 + αw1 ) = 0 yields w2 = (1 − α)w1 . 23.4 The total amount of the public good with k contributors must satisfy w G G=α + . k k Solving for G, we have G = αw/(k − α). As k increases, the amount of wealth becomes more equally distributed and the amount of the privately provided public good decreases. 23.5 The allocation is not in general Pareto eﬃcient, since for some patterns of preferences some of the private good must be thrown away. However, the amount of the public good provided will be the Pareto eﬃcient amount: 1 unit if i ri > c, and 0 units otherwise. ![Image 52: Ch. 24 EXTERNALITIES 49 23.6.a max ai ln(G−i + gi ) + wi − gi gi such that gi ≥ 0. 23.6.b The ﬁrst-order condition for an interior solution is ai = 1, G or G = ai . Obviously, the only agent who will give a positive amount is the one with the maximum ai . 23.6.c Everyone will free ride except for the agent with the maximum ai . 23.6.d Since all utility functions are quasilinear, a Pareto eﬃcient amount of the public good can be found by maximizing the sum of the utilities: n ai ln G − G, i=1 which implies G∗ = n i=1 ai . Chapter 24. Externalities 24.1.a Agent 1’s utility maximization problem is max u1 (x1 ) − p(x1 , x2)c1 , x1 while the social problem is max u1 (x1 ) + u2 (x2 ) − p(x1 , x2)[c1 + c2 ]. x1 ,x2 Since agent 1 ignores the cost he imposes on agent 2, he will generally choose too large a value of x1 . 24.1.b By inspection of the social problem and the private problem, agent 1 should be charged a ﬁne t1 = c2 . 24.1.c If the optimal ﬁnes are being used, then the total costs born by the agents in the case of an accident are 2[c1 + c2 ], which is simply twice the total cost of the accident. 24.1.d Agent 1’s objective function is (1 − p(x1 , x2 ))u1 (x1 ) − p(x1 , x2)c1 . This can also be written as u1 (x1 ) − p(x1 , x2 )[u1 (x1 ) + c1 ]. This is just the form of the previous objective function with u1 (x1 ) + c1 replacing c1 . Hence the optimal ﬁne for agent 1 is t1 = u2 (x2 ) + c2 . ]( Image 53: 50 ANSWERS Chapter 25. Information 25.1 By construction we know that f(u(s)) ≡ s. Diﬀerentiating one time shows that f (u)u (s) = 1. Since u (s) > 0, we must have f (u) > 0. Diﬀerentiating again, we have f (u)u (s) + f (u)u (s)2 = 0. Using the sign assumptions on u (s), we see that f (u) > 0. 25.2 According to the envelope theorem, ∂V /∂ca = λ + µ and ∂V /∂cb = µ. Thus, the sensitivity of the payment scheme to the likelihood ratio, µ, depends on how big an eﬀect an increase in cb would have on the principal. 25.3 In this case it is just as costly to undertake the action preferred by the principal as to undertake the alternative action. Hence, the incentive constraint will not be binding, which implies µ = 0. It follows that s(xi ) is constant. 25.4 If cb decreases, the original incentive scheme (si ) will still be feasible. Hence, an optimal incentive scheme must do at least as well as the original scheme. 25.5 In this case the maximization problem takes the form n max (xi − si )πib i=1 n such that si πib − cb ≥ u i=1 n n si πib − cb ≥ si πia − ca . i=1 i=1 Assuming that the participation constraint is binding, and ignoring the incentive-compatibility constraint for a moment, we can substitute into the objective function to write m max si πib − cb − u. i=1 Hence, the principal will choose the action that maximizes expected output minus (the agent’s) costs, which is the ﬁrst-best outcome. We can satisfy the incentive-compatibility constraint by choosing si = xi + F , and choose F so that the participation constraint is satisﬁed. Image 54: Ch. 25 INFORMATION 51 25.6 The participation constraints become st − c(xt ) ≥ ut , which we can write as st − (c(xt ) + ut ). Deﬁne ct (x) = c(x) + ut , and proceed as in the text. Note that the marginal costs of each type are the same, which adds an extra case to the analysis. 25.7 Since c2 (x) > c1 (x), we must have x2 x2 c2 (x) dx > c1 (x) dx. x1 x1 The result now follows from the Fundamental Theorem of Calculus. 25.8 The indiﬀerence curves take the form u1 = s−c1 (x) and u2 = s−c2 (x). Write these as s = u1 + c1 (x) and s = u2 + c2 (x). The diﬀerence between these two functions is d(x) = u2 − u1 + c2 (x) − c1 (x), and the derivative of this diﬀerence is d (x) = c2 (x) − c1 (x) > 0. Since the diﬀerence function is a monotonic function, it can hit zero at most once. 25.9 For only low-cost workers to be employed, there must be no proﬁtable contract that appeals to the high-cost workers. The most proﬁtable con- tract to a high-cost worker maximizes x2 − x2 , which implies x∗ = 1/2. 2 2 The cost of this to the worker is (1/2)2 = 1/4. For the worker to ﬁnd this acceptable, s2 − 1/4 ≥ u2 , or s2 = u2 + 1/4. For the ﬁrm to make a proﬁt, x∗ ≥ s2 . Hence we have 1/2 ≥ u2 + 1/4, or u2 ≤ 1/4. 2 25.10.a The professor must pay s = x2 /2 to get the assistant to work x hours. Her payoﬀ will be x − x2 /2. This is maximized where x = 1. 25.10.b The TA must get his reservation utility when he chooses the optimal x. This means that s − x2 /2 = s − 1/2 = 0, so s = 1/2. 25.10.c The best the professor can do is to get Mr. A to work 1 hour and have a utility of zero. Mr. A will work up to the point where he maximizes ax + b − x2 /2. Using calculus, we ﬁnd that Mr. A will choose x = a. Therefore he will work one hour if a = 1. Then his utility will be 0 if b = −1. 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Núcleo byJosé Antonio PAYANO YALE Cuaderno CLEIN 2013 byJosé Antonio PAYANO YALE Descubre Perú - Presencia Milenaria byJosé Antonio PAYANO YALE Laboratorio tecnológico byJosé Antonio PAYANO YALE Fisicoquimica y operaciones unitarias byJosé Antonio PAYANO YALE Microeconomic analysis 3d hal varian byJosé Antonio PAYANO YALE Capitulo 18 Sears byJosé Antonio PAYANO YALE Capitulo 36 Sears byJosé Antonio PAYANO YALE Capitulo 34 Sears byJosé Antonio PAYANO YALE Apéndice-Respuestas byJosé Antonio PAYANO YALE Capitulo 33 Sears byJosé Antonio PAYANO YALE Capitulo 31 Sears byJosé Antonio PAYANO YALE Capitulo 30 Sears byJosé Antonio PAYANO YALE Capitulo 32 Sears byJosé Antonio PAYANO YALE Capitulo 37 Sears byJosé Antonio PAYANO YALE Capitulo 35 Sears byJosé Antonio PAYANO YALE Capitulo 12 Sears byJosé Antonio PAYANO YALE Capitulo 04 Sears byJosé Antonio PAYANO YALE Capitulo 07 Sears byJosé Antonio PAYANO YALE Capitulo 23 Sears byJosé Antonio PAYANO YALE Varian, microeconomic analysis, solution book Answers to Exercises Microeconomic Analysis Third Edition Hal R. Varian University of California at Berkeley W. W. Norton & Company • New York • London Copyright c 1992,1984, 1978 by W. W. Norton & Company, Inc. All rights reserved Printed in the United States of America THIRD EDITION 0-393-96282-2 W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, N.Y. 10110 W. W. Norton Ltd., 10 Coptic Street, London WC1A 1PU 234567890 ANSWERS Chapter 1. Technology 1.1False. There are many counterexamples. Consider the technology generated by a production function f(x) = x2 . The production set is Y = {(y, −x) : y ≤ x2 } which is certainly not convex, but the input re- √ quirement set is V (y) = {x : x ≥ y} which is a convex set. 1.2 It doesn’t change. 1.3 1 = a and 2 = b. 1.4 Let y(t) = f(tx). Then n dy ∂f(x) = xi , dt ∂xi i=1 so that n 1 dy 1 ∂f(x) = xi . y dt f(x) ∂xi i=1 1.5 Substitute txi for i = 1, 2 to get 1 1 f(tx1 , tx2 ) = [(tx1 )ρ + (tx2 )ρ ] ρ = t[xρ + xρ ] ρ = tf(x1 , x2 ). 1 2 This implies that the CES function exhibits constant returns to scale and hence has an elasticity of scale of 1. 1.6 This is half true: if g (x) > 0, then the function must be strictly increasing, but the converse is not true. Consider, for example, the function g(x) = x3 . This is strictly increasing, but g (0) = 0. 1.7 Let f(x) = g(h(x)) and suppose that g(h(x)) = g(h(x )). Since g is monotonic, it follows that h(x) = h(x ). Now g(h(tx)) = g(th(x)) and g(h(tx )) = g(th(x )) which gives us the required result. 1.8 A homothetic function can be written as g(h(x)) where h(x) is ho- mogeneous of degree 1. Hence the TRS of a homothetic function has the 2 ANSWERS form g (h(x)) ∂h ∂h ∂x1 ∂x1 = . g (h(x)) ∂h ∂h ∂x2 ∂x2 That is, the TRS of a homothetic function is just the TRS of the un- derlying homogeneous function. But we already know that the TRS of a homogeneous function has the required property. 1.9 Note that we can write 1 1 a1 a2 ρ (a1 + a2 ) ρ xρ + xρ . a1 + a2 1 a1 + a2 2 1 Now simply deﬁne b = a1 /(a1 + a2 ) and A = (a1 + a2 ) ρ . 1.10 To prove convexity, we must show that for all y and y in Y and 0 ≤ t ≤ 1, we must have ty + (1 − t)y in Y . But divisibility implies that ty and (1 − t)y are in Y , and additivity implies that their sum is in Y . To show constant returns to scale, we must show that if y is in Y , and s > 0, we must have sy in Y . Given any s > 0, let n be a nonnegative integer such that n ≥ s ≥ n − 1. By additivity, ny is in Y ; since s/n ≤ 1, divisibility implies (s/n)ny = sy is in Y . 1.11.a This is closed and nonempty for all y > 0 (if we allow inputs to be negative). The isoquants look just like the Leontief technology except we are measuring output in units of log y rather than y. Hence, the shape of the isoquants will be the same. It follows that the technology is monotonic and convex. 1.11.b This is nonempty but not closed. It is monotonic and convex. 1.11.c This is regular. The derivatives of f(x1 , x2) are both positive so the technology is monotonic. For the isoquant to be convex to the origin, it is suﬃcient (but not necessary) that the production function is concave. To check this, form a matrix using the second derivatives of the production function, and see if it is negative semideﬁnite. The ﬁrst principal minor of the Hessian must have a negative determinant, and the second principal minor must have a nonnegative determinant. ∂ 2 f(x) 1 −3 1 ∂ 2 f(x) 1 −1 − 1 2 = − x1 2 x2 2 = x12 x2 2 ∂x1 4 ∂x1 ∂x2 4 ∂ 2 f(x) 1 1 −3 2 = − x1 x22 2 ∂x2 4 Ch. 2 PROFITMAXIMIZATION 3 −3/2 1/2 1 −1/2 −1/2 − 1 x1 x2 4 4 x1 x2 Hessian = 1 −1/2 −1/2 1 1/2 −3/2 4 x1 x2 − 4 x1 x2 1 −3/2 1/2 D1 = − x1 x2 < 0 4 1 −1 −1 1 D2 = x x − x−1 x−1 = 0. 16 1 2 16 1 2 So the input requirement set is convex. 1.11.d This is regular, monotonic, and convex. 1.11.e This is nonempty, but there is no way to produce any y > 1. It is monotonic and weakly convex. 1.11.f This is regular. To check monotonicity, write down the production √ function f(x) = ax1 − x1 x2 + bx2 and compute ∂f(x) 1 −1/2 1/2 = a − x1 x2 . ∂x1 2 1 x2 This is positive only if a > 2 x1 , thus the input requirement set is not always monotonic. Looking at the Hessian of f, its determinant is zero, and the determinant of the ﬁrst principal minor is positive. Therefore f is not concave. This alone is not suﬃcient to show that the input requirement sets are not convex. But we can say even more: f is convex; therefore, all sets of the form √ {x1 , x2 : ax1 − x1 x2 + bx2 ≤ y} for all choices of y are convex. Except for the border points this is just the complement of the input requirement sets we are interested in (the inequality sign goes in the wrong direction). As complements of convex sets (such that the border line is not a straight line) our input requirement sets can therefore not be themselves convex. 1.11.g This function is the successive application of a linear and a Leontief function, so it has all of the properties possessed by these two types of functions, including being regular, monotonic, and convex. Chapter 2. Profit Maximization 4 ANSWERS 2.1 Forproﬁt maximization, the Kuhn-Tucker theorem requires the follow- ing three inequalities to hold ∂f(x∗ ) p − wj x∗ = 0, j ∂xj ∂f(x∗ ) p − wj ≤ 0, ∂xj x∗ ≥ 0. j Note that if x∗ > 0, then we must have wj /p = ∂f(x∗ )/∂xj . j 2.2 Suppose that x is a proﬁt-maximizing bundle with positive proﬁts π(x ) > 0. Since f(tx ) > tf(x ), for t > 1, we have π(tx ) = pf(tx ) − twx > t(pf(x ) − wx ) > tπ(x ) > π(x ). Therefore, x could not possibly be a proﬁt-maximizing bundle. 2.3 In the text the supply function and the factor demands were computed for this technology. Using those results, the proﬁt function is given by a 1 w a−1 w a−1 π(p, w) = p −w . ap ap To prove homogeneity, note that a 1 w a−1 w a−1 π(tp, tw) = tp − tw = tπ(p, w), ap ap which implies that π(p, w) is a homogeneous function of degree 1. Before computing the Hessian matrix, factor the proﬁt function in the following way: 1 a a 1 1 a π(p, w) = p 1−a w a−1 a 1−a − a 1−a = p 1−a w a−1 φ(a), where φ(a) is strictly positive for 0 < a < 1. The Hessian matrix can now be written as ∂ 2 π(p,w) ∂ 2π(p,w) 2 ∂p2 ∂p∂w D π(p, ω) = ∂ 2 π(p,w) ∂ 2π(p,w) ∂w∂p ∂w 2  2a−1 a a 1  a (1−a)2 p 1−a w a−1 − (1−a)2 p 1−a w a−1 a     =  φ(a).   a 1 1 2−a − (1−a)2 p 1−a w a−1 a a (1−a)2 p 1−a w a−1 Ch. 2 PROFITMAXIMIZATION 5 The principal minors of this matrix are a 2a−1 a p 1−a w a−1 φ(a) > 0 (1 − a)2 and 0. Therefore, the Hessian is a positive semideﬁnite matrix, which implies that π(p, w) is convex in (p, w). 2.4 By proﬁt maximization, we have ∂f ∂x1 w1 \|T RS\| = = . ∂f w2 ∂x2 Now, note that ln(w2 x2 /w1 x1 ) = −(ln(w1 /w2 ) + ln(x1 /x2 )). Therefore, d ln(w2 x2 /w1 x1 ) d ln(w1 /w2 ) d ln \|T RS\| = −1= − 1 = 1/σ − 1. d ln(x1 /x2 ) d ln(x2 /x1 ) d ln(x2 /x1 ) 2.5 From the previous exercise, we know that ln(w2 x2 /w1 x1 ) = ln(w2 /w1 ) + ln(x2 /x1 ), Diﬀerentiating, we get d ln(w2 x2 /w1 x1 ) d ln(x2 /x1 ) =1− = 1 − σ. d ln(w2 /w1 ) d ln \|T RS\| 2.6 We know from the text that Y O ⊃ Y ⊃ Y I. Hence for any p, the maximum of py over Y O must be larger than the maximum over Y , and this in turn must be larger than the maximum over Y I. 2.7.a We want to maximize 20x − x2 − wx. The ﬁrst-order condition is 20 − 2x − w = 0. 2.7.b For the optimal x to be zero, the derivative of proﬁt with respect to x must be nonpositive at x = 0: 20 − 2x − w < 0 when x = 0, or w ≥ 20. 2.7.c The optimal x will be 10 when w = 0. 2.7.d The factor demand function is x = 10 − w/2, or, to be more precise, x = max{10 − w/2, 0}. 6 ANSWERS 2.7.e Proﬁtsas a function of output are 20x − x2 − wx = [20 − w − x]x. Substitute x = 10 − w/2 to ﬁnd w 2 π(w) = 10 − . 2 2.7.f The derivative of proﬁt with respect to w is −(10 − w/2), which is, of course, the negative of the factor demand. Chapter 3. Profit Function 3.1.a Since the proﬁt function is convex and a decreasing function of the factor prices, we know that φi(wi ) ≤ 0 and φi (wi) ≥ 0. 3.1.b It is zero. 3.1.c The demand for factor i is only a function of the ith price. Therefore the marginal product of factor i can only depend on the amount of factor i. It follows that f(x1 , x2 ) = g1 (x1 ) + g2 (x2 ). 3.2 The ﬁrst-order conditions are p/x = w, which gives us the demand function x = p/w and the supply function y = ln(p/w). The proﬁts from operating at this point are p ln(p/w) − p. Since the ﬁrm can al- ways choose x = 0 and make zero proﬁts, the proﬁt function becomes π(p, w) = max{p ln(p/w) − p, 0}. 3.3 The ﬁrst-order conditions are p a1 − w1 = 0 x1 p a2 − w2 = 0, x2 which can easily be solved for the factor demand functions. Substituting into the objective function gives us the proﬁt function. 3.4 The ﬁrst-order conditions are pa1 x1 1 −1 xa2 − w1 = 0 a 2 pa2 x2 2 −1 xa1 − w2 = 0, a 1 which can easily be solved for the factor demands. Substituting into the objective function gives us the proﬁt function for this technology. In order Ch. 4 COSTMINIMIZATION 7 for this to be meaningful, the technology must exhibit decreasing returns to scale, so a1 + a2 < 1. 3.5 If wi is strictly positive, the ﬁrm will never use more of factor i than it needs to, which implies x1 = x2 . Hence the proﬁt maximization problem can be written as max pxa − w1 x1 − w2 x2 . 1 The ﬁrst-order condition is pax1 − (w1 + w2 ) = 0. a−1 The factor demand function and the proﬁt function are the same as if the production function were f(x) = xa , but the factor price is w1 + w2 rather than w. In order for a maximum to exist, a < 1. Chapter 4. Cost Minimization 4.1 Let x∗ be a proﬁt-maximizing input vector for prices (p, w). This means that x∗ must satisfy pf(x∗ ) − wx∗ ≥ pf(x) − wx for all permissible x. Assume that x∗ does not minimize cost for the output f(x∗ ); i.e., there exists a vector x∗∗ such that f(x∗∗ ) ≥ f(x∗ ) and w(x∗∗ − x∗ ) < 0. But then the proﬁts achieved with x∗∗ must be greater than those achieved with x∗ : pf(x∗∗ ) − wx∗∗ ≥ pf(x∗ ) − wx∗∗ > pf(x∗ ) − wx∗ , which contradicts the assumption that x∗ was proﬁt-maximizing. 4.2 The complete set of conditions turns out to be ∂f(x∗ ) t − wj x∗ = 0, j ∂xj ∂f(x∗ ) t − wj ≤ 0, ∂xj x∗ ≥ 0, j (y − f(x∗ )) t = 0, y − f(x∗ ) ≤ 0, t ≥ 0. If, for instance, we have x∗ > 0 and x∗ = 0, the above conditions imply i j ∂f(x∗ ) ∂xi wi ∗ ≥ . ∂f(x ) wj ∂xj 8 ANSWERS This meansthat it would decrease cost to substitute xi for xj , but since there is no xj used, this is not possible. If we have interior solutions for both xi and xj , equality must hold. 4.3 Following the logic of the previous exercise, we equate marginal costs to ﬁnd y1 = 1. We also know y1 + y2 = y, so we can combine these two equations to get y2 = y −1. It appears that the cost function is c(y) = 1/2+y −1 = y −1/2. However, on reﬂection this can’t be right: it is obviously better to produce everything in plant 1 if y1 < 1. As it happens, we have ignored the implicit constraint that y2 ≥ 0. The actual cost function is y2 /2 if y < 1 c(y) = y − 1/2 if y > 1. 4.4 According to the text, we can write the cost function for the ﬁrst plant as c1 (y) = Ay and for the second plant as c2 (y) = By, where A and B depend on a, b, w1 , and w2 . It follows from the form of the cost functions that c(y) = min{A, B}y. 4.5 The cost of using activity a is a1 w1 +a2 w2 , and the cost of using activity b is b1 w1 + b2 w2 . The ﬁrm will use whichever is cheaper, so c(w1 , w2 , y) = y min{a1 w1 + a2 w2 , b1 w1 + b2 w2 }. The demand function for factor 1, for example, is given by   a1 y  if a1 w1 + a2 w2 < b1 w1 + b2 w2 b1 y if a1 w1 + a2 w2 > b1 w1 + b2 w2 x1 =   any amount between a1 y and b1 y otherwise. The cost function will not be diﬀerentiable when a1 w 1 + a2 w 2 = b 1 w 1 + b 2 w 2 . 4.6 By the now standard argument, √ √ c(y) = min{4 y1 + 2 y2 : y1 + y2 ≥ y}. It is tempting to set M C1 (y1 ) = M C2 (y2 ) to ﬁnd that y1 = y/5 and y2 = 4y/5. However, if you think about it a minute you will see that this Ch. 5 COSTFUNCTION 9 doesn’t make sense—you are producing more output in the plant with the higher costs! It turns out that this corresponds to a constrained maximum and not to the desired minimum. Check the second-order conditions to verify this. Since the cost function is concave, rather than convex, the optimal solu- tion will always occur at a boundary. That is, you will produce all output √ at the cheaper plant so c(y) = 2 y. 4.7 No, the data violate WACM. It costs 40 to produce 100 units of output, but at the same prices it would only cost 38 to produce 110 units of output. 4.8 Set up the minimization problem min x1 + x2 x1 x2 = y. Substitute to get the unconstrained minimization problem min x1 + y/x1 . The ﬁrst-order condition is 1 − y/x2 , 1 √ √ which implies x1 = y. By symmetry, x2 = y. We are given that √ √ 2 y = 4, so y = 2, from which it follows that y = 4. Chapter 5. Cost Function 5.1 The ﬁrm wants to minimize the cost of producing a given level of output: 2 2 c(y) = min y1 + y2 y1 ,y2 such that y1 + y2 = y. The solution has y1 = y2 = y/2. Substituting into the objective function yields c(y) = (y/2)2 + (y/2)2 = y2 /2. 5.2 The ﬁrst-order conditions are 6y1 = 2y2 , or y2 = 3y1 . We also require y1 + y2 = y. Solving these two equations in two unknowns yields y1 = y/4 and y2 = 3y/4. The cost function is 2 y 2 3y 3y2 c(y) = 3 + = . 4 4 4 10 ANSWERS 5.3 Considerthe ﬁrst technique. If this is used, then we need to have 2x1 + x2 = y. Since this is linear, the ﬁrm will typically specialize and set x2 = y or x1 = y/2 depending on which is cheaper. Hence the cost function for this technique is y min{w1 /2, w2 }. Similarly, the cost function for the other technique is y min{w3 , w4/2}. Since both techniques must be used to produce y units of output, c(w1 , w2, y) = y [min{w1 /2, w2} + min{w3 , w4 /2}] . 5.4 The easiest way to answer this question is to sketch an isoquant. First draw the line 2x1 + x2 = y and then the line x1 + 2x2 = y. The isoquant is the upper northeast boundary of this “cross.” The slope is −2 to the left of the diagonal and −1/2 to the right of the diagonal. This means that when w1 /w2 < 1/2, we have x1 = 0 and x2 = y. When w1 /w2 < 1/2, we have x1 = y and x2 = 0. Finally, when 2 > w1 /w2 > 1/2, we have x1 = x2 = y/3. The cost function is then c(w1 , w2 , y) = min{w1 , w2 , (w1 + w2 )/3}y. 5.5 The input requirement set is not convex. Since y = max{x1 , x2 }, the ﬁrm will use whichever factor is cheaper; hence the cost function is c(w1 , w2 , y) = min{w1 , w2}y. The factor demand function for factor 1 has the form y if w1 < w2 x1 = either 0 or y if w1 = w2 . 0 if w1 > w2 5.6 We have a = 1/2 and c = −1/2 by homogeneity, and b = 3 since ∂x1 /∂w2 = ∂x2 /∂w1 . 5.7 Set up the minimization problem min x1 + x2 x1 x2 = y. Substitute to get the unconstrained minimization problem min x1 + y/x1 . The ﬁrst-order condition is 1 − y/x2 , 1 √ √ which implies x1 = y. By symmetry, x2 = y. We are given that √ √ 2 y = 4, so y = 2, from which it follows that y = 4. Ch. 5 COSTFUNCTION 11 5.8 If p = 2, the ﬁrm will produce 1 unit of output. If p = 1, the ﬁrst- order condition suggests y = 1/2, but this yields negative proﬁts. The ﬁrm can get zero proﬁts by choosing y = 0. The proﬁt function is π(p) = max{p2 /4 − 1, 0}. 5.9.a dπ/dα = py > 0. 5.9.b dy/dα = p/c (y) > 0. 5.9.c p (α) = n[y + αp/c ]/[D (p) − nα/c ] < 0. 5.10 Let y(p, w) be the supply function. Totally diﬀerentiating, we have n n n ∂y(p, w) ∂xi (p, w) ∂xi (w, y) ∂y(p, w) dy = dwi = − dwi = − dwi. ∂wi ∂p ∂y ∂p i=1 i=1 i=1 The ﬁrst equality is a deﬁnition; the second uses the symmetry of the substitution matrix; the third uses the chain rule and the fact that the unconditional factor demand, xi (p, w), and the conditional factor demand, xi (w, y), satisfy the identity xi(w, y(p, w)) = xi (p, w). The last expression on the right shows that if there are no inferior factors then the output of the ﬁrm must increase. 5.11.a x = (1, 1, 0, 0). 5.11.b min{w1 + w2 , w3 + w4 }y. 5.11.c Constant returns to scale. 5.11.d x = (1, 0, 1, 0). 5.11.e c(w, y) = [min{w1 , w2 } + min{w3 , w4 }]y. 5.11.f Constant. 5.12.a The diagram is the same as the diagram for an inferior good in consumer theory. 5.12.b If the technology is CRS, then conditional factor demands take the form xi(w, 1)y. Hence the derivative of a factor demand function with respect to output is xi(w) ≥ 0. 5.12.c The hypothesis can be written as ∂c(w, y)2 /∂y∂wi < 0. But ∂c(w, y)2 /∂y∂wi = ∂c(w, y)2 /∂wi ∂y = ∂xi (w, y)/∂y. 12 ANSWERS 5.13.a Factordemand curves slope downward, so the demand for unskilled workers must decrease when their wage increases. 5.13.b We are given that ∂l/∂p < 0. But by duality, ∂l/∂p = −∂ 2 π/∂p∂w = −∂ 2 π/∂w∂p = −∂y/∂w. It follows that ∂y/∂w > 0. 5.14 Take a total derivative of the cost function to get: n ∂c ∂c dc = dwi + dy. ∂wi ∂y i=1 It follows that dc − n ∂c dw ∂c i=1 ∂wi i = . ∂y dy Now substitute the ﬁrst diﬀerences for the dy, dc, dwi terms and you’re done. 5.15 By the linearity of the function, we know we will use either x1 , or a combination of x2 and x3 to produce y. By the properties of the Leontief function, we know that if we use x2 and x3 to produce y, we must use 3 units of both x2 and x3 to produce one unit of y. Thus, if the cost of using one unit of x1 is less than the cost of using one unit of both x2 and x3 , then we will use only x1 , and conversely. The conditional factor demands can be written as: 3y if w1 < w2 + w3 x1 = 0 if w1 > w2 + w3 0 if w1 < w2 + w3 x2 = 3y if w1 > w2 + w3 0 if w1 < w2 + w3 x3 = 3y if w1 > w2 + w3 if w1 = w2 +w3 , then any bundle (x1 , x2 , x3 ) with x2 = x3 and x1 +x2 = 3y (or x1 + x3 = 3y) minimizes cost. The cost function is c(w, y) = 3y min(w1 , w2 + w3 ). 5.16.a Homogeneous: c(tw, y) = y1/2 (tw1 tw2 )3/4 = t3/2 (y1/2 (w1 w2 )3/4 ) = t3/2 c(w, y) No. Ch. 5 COSTFUNCTION 13 Monotone: ∂c 3 −1/4 3/4 ∂c 3 3/4 −1/4 = y1/2 w1 w2 > 0 = y1/2 w1 w2 >0 Yes. ∂w1 4 ∂w2 4 Concave: −5/4 3/4 9 1/2 −1/4 −1/4 − 16 y1/2 w1 w2 3 16 y w1 w2 Hessian = 9 1/2 −1/4 −1/4 3 1/2 3/4 −5/4 16 y w1 w2 − 16 y w1 w2 \|H1 \| < 0 9 −1/2 −1/2 81 −1/2 −1/2 \|H2 \| = yw w2 − yw w2 256 1 256 1 72 y =− √ < 0 No 256 w1 w2 Continuous: Yes 5.16.b Homogeneous: √ c(tw, y) = y(tw1 + tw1 tw2 + tw2 ) √ = ty(w1 + w1 w2 + w2 ) = tc(y, w) Yes Monotone: ∂c 1 w2 ∂c 1 w1 =y 1+ >0 =y 1+ >0 Yes ∂w1 2 w1 ∂w2 2 w2 Concave: 1/2 −3/2 −1/2 −1/2 − 1 yw2 w1 4 1 4 yw2 w1 H= −1/2 −1/2 −3/2 1/2 1 4 yw2 w1 − 4 yw2 w1 1 \|H1 \| < 0 1 1 \|H2 \| = yw−1 w−1 − yw2 w1 = 0 Yes −1 −1 16 2 1 16 Continuous: Yes 14 ANSWERSProduction Function: 1 w2 x1 (w, y) = y 1 + (1) 2 w1 1 w1 x2 (w, y) = y 1 + (2) 2 w2 Rearranging these equations: y w2 x1 − y = (1 ) 2 w1 y w1 x2 − y = (2 ) 2 w2 2 Multiply (1 ) and (2 ): (x1 − y)(x2 − y) = y . This is a quadratic equation 4 which gives y = 2 (x2 + x1 ) ± 2 x2 + x2 + 2 − x1 x2 . 3 3 1 2 5.16.c Homogeneous: c(tw, y) = y(tw1 e−tw1 + tw2 ) = ty(w1 e−tw1 + w2 ) = tc(w, y) No Monotone: ∂c = y(−w1 e−w1 + e−w1 ) = ye−w1 (1 − w1 ) ∂w1 This is positive only if w1 < 1. ∂c = y > 0 No ∂w2 Concave: y(w1 − 2)e−w1 0 H= 0 0 \|H1 \| = y(w1 − 2)e−w1 This is less than zero only if w1 < 2. \|H2 \| = 0 No Continuous: Yes Ch. 5 COSTFUNCTION 15 5.16.d Homogeneous: √ c(tw, y) = y(tw1 − tw1 tw2 + tw2 √ = ty(w1 − w1 w2 + w2 ) = tc(w, y) Yes Monotone: ∂c 1 w2 = y(1 − ) ∂w1 2 w1 1 w2 This is greater than 0 only if 1 > 2 w1 ∂c 1 w1 = y(1 − ) ∂w2 2 w2 w2 This is greater than 0 only if 2 > w1 1 √ √ w2 > w1 (by symmetry) 2 w1 > w2 4 or 1 w1 < 4w2 w1 > w2 4 Monotone only if 1 w2 < w1 < 4w2 . No. 4 Concave: −3/2 1/2 −1/2 −1/2 1 4 yw1 w2 − 1 yw1 w2 4 H= −1/2 −1/2 1/2 −1/2 − 4 yw1 w2 1 1 4 yw1 w2 1 −3/2 1/2 \|H1 \| = yw1 w2 > 0 4 \|H2 \| = 0 No (it is convex) Continuous: Yes 5.16.e Homogeneous: 1√ c(tw, y) = (y + tw1 tw2 ) y = tc(y, w) Yes 16 ANSWERSMonotone in w: ∂c 1 1 w2 ∂c 1 1 w1 = (y + ) >0 = (y + ) >0 Yes ∂w1 2 y w1 ∂w2 2 2 w2 Concave: −3/2 1/2 −1/2 −1/2 − 1 (y + y w1 4 1 w2 1 1 4 (y + y )w1 w2 H= −1/2 −1/2 1/2 −3/2 But not in y! 1 1 4 (y, y )w1 w2 − 4 (y + y )w1 w2 1 1 \|H1\| < 0 Yes \|H2\| = 0 Continuous: Not for y = 0. √ 5.17.a y = ax1 + bx2 5.17.b Note that this function is exactly like a linear function, except that the linear combination of x1 and x2 will produce y2 , rather than just y. So, we know that if x1 is relatively cheaper, we will use all x1 and no x2 , and conversely. 5.17.c The cost function is c(w, y) = y2 min( w1 , w2 ). a b Chapter 6. Duality 6.1 The production function is f(x1 , x2 ) = x1 + x2 . The conditional factor demands have the form  y if wi < wj xi = 0 if wi > wj  any amount between 0 and y if wi = wj . 6.2 The conditional factor demands can be found by diﬀerentiating. They are x1 (w1 , w2 , y) = x2 (w1 , w2, y) = y. The production function is f(x1 , x2 ) = min{x1 , x2 }. 6.3 The cost function must be increasing in both prices, so a and b are both nonnegative. The cost function must be concave in both prices, so a and b are both less than 1. Finally, the cost function must be homogeneous of degree 1, so a = 1 − b. Ch. 8 CHOICE 17 Chapter 7. Utility Maximization 7.1 The preferences exhibit local nonsatiation, except at (0, 0). The con- sumer will choose this consumption point when faced with positive prices. 7.2 The demand function is   m/p1 if p1 < p2 x1 = any x1 and x2 such that p1 x2 + p2 x2 = m if p1 = p2  0 if p1 > p2 The indirect utility function is v(p1 , p2 , m) = max{m/p1 , m/p2 }, and the expenditure function is e(p1 , p2 , u) = u min{p1 , p2}. 7.3 The expenditure function is e(p1 , p2 , u) = u min{p1 , p2}. The utility function is u(x1 , x2 ) = x1 + x2 (or any monotonic transformation), and the demand function is   m/p1 if p1 < p2 x1 = any x1 and x2 such that p1 x1 + p2 x2 = m if p1 = p2  0 if p1 > p2 7.4.a Demand functions are x1 = m/(p1 + p2 ), x2 = m/(p1 + p2 ). 7.4.b e(p1 , p2 , u) = (p1 + p + 2)u 7.4.c u(x1 , x2 ) = min{x1 , x2} 7.5.a Quasilinear preferences. 7.5.b Less than u(1). 7.5.c v(p1 , p2 , m) = max{u(1) − p1 + m, m} 7.6.a Homothetic. 7.6.b e(p, u) = u/A(p) 7.6.c µ(p; q, m) = mA(q)/A(p) 7.6.d It will be the same, since this is just a monotonic transformation. Chapter 8. Choice 8.1 We know that xj (p, m) ≡ hj (p, v(p, m)) ≡ ∂e(p, v(p, m))/∂pj . (0.1) 18 ANSWERS (Note thatthe partial derivative is taken with respect to the ﬁrst occurrence of pj .) Diﬀerentiating equation (0.1) with respect to m gives us ∂xj ∂ 2 e(p, v(p, m)) ∂v(p, m) = . ∂m ∂pj ∂u ∂m Since the marginal utility of income, ∂v/∂m, must be positive, the result follows. 8.2 The Cobb-Douglas demand system with two goods has the form a1 m x1 = p1 a2 m x2 = p2 where a1 + a2 = 1. The substitution matrix is −a1 mp−2 − a2 mp−2 1 1 1 −a1 a2 mp−1 p−1 1 2 . −a1 a2 mp−1 p−1 2 1 2 −a2 mp−2 − a2 mp−2 2 2 2 This is clearly symmetric and negative deﬁnite. 8.3 The equation is dµ/dt = at + bµ + c. The indirect money metric utility function is c a c c a aq µ(q, p, m) = eb(q−p) m + + 2+ p − − 2− . b b b b b b 8.4 The demand function can be written as x = ec+ap+bm . The integrability equation is dµ = eat+bµ+c . dt Write this as dµ e−bµ = ec eat . dt Integrating both sides of this equation between p and q, we have e−bµ q ec eat q − = . b p a p Evaluating the integrals, we have bec ap ebµ(q;p,m) = e−bm − [e − eaq ] . a Ch. 8 CHOICE 19 8.5 Write the Lagrangian 3 L(x, λ) = ln x1 + ln x2 − λ(3x1 + 4x2 − 100). 2 (Be sure you understand why we can transform u this way.) Now, equating the derivatives with respect to x1 , x2 , and λ to zero, we get three equations in three unknowns 3 = 3λ, 2x1 1 = 4λ, x2 3x1 + 4x2 = 100. Solving, we get x1 (3, 4, 100) = 20, and x2 (3, 4, 100) = 10. Note that if you are going to interpret the Lagrange multiplier as the marginal utility of income, you must be explicit as to which utility function you are referring to. Thus, the marginal utility of income can be measured in original ‘utils’ or in ‘ln utils’. Let u∗ = ln u and, correspondingly, v∗ = ln v; then ∂v(p, m) ∂v∗ (p, m) ∂m µ λ= = = , ∂m v(p, m) v(p, m) where µ denotes the Lagrange multiplier in the Lagrangian 3 L(x, µ) = x1 x2 − µ(3x1 + 4x2 − 100). 2 3 20 2 1 Check that in this problem we’d get µ = 4 ,λ= 40 , and v(3, 4, 100) = 3 20 2 10. 8.6 The Lagrangian for the utility maximization problem is 1 1 L(x, λ) = x1 x2 − λ(p1 x1 + p2 x2 − m), 2 3 taking derivatives, 1 −1 3 1 x1 2 x2 = λp1 , 2 1 2 −2 1 x1 x2 3 = λp2 , 3 p1 x1 + p2 x2 = m. Solving, we get 3m 2m x1 (p, m) = , x2 (p, m) = . 5 p1 5 p2 20 ANSWERS Plugging thesedemands into the utility function, we get the indirect utility function 1 1 1 1 5 3m 2 2m 3 m 6 3 2 2 3 v(p, m) = U (x(p, m)) = = . 5 p1 5 p2 5 p1 p2 Rewrite the above expression replacing v(p, m) by u and m by e(p, u). Then solve it for e(·) to get 3 2 p1 5 p2 5 6 e(p, u) = 5 u5 . 3 2 Finally, since hi = ∂e/∂pi , the Hicksian demands are p1 −2 5 p2 2 5 6 h1 (p, u) = u5 , 3 2 and p1 3 5 p2 −3 5 6 h2 (p, u) = u5 . 3 2 8.7 Instead of starting from the utility maximization problem, let’s now start from the expenditure minimization problem. The Lagrangian is L(x, µ) = p1 x1 + p2 x2 − µ((x1 − α1 )β1 (x2 − α2 )β2 − u); the ﬁrst-order conditions are p1 = µβ1 (x1 − α1 )β1 −1 (x2 − α2 )β2 , p2 = µβ2 (x1 − α1 )β1 (x2 − α2 )β2 −1 , (x1 − α1 )β1 (x2 − α2 )β2 = u. Divide the ﬁrst equation by the second p1 β2 x2 − α 2 = , p2 β1 x1 − α 1 using the last equation 1 x2 − α2 = (x1 − α1 )−β1 u β2 ; substituting and solving, β2 p2 β1 β1 β1 +β2 h1 (p, u) = α1 + u 2 , p1 β2 Ch. 8 CHOICE 21 and β1 p1 β2 β1 β1 +β2 h2 (p, u) = α2 + u 1 . p2 β1 Verify that 1 β2 β1 β1 +β2 ∂h1 (p, m) u β1 β2 ∂h2 (p, m) = = . ∂p2 β1 + β2 p1 p2 ∂p1 The expenditure function is  β2   β1  p2 β1 β1 β1 +β2 p1 β2 β1 β1 +β2 e(p, u) = p1 α1 + u 2  + p2 α2 + u 1 . p1 β2 p2 β1 Solving for u, we get the indirect utility function v(p, m) = β1 β2 β1 m − α 2 p2 β2 m − α 1 p1 − α1 − α2 . β1 + β2 p1 β1 + β2 p2 By Roy’s law we get the Marshallian demands 1 m − α 1 p1 x1 (p, m) = β1 α2 + β2 , β1 + β2 p2 and 1 m − α 2 p2 x2 (p, m) = β2 α1 + β1 . β1 + β2 p1 8.8 Easy—a monotonic transformation of utility doesn’t change anything about observed behavior. 8.9 By deﬁnition, the Marshallian demands x(p, m) maximize φ(x) subject to px = m. We claim that they also maximize ψ(φ(x)) subject to the same budget constraint. Suppose not. Then, there would exist some other choice x such that ψ(φ(x )) > ψ(φ(x(p, m))) and px = m. But since applying the transformation ψ−1 (·) to both sides of the inequality will preserve it, we would have φ(x ) > φ(x(p, m)) and px = m, which contradicts our initial assumption that x(p, m) maximized φ(x) subject to px = m. Therefore x(p, m) = x∗ (p, m). (Check that the reverse proposition also holds—i.e., the choice that maximizes u∗ also maximizes u when the the same budget constraint has to be veriﬁed in both cases.) v∗ (p, m) = ψ(φ(x∗ (p, m))) = ψ(φ(x(p, m)) = ψ(v(p, m)), 22 ANSWERS the ﬁrstand last equalities hold by deﬁnition and the middle one by our previous result; now e∗ (p, u∗) = min{px : ψ(φ(x)) = u∗ } = min{px : φ(x) = ψ−1 (u∗ )} = e(p, ψ−1 (u∗ )); again, we’re using deﬁnitions at both ends and the properties of ψ(·) — namely that the inverse is well deﬁned since ψ(·) is monotonic— to get the middle equality; ﬁnally using deﬁnitions and substitutions as often as needed we get h∗(p, u∗) = x∗ (p, e∗ (p, u∗)) = x(p, e∗ (p, u∗)) = x(p, e(p, ψ−1 (u∗ ))) = h(p, ψ−1(u∗ )). 8.10.a Diﬀerentiate the identity hj (p, u) ≡ xj (p, e(p, u)) with respect to pi to get ∂hj (p, u) ∂xj (p, m) ∂xj (p, e(p, u)) ∂e(p, u) = + . ∂pi ∂pi ∂m ∂pi We must be careful with this last term. Look at the expenditure minimiza- tion problem e(p, u) = min{p(x − x) : u(x) = u}. By the envelope theorem, we have ∂e(p, u) = hi (p, u) − xi = xi (p, e(p, u)) − xi . ∂pi Therefore, we have ∂hj (p, u) ∂xj (p, m) ∂xj (p, e(p, u)) = + (xi (p, m) − xi ), ∂pi ∂pi ∂m and reorganizing we get the Slutsky equation ∂xj (p, m) ∂hj (p, u) ∂xj (p, e(p, u)) = + (xi − xi(p, m)). ∂pi ∂pi ∂m 8.10.b Draw a diagram, play with it and verify that Dave is better oﬀ when p2 goes down and worse oﬀ when p1 goes down. Just look at the sets of allocations that are strictly better or worse than the original choice—i.e., the sets SB(x) = {z : z x} and SW (x) = {z : z x}. When p1 goes down the new budget set is contained in SW (x), while when p2 goes down there’s a region of the new budget set that lies in SB(x). Ch. 8 CHOICE 23 8.10.c The rate of return—also known as “own rate of interest”—on good x is (p1 /p2 ) − 1 8.11 No, because his demand behavior violates GARP. When prices are (2, 4) he spends 10. At these prices he could aﬀord the bundle (2, 1), but rejects it; therefore, (1, 2) (2, 1). When prices are (6, 3) he spends 15. At these prices he could aﬀord the bundle (1, 2) but rejects it; therefore, (2, 1) (1, 2). 8.12 Inverting, we have e(p, u) = u/f(p). Substituting, we have µ(p; q, y) = v(q, y)/f(p) = f(q)y/f(p). 8.13.a Draw the lines x2 + 2x1 = 20 and x1 + 2x2 = 20. The indiﬀerence curve is the northeast boundary of this X. 8.13.b The slope of a budget line is −p1 /p2 . If the budget line is steeper than 2, x1 = 0. Hence the condition is p1 /p2 > 2. 8.13.c Similarly, if the budget line is ﬂatter than 1/2, x2 will equal 0, so the condition is p1 /p2 < 1/2. 8.13.d If the optimum is unique, it must occur where x2 − 2x1 = x1 − 2x2 . This implies that x1 = x2 , so that x1 /x2 = 1. α 8.14.a This is an ordinary Cobb-Douglas demand: S1 = α+β+γ Y and β S2 = α+β+γ Y . 8.14.b In this case the utility function becomes U (C, S1 , L) = S1 Lβ C γ . α The L term is just a constant, so applying the standard Cobb-Douglas α formula S1 = α+γ Y . s 8.15 Use Slutsky’s equation to write: ∂L = ∂L + (L − L) ∂L . Note that ∂w ∂w ∂m the substitution eﬀect is always negative, (L − L) is always positive, and hence if leisure is inferior, ∂L is necessarily negative. Thus the slope of ∂w the labor supply curve is positive. 8.16.a True. With the grant, the consumer will maximize u(x1 , x2 ) subject to x1 + x2 ≤ m + g1 and x1 ≥ g1 . We know that when he maximizes his utility subject to x1 + x2 ≤ m, he chooses x∗ ≥ g1 . Since x1 is a normal 1 good, the amount of good 1 that he will choose if given an unconstrained grant of g1 is some number x1 > x∗ ≥ g1 . Since this choice satisﬁes the 1 constraint x1 ≥ g1 , it is also the choice he would make when forced to spend g1 on good 1. 24 ANSWERS 8.16.b False.Suppose for example that g1 = x∗ . Then if he gets an 1 unconstrained grant of g1 , since good 1 is inferior, he will choose to reduce his consumption to less than x∗ = g1 . But with the constrained grant, he 1 must consume at least g1 units of good 1. Incidentally, he will accept the grant, since with the grant he can always consume at least as much of both goods as without the grant. 8.16.c If he got an unconstrained grant of g1 , he would spend (48 + g1 )/4 on good 1. This is exactly what he will spend if g1 ≤ (48 + g1 )4. But if g1 > (48 + g1 )/4, he will spend g1 on good 1. The curve therefore has slope 1/4 if g1 < 16 and slope 1 if g1 > 16. Kink is at g1 = 16. Chapter 9. Demand 9.1 If preferences are homothetic, demand functions are linear in income, so we can write xi (p)m and xj (p)m. Applying Slutsky symmetry, we have ∂xi (p) ∂xj + xi(p)xj (p)m = + xj (p)xi (p)m. ∂pj ∂pi Subtracting xi (p)xj (p)m from each side of the equation establishes the result. 9.2 Note that p is the relative price of good x with respect to the other good which we’ll call z. Also, let m be income measured in units of z. Thus the consumer’s budget constraint is px + z = m. How do you know that there must be another good around? From dµ(p; q, m) = a + bp, dp we ﬁnd bp2 µ(p; q, m) = ap + + C. 2 Here C is a constant of integration. Since µ(q; q, m) = m, note that bq 2 C = m − aq − ; 2 therefore, bp2 bq 2 µ(p; q, m) = ap + + m − aq − . 2 2 A money metric utility function behaves like an indirect utility function with respect to q and m when holding p ﬁxed. Therefore, an indirect utility function consistent with the demand function given above is bq 2 v(q, m) = m − aq − . 2 Ch. 9 DEMAND 25 We can drop the terms in p since they are just constants. (Use Roy’s identity to check that this indirect utility is indeed consistent with our original demand function.) To get the direct utility function∗ we must solve bq 2 u(x, z) = min{v(q, m) : qx + z = m} = min{m − aq − : qx + z = m}; q q 2 use the budget constraint to eliminate m from the objective function and get the optimal value of q x−a q∗ = . b Thus (x − a)2 u(x, z) = z + . 2b This is, of course, a quasilinear utility function. On the back of an envelope, solve maxx,z {u(x, z) : px + z = m} and check that you get precisely the original demand function for x. What’s the demand for z? 9.3 We have to solve dµ(p; q, m) = a + bp + cµ(p; q, m). dp The homogeneous part has a solution of the form Aecp . A particular solu- tion to the nonhomogeneous equation is given by (a + bp)c + b µ=− . c2 Therefore the general solution to the diﬀerential equation is given by (a + bp)c + b µ(p; q, m) = Aecp − . c2 Since µ(q; q, m) = m we get (a + bq)c + b (a + bp)c + b µ(p; q, m) = m + ec(p−q) − . c2 c2 Hence, the indirect utility function is (a + bq)c + b v(q, m) = m + e−cq . c2 ∗ Strictly speaking, we should be saying “a utility function consistent with the given demand,” but we’ll just say “the utility function” with the understanding that any monotonic transformation of it would also generate the same demand function. 26 ANSWERS (Verify thatusing Roy’s identity we get the original demand function.) To get the direct utility function, we must solve (a + bq)c + b min m+ e−cq q c2 such that qx + z = m. The optimal value is given by x − cz − a q∗ = , b + cx which implies that b + cx ac − cx + c2 z u(x, z) = exp . c2 b + cx (Again, substitute z by m−px above, equate the derivative of the resulting expression with respect to x to zero, solve for x and recover the original demand function.) 9.4 Now the budget constraint is given by z + p1 x1 + p2 x2 = m. The symmetry of the substitution eﬀects implies ∂x1 ∂x2 = =⇒ b12 = b21 . ∂p2 ∂p1 The negative semideﬁniteness of the substitution matrix implies b1 < 0 and b1 b2 − b2 > 0. (Prove that these two conditions together imply that b2 < 0 must also hold.) We have to solve the following system of partial diﬀerential equations ∂µ(p; q, m) = a1 + b1 p1 + bp2 , ∂p1 and ∂µ(p; q, m) = a2 + bp1 + b2 p2 . ∂p2 The ﬁrst equation implies b1 2 µ(p; q, m) = a1 p1 + p + bp1 p2 + C1 , 2 1 where C1 is a constant of integration. The second implies b2 2 µ(p; q, m) = a2 p2 + p + bp1 p2 + C2 . 2 2 Ch. 9 DEMAND 27 Therefore, we must have b1 2 b2 µ(p; q, m) = a1 p1 + p + bp1 p2 + a2 p2 + p2 + C 2 1 2 2 a 1 b b p1 = [p1 , p2 ] 1 + [p1 , p2 ] 1 + C. a2 2 b b2 p2 Using µ(q; q, m) = m, we have b1 2 µ(p; q, m) = m + a1 (p1 − q1 ) + (p − q1 ) + b(p1 p2 − q1 q2 ) 2 2 1 b2 + a2 (p2 − q2 ) + (p2 − q2 ). 2 2 2 The indirect utility function is given by b1 2 b2 2 v(q, m) = m − a1 q1 − q1 − bq1 q2 − a2 q2 − q2 2 2 a1 1 b1 b q1 = m − [q1 , q2 ] − [q1 , q2] . a2 2 b b2 q2 9.5 To get the direct utility function we must solve u(x, z) = min{v(q, m) : z + q1 x1 + q2 x2 = m}. q After a few minutes of algebraic fun, we get ∗ b2 (x1 − a1 ) − b(x2 − a2 ) q1 = , b1 b2 − b2 and ∗ b1 (x2 − a2 ) − b(x1 − a1 ) q2 = . b1 b2 − b2 Substituting these values back into v(·), we get u(x, z) b2 (x1 − a1 )2 + b1 (x2 − a2 )2 b(a1 x2 + a2 x1 − x1 x2 − a1 a2 ) =z+ + 2(b1 b2 − b2 ) b1 b2 − b2 1 b −b x 1 − a1 =z+ [x1 − a1 , x2 − a2 ] 2 . 2(b1 b2 − b2 ) −b b1 x 2 − a2 9.6 Write the indirect utility function as v(p) = v(q/m) and diﬀerentiate with respect to qi and m: ∂v ∂v 1 = ∂qi ∂pi m k ∂v ∂v qi =− ∂m i=1 ∂pi m2 k ∂v 1 =− pi . i=1 ∂pi m 28 ANSWERS Dividing ∂vby ∂v yields the result. ∂qi ∂m 9.7 The function is weakly separable, and the subutility for the z-good b c consumption is z2 z3 . The conditional demands for the z-goods are Cobb- Douglas demands: b mz z1 = b + c p2 c mz z2 = . b + c p3 9.8.a dµ dp = a − bp + cµ 9.8.b µ(q, q, y) ≡ y 9.9.a The function V (x, y) = min{x, y}, and U (V, z) = V + z. 9.9.b The demand function for the z-good is z = m/pz if pz < px + py . If pz > px + py , then the demand for the x-good and the y-good is given by x = y = m/(px + py ). If pz = px + py , then take any convex combination of these demands. 9.9.c The indirect utility function is m m v(px , py , pz , m) = max , . px + py pz 9.11.a There are a variety of ways to solve this problem. The easiest is to solve for the indirect utility function to get v1 (p1 , p2 , m1 ) = m1 (p1 p2 )−1/2 . Now use Roy’s identity to calculate: 1 m1 x1 = 2 p1 1 m1 x2 = . 2 p2 Note that these are Cobb-Douglas demands. Recognizing that person 2 has Cobb-Douglas utility, we can write down the demands immediately: 3 m2 x1 = 3 + a p1 a m2 x2 = . 3 + a p2 9.11.b We must have the marginal propensity to consume each good the same for each consumer. This means that 1 3 = , 2 3+a which implies that a = 3. Ch. 11 UNCERTAINTY 29 Chapter 10. Consumers’ Surplus 10.1 We saw that in this case the indirect utility function takes the form v(p)+m. Hence the expenditure function takes the form e(p, u) = u−v(p). The expenditure function is necessarily a concave function of prices, which implies that v(p) is a convex function. 10.2 Ellsworth’s demand functions for the x-good and the y-good take the form 150 x=y= px + py . Plugging this into the utility function, we ﬁnd that the indirect utility function takes the form 150 v(px , py , 150) = . px + py Hence A is the solution to 150 − A 150 = 1+1 1+2 and B is the solution to 150 150 + B = . 1+1 1+2 Solving, we have A = 50 and B = 75. Chapter 11. Uncertainty 11.1 The proof of Pratt’s theorem established that 1 π(t) ≈ r(w)σ2 t2 . 2 But the σ2 t2 is simply the variance of the gamble t˜. 11.2 If risk aversion is constant, we must solve the diﬀerential equation u (x)/u (x) = −r. The answer is u(x) = −e−rx , or any aﬃne transforma- tion of this. If relative risk aversion is constant, the diﬀerential equation is u (x)x/u (x) = −r. The solution to this is u(x) = x1−r /(1 − r) for r = 1 and u(x) = ln x for r = 1. 11.3 We have seen that investment in a risky asset will be independent of wealth if risk aversion is constant. In an earlier problem, we’ve seen that 30 ANSWERS constant absoluterisk aversion implies that the utility function takes the form u(w) = −e−rw . 11.4 Marginal utility is u (w) = 1 − 2bw; when w is large enough this is a negative number. Absolute risk aversion is 2b/(1 − 2bw). This is an increasing function of wealth. 11.5.a The probability of heads occurring for the ﬁrst time on the j th toss ∞ is (1 − p)j−1p. Hence the expected value of the bet is j=1 (1 − p)j−1p2j = ∞ −j j ∞ j=1 2 2 = j=1 1 = ∞. 11.5.b The expected utility is ∞ ∞ (1 − p)j−1 p ln(2j ) = p ln(2) j(1 − p)j−1 . j=1 j=1 11.5.c By standard summation formulas: ∞ 1 (1 − p)j = . p j=0 Diﬀerentiate both sides of this expression with respect to p to obtain ∞ 1 j(1 − p)j−1 = . p2 j=1 Therefore, ∞ ln(2) p ln(2) j(1 − p)j−1 = . p j=1 11.5.d In order to solve for the amount of money required, we equate the utility of participating in the gamble with the utility of not participating. This gives us: ln(2) ln(w0 ) = , p Now simply solve this equation for w0 to ﬁnd w0 = eln(2)/p. 11.6.a Note that ∞ 2 1 1 s−µ E[u(R)] = u(s) √ exp − ds = φ(µ, σ2 ). −∞ σ 2π 2 σ Ch. 11 UNCERTAINTY 31 11.6.b Normalize u(·) such that u(µ) = 0. Diﬀerentiating, we have ∞ ∂E[u(R)] 1 = 2 u(s)(s − µ)f(s)ds > 0, ∂µ σ −∞ since the terms [u(s)(s − µ)] and f(s) are positive for all s. 11.6.c Now we have ∞ ∂E[u(R)] 1 = 3 u(s)((s − µ)2 − σ2 )f(s) ds ∂σ2 σ −∞ ∞ 1 < 3 u (µ)(s − µ)((s − µ)2 − σ2 )f(s)ds σ −∞ ∞ ∞ u (µ) = (s − µ)3 f(s)ds − σ2 (s − µ)f(s)ds σ3 −∞ −∞ = 0. The ﬁrst inequality follows from the concavity of u(·) and the normaliza- tion imposed; the last equality follows from the fact that R is normally distributed and, hence, E[(R − E[R])k ] = 0 for k odd. 11.7 Risk aversion implies a concave utility function. Denote by α ∈ [0, 1] the proportion of the initial wealth invested in asset 1. We have E[u(αw0 (1 + R1 ) + (1 − α)w0 (1 + R2 ))] = u(αw0 (1 + r1 ) + (1 − α)w0 (1 + r2 ))f(r1 )f(r2 )dr1 dr2 > [αu(w0 (1 + r1 )) + (1 − α)u(w0 (1 + r2 ))]f(r1 )f(r2 )dr1 dr2 = u(w0 (1 + r1 ))f(r1 )dr1 = u(w0 (1 + r2 ))f(r2 )dr2 = = E[u(w0 (1 + R1 ))] = E[u(w0 (1 + R2 ))]. The inequality follows from the concavity of u(·). For part b, proceed as before reversing the inequality since now u(·) is convex. 11.8.a Start by expanding both sides of E[u(w − πu )] = E[u(w + ˜)]: ˜ ˜ E[u(w − πu )] = pu(w1 − πu ) + (1 − p)u(w2 − πu ) ˜ ≈ p(u(w1 ) − u (w1 )πu ) + (1 − p)(u(w2 ) − u (w2 )πu ); p E[u(w + ˜)] = ˜ (u(w1 − ) + u(w1 + )) + (1 − p)u(w2 ) 2 u (w1 ) 2 ≈ p u(w1 ) + + (1 − p)u(w2 ). 2 32 ANSWERS Combining, weobtain 1 −(pu (w1 ) + (1 − p)u (w2 ))πu ≈ pu (w1 ) 2 , 2 or − 1 pu (w1 ) 2 πu ≈ 2 . pu (w1 ) + (1 − p)u (w2 ) 11.8.b For these utility functions, the Arrow-Pratt measures are −u /u = a, and −v /v = b. 11.8.c We are given a > b and we want to show that a value of (w1 − w2 ) large enough will eventually imply πv > πu , thus we want to get ae−aw1 be−bw1 < −bw1 ; pe−aw1 + (1 − p)e−aw2 pe + (1 − p)e−bw2 cross-multiplying we get ape−w1 (a+b) + a(1 − p)e−(aw1 +bw2 ) < bpe−w1 (a+b) + b(1 − p)e−(aw1 +bw2 ) , which implies p (a − b) < bea(w1 −w2 ) − aeb(w1 −w2 ) . 1−p The derivative of the RHS of this last inequality with respect to w1 − w2 is ab ea(w1 −w2 ) − eb(w1 −w2 ) > 0 whenever w1 > w2 ; the LHS does not depend on w1 or w2 . Therefore, this inequality will eventually hold for (w1 − w2 ) large enough. According to the Arrow-Pratt measure, u exhibits a higher degree of risk aversion than v. We’ve shown that v could imply a higher risk premium than u to avoid a fair lottery provided there’s an additional risk “big” enough. In this case, higher risk premium would no longer be synonymous with higher absolute risk aversion. 