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3300	https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-4-techniques-of-integration/part-a-trigonometric-powers-trigonometric-substitution-and-completing-the-square/session-73-completing-the-square/	Session 73: Completing the Square \| Single Variable Calculus \| Mathematics \| MIT OpenCourseWare Browse Course Material Syllabus 1. Differentiation Part A: Definition and Basic Rules Part B: Implicit Differentiation and Inverse Functions Exam 1 2. Applications of Differentiation Part A: Approximation and Curve Sketching Part B: Optimization, Related Rates and Newton's Method Part C: Mean Value Theorem, Antiderivatives and Differential Equa Exam 2 3. The Definite Integral and its Applications Part A: Definition of the Definite Integral and First Fundamental Part B: Second Fundamental Theorem, Areas, Volumes Part C: Average Value, Probability and Numerical Integration Exam 3 4. Techniques of Integration Part A: Trigonometric Powers, Trigonometric Substitution and Com Part B: Partial Fractions, Integration by Parts, Arc Length, and Part C: Parametric Equations and Polar Coordinates Exam 4 5. Exploring the Infinite Part A: L'Hospital's Rule and Improper Integrals Part B: Taylor Series Final Exam Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types grading Exams with Solutions notes Lecture Notes theaters Lecture Videos assignment_turned_in Problem Sets with Solutions laptop_windows Simulations theaters Problem-solving Videos Download Course menu search Give Now About OCW Help & Faqs Contact Us searchGIVE NOWabout ocwhelp & faqscontact us 18.01SC \| Fall 2010 \| Undergraduate Single Variable Calculus Menu More Info Syllabus 1. Differentiation Part A: Definition and Basic Rules Part B: Implicit Differentiation and Inverse Functions Exam 1 2. Applications of Differentiation Part A: Approximation and Curve Sketching Part B: Optimization, Related Rates and Newton's Method Part C: Mean Value Theorem, Antiderivatives and Differential Equa Exam 2 3. The Definite Integral and its Applications Part A: Definition of the Definite Integral and First Fundamental Part B: Second Fundamental Theorem, Areas, Volumes Part C: Average Value, Probability and Numerical Integration Exam 3 4. Techniques of Integration Part A: Trigonometric Powers, Trigonometric Substitution and Com Part B: Partial Fractions, Integration by Parts, Arc Length, and Part C: Parametric Equations and Polar Coordinates Exam 4 5. Exploring the Infinite Part A: L'Hospital's Rule and Improper Integrals Part B: Taylor Series Final Exam Part A: Trigonometric Powers, Trigonometric Substitution and Completing the Square Session 73: Completing the Square « Previous \| Next » Overview For integrals involving the square root of some more general quadratic function, complete the square and then use trig substitution. Lecture Video and Notes Video Excerpts Clip 1: Completing the Square Recitation Video Integration by Completing the Square Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate Chapters Chapters Descriptions descriptions off, selected Captions captions and subtitles off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. View video page Download video Download transcript Worked Example Complete the Square Problem (PDF) Solution (PDF) « Previous \| Next » Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2010 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types grading Exams with Solutions notes Lecture Notes theaters Lecture Videos assignment_turned_in Problem Sets with Solutions laptop_windows Simulations theaters Problem-solving Videos Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Accessibility Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare close Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Stay Here Continue
3301	https://www.youtube.com/watch?v=LY6-YvdJMJA	Calculus 4.9 Antiderivatives Asher Roberts 13400 subscribers 63 likes Description 6810 views Posted: 29 Aug 2020 My notes are available at (so you can write along with me). Calculus: Early Transcendentals 8th Edition by James Stewart 2 comments Transcript: section 4.9 anti-derivatives a function f big f is called an anti-derivative of f little f on an interval i if f prime equals f for all x in the interval i so we're saying is you go backwards if you have a function and its derivative is another function then the original function that you were taking the derivative of to get to the new function is called an antiderivative of the new function so uh if f is an antiderivative of f on interval i the most general anti-derivative is capital f of x plus c where c is some arbitrary constant this follows pretty quickly to the corollary to the mean value theorem we did you might remember that we said that um if a function f prime is the derivative of f and we also have the g prime is the derivative of f then we said that f prime and g prime had to differ by a constant this was the theorem this is the corollary to the mean value theorem that we proved in section 4.2 we said that g had to equal f of x plus c which should make sense because if you take the derivative of both sides you get g prime equals f prime derivatives the constant is just zero so um we can see that in general if we take the derivative of any function plus a constant the constant gets eliminated we get the derivative that we wanted so we can add anything to it for example if i had x squared and i want the derivative that's 2x but if i have x squared plus 10 the derivative is still 2x x squared plus 10 plus i don't know 11 x squared plus derivative would still be 2x so it doesn't matter what constant we add if we want a general antiderivative we have to just call this entire possibility a constant c in general because it could equal anything because who knows what was eliminated to get you back to zero let's do an example let's try finding the most general anti-derivative of each of the following functions so we'll start with f of x equals sine x so we need a function whose derivative is sine well the derivative of cosine is minus sine right so the derivative of minus cosine would equal sine derivative of the cosine part is minus sign then the minus cancels so this means that the most general anti-derivative capital f will equal minus cosine plus some constant because when we take the derivative to test this thing we get sine plus zero so that works let's look at uh anti-derivative of f of x equals 1 over x let's see we know that the derivative of ln of x is 1 over x so we could say okay the anti-derivative of 1 over x is l and x plus c but actually that's not the most general anti-derivative because this only applies to positive values of x but remember we have another function whose derivative is also 1 over x and that's the absolute value of x when we take the natural log ln absolute value of x also has a derivative of 1 over x and that applies to a lot more x values that applies to every possible x value that's not zero so we can plug in negatives over there too so that means that the most general anti-derivative of one over x is oops ln of absolute value of x plus c which we can write as a piecewise function ln x plus some constant and ln of minus x plus some constant in this case it's if x is positive and the other one if x is negative however this is still not completely general there's no reason to think these two constants have to be the same so i could say that this is one constant and then this is a second constant that's possibly different so i make one of them c1 and one of them c2 now that's the most general you could possibly be let's look at our power function f of x equals x to the n assuming that n is not minus 1 because if it was then we would just be over there so we need a derivative of something to be x to the n well by the power rule remember you reduce the number each time so you might think okay let's take the derivative of x to the n plus one because then it'll be reduced by one and give you x n and that's true if you take the derivative of um x n plus one it will be reduced and it will give you x n however you also have to multiply by the power when you're taking the derivative so you would have n plus 1 times x to the n and there's no n plus 1 in front over there so we need to make sure that our function divides by n plus 1 so that it cancels when we take the derivative now notice if i take the derivative of this guy n plus 1 hops out cancels with n plus 1 on the bottom and then the power goes down by one so that's perfect this derivative is equal to n plus one times x to the n over n plus 1 which cancels and leaves us x to the n so that's our most general anti-derivative so now we have a way of finding anti-derivatives of power functions in general and our way is that we increase the power by 1 and divide by the new power which should make sense because your power rule says first you multiply and then you subtract 1 for your exponent so if you want to go backwards first you should add one and then you should divide instead of multiply so we add one and then divide instead of multiplying and subtracting let's find all functions g such that g prime is this in other words let's find the anti-derivative of g prime okay so let's simplify this a little bit and by simplify i mean let's split this up so we'll write this as four times sine x plus two x to the fifth over x minus square root of x over x that becomes four times sine x plus two x to the fourth minus one over x well let's convert that all to powers so that we can use our reverse power rule so 4 sine x plus 2x to the fourth minus x to the minus one-half that means well we already got the anti-derivative of sine we said it's minus cosine so this is four times minus cosine and then 2x to the fourth well x to the fourth we increase the exponent and then divide so it's going to be plus 2 times x to the fifth over 5. and then again we increase the exponent by one so this becomes x to the one over two and then divide by half and we have our arbitrary constant simplifying we get minus four times cosine plus two-fifths x to the fifth minus two rad x plus c let's find f if f prime is equal to e to the x plus 20 times one plus x squared to the minus one and f of zero is minus two so let's start by writing out our derivative we've got f prime equal to e to the x plus 20 times 1 plus sorry 1 over 1 plus x squared notice that we're not writing this as capital f we're writing it as f prime just because they gave it to us that way if they don't give it to us in general and we're looking at an anti-derivative of f and that's what we're starting with then we say that um f will be the derivative and big f is the anti-derivative but if we're starting with the prime notation then we just keep using whatever they gave us and then the antiderivative will just be the f without the prime just like in the previous example or it was g without the prime so taking the antiderivative we get that f is equal to well let's see e to the x we could differentiate that a million times over e to the x to be derivative e to the x e to the x e to the x e to the x so anti-derivative will also be e to the x because if i take the derivative i get the e to the x uh 1 over 1 plus x squared is a little bit more obscure remember that's the derivative of a function it is derivative of inverse tangent so this is plus 20 times tan inverse because if you take the derivative you get e to the x and you get one over one plus x squared let's look at f of zero we'll plug in zero we get e to the zero plus twenty times tan inverse of zero plus c and we know that this has to equal minus two because f of zero is minus two okay e to the zero is one tan inverse of zero well that's when sine is zero which is zero so this implies that our constant is equal to -2 minus the e to the 0 is 1 so it equals minus 3. so now we have our particular function for this example a to the x plus 20 tan inverse x instead of plus c we know what c is so it's minus 3. let's find f if f double prime is 12x squared plus 6x minus 4 f of 0 is 4 and f of 1 is 1. notice in the previous example we had f prime so they only gave us one f 1 y value and in this example we have double prime so we need 2 in order to solve this thing uniquely so let's start with our derivative f prime remember that's the antiderivative of f double prime because its derivative is f double prime so that's 12 times well instead of x squared it'll be x cubed we increase the power by one and divide by our new power so plus six times x squared over two minus four times well let's see the derivative of something to get four there would have to be an x there if we take the derivative we lose the x and just get four and we have our arbitrary constant simplifying we get four x cubed plus three x squared minus four x plus c taking another anti-derivative we get four times x to the fourth over four plus three times x to the third over three minus four times x squared over two plus well you take the anti-derivative of a constant like this one you just add an x so when we take the anti-derivative of c we'll keep c and just throw an x in there however we just took an antiderivative so it's possible that there was some other constant let's say d if you take the derivative of this well all of these it's pretty clear that they work out via the power rule to give you all of these and similarly if you take the derivative of this guy it's a constant times x so it leaves you with just a constant and then taking the derivative of this guy just leaves you with zero simplifying we get that this equals x to the fourth plus x to the third minus 2x squared plus cx plus d so now let's use our f of zero they told us f of zero was uh four but if we plug in zero this is gone this is gone this is gone this is gone so we just get zero plus d that must equal four because f of zero is four okay so that means that d must equal four so now we have one of the constants solve for let's use our other y value to get the other constant f of one they said was uh one well if we plug in one for each of these guys then we get uh one plus one minus two plus c times one which is c plus d but we said d was 4 and we know that that has to equal f of 1 which is 1. okay solving for c this implies that c is equal to minus 3. so now we have our function f f must be equal to x to the fourth plus x cubed