11.9 Initially the person has expected utility of 1√ 1√ 4 + 12 + 4 + 0 = 3. 2 2 If he sells his ticket for price p, he needs to get at least this utility. To ﬁnd the breakeven price we write the equation 4 + p = 3. Ch. 13 COMPETITIVEMARKETS 33 Solving, we have p = 5. 11.10 The utility maximization problem is max π ln(w+x)+(1−π) ln(w−x). The ﬁrst-order condition is π 1−π = , w+x w−x which gives us x = w(2π − 1). If π = 1/2, x = 0. 11.11 We want to solve the equation p 1−p 1 + = . w1 w2 w After some manipulation we have w1 w2 w= . pw2 + (1 − p)w1 11.12.a max αpx x such that x is in X. 11.12.b v(p, α) has the same form as the proﬁt function. 11.12.c Just mimic the proof used for the proﬁt function. 11.12.d It must be monotonic and convex, just as in the case of the proﬁt function. Chapter 13. Competitive Markets 13.1 The ﬁrst derivative of welfare is v (p) + π (p) = 0. Applying Roy’s law and Hotelling’s lemma, we have −x(p) + y(p) = 0, which is simply the condition that demand equals supply. The second derivative of this welfare measure is −x (p)+y (p) which is clearly positive; hence, we have a welfare minimum rather than a welfare maximum. The intuition behind this is that at any price other than the equilibrium price, the ﬁrm wants to supply a diﬀerent amount than the consumer wants to demand; hence, the “welfare” associated with all prices other than the equilibrium price is not attainable. 13.2 By Hotelling’s law we know that ∂π(p, w)/∂p = y(p); therefore, p1 y(p)dp = π(p1 , w) − π(p0 , w). p0 34 ANSWERS 13.3.a Theaverage cost curve is just c(w, y) y2 + 1 y2 + 2 = w1 + w2 . y y y You should verify that it is a convex function that has a unique minimum at w1 /w2 + 2 ym = . w1 /w2 + 1 The derivative of ym with respect to w1 /w2 is negative, so the minimum of the average cost shifts to the left (right) as w1 /w2 increases (decreases). √ In fact it converges to 1 as the ratio approaches ∞ and to 2 as it goes down to 0. 13.3.b The marginal cost is ∂c(w, y) = 2y(w1 + w2 ), ∂y so short-run supply schedule is given by p y(p) = . 2(w1 + w2 ) 13.3.c The long-run supply curve is Y (p) = arbitrarily large amount if p > 2ym (w1 + w2 ) 0 otherwise. 13.3.d From the cost function we have that x1 = y2 + 1 and x2 = y2 + 2. Also, we see that x1 and x2 are not substitutes at any degree. Therefore, the input requirement set for an individual ﬁrm is √ √ V (y) = (x1 , x2) ∈ [1, ∞) × [2, ∞): y ≤ min x1 − 1, x2 − 2 . 13.4.a y(p) = p/2 13.4.b Y (p) = 50p 13.4.c Equating D(p) = Y (p), we get p∗ = 2 and y∗ = 1 (Y ∗ = 100). 13.4.d The equilibrium rent on land r must equal the diﬀerence between each ﬁrm revenues and labor costs at the competitive equilibrium. There- fore, r = 2 − 1 = 1. Ch. 13 COMPETITIVEMARKETS 35 13.5.a In order to oﬀset the output subsidy, the U.S. should choose a tax of the same size as the subsidy; that is, choose t(s) = s. 13.5.b In the case of the capital subsidy, the producers receive p − t(s). If y∗ is to remain optimal, we must have p − t(s) = ∂c(w, r − s, y∗ )/∂y. 13.5.c Diﬀerentiate the above expression to get ∂ 2 c(w, r − s, y∗ ) ∂K(w, r − s, y∗ ) t (s) = = . ∂y∂r ∂y 13.5.d Since K(w, r − s, y) = K(w, r − s, 1)y, the formula reduces to t (s) = K(w, r − s, 1). 13.5.e In this case ∂K/∂y < 0 so that an increase in the subsidy rate implies a decrease in the tariﬀ. 13.6 Each ﬁrm that has marginal cost less than 25 will produce to capacity. What about the ﬁrm that has marginal cost equal to 25? If it produces a positive amount, it will just cover its variable cost, but lose the quasiﬁxed cost. Hence it prefers to stay out of business. This means that there will be 24 ﬁrms in the market, each producing 12 units of output, giving a total supply of 288. 13.7.a ym = 500 13.7.b p = 5 13.7.c yc = 50 × 5 = 250 13.8.a Price equals marginal cost gives us p = y, so Y = p + p = 2p. 13.8.b Set demand equal to supply 90−p = 2p to ﬁnd p∗ = 30 and Y ∗ = 60. 13.8.c Let p be the price paid by consumers. Then the domestic ﬁrms receive a price of p and the foreign ﬁrms receive a price of p − 3. Demand equals supply gives us 90 − p = p + [p − 3]. Solving we have p∗ = 31. 13.8.d The supply of umbrellas by domestic ﬁrms is 31 and by foreign ﬁrms is 28. 36 ANSWERS Chapter 14.Monopoly 14.1 The proﬁt-maximizing level of output is 5 units. If the monopolist only has 4 to sell, then it would ﬁnd it most proﬁtable to charge a price of 6. This is the same as the competitive solution. If, however, the monopolist had 6 units to sell, it would be most proﬁtable to dispose of one unit and only sell 5 units at a price of 5. 14.2 The monopolist has zero marginal costs up until 7 units of output and inﬁnite marginal costs for any output greater than 7 units. The proﬁt- maximizing price is 5 and the proﬁts are 25. 14.3 For this constant elasticity demand function revenue is constant at 10, regardless of the level of output. Hence output should be as small as possible—that is, a proﬁt-maximizing level of output doesn’t exist. 14.4 According to the formula given in the text, we must have 1 = 1, 2 + yp (y)/p (y) or yp (y) = −p (y). The required inverse demand function is the solution to this diﬀerential equation. It turns out that it is given by p(y) = a − b ln x. The direct demand function then takes the form ln x = a/b − p/b, which is sometimes called a semilog demand function. 14.5 The monopolist’s proﬁt maximization problem is max p(y, t)y − cy. y The ﬁrst-order condition for this problem is ∂p(y, t) p(y, t) + y − c = 0. ∂y According to the standard comparative statics calculations, the sign of dy/dt is the same as the sign of the derivative of the ﬁrst-order expression with respect to t. That is, dy ∂p ∂p2 sign = sign + y. dt ∂t ∂y∂t For the special case p(y, t) = a(p) + b(t), the second term on the right-hand side is zero. Ch. 14 MONOPOLY 37 14.6 There is no proﬁt-maximizing level of output since the elasticity of demand is constant at −1. This means that revenue is independent of output, so reductions in output will lower cost but have no eﬀect on revenue. 14.7 Since the elasticity of demand is −1, revenues are constant at any price less than or equal to 20. Marginal costs are constant at c so the monopolist will want to produce the smallest possible output. This will happen when p = 20, which implies y = 1/2. 14.8 For this to occur, the derivative of consumer’s surplus with respect to quality must be zero. Hence ∂u/∂q − ∂p/∂qx ≡ 0. Substituting for the deﬁnition of the inverse demand function, this means that we must have ∂u/∂q ≡ x∂ 2 u/∂x∂q. It is easy to verify that this implies that u(x, q) = f(q)x. 14.9 The integral to evaluate is x x ∂ 2 u(z, q) ∂p(x, q) dz < dz. 0 ∂z∂q 0 ∂q Carrying out the integration gives ∂u(x, q) ∂p(x, q) < x, ∂q ∂q which is what is required. 14.10 If the ﬁrm produces x units of output which it sells at price p(x), then the most that it can charge for entry is the consumer’s surplus, u(x)−p(x)x. Once the consumer has chosen to enter, the ﬁrm makes a proﬁt of p(x) − c on each unit of output purchased. Thus the proﬁt maximization problem of the ﬁrm is max u(x) − p(x)x + (p(x) − c(x))x = u(x) − c(x). x It follows that the monopolist will choose the eﬃcient level of output where u (x) = c (x). The entry fee is set equal to the consumer’s surplus. 14.11 The ﬁgure depicts the situation where the monopolist has reduced the price to the point where the marginal beneﬁt from further reductions just balance the marginal cost. This is the point where p2 = 2p1 . If the high-demand consumer’s inverse demand curve is always greater than twice the low-demand consumer’s inverse demand curve, this condition cannot be satisﬁed and the low-demand consumer will be pushed to a zero level of consumption. 38 ANSWERS 14.12 AreaB is what the monopolist would gain by selling only to the high-demand consumer. Area A is what the monopolist would lose by doing this. 14.13 This is equivalent to the price discrimination problem with x = q and wt = rt . All of the results derived there translate on a one-to-one basis; e.g., the consumer who values quality the more highly ends up consuming the socially optimal amount, etc. 14.14 The maximization problem is maxp py(p) − c(y(p)). Diﬀerentiating, we have py (p) + y(p) − c (y)y (p) = 0. This can also be written as p + y(p)/y (p) − c (y) = 0, or p[1 + 1/ ] = c (y). 14.15 Under the ad valorem tax we have 1 (1 − τ )PD = 1+ c. Under the output tax we have 1 PD − t = 1 + . Solve each equation for PD , set the results equal to each other, and solve for t to ﬁnd τ kc 1 t= k= 1−τ 1+ 1 14.16.a The monopolist’s proﬁt maximization problem is max p(y, t)y − cy. y The ﬁrst-order condition for this problem is ∂p(y, t) p(y, t) + y − c = 0. ∂y According to the standard comparative statics calculations, the sign of dy/dt is the same as the sign of the derivative of the ﬁrst-order expression with respect to t. That is, dy ∂p ∂p2 sign = sign + y. dt ∂t ∂y∂t Ch. 14 MONOPOLY 39 14.16.b For the special case p(y, t) = a(y) + b(t), the second term on the right-hand side is zero, so that ∂p/∂t = ∂b/∂t. 14.17.a Diﬀerentiating the ﬁrst-order conditions in the usual way gives ∂x1 1 = <0 ∂t1 p1 − c1 ∂x2 1 = < 0. ∂t2 2p2 + p2 x2 − c2 14.17.b The appropriate welfare function is W = u1 (x1 )+u2 (x2 )−c1 (x1 )− c2 (x2 ). The total diﬀerential is dW = (u1 − c1 )dx1 + (u2 − c2 )dx2 . 14.17.c Somewhat surprisingly, we should tax the competitive industry and subsidize the monopoly! To see this, combine the answers to the ﬁrst two questions to get the change in welfare from a tax policy (t1 , t2 ). dx1 dx2 dW = (p1 − c1 ) dt1 + (p2 − c2 ) dt2 . dt1 dt2 The change in welfare from a small tax or subsidy on the competitive in- dustry is zero, since price equals marginal cost. But for the monopolized industry, price exceeds marginal cost, so we want the last term to be posi- tive. But this can only happen if dt2 is negative—i.e., we subsidize industry 2. 14.18.a The proﬁt maximization problem is max r1 + r2 such that a1 x1 − r1 ≥ 0 a2 x 2 − r 2 ≥0 a1 x 1 − r 1 ≥ a1 x 2 − r 2 a2 x2 − r2 ≥ a2 x 1 − r 1 x1 + x2 ≤ 10. 14.18.b The binding constraints will be a1 x1 = r1 and a2 x2 −r2 = a2 x1 −r1 , and x1 + x2 = 10. 14.18.c The expression is a2 x2 + (2a1 − a2 )x1 . 40 ANSWERS 14.18.d Formally,our problem is to solve max a2 x2 + (2a1 − a2 )x1 subject to the constraint that x1 + x2 = 10. Solve the constraint for x2 = 10 − x1 and substitute into the objective function to get the problem max 10a2 + 2(a1 − a2 )x1 . x1 Since a2 > a1 the coeﬃcient on the second term is negative, which means that x∗ = 0 and, therefore, x∗ = 10. Since x∗ = 10, we must have r2 = 1 2 2 ∗ ∗ ∗ 10a2 . Since x1 = 0, we must have r1 = 0. 14.19.a The proﬁt-maximizing choices of p1 and p2 are p1 = a1 /2b1 p2 = a2 /2b2 . These will be equal when a1 /b1 = a2 /b2 . 14.19.b We must have p1 (1 − 1/b1 ) = c = p2 (1 − 1/b2 ). Hence p1 = p2 if and only if b1 = b2 . 14.20.a The ﬁrst-order condition is (1 − t)[p(x) + p (x)x] = c (x), or p(x) + p (x)x = c (x)/(1 − t). This expression shows that the revenue tax is equivalent to an increase in the cost function, which can easily be shown to reduce output. 14.20.b The consumer’s maximization problem is maxx u(x) − m − px + tpx = maxx u(x) − m − (1 − t)px. Hence the inverse demand function satisﬁes u (x) − (1 − t)p(x), or p(x) = u (x)/(1 − t). 14.20.c Substituting the inverse demand function into the monopolist’s ob- jective function, we have (1 − t)p(x)x − c(x) = (1 − t)u (x)x/(1 − t) − c(x) = u (x)x − c(x). Since this is independent of the tax rate, the monopolist’s behavior is the same with or without the tax. 14.21 Under the ad valorem tax we have 1 (1 − τ )PD = 1+ c. Under the output tax we have 1 PD − t = 1 + . Ch. 15 GAMETHEORY 41 Solve each equation for PD , set the results equal to each other, and solve for t to ﬁnd τ kc 1 t= k= 1−τ 1+ 1 14.22.a Note that his revenue is equal to 100 for any price less than or equal to 20. Hence the monopolist will want to produce as little output as possible in order to keep its costs down. Setting p = 20 and solving for demand, we ﬁnd that D(20) = 5. 14.22.b They should set price equal to marginal cost, so p = 1. 14.22.c D(1) = 100. 14.23.a If c < 1, then proﬁts are maximized at p = 3/2 + c/2 and the monopolist sells to both types of consumers. The best he can do if he sells only to Type A consumers is to sell at a price of 2 + c/2. He will do this if c ≥ 1. 14.23.b If a consumer has utility ax1 −x2 /2+x2, then she will choose to pay 1 k if (a−p)2 /2 > k. If she buys, she will buy a−p units. So if k < (2−p)2 /2, then demand is N (4 − p) + N (2 − p). If (2 − p)2 < k < (4 − p)2 /2, then demand is N (4 − p). If k > (4 − p)2 /2, then demand is zero. 14.23.c Set p = c and k = (4 − c)2 /2. The proﬁt will be N (4 − c)2 /2. 14.23.d In this case, if both types of consumers buy the good, then the proﬁt-maximizing prices will have the Type B consumers just indiﬀerent between buying and not buying. Therefore k = (2 − p)2 /2. Total proﬁts will then be N ((6 − 2p)(p − c) + (2 − p)2 /2). This is maximized when p = 2(c + 2)/3. Chapter 15. Game Theory 15.1 There are no pure strategy equilibria and the unique mixed strategy equilibrium is for each player to choose Head or Tails with probability 1/2. 15.2 Simply note that the dominant strategy on the last move is to defect. Given that this is so, the dominant strategy on the next to the last move is to defect, and so on. 15.3 The unique equilibrium that remains after eliminating weakly domi- nant strategies is (Bottom, Right). 15.4 Since each player bids v/2, he has probability v of getting the item, giving him an expected payoﬀ of v2 /2. 42 ANSWERS 15.5.a a≥ e, c ≥ g, b ≥ d, f ≥ h 15.5.b Only a ≥ e, b ≥ d. 15.5.c Yes. 15.6.a There are two pure strategy equilibria, (Swerve, Stay) and (Stay, Swerve). 15.6.b There is one mixed strategy equilibrium in which each player chooses Stay with probability .25. 15.6.c This is 1 − .252 = .9375.. 15.7 If one player defects, he receives a payoﬀ of πd this period and πc forever after. In order for the punishment strategy to be an equilibrium the payoﬀs must satisfy πc πj πd + ≤ πj + . r r Rearranging, we ﬁnd πj − πc r≤ πd − πj . 15.8.a Bottom. 15.8.b Middle. 15.8.c Right. 15.8.d If we eliminate Right, then Row is indiﬀerent between his two re- maining strategies. 15.9.a (Top, Left) and (Bottom, Right) are both equilibria. 15.9.b Yes. (Top, Left) dominates (Bottom, Right). 15.9.c Yes. 15.9.d (Top, Left). Chapter 16. Oligopoly 16.1 The Bertrand equilibrium has price equal to the lowest marginal cost, c1 , as does the competitive equilibrium. Ch. 16 OLIGOPOLY 43 16.2 ∂F (p, u)/∂u = 1 − r/p. Since r is the largest possible price, this expression will be nonpositive. Hence, increasing the ratio of uninformed consumers decreases the probability that low prices will be charged, and increases the probability that high prices will be charged. 16.3 Let δ = β1 β2 − γ 2 . Then by direct calculation: ai = (αi βj − αj γ)/δ, bi = βj /δ, and c = γ/δ. 16.4 The calculations are straightforward and may be found in Singh & Vives (1984). Let ∆ = 4β1 β2 − γ 2 , and D = 4b1 b2 − c2 . Then it turns out that pc − pb = αiγ 2 /∆ and qi − qi = aic2 /D, where superscripts refer to i i b c Bertrand and Cournot. 16.5 The argument is analogous to the argument given on page 297. 16.6 The problem is that the thought experiment is phrased wrong. Firms in a competitive market would like to reduce joint output, not increase it. A conjectural variation of −1 means that when one ﬁrm reduces its output by one unit, it believes that the other ﬁrm will increase its output by one unit, thereby keeping joint output—and the market price—unchanged. 16.7 In a cartel the ﬁrms must equate the marginal costs. Due to the assumption about marginal costs, such an equality can only be established when y1 > y2 . 16.8 Constant market share means that y1 /(y1 + y2 ) = 1/2, or y1 = y2 . Hence the conjectural variation is 1. We have seen that the conjectural variation that supports the cartel solution is y2 /y1 . In the case of identical ﬁrms, this is equal to 1. Hence, if each ﬁrm believes that the other will attempt to maintain a constant market share, the collusive outcome is “stable.” 16.9 In the Prisoner’s Dilemma, (Defect, Defect) is a dominant strategy equilibrium. In the Cournot game, the Cournot equilibrium is only a Nash equilibrium. 16.10.a Y = 100 16.10.b y1 = (100 − y2 )/2 16.10.c y = 100/3) 16.10.d Y = 50 16.10.e y1 = 25, y2 = 50 16.11.a P (Y ) + P (Y )yi = c + ti 44 ANSWERS 16.11.b Sumthe ﬁrst order conditions to get nP (Y ) + P (Y )Y = nc + n i=1 ti , and note that industry output Y can only depend on the sum of the taxes. 16.11.c Since total output doesn’t change, ∆yi must satisfy P (Y ) + P (Y )[yi + ∆yi] = c + ti + ∆ti . Using the original ﬁrst order condition, this becomes P (Y )∆yi = ∆ti, or ∆yi = ∆ti /P (Y ). 16.12.a y = p 16.12.b y = 50p 16.12.c Dm (p) = 1000 − 100p 16.12.d ym = 500 16.12.e p = 5 16.12.f yc = 50 × 5 = 250 16.12.g Y = ym + yc = 750. Chapter 17. Exchange 17.1 In the proof of the theorem, we established that x∗ ∼i xi . If x∗ and xi i i were distinct, a convex combination of the two bundles would be feasible and strictly preferred by every agent. This contradicts the assumption that x∗ is Pareto eﬃcient. 17.2 The easiest example is to use Leontief indiﬀerence curves so that there are an inﬁnite number of prices that support a given optimum. 17.3 Agent 2 holds zero of good 2. 17.4 x1 = ay/p1 = ap2 /p1 , x1 = x2 so from budget constraint, (p1 + A B B p2 )x1 = p1 , so x1 = p1 /(p1 + p2 ). Choose p1 = 1 an numeraire and solve B B ap2 + 1/(1 + p2 ) = 1. 17.5 There is no way to make one person better oﬀ without hurting someone else. 17.6 x1 = ay1 /p1 , x2 = by2 /p1 1 y1 = y2 = p1 + p2 . Solve x1 + x1 = 2. 1 2 Ch. 18 PRODUCTION 45 17.7 The Slutsky equation for consumer i is ∂xi ∂hi = . ∂pj ∂pj 17.8 The strong Pareto set consists of 2 allocations: in one person A gets all of good 1 and person B gets all of good 2. The other Pareto eﬃcient allocation is exactly the reverse of this. The weak Pareto set consists of all allocations where one of the consumers has 1 unit of good 1 and the other consumer has at least 1 unit of good units of good 2. 17.9 In equilibrium we must have p2 /p1 = x2 /x1 = 5/10 = 1/2. 3 3 17.10 Note that the application of Walras’ law in the proof still works. 17.11.a The diagram is omitted. 17.11.b We must have p1 = p2 . 17.11.c The equilibrium allocation must give one agent all of one good and the other agent all of the other good. Chapter 18. Production 18.1.a Consider the following two possibilities. (i) Land is in excess supply. (ii) All land is used. If land is in excess supply, then the price of land is zero. Constant returns requires zero proﬁts in both the apple and the bandanna industry. This means that pA = pB = 1 in equilibrium. Every consumer will have income of 15. Each will choose to consume 15c units of apples and 15(1 − c) units of bandannas. Total demand for land will be 15cN . Total demand for labor will be 15N . There will be excess supply of land if c < 2/3. So if c < 2/3, this is a competitive equilibrium. If all land is used, then the total outputs must be 10 units of apples and 5 units of bandannas. The price of bandannas must equal the wage which is 1. The price of apples will be 1 + r where r is the price of land. Since preferences are homothetic and identical, it will have to be that each person consumes twice as many apples as bandannas. People will want to consume c twice as much apples as bandannas if pA /pB = (1−c) (1/2). Then it also must be that in equilibrium, r = (pA /pB ) − 1 ≥ 0. This last inequality will hold if and only if c ≥ 2/3. This characterizes equilibrium for c ≥ 2/3. 18.1.b For c < 2/3. 18.1.c For c < 2/3. 46 ANSWERS 18.2.a Letthe price of oil be 1. Then the zero-proﬁt condition implies that pg 2x − x = 0. This means that pg = 1/2. A similar argument shows that pb = 1/3. 18.2.b Both utility functions are Cobb-Douglas, and each consumer has an endowment worth 10. From this we can easily calculate that xg = 8, 1 xb = 18, xg = 10, xb = 15. 1 2 2 18.2.c To make 18 guns, ﬁrm 1 needs 9 barrels of oil. To make 33 units of butter, ﬁrm 2 needs 11 barrels of oil. Chapter 19. Time 19.1 See Ingersoll (1987), page 238. 19.2.a Apartments will be proﬁtable to construct as long as the present value of the stream of rents is at least as large as the cost of construction. In equations: (1 + π)p p+ ≥ c. 1+r In equilibrium, this condition must be satisﬁed as an equality, so that 1+r p= c. 2+r+π 19.2.b Now the condition becomes 1+r p= c. 2 + r + 3π 4 19.2.c Draw the ﬁrst period demand curve and subtract oﬀ the K rent controlled apartments to get the residual demand for new apartments. Look for the intersection of this curve with the two ﬂat marginal cost curves derived above. 19.2.d Fewer. 19.2.e The equilibrium price of new apartments will be higher. Ch. 22 WELFARE 47 Chapter 20. Asset Markets 20.1 The easiest way to show this is to write the ﬁrst-order conditions as ˜ ˜ ˜ Eu (C)Ra = Eu (C)R0 ˜ ˜ ˜ Eu (C)Rb = Eu (C)R0 and subtract. 20.2 Dividing both sides of the equation by pa and using the deﬁnition ˜ ˜ Ra = Va /pa , we have ˜ ˜ Ra = R0 − R0 cov(F (C), Ra ). Chapter 21. Equilibrium Analysis 21.1 The core is simply the initial endowment. 21.2 Since the income eﬀects are zero, the matrix of derivatives of the Marshallian demand function is equal to the matrix of derivatives of the Hicksian demand function. It follows from the discussion in the text that the index of every equilibrium must be +1, which means there can be only one equilibrium. 21.3 Diﬀerentiating V (p), we have dV (p) = −2z(p)Dz(p)p˙ dt = −2z(p)Dz(p)Dz(p)−1z(p) = −2z(p)z(p) < 0. Chapter 22. Welfare 22.1 We have the equation k ∂hj θxi = tj . ∂pi j=1 Multiply both sides of this equation by ti and sum to get k k ∂hj θR = θ tixi = ti tj . i j=1 i=1 ∂pi 48 ANSWERS The right-handside of expression is nonpositive (and typically negative) since the Slutsky matrix is negative semideﬁnite. Hence θ has the same sign as R. 22.2 The problem is max v(p, m) k such that (pi − ci )xi(pi ) = F. i=1 This is almost the same as the optimal tax problem, where pi − ci plays the role of ti. Applying the inverse elasticity rule gives us the result. Chapter 23. Public goods 23.1 Suppose that it is eﬃcient to provide the public good together, but neither agent wants to provide it alone. Then any set of bids such that b1 + b2 = c and bi ≤ ri is an equilibrium to the game. However, there are also many ineﬃcient equilibria, such as b1 = b2 = 0. 23.2 If utility is homothetic, the the consumption of each good will be proportional to wealth. Let the demand function for the public good be given by ai fi (w) = w. 1 + ai Then the equilibrium amount of the public good is the same as in the Cobb-Douglas example given in the text. 23.3 Agent 1 will contribute g1 = αw1 . Agent 2’s reaction function is f2 (w2 + g1 ) = max{α(w2 + g1 ) − g1 , 0}. Solving f2 (w2 + αw1 ) = 0 yields w2 = (1 − α)w1 . 23.4 The total amount of the public good with k contributors must satisfy w G G=α + . k k Solving for G, we have G = αw/(k − α). As k increases, the amount of wealth becomes more equally distributed and the amount of the privately provided public good decreases. 23.5 The allocation is not in general Pareto eﬃcient, since for some patterns of preferences some of the private good must be thrown away. However, the amount of the public good provided will be the Pareto eﬃcient amount: 1 unit if i ri > c, and 0 units otherwise. Ch. 24 EXTERNALITIES 49 23.6.a max ai ln(G−i + gi ) + wi − gi gi such that gi ≥ 0. 23.6.b The ﬁrst-order condition for an interior solution is ai = 1, G or G = ai . Obviously, the only agent who will give a positive amount is the one with the maximum ai . 23.6.c Everyone will free ride except for the agent with the maximum ai . 23.6.d Since all utility functions are quasilinear, a Pareto eﬃcient amount of the public good can be found by maximizing the sum of the utilities: n ai ln G − G, i=1 which implies G∗ = n i=1 ai . Chapter 24. Externalities 24.1.a Agent 1’s utility maximization problem is max u1 (x1 ) − p(x1 , x2)c1 , x1 while the social problem is max u1 (x1 ) + u2 (x2 ) − p(x1 , x2)[c1 + c2 ]. x1 ,x2 Since agent 1 ignores the cost he imposes on agent 2, he will generally choose too large a value of x1 . 24.1.b By inspection of the social problem and the private problem, agent 1 should be charged a ﬁne t1 = c2 . 24.1.c If the optimal ﬁnes are being used, then the total costs born by the agents in the case of an accident are 2[c1 + c2 ], which is simply twice the total cost of the accident. 24.1.d Agent 1’s objective function is (1 − p(x1 , x2 ))u1 (x1 ) − p(x1 , x2)c1 . This can also be written as u1 (x1 ) − p(x1 , x2 )[u1 (x1 ) + c1 ]. This is just the form of the previous objective function with u1 (x1 ) + c1 replacing c1 . Hence the optimal ﬁne for agent 1 is t1 = u2 (x2 ) + c2 . 50 ANSWERS Chapter 25.Information 25.1 By construction we know that f(u(s)) ≡ s. Diﬀerentiating one time shows that f (u)u (s) = 1. Since u (s) > 0, we must have f (u) > 0. Diﬀerentiating again, we have f (u)u (s) + f (u)u (s)2 = 0. Using the sign assumptions on u (s), we see that f (u) > 0. 25.2 According to the envelope theorem, ∂V /∂ca = λ + µ and ∂V /∂cb = µ. Thus, the sensitivity of the payment scheme to the likelihood ratio, µ, depends on how big an eﬀect an increase in cb would have on the principal. 25.3 In this case it is just as costly to undertake the action preferred by the principal as to undertake the alternative action. Hence, the incentive constraint will not be binding, which implies µ = 0. It follows that s(xi ) is constant. 25.4 If cb decreases, the original incentive scheme (si ) will still be feasible. Hence, an optimal incentive scheme must do at least as well as the original scheme. 25.5 In this case the maximization problem takes the form n max (xi − si )πib i=1 n such that si πib − cb ≥ u i=1 n n si πib − cb ≥ si πia − ca . i=1 i=1 Assuming that the participation constraint is binding, and ignoring the incentive-compatibility constraint for a moment, we can substitute into the objective function to write m max si πib − cb − u. i=1 Hence, the principal will choose the action that maximizes expected output minus (the agent’s) costs, which is the ﬁrst-best outcome. We can satisfy the incentive-compatibility constraint by choosing si = xi + F , and choose F so that the participation constraint is satisﬁed. Ch. 25 INFORMATION 51 25.6 The participation constraints become st − c(xt ) ≥ ut , which we can write as st − (c(xt ) + ut ). Deﬁne ct (x) = c(x) + ut , and proceed as in the text. Note that the marginal costs of each type are the same, which adds an extra case to the analysis. 25.7 Since c2 (x) > c1 (x), we must have x2 x2 c2 (x) dx > c1 (x) dx. x1 x1 The result now follows from the Fundamental Theorem of Calculus. 25.8 The indiﬀerence curves take the form u1 = s−c1 (x) and u2 = s−c2 (x). Write these as s = u1 + c1 (x) and s = u2 + c2 (x). The diﬀerence between these two functions is d(x) = u2 − u1 + c2 (x) − c1 (x), and the derivative of this diﬀerence is d (x) = c2 (x) − c1 (x) > 0. Since the diﬀerence function is a monotonic function, it can hit zero at most once. 25.9 For only low-cost workers to be employed, there must be no proﬁtable contract that appeals to the high-cost workers. The most proﬁtable con- tract to a high-cost worker maximizes x2 − x2 , which implies x∗ = 1/2. 2 2 The cost of this to the worker is (1/2)2 = 1/4. For the worker to ﬁnd this acceptable, s2 − 1/4 ≥ u2 , or s2 = u2 + 1/4. For the ﬁrm to make a proﬁt, x∗ ≥ s2 . Hence we have 1/2 ≥ u2 + 1/4, or u2 ≤ 1/4. 2 25.10.a The professor must pay s = x2 /2 to get the assistant to work x hours. Her payoﬀ will be x − x2 /2. This is maximized where x = 1. 25.10.b The TA must get his reservation utility when he chooses the optimal x. This means that s − x2 /2 = s − 1/2 = 0, so s = 1/2. 25.10.c The best the professor can do is to get Mr. A to work 1 hour and have a utility of zero. Mr. A will work up to the point where he maximizes ax + b − x2 /2. Using calculus, we ﬁnd that Mr. A will choose x = a. Therefore he will work one hour if a = 1. Then his utility will be 0 if b = −1. Download AboutSupportTermsPrivacyCopyrightCookie PreferencesDo not sell or share my personal information English © 2025 Slideshare from Scribd
14595	https://josr-online.biomedcentral.com/articles/10.1186/s13018-020-01773-9	Assessment of grip-motion characteristics in carpal tunnel syndrome patients using a novel finger grip dynamometer system \| Journal of Orthopaedic Surgery and Research \| Full Text Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Advertisement Search Explore journals Get published About BMC Login Menu Explore journals Get published About BMC Login Search all BMC articles Search Journal of Orthopaedic Surgery and Research Home About Articles Submission Guidelines Submit manuscript Assessment of grip-motion characteristics in carpal tunnel syndrome patients using a novel finger grip dynamometer system Download PDF Download PDF Research article Open access Published: 06 July 2020 Assessment of grip-motion characteristics in carpal tunnel syndrome patients using a novel finger grip dynamometer system Toru Sasaki1, Koji Makino2, Akimoto Nimura3, Shiro Suzuki3, Tomoyuki Kuroiwa1, Takafumi Koyama1, Atsushi Okawa1, Hidetsugu Terada4& … Koji FujitaORCID: orcid.org/0000-0003-3733-01881 Show authors Journal of Orthopaedic Surgery and Researchvolume 15, Article number:245 (2020) Cite this article 2701 Accesses 10 Citations Metrics details Abstract Background Grip strength measurement is widely used in daily medical practice, and it has been reported that the grip strength decreases in patients with carpal tunnel syndrome (CTS). However, conventional grip dynamometers evaluate only the maximum power of total grip strength and cannot measure the time course of grip motion. In this report, we aimed to determine the grip characteristics of CTS patients by measuring the time course of each finger’s grip motion and to analyze the relationship between finger grip strength and subjective symptoms using this new grip system. Methods The grip strength of each finger was measured using the new grip system that has four pressure sensors on the grip parts of each finger of the Smedley grip dynamometer. We analyzed the time course of grip motion and relationship between finger grip strength and subjective symptoms in 104 volunteer and 51 CTS hands. The Japanese Society for Surgery of the Hand version of the Carpal Tunnel Syndrome Instrument (CTSI-JSSH) and the Disability of Arm, Shoulder, and Hand questionnaire (DASH) were used as subjective evaluation scores. Results In the CTS group, the grip time with the index, middle, and ring fingers was longer, and the time at which strength was lost after reaching the maximum was earlier. Patients with severe subjective symptoms tended to not use the index and middle fingers during grip motion. Conclusions This new system that measures each finger’s grip strength at one time and record the time course of grip motion could quantify a patient’s symptoms easily and objectively, which may contribute to the evaluation of hand function. Background Grip strength measurement is widely used in daily medical practice because it is a non-invasive, quick, and inexpensive way of assessing muscle strength. In the field of hand surgery, grip strength may decrease due to pain and neuropathy; thus, it is important to measure the grip strength to evaluate hand function [1,2,3]. In particular, it has been shown that grip strength is decreased in patients with carpal tunnel syndrome (CTS) [4, 5], which is a clinical syndrome of numbness, pain, and disorder of thumb opposition associated with localized compression of the median nerve at the wrist [6,7,8,9]. However, the grip characteristics of patients with CTS are still unclear, and the mechanism underlying diminished grip strength in those with damaged sensory and motor fibers of the median nerve needs to be investigated. A conventional grip dynamometer, which has been used in many studies that have demonstrated weakened grip strength in patients with CTS, evaluates only the maximum power in the total grip strength of all fingers. Moreover, there have been no reports about the time course of grip motion, such as the time from the beginning of grip motion to maximum power and the time from maximum power to loss of grip strength. Thus, it was necessary to develop a grip dynamometer that can measure the time course to clarify the characteristics of grip motion in CTS patients. Although the Japanese Society for Surgery of the Hand version of the Carpal Tunnel Syndrome Instrument (CTSI-JSSH) and the Disability of Arm, Shoulder, and Hand questionnaire (DASH) are used as subjective evaluation scores in CTS patients, and the usefulness of these scores as indicators of patient’s symptom has been reported , there is no report on the correlation between subjective symptoms of CTS and grip strength. To evaluate the subjective symptoms of patients with only distal median nerve dysfunction using a grip dynamometer, we needed a new system for measuring the grip strength of each finger, since it is difficult to accurately evaluate these symptoms using the total grip strength of all fingers. Therefore, we developed a new grip system that could measure the grip strength of each finger in a single instance to record the time course and analyze the grip motion of healthy volunteers . In this report, to understand the pathophysiology of CTS, we aimed to analyze the grip characteristics of CTS patients by measuring the time course of the grip motion, and we sought to determine the relationship between finger grip strength and subjective symptoms using this new grip system. Methods This study was approved by the institutional review board of our institution, and we obtained informed consent from all patients. Participants We included 51 hands of 30 patients with CTS (bilateral CTS—21, unilateral CTS—9) who were diagnosed by hand surgeons (CTS group) and 104 hands of 52 healthy volunteers (control group). The inclusion criteria for this study were as follows: clinical symptoms of CTS (numbness, tingling, and pain), positive examination for CTS including a positive Phalen’s sign and Tinel’s sign, and an abnormal nerve conduction study (NCS) with sensory nerve conduction velocity (SCV) of ≤ 44 m/s or abductor pollicis brevis–distal latency (APB-DL) of > 4.0 ms. Patients who continued to have symptoms and an abnormal NCS after carpal tunnel release were included. Patients were excluded from this study if they had compressive neuropathy in the ipsilateral arm or peripheral polyneuropathy, cervical disease, DeQuervein syndrome, or trigger finger, had a history of a distal radial fracture, were pregnant, and were non-Japanese speaking. The NCS results were classified according to the Bland classification, which is based on the electrophysiological severity . For the control group, the inclusion criteria included participants who had no symptoms or previous history of neuropathies or trauma to the upper limbs, and they were age-matched to the participants in the CTS group. Measurement The grip strength of each finger was measured using the new grip system developed in our institution. The apparatus used and measurement posture were the same as those in the previous report, and data regarding the healthy volunteers have already been reported . The structure of this system was the same as that in the previous report and is shown in Fig. 1. This system consists of three units, which are as follows: a measurement unit with four pressure sensors on the grip parts of each finger (index, middle, ring, and little fingers) of the Smedley grip dynamometer (Fig. 1a), a wireless communication unit that sends the data to the display every 10 ms, and a display unit that shows each finger’s grip strength in real time (Fig. 1b). When the patients grip this system, the four sensors measure the voltage values and send the data to the display through a wireless communication unit. The voltage values are converted to grip strength, and each finger’s grip strength is displayed and saved on the computer in real time. Fig. 1 The finger grip dynamometer system. a Measurement unit: four pressure sensors on the grip parts of each finger (index, middle, ring, and little fingers) of the Smedley grip dynamometer. The four sensors measure the voltage values and send the data to the display through a wireless communication unit. b Display unit: the display converts measuring date to grip strength and shows date in real time Full size image The right and left finger grip strengths were measured twice [13, 14] using this system while the patient was standing with the shoulders adducted, elbows in the straight position, and forearms and wrists in the neutral position. The average of the two trials was calculated. The patients were instructed to hold the device as hard and as early as possible and to release it when the power reached the maximum. Statistical analysis Data regarding age, CTSI-JSSH, DASH, and measured grip strength are presented as medians with interquartile ranges. The Student t test, Fisher exact test, and Mann-Whitney U tests were used to compare differences in nominal and nonparametric variables. Based on previous studies, we considered a 3.0-kg decrease in the grip strength as clinically meaningful when comparing the difference in grip strength between the CTS patients and controls [15, 16]. We calculated that, with a sample of 88 participants (44 participants per group), the study would have 80% power to detect a 3.0-kg mean decrease in the grip strength, with a type 1 error of 5%. For the power analysis, we used a standard deviation of 5.0 in the grip strength using data reported in previous studies . To evaluate the characteristics of grip motion in the CTS patients, we performed the following types of analysis. First, to compare the time course of finger grip strength between the CTS and control groups, the grip time at which the maximum grip strength was reached was analyzed. We calculated the time from reaching 20% of the maximum strength to reaching the maximum strength [18, 19]. We compared the grip time between the CTS and control groups using t tests. In addition, we analyzed the mechanism underlying the diminished grip strength in CTS patients. The ratio of the maximum finger grip strength to real time finger grip strength was considered to be the ΔGrip strength. We calculated ΔGrip strength from − 1.6 to + 3.2 s of the time when the maximum grip strength was reached. At each time point, the average of ΔGrip strength of the CTS and control groups was calculated and visualized graphically for each finger. Since the absolute values of each finger’s grip strength were not being compared, differences in gender and individual physique have less of an influence on the grip time and ΔGrip strength. Second, to evaluate the relationship between subjective symptoms and finger grip strength in CTS patients, a principal component analysis (PCA) of each finger’s grip strength and subjective symptom scores was performed in the CTS group. PCA is a multivariate statistical technique applied to systematically reduce the number of dimensions [20,21,22]. This method extracts the important information from the data to represent these data as a set of new orthogonal variables called principal components and to display the pattern similarity between the observations and the variables as points in maps. PCA results show that factors with arrows pointing in the same direction have a tendency to be correlated, while factors with scattered arrows have little correlation. In the PCA, we set the grip strength of the individual’s largest finger (index, middle, ring, or little finger) to 1.0 and calculated the ratio of the other fingers to that of the largest finger. PCA of their finger grip strength ratios and subjective scores was performed. For this reason, our results compare of the ratio of forces per finger; thus, it was not affected by differences in individual physique. The subjective symptom score was obtained using the CTSI-JSSH symptom severity scale (CTSI-SS), CTSI-JSSH functional condition scale (CTSI-FS), and DASH. In order to investigate the characteristics of body size, we performed PCA on body mass index (BMI) and finger grip strength. Results The participants’ demographic features are presented in Table 1. Table 1 Characteristics of the participants in the CTS and control groups Full size table Time course of each finger’s grip strength Grip time with the index, middle, and ring fingers was longer in the CTS group than in the control group (Fig. 2). However, there was no difference in grip time with the little finger between the two groups. In particular, a significant difference (P = 0.01) was observed in the grip time of the middle finger. Fig. 2 Grip time of each finger’s grip strength. Grip time with index, middle, and ring fingers is longer in the CTS patients than in the controls. Statistical significance was determined by using the Student t test. In the grip time with the middle finger, a significant difference is observed between the CTS patients and controls (P = 0.01) Full size image Furthermore, in the time course of ΔGrip strength of the middle and ring fingers, the CTS group showed an early loss of strength after reaching the maximum grip strength than the control group (Fig. 3). In the time course of the little finger, the CTS group tended to release later than the control group. For the index finger, there was no difference between the CTS and control groups. Fig. 3 The time to loss of grip strength. The time at which grip strength was lost in the middle and ring fingers was earlier in the CTS patients than in the controls Full size image Principal component analysis The results of the PCA of the subjective symptom score and each finger’s grip strength in the CTS group were examined. In the PCA of each finger’s grip strength and CTSI-SS, CTSI-SS was drawn between the ring and little fingers and showed a strong correlation with the ring finger (Fig. 4). In the PCA of each finger’s grip strength and CTSI-FS, CTSI-FS was drawn between the ring and little fingers and showed a strong correlation with the little finger, similar to the PCA of DASH (Figs. 5 and 6). In the PCA of each finger’s grip strength and BMI, the arrows of little finger grip strength and BMI pointed same direction (Fig. 7). Fig. 4 Result of the PCA of each finger’s grip strength and CTSI-SS Full size image Fig. 5 Result of the PCA of each finger’s grip strength and CTSI-FS Full size image Fig. 6 Result of the PCA of each finger’s grip strength and DASH Full size image Fig. 7 Result of the PCA of each finger’s grip strength and BMI Full size image Discussion In this study, we analyzed the characteristics of grip motion in CTS patients and examined the relationship between grip strength and subjective symptoms (CTSI-JSSH and DASH) using a new grip system that can measure each finger’s grip strength at a single instance and record the time course of grip motion. There have been no reports focused on the time course of grip motion, such as the time from the beginning of grip motion to the maximum power, and the time from maximum power to loss of grip strength. Our system could show the characteristics of CTS patients in whom the grip time of the index, middle, and ring fingers was long, and grip strength was lost early after it reached the maximum. In the little finger, the CTS group tended to release later than the control group. This may be due to the fact that the middle and ring fingers are released early in the CTS group and rely on the little finger for gripping. From these results, we may be able to quantitatively evaluate sensory and motor nerve disorders, which may affect the hand function of CTS patients. Furthermore, a grip system was previously used to measure the maximum grip strength of each finger of healthy hands with anesthesia to the median nerve . In that report, grip strength of the middle, ring, and little fingers decreased after anesthesia, which was not consistent with the pathophysiology of CTS; thus, this method, using anesthesia, was deemed inadequate to accurately evaluate the hand function of CTS patients. Conversely, in our study, the symptom severity correlated with the tendency of patients to not use the index and middle fingers during grip motion. These results were consistent with the pathophysiology of CTS. A previous study using a conventional grip dynamometer reported that no correlation was found between grip strength and severity of CTS . It was also demonstrated that grip motion requires a synergistic function of the intrinsic and extrinsic muscles of the hand and does not use the muscles affected by CTS . Thus, the conventional grip dynamometer may not be accurate enough to evaluate the function of the hands with CTS. Conversely, in our PCA for CTSI-FS, CTSI-SS, DASH, and finger grip strength, the arrow was shown between the ring and little finger. In the PCA, factors with arrows pointing in the same direction have a tendency to be correlated. These results show that patients with severe subjective symptoms, as well as high CTSI-FS, CTSI-SS, and DASH scores, tended to not use the index and middle fingers during grip motion. These results proved that the function of the index and middle fingers worsens when the sensory nerve fibers of the digital nerve (index, middle, and ring fingers) and motor nerve fibers of the first and second lumbricals are damaged at the carpal tunnel. Thus, the finger grip dynamometer system could become a useful tool for evaluating the severity of CTS. It is not clear why patients with a high BMI are more likely to hold their little finger. In a study using a conventional grip strength dynamometer, people with a higher BMI were found to have greater grip strength ; thus, people with stronger grip strength may tend to grip on the ulnar side. We also analyzed the SCV and APB-DL of the NCS as an objective evaluation. Although PCA was performed on SCV, APB-DL, and finger grip strength, none showed a correlation between electrophysiological severity and finger grip strength (data not shown). This study has several limitations. First, the CTS and control groups matched in age but not in sex. Although the comparison of the absolute value of the total grip strength and each finger’s grip strength is inadequate, the analysis of time course and the PCA were not evaluations of the absolute value of grip strength and were considered to be useful results. Second, because DASH is not a questionnaire that asked for answers on the left and right hand separately, DASH results do not necessarily reflect the subjective symptoms of hands measured by finger grip strength. CTSI is a questionnaire that asks for answers on the left and right hands separately. Conclusion We developed a new grip system that can measure each finger’s grip strength at a single instance and record the time course of grip motion. Our system could show the characteristics of CTS patients in whom the grip time of the index, middle, and ring fingers was long, and grip strength was lost early after it reached the maximum. Furthermore, we could show that patients with severe subjective symptoms tended to not use the index and middle fingers during grip motion. From these results, our system could quantify the patient’s symptoms easily and objectively. To measure each finger’s grip strength in real time is an unprecedented method for characterizing the grip motion of patients with various diseases and understanding the pathophysiology of diseases. In the future, we would like to evaluate whether this finger grip dynamometer system is useful for diagnosis. Availability of data and materials The datasets used and/or analyzed during the current study are available from the corresponding author on reasonable request. Abbreviations ABP-DL: Abductor pollicis brevis-distal latency CTS: Carpal tunnel syndrome CTSI-FS: CTSI-JSSH functional condition scale CTSI-JHSS: Japanese Society for Surgery of the Hand version of the Carpal Tunnel Syndrome DASH Disability of Arm, Shoulder, and Hand questionnaire CTSI-SS: CTSI-JSSH symptom severity scale NCS: Nerve conduction study PCA: Principal component analysis SCV: Sensory conduction velocity References Bot AGJ, Mulders MAM, Fostvedt S, Ring D. Determinants of grip strength in healthy subjects compared to that in patients recovering from a distal radius fracture. J Hand Surg Am. 2012;37:1874–80. ArticleGoogle Scholar Czitrom AA, Lister GD. Measurement of grip strength in the diagnosis of wrist pain. J Hand Surg Am. 1988, 9:13a16. Edmunds JO. Current concepts of the anatomy of the thumb trapeziometacarpal joint. J Hand Surg Am. 2011;36:170–82. ArticleGoogle Scholar Baker NA, Moehling KK, Desai AR, Gustafson NP. Effect of carpal tunnel syndrome on grip and pinch strength compared with sex- and age-matched normative data. Arthritis Care Res. 2013;65:2041–5. ArticleGoogle Scholar Yucel H. Factors affecting symptoms and functionality of patients with carpal tunnel syndrome: a retrospective study. J Phys Ther Sci. 2015;27:1097–101. ArticleGoogle Scholar Fujita K, Kimori K, Nimura A, Okawa A, Ikuta Y. MRI analysis of carpal tunnel syndrome in hemodialysis patients versus non-hemodialysis patients: a multicenter case-control study. J Orthop Surg Res. 2019;14:91. ArticleGoogle Scholar Kuroiwa T, Fujita K, Nimura A, Miyamoto T, Sasaki T, Okawa A. A new method of measuring the thumb pronation and palmar abduction angles during opposition movement using a three-axis gyroscope. J Orthop Surg Res. 2018;13:288. ArticleGoogle Scholar Kuroiwa T, Nimura A, Suzuki S, Sasaki T, Okawa A, Fujita K. Measurement of thumb pronation and palmar abduction angles with a small motion sensor: a comparison with Kapandji scores. J Hand Surg Eur Vol. 2019;44:728–33. ArticleGoogle Scholar Werner RA, Andary M. Electrodiagnostic evaluation of carpal tunnel syndrome. Muscle Nerve. 