minus 2x squared minus 3 times x because c was minus 3 plus 4 because d was plus 4. okay say we have the graph of a function f let's see if we could find an antiderivative graph well we are given that f big f of zero is two so how about we start there so that's two that could be our first point why not then it looks like from zero to one our derivative is negative if our derivative is negative it's underneath the x-axis then that means our function is going down so up until 1 our function is decreasing if we go from 1 to 3 then our derivative is positive so whenever our derivative is above the x-axis that means our function is increasing so we should be going up until we hit let's see which is two and then three and then it looks like we should start going down so let me erase this a little bit because we should start going down once we hit three so it looks like i should stay up there a little bit more and then oop and then maybe start going down once we hit three let's see if i can get put these in first maybe it'll be easier to draw okay and then from three on it's negative so it should be decreasing forever more but see how it gets closer and closer to zero so it gets less and less negative so our function kind of levels off towards the end and over here is one uh this looks like it's probably pretty good we could even confirm this by taking a look at the second derivative if we look over here at these tangent lines you can see that the slope of the tangent line goes from positive to negative so that means the second derivative changes from positive to negative over here so it looks like we have an inflection point over here and an inflection point over here which makes sense we have inflection points over here and inflection points over here so it looks like this is a pretty good graph for our anti-derivative a particle moves in a straight line and has acceleration given by a of t equals six t plus four its initial velocity is v of zero equals negative six centimeters per second and its initial displacement is s of zero equals nine centimeters let's find its position function s of t okay let's start with the acceleration remember acceleration is the derivative of velocity so v prime of t is a of t and they told us that that's six t plus four so since we're starting with a derivative we can get back to velocity by doing an anti-derivative so we could say that the velocity is the anti-derivative of acceleration so that's 6 times t squared over 2 increase the power divide by it plus 4t plus c which simplifies to 3t squared plus 4t plus c well it would be great if we knew what the c was it turns out we can now though they told us that v of zero was minus six so let's use that information v of zero well if i plug in zero this is zero this is zero so i'm left with only c so i get that v of zero c and i know that that equals minus six because v of zero is minus six so now i know that c is minus six okay great i can rewrite v of t now getting rid of that constant because i know exactly what that constant is i can replace it with six minus six so i have three t squared plus four t minus 6. in the same way that we started with acceleration because that's the derivative of velocity we can now start with the derivative of position because velocity is the derivative of position so taking the anti-derivative velocity will give us the position function so that's 3 times t cubed over 3 plus 4 times t squared over two minus six t plus another constant we already use c so let's use d simplifying we get t cubed plus two t squared minus 6t plus d so let's use the fact that s of 0 they told us was 9. so s of 0 if we plug in 0 this is gone this is gone this is gone we just have d so that must be equal to nine so that means that d must be equal to nine so now we know what our constant is s of t must be equal to t cubed plus two t squared minus six t plus nine and that's our position function say a ball is thrown upward with a speed of 48 feet per second from the edge of a cliff 432 feet above the ground let's find its height above the ground t seconds later when does it reach its maximum height when does it hit the ground in order to do this we'll assume that uh downward acceleration g is constant and its value is about 9.8 meters per second squared or 32 feet per second squared okay we want to find its height above the ground so we want to find the uh position function let's assume that going up is positive and going down is negative so that means that if we start with our acceleration function a of t that's the derivative of velocity with respect to time and that'll be minus 32 because it's accelerating downward at 32 feet per second squared let's uh anti-derive this to get velocity so v of d would have to be minus 32 times t plus some constant but we know that the initial velocity it said was 48 feet per second because that was our upward speed that the ball was thrown at so that means that if we plug in 0 over here we get 0 plus c would have to be equal to 48 so now we know that c is 48 so we can rewrite our velocity function without the arbitrary constant and with the specific constant of 48 this tells us what our maximum height is the maximum height would have to be when the velocity is zero right that's when we have a maximum because velocity is the derivative of position so when the derivative is zero that's when you have a maximum when the velocity is 0 that tells us that t would have to be 1.5 seconds let's do another anti-derivative now let's get position so that'll be the antiderivative velocity so that's minus 16 t squared because it's t squared over 2 and then half times 32 is minus 16 so plus 48 t and a new constant d again they gave us an initial value they told us that it's thrown upward from the edge of the cliff 432 feet above the ground so that means that if we plug in 0 0 plus d must be equal to 432 so that means d must be 432 so now we have our position function all filled in minus 16 t squared plus 48 t plus 432 well the ball hits the ground when the y value is zero so that's when the position function is zero [Music] so s of t needs to equal 0 for this thing to hit the ground let's see if we can figure out what t is okay we have minus 16 t squared plus 48 t plus 432 i want to know when that equals zero we can divide by minus 16 and get that this is t squared minus three t minus 27 equal to 0 so they both have the same root so now we'll use the quadratic formula so t is going to be equal to 3 plus or minus the square root of 3 times radical 13 over two so that's when the ball hits the ground ball will hit the ground after 3 times 1 plus rad 13 over 2 seconds which is about 6.9 seconds notice we reject that negative value because we're not time travelers
3302	https://www.britannica.com/science/smooth-endoplasmic-reticulum	SUBSCRIBE Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos Related Topics: : endoplasmic reticulum : sarcoplasmic reticulum : terminal cisterna See all related content smooth endoplasmic reticulum (SER), meshwork of fine disklike tubular membrane vesicles, part of a continuous membrane organelle within the cytoplasm of eukaryotic cells, that is involved in the synthesis and storage of lipids, including cholesterol and phospholipids, which are used in the production of new cellular membrane. The smooth endoplasmic reticulum (SER) is distinguished from the rough endoplasmic reticulum (RER), the other basic type of endoplasmic reticulum, by its lack of ribosomes, which are protein-synthesizing particles that can be found attached to the outer surface of the RER to give the membrane its “rough” appearance. SER occurs both in animal and in plant cells. The function of the SER can vary, depending on cell type. In some cells, such as those of the adrenal gland and certain other endocrine glands, it plays a key role in the synthesis of steroid hormones from cholesterol. In the liver, enzymes in the SER catalyze reactions that render drugs, metabolic wastes, and harmful chemicals water-soluble, thereby contributing to their detoxification, or removal, from the body. The SER also plays a role in the conversion of glycogen to glucose, with glucose-6-phosphatase, an enzyme present in SER, catalyzing the final step in glucose production in the liver. In skeletal muscle cells, SER occurs as a specialized membrane structure known as the sarcoplasmic reticulum. The sarcoplasmic reticulum is a critical storage site for calcium ions, taking up the ions from the cytoplasm. It also releases calcium ions when the muscle cell is triggered by nerve stimuli, resulting in muscle contraction. In this way, the sarcoplasmic reticulum helps regulate calcium ion concentrations in the cytoplasm of skeletal muscle cells. The sarcoplasmic reticulum is also found in smooth muscle cells, though in a more loosely organized form than in skeletal muscle. More From Britannica cell: The smooth endoplasmic reticulum Kara Rogers calcium Table of Contents Introduction Occurrence, properties, and uses Compounds References & Edit History Quick Facts & Related Topics Images, Videos & Interactives For Students calcium summary Quizzes 118 Names and Symbols of the Periodic Table Quiz Facts You Should Know: The Periodic Table Quiz Related Questions Is mathematics a physical science? calcium chemical element Also known as: Ca Written by Timothy P. Hanusa Professor, Department of Chemistry, Vanderbilt University, Nashville, Tennessee. Timothy P. Hanusa Fact-checked by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Last Updated: • Article History Key People: : Sir Humphry Davy : Johann Wolfgang Döbereiner Related Topics: : chemical element : human nutrition : alkaline-earth metal : calcium deficiency On the Web: : Open Library Publishing Platform - Human Nutrition - Calcium (Aug. 23, 2025) See all related content Top Questions What is calcium? What is the chemical symbol of calcium and its atomic number on the periodic table? What are some common uses of calcium in everyday life? How is calcium important for human health and nutrition? What are the natural sources of calcium in the diet? How does calcium function in the process of bone formation and maintenance? In what ways does calcium interact with other elements or compounds in chemical reactions? How is calcium extracted and processed for use in industrial applications? calcium (Ca), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table. It is the most abundant metallic element in the human body and the fifth most abundant element in Earth’s crust. Element Properties \| atomic number \| 20 \| \| atomic weight \| 40.078 \| \| melting point \| 842 °C (1,548 °F) \| \| boiling point \| 1,484 °C (2,703 °F) \| \| specific gravity \| 1.55 (20 °C, or 68 °F) \| \| oxidation state \| +2 \| \| electron configuration \| 1s22s22p63s23p64s2 \| Occurrence, properties, and uses Calcium does not occur naturally in the free state, but compounds of the element are widely distributed. One calcium compound, lime (calcium oxide, CaO) was extensively used by the ancients. The silvery, rather soft, lightweight metal itself was first isolated (1808) by Sir Humphry Davy after distilling mercury from an amalgam formed by electrolyzing a mixture of lime and mercuric oxide. The name for the element was taken from the Latin word for lime, calx. Calcium constitutes 3.64 percent of Earth’s crust and 8 percent of the Moon’s crust, and its cosmic abundance is estimated at 4.9 × 104 atoms (on a scale where the abundance of silicon is 106 atoms). As calcite (calcium carbonate), it occurs on Earth in limestone, chalk, marble, dolomite, eggshells, pearls, coral, stalactites, stalagmites, and the shells of many marine animals. Calcium carbonate deposits dissolve in water that contains carbon dioxide to form calcium bicarbonate, Ca(HCO3)2. This process frequently results in the formation of caves and may reverse to deposit limestone as stalactites and stalagmites. As calcium hydroxyl phosphate, it is the principal inorganic constituent of teeth and bones and occurs as the mineral apatite. As calcium fluoride, it occurs as fluorite, or fluorspar. And as calcium sulfate, it occurs as anhydrite. Calcium is found in many other minerals, such as aragonite (a type of calcium carbonate) and gypsum (another form of calcium sulfate), and in many feldspars and zeolites. It is also found in a large number of silicates and aluminosilicates, in salt deposits, and in natural waters, including the sea. Formerly produced by electrolysis of anhydrous calcium chloride, pure calcium metal is now made commercially by heating lime with aluminum. The metal reacts slowly with oxygen, water vapour, and nitrogen of the air to form a yellow coating of the oxide, hydroxide, and nitride. It burns in air or pure oxygen to form the oxide and reacts rapidly with warm water (and more slowly with cold water) to produce hydrogen gas and calcium hydroxide. On heating, calcium reacts with hydrogen, halogens, boron, sulfur, carbon, and phosphorus. Although it compares favourably with sodium as a reducing agent, calcium is more expensive and less reactive than the latter. In many deoxidizing, reducing, and degasifying applications, however, calcium is preferred because of its lower volatility and is used to prepare chromium, thorium, uranium, zirconium, and other metals from their oxides. The metal itself is used as an alloying agent for aluminum, copper, lead, magnesium, and other base metals; as a deoxidizer for certain high-temperature alloys; and as a getter in electron tubes. Small percentages of calcium are used in many alloys for special purposes. Alloyed with lead (0.04 percent calcium), for example, it is employed as sheaths for telephone cables and as grids for storage batteries of the stationary type. When added to magnesium-based alloys in amounts from 0.4 to 1 percent, it improves the resistance of degradable orthopedic implants to biological fluids, permitting tissues to heal fully before the implants lose their structural integrity. Britannica Quiz 118 Names and Symbols of the Periodic Table Quiz Naturally occurring calcium consists of a mixture of six isotopes: calcium-40 (96.94 percent), calcium-44 (2.09 percent), calcium-42 (0.65 percent), and, in smaller proportions, calcium-48, calcium-43, and calcium-46. Calcium-48 undergoes double beta decay with a half-life of roughly 4 × 1019 years, so it is stable for all practical purposes. It is particularly neutron-rich and is used in the synthesis of new heavy nuclei in particle accelerators. The radioactive isotope calcium-41 occurs in trace quantities on Earth through the natural bombardment of calcium-40 by neutrons in cosmic rays. Calcium is essential to both plant and animal life and is broadly employed as a signal transducer, enzyme cofactor, and structural element (e.g., cell membranes, bones, and teeth). A large number of living organisms concentrate calcium in their shells or skeletons, and in higher animals calcium is the most abundant inorganic element. Many important carbonate and phosphate deposits owe their origin to living organisms. Access for the whole family! Bundle Britannica Premium and Kids for the ultimate resource destination. Subscribe The human body is 2 percent calcium. Major sources of calcium in the human diet are milk, milk products, fish, and green leafy vegetables. The bone disease rickets occurs when a lack of vitamin D impairs the absorption of calcium from the gastrointestinal tract into the extracellular fluids. The disease especially affects infants and children. Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Rogers, Kara. "smooth endoplasmic reticulum". Encyclopedia Britannica, 3 Jul. 2025, Accessed 29 August 2025. Share Share to social media Facebook X External Websites BMC Molecular and Cell Biology - Direct observation of molecular arrays in the organized smooth endoplasmic reticulum British Society for Cell Biology - Endoplasmic Reticulum (Rough and Smooth) Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. print Print Please select which sections you would like to print: verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Hanusa, Timothy P.. "calcium". Encyclopedia Britannica, 23 Aug. 2025, Accessed 29 August 2025. Share Share to social media Facebook X URL External Websites National Center for Biotechnology Information - Calcium Royal Society of Chemistry - Periodic Table - Calcium The Nemours Foundation - For Teens - Calcium Oregon State University - Linus Pauling Institute - Calcium Lenntech - Calcium - Ca Open Library Publishing Platform - Human Nutrition - Calcium U.S. Department of Health & Human Services - National Institutes of Health - Calcium Medicine LibreTexts - Calcium Chemicool - Calcium University of Rochester Medical Center - Health Encyclopedia - Calcium WebMD - Calcium Live Science - Facts about Calcium Britannica Websites Articles from Britannica Encyclopedias for elementary and high school students. calcium - Children's Encyclopedia (Ages 8-11) calcium - Student Encyclopedia (Ages 11 and up)
3303	https://math.stackexchange.com/questions/2362317/determining-a-plane-vector-algebra	Determining a plane - vector algebra - 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 Determining a plane - vector algebra Ask Question Asked 8 years, 2 months ago Modified8 years, 2 months ago Viewed 183 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am a bit confused and probably mixing some terms, and I wish to ask for your assistance in putting things in the right place. There are 4 ways to determine a plane: 3 points a point and a line 2 intersecting lines 2 parallel lines I have a problem understanding the 4th way. If two lines are parallel, aren't they linearly dependent ? If so, how can they be a basis for this plane ? Can they still span it if they are parallel ? This equation: x=(1,2,-3)+t(1,-2,4)+s(-2,4,-8) does not represent a plane. This is confusing, I mean, (1,-2-4) and (-2,4,-8) are parallel vectors, right ? They are linearly dependent, so according to the definition above they should determine a plane. I know that a set of vectors in R 3 R 3 span the entire space if every vector can be written as a linear combination of the vectors. In addition if they are linearly independent they are also a basis. The elementary basis contains the vectors (1,0,0), (0,1,0) and (0,0,1). Now I am trying to translate this logic to planes in R 3 R 3, and to connect it to the parametric representation of planes and lines. Any explanations will be most appreciated ! Thank you ! linear-algebra vectors Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jul 18, 2017 at 5:47 user3275222user3275222 623 7 7 silver badges 18 18 bronze badges 2 1 there is other way: a point and a vector, an orthogonal vector to the plane.user173262 –user173262 2017-07-18 05:56:27 +00:00 Commented Jul 18, 2017 at 5:56 It seems that you’re thinking of parallel vectors, that is, of lines through the origin, in which case parallel lines are coincident. The lines in the methods that you list are not so restricted. E.g., the lines x=0,z=0 x=0,z=0 and x=1,z=0 x=1,z=0 uniquely determine the x x-y y plane.amd –amd 2017-07-18 06:34:39 +00:00 Commented Jul 18, 2017 at 6:34 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. In order to find the plane of 2 distinct parallel lines l 1 l 1 and l 2 l 2, take a point P P on one of the lines, for example l 2 l 2, and then determine the plane given by P P and l 1 l 1 (case 2). Note that P∉l 1 P∉l 1. Show that this plane contains the whole line l 2 l 2. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 18, 2017 at 5:57 Robert ZRobert Z 148k 12 12 gold badges 110 110 silver badges 193 193 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The parallel lines need to be distinct as well as parallel. The equations x(t)y(s)=(1,2,−3)+t(1,−2,4)=(1,2,−3)+s(2,−4,8)x(t)=(1,2,−3)+t(1,−2,4)y(s)=(1,2,−3)+s(2,−4,8) describe the same line, and cannot be used to uniquely describe any of the infinitely many planes that contain the line. But, as soon as you make the lines distinct, by changing appropriately the initial vector, e.g. x(t)y(s)=(1,2,−3)+t(1,−2,4)=(0,1,0)+s(2,−4,8),x(t)=(1,2,−3)+t(1,−2,4)y(s)=(0,1,0)+s(2,−4,8), then the plane described is unique. Any direction vector from one point in the first line to another point in the second line, will not be parallel to the two parallel direction vectors. Thus, we will have two linearly independent directions, which in conjunction with any point from either line, will be enough to define a plane. Taking the above example, we can form a new direction vector (1,2,−3)−(0,1,0)=(1,1,−3)(1,2,−3)−(0,1,0)=(1,1,−3). Then, our plane has a vector equation, p(s,t)=(0,1,0)+t(1,−2,4)+s(1,1,−3)p(s,t)=(0,1,0)+t(1,−2,4)+s(1,1,−3) By constantly setting s s to 0 0 or 1 1, we recover the two lines that formed the plane. Hope that helps. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 18, 2017 at 5:56 Theo BenditTheo Bendit 53.7k 5 5 gold badges 55 55 silver badges 108 108 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 vectors See similar questions with these tags. 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3304	https://www.bbc.co.uk/bitesize/articles/zy7xs82	Bar charts - KS3 Maths - BBC Bitesize BBC Homepage Skip to content Accessibility Help Sign in Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds More menu More menu Search Bitesize Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Close menu Bitesize Menu Home Learn Study support Careers Teachers Parents Trending My Bitesize More England Early years KS1 KS2 KS3 GCSE Functional Skills Northern Ireland Foundation Stage KS1 KS2 KS3 GCSE Scotland Early Level 1st Level 2nd Level 3rd Level 4th Level National 4 National 5 Higher Core Skills An Tràth Ìre A' Chiad Ìre An Dàrna Ìre 3mh ìre 4mh ìre Nàiseanta 4 Nàiseanta 5 Àrd Ìre Wales Foundation Phase KS2 KS3 GCSE WBQ Essential Skills Cyfnod Sylfaen CA2 CA3 CBC TGAU International KS3 IGCSE More from Bitesize About us All subjects All levels Primary games Secondary games KS3 Bar charts Part ofMathsRepresenting data Save to My Bitesize Save to My Bitesize Saving Saved Removing Remove from My Bitesize Save to My Bitesize close panel Jump to Key points Creating a bar chart Examples Questions Practise working out bar charts Quiz Real-life maths Key points Image caption, Bar graphs are used to show sets of data in more specific categories. A bar chart is a type of graph used to represent a non-numerical close non-numerical Data that is not numbers, eg, words. or discrete close discrete Data that takes specific values, often whole numbers, eg, the number of siblings, shoe sizes. set of data. It uses different height rectangles, or bars, to represent the frequency close frequency The number of times something occurs. of each category. It can be used to make comparisons between the categories in a set of data. The frequency is usually represented on the vertical axis close vertical axis The line on a graph that runs vertically (up-down) from the origin.. The labelling on the horizontal axis close horizontal axis The line on a graph that runs horizontally (left-right) from the origin. depends on what data the bar chart is representing. A bar chart should have gaps between the bars. Image caption, Bar graphs are used to show sets of data in more specific categories. Back to top Creating a bar chart To create a bar chart data is required. The data often comes in the form of a table. To create a bar chart: Look for the largest frequency in your table. Draw a vertical axis close vertical axis The line on a graph that runs vertically (up-down) from the origin. on your square paper or graph paper. Choose an appropriate scale for this axis and label your axis up to the largest frequency. Look at how many categories are needed for the horizontal axis. Draw and label the horizontal axis close horizontal axis The line on a graph that runs horizontally (left-right) from the origin. , remembering to leave spaces for the gaps between the bars. Draw each bar the correct height, based on the frequencies. Check you have labelled each axis correctly and give your bar chart a title. Examples Image gallerySkip image gallery 1. Image caption, 32 students were asked which season was their favourite. The table shows the results. Construct a bar chart that represents these results. 2. Image caption, To construct a bar chart first identify the largest frequency. The largest frequency is 12. The vertical axis must go up to at least 12. 3. Image caption, Draw the vertical axis and number it in regular intervals. In this example, the axis is increasing in increments of two. Label the axis as frequency. 4. Image caption, The horizontal axis requires four categories. Each bar must be the same width and gaps between the bars need to be included. Label each bar and the axis. 5. Image caption, Draw each bar the correct height based on the frequency. The bar for spring has a frequency of 10. The bar for summer has a frequency of 12. Using this axis requires the height of the bars showing odd numbers to be exactly between the even numbers along the vertical axis. The bar for autumn has a frequency of 3, putting it between 2 and 4 . The bar for winter has a frequency of 7, putting it between 6 and 8. 6. Image caption, Finally give the graph an appropriate title. 7. Image caption, It is possible to read data and make comparisons using a bar chart. This type of graph, called a dual bar chart, shows the eye colour of year 7 and year 8 students. 8. Image caption, Using the vertical axis, the frequency of each bar can be read. The axis is going up in increments of 10. Between each multiple of 10 is ten subdivisions. Each subdivision is worth 1. There are 22 students in Year 7 with green eyes. 9. Image caption, By adding the frequencies of the blue bars, which represent Year 7, the total number of students in that year group can be calculated. 22 + 34 + 20 = 76. There are 76 students in Year 7. 10. Image caption, By adding together the frequencies of two bars for each eye colour, the most frequent eye colour, the mode, can be found. Green eyes = 22 + 16 = 38. Blue eyes = 34 + 28 = 62. Brown eyes = 20 + 24 = 44. The modal eye colour for Year 7 and 8 students is blue. 1 of 10 Previous image Next image Slide 1 of 10, Example one. An image of a table. The table has two rows and five columns. The first row is labelled, season, and is populated with, spring, summer, autumn, and winter. The second row is labelled, frequency, and is populated with the numbers, ten, twelve, three, and seven. The cells for the labels are coloured purple., 32 students were asked which season was their favourite. The table shows the results. Construct a bar chart that represents these results. End of image gallery Questions This bar chart shows student’s favourite colour. How many students were asked what their favourite colour was? Show answer Hide answer Number of students asked = 6 + 9 + 2 + 3 + 6 = 26 Twenty six students were asked their favourite colour. This compound bar chart shows the outcome of fixtures for four football teams for a season. What percentage of games did Torquay United draw? Show answer Hide answer Each square in the bar represents 10% The bar for draw, for Torquay United is two squares high. This is equivalent to 20% Back to top Practise working out bar charts Practise working out bar charts with this quiz. You may need a pen and paper to help you with your answers. Quiz Back to top Real-life maths Image caption, Scientists can use bar graphs to sort and analyse data about the environment and climate. A geographer close geographer A scientist who studies both the Earth's natural environment and human society. may use climate graphs to study changes in weather patterns in regions of the world. One type of climate graph shows precipitation or rainfall. This takes the form of a bar chart with the rainfall for each month being represent as a bar. The height of the bar indicates the quantity of a rainfall during that month. Image caption, Scientists can use bar graphs to sort and analyse data about the environment and climate. Back to top Play Sudoku with BBC Bitesize! Every weekday we release brand new easy, medium and hard Sudoku puzzles. Perfect for testing your skill with numbers and logic. Back to top More on Representing data Find out more by working through a topic Line graphs count 2 of 6 Pictograms count 3 of 6 Pie charts count 4 of 6 Frequency diagrams and frequency polygons count 5 of 6 Language: Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Terms of Use About the BBC Privacy Policy Cookies Accessibility Help Parental Guidance Contact the BBC BBC emails for you Advertise with us Do not share or sell my info Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking.