2011;44:597–607. ArticleGoogle Scholar Uchiyama S, Imaeda T, Toh S, Kusunose K, Sawaizumi T, Wada T, et al. Comparison of responsiveness of the japanese society for surgery of the hand version of the carpal tunnel syndrome instrument to surgical treatment with dash, sf-36, and physical findings. J Orthop Sci. 2007;12:249–53. ArticleCASGoogle Scholar Makino K: Development of a dynamometer to measure grip strength of each finger. 2018 IEEE 27th International Symposium on Industrial Electronics. Bland JD. A neurophysiological grading scale for carpal tunnel syndrome. Muscle Nerve . 2000;23:1280–3. ArticleCASGoogle Scholar Coldham F, Lewis J, Lee H. The reliability of one vs. Three grip trials in symptomatic and asymptomatic subjects. Journal of Hand Therapy. 2006;19:318–27. ArticleGoogle Scholar Watanabe T, Owashi K, Kanauchi Y, Mura N, Takahara M, Ogino T. The short-term reliability of grip strength measurement and the effects of posture and grip span. J Hand Surg-Am. 2005:30a:603–9. Jeong S, Kim J. Prospective association of handgrip strength with risk of new-onset cognitive dysfunction in korean adults: a 6-year national cohort study. Tohoku J Exp Med. 2018;244:83–91. ArticleGoogle Scholar Wolny T, Linek P, Saulicz E. Assessment of manual dysfunction in occupationally active women with carpal tunnel syndrome. Int J Occup Med Environ Health. 2019;32:185–96. PubMedGoogle Scholar Atkinson HH, Rapp SR, Williamson JD, Lovato J, Absher JR, Gass M, et al. The relationship between cognitive function and physical performance in older women: results from the women's health initiative memory study. J Gerontol A Biol Sci Med Sci. 2010;65:300–6. ArticleGoogle Scholar Merletti R, Holobar A, Farina D. Analysis of motor units with high-density surface electromyography. J Electromyogr Kinesiol. 2008;18:879–90. ArticleGoogle Scholar Peng Y, He J, Yao B, Li S, Zhou P, Zhang Y. Motor unit number estimation based on high-density surface electromyography decomposition. Clin Neurophysiol. 2016;127:3059–65. ArticleGoogle Scholar Gray RJ, Thom M, Riddle M, Suh N, Burkhart T, Lalone E. Image-based comparison between the bilateral symmetry of the distal radii through established measures. J Hand Surg Am. 2019;44:966–72. ArticleGoogle Scholar Luria S, Schwarcz Y, Wollstein R, Emelife P, Zinger G, Peleg E. 3-dimensional analysis of scaphoid fracture angle morphology. J Hand Surg Am. 2015;40:508–14. ArticleGoogle Scholar Sollaccio DR, Navo P, Ghiassi A, Orr CM, Patel BA, Lewton KL. Evaluation of articular surface similarity of hemi-hamate grafts and proximal middle phalanx morphology: a 3D geometric morphometric approach. J Hand Surg Am. 2019;44:121–8. ArticleGoogle Scholar Kozin SH, Porter S, Clark P, Thoder JJ. The contribution of the intrinsic muscles to grip and pinch strength. J Hand Surg Am. 1999;24:64–72. ArticleCASGoogle Scholar Atalay NS, Sarsan A, Akkaya N, Yildiz N, Topuz O. The impact of disease severity in carpal tunnel syndrome on grip strength, pinch strength, fine motor skill and depression. J Phys Ther Sci. 2011;23:115–8. ArticleGoogle Scholar Geere J, Chester R, Kale S, Jerosch-Herold C. Power grip, pinch grip, manual muscle testing or thenar atrophy - which should be assessed as a motor outcome after carpal tunnel decompression? A systematic review. BMC Musculoskelet Disord. 2007;8:114. ArticleGoogle Scholar Jeong SM, Choi S, Kim K, Kim SM, Kim S, Park SM. Association among handgrip strength, body mass index and decline in cognitive function among the elderly women. Bmc Geriatr. 2018;18. Download references Acknowledgements We would like to thank Editage (www.editage.jp) for the English-language editing. Funding This work was supported by JST AIP-PRISM Grant Number JPMJCR18Y2, Japan. Author information Authors and Affiliations Department of Orthopaedic and Spinal Surgery, Graduate School of Medical and Dental Sciences, Tokyo Medical and Dental University, 1-5-45, Yushima, Bunkyo-ku, Tokyo, 113-8519, Japan Toru Sasaki,Tomoyuki Kuroiwa,Takafumi Koyama,Atsushi Okawa&Koji Fujita Center for Creative Technology, University of Yamanashi, 4-3-11, Takeda, Kofu, Yamanashi, Japan Koji Makino Department of Functional Joint Anatomy, Graduate School of Medical and Dental Sciences, Tokyo Medical and Dental University, 1-5-45, Yushima, Bunkyo-ku, Tokyo, 113-8519, Japan Akimoto Nimura&Shiro Suzuki Department of Mechatronics, University of Yamanashi, 4-3-11, Takeda, Kofu, Yamanashi, Japan Hidetsugu Terada Authors 1. Toru SasakiView author publications Search author on:PubMedGoogle Scholar 2. Koji MakinoView author publications Search author on:PubMedGoogle Scholar 3. Akimoto NimuraView author publications Search author on:PubMedGoogle Scholar 4. Shiro SuzukiView author publications Search author on:PubMedGoogle Scholar 5. Tomoyuki KuroiwaView author publications Search author on:PubMedGoogle Scholar 6. Takafumi KoyamaView author publications Search author on:PubMedGoogle Scholar 7. Atsushi OkawaView author publications Search author on:PubMedGoogle Scholar 8. Hidetsugu TeradaView author publications Search author on:PubMedGoogle Scholar 9. Koji FujitaView author publications Search author on:PubMedGoogle Scholar Contributions T.S. and K.M. performed all experiments and analyses and prepared the first draft of the paper. K.F. designed the study, collected patient data, and supervised the project. He is the guarantor. A.N., S.S., T.Kuroiwa, and T.Koyama collected the patients’ data and provided advice about the experimental conditions. A.O. and H.T. provided advice about the experimental conditions. All authors read and approved the final manuscript. Corresponding author Correspondence to Koji Fujita. Ethics declarations Ethics approval and consent to participate This comparative study was approved by the Tokyo Medical and Dental University institutional review board (M2017-123). Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. 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Copy to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Carpal tunnel syndrome Grip dynamometer Grip motion Principal component analysis Download PDF Sections Figures References Abstract Background Methods Results Discussion Conclusion Availability of data and materials Abbreviations References Acknowledgements Funding Author information Ethics declarations Additional information Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Fig. 4 View in articleFull size image Fig. 5 View in articleFull size image Fig. 6 View in articleFull size image Fig. 7 View in articleFull size image Bot AGJ, Mulders MAM, Fostvedt S, Ring D. Determinants of grip strength in healthy subjects compared to that in patients recovering from a distal radius fracture. J Hand Surg Am. 2012;37:1874–80. ArticleGoogle Scholar Czitrom AA, Lister GD. Measurement of grip strength in the diagnosis of wrist pain. J Hand Surg Am. 1988, 9:13a16. Edmunds JO. Current concepts of the anatomy of the thumb trapeziometacarpal joint. J Hand Surg Am. 2011;36:170–82. ArticleGoogle Scholar Baker NA, Moehling KK, Desai AR, Gustafson NP. Effect of carpal tunnel syndrome on grip and pinch strength compared with sex- and age-matched normative data. Arthritis Care Res. 2013;65:2041–5. ArticleGoogle Scholar Yucel H. Factors affecting symptoms and functionality of patients with carpal tunnel syndrome: a retrospective study. J Phys Ther Sci. 2015;27:1097–101. ArticleGoogle Scholar Fujita K, Kimori K, Nimura A, Okawa A, Ikuta Y. MRI analysis of carpal tunnel syndrome in hemodialysis patients versus non-hemodialysis patients: a multicenter case-control study. J Orthop Surg Res. 2019;14:91. ArticleGoogle Scholar Kuroiwa T, Fujita K, Nimura A, Miyamoto T, Sasaki T, Okawa A. A new method of measuring the thumb pronation and palmar abduction angles during opposition movement using a three-axis gyroscope. J Orthop Surg Res. 2018;13:288. ArticleGoogle Scholar Kuroiwa T, Nimura A, Suzuki S, Sasaki T, Okawa A, Fujita K. Measurement of thumb pronation and palmar abduction angles with a small motion sensor: a comparison with Kapandji scores. J Hand Surg Eur Vol. 2019;44:728–33. ArticleGoogle Scholar Werner RA, Andary M. Electrodiagnostic evaluation of carpal tunnel syndrome. Muscle Nerve. 2011;44:597–607. ArticleGoogle Scholar Uchiyama S, Imaeda T, Toh S, Kusunose K, Sawaizumi T, Wada T, et al. Comparison of responsiveness of the japanese society for surgery of the hand version of the carpal tunnel syndrome instrument to surgical treatment with dash, sf-36, and physical findings. J Orthop Sci. 2007;12:249–53. ArticleCASGoogle Scholar Makino K: Development of a dynamometer to measure grip strength of each finger. 2018 IEEE 27th International Symposium on Industrial Electronics. Bland JD. A neurophysiological grading scale for carpal tunnel syndrome. Muscle Nerve . 2000;23:1280–3. ArticleCASGoogle Scholar Coldham F, Lewis J, Lee H. The reliability of one vs. Three grip trials in symptomatic and asymptomatic subjects. Journal of Hand Therapy. 2006;19:318–27. ArticleGoogle Scholar Watanabe T, Owashi K, Kanauchi Y, Mura N, Takahara M, Ogino T. The short-term reliability of grip strength measurement and the effects of posture and grip span. J Hand Surg-Am. 2005:30a:603–9. Jeong S, Kim J. Prospective association of handgrip strength with risk of new-onset cognitive dysfunction in korean adults: a 6-year national cohort study. Tohoku J Exp Med. 2018;244:83–91. ArticleGoogle Scholar Wolny T, Linek P, Saulicz E. Assessment of manual dysfunction in occupationally active women with carpal tunnel syndrome. Int J Occup Med Environ Health. 2019;32:185–96. PubMedGoogle Scholar Atkinson HH, Rapp SR, Williamson JD, Lovato J, Absher JR, Gass M, et al. The relationship between cognitive function and physical performance in older women: results from the women's health initiative memory study. J Gerontol A Biol Sci Med Sci. 2010;65:300–6. ArticleGoogle Scholar Merletti R, Holobar A, Farina D. Analysis of motor units with high-density surface electromyography. J Electromyogr Kinesiol. 2008;18:879–90. ArticleGoogle Scholar Peng Y, He J, Yao B, Li S, Zhou P, Zhang Y. Motor unit number estimation based on high-density surface electromyography decomposition. Clin Neurophysiol. 2016;127:3059–65. ArticleGoogle Scholar Gray RJ, Thom M, Riddle M, Suh N, Burkhart T, Lalone E. Image-based comparison between the bilateral symmetry of the distal radii through established measures. J Hand Surg Am. 2019;44:966–72. ArticleGoogle Scholar Luria S, Schwarcz Y, Wollstein R, Emelife P, Zinger G, Peleg E. 3-dimensional analysis of scaphoid fracture angle morphology. J Hand Surg Am. 2015;40:508–14. ArticleGoogle Scholar Sollaccio DR, Navo P, Ghiassi A, Orr CM, Patel BA, Lewton KL. Evaluation of articular surface similarity of hemi-hamate grafts and proximal middle phalanx morphology: a 3D geometric morphometric approach. J Hand Surg Am. 2019;44:121–8. ArticleGoogle Scholar Kozin SH, Porter S, Clark P, Thoder JJ. The contribution of the intrinsic muscles to grip and pinch strength. J Hand Surg Am. 1999;24:64–72. ArticleCASGoogle Scholar Atalay NS, Sarsan A, Akkaya N, Yildiz N, Topuz O. The impact of disease severity in carpal tunnel syndrome on grip strength, pinch strength, fine motor skill and depression. J Phys Ther Sci. 2011;23:115–8. ArticleGoogle Scholar Geere J, Chester R, Kale S, Jerosch-Herold C. Power grip, pinch grip, manual muscle testing or thenar atrophy - which should be assessed as a motor outcome after carpal tunnel decompression? A systematic review. BMC Musculoskelet Disord. 2007;8:114. ArticleGoogle Scholar Jeong SM, Choi S, Kim K, Kim SM, Kim S, Park SM. Association among handgrip strength, body mass index and decline in cognitive function among the elderly women. Bmc Geriatr. 2018;18. Journal of Orthopaedic Surgery and Research ISSN: 1749-799X Contact us Submission enquiries: journalsubmissions@springernature.com Read more on our blogs Receive BMC newsletters Manage article alerts Language editing for authors Scientific editing for authors Policies Accessibility Press center Support and Contact Leave feedback Careers Follow BMC BMC Twitter page BMC Facebook page BMC Weibo page By using this website, you agree to our Terms and Conditions, Your US state privacy rights, Privacy statement and Cookies policy. 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14596	https://www.youtube.com/watch?v=uM9qPqhmE_w	Finding the "critical points" in a quadratic equation Wyzant 5950 subscribers Description 20 views Posted: 14 Jul 2024 View full question and answer details: Question: Find the critical points of the linear equation y = 4x^3 - 3x So to be clear, when I state critical point, I want the maxima, minima, and any zeros in the domain of the function. Answered By: Charles W. Middle School Math Teacher who Loves Seeing Students Succeed More information: Written Explanation: I wanted to make sure this was posted because in my session with one of my students, we had to stop mid-explanation due to a professional reason, so lets get to this:Quadratic linear equations at some point cross the x and y axis, and quite simply its good to know a few key points: all x-intercepts including origin, the points it will reach before the domain (x values) result in a turn down/up (these are called the maxima and the minima).To find the zeros first, we will use the original equation (y) and follow the steps below:Set the equation equal to zerofactor out any x's from the termsmake conclusions about the terms remaining by setting each set of terms equal to zeroStep 1 is done for us.y=4x^3 - 3xStep 2:y=4x^3 - 3x - we have an x in each term on the right side of the equationy = x(4x^2 - 3) - now we have two terms that include an operation with x. (x) and (4x2-3)Step 3:Operationally this lets find what would make y in this linear equation equal 0.It could be 0 =› y=(0)(4x2 - 3) would make the final product zeroIt could be making the difference between (4x2 - 3) equal zero.so we do the following:4x2 -3 = 04x2 = 3x2 = 3/4x = ±√(3)/2+√(3)/2 and -√(3)/2 can both result in that part of the equation equal zero.Then we apply what we know about these equations....the amount of maxima and minima in quadratic equations is a ((highest power exponent)-1). In this equation, it would be 2.So x could be +√(3)/2, to make y = zero, and are CRITICAL POINTS in the domain.These are the Zeros in the equation.And the minimum x could equal because it is a quadratic equation with x raised to the 3rd power is -∞, and the maximum it could be is +∞.Final answer pt 1 of 2:The critical points in this problem are ,(0,-√(3)/2),(0,0), (0,+√(3)/2, +∞) per defining the domain. If the question needs the maxima or minima of the equation we can continue using original equation,y, and the first derivative of the original equation, y'.Other textbooks and such define the cri... See full answer: About: Wyzant Ask an Expert offers free answers to your toughest academic and professional questions from over 65,000 verified experts. It’s trusted by millions of students each month with the majority of questions receiving an answer within 1 hour of being asked. If you ever need more than just an answer, Wyzant also offers personalized 1-on-1 sessions with experts that will work with you to help you understand whatever you’re trying to learn. Ask your own question for free: Find a tutor for a 1-on-1 session: Subscribe to Wyzant on YouTube: Transcript: [Music] [Applause] [Music] all right finally got everything formatted the right way and we got that set set up here I'm going to cover this final answer so you don't see it yet but we got to go through the steps so know how to get it they're usually looking for critical points which are the X intercepts if it crosses origin which would be zero X intercepts would be any point we would have some number comma zero as a coordinate for it and any other places specifically where this quadratic equation would turn up well turn down or turned up at that point well one while the X may keep continue to increase the Y would go the Y value rather for that coordinate will go down or go from that point go up all right so to do this we kind of start with the first step which have already dumb Force um setting it equal to zero uh no looking at the first step and then trying to factor out any X's from that x side if you just have it set to set equal to a y and that would result in a x is m x times 4x^2 - 3 = y so it's pass to X origin and then we have 4x^2 is equal to no minus three is equal to zero so effectively saying now we know this one's going to give us zero because anything times that other side will give us zero to me X so we have one point here that's represented by Z 0 z0 the next one would be be based on this term we said this term alone equal to zero and we have 4x^2 = 3 x^2 is equal to 3 over/ 4 find the square root of that entire fraction gives the square root of 3 plus or and minus uh two right which really means you have two numbers here represented by one a negative otk of 3 over two and a positive3 over two so these are the two x coordinates the X X intercepts in addition to origin in addition to origin because origin is the X and the y coordinate a y intercept all right now here's how we find the maximum and minimum we're going to take uh the first derivative of the original equation Y Prime find the zeros find the zeros and plug those back into the original equation into the original equation all right so we have y is equal to 4X Cub 4X Cub - 3x y Prime applying our understanding of derivatives is 12 x^2 - 3 we said that equal to equal to zero therefore it be 12x = 3 x² 12x2 3 X2 is = to 3 over 12al 1/4 x is equal to2 or x equal to negative one2 being the possible scenarios kind of showing right here right so we plug these two values separately into the equation in separately into the original equation Y not y Prime and we have that that equation is 4X Cub minus 3x once again we just pluged in that 1/2 here and a negative one2 there respectfully after we respectfully after we solve through we have a negative one we plug this in for the Y and we have a one on this end for the Y so that gives us the two Maxima and Minima or the places where there's a turn in the graph turn in the graph right so we got this point this point this point this point as well as what this represents which would be infinity and this one represent as which would be Infinity all right so the critical points that I see and how I've been taught to kind of look at it is here right The Final Answer negative infinity and positive Infinity are included in this graph under there it's a parab well it's a it's a quadratic equation so they go on forever and ever and ever and ever we have a negative s < TK of two zero right here we have zero we have a positive < TK of 3 over two which equal zero right here our Maxima or place where we turn would be at negative one2 one one and we have another one here where we have a one half negative one right where turns back up and these obviously consider the critical points in this particular uh graph as I said this one I it was a relatively vague writing of it U we had to use the derivative as well I do want to reinforce that it's key that you do understand how to to find the first derivative um for numbers that for terms rather in any equation but other than that this is pretty simple so um I I think pretty simple but rather pretty direct um want to make sure this is clear and if you have any questions please leave comments please make sure uh we go over this um and thanks for letting me know and once again thanks for your time [Music]
14597	https://nhts.ornl.gov/assets/2017_nhts_summary_travel_trends.pdf	.0 Summary of Travel Trends 2017 National Household Travel Survey 2017 National Household Travel Survey Technical Report Documentation Page 1. Report No. FHWA-PL-18-019 2. Government Accession No. N/A 3. Recipient's Catalog No. N/A 4. Title and Subtitle Summary of Travel Trends: 2017 National Household Travel Survey 5. Report Date July 2018 6. Performing Organization Code 3301 7. Author(s) N. McGuckin and A. Fucci 8. Performing Organization Report No. N/A 9. Performing Organization Name and Address Westat 1600 Research Blvd. Rockville, MD 10. Work Unit No. (TRAIS) N/A 11. Contract or Grant No. DTFH6114F00113 13. Type of Report and Period Covered Trends in travel behavior, 1969-2017 12. Sponsoring Agency Name and Address Federal Highway Administration Office of Policy and Governmental Affairs Daniel E. Jenkins, PE, Travel Behavior Data Program Manager 1200 New Jersey Avenue, SE Washington DC 20590 14. Sponsoring Agency Code FHWA/HPP1-30 15. Supplementary Notes 16. Abstract The 2017 National Household Travel Survey (NHTS) provides an inventory of daily travel in the US and its major Census Divisions and add-on areas. It is the only source of national-level statistics on personal travel in the US. The survey series (conducted since 1969) includes demographic data on households, people, vehicles, and detailed information on daily travel by all modes of transportation and for all purposes. NHTS survey data are collected from a sample of households and expanded to provide estimates of trips and miles of travel by travel mode, trip purpose, and other important attributes. When combined with historical data from the earlier surveys (1969, 1977, 1983, 1990, and 1995 NPTS and the 2001 NHTS, 2009 NHTS, and 2017 NHTS) these data serve as a rich source of information on the trends in travel over time. This report summarizes trends in household and personal travel patterns, including information on changes to the household-based vehicle fleet and commuting patterns. The report begins with a summary of the changes in the population, demographics, and related travel. Next, travel trends are examined at the household level, including differences between different areas of the US and by household income, for example. Next, changes in travel are summarized at the person-level, including trips by purpose and miles of travel by age and sex. Following sections detail changes in vehicle availability and usage, commute travel patterns, temporal distribution, and the travel of special populations. The 2017 NHTS was conducted with major changes in sampling strategy (an address-based sample compared to previous land-line random-digit sample) and methodology (Web-based self-reports compared to previous computer-aided interviewing). These and other critical changes are summarized here in Appendix A and in the data documentation at Researchers and data users are cautioned to do their best to assess how the change in methods may affect their estimates and to caution their readers about these critical changes in the data series. 17. Key Words National Household Travel Survey, NHTS, US travel, daily travel, means of travel, mode of travel, trip purpose, VMT, PMT, demographics and travel, vehicle use trends, fleet changes, vehicle availability, commute data, commute patterns, travel purpose, travel patterns by age, travel patterns by gender 18. Distribution Statement No restrictions 19. Security Classif. (of this report) Unclassified 20. Security Classif. (of this page) Unclassified 21. No. of Pages 148 22. Price N/A Form DOT F 1700.7 (8-72) Reproduction of completed page authorized 2017 National Household Travel Survey Notice This document is disseminated under the sponsorship of the U.S. Department of Transportation in the interest of information exchange. The U.S. Government assumes no liability for use of the information contained in this document. This report does not constitute a standard, specification, or regulation. Quality Assurance Statement The Federal Highway Administration (FHWA) provides high-quality information to serve government agencies, industry, and the public in a manner that promotes understanding. High standards are applied to ensure and maximize the quality, objectivity, utility, and integrity of the information. FHWA periodically reviews quality issues and adjusts its programs and processes to ensure continuous quality improvement. The most recent assessment of the NHTS program is here: In 2017, the NHTS underwent a major change in survey methodology. The most impactful changes are 1) using an address-based sample rather than an RDD land-line sample, and 2) moving from an interviewer assisted telephone surveys (CATI) to a self-completed web-based survey. These changes made the 2017 NHTS a better sample survey, with better coverage of US households and lower respondent burden. In addition, the method of obtaining trip length used a Google API shortest path route between a geocoded origin and destination whereas previous NHTS used the respondent’s estimate of trip length for each trip. These changes may have impacted the number of reported trips, including incidental trips, and the estimate of trip lengths, which in turn impact VMT and PMT estimates. The change in methods may have measurable impacts on many of the survey estimates, and unknown impacts yet to be identified. Some of the measured impacts of methods changes in 2017 are outlined in Appendix A. Users should take into account the impacts identified here and do further analysis of their own to assess the best use of the data series for any specific application. The data presented here are based on a sample of the population, and so is subject to sampling error. Sampling error is the calculated statistical imprecision due to interviewing a random sample instead of the entire population. The margin of error provides an estimate of how much the results of the sample may differ due to chance when compared to what would have been found if the entire population was interviewed. For the 2017 data the margin of error is added to and subtracted from the point estimate to provide the range for each estimate. Sampling error is the only error that can be quantified, but there are other errors to which surveys are susceptible. Please read ‘Reliability of the Estimates’ in Chapter 1 for more details. 2017 National Household Travel Survey TABLE OF CONTENTS iv TABLE OF CONTENTS TABLE OF CONTENTS ............................................................................................................. iv LIST OF TABLES ........................................................................................................................ v LIST OF FIGURES ................................................................................................................... vii SUMMARY OF TRAVEL TRENDS: ........................................................................................... 1 2017 National Household Travel Survey .................................................................................... 1 1.0 INTRODUCTION AND RELIABILITY OF THE ESTIMATES ............................................ 1 2.0 OVERVIEW ..................................................................................................................... 5 3.0 HOUSEHOLD TRAVEL ..................................................................................................15 4.0 PERSON TRAVEL .........................................................................................................26 5.0 PRIVATE VEHICLE TRAVEL .........................................................................................56 6.0 VEHICLE USE AND AVAILABILITY ...............................................................................59 7.0 COMMUTE TRAVEL PATTERNS ..................................................................................75 8.0 TEMPORAL DISTRIBUTION ..........................................................................................86 9.0 SPECIAL POPULATIONS ..............................................................................................90 APPENDIX A: CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES ............................................................................................................ A-1 APPENDIX B: KEY CHANGES ............................................................................................... B-1 APPENDIX C: TRAVEL CONCEPTS AND GLOSSARY OF TERMS ...................................... C-1 2017 National Household Travel Survey LIST OF TABLES v LIST OF TABLES Table 1a. Summary Statistics on Demographic Characteristics: Households ............................ 5 Table 1b. Summary Statistics on Demographic Characteristics: Persons .................................. 6 Table 1c. Summary Statistics on Demographic Characteristics: Drivers and Workers ............... 7 Table 1d. Summary Statistics on Demographic Characteristics and Total Travel ...................... 8 Table 2a. Major Travel Indicators by Survey Year ..................................................................... 9 Table 2b. Major Travel Indicators by Survey Region.................................................................10 Table 3a. Summary of Household Travel Statistics ...................................................................12 Table 3b. Summary of Person Travel Statistics ........................................................................13 Table 4. Comparison of Survey Variables with Other Sources (Numbers in Thousands, Except VMT [millions]) ..........................................................................................................................14 Table 5a. Trends in the Average Annual Person Miles of Travel per Household by Trip Purpose .................................................................................................................................................16 Table 5b. Trends in the Average Person Trip Length by Trip Purpose ......................................17 Table 5c. Trends in the Average Annual Person Trips per Household by Trip Purpose ............18 Table 6a. Trends in the Average Annual Vehicle Miles of Travel by Selected Trip Purposes ....19 Table 6b. Trends in the Average Trip Length by Selected Trip Purposes .................................20 Table 6c. Trends in the Average Annual Vehicle Trips per Household by Selected Trip Purposes ...................................................................................................................................22 Table 7. Trends in the Average Annual Person Trips per Household by Mode of Transportation and MSA Size ...........................................................................................................................23 Table 8. Trends in the Number of Annual Person Trips per Household by Household Income .25 Table 9a. Trends in the Annual Number (millions) of Person Trips by Mode of Transportation and Trip Purpose ......................................................................................................................27 Table 9b. Trends in the Percent of Person Trips by Mode of Transportation and Trip Purpose (Millions) ...................................................................................................................................30 Table 10a. Trends in the Annual Number of Person Trips per Person by Trip Purpose and Gender ......................................................................................................................................32 Table 10b. Trends in the Percent of Person Trips per Person by Trip Purpose and Gender .....35 Table 11. Trends in the Daily Trip Rates and Person Miles of Travel per Person by Trip Purpose .................................................................................................................................................40 Table 12. Trends in the Distribution of Daily Person Miles of Travel per Person by Mode of Transportation and Trip Purpose ...............................................................................................43 Table 13. Trends in the Average Daily Person Trips per Person by Age and Gender ...............48 Table 14. Trends in the Average Daily Person Miles of Travel per Person by Age and Gender 50 2017 National Household Travel Survey LIST OF TABLES vi Table 15. Trends in the Average Time Spent Driving a Private Vehicle in a Typical Day by MSA Size (minutes) ...........................................................................................................................56 Table 16. Average Vehicle Occupancy for Selected Trip Purposes ..........................................58 Table 17. Trends in the Number and Percent of Households by Availability of Household Vehicles (Thousands) ...............................................................................................................59 Table 18. Trends in the Distribution of Households by Household Vehicle Availability and Population Density ....................................................................................................................62 Table 19. Trends in the Percent of Households Without a Vehicle Within MSA Size Group ......65 Table 20. Household-Based Vehicle Distribution and Average Vehicle Age by Vehicle Type ...66 Table 21. Trends in the Distribution of Household-Based Vehicles by Vehicle Age and Vehicle Type (Percent) ..........................................................................................................................69 Table 22. Trends in the Average Annual Miles per Vehicle by Vehicle Age (Vehicle Owner's Estimate) ...................................................................................................................................72 Table 23a. Trends in the Average Annual Miles per Licensed Driver-by-Driver Age (Self-Estimate) ...................................................................................................................................73 Table 23b. Trends in the Average Annual Miles per Licensed Driver-by-Driver Age and Gender (Self-Estimate) ..........................................................................................................................74 Table 24. Trends in Commute Trips and Vehicle Miles in Commute .........................................75 Table 25. Trends in the Distribution of Workers by Usual Commute Mode (Percent of Workers) .................................................................................................................................................77 Table 26. Usual Commute Mode to Work vs Actual Commute Mode on Travel Day .................78 Table 27. Trends in General Commute Patterns by Mode of Transportation .............................79 Table 28. Trends in Average Commute Speed by MSA Size (Miles per Hour) 1977, 1983, 1990, 1995 NPTS, and 2001, 2009, and 2017 NHTS..........................................................................82 Table 29. Trends in the Distribution of Person Trips by Start Time of Trip ................................86 Table 30. Trends in Travel Characteristics for Weekday vs. Weekend .....................................88 Table 31. Daily Travel Statistics of People 65 and Older ..........................................................91 Table 32a. Selected Data for Older Persons .............................................................................92 Table 32b. Selected Data for Older Men...................................................................................93 Table 32c. Selected Data for Older Women .............................................................................94 Table 33. Vehicle Miles of Travel (VMT) Trends for Younger People by Urban or Rural Household Location ..................................................................................................................96 Table 34. Travel Characteristics of People in Urban and Rural Areas, 2017 NHTS ..................99 Table 35. Average Number of On-Line Purchases and Deliveries to U.S. Households in the Last Month ...................................................................................................................................... 100 Table 36. Characteristics of Users of Transportation Network Companies (Uber/Lyft), 2017 NHTS ...................................................................................................................................... 101 2017 National Household Travel Survey LIST OF FIGURES vii LIST OF FIGURES Figure 1. Changes in Summary Statistics on Demographics and Total Travel ..........................11 Figure 2. Trends in the Distribution of Person Trips per Person by Gender and Trip Purpose ..38 Figure 3a. Daily Trip Rates per Person by Trip Purpose ...........................................................41 Figure 3b. Daily Person Miles of Travel per Person by Trip Purpose ........................................41 Figure 4. Trends in the Average Daily Person Trips by Age .....................................................49 Figure 5. Average Daily Person Miles of Travel by Gender, 1983, 1990, 1995 NPTS and 2001, 2009, and 2017 NHTS ..............................................................................................................52 Figure 6. Average Daily Person Miles of Travel by Age Group 1995 NPTS and 2001, 2009, and 2017 NHTS ...............................................................................................................................53 Figure 7. Trends in the Time Spent in a Vehicle by Age Group (Minutes per Day) ...................55 Figure 8. Average Time Spent Driving and Miles Traveled by MSA Size ..................................57 Figure 9. Household Distribution by Number of Household Vehicles ........................................61 Figure 10. Distribution of the Number of U.S. Households by Vehicle Ownership and Population Density, 2017 NHTS (Millions) ..................................................................................................64 Figure 11. Trends in the Number of Household-Based Vehicles by Type (Millions) ..................68 Figure 12. Distribution of Household-Based Vehicles Two Years old or Newer by Vehicle Type (Percent) ...................................................................................................................................71 Figure 13. Trends in the Distribution of Workers by Usual Commute Mode (Percent of Workers) .................................................................................................................................................76 Figure 14. Trends in Average Commute Speeds by MSA Size (All Modes) ..............................84 Figure 15. Distribution of Vehicle Trips by Trip Purpose and Start Time of Trip, 2017 NHTS ....87 2017 National Household Travel Survey INTRODUCTION AND RELIABILITY OF THE ESTIMATES 1 SUMMARY OF TRAVEL TRENDS: 2017 National Household Travel Survey 1.0 INTRODUCTION AND RELIABILITY OF THE ESTIMATES Policymakers rely on transportation statistics, including data on personal travel behavior, to formulate strategic transportation policies and to improve the safety and efficiency of the U.S. transportation system. Policymakers, individual state Department of Transportation (DOTs), metropolitan planning organizations, industry professionals, and academic researchers use the data to gauge the extent and patterns of travel, plan new investments, and better understand the implications of travel trends on the nation’s transportation infrastructure. To address these data needs, the U. S. Department of Transportation (USDOT) initiated an effort in 1969 to collect detailed data on personal travel. The 1969 survey was the first Nationwide Personal Transportation Survey (NPTS). The survey was conducted again in 1977, 1983, 1990, and 1995. In 2001, the survey was expanded by integrating the Federal Highway Administration (FHWA) managed NPTS and the Bureau of Transportation Statistics-sponsored American Travel Survey (ATS), and the survey was re-named the National Household Travel Survey (NHTS). The NHTS was conducted without the long-distance component again in 2009 and 2017. The recent evaluation of the NHTS data program found that NHTS data are used extensively to inform policy initiatives, provide context for decision-making, and benchmark progress for policies and programs.1 More directly, NHTS data are used as inputs to statistical analyses and models related to health, energy, air quality, and mobility. At the state and local levels, NHTS has its greatest impact in developing, calibrating, or validating travel demand models that are used to inform transportation planning and project selection. The 2017 NHTS is the most recent national inventory of daily travel, and the authoritative source on the travel behavior of the American public. The NPTS/NHTS data series is the only source of national travel behavior data that tracks trends in personal and household travel. The survey gathers trip-related data, such as mode of transportation, duration, distance, and purpose of trip, and links the travel-related information to demographic, geographic, and economic data for analysis purposes. The 2017 NHTS is a nationally representative survey of travel behavior conducted from April 2016 through April 2017. The 2017 survey is the latest in the series and updates information gathered in the NPTS conducted in 1969, 1977, 1983, 1990, and 1995, and the NHTS conducted in 2001 and 2009. The 2017 NHTS includes samples added by 13 state and local planning agencies from around the country, plus the core national sample. 1 Federal Highway Administration Research and Technology Evaluation: National Household Travel Survey Program Final Report, Publication Number: FHWA-HRT-16-082, Date: August 2017: 2017 National Household Travel Survey INTRODUCTION AND RELIABILITY OF THE ESTIMATES 2 During the survey period, researchers collected data from roughly 130,000 households, which were sampled based on postal address lists, and 275,000 persons in the United States. They mailed sampled households a survey form with a small incentive and asked them to join the survey by either logging onto the website or mailing the form back. Each participating household reported all travel by household members on a randomly assigned 24-hour single “travel day.” They assigned travel days for all 7 days of the week, including all holidays. Weighting reflected the day of week and month of travel to allow comparisons of weekdays or seasons. This report uses 2017 NHTS data to highlight travel trends over the entire survey series: almost 50 years of travel data for the United States. There are nine chapters, with each chapter representing a topic in travel behavior. The first section of statistical data focuses on demographic trends of households, persons, vehicles, and workers. The next chapter provides statistical data on overall household travel. Subsequent sections of this report present person travel, private vehicle travel, vehicle use, and commute travel patterns. The final chapter highlights travel behavior of special populations and some new data elements from the 2017 NHTS. The research findings in this report do not include a detailed analysis of the 2017 NHTS data set in its entirety but provide a very short overview of available data. Of course, this report relies on the work of previous authors and reproduces the analysis done as part of the previous reports. The first Summary of Travel Trends was a pamphlet produced for the 1983 NPTS by Comsis. In 1995 and 2001, Oak Ridge (ORNL) produced the trends report after retrieving the 1977 archived data. In 2009, the FHWA produced the report with Travel Behavior Analysts, and FHWA produced the current report with Travel Behavior Analysts and Westat. All errors are the responsibility of the authors. 1.1. CHANGES IN THE NHTS DATA COLLECTION METHOD In 2017, the NHTS underwent a major change in survey methodology. The most impactful changes are 1) using an address-based sample rather than a random digit dialing (RDD) landline telephone sample, and 2) moving from primarily an interviewer-led computer-assisted telephone interviewing (CATI) to a self-completed web-based survey with CATI as an alternative. With these changes, the 2017 NHTS sample had better coverage of U.S. households as it included households without landline telephones. The design reduced coverage bias and respondent burden. In addition, the method of obtaining trip length used a Google API (application programming interfaces) shortest path route between a geocoded origin and destination whereas previous NHTS’ used the respondent’s estimate of trip length for each trip. These changes may have impacted the number of reported trips, including incidental trips, and the estimate of trip lengths, which in turn impact vehicle miles of travel (VMT) and person miles of travel (PMT) estimates. The change in methods may have measurable impacts on many of the survey estimates, and unknown impacts that not yet identified. Appendix A outlines some of the measured impacts of methods changes in 2017. Users should consider the impacts identified here and do further analysis of their own to assess the best use of the data series for any specific application. 