3305	https://www.out-class.org/blogs/gravitational-potential-energy	Gravitational Potential Energy: Basics and Formulas by Taha Cheema Gravitational Potential Energy: A Brief Overview Gravitational potential energy is a common term in IGCSE and O Level Physics and often comes up when discussing energy transformations. But what is gravitational potential energy and how is it a useful quantity in Physics? Out-Class has prepared this succinct guide to help you get an overview! What is Gravitational Potential Energy Gravitational potential energy refers to the energy an object possesses due to its position in a gravitational field. In simple terms, a planet exerts a force on objects, pulling them towards its center. But if you raise an object above the surface, it now has a “potential” to fall back and do some kind of work. An Example of Gravitational Potential Energy Suppose we lift a block of iron a meter above the ground. We also place a nail on a wooden plank below it. The iron block has some “potential” energy as it has been lifted against the pull of gravity which can allow it to do some work. If we release the block, it will fall and drive the nail into the wooden plank, utilizing its gravitational potential energy. A Simple Rule The higher the object is raised above the surface, the higher the gravitational energy it possesses. This should be an intuitive idea (think about it)! Gravitational Potential Energy Formula The formula is as follows: U = mgh Where: U is the gravitational potential energy m is the mass of the object whose potential energy we are calculating g is the acceleration due to gravity (9.81 ms-2 near the Earth's surface) h is the height the object is raised above a zero reference point. The SI unit of potential energy is J. Energy Transformations As an object falls into a gravitational field, it gradually loses its gravitational potential energy, but this energy doesn’t just disappear. It gets converted to the kinetic energy of the falling object (as it gains speed). Think about these energy transformations in a pendulum! FAQs Q. What is gravitational potential energy? Gravitational potential energy refers to the energy an object possesses due to its position in a gravitational field. It is the energy associated with the gravitational force acting on the object. Q. How is gravitational potential energy useful in Physics? Gravitational potential energy helps in understanding energy transformations, especially when dealing with objects in a gravitational field. It is crucial in analyzing systems like pendulums, falling objects, and potential work done by gravity. Q. Can you explain gravitational potential energy with an example? Sure! If you lift an object, such as a block of iron, above the ground, it gains gravitational potential energy because it can do work when it falls back due to gravity. For instance, dropping the block can drive a nail into a wooden plank, utilizing its stored energy. Q. What factors affect gravitational potential energy? Gravitational potential energy depends on the mass of the object, the acceleration due to gravity (which is approximately 9.81 m/s² near the Earth's surface), and the height the object is raised above a reference point. Q. What is the formula for gravitational potential energy? The formula for gravitational potential energy (U) is U = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above a reference point. Q. How does gravitational potential energy transform into kinetic energy? When an object falls into a gravitational field, it loses gravitational potential energy, which is converted into kinetic energy as the object gains speed. This transformation of energy is fundamental in understanding the dynamics of falling objects and systems like pendulums. Q. What is the SI unit of gravitational potential energy? The SI unit of gravitational potential energy is the joule (J), which is the same unit used for energy in general. Q. Is gravitational potential energy always positive? Gravitational potential energy can be positive or negative, depending on the reference point chosen. For example, if the reference point is at ground level, lifting an object above the ground increases its potential energy (positive value), while lowering it below the ground level decreases its potential energy (negative value). If you want to dive right in, let's start with selecting the course you want What Course Are You Interested In? Choose from the list Explore
3306	https://webspace.ship.edu/AMRashed/123_Labs/123_Lab_PDF/05Coefficients_of_Friction.pdf	1 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] Kinetic Friction and Coefficients of Friction Pre-lab questions 1. What is the goal of this experiment? What physics and general science concepts does this activity demonstrate to the student? 2. What is the mathematical form of kinetic friction? 3. Should the friction coefficient depend on the contact area? The goal of the experiment is to study the nature of sliding friction and changes in friction force with differing surfaces in contact. Introduction The friction force is the resistive force encountered when one surface slides over a second surface. When viewed close up, the detailed nature of friction forces is quite complex. However, a simple linear model serves most everyday purposes in describing the action of friction for us. As a resistive force, friction always opposes the direction of motion of an object. Since sliding friction only exists between two surfaces in contact, it also depends on the force (called the normal force) that pushes the surfaces against each other. And, the force of friction depends on the nature of the surfaces in contact, their composition and surface roughness. The empirical model for contact friction says the maximum friction force Ff = N, where  is the coefficient of friction (between the pair of surfaces) and N is the normal force. (Normal is used here in the mathematical sense of perpendicular. The normal force is perpendicular to the surfaces in contact.) In the diagram, W1 = m1g is the downward force due to gravity that is being balanced by the upward normal force, N. T is the applied force trying to move block m1, while Ff is the friction force resisting that tendency toward motion. If the magnitude of the applied force T is greater than that of the friction force Ff, the block m1 will accelerate. If T = Ff, then the net force on the block m1 is zero and it will move at a constant speed. There are two types of frictional forces -- static friction when an object is at rest and kinetic friction when an object slides. From everyday life, we know it is usually harder (requires a larger force) to get an object to begin to slide that it does to keep it in motion. Since both are related to the same normal force, the coefficient of static friction s can be greater than the coefficient of kinetic friction, k. While Ff = kN, for sliding (kinetic) friction, Ff ≤ sN, for static friction. That is, static friction supplies just enough resistive force to cancel out the applied force trying to move an object – up to its maximum value Ff = sN. Once that maximum value is reached, the object begins to slide and kinetic friction takes over. 2 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] Equipment: Pasco Dynamics System, acoustic motion sensor, force sensor, assorted friction trays, motorized cart, compact masses, string, bubble level, balance scale. Experiment The Friction Trays have different materials on their bottom surface. The Motorized Cart (see Fig. 1) is used to pull the trays in a controlled manner along the track, as the Force Sensor directly measures the frictional force. All parameters are investigated, including speed, normal force, and surface area. Figure 1: Measuring Sliding Friction Setup 1. Set up the track as shown in Figure 1, and use the bubble level and the adjustable feet to level the track. 2. Connect the Motion Sensor to the PASPORT input P1, and attach it to the track. Make sure the switch on the top of the Motion Sensor is set to “cart”. 3. Connect the Force Sensor to the PASPORT analog input. Place the Cart/Force Sensor 3 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] assembly on the track. Change the sample rate to 10 Hz (at bottom of screen). Open the properties for the Force Sensor in the Data Summary and check the Change Sign box . 4. Connect the Motorized Cart power cord to Output #1, and make sure the cart is on the OFF/EXTERNAL setting. With the DC Voltage set to 2 volts, turn the output on and off to ensure that the cart is working. You can open and close the output window by clicking on the Signal Generator icon in the tool palette (see Fig. 2). 5. Connect the Friction Trays to the Force Sensor using string as shown in Figure 3. The lower tray should have the black felt surface. You must always pull with two carts stacked, so that the Force Sensor pulls level. For the top tray, use one of the trays with the white plastic on the bottom. The lower end of the string is just looped around the tray hook: This makes it easier to swap out the lower tray to change the surface. 6. Place both silver cart masses in the upper friction tray. If the Motorized Cart has trouble pulling the load, add the black Compact Masses to the cart in front of the Force Sensor. If you still have trouble, remove one of the silver masses from the tray. Figure 3: Towing the Friction Carts 7. Make a graph of velocity vs. time, add a plot area ( ) and put force in the second plot area. Select a Quick-Calc for the velocity (-v) to change the sign of the velocity so it will be positive. Click on the velocity units of the graph and select cm/s. Procedure: Measuring Speed and Force Figure 2: Signal Generator 4 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] 1. Position the cart and trays at the far end of the track, opposite the motion sensor. Note this spot as the starting position for all the runs. 2. Open the Signal Generator window, and set the voltage for 1.0 volt DC. The output should be set on Auto, which will automatically turn on the cart when you start recording data, and stop the cart when data collection is ended. 3. Click on Record. Stop recording before the cart hits the motion sensor. 4. If the Motorized Cart can't pull the load, increase the voltage to 1.5 volts. 5. Adjust the Motion Sensor if necessary to get good velocity data. 6. The force data will be noisy, but you will be able to get an average reading. Try using the Smoothing tool ( ) in the Graph tool palette. The trays will move at a relatively constant speed when the string tension (measured by the force sensor) equals the friction force. Changing Speed: 1. Using the control at the bottom of the screen, change the sample rate of all the sensors to a common rate of 20 Hz. Figure 4: Create mixed-input data table. 5 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] 2. Create a table with three columns, as shown in Fig. 4: In the first column, create a User-Entered data set called Voltage (units of V); In the second column, create a Run-Tracked User-Entered data set called Speed (units of cm/s) ); In the third column, create a Run-Tracked User-Entered data set called Friction (units of N). 3. Use the Coordinates tool ( ) to measure the speed and the frictional force. Pick a time that has reasonably good data for both. 4. Record your values in the table. 5. Repeat for voltages of 1.5 V, 2.0 V, 2.5 V, etc. up to 5.0 volts. 6. Create a graph of Friction vs. Speed. What trend can you see in the data? How does the frictional force depend on the speed? 7. Take the ratio of the fastest speed/ slowest speed. By what factor did you vary the speed? By what factor did the resulting frictional force change? Would you say that the friction changed a little or a lot? 8. Does your data support the concept that it is a useful approximation to assume that sliding friction is independent of speed? Changing Normal Force: 1. Create a table with four columns, as shown in Fig. 5: In the first column, create a Run-Tracked User-Entered data set called Tray (unitless); In the second column, create a User-Entered data set called Mass (units of kg); In the third column, create a calculation (units of Newtons) Normal Force = [mass (kg), ▼][Accel due to gravity (m/s²)] (Note: When you type [ in the calculation box it will bring up the selection box allowing you to choose data columns and constants.); In the fourth column, select Friction. 2. Replace the lower felt tray with a tray that has a white plastic surface. Start with both trays empty, and use the balance to determine the mass. Calculate the combined weight, and enter this as the normal force in the table below. 3. Set the voltage output to 1.5 volts. Click on Record. Stop recording before the cart hits the motion sensor. 4. Using the graph, note the approximate speed and, if necessary, adjust the voltage in later runs to keep the speed the same. 5. Record the frictional force in the table. Add the 4 mass bars (one at a time) and repeat. You should end up with 5 values. 6 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] 6. Create a graph of Friction vs. Normal Force1. What trend can you see in the data? Try a linear Curve Fit. Is your data linear? 7. Most text books make the assumption that Ff = μN, where Ff = friction, N= normal force, and μ = the friction coefficient. Does your data support this assumption? What are the units for μ? Changing Surface Material: 1. Create a table with four columns: Figure 5: Normal force vs. friction data table 7 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] a. In the first column, create a User-Entered data set called Cart Material (unitless); b. In the second column, create a User-Entered data set called Normal Force (units of N); c. In the third column, create a User-Entered data set called Frictional Force (units of N); d. In the fourth column, create a calculation (unitless) μ = [Frictional Force (N)]/[Normal Force (N)] 2. Remove all but one of the mass bars. Click on Record, and use the graph to measure the frictional force. 3. Coefficients are always between two surfaces. What is the other surface? 4. Replace the lower plastic tray with a tray with black felt and repeat. Replace the lower felt tray with a tray with brown cork material and repeat. 5. Which material has the largest coefficient? Which has the least? 6. What type of friction are you measuring: Static or Sliding (Kinetic) friction? Changing Area: 1. Connect the trays as shown in Figure 3. The bottom two trays should both be the white plastic. Place one of the silver mass bars in each of the two upper trays, and perform the pull test as before. 2. Measure the force from the graph and use equation (1) to calculate the coefficient of friction for the white plastic material. 3. How does this value (with twice the surface area) compare to your previous values? Most text books assume that sliding friction is independent of surface area. Do you agree? Figure 6: Changing Surface Area 8 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] Static Friction: 1. Increase the sample rate of the force sensor to 2 kHz (the maximum). 2. On the graph, change the negative velocity to the Position. 3. Remove the trailing two trays, returning to the original set-up pulling only two trays. The lower tray should be cork. Place both silver masses in the upper tray, and both black masses on the Motorized Cart in front of the Force Sensor. If the cart won't pull the trays, remove one of the silver masses. 4. Open the Signal Generator (see Fig. 7) and set the waveform to Positive Up Ramp Wave. Set the frequency to 0.1 Hz, the phase shift to 180º, and the amplitude to 1.5V. The frequency generator is now set to ramp up its voltage, to slowly increase the pulling force over a 10 second period. If the trays don't break free and move in that time, increase the voltage amplitude. 5. Click on Record. As soon as the tray starts to slide, click on stop. 6. Examine your data on the graph. You can see from the position data where the tray starts to slide, and the value of the sliding friction force after this point. In the time before this, Figure 7: Setting the signal generator 9 Lab experiment #5 [Based on PASCO lab manual 23 Coefficients of Friction, written by Jon Hanks] what does the force data look like? 7. Most materials show a value for static friction slightly larger than for sliding (kinetic) friction. Does your data support this? What is your largest value for the Static coefficient?
3307	https://pubmed.ncbi.nlm.nih.gov/31920114/	Switching Pharmacological Treatment in Wilson Disease: Case Report and Recommendations - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Switching Pharmacological Treatment in Wilson Disease: Case Report and Recommendations Marcia Leung1,Jaimie Wu Lanzafame1,Valentina Medici1 Affiliations Expand Affiliation 1 University of California Davis, Sacramento, CA, USA. PMID: 31920114 PMCID: PMC6956597 DOI: 10.1177/2324709619896876 Item in Clipboard Case Reports Switching Pharmacological Treatment in Wilson Disease: Case Report and Recommendations Marcia Leung et al. J Investig Med High Impact Case Rep.2020 Jan-Dec. Show details Display options Display options Format J Investig Med High Impact Case Rep Actions Search in PubMed Search in NLM Catalog Add to Search . 2020 Jan-Dec:8:2324709619896876. doi: 10.1177/2324709619896876. Authors Marcia Leung1,Jaimie Wu Lanzafame1,Valentina Medici1 Affiliation 1 University of California Davis, Sacramento, CA, USA. PMID: 31920114 PMCID: PMC6956597 DOI: 10.1177/2324709619896876 Item in Clipboard Full text links Cite Display options Display options Format Abstract Background. Available treatments for Wilson disease (WD) prevent longterm complications of copper accumulation. Current anti-copper agents include zinc salts, penicillamine, and trientine. Patients with WD may switch between the agents for a number of reasons. Due to the different mechanisms of action between the copper chelators and zinc salts, transitioning could require a period of overlap and increased monitoring. There are no large studies that investigate the best transition strategies between agents. In this article, we review the treatments for WD and how to monitor for treatment efficacy. Case Summary. The patient had been diagnosed with WD for over 20 years prior to establishing care in our Hepatology Clinic. During his initial course, he was transitioned from penicillamine to zinc due to evidence suggesting penicillamine had greater adverse effects in the long term. Later, he was switched to trientine. His liver enzymes and 24-hour urine copper were monitored. During these years, he intermittently had some financial hardship, requiring him to be on penicillamine rather than trientine. He also had developed acute kidney injury. Overall, his liver disease remained under control and he never had signs of decompensated cirrhosis, but had fluctuations of liver enzymes over the years. Conclusion. Anti-copper treatment for WD has to be tailored to medication side effects profile, patient's chronic and emerging comorbidities, as well as costs. Transitioning regimens is often challenging, and it requires closer monitoring, with no predictors of response. Keywords: Wilson disease; penicillamine; trientine; zinc acetate. PubMed Disclaimer Conflict of interest statement Declaration of Conflicting Interests: The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article. Figures Figure 1. ALT and AST levels trends.… Figure 1. ALT and AST levels trends. Abbreviations: ALT, alanine transaminase; AST, aspartate transaminase; BID,… Figure 1. ALT and AST levels trends. Abbreviations: ALT, alanine transaminase; AST, aspartate transaminase; BID, 2 times a day; TID, 3 times a day; QID, 4 times a day. Figure 2. Twenty-four-hour urine copper trends. Abbreviations:… Figure 2. Twenty-four-hour urine copper trends. Abbreviations: BID, 2 times a day; TID, 3 times… Figure 2. Twenty-four-hour urine copper trends. Abbreviations: BID, 2 times a day; TID, 3 times a day; QID, 4 times a day. See this image and copyright information in PMC Similar articles Efficacy and safety of oral chelators in treatment of patients with Wilson disease.Weiss KH, Thurik F, Gotthardt DN, Schäfer M, Teufel U, Wiegand F, Merle U, Ferenci-Foerster D, Maieron A, Stauber R, Zoller H, Schmidt HH, Reuner U, Hefter H, Trocello JM, Houwen RH, Ferenci P, Stremmel W; EUROWILSON Consortium.Weiss KH, et al.Clin Gastroenterol Hepatol. 2013 Aug;11(8):1028-35.e1-2. doi: 10.1016/j.cgh.2013.03.012. Epub 2013 Mar 28.Clin Gastroenterol Hepatol. 2013.PMID: 23542331 Efficacy and safety of D-penicillamine, trientine, and zinc in pediatric Wilson disease patients.Lee EJ, Woo MH, Moon JS, Ko JS.Lee EJ, et al.Orphanet J Rare Dis. 2024 Jul 9;19(1):261. doi: 10.1186/s13023-024-03271-1.Orphanet J Rare Dis. 2024.PMID: 38982450 Free PMC article. Comparative effectiveness of common therapies for Wilson disease: A systematic review and meta-analysis of controlled studies.Appenzeller-Herzog C, Mathes T, Heeres MLS, Weiss KH, Houwen RHJ, Ewald H.Appenzeller-Herzog C, et al.Liver Int. 2019 Nov;39(11):2136-2152. doi: 10.1111/liv.14179. Epub 2019 Jul 10.Liver Int. 2019.PMID: 31206982 Trientine induced colitis during therapy for Wilson disease: a case report and review of the literature.Boga S, Jain D, Schilsky ML.Boga S, et al.BMC Pharmacol Toxicol. 2015 Nov 20;16:30. doi: 10.1186/s40360-015-0031-z.BMC Pharmacol Toxicol. 2015.PMID: 26589720 Free PMC article.Review. Discontinuation of penicillamine in the absence of alternative orphan drugs (trientine-zinc): a case of decompensated liver cirrhosis in Wilson's disease.Ping CC, Hassan Y, Aziz NA, Ghazali R, Awaisu A.Ping CC, et al.J Clin Pharm Ther. 2007 Feb;32(1):101-7. doi: 10.1111/j.1365-2710.2007.00794.x.J Clin Pharm Ther. 2007.PMID: 17286794 See all similar articles Cited by The Present and Future Challenges of Wilson's Disease Diagnosis and Treatment.Leung M, Aronowitz PB, Medici V.Leung M, et al.Clin Liver Dis (Hoboken). 2021 May 1;17(4):267-270. doi: 10.1002/cld.1041. eCollection 2021 Apr.Clin Liver Dis (Hoboken). 2021.PMID: 33968387 Free PMC article.Review.No abstract available. Seven decades of clinical experience with Wilson's disease: Report from the national reference centre in Poland.Członkowska A, Niewada M, Litwin T, Kraiński Ł, Skowrońska M, Piechal A, Antos A, Misztal M, Khanna I, Kurkowska-Jastrzębska I.Członkowska A, et al.Eur J Neurol. 2024 Nov;31(11):e15646. doi: 10.1111/ene.15646. Epub 2022 Dec 11.Eur J Neurol. 2024.PMID: 36427277 Free PMC article. References Frydman M. Genetic aspects of Wilson’s disease. J Gastroenterol Hepatol. 1990;5:483-490. doi:10.1111/j.1440-1746.1990.tb01427.x - DOI - PubMed Wiggelinkhuizen M, Tilanus ME, Bollen CW, Houwen RH. Systematic review: clinical efficacy of chelator agents and zinc in the initial treatment of Wilson disease. Aliment Pharmacol Ther. 2009;29:947-958. doi:10.1111/j.1365-2036.2009.03959.x - DOI - PubMed Coffey AJ, Durkie M, Hague S, et al. A genetic study of Wilson’s disease in the United Kingdom. Brain. 2013;136(pt 5):1476-1487. doi:10.1093/brain/awt035 - DOI - PMC - PubMed Gao J, Brackley S, Mann JP. The global prevalence of Wilson disease from next-generation sequencing data. Genet Med. 2019;21:1155-1163. doi:10.1038/s41436-018-0309-9 - DOI - PubMed Tao TY, Gitlin JD. Hepatic copper metabolism: insights from genetic disease. Hepatology. 2003;37:1241-1247. doi:10.1053/jhep.2003.50281 - DOI - PubMed Show all 41 references Publication types Case Reports Actions Search in PubMed Search in MeSH Add to Search Research Support, N.I.H., Extramural Actions Search in PubMed Search in MeSH Add to Search MeSH terms Acute Kidney Injury / chemically induced Actions Search in PubMed Search in MeSH Add to Search Adult Actions Search in PubMed Search in MeSH Add to Search Chelating Agents / adverse effects Actions Search in PubMed Search in MeSH Add to Search Chelating Agents / economics Actions Search in PubMed Search in MeSH Add to Search Copper / urine Actions Search in PubMed Search in MeSH Add to Search Hepatolenticular Degeneration / drug therapy Actions Search in PubMed Search in MeSH Add to Search Hepatolenticular Degeneration / urine Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Liver / physiopathology Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Penicillamine / adverse effects Actions Search in PubMed Search in MeSH Add to Search Treatment Outcome Actions Search in PubMed Search in MeSH Add to Search Trientine / adverse effects Actions Search in PubMed Search in MeSH Add to Search Zinc / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Substances Chelating Agents Actions Search in PubMed Search in MeSH Add to Search Copper Actions Search in PubMed Search in MeSH Add to Search Penicillamine Actions Search in PubMed Search in MeSH Add to Search Zinc Actions Search in PubMed Search in MeSH Add to Search Trientine Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound (MeSH Keyword) Grants and funding R01 DK104770/DK/NIDDK NIH HHS/United States LinkOut - more resources Full Text Sources Europe PubMed Central PubMed Central Medical MedlinePlus Health Information Full text links[x] Free PMC article [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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3308	https://resources.system-analysis.cadence.com/blog/msa2023-all-about-the-heat-flux-equation	All About the Heat Flux Equation \| System Analysis Blog \| Cadence Skip to main content ### System Analysis PCB Design & Analysis Email I have read/accept theCadence Terms of Use and Privacy Policy. 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The heat flux in a system is dependent on the temperature gradient and heat transfer coefficient. Depending on the medium, heat transfer mechanisms are classified into conduction, convection, or radiation The demand for energy encourages us to explore opportunities in renewable energy sources. Among non-conventional energy sources, solar energy is critical due to its abundance. Solar power is used for utility power generation as well as heating. Concentrated solar power plants meet today’s energy requirements. In the design of concentrated solar power plants, especially solar receivers, heat flux and temperature are two primary design parameters. Understanding heat flux aids in determining the efficiency of solar receivers. In most thermodynamic applications, heat flux is a fundamental quantity of importance, as it influences efficiency and performance. Theoretically, the heat flux equation is utilized for calculating heat flux. However, in practicality, a range of calorimeters, gauges, and radiometers are used. Let’s explore heat flux and its equations. Heat Flow and Heat Transfer Mechanisms In a given system, heat flows from one point to another only if there is a temperature difference between them. The heat flows from a warmer location to a colder location. Heat flow occurs only if there is a medium for the heat to travel between points with different temperatures. The heat flow or heat transfer phenomenon is complex and multi-dimensional. Depending on the medium or the set of media where there is a temperature gradient, heat transfer mechanisms can be classified into: Conduction -In conduction, heat flow takes place through solid materials. Convection -When heat flows through gases and liquids, the heat transfer mechanism is called convection. Radiation -When electromagnetic waves carry heat energy, it forms the radiation mechanism of heat transfer. In the above heat transfer mechanisms, heat is transferred from one point to another through a medium. The rate of heat energy transferred gives the idea of heat flux in conduction, convection, and radiation. What Is Heat Flux? Heat flux is the amount of heat energy transferred through a surface in a unit area in unit time. The heat flux can be the amount of heat transferred from or dissipated on the surface of consideration. Heat flux is also known as