2017 National Household Travel Survey INTRODUCTION AND RELIABILITY OF THE ESTIMATES 3 1.2. RELIABILITY OF THE ESTIMATES (SOURCE AND ACCURACY) An estimate based on a sample survey has two types of error — sampling error and nonsampling error. The estimated standard errors provided approximate the true sampling errors. They do incorporate the effect of some nonsampling errors in response and enumeration, but do not account for any systematic biases in the data. Nonsampling error. The full extent of nonsampling error is unknown, but special studies have quantified some sources of nonsampling error. Some sources of nonsampling errors in surveys include the inability to obtain information about all persons in the sample, differences in the interpretation of questions, inability or unwillingness of respondents to provide correct information, inability of respondents to recall information, errors made in collecting and processing the data, errors made in estimating values for missing data, and failure to represent all sample households and all persons within sample households (undercoverage). In a national sample such as that used for the NHTS, undercoverage can occur when households reside in very newly constructed homes whose addresses are not yet available on the sampling frame, households have simplified addresses (e.g., John Doe, Anytown, MD 12345), or the household respondent either accidentally or purposely does not report all the people living in the household. The weighting process adjusts for some nonresponse and matches independent age-sex-race-ethnicity population controls, which partially corrects for the biases due to survey undercoverage. However, biases exist in the estimates to the extent that missed persons in missed households or missed persons in interviewed households have travel characteristics different from those of interviewed persons in the same age-sex-race-origin group. Sampling error. When a portion of the population is surveyed, rather than the entire population, estimates differ from the true population values that they represent. This difference, or sampling error, occurs by chance, and variability is measured by the standard error of the estimate. The standard error is the margin of error (MOE), which is the half-confidence interval at the 95% confidence level. Sample estimates from a given survey design are unbiased when an average of the estimates from all possible samples would yield, hypothetically, the true population value. In this case, the sample estimate and its margin of error can be used to construct approximate confidence intervals, or ranges of values that include the true population value with known probabilities. The margin of error in this document is at the 95 percent confidence level. To construct the bounds of the margin of error—that is, a high estimate and a low estimate—the MOE shown in tables is added to and subtracted from the estimate given. For example, if the estimate is 500 and the margin of error is 2, then in 95 repeated samples the estimates obtained would fall between 498 and 502; therefore, if the survey were conducted 100 times with the same protocols, 95 percent of the time the true population estimate would fall between 498 and 502. It is important to determine the significant differences from those estimates that are a product of the known sample error when analyzing these data. When comparing values, if the ranges of two estimates overlap, then there is no significant difference in the estimated values. 2017 National Household Travel Survey INTRODUCTION AND RELIABILITY OF THE ESTIMATES 4 Users should be cautious when computing estimates for smaller population groups, such as specific geographies, groups of people, or even less common forms of transportation, like bicycle, Uber/Lyft, or even transit. While the weights support a large variety of travel-related estimates, caution should be taken for estimates generated from a small number of responding households or persons. Computing the confidence interval or MOE is especially important for such analyses to ascertain whether any apparent nominal differences are actually statistically different. On the other hand, the NHTS sample can produce robust estimates of major travel indicators at census region or division (as shown in Table 2b) or by Metropolitan Area size (as shown in Table 28), and for specific groups of travelers (see Section 9 on Travel by Special Populations). Using the data appropriately is the responsibility of the analyst. The data trends shown here are just a small sample of the analysis possible with the NHTS data, and each of the topics presented could be the subject of a more in-depth and stringent analysis. Public-use national data from the 2017 NHTS is available for download and for on-line analysis on the NHTS website ( Weights and replicates are included for each of the data files. Weights match the sample of households and persons to the population for demographic characteristics and geographic levels. Use replicate weights to calculate the MOE of each estimate. 2017 National Household Travel Survey OVERVIEW 5 2.0 OVERVIEW Tables 1a through 1d present summary statistics on key demographic characteristics by survey year. For years 2009 and 2017, the MOEs are also included. There was a major change in the method used to collect trip distance in 2017 that impacts the estimates of PMT, VMT, and average person and vehicle trip lengths. In 2017, the NHTS calculated trip length using the shortest path routes between geocoded origins and destinations. Previous surveys used self-reported distances. As a result of the change in method, the 2017 original estimates of VMT and PMT may not be directly comparable with previous years. The 2017 trip distance is adjusted to be more comparable, shown as “adj.” in this document. See Appendix A for further details. Table 1a. Summary Statistics on Demographic Characteristics: Households Households (thousands) Survey Year All 1 person 2 persons 3 persons 4+ persons 1969 62,504 10,980 18,448 10,746 22,330 1977 75,412 16,214 22,925 13,046 23,227 1983 85,371 19,354 27,169 14,756 24,092 1990 (adj) 93,347 22,999 30,114 16,128 24,106 1995 98,990 24,732 31,834 16,827 25,597 2001 107,365 27,718 35,032 17,749 26,867 2009 113,101 31,741 37,728 18,104 25,528 2009 MOE - 106 135 257 243 2017 118,208 32,952 40,056 18,521 26,679 2017 MOE - - - 97 97 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones, the cell-phone only (CPO) households. • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey OVERVIEW 6 Table 1b. Summary Statistics on Demographic Characteristics: Persons Persons (thousands) Survey Year All Under 16 16-19 20-34 35-64 65+ 1969 197,213 60,100 14,598 40,060 62,982 19,473 1977 213,141 54,958 16,552 52,252 66,988 22,391 1983 229,453 53,682 15,268 60,788 75,353 24,362 1990 (adj) 239,416 54,303 13,851 59,517 82,480 26,955 1995 259,994 61,411 14,074 59,494 93,766 31,249 2001 277,203 44,985 14,296 57,680 103,296 32,884 2009 283,054 44,724 19,414 50,844 129,202 38,870 2009 MOE - 441 743 1,089 874 0 2017 321,419 45,498 17,755 64,339 126,350 47,657 2017 MOE 0 756 945 954 985 0 Persons (thousands) Survey Year All 16+ All Male All Male 16+ All Female All Female 16+ All 5+ 1969 137,113 94,465 66,652 102,748 73,526 NA 1977 158,183 102,521 74,542 110,620 83,721 198,434 1983 175,771 111,514 83,645 117,939 92,080 212,932 1990 (adj) 182,803 114,441 86,432 124,975 96,371 222,101 1995 198,583 126,553 95,627 133,441 102,956 241,675 2001 208,155 125,321 100,308 132,240 107,847 257,560 2009 238,330 139,257 116,421 143,797 121,908 283,054 2009 MOE 441 81 338 81 338 0 2017 256,101 148,039 124,903 153,560 131,198 321,419 2017 MOE 756 0 471 0 397 0 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey OVERVIEW 7 Table 1c. Summary Statistics on Demographic Characteristics: Drivers and Workers Survey Year Drivers (thousands) Workers (thousands) All Male Female All Male Female 1969 102,986 57,981 45,005 75,758 48,487 27,271 1977 127,552 66,199 61,353 93,019 55,625 37,394 1983 147,015 75,639 71,376 103,244 58,849 44,395 1990 (adj) 163,025 80,289 82,707 118,343 63,996 54,334 1995 176,330 88,480 87,851 131,697 71,105 60,593 2001 190,425 94,651 95,773 145,272 78,264 67,007 2009 212,309 106,813 105,496 151,373 81,939 69,434 2009 MOE 959 709 631 893 769 728 2017 223,277 111,163 112,114 156,988 83,589 73,399 2017 MOE 827 588 963 1,012 495 859 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey OVERVIEW 8 Table 1d. Summary Statistics on Demographic Characteristics and Total Travel Travel Characteristics Survey Year Household Vehicles (thousands) Household Vehicle Trips (millions) Household Vehicle Miles of Travel (VMT in millions) Person Trips (millions) Person Miles of Travel (PMT in millions) 1969 72,500 87,284 775,940 145,146 1,404,137 1977 120,098 108,826 907,603 211,778 1,879,215 1983 143,714 126,874 1,002,139 224,385 1,946,662 1990 (adj) 165,221 193,916 1,695,290 304,471 2,829,936 1995 176,067 229,745 2,068,368 378,930 3,411,122 2001 201,308 233,030 2,274,769 384,485 3,783,979 2009 210,778 233,849 2,245,111 392,023 3,732,791 2009 MOE 918 2,381 56,157 3,644 141,396 2017 222,579 220,430 2,105,882 371,152 3,970,287 2017 MOE 917 2,561 88,113 4,395 150,877 2017 (adj) - - 2,321,820 - 4,291,150 2017 (adj) MOE - - 98,064 - 155,470 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Household VMT and PMT "adjusted” includes estimates of miles in all vehicles, including “18” Rental Car. • In 1969, household vehicles did not include pickups or other light trucks. 2017 National Household Travel Survey OVERVIEW 9 The 2017 NHTS obtained larger households with more workers compared to the 2009 survey, possibly because the 2017 address-based sample included about 45 percent cell phone only (CPO) households, which are more likely younger and working. CPO households were not included in the sample in 2009 (see Appendix B). The data series in Tables 2a and 2b show that over the last five decades, American households acquired more vehicles and drivers. In the United States in 1969, there were as many vehicles as workers. By 1990 and continuing to the present, there are as many vehicles as drivers. As average household size has stabilized, average vehicles per household, licensed drivers per household, and workers per household have all remained rather stable over the last decade or so. There are important differences between the census regions listed in Table 2b (the states in each census region are listed in Appendix C). The West continues to have the highest household size, vehicle ownership, and driver rates in the country. The Midwest has smaller households on average, and fewer workers per household. The Northeast has fewer vehicles and drivers per household. Table 2a. Major Travel Indicators by Survey Year Major Travel Indicators by Year Travel Indicator 1969 1977 1983 1990 1995 2001 2009 2017 Persons per Household 3.16 2.83 2.69 2.56 2.63 2.58 2.50 2.55 Vehicles per Household 1.16 1.59 1.68 1.77 1.78 1.89 1.86 1.88 Licensed drivers per Household 1.65 1.69 1.72 1.75 1.78 1.77 1.88 1.89 Vehicles per Licensed Driver 0.70 0.94 0.98 1.01 1.00 1.06 0.99 1.00 Workers per Household 1.21 1.23 1.21 1.27 1.33 1.35 1.34 1.33 Vehicles per Worker 0.96 1.29 1.39 1.40 1.34 1.39 1.39 1.42 Note: • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. 2017 National Household Travel Survey OVERVIEW 10 Table 2b. Major Travel Indicators by Survey Region Major Travel Indicators by Region Census Region Persons per Household Vehicles per Household Drivers per Household Vehicles per Driver Workers per Household Vehicles per Worker ALL (1) 2.55 1.88 1.89 1.00 1.33 1.42 Northeast 2.53 1.63 1.79 0.91 1.34 1.22 Midwest 2.42 1.96 1.83 1.07 1.29 1.52 South 2.56 1.90 1.91 0.99 1.31 1.45 West 2.70 1.98 1.98 1.00 1.38 1.43 Note: • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. During the past four decades, the growth in the number of workers and drivers has far outpaced the growth in the number of households and persons. However, as shown in Figure 1, the growth in the number of vehicles has outpaced all other indicators. Since 1969, the annual rate of increase in the number of personal vehicles was almost 1½ times the annual rate of increase in the number of drivers. 2017 National Household Travel Survey OVERVIEW 11 Figure 1. Changes in Summary Statistics on Demographics and Total Travel Note: • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. The data series indicates that the per capita growth in travel that the United States experienced over the last four decades may be changing. Statistically, of the 10 estimates of major travel indicators shown in Tables 3a and 3b, 7 are lower than the 2001 estimates and the remainder are statistically the same (within the confidence interval). Importantly, the number of reported person- and vehicle-trips per person is statistically lower in 2017 than in 2009, which is statistically lower than 2001. The estimates of travel for U.S. households show significant changes in trip-making. The estimates of person and vehicle trips per household are lower in 2017 than 2009, which in turn was lower than the 2001 estimates. As mentioned earlier, there was a major change in the method used to collect trip distance in 2017 that impacts the estimates of PMT, VMT, and Average Person and Vehicle Trip Lengths. In 2017, the NHTS calculated trip length using the shortest path routes between geocoded origins and destinations. Previous surveys used self-reported distances. 1.0 1.5 2.0 2.5 3.0 3.5 1969 1977 1983 1990 1995 2001 2009 2017 Indexed (1969=1.0) Vehicles Drivers Workers Households Persons 2017 National Household Travel Survey OVERVIEW 12 As a result of the change in method, the original estimates of VMT and PMT may not be directly comparable with previous years. See Appendix A for further details. Table 3a. Summary of Household Travel Statistics Household Statistics Survey Year: Daily Person Trips per Household Daily PMT per Household Daily Vehicle Trips per Household Daily VMT per Household 1969 6.36 61.55 3.83 34.01 1977 7.69 68.27 3.95 32.97 1983 7.20 62.47 4.07 32.16 1990 8.94 83.06 5.69 49.76 1995 10.49 94.41 6.36 57.25 2001 9.66 95.24 5.95 58.05 2009 9.50 90.42 5.66 54.38 2009 MOE 0.09 3.38 0.06 1.34 2017 orig. 8.60 92.02 5.11 48.81 2017 orig. MOE 0.10 3.50 0.06 2.04 2017 adj. 99.46 53.81 2017 adj. MOE 3.60 2.27 Note: • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here excludes them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Household VMT and PMT "adjusted” includes estimates of miles in all vehicles, including “18” Rental Car. 2017 National Household Travel Survey OVERVIEW 13 Table 3b. Summary of Person Travel Statistics Survey Year: Person Statistics Daily Person Trips per Person Daily PMT per Person Daily Vehicle Trips per Driver Daily VMT per Driver Average Person Trip Length (miles) Average Vehicle Trip Length (miles) 1969 2.02 19.51 2.32 20.64 9.67 8.89 1977 2.92 25.95 2.34 19.49 8.87 8.34 1983 2.89 25.05 2.36 18.68 8.68 7.90 1990 3.76 34.91 3.26 28.49 9.47 8.85 1995 4.30 38.67 3.57 32.14 9.13 9.06 2001 4.09 36.89 3.35 32.73 10.04 9.87 2009 3.79 36.13 3.02 28.97 9.75 9.72 2009 MOE 0.03 1.35 0.03 0.71 0.36 0.22 2017 orig. 3.37 36.07 2.70 25.84 10.70 9.55 2017 orig. MOE 0.04 1.47 0.03 1.04 0.40 0.37 2017 adj. 38.98 28.49 11.57 10.53 2017 adj. MOE 1.41 1.16 0.41 0.42 Note: • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here excludes them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Household VMT and PMT "adjusted” includes estimates of miles in all vehicles, including “18” Rental Car. 2017 National Household Travel Survey OVERVIEW 14 Table 4 compares key survey variables for each NPTS survey with external sources. Table 4. Comparison of Survey Variables with Other Sources (Numbers in Thousands, Except VMT [millions]) Category Households Population Licensed Drivers Workers Vehicles VMT 1969 Other Sources 61,806 199,145 108,306 89,174 1969 NPTS 62,504 197,213 102,986 72,500 1977 Other Sources 74,142 218,106 138,121 132,155 1977 NPTS 75,412 213,141 127,552 120,098 1983 Other Sources 83,918 232,086 154,389 152,070 1,652,788 1983 NPTS 85,371 229,453 147,015 143,714 1,002,139 1990 Other Sources 91,947 247,826 167,015 125,840 172,902 2,144,362 1990 NPTS 93,347 239,416 163,025 118,343 165,221 1,695,290 1995 Other Sources 97,386 261,538 176,628 132,300 180,735 2,139,307 1995 NPTS 98,990 259,994 176,330 131,697 176,067 2,068,368 2001 Other Sources 108,209 285,318 191,276 143,730 205,551 2,494,951 2001 NHTS 107,365 277,203 186,280 142,850 202,586 2,274,769 2009 Other Sources 117,181 307,007 208,321 154,140 231,490 2,562,305 2009 NHTS 112,520 299,802 211,270 151,370 216,056 2,245,111 2017 Other Sources 118,208 321,419 218,084 151,144 231,490 2,638,583 2017 NHTS 118,208 321,419 223,277 156,988 222,579 2,105,882 2017 NHTS (adj) 2,431,558 Note: Please see previous Summary of Travel Trends publications for the sources used for comparisons to prior surveys. Other Sources for 2017 Comparisons: Households - Census QuickFacts Table US Households 2012-2016 Population - Population in Occupied Housing Units, estimate 2016 Drivers - 2015 estimate from Highway Statistics Table DL-22 Workers - Source: 2016 American Community Survey 1-year estimate, Table B18120 Vehicles and VMT - Light Duty Vehicles (short WB) plus Motorcycles plus (based on the 2002 VIUS) 85.6% of Light Duty Vehicles with wheelbases (WB) larger than 121 inches) 2017 National Household Travel Survey HOUSEHOLD TRAVEL 15 3.0 HOUSEHOLD TRAVEL Overall, households generated about the same person miles of travel in 2017 (Table 5a) compared to the 2009 estimate, but fewer person trips (Table 5c). The person miles of travel– overall and for most trip purposes–were statistically the same between 2009 for both the original and adjusted estimates for 2017. The exception was person miles of travel for social and recreational purposes, which were significantly lower in the original 2017 estimate. The fact that the number of reported trips is lower while the total miles of travel is about the same as previous surveys could be an artifact of the shift away from interviewer-aided surveys to self-reported travel on the web. Without the aid of an interviewer, people may forget to report incidental stops and other short trips that impact the estimate of trips more than the estimate of miles of travel. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 16 Table 5a. Trends in the Average Annual Person Miles of Travel per Household by Trip Purpose Average Annual PMT per Household Trip Purpose All Purposes To / From Work Work Related Business Shopping Other Family / Personal Errands School / Church Social / Recreation Other 1983 22,802 4,586 1,354 2,567 3,311 1,522 8,964 500 1990 30,316 5,637 1,043 3,343 7,167 1,599 11,308 214 1995 34,459 7,740 1,987 4,659 7,381 1,973 10,571 131 2001 35,244 6,706 2,987 4,887 6,671 2,060 10,586 1,216 2009 33,004 6,256 2,078 4,620 5,134 2,049 9,989 2,878 2009 MOE 1,235.1 170.1 247.2 181.4 222.8 123.0 585.8 864.6 2017 Orig. 33,587 6,259 1,326 4,122 4,469 2,189 8,964 6,260 2017 Orig. MOE 1,276.2 204.6 326.0 343.3 253.6 394.0 362.3 971.4 2017 Adj. 36,302 6,678 1,399 4,578 4,939 2,396 9,883 6,429 2017 Adj. MOE 1,315.0 217.3 330.0 378.2 280.0 437.8 386.8 960.7 Note • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 17 Table 5b. Trends in the Average Person Trip Length by Trip Purpose Trip Purpose Average Person Trip Length (miles) All Purposes To / From Work Work Related Business Shopping Other Family / Personal Errands School / Church Social / Recreation Other 1983 8.7 8.5 21.8 5.4 7.3 4.9 12.3 8.2 1990 9.5 10.7 28.2 5.4 8.6 5.4 13.2 10.3 1995 9.1 11.6 20.3 6.1 7.6 6.0 11.3 22.8 2001 10.0 12.1 28.3 7.0 7.8 6.0 11.4 43.1 2009 9.7 11.8 20.0 6.5 7.0 6.3 10.7 51.5 2009 MOE 0.4 0.3 2.0 0.2 0.3 0.3 0.6 14.5 2017 Orig. 10.7 11.5 25.9 7.1 7.1 6.4 10.4 49.1 2017 Orig. MOE 0.4 0.3 6.4 0.5 0.3 1.2 0.5 7.3 2017 Adj. 11.6 12.2 27.4 7.9 7.9 7.0 11.4 50.4 2017 Adj. MOE 0.4 0.4 6.5 0.6 0.3 1.4 0.5 7.2 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. While the 2017 estimates of the number of person trips for work and school/church are statistically the same as in 2009 and 2001, the 2017 survey shows a significant decrease in the number of person trips for three major purposes: shopping, family and personal errands, and social and recreational travel. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 18 There may also be a change in trip-making for shopping, family errands, and social and recreational travel. This is a large, catch-all category of purposes that may be affected by changes in on-line shopping and other electronic communication. Further research into the specific and detailed trends of changes in trip-making by purpose, including changes in trip- chaining, would be useful. Table 5c. Trends in the Average Annual Person Trips per Household by Trip Purpose Trip Purpose Average Annual Person Trips per Household All Purposes To / From Work Work Related Business Shopping Other Family / Personal Errands School / Church Social / Recreation Other 1983 2,628 537 62 474 456 310 728 61 1990 3,262 539 38 630 854 304 874 22 1995 3,828 676 100 775 981 337 953 6 2001 3,581 565 109 707 863 351 952 30 2009 3,466 541 106 725 748 333 952 61 2009 MOE 31.8 7.9 7.4 14.6 13.9 9.8 14.1 4.1 2017 Orig. 3,140 546 51 580 628 341 866 128 2017 Orig. MOE 37.2 11.3 3.5 14.1 13.8 8.1 22.0 3.1 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. Tables 6a and 6b display trends in the average annual vehicle miles of travel and average trip length by select trip purposes. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 19 The original (unadjusted) 2017 estimates of overall VMT per household is statistically lower than 2009, while the adjusted estimate is about the same—within the margin of error of the 2009 estimate. While nominally lower, the VMT per household for shopping is within range of the earlier estimates. However, the estimates of VMT per household in 2017 for errands and social/recreational travel are statistically lower than the 2001 estimates for the same purposes. Using the adjusted estimates of vehicle miles of travel increases the estimate of VMT per household to be about the same as the 2009 estimates (within the margin of error) overall and for all trip purposes. For more information on the trip length adjustment, see Appendix A. Table 6a. Trends in the Average Annual Vehicle Miles of Travel by Selected Trip Purposes Trip Purpose Average Annual VMT per Household All Purposes To / From Work Shopping Other Family / Personal Errands Social / Recreation 1969 12,423 4,183 929 1,270 4,094 1977 12,036 3,815 1,336 1,444 3,286 1983 11,739 3,538 1,567 1,816 3,534 1990 18,161 4,853 2,178 4,250 5,359 1995 20,895 6,492 2,807 4,307 4,764 2001 21,187 5,724 3,062 3,956 5,186 2009 19,850 5,513 2,979 3,515 4,842 2009 MOE 490.5 146.7 95.9 120.1 257.8 2017 Original 17,815 5,379 2,618 2,982 4,327 2017 Orig. MOE 745.4 192.3 304.3 217.0 182.3 2017 Adjusted 19,641.8 5,773.9 2,919.9 3,325.2 4,825.5 2017 Adj. MOE 829.6 206.5 339.3 241.9 203.2 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 20 Table 6b. Trends in the Average Trip Length by Selected Trip Purposes Trip Purpose: Average Vehicle Trip Length (miles) All Purposes To / From Work Shopping Other Family / Personal Errands Social / Recreation 1969 8.9 9.4 4.4 6.5 13.1 1977 8.4 9.0 5.0 6.7 10.3 1983 7.9 8.6 5.3 6.7 10.6 1990 8.9 11.0 5.1 7.4 11.8 1995 9.1 11.8 5.6 6.9 11.2 2001 9.9 12.1 6.7 7.5 11.9 2009 9.7 12.2 6.4 7.1 11.2 2009 MOE 0.2 0.3 0.2 0.2 0.6 2017 Original 9.6 12.0 7.0 6.9 10.6 2017 Orig. MOE 0.4 0.4 0.8 0.4 0.4 2017 Adjusted 10.5 12.8 7.9 7.7 11.8 2017 Adj. MOE 0.4 0.4 0.8 0.4 0.4 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 21 Following the trends in person trips, in 2017, a typical household generated significantly fewer vehicle trips than in 2009 (Table 6c). While the 2017 estimates of the number of vehicle trips for work and school/church are statistically the same as in 2009 and 2001, the 2017 survey shows a significant decrease in the number of vehicle trips for three major purposes: shopping, family and personal errands, and social and recreational travel. The original estimates of vehicle miles overall and for most purposes (except commuting) are statistically lower in 2017 compared to 2009. The adjustment for vehicle miles of travel brings the estimates into the same range as the 2009 estimates (within the margin of error). For more information on the trip length adjustment, see Appendix A. The fact that the number of reported vehicle trips is lower while the total (adjusted) vehicle miles of travel (Table 6a) is about the same as previous surveys could be an artifact of the shift away from interviewer-aided surveys to self-reported travel on the web. Without the aid of an interviewer, people may forget to report incidental stops and other short trips that impact the estimate of trips more than the estimate of miles of travel. However, there may also be a change in trip-making for shopping, family errands, and social and recreational travel. This is a large, catch-all category of purposes that may be affected by changes in on-line shopping and other electronic communication. Further research into the specific and detailed trends of changes in trip-making by purpose, including trip-chaining, would be enlightening. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 22 Table 6c. Trends in the Average Annual Vehicle Trips per Household by Selected Trip Purposes Trip Purpose Average Annual Vehicle Trips per Household All Purposes To / From Work Shopping Other Family / Personal Errands Social / Recreation 1969 1,396 445 213 195 312 1977 1,442 423 268 215 320 1983 1,486 414 297 272 335 1990 2,077 448 431 579 460 1995 2,321 553 501 626 427 2001 2,171 479 459 537 441 2009 2,068 457 468 500 436 2009 MOE 20.8 7.8 9.2 9.2 8.4 2017 Original 1,865 450 372 434 410 2017 Orig. MOE 21.7 9.6 10.2 11.0 10.6 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. Table 7 displays the trends in average annual person trips per household by mode of transportation and metropolitan statistical area (MSA) size. Future surveys will tell if there is a shift to using public transit instead of private vehicles. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 23 Table 7. Trends in the Average Annual Person Trips per Household by Mode of Transportation and MSA Size MSA Size 1977 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE Private Vehicle ALL 2,351 2,152 2,861 3,307 3,090 2,892 30 2,592 30 Not in MSA 2,436 2,322 2,837 3,492 3,076 2,898 72 2,623 81 Less than 250,000 2,517 2,375 3,090 3,503 3,304 2,980 118 2,620 123 250,000 - 499,999 2,574 2,443 3,014 3,472 3,251 2,950 141 2,718 122 500,000 - 999,999 2,628 2,140 2,957 3,509 3,348 3,020 144 2,698 73 1,000,000 - 2,999,999 2,366 2,031 2,986 3,354 3,174 2,951 74 2,678 89 3,000,000 and above 1,785 1,691 2,649 3,075 2,911 2,793 50 2,446 37 Public Transit ALL 73 60 58 67 58 66 4 80 4 Not in MSA 22 11 14 9 6 4 2 6 2 Less than 250,000 47 17 30 23 12 14 8 33 8 250,000 - 499,999 44 23 22 18 18 15 7 34 12 500,000 - 999,999 58 48 33 33 11 41 17 42 9 1,000,000 - 2,999,999 86 67 52 37 36 39 8 50 9 3,000,000 and above 189 181 124 137 128 148 11 170 8 Walk ALL 261 226 234 205 309 362 13 329 14 Not in MSA 199 211 175 134 221 239 17 204 36 Less than 250,000 241 280 212 138 248 270 48 217 18 250,000 - 499,999 206 199 203 152 251 268 23 228 33 500,000 - 999,999 256 184 161 138 224 314 52 274 20 1,000,000 - 2,999,999 295 179 207 162 275 313 20 303 26 3,000,000 and above 396 330 337 301 423 514 29 479 16 ALL Modes ALL 2,808 2,628 3,262 3,828 3,581 3,466 32 3,140 37 Not in MSA 2,800 2,766 3,151 3,878 3,435 3,275 77 2,966 85 Less than 250,000 2,944 2,889 3,450 3,926 3,678 3,395 128 2,984 128 250,000 - 499,999 2,945 2,891 3,340 3,894 3,645 3,356 144 3,103 128 500,000 - 999,999 3,049 2,542 3,252 3,916 3,692 3,529 151 3,141 79 1,000,000 - 2,999,999 2,861 2,463 3,344 3,795 3,602 3,446 78 3,178 100 3,000,000 and above 2,459 2,326 3,213 3,765 3,593 3,614 55 3,246 43 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 24 • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Rural, Not in MSA" includes only full counties designated as rural. There may also be rural pockets included within MSA boundaries. • The population size groups for 1977 - 1983 NPTS are MSA size groups. 1990 - 2001 are MSA size groups. 2009 - 2017 are Consolidated Metropolitan Statistical Area (CMSA) size groups. • Changes in walk trips throughout the data series could be a result, at least in part, to questionnaire changes: Recent NHTS surveys explicitly prompt respondents to include walk and bike trips, which was not the case in prior surveys. The 2017 NHTS changed the definition of a trip to allow walk and bike trips to and from hone (loop trips). • Public transit includes local bus, commuter bus, commuter train, subway, trolley, and streetcar. The data series in Table 8 shows that more income is related to more travel. The households in the highest income group annually produce 80 percent more person trips compared to households in the lowest income group. The income categories in 2017 changed slightly from the 2009 and earlier surveys. The data here are shown in 2017 current dollars The 2009 and earlier surveys were conducted with a telephone sample (landline only) which excluded CPO households. This was especially an issue in 2009, when an estimated 25 percent of all US households did not have a landline. Therefore, the 2009 sample may have under coverage of households with lower income. Care should be taken in interpreting trends of estimates that might be correlated to telephone ownership, such as household income. 2017 National Household Travel Survey HOUSEHOLD TRAVEL 25 Table 8. Trends in the Number of Annual Person Trips per Household by Household Income Income 1990 1995 2001 2009 2009 MOE 2017 2017 MOE ALL 3,262 3,828 3,793 3,466 31 3,140 37 Less than $15,000 2,298 2,525 2,272 2,200 99 2,214 112 $15,000 to $24,999 3,072 3,263 3,028 2,616 102 2,477 146 $25,000 to $34,999 3,685 3,914 3,411 3,018 112 2,756 94 $35,000 to $49,999 4,214 4,483 4,015 3,278 110 2,979 134 $50,000 to $74,999 4,549 4,710 4,761 3,967 100 3,172 81 $75,000 to $99,999 4,537 4,910 5,214 4,504 112 3,487 90 $100,000 and over - 4,723 5,253 4,947 117 4,033 105 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 NHTS asked income in different categories than previous surveys, therefore this table will not match the Summary of Travel Trends 2009 and earlier • In 1990 the highest income group was $80,000 and above • Incomes for 1983, 1990, adjusted 1990, and 1995 have been adjusted to 2001 dollars: 2017 National Household Travel Survey PERSON TRAVEL 26 4.0 PERSON TRAVEL In 2017, the overall number of reported trips by private vehicle was significantly lower than the 2009 estimate. However, the declines were not equal across all purposes. For example, the estimate for the number of vehicle commutes and vehicle trips to school and church were statistically the same in 2017 compared to 2009 and previous years (within the margin of error). However, the reported total number of vehicle trips for shopping and errands was nominally closer to the 1990 estimate than any intervening year and a significant decline from the 2009 estimate. On the other hand, the overall number of transit trips reported was significantly higher than the 2009 estimate, fueled by the significant increase in the number of reported commutes on transit. The estimate for the number of transit trips for all other purposes was statistically the same in 2017 compared to 2009. The total number of walk trips reported was statistically within the margin of error of the 2009 estimate. The definition of a reported walk trip changed slightly to allow trips that begin and end at home, like walks for exercise. This change in definition impacts the total estimate of walks and requires more investigation. But it should be noted that the common thread is an overall decline in reported trips for shopping and errands. This category of trip purposes is a large, catch-all category of trip-making that may be affected by many competing factors: For example, some of the difference in reported trips in 2017 NHTS may be a result of moving to a self-completed questionnaire compared to interview-assisted in previous surveys. Interviewers are trained to prompt for short stops and under-reported trips. There may also be changes in trip-making for shopping and errands related to on-line purchasing. Other demographic trends, such as shifts in the percentage of households with children, may also be a factor. It would be helpful to conduct further research into the specific and detailed trends of changes in trip-making by purpose, including trip-chaining. The Table 9 series displays these findings. 2017 National Household Travel Survey PERSON TRAVEL 27 Table 9a. Trends in the Annual Number (millions) of Person Trips by Mode of Transportation and Trip Purpose Category Private Vehicle To/ From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other 1990 45,856 3,178 128,368 17,545 70,382 1,629 1995 60,740 8,835 156,065 22,436 78,809 470 2001 56,054 10,648 153,270 26,861 82,437 2,147 2009 55,969 10,525 146,158 26,654 82,887 4,925 2009 MOE 941.4 767.1 2487.7 968.2 1583.2 304.1 2017 56,981 4,844 126,268 28,427 78,890 10,988 2017 MOE 1276.6 272.7 1343.8 990.0 2262.4 400.8 Category Public Transit To/ From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other 1990 1,992 92 1,318 1,076 946 35 1995 2,328 123 2,000 826 1,350 11 2001 2,271 213 1,776 800 989 134 2009 2,247 264 2,344 829 1,426 409 2009 MOE 254.2 93.7 264.7 131.8 195.0 114.5 2017 3,537 208 2,586 1,009 1,618 487 2017 MOE 214.3 72.2 198.5 182.0 131.0 82.5 2017 National Household Travel Survey PERSON TRAVEL 28 Table 9a. Trends in the Annual Number (millions) of Person Trips by Mode of Transportation and Trip Purpose (continued) Category Walk To/ From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other 1990 1,999 154 7,722 3,649 8,090 265 1995 1,510 240 8,756 2,925 6,845 47 2001 1,715 487 11,936 3,630 14,824 507 2009 1,854 684 15,174 3,542 18,833 874 2009 MOE 230.4 136.1 818.7 479.4 768.4 157.6 2017 2,523 510 11,496 4,146 18,483 1,790 2017 MOE 258.3 68.2 680.0 459.5 724.0 122.3 Category Other To/ From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other 1990 428 95 1,087 6,086 2,098 73 1995 887 417 1,768 6,035 2,954 37 2001 584 317 1,468 6,351 3,829 394 2009 1,144 469 2,859 6,651 4,576 725 2009 MOE 166.1 169.2 337.3 413.1 387.4 135.1 2017 1,540 486 2,404 6,721 3,330 1,873 2017 MOE 184.4 139.6 296.1 294.8 309.0 274.4 2017 National Household Travel Survey PERSON TRAVEL 29 Table 9a. Trends in the Annual Number (millions) of Person Trips by Mode of Transportation and Trip Purpose (continued) Category Total To/ From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other 1990 50,314 3,529 138,559 28,397 81,575 2,014 1995 66,901 9,860 173,764 33,355 94,362 623 2001 60,690 11,676 168,560 37,671 102,165 3,198 2009 61,214 11,943 166,535 37,676 107,722 6,933 2009 MOE 901.9 849.2 2536.5 1119.2 1617.9 468.3 2017 64,582 6,048 142,754 40,303 102,327 15,139 2017 MOE 1333.0 409.3 1469.3 955.6 2605.5 362.8 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Other" trip purpose includes trips for work-related business and trips not categorized. 2017 National Household Travel Survey PERSON TRAVEL 30 Table 9b. Trends in the Percent of Person Trips by Mode of Transportation and Trip Purpose (Millions) Year Private Vehicle To/From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other Total 1990 91.2% 90.3% 92.7% 61.9% 86.3% 81.4% 87.8% 1995 92.8% 91.9% 92.6% 69.6% 87.6% 83.2% 89.3% 2001 92.4% 91.2% 90.9% 71.3% 80.7% 67.2% 86.3% 2009 91.4% 88.1% 87.8% 70.7% 76.9% 71.0% 83.4% 2017 88.2% 80.1% 88.5% 70.5% 77.1% 72.6% 82.6% Year Public Transit To/From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other Total 1990 4.0% 2.6% 1.0% 3.8% 1.2% 1.7% 1.8% 1995 3.6% 1.3% 1.2% 2.6% 1.5% 1.9% 1.8% 2001 3.7% 1.8% 1.1% 2.1% 1.0% 4.2% 1.6% 2009 3.7% 2.2% 1.4% 2.2% 1.3% 5.9% 1.9% 2017 5.5% 3.4% 1.8% 2.5% 1.6% 3.2% 2.5% Year Walk To/From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other Total 1990 4.0% 4.4% 5.6% 12.8% 9.9% 13.2% 7.2% 1995 2.3% 2.4% 5.0% 8.8% 7.3% 7.6% 5.4% 2001 2.8% 4.2% 7.1% 9.6% 14.5% 15.9% 8.6% 2009 3.0% 5.7% 9.1% 9.4% 17.5% 12.6% 10.4% 2017 3.9% 8.4% 8.1% 10.3% 18.1% 11.8% 10.5% 2017 National Household Travel Survey PERSON TRAVEL 31 Table 9b. Trends in the Percent of Person Trips by Mode of Transportation and Trip Purpose (Millions) (continued) Year Other To/From Work Work-Related Business Shopping and Errands School or Church Social and Recreational Other Total 1990 0.8% 2.7% 0.8% 21.4% 2.6% 3.6% 3.2% 1995 1.3% 4.2% 1.0% 18.1% 3.1% 6.0% 3.2% 2001 1.0% 2.7% 0.9% 16.9% 3.7% 12.3% 3.4% 2009 1.9% 3.9% 1.7% 17.7% 4.2% 10.5% 4.2% 2017 2.4% 8.0% 1.7% 16.7% 3.3% 12.4% 4.4% Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. . • Changes in walk trips throughout the data series could be a result, at least in part, to questionnaire changes: Recent NHTS surveys explicitly prompt respondents to include walk and bike trips, which was not the case in prior surveys. The 2017 NHTS changed the definition of a trip to allow walk and bike trips to and from hone (loop trips). • "Other" trip purpose includes trips for work-related business and trips not categorized. 2017 National Household Travel Survey PERSON TRAVEL 32 The most striking gender difference in travel behavior is in the difference in the number of household-supporting trips taken by men and women. Traditionally, women make many more trips for shopping and errands compared to men. Table 10a shows that these gender differences persist in the 2017 data. In the 2017 NHTS, women reported making more trips overall than men and more trips for shopping and family errands compared to men. On the other hand, men reported more trips than women for work and for work-related business. Men and women reported about the same number of social and recreational trips (within the margin of error). Continuing trends noted previously, both men and women took fewer trips on average in 2017 compared to the estimates for 2009 and 2001 (Table 10b). Men and women reported about 11 percent fewer trips in 2017 compared to 2009. Nearly all the decline in trip-making came from declines in the estimate of trips for shopping and errands. Table 10a. Trends in the Annual Number of Person Trips per Person by Trip Purpose and Gender Category All 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 1,371 1,568 1,469 1,385 16.1 1,231 15.7 To or From Work 210 257 219 216 4.7 214 4.7 Work Related Business 15 38 42 42 3.9 20 1.5 Shopping and Errands 579 668 608 588 11.4 473 5.2 School/Church 119 128 136 133 4.9 134 3.4 Social and Recreational 341 363 369 381 7.5 339 8.6 Other 8 2 12 24 2.2 50 1.2 2017 National Household Travel Survey PERSON TRAVEL 33 Table 10a. Trends in the Annual Number of Person Trips per Person by Trip Purpose and Gender (continued) Men 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 1,339 1,579 1,491 1,368 15.7 1,210 23.2 To or From Work 259 327 273 241 4.6 240 6.9 Work Related Business 21 60 66 58 5.2 25 2.4 Shopping and Errands 549 648 590 529 10.7 420 9.9 School/Church 123 134 141 128 5.3 132 4.3 Social and Recreational 377 406 405 386 7.9 335 10.9 Other 9 2 13 26 2.4 58 2.4 Women 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 1,401 1,558 1,494 1,401 16.4 1,251 16.0 To or From Work 197 229 200 193 4.7 189 4.5 Work Related Business 11 23 25 27 2.6 15 0.8 Shopping and Errands 693 786 715 646 12.1 525 10.4 School/Church 132 141 151 138 4.5 135 3.9 Social and Recreational 358 375 389 375 7.2 344 9.3 Other 9 3 12 23 2.0 42 2.5 2017 National Household Travel Survey PERSON TRAVEL 34 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Other" trip purpose includes trips for work-related business and trips not categorized. 2017 National Household Travel Survey PERSON TRAVEL 35 Table 10b. Trends in the Percent of Person Trips per Person by Trip Purpose and Gender Category All 1990 1995 2001 2009 2017 TOTAL 100% 100% 100% 100% 100% To or From Work 15.3% 16.4% 14.9% 15.6% 17.4% Work Related Business 1.1% 2.4% 2.9% 3.0% 1.6% Shopping and Errands 42.2% 42.6% 41.4% 42.5% 38.4% School/Church 8.7% 8.2% 9.2% 9.6% 10.9% Social and Recreational 24.9% 23.1% 25.1% 27.5% 27.5% Other 0.6% 0.2% 0.8% 1.8% 4.1% Category Men 1990 1995 2001 2009 2017 TOTAL 100% 100% 100% 100% 100% To or From Work 19.3% 20.7% 18.3% 17.6% 19.8% Work Related Business 1.6% 3.8% 4.4% 4.2% 2.1% Shopping and Errands 41.0% 41.0% 39.6% 38.7% 34.7% School/Church 9.2% 8.5% 9.5% 9.4% 10.9% Social and Recreational 28.2% 25.7% 27.2% 28.2% 27.7% Other 0.7% 0.1% 0.9% 1.9% 4.8% 2017 National Household Travel Survey PERSON TRAVEL 36 Table 10b. Trends in the Percent of Person Trips per Person by Trip Purpose and Gender (continued) Category Women 1990 1995 2001 2009 2017 TOTAL 100% 100% 100% 100% 100% To or From Work 14.1% 14.7% 13.4% 13.8% 15.1% Work Related Business 0.8% 1.5% 1.7% 1.9% 1.2% Shopping and Errands 49.5% 50.4% 47.9% 46.1% 42.0% School/Church 9.4% 9.1% 10.1% 9.9% 10.9% Social and Recreational 25.6% 24.1% 26.0% 26.8% 27.5% Other 0.6% 0.2% 0.8% 1.6% 3.4% Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Other" trip purpose includes trips for work-related business and trips not categorized. 2017 National Household Travel Survey PERSON TRAVEL 37 Figure 2 shows the estimate of the number of annual person trips by purpose for men and women from 1990 to 2017. The decline in the total number of trips per person since 1995 appears to be mostly due to declines in the estimate of trips for shopping and errands. Interestingly, both men and women report about one-third fewer trips for shopping and errands in 2017 compared to 1995. However, in 2017, women still reported making about 25 percent more shopping and errand trips than men. The category of trip purposes called “shopping and errands” is a large, catch-all category of purposes that may be affected by the change in methods (e.g., self-reports on the web may under-report incidental stops) and may also be affected by increases in online shopping as well as shifts in the number of households with children. It would be enlightening to conduct further research into the specific and detailed changes in trip-making by purpose, including trip-chaining. 2017 National Household Travel Survey PERSON TRAVEL 38 Figure 2. Trends in the Distribution of Person Trips per Person by Gender and Trip Purpose Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Other" trip purpose includes trips for work-related business and trips not categorized. 0 200 400 600 800 1000 1200 1400 1600 1800 1990 1995 2001 2009 2017 1990 1995 2001 2009 2017 Men Women Annual Person Trips per Person Shopping and Errands Social and Recreational School/Church To or From Work 2017 National Household Travel Survey PERSON TRAVEL 39 In 2017, the person trip rates overall were lower than the 2009 estimates (Table 11). It is interesting to note that not all trip purposes declined at the same rate. For example, the estimate for the number of trips to and from work and trips to school and church were statistically the same in 2017 compared to 2009 and previous years. The majority of the decline in trip-making came from lower estimates for daily trips for shopping and family errands. The estimate for the number of daily trips for shopping and errands declined from 1.61 in 2009 to 1.31 in 2017. This follows a decline from 2001-2009 (from 1.79 to 1.61), which follows a decline from 1995-2001 from 1.97 to 1.79). This is a large, catch-all category of purposes that may be affected by the change in methods (e.g., self-reports on the web may under-report incidental stops) and may also be affected by changes in online shopping as well as shifts in the number of households with children. It would be enlightening to conduct further research into the specific and detailed trends of changes in trip-making by purpose, including trip-chaining. In terms of miles of travel, the results are also mixed. The average daily miles travelled for work, school, and church were statistically lower for all purposes when measured via the shortest path. However, with the adjusted factors applied, the average daily miles were significantly higher for shopping and errands and for social and recreational travel in 2017 compared to 2009. Details about the mileage estimate obtained in the 2017 NHTS is in Appendix A. 2017 National Household Travel Survey PERSON TRAVEL 40 Table 11. Trends in the Daily Trip Rates and Person Miles of Travel per Person by Trip Purpose Survey Year Total To / From Work Shopping / Errands School / Church Social / Recreation Person Trips per Day 1977 2.92 0.57 0.91 0.35 0.71 1983 2.89 0.59 1.02 0.34 0.8 1990 3.76 0.62 1.71 0.35 1.01 1995 4.30 0.76 1.97 0.38 1.07 2001 4.09 0.65 1.79 0.4 1.09 2009 3.79 0.59 1.61 0.36 1.04 2009 MOE 0.03 0.01 0.02 0.01 0.02 2017 3.37 0.59 1.30 0.37 0.93 2017 MOE 0.04 0.01 0.01 0.01 0.02 Survey Year Total To / From Work Shopping / Errands School / Church Social / Recreation Person Miles per Day 1977 25.95 5.16 5.68 1.61 7.81 1983 25.05 5.04 6.46 1.67 9.85 1990 34.91 6.49 12.1 1.84 13.02 1995 38.67 8.69 13.51 2.21 11.86 2001 40.25 7.66 13.2 2.35 12.09 2009 36.13 6.85 10.68 2.24 10.93 2009 MOE 1.35 0.19 0.31 0.13 0.64 2017 Orig. 36.07 6.72 9.22 2.35 9.63 2017 Orig. MOE 1.47 0.24 0.50 0.45 0.39 2017 Adj 38.98 7.17 10.22 2.57 10.61 2017 Adj. MOE 1.41 0.23 0.51 0.47 0.42 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. • "Other" trip purpose includes trips for work-related business and trips not categorized. • Trip rates are calculated including travelers and non-travelers, resulting in travel estimates per-capita. 2017 National Household Travel Survey PERSON TRAVEL 41 Figures 3a and 3b and Tables 12 and 13 display daily trip and person rates and person miles of travel and show a decline in overall trip-making. Figure 3a. Daily Trip Rates per Person by Trip Purpose . Figure 3b. Daily Person Miles of Travel per Person by Trip Purpose 0.6 0.8 0.7 0.6 0.6 1.7 2.0 1.8 1.6 1.3 0.4 0.4 0.4 0.4 0.4 1.0 1.1 1.1 1.0 0.9 1990 1995 2001 2009 2017 Other Social and Recreational School/Church Shopping and Errands To or From Work 4.3 4.1 3.8 3.4 x.x Daily Trip Rate Estimate 3.8 6.5 8.7 7.7 6.9 6.7 7.2 12.1 13.5 13.2 10.7 9.2 10.2 13.0 11.9 12.1 10.9 9.6 10.6 1.5 2.4 4.8 5.4 8.2 8.4 1990 1995 2001 2009 2017 Orig 2017 Adj Other Social and Recreational School/Church Shopping and Errands To or From Work 34.9 38.7 36.1 x.x Daily Person Miles Estimate 40.3 36.1 39.0 2017 National Household Travel Survey PERSON TRAVEL 42 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. • "Other" trip purpose includes trips for work-related business and trips not categorized. 