thermal flux, heat flow density, heat flux density, or heat flow rate intensity. Heat flux is evaluated based on two fundamental quantities: The amount of heat transfer per unit area (Q) The area where the heat transfer takes place (A) (Alt text: Generalized heat flux equation) Since the heat flux is based on these two quantities, it is considered a derived quantity. Unit of Heat Flux Density The amount of heat energy transferred, or the heat transfer rate, can be measured in Joule per second or Watt. Heat flux can be calculated as the heat transfer rate per unit area, otherwise known as the heat flux density. Heat flux density is measured in SI unit-Watts per square meter (W/m2). The heat flux is a vector quantity with both magnitude and direction. Factors Influencing Heat Flux The heat flux in a system is dependent on: Temperature difference -The temperature difference or gradient is necessary for any heat transfer to take place. The heat flux shares a direct relationship with the temperature gradient. As the temperature gradient increases, the heat flux magnitude increases. Thermal transfer coefficient or heat transfer coefficient -The thermal transfer coefficient is introduced through Newton’s law of cooling. According to this law, the heat flux associated with a surface is linearly related to the temperature gradient. The proportionality constant linking heat flux and temperature gradient is called the heat transfer coefficient. Heat Flux Equation The heat flux equation can be obtained from the law of thermal conduction or the law of thermal conductivity, popularly known as Fourier’s law. The law is also referred to as the law of heat conduction. According to Fourier’s law, the heat flux is directly proportional to the thermal or temperature gradient. Mathematically, the heat flux equation can be expressed as: q is the heat flux Q is the heat transfer rate A is the area of the cross-section of the surface T is the temperature gradient K is the heat transfer coefficient Heat Flux Equation for Convection and Radiation Heat Transfer The heat flux equation for conduction heat transfer can be utilized for convection, provided the convective heat transfer coefficient is used in the place of constant K. To determine the heat flux in radiative heat transfer, the equation is given by the Stefan-Boltzmann law. The heat flux equation for radiation heat transfer is: σ is the Stefan-Boltzmann constant ϵ is the emissivity T is the temperature in (K) The calculation of heat flux is crucial in chemical processes, thermodynamic systems, and the aviation industry, among other things. Cadence’s CFD solver can support multi-dimensional heat flux problems. With the proper CFD tools, solving complex problems involving heat flux equations is a breeze. Subscribe to our newsletter for the latest CFD updates or browse Cadence’s suite of CFD software, including Fidelity and Fidelity Pointwise, to learn more about how Cadence has the solution for you. CFD SoftwareSubscribe to Our Newsletter About the Author With an industry-leading meshing approach and a robust host of solver and post-processing capabilities, Cadence Fidelity provides a comprehensive Computational Fluid Dynamics (CFD) workflow for applications including propulsion, aerodynamics, hydrodynamics, and combustion. 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3309	https://www.cs.arizona.edu/~alon/papers/regular-ire.pdf	On the Boundary of the Union of Planar Con v ex Sets J anos P ac h Mic ha Sharir y Abstract W e giv e t w o alternativ e pro ofs leading to dieren t generalizations of the follo wing theorem of [ ]. Giv en n con v ex sets in the plane, suc h that the b oundaries of eac h pair of sets cross at most t wice, then the b oundary of their union consists of at most n arcs. (An ar c is a connected piece of the b oundary of one of the sets.) In the generalizations w e allo w pairs of b oundaries to cross more than t wice. In tro duction Let C b e a collection of n non-degenerate con v ex sets (b o dies) in the plane, an y t w o of whic h ha v e at most a nite n um b er of b oundary p oin ts in common. W e also assume, for simplicit y , that no t w o b oundary curv es are tangen t to eac h other, and no three pass through the same p oin t. If t w o mem b ers of C ha v e exactly t w o b oundary p oin ts in common, then these p oin ts are called r e gular vertic es of the arrangemen t A(C ). All other in tersection p oin ts of the b oundary curv es are said to b e irr e gular. Let U = [C denote the union of all mem b ers of C . Let R and I denote the set of regular and irregular v ertices of A(C ), resp ectiv ely , lying on @ U , the b oundary of U . F urther, put V = R [ I . Clearly , jV j is equal to the n um b er of arcs that comp ose @ U . It w as sho wn in [ ] that if an y t w o mem b ers of C ha v e at most t w o b oundary p oin ts in common (i.e., if there are no irregular v ertices), then jRj = jV j n . In Section of this note, w e generalize this result as follo ws. Theorem With the ab ove notation, for any c ol le ction of n c onvex sets in the plane, we have jRj jI j + n : Actually , in [] the mem b ers of C w ere not required to b e con v ex, and it is v ery lik ely that Theorem also generalizes to that case. Whitesides and Zhao [ ] in tro duced the follo wing denition. A collection of closed Jordan curv es is called k -admissible if no t w o curv es touc h eac h other, an y t w o curv es Departmen t of Computer Science, Cit y College, CUNY, New Y ork, NY, USA, Couran t Institute of Mathematical Sciences, New Y ork Univ ersit y , New Y ork, NY 00, USA, and Hungarian Academ y of Sciences, Budap est, Hungary y Sc ho ol of Mathematical Sciences, T el Aviv Univ ersit y , T el Aviv , Israel, and Couran t Institute of Mathematical Sciences, New Y ork Univ ersit y , New Y ork, NY 00, USA in tersect in at most k p oin ts, and the in terior of no curv e disconnects the in terior of another. Clearly , w e can restrict our atten tion to the case when k is ev en. In Section , w e giv e a new pro of of the follo wing result of [ ], whic h pro vides y et another generalization of the ab o v e men tioned theorem of []. Theorem The numb er of vertic es on the b oundary of the union of the interiors of n Jor dan curves that form a k -admissible family, is at most k (n ). Pro of of Theorem First w e need some preparation. Let b e an orien ted con tin uous curv e in the plane. If at some p oin t w of , there is no unique tangen t line, then w is called a br e akp oint. W e sa y that is pie c ewise smo oth, if it has nitely man y breakp oin ts, and ev ery piece of b et w een t w o consecutiv e breakp oin ts is dieren tiable (including at its endp oin ts). Dene the total turning angle ( ) of a piecewise smo oth, orien ted curv e , as follo ws. If necessary , sub divide in to smaller dieren tiable orien ted arcs ; : : : ; m , suc h that eac h i is smo oth and an y t w o tangen ts to the same arc i , orien ted according to the orien tation of the curv e, dier in their orien tations b y less than . Let < ( i ) < + b e the smaller angle from the tangen t v ector at the starting p oin t of i to the tangen t v ector at the endp oin t of i , tak en with p ositiv e sign if the c hange is coun ter-clo c kwise and with negativ e sign otherwise. A t eac h p oin t w i separating t w o pieces, i and i+ , let (w i ) b e the smaller angle from the tangen t to i at w i to the tangen t to i+ at w i , with p ositiv e sign if and only if it is coun ter-clo c kwise. If w i is not a breakp oin t, then, b y construction, (w i ) = 0. Finally , let the total turning angle ( ) b e dened as the sum of ( i ) o v er all pieces i plus the sum of (w i ) o v er all v ertices w i . Eviden tly , this denition of the turning angle is indep enden t of the particular sub division of . ( i ) and (w i ) are called, resp ectiv ely , the turning angle of along the arc i and at the p oin t w i . The follo wing lemma summarizes the elemen tary prop erties of the total turning angle. W e omit the trivial pro of. Lemma Let b e a piecewise smo oth, orien ted curv e in the plane with total turning angle ( ). (i) If is a closed curv e, then ( ) is an in teger m ultiple of . (ii) If is a coun ter-clo c kwise (resp. clo c kwise) orien ted closed curv e whic h do es not in ter-sect itself, then ( ) = (resp. ). (iii) If in tersects itself at a p oin t w , then it can b e decomp osed in to t w o piecewise smo oth, orien ted curv es, 0 and ", ha ving the common breakp oin t w . (If is a closed curv e, then so are 0 and "; if is op en, then one of the t w o parts is op en and the other is closed.) In b oth cases, w e ha v e ( ) = ( 0 ) + ( "): W e refer to the last equalit y as the additivity pr op erty of the total turning angle. No w w e turn to the pro of of Theorem . W e can assume without loss of generalit y that ev ery mem b er of C is b ounded and its b oundary is smo oth. It is sucien t to establish the theorem in the case when U = [C is connected; otherwise, arguing for eac h comp onen t of U separately , w e obtain the stronger inequalit y jRj jI j + n k 0k k , where k (resp. k , k ) is the n um b er of connected comp onen ts of U formed b y one (resp. t w o, at least three) sets of C . A connected comp onen t H of the complemen t of U is called a hole. Let V (H ) denote the set of v ertices along the b oundary of a hole H . These v ertices divide the b oundary of H in to jV (H )j arcs, whic h form a set denoted b y (H ). The set of all arcs comp osing @ U will b e denoted b y ext = [ H (H ). Note that ev ery b ounded hole has at least three v ertices. The unique un b ounded hole ma y ha v e few er v ertices (zero or t w o), but then jV j . W e ma y therefore assume that ev ery hole has at least three v ertices, so the n um b er of holes is at most jV j=. Orien t the b oundary of ev ery c C in the coun ter-clo c kwise direction. Accordingly , ev ery (unit) tangen t v ector to c will b e orien ted so that c lies on its left-hand side. Consider no w t w o sets c; c 0 C whose b oundaries in tersect in exactly t w o p oin ts v and v 0 . (These are r e gular v ertices of the arrangemen t.) Then c \ c 0 is a lens-lik e region, whose b oundary is a coun ter-clo c kwise orien ted closed curv e cc 0 , with t w o breakp oin ts, v and v 0 . Denote the turning angle of cc 0 at v and v 0 b y a(v ) and a(v 0 ), resp ectiv ely . (Note that a(v ) is alw a ys p ositiv e.) Notice that the orien tation of the b oundary of an y b ounde d hole H is clo c kwise, and the orien tation of the b oundary of the unique unb ounde d hole is coun ter-clo c kwise. A t an y regular v ertex v on the b oundary of an y hole, the turning angle of the b oundary is a(v ). Th us, Lemma (ii) implies that for an y xed b ounded hole H , X v V (H ) a(v ) + X (H ) ( ) = ; and, for the unique un b ounded hole, the left-hand side is equal to . Adding all these equations, w e obtain X v R ( a(v )) X ext ( ) (jH j ) = jH j (jRj + jI j) : () Let in t denote the collection of maximal b oundary arcs of the sets in C , orien ted as ab o v e, that are con tained in the in terior of U . Lemma X v R ( a(v )) X in t ( ): () It is easy to see that Lemma implies Theorem . Indeed, the righ t-hand side of () is equal to n P ext ( ). Substituting () in to (), w e obtain jRj 0 @ n X ext ( ) A X ext ( ) (jRj + jI j) ; whic h yields that jRj jI j + n , as asserted. Pro of of Lemma : Let R denote the subset of those arcs in in t that ha v e at least one regular endp oin t. The union of R is decomp osed in to a collection of orien ted cycles and paths; the v ertices (breakp oin ts) of the cycles and the in ternal v ertices of the paths b elong to R, and the endp oin ts of the paths b elong to I . A: Let = v 0 v v k (v k = v 0 ) b e one of these orien ted cycles, with v ertices v 0 ; v ; : : : ; v k R. Let i denote the orien ted arc along connecting v i to v i , and let c i b e the set in C whose b oundary con tains i , for i = ; : : : ; k . W e tra v erse from v 0 , and consider the tangen ts to , orien ted in accordance with the orien tation of (so that the sets they are tangen t to lie on their left). By construction, as w e follo w these tangen ts, they k eep turning in the coun ter-clo c kwise (p ositiv e) direction, and this also holds at eac h v ertex of . See Figure ??. F or eac h i = ; : : : ; k , c ho ose a v ery small " > 0, and dra w a circle of radius " around eac h v ertex v i . Let v i and v + i denote an in tersection p oin ts of this circle with i and i+ , resp ectiv ely (with k + = ). Let 0 denote the closed curv e obtained from b y replacing the p ortion of i b et w een v + i and v i b y a straigh t-line segmen t, for ev ery i. Clearly , the total turning angle of 0 is equal to the total turning angle of . W e claim that 0 can b e decomp osed in to k p ositiv ely (i.e., coun ter-clo c kwise) orien ted lo ops at the v ertices v i and an orien ted closed p olygon = v 0 v v k ; (v k = v 0 ). This follo ws from the fact that v i and v i+ , the other endp oin ts of the arcs i and i+ , lie on dieren t sides of the line connecting the t w o regular in tersection p oin ts of the b oundaries of c i and c i+ . Consequen tly , if " is sucien tly small, then the segmen ts v + i v i and v + i v i+ m ust cross eac h other in a small neigh b orho o d of v i , at a p oin t denoted b y v i . The i-th lo op of 0 is its p ortion that starts and ends at v i . Th us, b y the additivit y the turning angle, ( 0 ) = k ( ) + ( ): By denition, at eac h v ertex of , the absolute v alue of the turning angle is at most (and the turning angle along its edges is 0). Consequen tly , ( ) = ( 0 ) k . (Actually , b y Lemma (i), the total turning angle of m ust b e a m ultiple of , so ( ) dk =e is also true.) B: Consider no w a path = v 0 v v k v k + with irregular endp oin ts and regular in ternal v ertices. Let i ; c i , for i = ; : : : ; k + , and v i ; v + i , for i = ; : : : ; k , denote the same en tities as for cycles (the previous case). W e also put v + 0 = v 0 and v k + = v k + . In exactly the same w a y as b efore, w e construct a curv e 0 from b y replacing with a straigh t-line segmen t the p ortion of i b et w een v + i and v i , for ev ery i = ; : : : ; k + . W e ha v e that ( ) ( 0 ) (w e turn more along from v 0 to v than b y going straigh t from v 0 to v and then turning at v un til w e are tangen t to , and similarl y at the other end of ; see Figure ?? ). No w, arguing as in the case of cycles, 0 is decomp osed in to k p ositiv ely orien ted lo ops and a p olygonal path . Again, the additivit y of the turning angle implies that ( ) ( 0 ) = k ( ) + ( ): Since at eac h in ternal v ertex of , the turning angle is b et w een and + , w e ha v e that the total turning angle of is at least k . C: If is a cycle, as in A, its total turning angle is k X i= a(v i ) + k X i= ( i ) k ; whic h implies that k X i= ( a(v i )) k X i= ( i ): If is a path, as in B, its total turning angle is k X i= a(v i ) + k + X i= ( i ) k ; whic h implies that k X i= ( a(v i )) k + X i= ( i ): W e no w add these inequalities, o v er all cycles and paths comp osing R , and obtain X v R ( a(v )) X R ( ) X in t ( ); as asserted. Remarks. () In [], w e pro v ed that jRj n , under the assumption that al l v ertices of A(C ) are regular. Theorem sho ws that the same b ound holds with the w eak er assumption that there are no irregular v ertices on the b oundary of U = [C . (Recall, ho w ev er, that the result in [ ] do es not require, as w e do, that all mem b ers of C b e con v ex.) () Note that Theorem also holds if I denotes the set of irregular v ertices of @ U that lie on b oundaries of holes that con tain at least one regular v ertex. () Supp ose that an y t w o mem b ers of C ha v e at most s (a constan t n um b er) of b oundary p oin ts in common. Ho w large can jRj b e? One can sho w that, ev en for s = , the maxim um p ossible v alue of jRj can b e (n = ). T o see this, tak e a set P of n p oin ts and a set L of n lines, so that there are (n = ) incidences b et w een P and L (see [ , Chap. ]). Replace eac h p oin t in P b y a disk of radius , for some sucien tly small > 0, and replace eac h line L b y a long rectangle whose width is and whose long b ottom edge is parallel to, lying ab o v e , and at distance 0 < from it. One can sho w that, for an appropriate c hoice of and 0 , the n um b er of in tersections b et w een an y disk and an y rectangle is at most t w o, that eac h incidence b et w een a p oin t of P and a line of L corresp onds to an in tersecting pair of a disk and a rectangle, and that eac h in tersection p oin t b et w een suc h a pair lies on the b oundary of the union. Hence, w e ha v e a collection of n disks and rectangles satisfying jRj = (n = ). Is this construction asymptotically b est p ossible? () It is not hard to see that the co ecien t of the term jI j in Theorem cannot b e replaced b y an y smaller constan t. T o see this, tak e n copies of a regular n-gon, sligh tly rotated around their common cen ter, and, for eac h original v ertex, clip the batc h of its copies with a small rectangle. This creates roughly n regular v ertices on the b oundary of the union of the resulting collection of n con v ex sets. On the other hand, jI j is ab out n . Pro of of Theorem Assume without loss of generalit y that ev ery curv e c has a p oin t p c that b elongs to the b oundary of U , the union of the in teriors of all family mem b ers. Let q b e one of the (at most k ) in tersection p oin ts of t w o curv es, c and c 0 . Connect p c to p c 0 b y an arc (edge'), going rst from p c to q in clo c kwise direction around c, and then follo wing the b oundary of c 0 in coun terclo c kwise direction to p c 0 . F or eac h pair c; c 0 of family mem b ers that con tribute an in tersection p oin t q to the b oundary of U , construct suc h an edge that connects p c to p c 0 via q , but do this for only one suc h p oin t q . The t w o pieces an edge consists of are called half-e dges. It is easy to sho w that an y t w o half-edges not inciden t to the same p oin t p c in tersect an ev en n um b er of times. Th us, these edges form a graph dra wing with the prop ert y that an y t w o edges not inciden t to the same v ertex p c in tersect an ev en n um b er of times. This implies that the underlying graph is planar (see [, Cor. .], and, since it has no m ultiple edges, the n um b er of its edges is at most n . The total n um b er of v ertices along the b oundary of U is ob viously at most k times larger than that. References [] K. Kedem, R. Livne, J. P ac h, and M. Sharir, On the union of Jordan regions and collision-free translational motion amidst p olygonal obstacles, Discr ete Comput. Ge om. ( ), {. [] L. Lo v asz, J. P ac h and M. Szegedy , On Con w a y's thrac kle conjecture, Pr o c. th A CM Symp. on Computational Ge ometry ( ), {. [] J. P ac h and P .K. Agarw al, Combinatorial Ge ometry, J. Wiley-In terscience, New Y ork, . [] S. Whitesides and R. Zhao. K -admissible collections of Jordan curv es and osets of circular arc gures. T e chnic al R ep ort SOCS 0.0, McGill Univ ersit y , Mon treal, 0.
3310	https://www.geeksforgeeks.org/aptitude/difference-between-modulo-and-modulus/	Difference between Modulo and Modulus - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Quantitiative Aptitude Logical Reasoning Verbal Ability Aptitude Quiz Quantitiative Aptitude Quiz Verbal Ability Quiz Aptitude For Placements Interview Corner Practice Sets Sign In ▲ Open In App Difference between Modulo and Modulus Last Updated : 23 Jul, 2025 Comments Improve Suggest changes 5 Likes Like Report In the world of Programming and Mathematics we often encounter the two terms "Modulo"and "Modulus".In programming we use the operator "%" to perform modulo of two numbers. It basically finds the remainder when a number xis divided by another number N. It is denoted by : x mod N where x : Dividend N : Divisor Some important links to find out more about concepts in Modular Arithmetic : Modular Arithmetic Euler’s Totient Function Compute n! under modulo p Wilson’s Theorem How to compute mod of a big number? Find value of y mod (2 raised to power x) Modulus Operator in C Modulus of Negative Numbers In this article we are going to see the difference between the two terms, modulo and modulus, used frequently in Modular Arithmetic. Modulo : It is basically an operator which is denoted by "mod"and, in programming, uses "%".It is a function which returns the remainder value when a number is divided by another number. For example - 12 mod 5 // 12 modulo 5 38 ≡ 14 (mod 12) // 38,14 are congruent modulo 12 Modulus : It is simply a noun which is the value N in the expression "x mod N". We abbreviate it as N is the modulus. For example - 12 mod 3 // 3 is the modulus In the previous example, note that thecongruent modulo basically means that the difference of the two numbers is an integer multiple of the modulus 12. The difference of 38 and 14 is 24 and 24 is double of 12. It is important to note that these two words can't be interchanged as it will completely change the meaning. It would be bogus to say that - 12 mod 10 // modulo is 10 38 ≡ 14 (mod 12) // 38,14 are congruent modulus In terms of English, we can conclude that modulo is a preposition,as prepositions are used before a noun to show some spatial and temporal relations of the noun modulus.The relation in modular arithmetic is the remainderwhen two numbers are added. 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Register Oct 05 Ace Arithmetic on the GMAT Focus Edition 11:00 AM IST 01:00 PM IST Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions. Save Now! Sep 22 Special Offer: Get 25% Off Target Test Prep GMAT Plans 12:00 PM EDT 11:59 PM EDT The Target Test Prep GMAT Flash Sale is LIVE! Get 25% off our game-changing course and save up to $450 today! Use code FLASH25 at checkout. This special offer expires on September 30, so grab your discount now! Back to Forum Create Topic Reply Two boys start cycling around a circular track in opposite directions rishabhmishra rishabhmishra Joined: 23 Sep 2016 Last visit: 16 Aug 2019 Posts: 181 Products: Posts: 181 Post URL08 Mar 2018, 04:17 Show timer 00:00 Start Timer Pause Timer Resume Timer Show Answer a 3% b 17% c 15% d 61% e 5% A B C D E Hide Show History My Mistake Official Answer and Stats are available only to registered users.Register/Login. Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)! Difficulty: 65% (hard) Question Stats: 61% (02:19) correct 39%(02:31) wrong based on 632 sessions History Date Time Result Not Attempted Yet Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show Hide Answer Official Answer Official Answer and Stats are available only to registered users.Register/Login. 