2017 National Household Travel Survey PERSON TRAVEL 43 Table 12. Trends in the Distribution of Daily Person Miles of Travel per Person by Mode of Transportation and Trip Purpose Category Private Vehicle 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 30.85 35.26 35.49 31.92 0.88 27.54 0.80 30.45 0.83 Percent 88.4% 91.2% 88.2% 88.3% 76.4% 78.1% To or From Work 6.15 8.09 7.11 6.47 0.17 6.13 0.21 6.58 0.22 Percent 17.6% 20.9% 17.7% 17.9% 17.0% 16.9% Work-Related Business 0.63 1.85 2.27 1.88 0.21 0.68 0.06 0.76 0.07 Percent 1.80% 4.78% 5.64% 5.20% 1.89% 1.95% Shopping and Errands 11.39 12.7 12.77 10.30 0.32 8.65 0.45 9.64 0.50 Percent 32.6% 32.8% 31.7% 28.5% 24.0% 24.7% School/Church 1.32 1.68 1.87 1.80 0.13 1.93 0.41 2.15 0.46 Percent 3.78% 4.34% 4.65% 4.98% 5.35% 5.52% Social and Recreational 11.12 10.83 11.01 9.98 0.52 8.57 0.42 9.56 0.47 Percent 31.9% 28.0% 27.4% 27.6% 23.8% 24.5% Other 0.23 0.10 0.36 1.49 0.35 1.58 0.20 1.76 0.22 Percent 0.66% 0.26% 0.89% 4.12% 4.38% 4.52% 2017 National Household Travel Survey PERSON TRAVEL 44 Table 12. Trends in the Distribution of Daily Person Miles of Travel per Person by Mode of Transportation and Trip Purpose (continued) Public Transit Category 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 0.74 0.82 0.47 0.53 0.11 0.94 0.10 0.94 0.10 Percent 2.1% 2.1% 1.2% 1.5% 2.6% 2.4% To or From Work 0.27 0.30 0.24 0.18 0.04 0.39 0.04 0.39 0.04 Percent 0.77% 0.78% 0.60% 0.50% 1.08% 1.00% Work-Related Business 0.01 0.02 0.01 0.02 0.01 0.06 0.05 0.06 0.05 Percent 0.03% 0.05% 0.02% 0.06% 0.17% 0.15% Shopping and Errands 0.14 0.19 0.10 0.10 0.02 0.17 0.02 0.17 0.02 Percent 0.40% 0.49% 0.25% 0.28% 0.47% 0.44% School/Church 0.12 0.07 0.04 0.05 0.01 0.07 0.01 0.07 0.01 Percent 0.34% 0.18% 0.10% 0.14% 0.19% 0.18% Social and Recreational 0.18 0.24 0.07 0.10 0.03 0.18 0.06 0.18 0.06 Percent 0.52% 0.62% 0.17% 0.28% 0.50% 0.46% Other 0.01 0.00 0.00 0.08 0.09 0.08 0.02 0.08 0.02 Percent 0.03% 0.00% 0.00% 0.22% 0.22% 0.21% 2017 National Household Travel Survey PERSON TRAVEL 45 Table 12. Trends in the Distribution of Daily Person Miles of Travel per Person by Mode of Transportation and Trip Purpose (continued) Category Other Means 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 3.31 2.2 4.10 3.68 0.96 7.58 1.24 7.58 1.24 Percent 9.5% 5.7% 10.2% 10.2% 21.0% 19.4% To or From Work 0.06 0.22 0.30 0.20 0.09 0.20 0.06 0.20 0.06 Percent 0.17% 0.57% 0.75% 0.55% 0.55% 0.51% Work-Related Business 0.56 0.34 1.12 0.38 0.15 0.69 0.32 0.69 0.32 Percent 1.60% 0.88% 2.78% 1.05% 1.91% 1.77% Shopping and Errands 0.57 0.49 0.32 0.28 0.04 0.41 0.16 0.41 0.16 Percent 1.63% 1.27% 0.80% 0.77% 1.14% 1.05% School/Church 0.40 0.44 0.44 0.40 0.03 0.35 0.05 0.35 0.05 Percent 1.15% 1.14% 1.09% 1.11% 0.97% 0.90% Social and Recreational 1.71 0.66 1.01 0.85 0.35 0.88 0.42 0.88 0.42 Percent 4.90% 1.71% 2.51% 2.35% 2.44% 2.26% Other 0.01 0.05 0.87 1.57 0.87 5.06 1.16 5.06 1.16 Percent 0.0% 0.1% 2.2% 4.3% 14.0% 13.0% 2017 National Household Travel Survey PERSON TRAVEL 46 Table 12. Trends in the Distribution of Daily Person Miles of Travel per Person by Mode of Transportation and Trip Purpose (continued) Total Category 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 34.91 38.67 40.25 36.13 1.35 36.07 1.37 38.98 1.41 Percent 100% 100% 100% 100% 100% 100% To or From Work 6.49 8.69 7.66 6.85 0.19 6.72 0.22 7.17 0.23 Percent 18.6% 22.5% 19.0% 19.0% 18.6% 18.4% Work-Related Business 1.20 2.23 3.41 2.28 0.27 1.42 0.35 1.5 0.35 Percent 3.44% 5.77% 8.47% 6.31% 3.94% 3.85% Shopping and Errands 12.10 13.51 13.2 10.68 0.31 9.22 0.46 10.22 0.51 Percent 34.7% 34.9% 32.8% 29.6% 25.6% 26.2% School/Church 1.84 2.21 2.35 2.24 0.13 2.35 0.42 2.57 0.47 Percent 5.27% 5.72% 5.84% 6.20% 6.52% 6.59% Social and Recreational 13.02 11.86 12.09 10.93 0.64 9.63 0.39 10.61 0.42 Percent 37.3% 30.7% 30.0% 30.3% 26.7% 27.2% Other 0.25 0.15 1.39 3.15 0.95 6.72 1.04 6.9 1.03 Percent 0.7% 0.4% 3.5% 8.7% 18.6% 17.7% Note • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. 2017 National Household Travel Survey PERSON TRAVEL 47 • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 2001, the mode "Bus" was divided into "Local Public Transit Bus," "Commuter Bus," "Charter/Tour Bus," and "City to City Bus." Only "Local Public Transit Bus" and "Commuter Bus" are included in public transit calculations. • Increases in walk trips between 2001 and 2017 are due, at least in part, to questionnaire changes: recent NHTS surveys explicitly ask respondents to include walk and bike trips, which was not the case in prior surveys. • In 2017, walk and bike trips were sometimes reported as Home-Home loops (single round trips). In prior surveys, “loop” trips were coded to the farthest destination and reported as two trips: outbound and return. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. • "Other" trip purpose includes trips for work-related business and trips not categorized. • Percentages are a percent of total daily person miles of travel. 2017 National Household Travel Survey PERSON TRAVEL 48 Table 13. Trends in the Average Daily Person Trips per Person by Age and Gender Age Total 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 2.9 3.8 4.3 4.1 3.8 0.03 3.4 0.04 Under 16 2.3 3.1 3.7 3.4 3.2 0.07 2.8 0.06 16 to 20 3.3 4.2 4.6 4.1 3.5 0.11 2.8 0.08 21 to 35 3.5 4.4 4.6 4.3 3.9 0.09 3.4 0.10 36 to 65 2.9 3.9 4.6 4.5 4.2 0.05 3.7 0.03 Over 65 1.8 2.4 3.4 3.4 3.2 0.07 3.2 0.04 Age Men 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 2.9 3.7 4.3 4.1 3.7 0.04 3.3 0.06 Under 16 2.3 3 3.7 3.5 3.2 0.09 2.8 0.07 16 to 20 3.2 4.2 4.6 4.0 3.3 0.13 2.8 0.13 21 to 35 3.4 4.2 4.5 4.2 3.7 0.11 3.2 0.10 36 to 65 2.9 3.7 4.6 4.4 4.1 0.06 3.6 0.06 Over 65 2.2 2.8 3.9 3.8 3.5 0.10 3.4 0.05 Age Women 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 2.9 3.8 4.3 4.1 3.8 0.04 3.4 0.04 Under 16 2.3 3.1 3.8 3.4 3.2 0.10 2.8 0.07 16 to 20 3.4 4.2 4.7 4.2 3.7 0.15 2.8 0.12 21 to 35 3.5 4.6 4.8 4.5 4.1 0.12 3.6 0.12 36 to 65 3 4.1 4.6 4.5 4.3 0.06 3.8 0.04 Over 65 1.5 2.2 3 3.1 2.9 0.09 3.0 0.06 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey PERSON TRAVEL 49 According to the 2017 NHTS estimates, all people younger than 65 reported significantly fewer trips in 2017 compared to 2009 (which was significantly lower than 2001, which was lower than 1995). Figure 4 shows that the 2017 estimate of person trips per person by age in these categories were lower than previous survey estimates, except for people aged 65 and older. The data show that the decrease in trip-making was similar for both men and women, with men's trip-making declining by 21 percent and women’s by 19 percent since 1995. Some of the difference in reported trips in 2017 NHTS may be a result of moving to a self-completed questionnaire, compared to interview-assisted in previous surveys. For example, interviewers are trained to prompt for short stops and under-reported trips. Other factors, such as shifts related to online shopping may affect these estimates. Changes in household structure and other demographic trends may also play a role. However, the trends over the last two decades clearly indicate that the NHTS estimates of overall trip-making are declining, with larger declines noted for younger people. Figure 4. Trends in the Average Daily Person Trips by Age Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Total Under 16 16 to 20 21 to 35 36 to 65 Over 65 Person Trips per Day 1990 1995 2001 2009 2017 2017 National Household Travel Survey PERSON TRAVEL 50 Table 14. Trends in the Average Daily Person Miles of Travel per Person by Age and Gender Age TOTAL 1983 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 25.1 34.9 38.7 40.2 36.1 1.4 36.1 1.4 39.0 1.4 Under 16 16.2 20.1 25 24.5 25.3 3.5 22.9 2.8 24.9 3.0 16 to 20 22.2 34.4 36.4 38.1 29.5 1.8 27.3 2.0 29.6 2.2 21 to 35 31.1 44.3 46 45.6 37.7 1.9 41.4 5.7 44.6 6.1 36 to 65 29.2 40.1 45.1 48.8 44.0 1.9 41.7 1.6 44.9 1.7 Over 65 12.0 18.4 24.4 27.5 24.0 1.2 30.1 2.8 32.8 2.8 Age Men 1983 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 27.7 38.0 43.9 45.0 40.9 2.1 39.5 1.4 42.5 1.5 Under 16 16.8 20.3 23.7 24.6 27.2 6.3 25.6 4.8 27.7 5.1 16 to 20 23.0 36.9 37.6 34.1 28.2 2.3 25.9 3.1 28.0 3.2 21 to 35 32.8 48.2 51.3 49.8 40.5 2.8 42.9 5.4 46.0 5.5 36 to 65 33.6 43.4 53.2 57.7 50.9 3.0 47.1 2.3 50.6 2.3 Over 65 14.8 22.5 31.7 32.9 30.5 1.9 33.8 4.0 36.8 4.2 2017 National Household Travel Survey PERSON TRAVEL 51 Table 14. Trends in the Average Daily Person Miles of Travel per Person by Age and Gender (continued) Age Women 1983 1990 1995 2001 2009 2009 MOE 2017 Orig. 2017 Orig. MOE 2017 Adj. 2017 Adj. MOE TOTAL 22.6 32.1 33.8 35.7 31.5 1.0 32.8 1.8 35.6 2.0 Under 16 15.4 19.9 26.2 24.4 23.3 2.7 20.2 1.7 22.1 1.8 16 to 20 21.5 32.2 35 42.5 31.0 2.8 28.8 2.7 31.3 2.8 21 to 35 29.5 40.7 40.8 41.5 35.0 2.3 39.8 6.8 43.2 7.4 36 to 65 25.2 37 37.5 40.4 37.0 1.6 36.4 1.5 39.5 1.6 Over 65 10.2 15.3 19.2 23.5 19.3 1.2 27.2 2.9 29.5 2.9 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. 2017 National Household Travel Survey PERSON TRAVEL 52 Overall, the unadjusted estimate of person miles per day in 2017 was 36.1 miles on average, nominally the same as the 2009 estimate. These miles are reported for all means of transportation and for all purposes and include people who traveled and those who did not. In 2017 (Figure 5), the unadjusted estimate for average daily miles for men was 39.5 miles per day, for women the estimate was 32.8 miles per day. These were statistically the same as the estimates in 2009 (within the margin of error). The adjusted estimates are higher for both men and women than the 2009 estimates. The adjusted estimates were 42.5 miles per day for men and 35.6 miles for women. See Appendix A for more details. Figure 5. Average Daily Person Miles of Travel by Gender, 1983, 1990, 1995 NPTS and 2001, 2009, and 2017 NHTS Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. 0 5 10 15 20 25 30 35 40 45 50 1983 1990 1995 2001 2009 2017 Orig 2017 Adj Daily Person Miles of Travel per Person Men Women 2017 National Household Travel Survey PERSON TRAVEL 53 The overall trends in person miles of travel (Figure 6) are not as significant as the changes in trip-making. The original estimate of person miles was exactly the same as the estimate in 2009 (36.1 miles per day), while the adjusted estimate is exactly the same as the 1995 estimate (38.7 miles per day). A notable trend is the increase in travel by people aged 65 and older. The 2017 estimates of daily miles of travel are higher than all previous surveys. For every other age group shown, the 2017 original estimate of person miles per person is within the margin of error of estimates from the earlier surveys. Figure 6. Average Daily Person Miles of Travel by Age Group 1995 NPTS and 2001, 2009, and 2017 NHTS Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Total Under 16 16 to 20 21 to 35 36 to 65 Over 65 Daily Person Miles of Travel per Person 1983 1990 1995 2001 2009 2017 Orig. 2017 Adj. 2017 National Household Travel Survey PERSON TRAVEL 54 Including people who drive and those who are passengers in vehicles, the average American in 2017 spends just under 1 hour a day in a vehicle—58.6 minutes per capita—as a driver or passenger (Figure 7). This estimate is 4 percent lower (2.7 minutes) compared to the 2009 estimate, and the difference is statistically significant. People in their prime working and commuting years, ages 36-55, spend the most amount of time in a vehicle while children under the age of 16 spend the least amount of time in a vehicle. In the 2017 NHTS, only people aged 16-20 have a significant decrease in time spent in a vehicle as a passenger or driver. All other age groups have estimates that fall within the margin of error. 2017 National Household Travel Survey PERSON TRAVEL 55 Figure 7. Trends in the Time Spent in a Vehicle by Age Group (Minutes per Day) Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The 2017 estimates of vehicle trip length have an adjusted value to account for different methods in trip length reporting, see Appendix A. 0 10 20 30 40 50 60 70 80 90 Total 5-16 16-20 21-35 36-55 56-65 Over 65 Minutes per Day in a Vehicle as Driver or Passenger 1995 2001 2009 2017 2017 National Household Travel Survey PRIVATE VEHICLE TRAVEL 56 5.0 PRIVATE VEHICLE TRAVEL In Table 15, researchers calculated the average amount of time spent driving using two different methods: (1) by including all drivers, even those who did not drive a private vehicle on the designated travel day, and (2) by excluding any drivers who did not drive on the designated travel day. In 2017, while the nominal estimates were slightly lower than 2009, they were significantly lower than the 2001 estimates. That is, the estimate of the time spent driving for all drivers (including those who drove and those who did not) did not change between 2009 and 2017 (were within the margin of error); the 2017 estimate was significantly lower than the 2001 estimate. However, looking at people who reported driving on the travel day, the estimate of time spent driving was significantly higher in 2017 compared to 2009. The increase in reported time driving on travel day was notably higher for drivers in metro areas of 1-3 million in population. Table 15. Trends in the Average Time Spent Driving a Private Vehicle in a Typical Day by MSA Size (minutes) MSA Size All Drivers 1990 1995 2001 2009 2009 MOE 2017 2017 MOE ALL 49.35 56.28 62.32 56.09 0.71 55.62 0.80 Rural, Not in MSA 48.85 56.47 61.83 55.87 1.80 54.08 1.15 < 250,000 48.36 53.98 60.22 55.01 4.02 52.45 1.36 250,000 to 499,999 47.82 55.96 59.63 54.79 2.68 52.49 3.11 500,000 to 999,999 50.20 56.91 62.59 55.21 2.36 55.07 1.42 1 million to 2.9 million 50.61 56.48 62.89 56.20 1.76 58.37 1.73 3 million+ 49.38 56.49 63.29 56.85 1.15 56.49 1.01 MSA Size Only Persons Who Drove 1990 1995 2001 2009 2009 MOE 2017 2017 MOE ALL 71.88 73.24 81.35 76.37 0.87 78.91 0.90 Rural, Not in MSA 69.20 72.96 81.74 76.28 2.13 78.45 2.14 < 250,000 67.94 69.35 76.40 73.30 4.75 72.69 1.79 250,000 to 499,999 71.66 71.72 76.50 72.55 3.42 72.94 3.33 500,000 to 999,999 72.42 73.35 79.34 73.57 2.86 76.55 1.62 1 million to 2.9 million 74.38 72.19 79.55 73.64 1.96 79.19 1.67 3 million+ 71.08 75.02 85.12 80.48 1.34 83.22 1.49 2017 National Household Travel Survey PRIVATE VEHICLE TRAVEL 57 Figure 8 displays the trends in driving by American households in minutes and miles by MSA size for the 2001, 2009 and 2017 surveys. Figure 8. Average Time Spent Driving and Miles Traveled by MSA Size Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here excludes them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • "Rural, Not in MSA" includes only full counties designated as rural. There may also be rural pockets included within MSA boundaries. 0 10 20 30 40 50 60 70 Rural, Not in a MSA < 250M 250K - < 500K 500K - < 1M 1M - < 3M 3M+ Average Minutes or Miles 2001 Minutes 2001 Miles 2009 Minutes 2009 Miles 2017 Minutes 2017 Miles 2017 National Household Travel Survey PRIVATE VEHICLE TRAVEL 58 Since about 1990, the vehicle occupancy estimates, measured as person miles per vehicle mile, seems to have stayed about the same (Table 16). While there are small nominal differences between the 2017 and earlier estimates, these differences are all within the margins of error. Table 16. Average Vehicle Occupancy for Selected Trip Purposes (Person Mile per Vehicle Mile) Survey Year Trip Purpose To / From Work Shopping Other Family / Personal Errands Social / Recreation All Purposes 1977 1.30 2.10 2.00 2.40 1.90 1983 1.29 1.79 1.81 2.12 1.75 1990 1.14 1.71 1.84 2.08 1.64 1995 1.14 1.74 1.78 2.04 1.59 2001 1.14 1.79 1.83 2.03 1.63 2009 1.13 1.78 1.84 2.20 1.67 2009 MOE 0.05 0.78 0.84 1.20 0.67 2017 1.18 1.82 1.82 2.10 1.67 2017 MOE 0.01 0.05 0.13 0.04 0.04 Note • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • “Other Family/Personal Errands” includes trips such as to the post office, dry cleaners, or library. • All Purposes includes other trip purposes not shown, such as trips to school, church, doctor, dentist, and work-related business trips. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 59 6.0 VEHICLE USE AND AVAILABILITY As displayed in Table 17, two thirds of the households in the United States have one or two vehicles available, according to the 2017 NHTS. Statistically, the number of households with zero vehicles or two vehicles remained about the same. On the other hand, the number of households with one vehicle and three or more vehicles were significantly higher in 2017 compared to the 2009 estimates. The estimate of the number of households with three or more vehicles rose significantly between 2009 and 2017, from 25.7 million households to 28.9 million households in 2017. Table 17. Trends in the Number and Percent of Households by Availability of Household Vehicles (Thousands) Survey Year No Vehicle One Vehicle Two Vehicles Three or More Vehicles ALL Vehicles Per Household 1969 12,876 30,252 16,501 2,875 62,504 1.16 1977 11,538 26,092 25,942 11,840 75,412 1.59 1983 11,548 28,780 28,632 16,411 85,371 1.68 1990 8,573 30,654 35,872 18,248 93,347 1.77 1995 7,989 32,064 40,024 18,914 98,990 1.78 2001 8,716 33,757 39,938 24,955 107,365 1.89 2009 9,828 36,509 41,077 25,688 113,101 1.86 2009 MOE 49 302 274 270 0 0.01 2017 10,567 39,648 39,125 28,869 118,208 1.88 2017 MOE 0 0 272 272 0 0.01 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 60 Table 17. Trends in the Number and Percent of Households by Availability of Household Vehicles (Thousands) (continued) Percent No Vehicle One Vehicle Two Vehicles Three or More Vehicles ALL 1969 20.6% 48.4% 26.4% 4.6% 100% 1977 15.3% 34.6% 34.4% 15.7% 100% 1983 13.5% 33.7% 33.5% 19.2% 100% 1990 9.2% 32.8% 38.4% 19.6% 100% 1995 8.1% 32.4% 40.4% 19.1% 100% 2001 8.1% 31.4% 37.2% 23.2% 100% 2009 8.7% 32.3% 36.3% 22.7% 100% 2017 8.9% 33.5% 33.1% 24.4% 100% Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. • Standard error of the estimate is too small to show. • No Vehicle and One Vehicle categories were used as controls in calibrating the weights according to the weighting plan and should have nearly no variance in the replicate weights, resulting in standard errors close to 0. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 61 Out of the 120 million households in the United States, about 10.5 million are without a vehicle, according to the 2017 NHTS (Figure 9). The number of households with zero vehicles available remained statistically the same in 2017 (within the margin of error of the 2009 estimate). On the other hand, since 1969 the number of households that owned three or more vehicles has grown by tenfold—from 2.9 million to nearly 29 million. The percentage of households with three or more vehicles has gone from 5 percent to nearly a quarter of all U.S. households. Figure 9. Household Distribution by Number of Household Vehicles Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more and CPO urban households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 0 20,000 40,000 60,000 80,000 100,000 120,000 140,000 1969 1977 1983 1990 1995 2001 2009 2017 Number of Households (thousands) Three or More Vehicles Two Vehicles One Vehicle No Vehicle 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 62 Table 18 shows the traditional correlation between high population density and the percentage of households with fewer or no vehicles in the NHTS data series. Over a quarter (26.8%) of the households in areas with a population density greater than 10,000 per square mile did not own a vehicle in 2017 and 30.7 percent owned two or more vehicles. On the other hand, only 4.3 percent of the households in the least densely populated areas did not own a vehicle in 2017 and almost 70 percent (68.3%) owned two or more vehicles. Table 18. Trends in the Distribution of Households by Household Vehicle Availability and Population Density Population Density Survey Year Household Vehicle Availability ALL No Vehicle One Vehicle Two or more Vehicles Less than 2,000 People per Square Mile 1990 100.0% 6.1% 30.4% 63.5% 1995 100.0% 3.6% 26.6% 69.8% 2001 100.0% 3.8% 25.8% 70.5% 2009 100.0% 4.4% 26.8% 68.8% 2009 MOE - 0.41 0.76 0.89 2017 100.0% 4.3% 27.4% 68.3% 2017 MOE - 0.31 0.42 0.56 2,000 to 4,000 People per Square Mile 1990 100.0% 7.6% 33.4% 59.0% 1995 100.0% 5.8% 33.3% 60.8% 2001 100.0% 5.8% 32.8% 61.4% 2009 100.0% 6.4% 34.1% 59.5% 2009 MOE - 0.84 1.47 1.66 2017 100.0% 6.7% 35.6% 57.7% 2017 MOE - 0.61 0.86 0.89 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 63 Table 18. Trends in the Distribution of Households by Household Vehicle Availability and Population Density (continued) Population Density Survey Year Household Vehicle Availability ALL No Vehicle One Vehicle Two or more Vehicles 4,000 to 10,000 People per Square Mile 1990 100.0% 10.9% 38.2% 50.9% 1995 100.0% 7.7% 37.2% 55.1% 2001 100.0% 8.1% 36.3% 55.6% 2009 100.0% 8.4% 37.5% 54.1% 2009 MOE - 0.73 1.36 1.34 2017 100.0% 9.3% 38.1% 52.7% 2017 MOE - 0.79 1.23 1.39 10,000 or more People per Square Mile 1990 100.0% 35.1% 40.0% 24.9% 1995 100.0% 27.4% 41.8% 30.8% 2001 100.0% 26.3% 40.3% 33.4% 2009 100.0% 28.4% 39.9% 31.7% 2009 MOE - 1.40 1.68 1.55 2017 100.0% 26.8% 42.5% 30.7% 2017 MOE - 1.13 1.32 1.20 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 64 Overall, most households in the United States—over 51 million or 43.4 percent of all—are in low-density areas with less than 2,000 people per square mile (Figure 10). An equal amount, another 51 million and 43.4 percent of all, are in areas with between 2,000 and 10,000 people per square mile. Only 13.2 percent of households are in very high-density areas of more than 10,000 people per square mile. In these denser urban areas, households are less likely to have two or more vehicles, and more likely to have fewer vehicles. Figure 10. Distribution of the Number of U.S. Households by Vehicle Ownership and Population Density, 2017 NHTS (Millions) Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 0 5 10 15 20 25 30 35 40 < 2,000 pop/sq mi. 2,000-4,000 4,000-10,000 10,000 or more Number of US Households (millions) No Vehicles One Vehicle Two or more vehicles 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 65 Table 19 shows that larger metro areas have higher proportions of households with no vehicles than smaller towns and rural areas. Overall, the proportion of households without a vehicle declined significantly from 1977 to 1995, and then—in some areas—experienced a small shift upward. The proportion of households without a vehicle available overall was 15.3 percent in 1977, and fell to 8.1 percent in 1995 and 2001, rising to 8.7 percent in 2009 and 8.9 percent in 2017. Table 19. Trends in the Percent of Households Without a Vehicle Within MSA Size Group Survey Year Metro Area Size Rural, Not in MSA Less than 250,000 250,000 to 499,999 500,000 to 999,999 1 to 2.9 million 3+ million ALL 1977 12.2% 13.7% 12.2% 14.0% 14.2% 26.1% 15.3% 1983 10.5% 10.1% 8.1% 14.3% 12.1% 25.4% 13.5% 1990 7.7% 8.6% 5.7% 8.4% 8.2% 12.4% 9.2% 1995 5.3% 4.8% 7.3% 6.3% 6.9% 11.2% 8.1% 2001 5.8% 5.8% 5.2% 7.0% 6.4% 11.9% 8.1% 2009 5.6% 6.3% 5.6% 8.3% 7.2% 12.6% 8.7% 2009 MOE 0.14 0.12 0.09 0.12 0.15 0.14 0.04 2017 6.8% 7.0% 5.8% 7.4% 7.4% 12.8% 8.9% 2017 MOE 0.07 0.05 0.05 0.04 0.11 0.12 0.00 Note: • Totals in all tables can include cases that were not included in any table subcategory.. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. • "Rural, Not in MSA" includes only full counties designated as rural. There may also be rural pockets included within MSA boundaries. • The population size groups for 1977 - 1983 NPTS are MSA Size Groups. 1990 - 2001 are MSA Size Groups. 2009 - 2017 are CMSA size groups. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 66 Table 20 shows vehicle in the household-based fleet by vehicle type and age. It shows how much the average vehicle has aged over the last decades. Figure 11 shows these trends in a pictorial format. The share of vans in the household vehicle fleet declined again in 2017—the percentage of vehicles classified as vans in 2017 (6.1%) was lower than the 2009 estimate (7.8%). On the other hand, the percentage of vehicles classified as SUVs continued to increase—as they have since the survey included a category for them in 1995. From just under 7 percent of all vehicles in 1995, SUVs grew to almost a quarter (23.7%) of all household vehicles in 2017. Continuing a long-standing trend, the household vehicle fleet continues to age. The most recent data shows the average vehicle owned by U.S. households is 10.3 years old, about 1 year older than the estimate in 2009. Auto, Van, SUV, and Pickups were significantly older in 2017 compared to the age estimate in 2009, and each of these vehicle types were significantly older in 2009 compared to 2001. Over the last 4 decades the U.S. fleet has aged almost 4 years—the average vehicle in the household fleet was 6.6 years old in 1977, compared to 10.27 years old in 2017. Table 20. Household-Based Vehicle Distribution and Average Vehicle Age by Vehicle Type Distribution of Vehicles by Vehicle Type Category 1977 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE TOTAL 100% 100% 100% 100% 100% 100% 0.00 100% 0.00 Auto 79.6% 75.9% 74.7% 64.3% 56.8% 49.9% 0.45 49.5% 0.44 Van 2.8% 3.6% 5.5% 7.8% 9.0% 8.2% 0.28 6.1% 0.28 Sport Utility NA NA NA 6.9% 12.1% 19.4% 0.35 23.7% 0.46 Pickup 12.8% 15.2% 17.2% 17.7% 18.4% 17.8% 0.29 15.9% 0.21 Other Truck 1.3% 1.5% 0.6% 0.4% 0.5% 0.4% 0.08 0.5% 0.10 RV/Motor Home 0.4% 0.5% 0.5% 0.5% 0.7% 0.5% 0.06 0.6% 0.07 Motorcycle/Moped 2.9% 3.1% 1.4% 0.9% 2.1% 3.3% 0.24 3.3% 0.14 Other 0.2% 0.2% 0.1% 0.1% 0.5% 0.3% 0.05 0.4% 0.04 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 67 Table 20. Household-Based Vehicle Distribution and Average Vehicle Age by Vehicle Type (continued) Average Vehicle Age Category 1977 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE All 6.60 7.60 7.71 8.33 8.87 9.38 0.10 10.27 0.12 Auto 6.40 7.20 7.61 8.24 8.98 9.57 0.11 10.10 0.18 Van 5.50 8.45 5.88 6.68 7.56 8.68 0.18 10.65 0.27 Sport Utility NA NA NA 6.56 6.44 7.09 0.15 8.34 0.13 Pickup 7.30 8.54 8.43 9.65 10.05 11.10 0.21 13.12 0.17 Other Truck 11.60 12.39 14.48 14.93 17.72 17.76 1.74 17.29 1.04 RV/Motor Home 4.50 10.69 10.44 13.21 13.49 15.46 1.47 15.77 1.29 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. • Totals do not include any unreported vehicle ages, but do include vehicle types such as motorcycle, RV, etc. that are not shown. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 68 Figure 11. Trends in the Number of Household-Based Vehicles by Type (Millions) Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 1977 1983 1990 1995 2001 2009 2017 Motorcycle / Moped Pickup Van SUV Auto 120 144 165 176 201 211 223 N=Total Household Vehicles (millions) 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 69 Over the last 4 decades, a striking feature of the household vehicle fleet is the increase in the number of years an average vehicle is operated (Table 21). In 1977, automobiles averaged 6.4 years of age while automobiles in 2017 averaged 10.1 years of age—an increase of 3.7 years on average. In 1995 (the first year SUVs were separately catalogued in the NHTS), Vans/SUV/Pickup Trucks were 8.3 years old on average. By 2017, they averaged 10.4 years—more than 2 years older. As a result, of the aging fleet, many older cars are in daily use. In 1977, about one out of six vehicles was 10 years old or older; by 2017, nearly half (48.5%) of the household-based fleet was 10 years old or more. Table 21. Trends in the Distribution of Household-Based Vehicles by Vehicle Age and Vehicle Type (Percent) Survey Year Vehicle Type Vehicle Age: 0 to 2 years 3 to 5 years 6 to 9 years 10 or more Total Average Age 1977 Auto 27.3% 30.4% 26.7% 15.6% 100.0% 6.4 Van/Pickup 29.9% 25.6% 21.1% 23.4% 100.0% 5.6 ALL 27.8% 29.6% 25.7% 16.9% 100.0% 6.6 1983 Auto 20.0% 28.0% 27.4% 24.6% 100.0% 7.2 Van/Pickup 16.6% 26.6% 25.0% 31.8% 100.0% 7.8 ALL 19.2% 27.6% 26.9% 26.3% 100.0% 7.6 1990 Auto 15.6% 27.7% 26.8% 29.9% 100.0% 7.6 Van/Pickup 19.7% 27.2% 20.9% 32.2% 100.0% 8.0 ALL 16.6% 27.5% 25.3% 30.6% 100.0% 7.7 1995 Auto 14.9% 21.7% 30.3% 33.1% 100.0% 8.2 Van/SUV/Pickup 19.2% 21.6% 25.5% 33.7% 100.0% 8.3 ALL 16.2% 21.5% 28.5% 33.8% 100.0% 8.3 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 70 Table 21. Trends in the Distribution of Household-Based Vehicles by Vehicle Age and Vehicle Type (Percent) (continued) Survey Year Vehicle Type Vehicle Age: 0 to 2 years 3 to 5 years 6 to 9 years 10 or more Total Average Age 2001 Auto 13.3% 20.4% 25.5% 40.9% 100.0% 9.0 Van/SUV/Pickup 18.6% 23.5% 22.6% 35.4% 100.0% 8.5 ALL 15.4% 21.5% 24.1% 39.0% 100.0% 8.9 2009 Auto 12.4% 19.7% 27.0% 40.9% 100.0% 9.6 Van/SUV/Pickup 12.8% 23.6% 27.1% 36.6% 100.0% 9.0 ALL 12.7% 21.6% 26.8% 38.9% 100.0% 9.4 2009 MOE Auto 0.49% 0.58% 0.70% 0.74% 0.00% 0.11 Van/SUV/Pickup 0.49% 0.60% 0.69% 0.66% 0.00% 0.11 ALL 0.36% 0.42% 0.49% 0.54% 0.00% 0.10 2017 Auto 12.2% 20.5% 20.8% 46.6% 100.0% 10.1 Van/SUV/Pickup 14.5% 17.5% 18.0% 50.0% 100.0% 10.4 ALL 13.2% 18.9% 19.4% 48.5% 100.0% 10.3 2017 MOE Auto 0.39% 0.60% 0.57% 0.91% 0.00% 0.18 Van/SUV/Pickup 0.50% 0.41% 0.74% 0.55% 0.00% 0.09 ALL 0.32% 0.45% 0.49% 0.62% 0.00% 0.12 Note: • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 71 Figure 12 shows that after cars, SUVs appear to be the most popular vehicle type among newer vehicles, according to the 2017 NHTS. Figure 12. Distribution of Household-Based Vehicles Two Years old or Newer by Vehicle Type (Percent) Note: • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. Van 4% Pickup 12% Other 2% Motorcycle 2% SUV 34% Car 46% 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 72 Based on vehicle owners’ estimates, an average U.S. vehicle was driven slightly more than 10,000 miles a year in 2017, statistically the same as in 2009 (Table 22). Overall, average miles per vehicle (from the owner's estimate) seems to have peaked in the 1990s. In the 2017 survey, it is lower than the estimates in 2001 for all vehicles in all age categories. Table 22. Trends in the Average Annual Miles per Vehicle by Vehicle Age (Vehicle Owner's Estimate) Survey Year Vehicle Age 0 to 2 years 3 to 5 years 6 to 9 years 10 or more years ALL 1969 15,700 11,200 9,700 6,500 11,600 1977 14,460 11,074 9,199 6,755 10,679 1983 15,292 11,902 9,253 7,023 10,315 1990 16,811 13,706 12,554 9,176 12,458 1995 16,092 14,004 12,608 8,758 12,226 2001 14,892 13,230 11,603 7,863 11,078 2009 13,851 12,042 10,741 7,401 10,088 2009 MOE 533 198 280 160 133 2017 13,065 12,582 11,432 7,812 10,164 2017 MOE 372 621 349 214 131 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 73 The annual miles shown in Table 23a and 23b are based on the driver's estimate of how many miles he or she drives (in all vehicles) in a year. Like other measures of vehicle travel, these estimates have also decreased significantly between 2009 and 2017. Drivers aged 20 to 54 estimated that in a year they drove significantly fewer miles than comparable age groups in 2009. The decrease in annual miles estimated by men drivers was significant for 20 to 54-year-olds, but not drivers 16-19 or those over 55. Women driver's estimates were statistically the same as in 2009 in all age groups (although the nominal estimate was lower in every age group). Table 23a. Trends in the Average Annual Miles per Licensed Driver-by-Driver Age (Self-Estimate) Survey Year Drivers 16 to 19 20 to 34 35 to 54 55 to 64 65+ ALL 1969 4,633 9,348 9,771 8,611 5,171 8,685 1977 5,662 11,063 11,539 9,196 5,475 10,006 1983 4,986 11,531 12,627 9,611 5,386 10,536 1990 8,485 14,776 14,836 11,436 7,084 13,125 1995 7,624 15,098 15,291 11,972 7,646 13,476 2001 7,331 15,650 15,627 13,177 7,684 13,827 2009 6,244 13,709 15,117 12,528 8,250 12,888 2009 MOE 540 615 321 387 346 204 2017 5,561 12,187 13,806 12,095 8,218 11,621 2017 MOE 383 466 294 267 223 169 2017 National Household Travel Survey VEHICLE USE AND AVAILABILITY 74 Table 23b. Trends in the Average Annual Miles per Licensed Driver-by-Driver Age and Gender (Self-Estimate) Survey Year Male Drivers 16 to 19 20 to 34 35 to 54 55 to 64 65+ ALL 1969 5,461 13,133 12,841 10,696 5,919 11,352 1977 7,045 15,222 16,097 12,455 6,795 13,397 1983 5,908 15,844 17,808 13,431 7,198 13,962 1990 9,543 18,310 18,871 15,224 9,162 16,536 1995 8,206 17,976 18,858 15,859 10,304 16,550 2001 8,228 18,634 19,287 16,883 10,163 16,946 2009 6,652 15,716 17,654 15,117 10,322 15,139 2009 MOE 633 1041 450 555 324 328 2017 5,893 13,291 15,705 14,717 9,974 13,393 2017 MOE 796 583 437 525 253 228 Survey Year Female Drivers 16 to 19 20 to 34 35 to 54 55 to 64 65+ ALL 1969 3,586 5,512 6,003 5,375 3,664 5,411 1977 4,036 6,571 6,534 5,097 3,572 5,940 1983 3,874 7,121 7,347 5,432 3,308 6,382 1990 7,387 11,174 10,539 7,211 4,750 9,528 1995 6,873 12,004 11,464 7,780 4,785 10,142 2001 6,106 12,266 11,590 8,795 4,803 10,267 2009 5,753 11,484 12,035 9,544 5,824 10,244 2009 MOE 881 472 381 407 646 213 2017 5,104 11,026 11,895 9,434 6,373 9,854 2017 MOE 610 562 389 200 237 241 Note: • Totals in all tables can include cases that were not included in any table subcategory. • In 1995, some drivers reported zero annual miles. These were changed to miles not reported. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1969, household vehicles did not include pickups or other light trucks. • SUVs were added as a vehicle class in the NHTS survey in 1995. • In 2009 the survey included Light Electric Vehicles (LEV) as a separate classification. • Motorcycle, moped, LEVs and “other” POV are excluded from the calculation of vehicle age. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 75 7.0 COMMUTE TRAVEL PATTERNS Table 24 shows that the estimate of the number of vehicle trips to and from work is about the same in 2017 compared to that of 2009 (within the margin of error). Although the estimate of total vehicle miles for commuting is nominally higher in 2017 compared to 2009, the differences are not significant. The total number of estimated workers has increased, while the annual commute vehicle trips per worker has remained virtually the same over many survey iterations, excepting the 1995 NPTS. Table 24. Trends in Commute Trips and Vehicle Miles in Commute Survey Year Commute Vehicle Trips (millions) Commute VMT (millions) Total VMT (millions) % Commute VMT of Total VMT Workers (thousands) Annual Commute Vehicle Trips per Worker 1969 27,844 260,716 775,940 33.60% 75,758 368 1977 31,886 287,710 907,603 31.70% 93,019 343 1983 35,271 301,644 1,002,139 30.10% 103,244 342 1990 41,792 453,042 1,695,290 26.72% 118,343 353 1995 54,782 642,610 2,068,368 31.07% 131,697 416 2001 51,395 614,548 2,274,797 27.02% 145,272 354 2009 51,699 623,479 2,245,112 27.77% 151,373 342 2009 MOE 897 16,794 56,158 - 893 - 2017 Orig. 53,154 635,792 2,105,882 30.19% 156,988 339 2017 Orig. MOE 1,131 22,741 88,132 - 1,012 - 2017 Adj. - 682,548 2,321,820 28.07% - - 2017 Adj. MOE - 24,399 98,080 - - - Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • Trip miles and travel times were calculated using actual trips to and from work as reported in the travel day file. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 76 • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. • Unlike the Census Journey-to-Work data, the NHTS does not include “work at home” in usual commute data. • “Other” includes travel modes not specifically cited, such as motorcycle, taxi, bike, truck, and other. Across many decades, the vast majority of workers have traveled to work in a privately-owned vehicle. However, in the 2017 NHTS the estimate of workers commuting by private vehicle is significantly lower (87.5% of workers) than the 2009 estimate (89.4% of workers) (Figure 13). Table 25 shows that the 2017 NHTS estimates 6.9 percent of workers use public transit as their usual means of travel to work, a significant increase from 2009 and previous estimates. Figure 13. Trends in the Distribution of Workers by Usual Commute Mode (Percent of Workers) Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Trip miles and travel times were calculated using actual trips to and from work as reported in the travel day file. • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. • Unlike the Census Journey-to-Work data, the NHTS does not include “work at home” in usual commute data. • “Other” includes travel modes not specifically cited, such as motorcycle, taxi, bike, truck, and other. • Public transit includes local bus, commuter bus, commuter train, subway, trolley, and streetcar. 0 10 20 30 40 50 60 70 80 90 100 1969 1977 1983 1990 1995 2001 2009 2017 Percent of Workers Private Vehicle Public Transit Walk Other 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 77 Table 25. Trends in the Distribution of Workers by Usual Commute Mode (Percent of Workers) Survey Year All Modes Private Vehicle Public Transit Walk Other 1969 100% 90.8 8.4 N/A 0.8 1977 100% 87.0 6.0 4.1 2.9 1983 100% 88.6 5.3 4.3 1.8 1990 100% 87.8 5.3 4.0 2.9 1995 100% 91.0 5.1 2.6 1.3 2001 100% 90.8 5.1 2.8 1.3 2009 100% 89.4 5.1 2.8 2.7 2009 MOE 0.52 0.41 0.34 0.25 2017 100% 87.5 6.9 2.9 2.7 2017 MOE 0.53 0.32 0.34 0.29 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. • Unlike the Census Journey-to-Work data, the NHTS does not include “work at home” in usual commute data. • “Other” includes travel modes not specifically cited, such as motorcycle, taxi, bike, truck, and other. • Public transit includes local bus, commuter bus, commuter train, subway, trolley, and streetcar. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 78 Interestingly, when comparing the report by the same respondents of how they “usually” commute and how they actually travelled to work on the travel day, some important differences emerge. For example, as shown in Table 26, driving alone has the highest mode loyalty—86.2 percent of workers who say they usually drive alone do so on the travel day. About 70 percent of commuters who usually travel by transit, walk, or bike report doing so on their travel day. When they do not use their usual mode, they are most likely to share a ride in a private auto. The percentage of workers on their assigned travel day who share a ride to work (including family members riding together) is 18.8 percent compared to the “usual” estimate of 11.0 percent. “Shared ride” does not include ride-hailing (such as Uber/Lyft, which is classified with “taxi” in the 2017 NHTS and would be in “Other”). The table does not show “Other” modes and excludes workers who did not report both a usual and actual mode to work (15% of all). Table 26. Usual Commute Mode to Work vs Actual Commute Mode on Travel Day 'Usual' Commute Mode On Travel Day Commuted by: Drove Alone Shared Ride Transit Walk Bike Usual Mode Share: Drove Alone 86.2% 12.8% 0.2% 0.6% 0.1% 76.2% Shared Ride 37.2% 60.2% 1.0% 1.2% 0.2% 11.0% Transit 4.8% 14.4% 70.8% 7.0% 0.8% 6.9% Walk 7.3% 18.2% 2.6% 69.8% 0.9% 2.9% Bike 8.1% 11.9% 3.4% 4.6% 70.3% 1.1% Actual Mode Share 71.0% 18.8% 5.2% 3.3% 1.0% Note: • Totals in all tables can include cases that were not included in any table subcategory. • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. • Table does not show “Other” modes of travel. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 79 Table 27 displays trends in average trip lengths, travel time, and speed for different modes of transportation. Table 27. Trends in General Commute Patterns by Mode of Transportation Survey Year All Modes Average Commute Trip Length (miles) Average Commute Travel Time (minutes) Average Commute Speed (miles per hour) 1977 9.06 19.23 34.72 1983 8.54 18.20 26.84 1990 10.65 19.60 33.35 1995 11.63 20.65 34.67 2001 12.11 23.32 32.23 2009 11.79 23.85 27.50 2009 MOE 0.29 0.35 0.33 2017 Orig. 11.46 26.58 23.42 2017 Orig. MOE 0.34 0.56 0.28 2017 Adj. 12.22 26.58 25.06 2017 Adj. MOE 0.36 0.56 0.29 Survey Year Private Vehicle Average Commute Trip Length (miles) Average Commute Travel Time (minutes) Average Commute Speed (miles per hour) 1977 9.61 18.95 37.50 1983 8.86 17.62 27.78 1990 11.02 19.05 31.49 1995 11.84 20.10 35.18 2001 12.10 22.49 32.27 2009 12.09 22.85 28.87 2009 MOE 0.25 0.34 0.31 2017 Orig. 11.84 25.01 25.22 2017 Orig. MOE 0.38 0.56 0.33 2017 Adj. 12.71 25.01 27.08 2017 Adj. MOE 0.41 0.56 0.35 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 80 Table 27. Trends in General Commute Patterns by Mode of Transportation (continued) Survey Year Public Transit Average Commute Trip Length (miles) Average Commute Travel Time (minutes) Average Commute Speed (miles per hour) 1977 7.48 37.59 12.58 1983 9.00 37.79 15.44 1990 12.75 41.10 18.02 1995 12.88 41.95 18.22 2001 11.73 55.50 12.96 2009 10.18 52.98 11.42 2009 MOE 1.54 4.19 0.99 2017 Orig. 12.09 58.11 11.63 2017 Orig. MOE 1.15 2.06 0.73 Survey Year Walk Average Commute Trip Length (miles) Average Commute Travel Time (minutes) Average Commute Speed (miles per hour) 1977 - - - 1983 - - - 1990 0.83 9.79 4.99 1995 0.74 10.86 3.58 2001 0.91 14.06 3.18 2009 0.98 16.15 4.77 2009 MOE 0.23 2.28 0.51 2017 Orig. 1.19 15.26 3.15 2017 Orig. MOE 0.73 1.59 0.18 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Trip miles and travel times were calculated using actual trips to and from work as reported in the travel day file. • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 81 • Average commute speed was calculated using only those trips with both trip mileage and travel time information present. • Average commute trip length was calculated using only those records with trip mileage information present. • Commute time for public transit includes total trip time, including access and egress. Wait time is not included. • Unlike the Census Journey-to-Work data, the NHTS does not include “work at home” in usual commute data. • Public transit includes local bus, commuter bus, commuter train, subway, trolley, and streetcar. 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 82 Table 28 shows the trends in the average speed of commutes in areas of different population sizes. On average, larger metro areas have slower speeds—both as a result of more congestion, but also more workers commuting by non-auto means of travel, like transit and walking. Table 28. Trends in Average Commute Speed by MSA Size (Miles per Hour) 1977, 1983, 1990, 1995 NPTS, and 2001, 2009, and 2017 NHTS MSA Size Rural, Not in MSA Less than 250,000 250,000 to 499,999 500,000 to 999,999 1 to 2.9 million 3 million and over All Modes (Including Private Vehicle) 1977 - 25.8 26.5 26.5 27.5 20.0 1983 28.9 25.6 26.3 27.3 27.4 24.8 1990 32.0 29.7 30.4 31.4 30.2 27.7 1995 31.2 28.9 30.0 30.4 29.9 28.4 2001 31.9 28.5 28.3 28.8 27.9 25.4 2009 31.6 27.6 27.6 28.1 27.8 24.7 2009 MOE 0.8 0.8 0.2 0.9 0.7 0.5 2017 Orig. 27.6 24.8 25.3 24.5 23.8 20.5 2017 Orig. MOE 0.7 0.9 1.3 0.6 0.6 0.3 2017 Adj. 29.6 26.6 27.1 26.2 25.4 21.9 2017 Adj. MOE 0.7 0.9 1.4 0.6 0.6 0.4 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2009 NHTS sample did not include households without landlines telephones (CPO households). 2017 National Household Travel Survey COMMUTE TRAVEL PATTERNS 83 • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • Trip miles and travel times were calculated using actual trips to and from work as reported in the travel day file. • The usual mode is defined as the means of transportation usually used to go to work in the week prior to the travel day. • Average commute speed was calculated using only those trips with both trip mileage and travel time information present. • Average commute trip length was calculated using only those records with trip mileage information present. • Commute time for public transit includes total trip time, including access and egress. Wait time is not included. • Unlike the Census Journey-to-Work data, the NHTS does not include “work at home” in usual commute data. • "Rural, Not in MSA" includes only full counties designated as rural. There may also be rural pockets included within MSA boundaries. • The population size groups for 1977 - 1983 NPTS are MSA Size Groups. 1990 - 2001 are MSA Size Groups. 2009 - 2017 are CMSA size groups. 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 84 Figure 14 shows that the average speed of commuting by all modes has declined in all metro areas, regardless of size. Since 1990, the largest metro areas have seen the greatest decline in commute speed. As mentioned earlier, trip distance was collected differently in the 2017 NHTS, which affects the trends in speed (see Appendix A). The 2017 adjusted values show higher speeds because the trip distance was adjusted to be more comparable to earlier surveys, while the reported time remained the same. Figure 14. Trends in Average Commute Speeds by MSA Size (All Modes) Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Trip miles and travel times were calculated using actual trips to and from work as reported in the travel day file. 0 5 10 15 20 25 30 35 Rural, Not in an MSA < 250K < 500K < 1M 1M- < 3M 3M+ Miles Per Hour for Commutes 1977 1983 1990 1995 2001 2009 2017 2017 Adj 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 85 • Average commute speed was calculated using only those trips with both trip mileage and travel time information present. • Average commute trip length was calculated using only those records with trip mileage information present. • Commute time for public transit includes total trip time, including access and egress. Wait time is not included. 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 86 8.0 TEMPORAL DISTRIBUTION Table 29 shows the percentage of person trips by time of day. The 2017 data shows a notable increase in the percentage of trips during the morning peak period (6-9 am). However, the distribution of trips by time of day has remained about the same for many decades. The 2017 survey data shows that almost half (47%) of all person trips start in the midday between 9 a.m. and 4 p.m., virtually the same as the estimates since 1995. Table 29. Trends in the Distribution of Person Trips by Start Time of Trip Time of Day 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE 10 p.m. - 1 a.m. 4.0 4.1 3.5 2.9 2.6 0.13 2.3 0.07 1 a.m. - 6 a.m. 3.3 1.8 1.7 1.8 1.8 0.08 1.9 0.12 6 a.m. - 9 a.m. 14.4 12.5 13.8 14.4 15.0 0.21 16.6 0.21 9 a.m. - 1 p.m. 23.4 20.6 24.2 24.6 24.8 0.29 25.4 0.35 1 p.m. - 4 p.m. 20.8 20.7 22.1 22.1 22.4 0.34 22.1 0.33 4 p.m. - 7 p.m. 21.2 22.9 23.0 22.3 22.6 0.29 22.1 0.26 7 p.m.- 10 p.m. 12.3 13.2 11.8 11.7 11.0 0.23 9.8 0.24 ALL 100.0 100.0 100.0 100.0 100.0 - 100.0 - Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 87 Figure 15 shows vehicle trips by time of day and purpose. The data show that the morning and evening peak periods include not just commutes, but shopping and family errands (which includes dropping children at school), and other non-work trips. These vehicle trips add to the total number of vehicles traveling during the peak periods. As expected, in 2017 most vehicle commutes started between 6 a.m. and 9 a.m. in the morning and between 4 p.m. and 7 p.m. More than half of vehicle trips for other purposes started between 9 a.m. and 4 p.m. Figure 15. Distribution of Vehicle Trips by Trip Purpose and Start Time of Trip, 2017 NHTS Note • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 0.0% 1.0% 2.0% 3.0% 4.0% 5.0% 6.0% 7.0% 8.0% 9.0% Midnight 1 2 3 4 5 6 7 8 9 10 11Noon 1 2 3 4 5 6 7 8 9 10 11 Percent of Vehicle Trips Start Hour Commute Shopping/Errands School/Church Social/Recreational Total 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 88 Table 30 displays trends for key travel characteristics for weekday and weekend travel. Table 30. Trends in Travel Characteristics for Weekday vs. Weekend Weekday Survey Year Vehicle Trips per Driver Percent Work Trips Percent Non-Work Trips VMT per Driver Average Vehicle Trip Length Average Time Spent Driving (in minutes) Person Trips per Person PMT per Person Average Person Trip Length 1990 3.4 28% 72% 28.5 8.5 50.7 3.8 32.6 9.5 1995 3.8 32% 68% 33.5 8.9 59.5 4.4 37.7 8.6 2001 3.6 31% 69% 34.4 9.8 64.8 4.2 39.4 9.6 2009 3.2 31% 69% 30.6 9.6 59.8 3.9 35.8 9.4 2009 MOE 0.0 0.58 0.58 0.9 0.3 0.8 0.0 1.3 0.3 2017 Orig 2.9 31% 69% 26.9 9.3 59.0 3.5 35.3 10.2 2017 Orig MOE 0.0 0.48 0.48 1.5 0.5 0.9 0.1 2.0 0.6 2017 Adj. - - - 29.6 10.3 - - 38.0 10.9 2017 Adj. MOE - - - 1.6 0.5 - - 2.1 0.6 Weekends Survey Year Vehicle Trips per Driver Percent Work Trips Percent Non-Work Trips VMT per Driver Average Vehicle Trip Length Average Time Spent Driving (in minutes) Person Trips per Person PMT per Person Average Person Trip Length 1990 2.9 10% 90% 28.4 10.0 46.1 3.6 40.6 11.5 1995 3.0 13% 88% 28.9 9.7 48.1 4.0 41.1 10.5 2001 2.9 11% 89% 28.7 10.2 52.4 3.9 42.3 11.2 2009 2.5 10% 90% 25.0 10.0 46.7 3.5 37.1 10.8 2009 MOE 0.1 0.65 0.65 1.1 0.5 1.3 0.1 3.3 1.0 2017 Orig 2.3 11% 89% 23.2 10.3 47.3 3.1 38.1 12.2 2017 Orig MOE 0.0 0.61 0.61 0.8 0.4 0.9 0.0 2.9 1.0 2017 Adj. - - - 25.7 11.4 - - 41.4 13.3 2017 Adj MOE - - - 0.9 0.5 - - 3.0 1.1 2017 National Household Travel Survey TEMPORAL DISTRIBUTION 89 Note • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Average time spent driving includes all drivers, even those who did not drive a private vehicle on the day in which the household was interviewed. • Average trip length is calculated using only those records with trip mileage information present. • "% Work Trips" also includes work-related business. 2017 National Household Travel Survey SPECIAL POPULATIONS 90 9.0 SPECIAL POPULATIONS Table 31 shows that the estimates of travel for people aged 65 and older is a mixed bag: While reported vehicle trips per driver are lower than 2009 estimates, person trips and person miles of travel both show increases for older individuals. On a daily basis, people aged 65 and older took significantly fewer vehicle trips per driver than the same age group in 2009, 2001, and 1995. This estimate includes all people who drive, whether they drove on the travel day or not. The original estimate of miles driven by drivers aged 65 and older in 2017 is statistically the same as in 2009, 2001, and 1995—meaning that there has been virtually no change in the estimates. The adjusted estimate for 2017 is significantly higher than the 2009 estimate. Likewise, the original estimate for the average vehicle- and person-trip length are statistically the same as in 2009, while the adjusted estimate is higher. However, the original and adjusted estimates for the daily PMT are significantly higher in 2017 than in 2009, but statistically the same as the estimate of PMT for people 65 and older in 2001. In addition, the reported number of person trips per person (including those who travel and those who do not) remains exactly the same as the 2009 estimate. Tables 32a, 32b, and 32c display additional characteristics for older persons. 2017 National Household Travel Survey SPECIAL POPULATIONS 91 Table 31. Daily Travel Statistics of People 65 and Older Daily Travel Statistics (65 and Older) 1983 1990 1995 2001 2009 2009 MOE 2017 2017 MOE Vehicle Trips per Driver 1.66 2.27 2.94 2.84 2.67 0.05 2.55 0.04 Percent Work Trips 10.2% 4.8% 8.5% 6.2% 10.6% 0.97 8.6% 0.69 Percent Non-Work Trips 89.8% 95.2% 91.5% 93.8% 89.4% 0.97 91.4% 0.69 VMT per Driver 9.80 14.83 19.56 21.13 19.69 0.75 20.21 1.21 2017 Adjusted VMT per Driver - - - - - - 22.47 1.35 Average Vehicle Trip Length 5.92 6.61 6.69 7.51 7.46 0.29 7.91 0.41 2017 Adjusted Vehicle Trip Length - - - - - - 8.80 0.45 Average Time Spent Driving (in minutes) - 30.83 42.89 49.11 46.37 1.26 48.29 1.48 Person Trips per Person 1.8 2.5 3.4 3.4 3.2 0.1 3.2 0.0 PMT per Person 12.2 19.9 25.2 28.0 25.0 1.2 31.6 2.6 2017 PMT per Person adj. - - - - - - 34.3 2.5 Average Person Trip Length 6.7 8.1 7.5 8.4 8.0 0.4 9.9 0.7 2017 Adjusted Person Trip Length - - - - - - 10.3 0.8 Note: • Totals in all tables can include cases that were not included in any table subcategory. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO. This and other methods changes in the data series are outlined in Appendix B. • Average time spent driving includes all drivers, even those who did not drive a private vehicle on the day in which the household was interviewed. • Average trip length is calculated using only those records with trip mileage information present. • "% Work Trips" also includes work-related business. 2017 National Household Travel Survey SPECIAL POPULATIONS 92 Table 32a. Selected Data for Older Persons Survey Year: Characteristic All All Age Groups 50 and Older 50-59 60-69 70-79 80 and older 2009 Percent Drivers 87.9% 93.7% 91.4% 83.0% 61.7% 2009 MOE 0.52 0.69 0.89 1.32 2.17 2017 Percent Drivers 87.3% 91.2% 89.5% 85.8% 63.5% 2017 MOE 0.40 0.38 0.41 1.52 2.23 2009 Vehicle Miles/Driver 26.83 31.51 27.63 18.77 12.04 2009 MOE 0.67 1.29 1.18 1.14 1.03 2017 Vehicle Miles/Driver Orig. 24.43 28.28 24.22 20.08 12.94 2017 MOE for Orig VMT/Driver 0.79 1.47 1.09 1.35 1.83 2017 Vehicle Miles/Driver Adj. 27.01 31.15 26.81 22.33 14.41 2017 MOE for Adj. VMT/Driver 0.88 1.63 1.21 1.50 2.04 2009 Percent with Zero Vehicles Available 7.7% 4.9% 6.8% 10.3% 17.6% 2009 MOE 0.40 0.54 0.92 1.28 1.84 2017 Percent with Zero Vehicles Available 7.7% 6.9% 7.7% 7.1% 12.6% 2017 MOE 0.40 0.32 0.74 0.93 1.30 2009 Percent Who Did Not Travel 17.3% 11.2% 14.9% 24.3% 38.0% 2009 MOE 0.60 0.71 0.94 1.58 2.91 2017 Percent Who Did Not Travel 19.7% 14.6% 18.4% 24.8% 37.3% 2017 MOE 0.59 1.09 0.71 0.89 2.50 2009 Percent with Disability 17.5% 10.9% 15.8% 22.6% 41.3% 2009 MOE 0.53 0.91 0.87 1.30 2.11 2017 Percent with Disability 13.9% 8.0% 11.2% 15.1% 48.9% 2017 MOE 0.40 0.71 0.82 1.06 1.43 2017 National Household Travel Survey SPECIAL POPULATIONS 93 Table 32b. Selected Data for Older Men Survey Year Characteristic Men All Men 50 and Older 50-59 60-69 70-79 80 and older 2009 Percent Drivers 93.2% 95.7% 95.1% 90.8% 77.4% 2009 MOE 0.50 0.67 0.94 1.42 2.76 2017 Percent Drivers 91.2% 92.4% 92.4% 91.6% 77.6% 2017 MOE 0.52 0.68 0.97 1.24 1.64 2009 Vehicle Miles/Driver 33.55 37.63 34.62 26.51 16.98 2009 MOE 1.14 1.76 2.25 2.18 2.20 2017 Vehicle Miles/Driver Orig. 30.06 33.85 30.14 25.72 16.42 2017 MOE for Orig VMT/Driver 1.24 2.03 2.21 2.06 2.64 2017 Vehicle Miles/Driver Adj. 33.22 37.26 33.36 28.61 18.28 2017 MOE for Adj. VMT/Driver 1.38 2.25 2.46 2.28 2.94 2009 Percent with Zero Vehicles Available 5.2% 4.5% 5.2% 5.4% 9.0% 2009 MOE 0.47 0.67 1.09 1.27 2.84 2017 Percent with Zero Vehicles Available 6.1% 6.5% 6.4% 4.5% 6.7% 2017 MOE 0.58 0.84 1.26 1.07 1.64 2009 Percent Who Did Not Travel 14.3% 10.8% 12.8% 18.3% 31.2% 2009 MOE 0.83 1.11 1.13 1.87 3.90 2017 Percent Who Did Not Travel 16.9% 13.2% 15.9% 21.7% 31.5% 2017 MOE 0.82 1.31 1.02 1.38 3.48 2009 Percent with Disability 14.4% 9.9% 13.5% 18.6% 34.2% 2009 MOE 0.71 1.19 1.26 1.70 2.95 2017 Percent with Disability 11.6% 7.1% 9.5% 12.7% 44.9% 2017 MOE 0.47 1.02 0.85 1.68 3.77 2017 National Household Travel Survey SPECIAL POPULATIONS 94 Table 32c. Selected Data for Older Women Survey Year Characteristic Women All Women 50 and Older 50-59 60-69 70-79 80 and older 2009 Percent Drivers 83.3% 91.8% 88.2% 77.1% 52.4% 2009 MOE 0.86 1.19 1.52 1.97 2.73 2017 Percent Drivers 83.8% 90.0% 87.0% 81.0% 54.3% 2017 MOE 0.93 0.78 1.18 2.30 3.82 2009 Vehicle Miles/Driver 20.33 25.29 20.92 11.84 7.76 2009 MOE 0.86 1.82 1.18 0.79 0.81 2017 Vehicle Miles/Driver Orig. 19.05 22.79 18.62 14.93 9.66 2017 MOE for Orig VMT/Driver 0.94 1.59 1.26 2.23 2.36 2017 Vehicle Miles/Driver Adj. 21.07 25.12 20.62 16.61 10.76 2017 MOE for Adj. VMT/Driver 1.38 1.75 1.39 2.49 2.63 2009 Percent with Zero Vehicles Available 9.9% 5.2% 8.3% 14.0% 22.7% 2009 MOE 0.61 0.94 1.18 1.90 2.23 2017 Percent with Zero Vehicles Available 9.1% 7.3% 8.9% 9.1% 16.4% 2017 MOE 0.82 0.63 1.26 1.19 2.34 2009 Percent Who Did Not Travel 20.0% 11.7% 16.7% 28.9% 42.1% 2009 MOE 0.84 0.83 1.40 2.14 3.42 2017 Percent Who Did Not Travel 22.1% 15.9% 20.6% 27.3% 41.1% 2017 MOE 0.69 1.25 0.80 1.33 3.01 2009 Percent with Disability 20.2% 11.8% 17.9% 25.7% 45.4% 2009 MOE 0.72 1.26 1.21 2.01 2.77 2017 Percent with Disability 15.9% 9.0% 12.7% 17.0% 51.5% 2017 MOE 0.62 0.89 1.41 1.57 1.97 2017 National Household Travel Survey SPECIAL POPULATIONS 95 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • Percent with Disability is based on respondents who answered that they had a temporary or permanent condition that makes it difficult for them to travel outside of the home. 2017 National Household Travel Survey SPECIAL POPULATIONS 96 Overall, younger drivers report driving fewer miles per capita (including drivers who drove on the travel day and those who did not) in 2017 compared to the trend data. However, the estimates for both the original and adjusted VMT in 2017 are statistically the same as the 2009 estimates across the board (within the margin of error) (Table 33). In urbanized areas, where the majority of the U.S. population lives, the declines in VMT per day are significant for 16-24 year old's compared to 2001 but not 2009. As the data series shows, VMT per driver in these age groups has not significantly declined between 2017 and 2009, but the estimates are statistically lower than 2001. Table 33. Vehicle Miles of Travel (VMT) Trends for Younger People by Urban or Rural Household Location Survey Year People in All Areas Daily VMT 16-24 25-34 35-44 45+ 1990 25.1 22.4 31.9 30.9 19.4 1995 28.5 22.6 33.5 34.6 25.0 2001 29.5 22.4 32.8 36.4 27.3 2009 25.8 17.4 26.8 32.5 25.2 2009 MOE 0.6 1.1 1.7 1.8 0.8 2017 Orig. 22.5 14.9 26.0 27.1 22.2 2017 Orig. MOE 1.0 1.2 4.8 1.8 0.7 2017 Adj. 24.8 16.4 28.6 29.8 24.6 2017 Adj. MOE 1.1 1.3 5.3 1.9 0.8 2017 National Household Travel Survey SPECIAL POPULATIONS 97 Table 33. Vehicle Miles of Travel (VMT) Trends for Younger People by Urban or Rural Household Location (continued) Survey Year People in Urban Areas Daily VMT 16-24 25-34 35-44 45+ 1990 22.4 20.2 28.5 27.4 17.0 1995 25.0 19.7 30.1 30.3 21.5 2001 27.3 20.9 30.7 33.3 25.0 2009 23.1 14.6 24.5 30.1 22.4 2009 MOE 0.7 1.0 2.0 2.1 0.8 2017 Orig. 20.8 13.3 25.0 25.2 20.2 2017 Orig. MOE 1.2 1.5 5.3 1.5 0.8 2017 Adj. 23.0 14.6 27.5 27.7 22.3 2017 Adj. MOE 1.4 1.7 5.9 1.6 0.8 Survey Year People in Rural Areas Daily VMT 16-24 25-34 35-44 45+ 1990 29.6 26.9 38.7 36.9 23.0 1995 34.6 28.2 40.1 41.6 30.8 2001 37.6 28.2 42.1 47.1 34.6 2009 34.2 25.8 34.6 40.5 34.2 2009 MOE 1.2 3.1 2.6 3.2 1.7 2017 Orig. 30.3 22.4 32.8 37.0 30.3 2017 Orig. MOE 0.8 3.1 3.0 4.4 1.6 2017 Adj. 33.5 24.6 36.1 40.7 33.4 2017 Adj. MOE 0.8 3.5 3.3 4.8 1.8 2017 National Household Travel Survey SPECIAL POPULATIONS 98 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • In 1995, VMT and vehicle trips with "To or From Work" as a trip purpose are believed to be overstated. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • “Rural” encompasses all territory not included within a Census Bureau classified urban area. 2017 National Household Travel Survey SPECIAL POPULATIONS 99 Table 34 shows select travel characteristics by urban and rural areas. Table 34. Travel Characteristics of People in Urban and Rural Areas, 2017 NHTS Characteristics Living in Urban Areas MOE Urban Living in Rural Areas MOE Rural Overall Percent (People 16 and older) 82.2% 0.52 17.8% 0.52 Percent Drivers 85.9% 0.54 91.9% 0.51 Percent Workers 62.0% 0.42 57.9% 0.37 Percent with Household Members Younger than 21 Years Old 42.5% 0.78 44.8% 0.35 Percent with Zero Vehicles Available 7.6% 0.22 2.3% 0.11 Percent Who Did Not Travel on Travel Day 16.3% 0.75 20.0% 0.16 Person Trips by Age Group Living in Urban Areas MOE Urban Living in Rural Areas MOE Rural All 16 and older 3.5 0.05 3.2 0.08 16-19 Years Old 2.8 0.14 2.7 0.20 20-34 3.4 0.09 3.2 0.17 35-54 3.9 0.07 3.5 0.10 55-64 3.6 0.06 3.4 0.20 65 and Older 3.2 0.05 3.0 0.10 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. • “Rural” encompasses all territory not included within a Census Bureau classified urban area. 2017 National Household Travel Survey SPECIAL POPULATIONS 100 One of the major new conveniences for U.S. households is online shopping and home delivery of many types of goods. The data series added a question about online purchases delivered to the home for the first time in the 2009 NHTS. The question also changed slightly in 2017. In 2009, the survey asked: “In the last month, how many of your online purchases were delivered to your home?”, while in 2017 the question was: “In the past 30 days, how many times did you purchase something online and have it delivered?" Assuming the answers are comparable, the estimate of the number of deliveries in an average month has doubled between the two survey time points (Table 35). The data indicates that online shopping is more prevalent in households with children, especially older teens and young adults (children aged 16-21). However, households with small children and those without children—including those headed by older individuals—had larger increases in the number of online purchases delivered to the household. Table 35. Average Number of On-Line Purchases and Deliveries to U.S. Households in the Last Month Household Type by Presence of Children 2009 NHTS 2017 NHTS Purchases Delivered to the Household 2009 MOE Purchases Delivered to the Household 2017 MOE All Households 2.4 0.1 4.9 0.1 Households Without Members <21 1.6 0.1 3.9 0.1 Households With Members Aged 5-15 3.7 0.2 6.9 0.1 Households With Members Aged 16-21 4.2 0.6 7.5 0.6 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • The 2009 NHTS was the first time data was collected on home deliveries from Internet shopping and on-line purchases. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey SPECIAL POPULATIONS 101 Table 36 displays select characteristics for users of transportation network companies. Table 36. Characteristics of Users of Transportation Network Companies (Uber/Lyft), 2017 NHTS Characteristic: Used Rideshare MOE All Others MOE Overall Percent (16 and older) 9.8% 0.44 90.2% 0.44 Percent Drivers 87.6% 0.36 86.9% 0.37 Percent Workers 81.3% 0.40 59.1% 0.37 Percent Urban 96.5% 0.41 80.6% 0.85 Percent with Household Members Younger than 21 Years Old 36.4% 0.20 43.6% 0.64 Percent with Zero Vehicles Available 12.3% 0.08 6.0% 0.20 Percent Who Did Not Travel on Travel Day 10.2% 0.12 17.7% 0.72 Person Trips by Age Group: All 16 and older 4.0 0.20 3.4 0.04 16-19 years old 3.2 0.48 2.8 0.11 20-34 3.9 0.17 3.3 0.08 35-54 4.1 0.17 3.8 0.06 55-64 4.1 0.25 3.6 0.06 65 and Older 3.9 0.35 3.2 0.04 Note: • Totals in all tables can include cases that were not included in any table subcategory, for instance people who did not report their age are included in the total persons, but not in any age category. • 1990 NPTS data were adjusted to make them more comparable with later surveys. • 2001 NHTS sample included children 0 to 4 in the survey. The data shown here exclude them to be comparable with other survey years. • 2009 NHTS sample did not include households without landlines telephones (CPO households). • 2017 NHTS sample was address-based and included more urban and CPO households. This and other methods changes in the data series are outlined in Appendix B. 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-1 APPENDIX A: CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-2 Introduction The 2017 National Household Travel Survey (NHTS) underwent a redesign of the survey methodology and sampling strategy. Although these improvements lowered respondent burden (web-based self-reports) and improved coverage (address-based sample selection), they make direct comparisons between the results of the 2017 NHTS and the 2009 and earlier surveys problematic. Any travel changes observed between the 2009 and 2017 surveys may reflect not only actual changes in travel during the period but also artifacts of differences in survey methodology and sampling, or some of both. That is, any changes observed between the 2009 and 2017 travel data are presumably attributable to: (1) Real changes in travel behavior, (2) Shift from using interviewer-assisted interviewers to web-based self-reports (about 70% of respondents reported via web), (3) Inclusion of households not sampled in 2009 (45% of completed households2 in 2017 are cell phone only [CPO]), and (4) Other improvements/changes in the 2017 survey methods. The first part of this document summarizes the potential impact of the changes in methods and sampling in the 2017 NHTS that will be the subject of on-going research. One specific change in the 2017 NHTS is an immediate and calculable impact on the survey estimates for trip distances. In the 2017 NHTS, researchers calculated trip distance via the shortest-path on the network from the geocoded origin of the trip to the geocoded destination. Previous surveys depended on the respondent to report the trip distance for each trip. The difference in trip distance reporting in 2017 NHTS impacts the estimation of average trip length by purpose and person miles of travel (PMT)/vehicle miles of travel (VMT) estimates for persons and households. The distance calculation estimates are in the second part of this document. This document has two parts: Part One presents a summary of a few of the important changes in methodology and protocols between the 2017 NHTS and earlier surveys (more detail is found in the User’s Guide here: Part Two describes an effort researchers made to quantify the impact of the change in trip distance reporting and to calculate simple adjustment factors to bring the 2017 more in line with earlier estimates and outside sources (Highway Performance Monitoring System (HPMS) VMT). 2 In 2009, a completed household was defined as having 50 percent of the adults complete the survey. In 2017 a completed household required 100 percent of household members 5 and older to have a completed survey. 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-3 The resulting “adjusted” estimates are displayed along with the original distance estimates in the tables in this report that include trip length, VMT, or PMT trends. Users of the data series should spend the time to understand how the changes in methodology and sampling in the 2017 NHTS might impact the estimates in their analyses. Researchers should include the necessary cautions to readers of their reports and findings. Part 1. Overview of Important Changes in Survey Methodology For major population estimates, the change in methodology and sampling had little effect, as shown in Table A-1 (a reprint of Table 4 in Section 2). The notable exception is the difference in the estimate of total household-based VMT from the NHTS 2017 and other sources (HPMS), which is discussed in Part Two. Table A-1 Comparison of NHTS 2017 to Other Sources (Thousands) Variable 2017 NHTS Other Sources Percent Difference: Other Sources/NHTS Households3 118,208 118,208 0% Population4 321,419 321,419 0% Drivers5 223,277 218,084 -2% Workers6 156,988 151,144 4% Vehicles7 222,579 231,490 4% VMT5 2,105,882 2,638,583 25% The population estimates match because researchers controlled them at the census division level during the weighting process. The weighting followed a similar protocol to the 2009 NHTS weighting process. This included the standard, best-practice methodology that is appropriate for any household survey, regardless of survey design or mode. The steps in weighting the survey data include: • Computing base weights as the inverse of the selection probability from each sampled unit (in the case of 2017 NHTS this was the household address), • Adjusting the base weights for eligibility and nonresponse, and 3 Households - Census QuickFacts Table US Households 2011-2015 4 Population - Population in Occupied Housing Units, estimate 2016 5Drivers - 2015 estimate from Highway Statistics Table DL-22 6 Workers - Source: Statista Civilian labor force in the United States from 1990 to 2016 (in millions) 7 Vehicles and VMT - Light Duty Vehicles (short WB) plus Motorcycles plus (based on the 2002 VIUS) 85.6% of Light Duty Vehicles with wheelbases (WB) larger than 121 inches 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-4 • Trimming and post-stratifying (or raking) to known reliable external data sources such as the Census. The 2017 NHTS data were raked by month and day of week, along with demographic characteristics such as age, sex, race/ethnicity, and worker status. (The User’s Guide provides more details on the weighting method: Researchers designed the 2017 NHTS to support state-, regional-, or city-level estimates only for areas that purchased additional samples (add-ons). The 2017 NHTS add-ons are: • Arizona • California • Dallas-Ft. Worth, Texas • Des Moines, Iowa • Georgia • Maryland • New York • North Carolina • South Carolina • Texas • Tulsa, Oklahoma • Waterloo, Iowa • Wisconsin The user is also cautioned not to attempt to estimate travel differences (e.g., between population groups, geographic areas, or between survey years) without calculating the confidence intervals to ensure statistically sound estimates. Sample Design and Address-based Sampling The random digit dialing (RDD) landline sample used in 2009 had coverage issues related to the growth in CPO households. In 2009, an estimated 25 percent of households nationwide did not have a landline, and these households were not included in the sample frame. To increase coverage the 2017 NHTS sample used an address-based sample frame, which included about 98 percent of U.S. households. About 45 percent of completed households in the 2017 NHTS are CPO (see definition of a completed household below). There are important demographic differences between people in CPO households compared to landline households. For example, the CPO respondents in the 2017 sample were more likely racial/ethnic minorities and younger than respondents in landline households. Mail-Out/Mail-Back Recruit The 2009 and earlier surveys mailed preliminary information to the sampled households but depended on a telephone interviewer to recruit the households into the study. In 2017, households at the sampled address received a recruitment package that they completed and returned by mail in order to be included in the survey. Definition of a Completed Household In 2017, 100 percent of household members aged 5 and older had to provide information relating to their travel on the assigned travel day in order for the household to be included in the survey. In previous (1995-2009) surveys, if 50 percent of adults 18 and older in the household provided information about their travel, the household was included in the survey. Therefore, in 2017, some larger households have more burden to complete the survey compared to smaller 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-5 households. It should be noted that the earliest NHTS surveys (1990 and earlier) accepted proxy reports from one household adult for all other household members. Web-Based Retrieval Questionnaire In 2017, the majority (70%) of respondents participated via the web-based questionnaire. Previous NHTS surveys were administered by computer-assisted telephone interviewing (CATI) only and used a trained interviewer to lead respondents through the survey. Interviewers were therefore available to answer respondent questions and probe responses where needed. In the 2017 survey, only 30 percent of respondents completed by CATI (these respondents either called in or were contacted via telephone). The mixed-mode nature of the 2017 NHTS resulted in different population groups utilizing different methods to complete the survey. The respondents who completed with an interviewer (CATI) were older, poorer, and on average less educated. A greater proportion of CATI respondents came from single-person households, households with no workers, rural households, and households with no vehicle or one vehicle. In contrast, people who reported via the web-based retrieval were younger, had higher income, and were more educated. Web-based respondents were more likely from larger households, more likely urban, with one or more workers, and had higher vehicle ownership. People aged 55 and older reporting via the web were almost twice as likely to be a worker and more likely to work at home compared to the same aged respondents who completed by CATI. The percentage of people reporting no travel also varied between the respondents completing via CATI or web. The data show that many more children under 16 (who all have their travel reported by proxy from a household adult) have no travel reports in the web-based format. At the other end of the age spectrum—people 65 and older—many fewer older respondents reported no travel on the web-based format. The differences in the proportion of people reporting no travel impacts the average trip rates. Changes in the Questionnaire The differences between the 2017 redesigned survey instrument and the 2009 instrument are in Table A-2. One difference was the use of a place-based reporting compared to trip-based. For example, in 2009 respondents were given the definition of a trip: “A trip is whenever you travel from one address to another.” In 2017, respondents were given the definition of a place: “A place is any location you go to, no matter how long you are there.” 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-6 Table A-2 Differences between 2009 and 2017 Travel Diary/Travel Log 2009 Diary Where did you go? What was the Location? What time did you start and end each trip? How did you travel? How far was it? (blocks or miles) 2017 Travel Log Where did you go next? What time did you arrive at this place? How did you get to this place? How many people went with you to this place? What time did you leave this place? What did you do at this place? Researchers changed the definition of a “trip” to allow reports of travel that began and ended at home (loop trips). This particularly influences walk and bike trends. In the 2017 NHTS, trips that began or ended at home were coded as a single trip. In 2009 and earlier surveys trips that began and ended at home were split into an outbound and inbound segment based on the farthest point. About 2 percent of trips were home-to-home loops. Most of these were walk and bicycle trips. Researchers asked additional walk and bike questions in the 2017 NHTS. 2009: Number of walk and bike trips. 2017: Number of walk and bike trips. Number of walk/bike trips for exercise. What keeps you from walking/biking more often? (>0 AND NOT PROXY) The 2017 NHTS also had additional trip prompts. 2009: Interviewer prompted respondent at the end of trip roster: So far, I have recorded {N} trip(s). Before we continue, did {you/SUBJECT} take any other walks, bike rides, or drives on {TRIPDATE}? Please include any other trips where {you/SUBJECT} used public transit or started and ended in the same place. 2017: The survey displayed a pop-up prompt after the places roster for respondents: - Did you include all places [$YOU] went on the assigned travel day, including short stops such as the dry cleaners or ATM? - Participants provided two options (I Need to Add a Place / I’m done). - Must select an option to advance. These changes in the questionnaire wording, and the change in trip definition, may have impacted travel estimates, especially for walk and bike trips. 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-7 Part 2: Calculation of Differences in Trip Distance Reports The 2017 NHTS collected trip distance based on the calculated shortest route between a valid geocoded origin and a valid address destination using an interface similar to Google maps. This marks a major change to the data series—previous surveys depended on the respondent to report the distance for each trip. The change in the calculation of trip distance impacts estimates of total PMT and VMT, as well as average person- and vehicle-trip lengths (including commute trip length). Analysts should use extreme caution in developing trends with these variables. Purpose of the Trip Distance Assessment To assess how these two measures of trip distance vary, the 2009 NHTS origin-destination data from the following add-on areas were geocoded and used to compute shortest distance paths using the same Google API used to compute trip distance in the 2017 NHTS: • California • Georgia • New York State • North Carolina • South Carolina • Texas • Wisconsin More than half a million (541,009) trips were assessed overall, including 352,565 vehicle (driver) trips (65%). Only vehicle trips were included in the analysis, because the 2017 estimate of VMT was lower than the HPMS estimate, and lower nominally compared to the 2001 and 2009 estimates. Researchers examined vehicle trips to understand how much the self-reported estimate differed from the calculated estimate by purpose. Interestingly, self-reported distances for work trips were closer to calculated (shortest-path) distances than self-reported distances for non-work purposes. Therefore, researchers analyzed work and non-work vehicle trips separately. The distribution of the difference between self-reported and calculated vehicle trip distances showed some extreme values—self-reported distances that were more than twice as long or twice as short compared to the calculated distance. Extreme values can have a big impact on the mean estimates. Researchers examined these outliers further by the trip characteristics. The reported trip distance for these outliers skewed toward very short trips, over half were trips with reported distances of less than one mile. The vehicle trips that had a difference between self-reported and calculated distance of +/- 100 percent as outliers were removed. With these outliers removed, the calculated distance in the 2009 dataset was shorter for both work and non-work trips (the raw data showed the opposite effect). Next, researchers applied the mean difference in vehicle trip length estimates between self-reported and calculated trip distance in 2009 to the 2017 data (the percentage difference applied to work and non-work trips separately). The adjustment raised the 2017 overall VMT 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-8 estimate by 10.3 percent. This brings the 2017 VMT adjusted estimate above the estimate for 2009—showing growth in VMT between the two survey years. They then compared the mean vehicle trip length—adjusted and original—to the estimates from previous surveys. The increases in average trip length were significant for most purposes (trips for shopping were nominally but not significantly longer). The overall difference was 7 percent for commute trips and 11 percent for trips of other purposes. The Summary of Travel Trends includes both the original and adjusted estimate, along with the margin of error, to let data users decide on the appropriate estimate for their particular use. Background and Context Though the “lower” estimate in 2017 for VMT is within the margin of error of the 2009 estimate and statistically the estimates for 2009 and 2017 VMT are not different (see Figure A-1), the total estimate of 2,105,882 million miles in 2017 was nominally 6 percent lower than the estimate in 2009. Importantly, other sources of VMT estimates show that total VMT had grown in the period between 2009 and 2017. HPMS estimates in 2015 (the most recent year available) were 3,095,373 million miles of vehicle travel. The 2017 estimate for passenger travel was only 68 percent of that total (compared to 76% in 2009 and 81% in 2001). The adjusted values for trip distance raises the nominal estimate of VMT above the nominal estimate for 2009 and within the margin of error of the 2001 estimate. Figure A-1 displays these estimates and the confidence limits at the 95 percent level. 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-9 Figure A-1 Estimates of VMT for 2001, 2009, and 2017 NHTS (original and adjusted) Method and Approach As shown in Figure A-2, the variation between reported and calculated distance was different for work and non-work trip purposes. Driver’s reports for commute trip lengths were close to the shortest path calculated distance—a plurality of work trips had reported distance within +/- 10% of the calculated trip distance. On the other hand, self-reported distance for non-work trips were not as close to the calculated distance. However, social and recreational, errands and shopping, and other purposes all had similar distributions. Therefore, going forward the purposes were categorized as “work” and “non-work”. 2001 VMT 2009 VMT 2017 VMT Orig. Distance 2017 VMT Adj. Distance High 2,491,535 2,301,276 2,197,419 2,419,900 Low 2,058,003 2,188,947 2,014,344 2,223,740 Estimate 2,274,769 2,245,111 2,105,882 2,321,820 - 500,000 1,000,000 1,500,000 2,000,000 2,500,000 3,000,000 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-10 Figure A-2 Difference in Trip Length by Purpose Examining the Distribution of the Data/Outliers Figure A-3 displays the mean unweighted trip distances from the 2009 self-reported and calculated distance estimates. The calculated distance is about 10 percent higher for work trips and about 20 percent higher for non-work (using (CALC_DIST-TRPMILES)/TRPMILES))8. Remember, these are the distance estimates from the Google API run as the shortest path at the time the respondent entered a valid (geocoded) origin and destination for the trip. 8 The NHTS uses negative values to code legitimate skip and unreported (-1, -8, -9), and these must be removed to calculate correct means. 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 100% Low or Lower 75-99% Low 50-75% Low 25-50% Low 10-25% Low Within 10% Either Way 10-25% High 25-50% High 50-75% High 75-99% High 100% High or Higher Percent of Trips To and From Work Fam/Pers Errands (Inc. Shop) Social/Rec (Inc. Visit) Work-Related Business School/Church Other 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-11 Figure A-3 Mean Distance for Work and Non-Work Trips by Two Methods: Uncapped The “average” or means in data such as these are very sensitive to the number of extreme values (outliers). The difference between reported and calculated miles skews to the right (shown in Figure A-4) —meaning that in most cases reported miles were higher than calculated miles. Few of the values were on the extreme edges of the distribution (reported distances were more or less than 100% of the calculated distance). 12.4 8.1 13.7 9.6 Work Non-Work Unweighted Reported Miles Unweighted Calculated Miles 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-12 Figure A-4 Distribution of the Percent Difference in Trip Length between Reported Miles and Calculated Miles 0% 5% 10% 15% 20% 25% 30% 100% or More Low 75-99% Low 50-75% Low 25-50% Low 10-25% Low Within 10% Either Way 10-25% High 25-50% High 50-75% High 75-99% High 100% or more High Percent of Trips Reported Miles are Lower than Calculated Miles<------->Reported Miles are Higher Than 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-13 After several univariate analyses, researchers identified the trips with a difference between reported and calculated miles of more than 100 percent as potential outliers. Table A-3 shows the original and final number of records and the logic used for each step. Table A-3 Number of Records Used In Analysis Category n Logic Geocoded Records 549,009 With Reported Miles 532,243 TRPMILES>0 Driver Trips 349,305 TRPMILES>0 and DRVR_FLG='01' Within Range 318,919 PCT_DIFF_MILES +/- 100% of Reported Miles Removed as Outliers 30,386 Outliers as a Percent of Driver Trips 8.7% The outliers skewed to the negative range, as shown in Figure A-5. The bottom graphic in Figure A-5 shows the distribution of trip records considered outliers. The blue bar across the bottom represents a frequency of “one”, with occasional spikes ranging from two to six reported trips with the same extreme difference between self-reported and calculated trip distance. Figure A-5 Distribution of Distance Outliers, 2009 NHTS 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-14 Figure A-6 shows the number of trip records in each bin. Note again that the outliers skew to the negative side: trips with self-reported distances that were less than half the calculated distance were 83 percent of all outliers (25,327 of 30,386). Altogether, the 30,386 total records with self-reported distances of more or less than 100 percent of the calculated distance represented 8.7 percent of driver trips in the analysis dataset. Figure A-6 Distribution and Frequency of Trips by Percent Difference between Reported and Calculated Miles Researchers examined the outliers further to identify the types of trips that had large differences between reported and calculated distances. Table A-4 shows some characteristics between the trips considered outliers (greater than +/-100% difference between reported and calculated trip distance) and all others. Households that 100% or More Low 75-99% Low 50-75% Low 25-50% Low 10-25% Low Within 10% Either Way 10-25% High 25-50% High 50-75% High 75-99% High 100% or More High Number of Trip Records 25,327 6,464 11,386 24,233 31,703 99,145 55,152 54,759 26,350 9,727 5,059 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-15 were rural and people who did not start their day at home were more likely to have trips that were considered outliers. Table A-4 Characteristics of Trips with Extreme Differences between Reported and Calculated Miles Characteristics of Trips with Extreme Difference Between Reported and Calculated Distance, 2009 Selected Areas (Driver Trips Only) Outliers Non-Outliers Reported by Proxy 12.5% 13.7% Household is Rural 39.4% 28.7% Purpose is Non-Work 88.8% 81.7% Trip was a Weekend Trip 22.0% 25.7% Person did Not Start the Travel Day at Home 8.4% 2.8% Of the outliers, fully half were under one mile in length (recall that only driver trips are included in this analysis). Overall, almost nine out of ten (88.3%) were for non-work purposes. Figure A-7 shows the distribution of the outliers by trip length and purpose. Figure A-7 Characteristics of Outliers Analysis of Trip Distance Figure A-8 shows the difference in the mean estimate of trip distance for the analysis areas in the 2009 NHTS for all reported vehicle trips (349,305 records), and for the same set of records with outliers removed (318,919 records). 0% 10% 20% 30% 40% 50% 60% Less Than 1 Mile 1-4.9 Miles 5-9.9 Miles 10-19.9 Miles 20-49.9 Miles 50 Miles of More Percent of Trips Work Non-Work 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-16 With the outliers removed, the relationship changed. With extreme values removed, the average trip distance using the shortest-path calculation is less than the average using reported miles. Table A-5 and Figure A-8 show the capped and uncapped values. Note that this calculation uses “calculated miles” as the base because it is common to both datasets. Table A-5 Percent Difference Between Calculated and Reported Miles Category A Category B Work Non-Work All Uncapped Reported Miles 12.41 8.08 8.87 Calculated Miles 13.71 9.61 10.36 Capped at 100% Diff Reported Miles 12.84 8.66 9.45 Calculated Miles 11.96 7.76 8.56 Difference 0.88 0.89 0.89 Percent Diff (Diff./calculated miles) 7.35% 11.51% 10.41% Figure A-8 Mean Distance for Work and Non-Work Trips: Raw Data and Outliers Removed 12.41 13.71 12.84 11.96 8.08 9.61 8.66 7.76 8.87 10.36 9.45 8.56 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 Reported Miles Calculated Miles Reported Miles Calculated Miles Uncapped Capped at 100% Diff Trip Miles Work Non-Work All 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-17 Testing the Effect on 2017 NHTS VMT Estimate Researchers tested the effect of adjusting the disaggregate trip miles (at the trip level) by these factors on the estimates of VMT for 2017. That is, the calculated trip miles in the 2017 NHTS trip file (vehicle trips) was adjusted at the trip level by a factor of 1.0735 for work trips and 1.1151 for non-work trips (based on the calculations in Table A-3). This adjustment to each vehicle trip distance was then weighted by the individual trip record weight (MILE_ADJWTTRDFIN) to obtain weighted total estimate of household-based VMT. In addition, they added a new mode of travel in 2017 NHTS (rental cars, including Car2Go and ZipCar)—to the estimate. The adjusted estimate of trip distance for vehicle trips added 10.3 percent to the total estimate for household-based VMT in 2017. Figure A-9 shows the 2001, 2009, 2017 original, and 2017 adjusted VMT estimates. Figure A-9 Trends in VMT Estimates, 2009, 2017 and 2017 Adjusted Trends in Trip Length Estimates by Purpose Researchers compared the adjusted vehicle trip length estimates to the original estimates in the 2017 NHTS and previous surveys for major trip purposes (see Figure A-10). For each major purpose category, the adjusted data are noticeably higher than the original estimates. (The data for this table is also shown in Table 6 of the 2017 Summary of Travel Trends). 2,274,769 2,245,111 2,105,881 2,321,820 2001 2009 2017 Orig. 2017 Adj. 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-18 Figure A-10 Trends in Mean Vehicle Trip Length by Purpose Researchers tested the mean trip lengths from the original distance measure and adjusted distance measure for significance. As shown in Figure A-11, the adjusted trip length estimates are significantly higher than previous estimates for commute trips, social/recreational trips, and overall. Shopping trips, while nominally longer (7.2 miles original to 8.0 miles adjusted), are not statistically different between 2009 and 2017. 0 2 4 6 8 10 12 14 All Purposes To or From Work Shopping Other Family/Personal Errands Social and Recreational Mean Vehicle Trip Length in Miles 1983 1990 1995 2001 2009 2017 Orig 2017 Adj 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-19 Figure A-11 Mean Vehicle Trip Length by Purpose with Confidence Intervals, 2001, 2009, 2017 Original and 2017 Adjusted Conclusion The 2017 NHTS obtained estimates of trip distance using a Google API shortest-path route distance between a geocoded origin and a geocoded destination. This is a major difference compared to previous surveys which depended on the driver’s estimate of trip distance for each reported trip. The impact of this change resulted in a low estimate of VMT in 2017, compared to previous estimates and other sources (HPMS). To assess the impact on the estimate of vehicle trip distance obtained by these two different methods, researchers calculated trip distances for a sub-set of 2009 (add-on) data from the geocoded origins and destinations using the same Google API method as that used in 2017. The analysis showed that the different methods of obtaining trip distance between 2017 NHTS and the earlier surveys resulted in a nominal decrease the estimates of vehicle trip lengths and VMT for the 2017 NHTS. The estimate of vehicle trip lengths from the two methods (self-reported and calculated) varied by trip purpose. Commuters who reported the trip distance to work (in the 2009 NHTS) were 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 2001 2009 2017 org. 2017 adj. 2001 2009 2017 org. 2017 adj. 2001 2009 2017 org. 2017 adj. 2001 2009 2017 org. 2017 adj. 2001 2009 2017 org. 2017 adj. Commute Shop Errands Soc/Rec All Mean Vehicle Trip Length (miles) 2017 National Household Travel Survey APPENDIX A – CHANGES IN SURVEY METHODOLOGY AND THE ADJUSTMENT OF TRIP LENGTH ESTIMATES A-20 closer to the calculated shortest-path distance obtained by Google API (within 7%). However, for other trip purposes, the self-reported distances were over 11 percent different compared to calculated shortest-path distances. Researchers developed an adjustment factor based on the percentage difference between calculated and self-reported vehicle trip distance for work and non-work purposes. They then adjusted the 2017 NHTS vehicle trip lengths by this factor for work and non-work trips. The adjusted estimates resulted in higher VMT estimates overall, and longer vehicle trip lengths for most purposes (shopping trips were nominally but not significantly longer after the adjustment). The 2017 NHTS Summary of Travel Trends report provides both the original and adjusted vehicle trip distance for the user. The adjusted mileage estimates for vehicle trips will also affect other estimates, such as PMT, and comparisons of trip length by mode. Therefore, including both the adjusted and original estimates in the Summary of Travel Trends documentation will offer the most flexibility to the NHTS user community. History of Adjusting NHTS Data As a reference, when the methods changed between the 1990 NPTS (which used a recall of “yesterday”) and the 1995 NPTS (which used a two-stage survey with a travel diary) the earlier survey was given an “adjustment” (in that case applied to the weights) to bring the trip reporting in line with the 1995 NPTS. The adjusted data were provided on the dataset and in the 1995 documentation along with the original estimate until 2001, when the Summary of Travel Trends dropped the original estimate for 1990 and only included the adjusted estimates. The documentation of the adjustment is found in the 1995 Summary of Travel Trends, Appendix 2 “Adjustment of the 1990 NPTS Data”: 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-1 APPENDIX B: KEY CHANGES 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-2 Key Changes in NHTS Survey Methodology and Content 1969 1977 1983 1990 1995 2001 2009 2017 Approximate Sample Size (Number of Households) 15,000 18,000 6,500 18,000 national and 4,300 add-on 21,000 national and 21,000 add-on 26,000 national and 40,000 add-ons 26,000 national and 125,000 add-on (Combined into single sample) 26,000 national and 104,000 add-on (Combined into a single sample) 1969 1977 1983 1990 1995 2001 2009 2017 Sample Selection Outgoing panels of Census Quarterly Housing Survey Outgoing panels of Census Current Population Survey Outgoing panels of Census Current Population Survey Random Digit Dialing (RDD) Telephone sample RDD Telephone sample RDD Telephone sample RDD Telephone sample Address-based sample 1969 1977 1983 1990 1995 2001 2009 2017 Interview Method In-home interview with some telephone follow-up In-home interview with some telephone follow-up In-home interview with some telephone follow-up One stage: computer-assisted telephone interviewing (CATI) recruit and recall of travel day Two stage: CATI recruit-mail out diary-CATI collection Two stage: CATI recruit-mail out diary-CATI collection Two stage: CATI recruit-mail out diary-CATI collection Two-stage: Mail-out recruit plus web-based self-report or CATI retrieval 1969 1977 1983 1990 1995 2001 2009 2017 Pre-Contact/Incentive Pre-Interview Letter Pre-Interview Letter Pre-Interview Letter Pre-Interview Letter Advance letter and $2 per person with diary Advance letter with $5 and $2 per person with diary Advance letter with $5 and $2 per person with diary Advance letter with $2 plus $5 per person plus $20 post-completion 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-3 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Diary ”Memory Jogger” None: Respondent recalled “yesterday” None: Respondent recalled “yesterday” None: Respondent recalled “yesterday” None: Respondent recalled “yesterday” Diary as a memory jogger Diary as a memory jogger Diary as a memory jogger Diary as a memory jogger 1969 1977 1983 1990 1995 2001 2009 2017 Trip Rostering to Reduce Item Nonresponse None None None None Full day trip rostering before collecting trip detail Full day trip rostering before collecting trip detail Full day trip rostering before collecting trip detail Full day trip rostering before collecting trip detail 1969 1977 1983 1990 1995 2001 2009 2017 Eligible Persons Household members aged 5 and older Household members aged 5 and older Household members aged 5 and older Household members aged 5 and older Household members aged 5 and older All household members Household members aged 5 and older Household members aged 5 and older 1969 1977 1983 1990 1995 2001 2009 2017 Usable Household Definition At least one adult member of the household At least one adult member of the household At least one adult member of the household At least one adult member of the household At least half the adult members of the household At least half the adult members of the household At least half the adult members of the household 100% of all household members aged 5 and older 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-4 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Proxy Rules An Adult household member reported all trips (excluding bike and walk trips) made by household members between the ages of 5 to 13 years An Adult household member reported all trips (excluding bike and walk trips) made by household members between the ages of 5 to 13 years An Adult household member reported all trips (excluding bike and walk trips) made by household members between the ages of 5 to 13 years An Adult household member reported all trips made by household members between the ages of 5 to 13 years. Adult proxy allowed Proxy reports required for 13 and under. Parental approval for 14- to 15- year olds. Adult proxy from diary allowed Proxy reports required for 13 and under. Parental approval for 14- to 15- year olds. Adult proxy from diary after 3 days Proxy reports required for 13 and under. Parental approval for 14- to 15-year olds. Adult proxy from diary after three days Whether travel day report was via a proxy was self-reported in the web-based retrieval. Proxy flag is carried on the person record 1969 1977 1983 1990 1995 2001 2009 2017 Travel Day Trip Definition Travel within a defined area (such as a strip mall or shopping mall) not counted Travel within a defined area (such as a strip mall or shopping mall) not counted Travel within a defined area (such as a strip mall or shopping mall) not counted Travel within a defined area (such as a strip mall or shopping mall) not counted Any stop from one address to another, including trips to change transportation mode Any trip from one address to another, mode changes not included (access and egress asked separately) Any trip from one address to another, mode changes not included (access and egress asked separately) Any trip from one address to another, including trips to change transportation mode 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-5 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Reporting Prompts None None Prompts to include walking and bike trips, to lunch, stopping at a gas station, etc. Prompts for forgotten trips Prompts for forgotten trips Prompts to include walk/bike trips and trips that started and ended in the same place Prompts to include walk/bike and trips that started and ended in the same place. Added prompts to include transit Prompts to include incidental trips/stops plus walk, bike rides and trips that started and ended in the same place 1969 1977 1983 1990 1995 2001 2009 2017 Walk and Bike Coding Collected walk and bike trips by respondents aged 14 and older Collected walk and bike trips by respondents aged 14 and older Collected walk and bike trips by respondents aged 14 and older Collected walk and bike trips by all respondents Collected walk and bike trips by all respondents Collected walk and bike by all respondents. Split home-to-home trips to geocode trip location Collected walk and bike by all respondents. Split home-to-home trips to geocode trip location Collected walk and bike by all respondents, allowed home-to-home trips (loop trips) 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-6 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Trip Verification (verifying joint trips reported by other household members) None None Manual checks across household member’s travel Interviewer instructed to check across household member’s travel CATI program checked across household members CATI program checked across household members CATI program checked across household members CATI and web-based systems checked across household members. Also checked as part of the QC and corrected with household recontact as necessary 1969 1977 1983 1990 1995 2001 2009 2017 Geocoding None None None None or limited manual coding Limited manual geocoding Extensive post-survey GIS-based geocoding Online real time geocoding during interview, followed by post processing GIS coding Real-time geocoding of each trip destination from a map interface. Shortest network-path distance calculated by Google between every geocoded origin and destination 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-7 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Weighting Raking to control totals Raking to control totals Nonresponse and noncoverage adjustments included in weight development Nonresponse and noncoverage adjustments included in weight development Raking to control totals, within household nonresponse adjustment Nonresponse adjustment, several stages of weighting, and trimming. Changes to the cells used for raking based on nonresponse follow-up survey Nonresponse adjustment, several stages of weighting, and trimming. Changes to the cells used for raking based on cell phone only sample Nonresponse adjustment, several stages of weighting, and trimming. Address-based sample weighted to geography. Raking variables consistent with 2009 1969 1977 1983 1990 1995 2001 2009 2017 Travel Day Trip Purpose There were 10 trip purposes plus “Other”, respondent selected the “Main" purpose of trip to code return home segment There were 21 trip purposes, respondent selected the “Main" purpose of trip to code return home segment There were 10 trip purposes plus “Other”, respondent selected the “Main" purpose of trip to code return home segment There were 10 trip purposes plus “Other”, respondent selected the “Main" purpose of trip to code return home segment There were 17 trip purposes plus “Other”, FHWA coded “Main" purpose for return home and included a separate tour file There were 36 trip purposes, FHWA coded “Main" purpose for return home and included a separate tour file There were 36 trip purposes, FHWA coded “Main" purpose for return home and included a separate tour file There were 19 purpose codes. FHWA coded “Main" purpose for return home trips 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-8 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Vehicle Detail Only included automobiles as household vehicles Included all motor vehicles in household: pickups, vans, motorcycles, etc. Included all motor vehicles in household: pickups, vans, motorcycles, etc. Included all motor vehicles in household: pickups, vans, motorcycles, etc. Coded SUVs separately, but not Hybrid or electric Coded SUVs separately, but not Hybrid or electric Coded Hybrid/alt fuel for all vehicle classes. Coded Light Electric Vehicles, but did not count them as household vehicles. Coded Hybrid/alt fuel for all vehicle classes. Coded Light Electric Vehicles, but did not count them as household vehicles. 1969 1977 1983 1990 1995 2001 2009 2017 Odometer Readings None None None None Two readings collected by contacting respondent by phone or mail Two readings collected multi-modal (Internet, mail, 800 number) One reading collected at time of interview One reading collected at time of interview 1969 1977 1983 1990 1995 2001 2009 2017 Long-Distance Component None Included 2-week travel period for trips of 75 miles or more Included 2-week travel period for trips of 75 miles or more Included 2-week travel period for trips of 75 miles or more Included 2-week travel period for trips of 75 miles or more Included 28-day travel period (long distance) None Some add-ons asked questions related to long-distance for their specific areas 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-9 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Other Notes NPTS and National Travel Survey (long distance) combined Major shift in methods from recall of travel day to two-stage survey with pre-mailed diary NPTS and American Travel Survey (long-distance) combined Major shifts in methods from RDD/CATI to address-based sample and web-based retrieval. See Appendix A and User's Guide for more detail 1969 1977 1983 1990 1995 2001 2009 2017 Add-Ons None None None New York MPO New York State Baltimore MPO California Arizona Connecticut Massachusetts Des Moines, IA MPO Florida California Indianapolis MPO Oklahoma and Tulsa, Oklahoma Hawaii Georgia Des Moines Area MPO Puget Sound Kentucky Indiana Georgia Lancaster, PA MPO Iowa Indian Nations Council of Governments New York State New York State Iowa Northland Regional COG Oahu HI MPO North Carolina Maryland Texas South Carolina New York State Wisconsin South Dakota North Carolina 2017 National Household Travel Survey APPENDIX B – KEY CHANGES B-10 Key Changes in NHTS Survey Methodology and Content (continued) 1969 1977 1983 1990 1995 2001 2009 2017 Tennessee South Carolina Texas Wisconsin Vermont Texas Virginia North Central Texas COG Wisconsin Chittenden County MPO Linn County RPC Maricopa Association of Governments Pima County MPO Piedmont Regional Transportation Omaha-Council Bluffs Metro Area Planning Agency 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-1 APPENDIX C: TRAVEL CONCEPTS AND GLOSSARY OF TERMS 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-2 Travel Concepts Person Trip A movement in the public space between two identifiable points. In 2017, NHTS trips that begin and end at home are included as one trip record and flagged as “loop” trips. These primarily include walks, jogs, and bike rides that in the past were divided into an outbound portion (geocoded to the farthest point) and an inbound portion. In 2017, the entire “loop” trip is included as one unit. Each record in the trip file represents one trip. For example, two household members traveling together in one car are counted as two person-trips. Three household members walking to the store together are counted as three person-trips. In 2017 NHTS, a jogger who leaves home and jogs around the neighborhood and back home is counted as one (loop) trip. Person Miles of The number of miles traveled by each person on a trip. Travel (PMT) For example, if two people traveling together take a 6-mile subway trip to the airport, that trip results in 12 person-miles of travel. A 4-mile van trip with a driver and four passengers counts as 16 person-miles of travel (4 people times 4 miles). Vehicle Trip A trip by a single privately-operated vehicle (POV) regardless of the number of persons in the vehicle. For example, two people traveling together in a car would be counted as one vehicle trip. Four people going to a restaurant in a van is considered one vehicle trip. Note: To be considered a vehicle trip in NHTS, the trip must have been made in a POV, namely a household-based car, van, sport utility vehicle (SUV), pickup truck, other truck, recreational vehicle, motorcycle or other POV. The vehicle does not need to belong to the household—in 2017 a category for rental cars was added to the mode list, and are included in estimates of private vehicle travel (including services like Car2Go and ZipCar). Trips made in other highway vehicles, such as buses, streetcars, taxis (including Uber/Lyft), and school buses are collected in the NHTS, but these are shown as person trips by those modes. The design of the NHTS is such that it does not serve as a source for vehicle trips in modes using other highway vehicles, because there is no way to trace the movement of these vehicles throughout the day. Those interested in vehicle trips by buses, taxis, etc., need to use a data source that relies on reports from the fleet operators of those vehicles. The National Transit Database of the Federal Transit Administration is one such source. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-3 Vehicle Miles of One vehicle mile of travel is the movement of one privately operated Travel (VMT) (POV) vehicle for one mile, regardless of the number of people in the vehicle. For example, when one person drives her car 12 miles to work, that equals 12 vehicle miles of travel. If two people travel 3 miles by pickup, that equals 3 vehicle miles of travel. The same definition of household vehicles is used. For NHTS data, vehicle miles are restricted to the same POVs as vehicle trips, that is a household-based car, van, SUV, pickup truck, other truck, recreational vehicle, or other POV, including rental car. Vehicle For NHTS data, vehicle occupancy is generally computed as person Occupancy miles of travel per vehicle mile (referred to as the travel method). Note that the other commonly used definition of vehicle occupancy is persons per vehicle trip (referred to as the trip method). Because longer trips often have higher occupancies, the distance-based method generally yields a higher rate than the trip-based method. The calculation of the distance-based method requires that trip distance be included in the record. In 2017, every geocoded origin and destination pair had a calculated shortest-path distance appended to the trip record. Some trips may be missing trip distance; therefore, vehicle occupancy using distance is calculated on a slightly smaller number of trips than the trip method. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-4 Glossary of Terms This glossary provides the most common terms used in this report and the NHTS survey, and definitions of those terms. These definitions are provided to assist the user in the interpretation of the NHTS data and tables in this report. Adult For NHTS, this is defined as a person 18 years or older. Census Region The U.S. Census Bureau divides the states into four regions and nine and Division divisions. Note that the divisions are wholly contained within a region (i.e., region lines do not split division lines). The regions and their component divisions are: Northeast Region: • New England Division: Connecticut, Maine, Massachusetts, New Hampshire, Rhode Island, Vermont • Middle Atlantic Division: New Jersey, New York, Pennsylvania Midwest Region: • East North Central Division: Illinois, Indiana, Michigan, Ohio, Wisconsin • West North Central Division: Iowa, Kansas, Minnesota, Missouri, Nebraska, North Dakota, South Dakota South Region: • South Atlantic Division: Delaware, Florida, Georgia, Maryland, North Carolina, South Carolina, Virginia, West Virginia • East South Central Division: Alabama, Kentucky, Mississippi, Tennessee • West South Central Division: Arkansas, Louisiana, Oklahoma, Texas West Region: • Mountain Division: Arizona, Colorado, Idaho, Montana, Nevada, New Mexico, Utah, Wyoming • Pacific Division: Alaska, California, Hawaii, Oregon, Washington Destination For travel day trips, the destination is the end-point of the reported trip. Driver A driver is a person who operates a motorized vehicle. NHTS does not specifically ask about license status. Employed A person is considered a worker/employed if they worked for pay, either (Worker) full time or part time, during the week before the interview. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-5 Education Level The number of years of regular schooling completed in graded public, private, or parochial schools, or in colleges, universities, or professional schools, whether day school or night school. Regular schooling advances a person toward an elementary or high school diploma, or a college, university, or professional school degree. Household A group of persons whose usual place of residence is a specific housing unit; these persons may or may not be related to each other. The total of all U.S. households represents the total civilian non-institutionalized population. Household Household income is the money earned by all family members in a Income household, including those temporarily absent. Annual income is the income earned 12 months preceding the interview. Household Household members include all people, whether present or temporarily Members absent, whose usual place of residence is in the sample unit. Household members also include people staying in the sample unit who have no other usual place of residence elsewhere and does not include anyone who usually lives somewhere else or is just visiting, such as a college student away at school. Household A household vehicle is a motorized vehicle that is owned, leased, rented Vehicle or company-owned and available to be used regularly by household members. Household vehicles include vehicles used solely for business purposes or business-owned vehicles, so long as they are driven home and can be used for the home to work trip, (e.g., taxicabs, police cars, etc.). Household vehicles include all vehicles that were owned or available for use by members of the household during the travel day, even though a vehicle may have been sold before the interview. Vehicles excluded from household vehicles are those that were not working and were not expected to be working, and vehicles that were purchased or received after the designated travel day. Means of A mode of travel used for going from one place (origin) to another Transportation (destination). A means of transportation includes private and public modes, as well as walking. The following transportation modes, grouped by major mode, are included in the NHTS data. Private Vehicle • Car: A privately owned and/or operated licensed motorized vehicle including cars and station wagons. Leased and rented cars are included if they are privately operated and not used for picking up passengers in return for fare. • Van: A privately owned and/or operated van or minivan designed to carry 5 to 13 passengers, or to haul cargo. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-6 • Sport utility vehicle: A privately owned and/or operated vehicle that is a hybrid of design elements from a van, a pickup truck and a station wagon. Examples include a Chevrolet Blazer, Ford Bronco, Jeep Cherokee, or Nissan Pathfinder. • Pickup truck: A pickup truck is a motorized vehicle, privately owned and/or operated, with an enclosed cab that usually accommodates two to three passengers, and an open cargo area in the rear. Later model pickups often have a back seat that allows for total seating of four to six passengers. Pickup trucks usually have the same size of wheel-base as a full-size station wagon. This category also includes pickups with campers. • Motorcycle/moped: This category includes large, medium, and small motorcycles and mopeds. Electric Bicycles are not included. • Golf cart/Segway: This category consists of self-powered small vehicles, generally light electric vehicles, and any two-wheeled motorized personal vehicle consisting of a platform for the feet mounted above an axle and an upright post surmounted by handles. • RV (motor home, ATV, snowmobile): An RV or motor home includes a self-powered recreational vehicle that is operated as a unit without being towed by another vehicle (e.g., a Winnebago motor home). This category includes all terrain vehicles and snowmobiles. Public Transportation • Public or commuter bus: This category includes buses that are part of transit systems, or a private service buses operating on a fixed schedule to serve commuters. • Subway/elevated/light rail/streetcar: Any transit service operated on a fixed rail or guide way system, vehicles that run on a fixed rail system powered by electricity obtained from an overhead power distribution system, and any other • Amtrak/commuter rail: This category includes all commuter trains and passenger trains. • City-to-city bus (Greyhound/Megabus): This category includes all passenger buses operating between population centers. • Paratransit/dial-a-ride: This category includes publicly operated on-call transit services for qualified individuals. Non-Motorized • Walk: This category includes walking and jogging. • Bicycle: This category includes bicycles of all speeds and sizes, including electric bikes. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-7 Other Modes: • Airplane includes commercial airplanes and smaller planes that are available for use by the public in exchange for a fare. Private and corporate planes and helicopters are also included. • Boat/ferry/water taxi: This includes travel by ships, cruise ships, passenger lines and ferries, sailboats, motorboats and yachts including water taxi. • Taxi/limo (including Uber/Lyft): This category includes the use of a mobility service by a passenger for fare, including traditional and ride-hailing services. The taxi category does not include rental cars if they are privately operated. • Private/charter/tour/shuttle bus: This includes privately operated large or shuttle buses that are operated for a fare. Metropolitan Geographic areas of more than 50,000 persons managed by the Office of Statistical Area Management and Budget to categorize official population estimates. (MSA) Counties and county equivalents are combined based on social and economic integration with its designated urban center. 2017 NHTS derived MSA variables using the 2010-2014 5-year American Community Survey B01003_001E variable. Margin of Error The 95 percent confidence interval of the estimate, calculated in this (MOE) report by multiplying a factor of 1.984 to the standard error of the estimate. Add and subtract the MOE to the estimate to determine the range of values that the statistic would fall into 95% of the time. Motorized Vehicle Motorized vehicles are all vehicles that are licensed for highway driving. Nationwide Personal Transportation The name of the national survey program responsible for data collected in Survey (NPTS) 1969, 1977, 1983, 1990, and 1995. Occupancy Occupancy is the number of persons, including driver and passenger(s) in a vehicle. NHTS occupancy rates are generally calculated as person miles divided by vehicle miles. See Vehicle Occupancy in Travel Concepts. Origin The starting point of a trip. Passenger For a specific trip, a passenger is any occupant of a motorized vehicle, other than the driver. Person Miles of PMT is a primary measure of person travel. When one person travels Travel (PMT) one mile, one person mile of travel results. Where 2 or more persons travel together in the same vehicle, each person makes the same number of person miles as the vehicle miles. Therefore, four persons traveling 5 miles in the same vehicle results in 20 person miles (4 x 5 = 20). 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-8 Person Trip A person trip is a trip by one or more persons in any mode of transportation. Each person is considered as making one person trip. For example, four persons traveling together in one auto are counted as four person trips. POV A privately-owned vehicle or privately-operated vehicle. Either way, the intent here is that this is not a vehicle available to the public for a fee, such as a bus, subway, taxi, etc. Travel Day A travel day is a 24-hour period from 4:00 a.m. to 3:59 a.m. designated as the reference period for studying trips and travel by members of a sampled household. Travel Day Trip A travel day trip is defined as any time the respondent went from one address to another by private motor vehicle, public transportation, bicycle, walking, or other means. Trip Purpose A trip purpose is the main reason that motivates a trip. In the 2017 NHTS survey, the number of trip purposes were reduced because of the move to self-reported travel on the web. For each trip, the origin and destination are on the file in generic terms, e.g. from work to shopping. There were 19 trip reasons that were on a pick-list for respondents to choose from, and the data were compiled into a legacy format (WHYTRP90) to match previous data from the NPTS/NHTS data series. These legacy purposes used in this report include trips to and from: ‘01' To and From Work (Commuting) ‘02' Work Related Business (meeting or trip) ‘03' Shopping ‘04' Family/Personal Errands (including drop-off/pickup, volunteer activities, and buying services such as cleaners, pet care, automotive care) ‘05' School/Church ‘06' Medical/Dental (any health care visit) ‘07' Vacation ‘08' Visit Friends and Family ‘10' Social/Recreational (exercise, movies, parks, museums and bars) ‘11', '98', '99' Other Urbanized Area An urbanized area consists of the built-up area surrounding a central core (or central city), with a population density of at least 1,000 persons per square mile. Urbanized areas do not follow jurisdictional boundaries thus it is common for the urbanized area boundary to divide a county. Vehicle In the 2017 NHTS, the term vehicle includes autos, passenger vans, sport utility vehicles, pickups and other light trucks, RVs, motorcycles and mopeds owned or available to the household. 2017 National Household Travel Survey APPENDIX C – TRAVEL CONCEPTS AND GLOSSARY OF TERMS C-9 Vehicle Miles of VMT is a unit to measure vehicle travel made by a private vehicle, such Travel (VMT) as an automobile, van, pickup truck, or motorcycle. Each mile traveled is counted as 1 vehicle mile regardless of the number of persons in the vehicle. Vehicle Vehicle occupancy is the number of persons, including driver and Occupancy passenger(s) in a vehicle; also includes persons who did not complete a whole trip. NHTS occupancy rates are generally calculated as person miles divided by vehicle miles. Vehicle Trip A trip by a single privately operated vehicle (POV) regardless of the number of persons in the vehicle. Vehicle Type The 2017 NHTS codes vehicles by make and model, and then generally into one of the following major vehicle types: 1. Automobile (including station wagon) 2. Van 3. Sport utility vehicle 4. Pickup truck (including pickup with camper) 5. Other truck 6. RV or motor home 7. Motorcycle 8. Other
14598	https://www.jewelersmutual.com/the-jewelry-box/sterling-silver-vs-silver	Skip to main content Click to view our statement on Accessibility Check Your Rate FOR SMART, STYLISH JEWELRY LOVERS EVERYWHERE The Jewelry Box Blog Sterling Silver vs Silver - How Do You Tell Them Apart? Posted by Matt Wodenka on Silver has been used for hundreds of years in a variety of ways, and sterling silver has long been a staple of the jewelry world. Yet, after being around for so many years, many questions still exist about both silver and sterling silver, including how they compare to each other, whether they tarnish and how to care for them. What is Sterling Silver? Sterling silver is a metal alloy typically made of 92.5% silver and 7.5% copper or nickel. Adding another harder metal like copper or nickel to silver allows the resulting sterling silver to be less susceptible to damage. It also has a brighter, shinier finish than pure silver. This shiny appearance is what many associate with “silver” (rather than pure silver’s grayish color) due to the heavy use of sterling silver in jewelry. Pros of sterling silver: Affordable: Sterling silver is generally more affordable than white gold or platinum. This makes it a great choice for everyday jewelry, especially for those on a budget. Versatile: Sterling silver comes in a wide variety of styles, from classic to modern. It can also be easily adorned with gemstones, pearls and other materials, making it a versatile choice for any occasion. Long-lasting shine: Tarnished sterling silver can be easily brightened up with a polishing cloth, restoring its finish to its original gleam. This makes it a great choice for jewelry that you might wear for many years. Cons of sterling silver: Prone to tarnishing: Although it is polishable, silver will tarnish when it comes in contact with air, moisture and chemicals. Not hypoallergenic: Sterling silver contains copper (and can contain nickel as mentioned above), which some people may have an adverse or allergic reaction to. Consider hypoallergenic metals such as surgical steel or titanium if you have sensitive skin. While sterling silver is a popular metal choice for jewelry, other silver jewelry options exist, including silver-filled and silver-plated jewelry. Learn more about the different options available by Exploring the Various Types of Silver for Jewelry. Sterling Silver vs. Silver Pure Silver is a metal that is abundant in nature. Like gold and copper, silver occurs naturally in ores (rocks containing silver) and is available with little processing. Pure silver is often called “fine silver” or “.999 silver” as it is made of 99.9% silver, with the remaining 0.1% being trace impurities. Ancient Greek, Egyptian and Roman civilizations widely used silver as currency. Because of its softness, silver was essentially useless in manufacturing and weaponry. But silver’s high malleability allows it to be easily hammered into thin sheets, which caught the eye of jewelry makers. Pros of pure silver: Easy to work with: Its malleability makes silver perfect for a wide variety of uses, including battery making, dentistry and much more. Highly conductive: Silver has the highest electrical and thermal conductivity of all metals, making it useful and versatile in technology and electronics. Antibacterial: Silver naturally has a low toxicity to humans and is often used for things like medical equipment and water purification processes. Cons of pure silver: Naturally soft. In its pure form, silver is too soft to serve as a solo option for metal jewelry without adding an alloy. Pure silver’s softness makes it difficult to mold into an intricate design that holds its shape. Expensive. Especially when compared to silver-alloy alternatives, fine silver will typically cost a bit more. What do these Numbers Mean on Sterling Silver? Sterling silver jewelry has several, authentic identification markings including: 925 925/100 92.5% pure Ster Sterling Sterling Silver Sterling silver is used across the spectrum of jewelry types due to its many positive features. However, some factors may be seen as downsides to sterling silver, too. Comparing Sterling Silver and Silver: How to Tell Them Apart By now, the primary similarities and differences between sterling silver and pure silver are fairly evident. Silver is soft and vulnerable in its raw, pure form, while sterling silver’s alloy blend makes it a harder, more resistant metal. Pure silver tends to be more gray, while sterling silver shines with a brighter, whiter brilliance, similar to white gold. Pure silver is hypoallergenic, but sterling silver’s blend of metals can create skin reactions in people with skin sensitivities. While both are susceptible to tarnish, fine silver (.999) will tarnish more slowly than sterling silver (.925) due to its higher concentration of silver. Pure silver is also typically more expensive than sterling silver. If you’re considering a jewelry purchase, you might want to weigh your options between white gold and silver. How to Tell if Sterling Silver is Real Sometimes “silver” jewelry is instead a cheaper silver that has been alloyed with a lower-quality metal. If you’re wondering whether your current or future sterling silver jewelry is real, here are a few tips and tests that can help you determine your answer: Check for the hallmark silver markings, such as “925,” as mentioned above. If no hallmarks are present, ask the seller for the contents of the jewelry. If you already own the piece, consult a professional jeweler. Silver is not magnetic. If your jewelry smells like a handful of change, it may be an imposter, as real silver does not omit that metallic money scent. Real silver jewelry tends to tarnish and turn black. Rub your silver jewelry with a white cloth. If no black residue appears on your cloth, it’s likely not silver. Learn more about spotting a silver fake with How to Tell if Silver Jewelry is Fake: 6 Tests to Try at Home. Does Sterling Silver Tarnish? What about Pure Silver? Yes, both pure silver and sterling silver can tarnish over time. Silver tarnish occurs when pure silver encounters sulfur-containing substances in the air, forming silver sulfide, which appears as black or gray tarnish. Sterling silver is an alloy that includes metals also susceptible to corrosion (such as copper or nickel). Sterling silver tarnishes at a faster rate than pure silver, due to its copper content, as copper is more reactive than pure silver. The color change is strong, and the surface of the alloy can become yellow, brown, or even a faint blue or black color. The word “tarnish” can often carry a negative connotation. However, some people enjoy the look of tarnished sterling silver, which can be a unique, tasteful touch to jewelry and home or office decor. How to Care for Sterling Silver Despite being prone to tarnish, sterling silver jewelry can maintain its beautiful shine with proper care and cleaning. Storing sterling silver pieces in separate containers or pouches can help prevent scratching caused by pieces clanking together in the same pouch. Be sure your sterling silver jewelry is not stored on a treated surface, such as stained or treated wood, as this can increase the tarnishing process. Once separated, store jewelry somewhere dry, away from direct sunlight. If excess moisture is a concern, consider adding silica packs or a piece of chalk in your jewelry box or jewelry pouches, as they help absorb moisture and chemicals. Lastly, gently wiping your jewelry with an anti-tarnish cloth after use and before storing can help stave off tarnishing. A great bonus of sterling silver is that its tarnish can be easily removed with a polishing cloth. However, for a thorough, deep cleaning of your silver jewelry, consider the following cleaning technique. From Werewolves and Spoons to the Silver Screen, Silver’s Shiny History That’s a lot of silver information to take in at one time. But, for those yearning for more, you’re in luck! Here are four fun, silver history tidbits: Silver and werewolves. In folklore and fiction, silver is a classic weapon against werewolves and other supernatural creatures. While silver has some antimicrobial properties, there's no scientific evidence it can ward off mythical beasts. Silverware. Many types of flatware are made with stainless steel, which makes it great for everyday use since it’s relatively affordable, durable and easy to clean. Flatware is also sometimes made from sterling silver, which is why it’s commonly referred to as silverware. However, this may be a pricier option and will require more frequent maintenance. Silver screen. In the early days of movie projection technology, industry innovators found that coating their movie screen surfaces with metallic paint increased the contrast of the image and reduced blurriness. While the term “silver screen” became synonymous with movie screens, the metal used to coat screens wasn’t necessarily silver at all. In fact, various shiny, reflective metals were used. And there you have it—your easy guide to silver and sterling silver. If you own or are hoping to purchase valuable silver jewelry, protect your investments by obtaining personal jewelry insurance through Jewelers Mutual instead of insuring your valuables through homeowners insurance. It’s an affordable option that covers loss, damage and disappearance without impacting your home insurance premiums. You can check your rate by clicking the button below. Get a Jewelry Insurance Quote Estimate your rate in less than one minute. Just tell us what you need to insure, how much it’s worth and your zip/postal code. CHECK YOUR RATE Already have a quote? Retrieve your saved quote or application. Coverage not available in Quebec Most Popular Posts 6 Ear Piercing Care Mistakes You're Probably Making 5 Questions to Ask Your Jeweler When Buying an Engagement Ring How to Clean Gold Jewelry the Right Way When and How to Make a Ring Smaller Without Resizing Using a Homemade Jewelry Cleaner? Avoid These 3 Most Popular Posts What to Know About Anti-Money Laundering Certification for Jewelers 5-Step Rooftop Burglary Procedure Criminals Use & What You Can Do to Stop Them USPS Shipping Options: What They Mean to Your Jewelry Business 5 Jewelry Store Promotion Ideas That Don't Cost a Dime The Difference Between Priority Mail Express® and Express Mail® Contact Us If you have a media-related question, please email us at mediainquiries@jminsure.com. About Jewelers Mutual Group Jewelers Mutual was founded in 1913 by a group of Wisconsin jewelers to meet their unique insurance needs. Later, consumers began putting their trust in Jewelers Mutual to protect their jewelry and the special memories each piece holds. Today, Jewelers Mutual continues to support and move the industry forward by listening to jewelers and consumers and offering products and services to meet their evolving needs. Beyond insurance, Jewelers Mutual’s powerful suite of innovative solutions and digital technology offerings help jewelers strengthen and grow their businesses, mitigate risk, and bring them closer to their customers. The Group insurers’ strong financial position is reflected in their 38 consecutive “A+ Superior” ratings from AM Best Company, as of November 2024. Policyholders of the Group insurers are members of Jewelers Mutual Holding Company. Jewelers Mutual is headquartered in Neenah, Wisconsin, with other Group offices in Dallas, Miami, and Raleigh, North Carolina. To learn more, visit JewelersMutual.com. Posts by Topic Anniversary & Wedding Appraisal Claims Cleaning & Care Coverage Earrings Gems Holidays Jewelry Industry Loss Prevention Necklaces Proposal & Engagement Rings Safety & Security Settings Tips Travel Trends Watches
14599	https://www.accessdata.fda.gov/drugsatfda_docs/label/2022/086069Orig1s026lbl.pdf	Each gram contains 0.1 mg estradiol in a nonliquefying base NDC 0430-3754-14 NET WT 42.5G Rx only This product also contains puriﬁed water, propylene glycol, stearyl alcohol, white ceresin wax, mono- and di-glycerides, hypromellose, sodium lauryl sulfate, methylparaben, edetate disodium, and t-butylhydroquinone. CAUTION: Keep this and all medications out of the reach of children. USUAL DOSAGE: See Package Insert for Full Prescribing Information. Store at room temperature. Protect from temperatures in excess of 40° C (104° F). Manufactured In Canada by: Contract Pharmaceuticals Limited Mississauga, Ontario, Canada L5N 6L6 Distributed by: Allergan USA, Inc. Irvine, CA 92612 © 2022 Allergan. All rights reserved. ESTRACE® is a registered trademark of Allergan Therapeutics LLC. 2 2 2 8 4 US 1 1 _ d ra ft 2009126 3 N 2 0430-3754-14 (b) (4) This label may not be the latest approved by FDA. For current labeling information, please visit NET WT 42.5G TUBE Calibrated Applicator Enclosed NDC 0430-3754-14 0430-3754-14 INSTRUCTIONS FOR USE OF APPLICATOR: 1. Wash and dry your hands well. Remove cap from the tube. 2. Screw threaded end of applicator onto the open nozzle of the tube until secure. Do not attach the plunger end of the applicator onto the open tube. 3. Gently squeeze tube from the bottom to push the prescribed amount of cream into the applicator. 4. Unscrew applicator from tube and replace cap on tube. Lie on your back with knees bent. Gently insert applicator deeply into your vagina and press the plunger downward to its original position. Remove applicator from your vagina. To cleanse: Pull plunger out from barrel. Wash with mild soap and warm water. DO NOT BOIL OR USE HOT WATER. NOTE: The number of doses per tube will vary with dosage requirements and patient handling. Usual Dosage: See enclosed Package Insert for Full Prescribing Information. CAUTION: Keep this and all medications out of the reach of children. Each gram contains 0.1 mg estradiol in a nonliquefying base. NDC 0430-3754-14 This product also contains purifed water, propylene glycol, stearyl alcohol, white ceresin wax, mono- and di-glycerides, hypromellose, sodium lauryl sulfate, methylparaben, edetate disodium, and t-butylhydroquinone. CALIBRATED APPLICATOR ENCLOSED Read Enclosed Information for the Patient GTIN 00304303754142 2009128 66229US11 _draft UNSCENTED NET WT 42.5G TUBE To report SUSPECTED ADVERSE REACTIONS, contact Allergan at 1-800-433-8871 or FDA at 1-800-FDA-1088 or www.fda.gov/medwatch. Store at room temperature. Protect from temperatures in excess of 40˚ C (104˚ F). Dispense With Insert. Manufactured In Canada by: Contract Pharmaceuticals Limited Mississauga, Ontario, Canada L5N 6L6 Distributed by: Allergan USA, Inc. Irvine, CA 92612 © 2022 Allergan. All rights reserved. ESTRACE® is a registered trademark of Allergan Therapeutics LLC. 3 N 2 0430-3754-14 (b) (4) This label may not be the latest approved by FDA. For current labeling information, please visit ESTRACE® Cream (estradiol vaginal cream, USP, 0.01%) WARNING: ENDOMETRIAL CANCER, CARDIOVASCULAR DISORDERS, BREAST CANCER and PROBABLE DEMENTIA Estrogen-Alone Therapy Endometrial Cancer There is an increased risk of endometrial cancer in a woman with a uterus who uses unopposed estrogens. Adding a progestin to estrogen therapy has been shown to reduce the risk of endometrial hyperplasia, which may be a precursor to endometrial cancer. Adequate diagnostic measures, including directed or random endometrial sampling when indicated, should be undertaken to rule out malignancy in postmenopausal women with undiagnosed persistent or recurring abnormal genital bleeding [see WARNINGS, Malignant Neoplasms, Endometrial Cancer]. Cardiovascular Disorders and Probable Dementia Estrogen-alone therapy should not be used for the prevention of cardiovascular disease or dementia [see CLINICAL STUDIES and WARNINGS, Cardiovascular Disorders, and Probable Dementia]. The Women’s Health Initiative (WHI) estrogen-alone substudy reported increased risks of stroke and deep vein thrombosis (DVT) in postmenopausal women (50 to 79 years of age) during 7.1 years of treatment with daily oral conjugated estrogens (CE) [0.625 mg]-alone, relative to placebo [see CLINICAL STUDIES and WARNINGS, Cardiovascular Disorders]. The WHI Memory Study (WHIMS) estrogen-alone ancillary study of WHI reported an increased risk of developing probable dementia in postmenopausal women 65 years of age or older during 5.2 years of treatment with daily CE (0.625 mg) -alone, relative to placebo. It is unknown whether this finding applies to younger postmenopausal women [see CLINICAL STUDIES and WARNINGS, Probable Dementia and PRECAUTIONS, Geriatric Use]. In the absence of comparable data, these risks should be assumed to be similar for other doses of CE and other dosage forms of estrogens. Estrogens with or without progestins should be prescribed at the lowest effective doses and for the shortest duration consistent with treatment goals and risks for the individual woman. Estrogen Plus Progestin Therapy Cardiovascular Disorders and Probable Dementia Estrogen plus progestin therapy should not be used for the prevention of cardiovascular disease or dementia [see CLINICAL STUDIES and WARNINGS, Cardiovascular Disorders, and Probable Dementia]. This label may not be the latest approved by FDA. For current labeling information, please visit The WHI estrogen plus progestin substudy reported increased risks of DVT, pulmonary embolism (PE), stroke and myocardial infarction (MI) in postmenopausal women (50 to 79 years of age) during 5.6 years of treatment with daily oral CE (0.625 mg) combined with medroxyprogesterone acetate (MPA) [2.5 mg], relative to placebo [see CLINICAL STUDIES and WARNINGS, Cardiovascular Disorders]. The WHIMS estrogen plus progestin ancillary study of the WHI reported an increased risk of developing probable dementia in postmenopausal women 65 years of age or older during 4 years of treatment with daily CE (0.625 mg) combined with MPA (2.5 mg), relative to placebo. It is unknown whether this finding applies to younger postmenopausal women [see CLINICAL STUDIES and WARNINGS, Probable Dementia and PRECAUTIONS, Geriatric Use]. Breast Cancer The WHI estrogen plus progestin substudy also demonstrated an increased risk of invasive breast cancer [see CLINICAL STUDIES and WARNINGS, Malignant Neoplasms, Breast Cancer]. In the absence of comparable data, these risks should be assumed to be similar for other doses of CE and MPA, and other combinations and dosage forms of estrogens and progestins. Estrogens with or without progestins should be prescribed at the lowest effective doses and for the shortest duration consistent with treatment goals and risks for the individual woman. DESCRIPTION Each gram of ESTRACE (estradiol vaginal cream, USP, 0.01%) contains 0.1 mg estradiol in a nonliquefying base containing purified water, propylene glycol, stearyl alcohol, white ceresin wax, mono- and di-glycerides, hypromellose 2208 (4000 cps), sodium lauryl sulfate, methylparaben, edetate di-sodium and tertiary-butylhydroquinone. Estradiol is chemically described as estra-1,3,5(10)-triene-3, 17(beta)-diol. It has an empirical formula of C18H24O2 and molecular weight of 272.37. The structural formula is: CLINICAL PHARMACOLOGY Endogenous estrogens are largely responsible for the development and maintenance of the female reproductive system and secondary sexual characteristics. Although circulating estrogens This label may not be the latest approved by FDA. For current labeling information, please visit exist in a dynamic equilibrium of metabolic interconversions, estradiol is the principal intracellular human estrogen and is substantially more potent than its metabolites, estrone and estriol at the receptor level. The primary source of estrogen in normally cycling adult women is the ovarian follicle, which secretes 70 to 500 mcg of estradiol daily, depending on the phase of the menstrual cycle. After menopause, most endogenous estrogen is produced by conversion of androstenedione, secreted by the adrenal cortex, to estrone by peripheral tissues. Thus, estrone and the sulfate conjugated form, estrone sulfate, are the most abundant circulating estrogens in postmenopausal women. Estrogens act through binding to nuclear receptors in estrogen-responsive tissues. To date, two estrogen receptors have been identified. These vary in proportion from tissue to tissue. Circulating estrogens modulate the pituitary secretion of the gonadotropins, luteinizing hormone (LH) and follicle stimulating hormone (FSH), through a negative feedback mechanism. Estrogens act to reduce the elevated levels of these hormones seen in postmenopausal women. Pharmacokinetics Absorption Estrogen drug products are absorbed through the skin, mucous membranes, and the gastrointestinal tract after release from the drug formulation. Distribution The distribution of exogenous estrogens is similar to that of endogenous estrogens. Estrogens are widely distributed in the body and are generally found in higher concentrations in the sex hormone target organs. Estrogens circulate in the blood largely bound to sex hormone binding globulin (SHBG) and albumin. Metabolism Exogenous estrogens are metabolized in the same manner as endogenous estrogens. Circulating estrogens exist in a dynamic equilibrium of metabolic interconversions. These transformations take place mainly in the liver. Estradiol is converted reversibly to estrone, and both can be converted to estriol, which is the major urinary metabolite. Estrogens also undergo enterohepatic recirculation via sulfate and glucuronide conjugation in the liver, biliary secretion of conjugates into the intestine, and hydrolysis in the gut followed by reabsorption. In postmenopausal women, a significant proportion of the circulating estrogens exist as sulfate conjugates, especially estrone sulfate, which serves as a circulating reservoir for the formation of more active estrogens. Excretion This label may not be the latest approved by FDA. For current labeling information, please visit Estradiol, estrone, and estriol are excreted in the urine along with glucuronide and sulfate conjugates. Special Populations No pharmacokinetic studies were conducted in special populations, including patients with renal or hepatic impairment. Drug Interactions In vitro and in vivo studies have shown that estrogens are metabolized partially by cytochrome P450 3A4 (CYP3A4). Therefore, inducers or inhibitors of CYP3A4 may affect estrogen drug metabolism. Inducers of CYP3A4 such as St. John’s Wort preparations (Hypericum perforatum), phenobarbital, carbamazepine, and rifampin may reduce plasma concentrations of estrogens, possibly resulting in a decrease in therapeutic effects and/or changes in the uterine bleeding profile. Inhibitors of CYP3A4 such as erythromycin, clarithromycin, ketoconazole, itraconazole, ritonavir and grapefruit juice may increase plasma concentrations of estrogens and may result in side effects. CLINICAL STUDIES Women’s Health Initiative Studies The WHI enrolled approximately 27,000 predominantly healthy postmenopausal women in two substudies to assess the risks and benefits of daily oral CE (0.625 mg)-alone or in combination with MPA (2.5 mg) compared to placebo in the prevention of certain chronic diseases. The primary endpoint was the incidence of coronary heart disease (CHD) (defined as nonfatal MI, silent MI and CHD death), with invasive breast cancer as the primary adverse outcome. A “global index” included the earliest occurrence of CHD, invasive breast cancer, stroke, PE, endometrial cancer (only in the CE plus MPA substudy), colorectal cancer, hip fracture, or death due to other cause. These substudies did not evaluate the effects of CE or CE plus MPA on menopausal symptoms. WHI Estrogen-Alone Substudy The WHI estrogen-alone substudy was stopped early because an increased risk of stroke was observed, and it was deemed that no further information would be obtained regarding the risks and benefits of estrogen-alone in predetermined primary endpoints. Results of the estrogen-alone substudy, which included 10,739 women (average 63 years of age, range 50 to 79; 75.3 percent White, 15.1 percent Black, 6.1 percent Hispanic, 3.6 percent Other), after an average follow-up of 7.1 years are presented in Table 1. TABLE 1 -Relative and Absolute Risk Seen in the Estrogen-Alone Substudy of WHIa This label may not be the latest approved by FDA. For current labeling information, please visit For those outcomes included in the WHI "global index" that reached statistical significance, the absolute excess risk per 10,000 women-years in the group treated with CE-alone was 12 more strokes, while the absolute risk reduction per 10,000 women-years was 7 fewer hip fractures1. The absolute excess risk of events included in the "global index" was a non-significant 5 events per 10,000 women-years. There was no difference between the groups in terms of all-cause mortality. No overall difference for primary CHD events (nonfatal MI, silent MI and CHD death) and invasive breast cancer incidence in women receiving CE-alone compared with placebo was reported in final centrally adjudicated results from the estrogen-alone substudy, after an average follow-up of 7.1 years. Centrally adjudicated results for stroke events from the estrogen-alone substudy, after an average follow-up of 7.1 years, reported no significant differences in distribution of stroke subtypes or severity, including fatal strokes, in women receiving CE-alone compared to This label may not be the latest approved by FDA. For current labeling information, please visit placebo. Estrogen-alone increased the risk for ischemic stroke, and this excess risk was present in all subgroups of women examined2. Timing of the initiation of estrogen-alone therapy relative to the start of menopause may affect the overall risk benefit profile. The WHI estrogen-alone substudy stratified by age showed in women 50 to 59 years of age a non-significant trend toward reduced risk for CHD [hazard ratio (HR) 0.63 (95 percent CI, 0.36-1.09)] and overall mortality [HR 0.71 (95 percent CI, 0.46– 1.11)]. WHI Estrogen Plus Progestin Substudy The WHI estrogen plus progestin substudy was also stopped early. According to the predefined stopping rule, after an average follow-up of 5.6 years of treatment, the increased risk of breast cancer and cardiovascular events exceeded the specified benefits included in the “global index.” The absolute excess risk of events included in the “global index” was 19 per 10,000 women-years. For those outcomes included in the WHI “global index” that reached statistical significance after 5.6 years of follow-up, the absolute excess risks per 10,000 women-years in the group treated with CE plus MPA were 7 more CHD events, 8 more strokes, 10 more PEs, and 8 more invasive breast cancers, while the absolute risk reductions per 10,000 women-years were 6 fewer colorectal cancers and 5 fewer hip fractures. Results of the CE plus MPA substudy, which included 16,608 women (average 63 years of age, range 50 to 79; 83.9 percent White, 6.8 percent Black, 5.4 percent Hispanic, 3.9 percent Other), are presented in Table 2. These results reflect centrally adjudicated data after an average follow- up of 5.6 years. TABLE 2 -Relative and Absolute Risk Seen in the Estrogen Plus Progestin Substudy of WHI at an Average of 5.6 Yearsa,b This label may not be the latest approved by FDA. For current labeling information, please visit Timing of the initiation of estrogen plus progestin therapy relative to the start of menopause may affect the overall risk benefit profile. The WHI estrogen plus progestin substudy stratified by age showed in women 50 to 59 years of age, a non-significant trend toward reduced risk for overall mortality [HR 0.69 (95 percent CI, 0.44-1.07)]. Women’s Health Initiative Memory Study The WHIMS estrogen-alone ancillary study of WHI enrolled 2,947 predominantly healthy hysterectomized postmenopausal women 65 to 79 years of age and older (45 percent were 65 to 69 years of age; 36 percent were 70 to 74 years of age; 19 percent were 75 years of age and older) to evaluate the effects of daily CE (0.625 mg)-alone on the incidence of probable dementia (primary outcome) compared to placebo. After an average follow-up of 5.2 years, the relative risk of probable dementia for CE-alone versus placebo was 1.49 (95 percent CI, 0.83-2.66). The absolute risk of probable dementia for CE-alone versus placebo was 37 versus 25 cases per 10,000 women-years. Probable dementia as defined in this study included Alzheimer’s disease (AD), vascular dementia (VaD) and mixed types (having features of both AD and VaD). The most common classification of probable dementia in the treatment group and the placebo group was AD. Since the ancillary study was This label may not be the latest approved by FDA. For current labeling information, please visit conducted in women 65 to 79 years of age, it is unknown whether these findings apply to younger postmenopausal women [see BOXED WARNINGS, WARNINGS, Probable Dementia and PRECAUTIONS, Geriatric Use]. The WHIMS estrogen plus progestin ancillary study of WHI enrolled 4,532 predominantly healthy postmenopausal women 65 years of age and older (47 percent were 65 to 69 years of age; 35 percent were 70 to 74 years; 18 percent were 75 years of age and older) to evaluate the effects of daily CE (0.625 mg) plus MPA (2.5 mg) on the incidence of probable dementia (primary outcome) compared to placebo. After an average follow-up of 4 years, the relative risk of probable dementia for CE plus MPA versus placebo was 2.05 (95 percent CI, 1.21-3.48). The absolute risk of probable dementia for CE plus MPA versus placebo was 45 versus 22 per 10,000 women-years. Probable dementia as defined in this study included AD, VaD and mixed types (having features of both AD and VaD). The most common classification of probable dementia in the treatment group and the placebo group was AD. Since the ancillary study was conducted in women 65 to 79 years of age, it is unknown whether these findings apply to younger postmenopausal women [see WARNINGS, Probable Dementia and PRECAUTIONS, Geriatric Use]. When data from the two populations were pooled as planned in the WHIMS protocol, the reported overall relative risk for probable dementia was 1.76 (95 percent CI, 1.19-2.60). Differences between groups became apparent in the first year of treatment. It is unknown whether these findings apply to younger postmenopausal women [see WARNINGS, Probable Dementia and PRECAUTIONS, Geriatric Use]. INDICATIONS AND USAGE ESTRACE (estradiol vaginal cream, USP, 0.01%) is indicated in the treatment of moderate to severe symptoms of vulvar and vaginal atrophy due to menopause. CONTRAINDICATIONS ESTRACE (estradiol vaginal cream, USP, 0.01%) should not be used in women with any of the following conditions: 1. Undiagnosed abnormal genital bleeding. 2. Known, suspected, or history of cancer of the breast. 3. Known or suspected estrogen-dependent neoplasia. 4. Active DVT, PE or history of these conditions. 5. Active arterial thromboembolic disease (for example, stroke, MI) or a history of these conditions. 6. Known anaphylactic reaction or angioedema to ESTRACE (estradiol vaginal cream, USP, 0.01%). 7. Known liver dysfunction or disease. 8. Known protein C, protein S, or antithrombin deficiency, or other known thrombophilic disorders. 9. Known or suspected pregnancy. This label may not be the latest approved by FDA. For current labeling information, please visit WARNINGS See BOXED WARNINGS. Systemic absorption may occur with the use of ESTRACE (estradiol vaginal cream, USP, 0.01%). The warnings, precautions, and adverse reactions associated with oral estrogen treatment should be taken into account. 1. Cardiovascular Disorders An increased risk of stroke and DVT has been reported with estrogen-alone therapy. An increased risk of PE, DVT, stroke and MI has been reported with estrogen plus progestin therapy. Should any of these occur or be suspected, estrogen with or without progestin therapy should be discontinued immediately. Risk factors for arterial vascular disease (e.g., hypertension, diabetes mellitus, tobacco use, hypercholesterolemia, and obesity) and/or venous thromboembolism (VTE) (e.g., personal history or family history of VTE, obesity, and systemic lupus erythematosus) should be managed appropriately. a. Stroke In the WHI estrogen-alone substudy, a statistically significant increased risk of stroke was reported in women 50 to 79 years of age receiving daily CE (0.625 mg)-alone compared to women in the same age group receiving placebo (45 versus 33 per 10,000 women-years). The increase in risk was demonstrated in year one and persisted [see CLINICAL STUDIES]. Should a stroke occur or be suspected, estrogen-alone therapy should be discontinued immediately. Subgroup analyses of women 50 to 59 years of age suggest no increased risk of stroke for those women receiving CE (0.625 mg)-alone versus those receiving placebo (18 versus 21 per 10,000 women-years)3. In the WHI estrogen plus progestin substudy, a statistically significant increased risk of stroke was reported in women 50 to 79 years of age receiving daily CE (0.625 mg) plus MPA (2.5 mg) compared to women in the same age group receiving placebo (33 versus 25 per 10,000 women- years) [see CLINICAL STUDIES]. The increase in risk was demonstrated after the first year and persisted3. Should a stroke occur or be suspected, estrogen plus progestin therapy should be discontinued immediately. b. Coronary Heart Disease In the WHI estrogen-alone substudy, no overall effect on coronary heart disease (CHD) events (defined as nonfatal MI, silent MI and CHD death) was reported in women receiving estrogen-alone compared to placebo4 [see CLINICAL STUDIES]. This label may not be the latest approved by FDA. For current labeling information, please visit Subgroup analyses of women 50 to 59 years of age suggest a statistically non-significant reduction in CHD events (CE [0.625 mg]-alone compared to placebo) in women with less than 10 years since menopause (8 versus 16 per 10,000 women-years) 3. In the WHI estrogen plus progestin substudy, there was a statistically non-significant increased risk of CHD events reported in women receiving daily CE (0.625 mg) plus MPA (2.5 mg) compared to women receiving placebo (41 versus 34 per 10,000 women-years) 3. An increase in relative risk was demonstrated in year 1, and a trend toward decreasing relative risk was reported in years 2 through 5 [see CLINICAL STUDIES]. In postmenopausal women with documented heart disease (n = 2,763), average age 66.7 years of age, in a controlled clinical trial of secondary prevention of cardiovascular disease (Heart and Estrogen/Progestin Replacement Study; HERS), treatment with daily CE (0.625 mg) plus MPA (2.5 mg) demonstrated no cardiovascular benefit. During an average follow-up of 4.1 years, treatment with CE plus MPA did not reduce the overall rate of CHD events in postmenopausal women with established coronary heart disease. There were more CHD events in the CE plus MPA-treated group than in the placebo group in year 1, but not during the subsequent years. Two thousand, three hundred and twenty one (2,321) women from the original HERS trial agreed to participate in an open label extension of HERS, HERS II. Average follow-up in HERS II was an additional 2.7 years, for a total of 6.8 years overall. Rates of CHD events were comparable among women in the CE plus MPA group and the placebo group in HERS, HERS II, and overall. c. Venous Thromboembolism In the WHI estrogen-alone substudy, the risk of VTE (DVT and PE) was increased for women receiving daily CE (0.625 mg)-alone compared to women receiving placebo (30 versus 22 per 10,000 women-years), although only the increased risk of DVT reached statistical significance (23 versus 15 per 10,000 women-years). The increase in VTE risk was demonstrated during the first 2 years5 [see CLINICAL STUDIES]. Should a VTE occur or be suspected, estrogen-alone therapy should be discontinued immediately. In the WHI estrogen plus progestin substudy, a statistically significant 2-fold greater rate of VTE was reported in women receiving daily CE (0.625 mg) plus MPA (2.5 mg) compared to women receiving placebo (35 versus 17 per 10,000 women-years). Statistically significant increases in risk for both DVT (26 versus 13 per 10,000 women-years) and PE (18 versus 8 per 10,000 women-years) were also demonstrated. The increase in VTE risk was observed during the first year and persisted6 [see CLINICAL STUDIES]. Should a VTE occur or be suspected, estrogen plus progestin therapy should be discontinued immediately. If feasible, estrogens should be discontinued at least 4 to 6 weeks before surgery of the type associated with an increased risk of thromboembolism, or during periods of prolonged immobilization. 2. Malignant Neoplasms a. Endometrial Cancer This label may not be the latest approved by FDA. For current labeling information, please visit An increased risk of endometrial cancer has been reported with the use of unopposed estrogen therapy in a woman with a uterus. The reported endometrial cancer risk among unopposed estrogen users is about 2- to 12-times greater than in non-users, and appears dependent on duration of treatment and on estrogen dose. Most studies show no significant increased risk associated with use of estrogens for less than one year. The greatest risk appears associated with prolonged use, with increased risks of 15- to 24-fold for five to ten years or more and this risk has been shown to persist for at least 8 to 15 years after estrogen therapy is discontinued. Clinical surveillance of all women using estrogen-alone or estrogen plus progestin therapy is important. Adequate diagnostic measures, including directed or random endometrial sampling when indicated, should be undertaken to rule out malignancy in postmenopausal women with undiagnosed persistent or recurring abnormal genital bleeding. There is no evidence that the use of natural estrogens results in a different endometrial risk profile than synthetic estrogens of equivalent estrogen dose. Adding a progestin to estrogen therapy has been shown to reduce the risk of endometrial hyperplasia, which may be a precursor to endometrial cancer. b. Breast Cancer The most important randomized clinical trial providing information about breast cancer in estrogen-alone users is the WHI substudy of daily CE (0.625 mg)-alone. In the WHI estrogen-alone substudy, after an average follow-up of 7.1 years, daily CE (0.625 mg)-alone was not associated with an increased risk of invasive breast cancer (relative risk [RR] 0.80) 7 [see CLINICAL STUDIES]. The most important randomized clinical trial providing information about breast cancer in estrogen plus progestin users is the WHI substudy of daily CE (0.625 mg) plus MPA (2.5 mg). After a mean follow-up of 5.6 years, the estrogen plus progestin substudy reported an increased risk of invasive breast cancer in women who took daily CE plus MPA. In this substudy, prior use of estrogen-alone or estrogen plus progestin therapy was reported by 26 percent of the women. The relative risk of invasive breast cancer was 1.24, and the absolute risk was 41 versus 33 cases per 10,000 women-years, for CE plus MPA compared with placebo. Among women who reported prior use of hormone therapy, the relative risk of invasive breast cancer was 1.86, and the absolute risk was 46 versus 25 cases per 10,000 women-years, for CE plus MPA compared with placebo. Among women who reported no prior use of hormone therapy, the relative risk of invasive breast cancer was 1.09, and the absolute risk was 40 versus 36 cases per 10,000 women- years, for CE plus MPA compared with placebo. In the same substudy, invasive breast cancers were larger, were more likely to be node positive, and were diagnosed at a more advanced stage in the CE (0.625 mg) plus MPA (2.5 mg) group compared with the placebo group. Metastatic disease was rare with no apparent difference between the two groups. Other prognostic factors such as histologic subtype, grade and hormone receptor status did not differ between the groups8 [see CLINICAL STUDIES]. Consistent with the WHI clinical trial, observational studies have also reported an increased risk of breast cancer for estrogen plus progestin therapy, and a smaller increased risk for estrogen- This label may not be the latest approved by FDA. For current labeling information, please visit alone therapy, after several years of use. The risk increased with duration of use, and appeared to return to baseline in about 5 years after stopping treatment (only the observational studies have substantial data on risk after stopping). Observational studies also suggest that the risk of breast cancer was greater, and became apparent earlier, with estrogen plus progestin therapy as compared to estrogen-alone therapy. However, these studies have not generally found significant variation in the risk of breast cancer among different estrogen plus progestin combinations, doses, or routes of administration. The use of estrogen-alone and estrogen plus progestin therapy has been reported to result in an increase in abnormal mammograms requiring further evaluation. All women should receive yearly breast examinations by a healthcare provider and perform monthly breast self-examinations. In addition, mammography examinations should be scheduled based on patient age, risk factors, and prior mammogram results. c. Ovarian Cancer The WHI estrogen plus progestin substudy reported a statistically non-significant increased risk of ovarian cancer. After an average follow-up of 5.6 years, the relative risk for ovarian cancer for CE plus MPA versus placebo was 1.58 (95 percent CI, 0.77-3.24). The absolute risk for CE plus MPA was 4 versus 3 cases per 10,000 women-years9. A meta-analysis of 17 prospective and 35 retrospective epidemiology studies found that women who used hormonal therapy for menopausal symptoms had an increased risk for ovarian cancer. The primary analysis, using case-control comparisons, included 12,110 cancer cases from the 17 prospective studies. The relative risks associated with current use of hormonal therapy was 1.41 (95% confidence interval [CI] 1.32 to 1.50); there was no difference in the risk estimates by duration of the exposure (less than 5 years [median of 3 years]vs. greater than 5 years [median of 10 years] of use before the cancer diagnosis). The relative risk associated with combined current and recent use (discontinued use within 5 years before cancer diagnosis) was 1.37 (95% CI 1.27-1.48), and the elevated risk was significant for both estrogen-alone and estrogen plus progestin products. The exact duration of hormone therapy use associated with an increased risk of ovarian cancer, however, is unknown. 3. Probable Dementia In the WHIMS estrogen-alone ancillary study of WHI, a population of 2,947 hysterectomized women 65 to 79 years of age was randomized to daily CE (0.625 mg)-alone or placebo. After an average follow-up of 5.2 years, 28 women in the estrogen-alone group and 19 women in the placebo group were diagnosed with probable dementia. The relative risk of probable dementia for CE-alone versus placebo was 1.49 (95 percent CI, 0.83-2.66). The absolute risk of probable dementia for CE-alone versus placebo was 37 versus 25 cases per 10,000 women-years 10 [see CLINICAL STUDIES and PRECAUTIONS, Geriatric Use]. In the WHIMS estrogen plus progestin ancillary study, a population of 4,532 postmenopausal women 65 to 79 years of age was randomized to daily CE (0.625 mg) plus MPA (2.5 mg) or placebo. This label may not be the latest approved by FDA. For current labeling information, please visit After an average follow-up of 4 years, 40 women in the CE plus MPA group and 21 women in the placebo group were diagnosed with probable dementia. The relative risk of probable dementia for CE plus MPA versus placebo was 2.05 (95 percent CI, 1.21-3.48). The absolute risk of probable dementia for CE plus MPA versus placebo was 45 versus 22 cases per 10,000 women-years10 [see CLINICAL STUDIES and PRECAUTIONS, Geriatric Use]. When data from the two populations in the WHIMS estrogen-alone and estrogen plus progestin ancillary studies were pooled as planned in the WHIMS protocol, the reported overall relative risk for probable dementia was 1.76 (95 percent CI, 1.19-2.60). Since both ancillary studies were conducted in women 65 to 79 years of age, it is unknown whether these findings apply to younger postmenopausal women 10 [see PRECAUTIONS, Geriatric Use]. 4. Gallbladder Disease A 2- to 4-fold increase in the risk of gallbladder disease requiring surgery in postmenopausal women receiving estrogens has been reported. 5. Hypercalcemia Estrogen administration may lead to severe hypercalcemia in patients with breast cancer and bone metastases. If hypercalcemia occurs, use of the drug should be stopped and appropriate measures taken to reduce the serum calcium level. 6. Visual Abnormalities Retinal vascular thrombosis has been reported in patients receiving estrogens. Discontinue medication pending examination if there is sudden partial or complete loss of vision, or a sudden onset of proptosis, diplopia, or migraine. If examination reveals papilledema or retinal vascular lesions, estrogens should be permanently discontinued. 7. Anaphylactic Reaction and Angioedema Cases of anaphylaxis, which develop within minutes to hours after taking orally-administered estrogen and require emergency medical management, have been reported in the postmarketing setting. Skin (hives, pruritis, swollen lips-tongue-face) and either respiratory tract (respiratory compromise) or gastrointestinal tract (abdominal pain, vomiting) involvement has been noted. Angioedema involving the tongue, larynx, face, hands and feet requiring medical intervention has occurred postmarketing in patients taking orally-administered estrogen. If angioedema involves the tongue, glottis, or larynx, airway obstruction may occur. Patients who develop an anaphylactic reaction with or without angioedema after treatment with oral estrogen should not receive oral estrogen again. 8. Hereditary Angioedema Exogenous estrogens may exacerbate symptoms of angioedema in women with hereditary angioedema. This label may not be the latest approved by FDA. For current labeling information, please visit PRECAUTIONS A. General 1. Addition of a Progestin When a Woman Has Not Had a Hysterectomy Studies of the addition of a progestin for 10 or more days of a cycle of estrogen administration, or daily with estrogen in a continuous regimen, have reported a lowered incidence of endometrial hyperplasia than would be induced by estrogen treatment alone. Endometrial hyperplasia may be a precursor to endometrial cancer. There are, however, possible risks that may be associated with the use of progestins with estrogens compared to estrogen-alone regimens. These include a possible increased risk of breast cancer. 2. Elevated Blood Pressure In a small number of case reports, substantial increases in blood pressure have been attributed to idiosyncratic reactions to estrogens. In a large, randomized, placebo-controlled clinical trial, a generalized effect of estrogens on blood pressure was not seen. 3. Hypertriglyceridemia In women with pre-existing hypertriglyceridemia, estrogen therapy may be associated with elevations of plasma triglycerides leading to pancreatitis and other complications. Consider discontinuation of treatment if pancreatitis occurs. 4. Hepatic Impairment and/or Past History of Cholestatic Jaundice Estrogens may be poorly metabolized in patients with impaired liver function. For women with a history of cholestatic jaundice associated with past estrogen use or with pregnancy, caution should be exercised and in the case of recurrence, medication should be discontinued. 5. Hypothyroidism Estrogen administration leads to increased thyroid-binding globulin (TBG) levels. Women with normal thyroid function can compensate for the increased TBG by making more thyroid hormone, thus maintaining free T4 and T3 serum concentrations in the normal range. Women dependent on thyroid hormone replacement therapy who are also receiving estrogens may require increased doses of their thyroid replacement therapy. These women should have their thyroid function monitored in order to maintain their free thyroid hormone levels in an acceptable range. 6. Fluid Retention Estrogens may cause some degree of fluid retention. Women with conditions that might be influenced by this factor, such as a cardiac or renal dysfunction, warrant careful observation when estrogen-alone is prescribed. 7. Hypocalcemia This label may not be the latest approved by FDA. For current labeling information, please visit Estrogen therapy should be used with caution in women with hypoparathyroidism as estrogen- induced hypocalcemia may occur. 8. Exacerbation of Endometriosis A few cases of malignant transformation of residual endometrial implants have been reported in women treated post-hysterectomy with estrogen alone therapy. For patients known to have residual endometriosis post-hysterectomy, the addition of progestin should be considered. 9. Exacerbation of Other Conditions Estrogen therapy may cause an exacerbation of asthma, diabetes mellitus, epilepsy, migraine or porphyria, systemic lupus erythematosus, and hepatic hemangiomas and should be used with caution in women with these conditions. B. Patient Information Physicians are advised to discuss the PATIENT INFORMATION leaflet with women for whom they prescribe ESTRACE (estradiol vaginal cream, USP, 0.01%). C. Laboratory Tests Serum FSH and estradiol levels have not been shown to be useful in the management of moderate to severe symptoms of vulvar and vaginal atrophy. D. Drug-Laboratory Test Interactions 1. Accelerated prothrombin time, partial thromboplastin time, and platelet aggregation time; increased platelet count; increased factors II, VII antigen, VIII antigen, VIII coagulant activity, IX, X, XII, VII-X complex, II-VII-X complex, and beta-thromboglobulin; decreased levels of anti-factor Xa and antithrombin III, decreased antithrombin III activity; increased levels of fibrinogen and fibrinogen activity; increased plasminogen antigen and activity. 2. Increased thyroid-binding globulin (TBG) leading to increased circulating total thyroid hormone levels, as measured by protein-bound iodine (PBI), T4 levels (by column or by radioimmunoassay) or T3 levels by radioimmunoassay. T3 resin uptake is decreased, reflecting the elevated TBG. Free T4 and free T3 concentrations are unaltered. Women on thyroid replacement therapy may require higher dose of thyroid hormone. 3. Other binding proteins may be elevated in serum, i.e., corticosteroid binding globulin (CBG), sex hormone-binding globulin (SHBG), leading to increased total circulating corticosteroids and sex steroids, respectively. Free hormone concentrations, such as testosterone and estradiol, may be decreased. Other plasma proteins may be increased (angiotensinogen/renin substrate, alpha-1-antitrypsin, ceruloplasmin). 4. Increased plasma high-density lipoprotein (HDL) and HDL2 cholesterol subfraction concentrations, reduced low-density lipoprotein (LDL) cholesterol concentration, increased triglycerides levels. This label may not be the latest approved by FDA. For current labeling information, please visit 5. Impaired glucose tolerance. E. Carcinogenesis, Mutagenesis, and Impairment of Fertility Long-term continuous administration of estrogen, with and without progestin, in women with and without a uterus, has shown an increased risk of endometrial cancer, breast cancer, and ovarian cancer. (See BOXED WARNINGS, WARNINGS and PRECAUTIONS.) Long term continuous administration of natural and synthetic estrogens in certain animal species increases the frequency of carcinomas of the breast, uterus, cervix, vagina, testis, and liver. F. Pregnancy ESTRACE should not be used during pregnancy [see CONTRAINDICATIONS]. There appears to be little or no increased risk of birth defects in children born to women who have used estrogens and progestins as an oral contraceptive inadvertently during early pregnancy. G. Nursing Mothers ESTRACE should not be used during lactation. Estrogen administration to nursing women has been shown to decrease the quantity and quality of the breast milk. Detectable amounts of estrogens have been identified in the milk of women receiving estrogen therapy. Caution should be exercised when ESTRACE is administered to a nursing woman. H. Pediatric Use ESTRACE Vaginal Cream is not indicated in children. Clinical studies have not been conducted in the pediatric population. I. Geriatric Use There have not been sufficient numbers of geriatric patients involved in studies utilizing ESTRACE to determine whether those over 65 years of age differ from younger subjects in their response to ESTRACE. The Women’s Health Initiative Study In the WHI estrogen-alone substudy (daily CE [0.625 mg]-alone versus placebo), there was a higher relative risk of stroke in women greater than 65 years of age [see CLINICAL STUDIES and WARNINGS]. In the WHI estrogen plus progestin substudy (daily CE [0.625 mg] plus MPA [2.5 mg] versus placebo), there was a higher relative risk of nonfatal stroke and invasive breast This label may not be the latest approved by FDA. For current labeling information, please visit cancer in women greater than 65 years of age [see CLINICAL STUDIES and WARNINGS]. The Women’s Health Initiative Memory Study In the WHIMS ancillary studies of postmenopausal women 65 to 79 years of age, there was an increased risk of developing probable dementia in women receiving estrogen-alone or estrogen plus progestin when compared to placebo [see CLINICAL STUDIES and WARNINGS]. Since both ancillary studies were conducted in women 65 to 79 years of age, it is unknown whether these findings apply to younger postmenopausal women 10 [see CLINICAL STUDIES and WARNINGS]. ADVERSE REACTIONS See BOXED WARNINGS, WARNINGS and PRECAUTIONS. Systemic absorption may occur with the use of ESTRACE. The warnings, precautions, and adverse reactions associated with oral estrogen treatment should be taken into account. The following adverse reactions have been reported with estrogen and/or progestin therapy. 1. Genitourinary System Abnormal uterine bleeding or spotting; dysmenorrhea or pelvic pain, increase in size of uterine leiomyomata; vaginitis, including vaginal candidiasis; change in cervical secretion; cystitis-like syndrome; application site reactions of vulvovaginal discomfort including burning and irritation; genital pruritus; ovarian cancer; endometrial hyperplasia; endometrial cancer. 2. Breasts Tenderness, enlargement, pain, nipple discharge, fibrocystic breast changes; breast cancer. 3. Cardiovascular Deep and superficial venous thrombosis; pulmonary embolism; myocardial infarction; stroke; increase in blood pressure. 4. Gastrointestinal Nausea, vomiting; abdominal cramps, bloating; increased incidence of gallbladder disease. 5. Skin Chloasma that may persist when drug is discontinued; loss of scalp hair; hirsutism; rash. 6. Eyes This label may not be the latest approved by FDA. For current labeling information, please visit Retinal vascular thrombosis, intolerance to contact lenses. 7. Central Nervous System Headache; migraine; dizziness; mental depression; nervousness; mood disturbances; irritability; dementia. 8. Miscellaneous Increase or decrease in weight; glucose intolerance; edema; arthralgias; leg cramps; changes in libido; urticaria; exacerbation of asthma; increased triglycerides; hypersensitivity. OVERDOSAGE Overdosage of estrogen may cause nausea, vomiting, breast tenderness, abdominal pain, drowsiness and fatigue, and withdrawal bleeding may occur in women. Treatment of overdose consists of discontinuation of ESTRACE therapy together with institution of appropriate symptomatic care. DOSAGE AND ADMINISTRATION Use of ESTRACE alone or in combination with a progestin, should be limited to the shortest duration consistent with treatment goals and risks for the individual woman. Postmenopausal women should reevaluate periodically as clinically appropriate to determine if treatment is still necessary. For treatment of vulvar and vaginal atrophy associated with the menopause, the lowest dose and regimen that will control symptoms should be chosen and medication should be discontinued as promptly as possible. For women who have a uterus, adequate diagnostic measures, including directed and random endometrial sampling when indicated, should be undertaken to rule out malignancy in cases of undiagnosed persistent or recurring abnormal genital bleeding. Usual Dosage: The usual dosage range is 2 to 4 g (marked on the applicator) daily for one or two weeks, then gradually reduced to one half initial dosage for a similar period. A maintenance dosage of 1 g, one to three times a week, may be used after restoration of the vaginal mucosa has been achieved. NOTE: The number of doses per tube will vary with dosage requirements and patient handling. HOW SUPPLIED ESTRACE (estradiol vaginal cream, USP, 0.01%). N 0430-3754-14: Tube containing 1 ½ oz (42.5 g) with a calibrated plastic applicator for delivery of 1, 2, 3, or 4 g. Smooth semi solid cream. White or off-white. This label may not be the latest approved by FDA. For current labeling information, please visit Store at room temperature 20° to 25°C (59° to 77°F). Protect from temperatures in excess of 40°C (104° F). Keep ESTRACE Vaginal Cream out of the reach of children. REFERENCES 1. Jackson RD, et al. Effects of Conjugated Equine Estrogen on Risk of Fractures and BMD in Postmenopausal Women With Hysterectomy: Results From the Women’s Health Initiative Randomized Trial. J Bone Miner Res. 2006;21:817-828. 2. Hendrix SL, et al. Effects of Conjugated Equine Estrogen on Stroke in the Women’s Health Initiative. Circulation. 2006;113:2425-2434. 3. Rossouw JE, et al. Postmenopausal Hormone Therapy and Risk of Cardiovascular Disease by Age and Years Since Menopause. JAMA. 2007;297:1465-1477. 4. Hsia J, et al. Conjugated Equine Estrogens and Coronary Heart Disease. Arch Int Med. 2006;166:357-365. 5. Curb JD, et al. Venous Thrombosis and Conjugated Equine Estrogen in Women Without a Uterus. Arch Int Med. 2006;166:772-780. 6. Cushman M, et al. Estrogen Plus Progestin and Risk of Venous Thrombosis. JAMA. 2004;292:1573-1580. 7. Stefanick ML, et al. Effects of Conjugated Equine Estrogens on Breast Cancer and Mammography Screening in Postmenopausal Women With Hysterectomy. JAMA. 2006;295:1647-1657. 8. Chlebowski RT, et al. Influence of Estrogen Plus Progestin on Breast Cancer and Mammography in Healthy Postmenopausal Women. JAMA. 2003;289:3234-3253. 9. Anderson GL, et al. Effects of Estrogen Plus Progestin on Gynecologic Cancers and Associated Diagnostic Procedures. JAMA. 2003;290:1739-1748. 10. Shumaker SA, et al. Conjugated Equine Estrogens and Incidence of Probable Dementia and Mild Cognitive Impairment in Postmenopausal Women. JAMA. 2004;291:29472958. Distributed by: Allergan USA, Inc. Madison, NJ 07940 © 2022 Allergan. All rights reserved. Allergan® and its design are trademarks of Allergan, Inc. ESTRACE® is a registered trademark of Allergan Pharmaceuticals International Limited. V3.0USPI3754 This label may not be the latest approved by FDA. For current labeling information, please visit PATIENT INFORMATION ESTRACE® CREAM (es-trays) (estradiol vaginal cream, USP, 0.01%) Read this Patient Information before you start using ESTRACE Vaginal Cream and each time you refill. There may be new information. This information does not take the place of talking with your healthcare provider about your menopausal symptoms or your treatment. What is the most important information I should know about ESTRACE Vaginal Cream (an estrogen hormone)? • Using estrogen-alone may increase your chance of getting cancer of the uterus (womb). Report any unusual vaginal bleeding right away while you are using ESTRACE Vaginal Cream. Vaginal bleeding after menopause may be a warning sign of cancer of the uterus (womb). Have your healthcare provider check any unusual vaginal bleeding to find out the cause. • Do not use estrogen-alone to prevent heart disease, heart attacks, strokes or dementia (decline in brain function). • Using estrogen-alone may increase your chances of getting strokes or blood clots. • Using estrogen-alone may increase your chance of getting dementia, based on a study of women - 65 years of age or older. • Do not use estrogens with progestins to prevent heart disease, heart attacks, strokes or dementia. • Using estrogens with progestins may increase your chances of getting heart attacks, strokes, breast cancer, or blood clots. • Using estrogens with progestins may increase your chance of getting dementia, based on a study of women 65 years of age or older. • Talk regularly with your healthcare provider about whether you still need treatment with ESTRACE Vaginal Cream. What is ESTRACE Vaginal Cream? ESTRACE Vaginal Cream is a prescription medicine that contains estradiol (an estrogen hormone). What is ESTRACE Vaginal Cream used for? ESTRACE Vaginal Cream is used after menopause to: • Treat moderate to severe menopausal changes in and around the vagina. Talk regularly with your healthcare provider about whether you still need treatment with ESTRACE. This label may not be the latest approved by FDA. For current labeling information, please visit Who should not use ESTRACE Vaginal Cream? Do not start using ESTRACE Vaginal Cream if you: ▪ have unusual vaginal bleeding. Vaginal bleeding after menopause may be a warning sign of cancer of the uterus (womb). Have your healthcare provider check any unusual bleeding to find out the cause. ▪ currently have or have had certain cancers. Estrogens may increase the chances of getting certain types of cancers, including cancer of the breast or uterus. If you have or have had cancer, talk with your healthcare provider about whether you should use ESTRACE Vaginal Cream. ▪ currently have or have had blood clots. ▪ had a stroke or heart attack. ▪ are allergic to ESTRACE Vaginal Cream or any of its ingredients. See the list of ingredients in ESTRACE Vaginal Cream at the end of this leaflet. ▪ currently have or have had liver problems. ▪ have been diagnosed with a bleeding disorder. ▪ think you may be pregnant. This label may not be the latest approved by FDA. For current labeling information, please visit Before you use ESTRACE Vaginal Cream, tell your healthcare provider about all of your medical conditions, including if you: ▪ have unusual vaginal bleeding. Vaginal bleeding after menopause may be a warning sign of cancer of the uterus (womb). Your healthcare provider should check any unusual vaginal bleeding to find out the cause. ▪ have any other medical conditions. Your healthcare provider may need to check you more carefully if you have certain conditions, such as asthma (wheezing), diabetes, epilepsy (seizures), migraine, endometriosis, lupus, problems with your heart, liver, thyroid, kidneys, or have high calcium levels in your blood. ▪ are going to have surgery or will be on bed rest. Your healthcare provider will let you know if you need to stop using ESTRACE Vaginal Cream. ▪ are breastfeeding. The hormone in ESTRACE Vaginal Cream can pass into your breast milk. Tell your healthcare provider about the medicines you take, including prescription and over-the-counter medicines, vitamins, and herbal supplements. Some medicines may affect how ESTRACE works. ESTRACE may also affect how your other medicines work. Keep a list of your medicines and show it to your healthcare provider and pharmacist when you get a new medicine. This label may not be the latest approved by FDA. For current labeling information, please visit How should I use ESTRACE Vaginal Cream? ESTRACE Vaginal Cream is a cream that you place in your vagina with the applicator provided with the cream. • Take the dose recommended by your healthcare provider and talk to them about how well that dose is working for you. • Use estrogens at the lowest dose possible for your treatment only as long as needed. Talk regularly (for example, every 3 to 6 months) with your healthcare provider about the dose you are using and whether you still need treatment with ESTRACE Vaginal Cream. Figure A Step 1 Wash and dry your hands well. Step 2. Remove the cap from the ESTRACE Vaginal Cream tube. (There is no seal on tube). Step 3. Hold the applicator as shown. Do not separate the plunger from applicator (Figure B). This label may not be the latest approved by FDA. For current labeling information, please visit Figure B Step 4. Screw threaded end of applicator onto the open nozzle of the ESTRACE Vaginal Cream tube until secure. Do not attach the plunger end of the applicator onto the open ESTRACE Vaginal Cream tube (see Figure C). This label may not be the latest approved by FDA. For current labeling information, please visit Figure C Step 5. Position upright to view the calibrated gram amounts. Step 6. Gently squeeze tube from the bottom to push the prescribed amount of ESTRACE Vaginal Cream into the applicator. As ESTRACE Vaginal Cream is squeezed out, the plunger will rise to indicate the amount of grams (see Figure D). This label may not be the latest approved by FDA. For current labeling information, please visit Figure D This Figure shows a 2 g dose for illustrative purposes only. Your prescribed dosage may vary. Step 7. Unscrew applicator from tube. Step 8. Replace cap on tube. Step 9 Lie on your back with knees bent. To deliver the medicine, gently insert applicator deeply into your vagina and press the plunger downward to its original position (see Figure E). This label may not be the latest approved by FDA. For current labeling information, please visit Figure E Step 10 Remove the applicator from your vagina. Step 11. To cleanse applicator: Pull plunger to remove it from barrel. Wash with mild soap and warm water (Do not boil or use hot water) (see Figure F). Figure F This label may not be the latest approved by FDA. For current labeling information, please visit What are the possible side effects of ESTRACE Vaginal Cream? Side effects are grouped by how serious they are and how often they happen when you are treated. Serious, but less common side effects include: • stroke • blood clots • breast cancer • dementia • high or low blood calcium • severe allergic reaction • high blood pressure • liver problems • fluid retention- this can make medical conditions worse in people with heart or kidney disease • worsening of endometriosis • changes in certain laboratory test results • heart disease (such as heart attack) • cancer of the lining of the uterus (womb) • cancer of the ovary • gallbladder disease • changes in vision • worsening of angioedema (swelling of face and tongue) • high triglyceride (fat) levels in your blood • low thyroid levels in your blood • worsening of other medical conditions • high blood sugar • enlargement of benign tumors of the uterus (“fibroids”) Call your healthcare provider right away if you get any of the following warning signs or any other unusual symptoms that concern you: • new breast lumps • unusual vaginal bleeding • changes in vision or speech • sudden new severe headaches • severe pains in your chest or legs with or without shortness of breath, weakness and fatigue • swollen lips, tongue or face Less serious, but common side effects include: • headache • breast pain • irregular vaginal bleeding or spotting • stomach or abdominal cramps, bloating This label may not be the latest approved by FDA. For current labeling information, please visit • nausea and vomiting • hair loss • fluid retention • vaginal yeast infection • reactions from inserting ESTRACE Vaginal Cream, such as vaginal burning, irritation, and itching These are not all the possible side effects of ESTRACE Vaginal Cream. For more information, ask your healthcare provider or pharmacist for advice about side effects. Call your doctor for medical advice about side effects. You may report side effects to FDA at 1-800-FDA-1088. How should I store Estrace Vaginal Cream? ▪ Store ESTRACE Vaginal Cream at room temperature between 59°F to 77°F (20°C to 25°C) ▪ Protect from temperatures above 104°F (40°C) ▪ Keep ESTRACE Vaginal Cream and all medicines out of the reach of children What can I do to lower my chances of a serious side effect with ESTRACE Vaginal Cream? • Talk with your healthcare provider regularly about whether you should continue using ESTRACE Vaginal Cream. • If you have a uterus, talk with your healthcare provider about whether the addition of a progestin is right for you. In general, the addition of a progestin is generally recommended for a woman with a uterus to reduce the chance of getting cancer of the uterus. •See your healthcare provider right away if you get vaginal bleeding while using ESTRACE Vaginal Cream. • Have a pelvic exam, breast exam and mammogram (breast X-ray) every year unless your healthcare provider tells you something else. If members of your family have had breast cancer or if you have ever had breast lumps or an abnormal mammogram, you may need to have breast exams more often. • If you have high blood pressure, high cholesterol (fat in the blood), diabetes, are overweight, or if you use tobacco, you may have higher chances for getting heart disease. Ask your healthcare provider for ways to lower your chances for getting heart disease. This label may not be the latest approved by FDA. For current labeling information, please visit General information about safe and effective use of ESTRACE Vaginal Cream Medicines are sometimes prescribed for purposes other than those listed in Patient Information leaflets. Do not use ESTRACE Vaginal Cream for conditions for which it was not prescribed. Do not give ESTRACE Vaginal Cream to other people, even if they have the same symptoms you have. It may harm them. You can ask your pharmacist or healthcare provider for information about ESTRACE Vaginal Cream that is written for health professionals. What are the ingredients in ESTRACE Vaginal Cream? Active ingredient: estradiol Inactive Ingredients: purified water, propylene glycol, stearyl alcohol, white ceresin wax, mono- and di- glycerides, hypromellose 2208 (4000 cps), sodium lauryl sulfate, methylparaben, edetate di- sodium and tertiary-butylhydroquinone. Distributed by: Allergan USA, Inc. Madison, NJ 07940 © 2022 Allergan. All rights reserved. Allergan® and its design are trademarks of Allergan, Inc. ESTRACE® is a registered trademark of Allergan Pharmaceuticals International Limited. V3.0PPI3754 This label may not be the latest approved by FDA. For current labeling information, please visit

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