3 Kudos Add Kudos 60 Bookmarks Bookmark this Post Most Helpful Reply generis generis Senior SC Moderator Joined: 22 May 2016 Last visit: 18 Jun 2022 Posts: 5,277 Expert Expert reply Posts: 5,277 Post URL11 Mar 2018, 16:09 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more Find combined rate per minute. (R + R). Find time required to cross paths one time (D/R). Finally, divide total time by time per crossing. Answer equals the # of times they meet. Convert seconds to minutes If time units are different, you can convert X seconds to one minute. Multiply the rate's number of seconds by N to get to 60 seconds. Multiply the numerator by N, too. Rate of Boy 1: (10 m 1 s e c∗60 60)=(600 m 60 s e c s)=600 m 1 m i n(10 m 1 s e c∗60 60)=(600 m 60 s e c s)=600 m 1 m i n Rate of Boy 2: (5 m 1 s e c)=(300 m 60 s e c s)=300 m 1 m i n(5 m 1 s e c)=(300 m 60 s e c s)=300 m 1 m i n Add rates When travelers move in opposite directions, whether towards or away from one another, add rates (speeds). Combined rate: 600 m 1 m i n+300 m 1 m i n=900 m 1 m i n 600 m 1 m i n+300 m 1 m i n=900 m 1 m i n Time it takes for them to cross paths once? R∗T=D R∗T=D, so T=D R T=D R T=400 m 900 m e t e r s m i n u t e=400 900 m i n s=4 9 m i n T=400 m 900 m e t e r s m i n u t e=400 900 m i n s=4 9 m i n for them to meet once Number of times they cross/meet? T o t a l M i n u t e s M i n u t e s P e r O n e M e e t=T o t a l M i n u t e s M i n u t e s P e r O n e M e e t= # of times they meet 40 4 9=40∗9 4=90 40 4 9=40∗9 4=90 times that they meet Answer D — The only thing more dangerous than ignorance is arrogance. Einstein — I stand with Ukraine: Donate to Help Ukraine! Signature Read More 8 Kudos Add Kudos 8 Bookmarks Bookmark this Post JeffTargetTestPrep JeffTargetTestPrep Target Test Prep Representative Joined: 04 Mar 2011 Last visit: 05 Jan 2024 Posts: 2,983 Status:Head GMAT Instructor Affiliations: Target Test Prep Expert Expert reply Posts: 2,983 Post URL12 Mar 2018, 10:08 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 Show more We can let t = the time they will meet for the first time: 10t + 5t = 400 15t = 400 t = 400/15 = 80/3 seconds Since 40 minutes = 40 x 60 = 2,400 seconds, they will meet 2,400/(80/3) = (2,400 x 3)/80 = 30 x 3 = 90 times. Answer: D Jeffrey Miller \| Head of GMAT Instruction \| Jeff@TargetTestPrep.com Flash Sale: Get 25% Off Our GMAT Plans \| Use Code:FLASH25 Try OnDemand GMAT, the Most Comprehensive Video Course Perfect 5-Star Ratingon GMAT Club See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews Signature Read More 5 Kudos Add Kudos 4 Bookmarks Bookmark this Post General Discussion rishabhmishra rishabhmishra Joined: 23 Sep 2016 Last visit: 16 Aug 2019 Posts: 181 Products: Posts: 181 Post URL08 Mar 2018, 04:26 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more SOLU lets first find out when they both will meet as we all know both cyclist are coming towards each other they will meet soon so total distance must be divided by sum of their speed. total distance will be 400 meters circumference of circle so 400/10+5= 400/15 =80/3 in every 80/3 seconds they will meet so in 40 min they will meet 4060(convert in seconds) we will divide total time with they took time to meet once so to identify how many times they met. 40603/80 =90 so OA IS D 4 Kudos Add Kudos 3 Bookmarks Bookmark this Post gracie gracie Joined: 07 Dec 2014 Last visit: 11 Oct 2020 Posts: 1,031 Posts: 1,031 Post URL08 Mar 2018, 12:25 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more 400m/15 combined mps=26 2/3 seconds until first crossing 40 min60 sec=2400 seconds 2400/(26 2/3)=90 crossings D 1 Kudos Add Kudos Bookmarks Bookmark this Post Archit3110 Archit3110 Major Poster Joined: 18 Aug 2017 Last visit: 28 Sep 2025 Posts: 8,425 Status:You learn more from failure than from success. Location: India Concentration: Sustainability, Marketing GMAT Focus 1:545 Q79 V79 DI73 GMAT Focus 2:645 Q83 V82 DI81 GPA: 4 WE:Marketing (Energy) GMAT Focus 2:645 Q83 V82 DI81 Posts: 8,425 Post URL28 Sep 2019, 04:16 total time taken by two cyclists ; 10+5 ; 15 mps 400/15 ; 80/3 seconds in 40 mins total circles ; 40 60 ; 2400 sec 2400/80 3 ; 90 times IMO D rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more Kudos Add Kudos Bookmarks Bookmark this Post Kinshook Kinshook Major Poster Joined: 03 Jun 2019 Last visit: 28 September 2025 Posts: 5,760 Location: India Schools:HBS '26CBS '26Wharton '26 GMAT 1:690 Q50 V34 WE:Engineering (Transportation) Products: Schools:HBS '26CBS '26Wharton '26 GMAT 1:690 Q50 V34 Posts: 5,760 Post URL31 Mar 2020, 11:13 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more Given: Two boys start cycling around a circular track in opposite directions at constant speeds. Asked: if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? 40 mins = 40 60 = 2400 seconds Time taken to meet each other = 400/15 = 26 10/15 = 26 2/3 seconds Number of times they met each other in 40 mins = 2400/(400/15) = 2415/4 = 615 = 90 times IMO D Kudos Add Kudos 1 Bookmarks Bookmark this Post AkshayC95 AkshayC95 Joined: 28 Jan 2018 Last visit: 23 Feb 2021 Posts: 77 Location: India Concentration: Finance, International Business GPA: 3.5 Products: Posts: 77 Post URL10 Jul 2020, 17:08 rishabhmishra Two boys start cycling around a circular track in opposite directions at constant speeds. if the circumference of the track is 400 meters, the faster boy cycles at 10 meters per second and the slower boy cycles at 5 meters per second, how many times would they have crossed each other after cycling for 40 minutes? A. 15 B. 30 C. 60 D. 90 E. 120 SOURCE EXPERTSGLOBAL Show more Speed of A= 10m/sec Speed of B= 5m/sec Distance=400m (One revolution) For A speed=distance/time 10m/sec=400/time Time=40sec Total Time Given 40mins(2400secs) No. of revolutions made by A in 40mins(2400sec) If it takes A 40secs for one revolution so in 40mins that is 2400secs A will make 2400secs/40sec= 60 revolutions For B Speed=distance/time 5m/sec=400/time time=80secs Total Time Given 40mins(2400) No. of revolutions made by B in 40mins(2400secs) If it takes B 80secs for one revolution then in 2400secs it will make 2400/80=30 revolutions. Adding no. of revolution of A and B= 60+30=90 revolutions in total. Please check and let me know If I am wrong, I tried on a few problems and I got the answer through this method. 2 Kudos Add Kudos Bookmarks Bookmark this Post Kritisood Kritisood Joined: 21 Feb 2017 Last visit: 19 Jul 2023 Posts: 492 Location: India GMAT 1:700 Q47 V39 Products: GMAT 1:700 Q47 V39 Posts: 492 Post URL10 Jul 2020, 23:37 total time they cycle for is 4060=2400 secs to find: 2400/how long they take to meet the first time Relative speed= 15 (since they are traveling in opp directions) total distance= 400 Time they meet = 400/15 = 80/3 answer = 2400/80 = 90 Kudos Add Kudos Bookmarks Bookmark this Post Mayfly Mayfly Joined: 03 Apr 2015 Last visit: 07 May 2021 Posts: 1 Posts: 1 Post URL04 Apr 2021, 08:52 D=SPEED X TIME 400x = 15 X 40 X 60 x=90 total distance = 400x total time = 40 X 60 total speed= relative speed added as in opposite direction. -> 15m/s Kudos Add Kudos Bookmarks Bookmark this Post Gemmie Gemmie Joined: 19 Dec 2021 Last visit: 29 Jul 2025 Posts: 499 Location: Viet Nam Concentration: Technology, Economics GMAT Focus 1:695 Q87 V84 DI83 GPA: 3.55 GMAT Focus 1:695 Q87 V84 DI83 Posts: 499 Post URL28 Aug 2024, 22:30 Total distance completed by both: (10+5)∗40∗60=36∗10 3(10+5)∗40∗60=36∗10 3 Total number of rounds completed by both: 36∗10 3 4∗10 2=90 36∗10 3 4∗10 2=90 Every time they together complete a round => they cross each other once => The 2 complete 90 rounds => cross each other 90 times Kudos Add Kudos Bookmarks Bookmark this Post ClaireCHEN ClaireCHEN Joined: 09 Jul 2024 Last visit: 15 Jan 2025 Posts: 25 Location: China Schools:Anderson '27Marshall '27 GMAT Focus 1:635 Q90 V77 DI77 GPA: 3.2 Schools:Anderson '27Marshall '27 GMAT Focus 1:635 Q90 V77 DI77 Posts: 25 Post URL29 Aug 2024, 05:51 The travel distance of the slower one within 40 mins is 12000m, then this q equals how many times they can meet within 12000m? Their 1st encounter happens at the distance of x, then x/5 = (400-x)/10 coz they should take the same amount of time, >>>x=400/3. Then within the 12000m they will encounter 12000/(400/3) times, which is 90 times. So D. Kudos Add Kudos Bookmarks Bookmark this Post einstein801 einstein801 Joined: 23 Jan 2024 Last visit: 18 Feb 2025 Posts: 176 Posts: 176 Post URL05 Dec 2024, 23:49 Hi MartyMurrayBunuel any idea how to tackle this question if they travel in the SAME direction? Kudos Add Kudos Bookmarks Bookmark this Post samarpan.g28 samarpan.g28 Joined: 08 Dec 2023 Last visit: 25 Sep 2025 Posts: 325 Location: India Concentration: General Management, Human Resources Schools:IIMA '26Schulich Sept '26Montreal '26HEC Sept '26 GPA: 8.88 WE:Engineering (Technology) Schools:IIMA '26Schulich Sept '26Montreal '26HEC Sept '26 Posts: 325 Post URL16 Dec 2024, 07:13 Hi einstein801, I am no expert but I believe this will answer your query. If you find the relative speed while they travel in the same direction, it will be 10-5=5m/s. We know the circumference is 400m, and time = distance/speed; thus, the time it takes for them to meet every time is 400/5=80 secs. Given that they travel for a total of 40 mins i.e. 4060=2400 secs. So, the total number of times they will meet is 2400/80=30. This is the answer. Or, you can do it this way too. The relative speed in the same direction is 10-5=5m/s. We know, distance=speedtime; so in 40 mins they cover 40605=12000 m distance. Given that the circumference is 400m, so the number of laps they cover i.e. the number of times they meet is 12000/400=30. I hope this helps. einstein801 Hi MartyMurrayBunuel any idea how to tackle this question if they travel in the SAME direction? Show more Kudos Add Kudos Bookmarks Bookmark this Post NEW TOPIC POST REPLY Question banks Downloads My Bookmarks Important topics Reviews Similar topics Similar Topic Author Kudos Replies Last Post Two trains run in opposite directions on a circular track. alex1233 by: alex1233 25 Nov 2024, 04:35 17 180 Two cars run in opposite directions on a circular track. Car A travels Bunuel by: Bunuel 29 Sep 2024, 05:43 16 207 Two people are walking around a circular track. They start at 8 A.M. 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3312	https://www.colonialsd.org/uploaded/Forms_and_Documents/Curriculum/Math/Integrated_Math/Blue_Unit_3/Compound_Growth_and_Decay.pdf	Compound Growth and Decay Example 1 Exponential Decay of the Form y = a(1 – r)t. RECREATION A particular chemical must be added to a swimming pool at regular intervals because it is released from the water into the air at the rate of 5% per hour. Sixteen ounces of the chemical is added at 8 am. At what time will three–fourths of this chemical be gone from the pool? Explore The problem gives the amount of chemical added to the pool and the rate at which the chemical is eliminated. It asks you to find the time it will take for three–fourths of the original amount of chemical to be eliminated from the pool. You then must add that time to 8 am to find the time of day. Plan Use the formula y = a(1 – r)t. Let t be the number of hours since adding the chemical. The initial amount a is 16 ounces, and the percent of decrease r is 0.05. The amount remaining y is one–fourth of 16 or 4. Solve y = a(1 – r)t Exponential decay formula 4 = 16(1 – 0.05)t Replace y with 4, a with 16, and r with 0.05. 0.25 = (0.95)t Divide each side by 16. log 0.25 = log (0.95)t Property of Equality for Logarithms log 0.25 = t log (0.95) Product Property for Logarithms log 0.25 log 0.95 = t Divide each side by log 0.95. 27.0268 ≈ t Use a calculator. Since 27.0268 ≈ 27, it will take approximately 27 hours for three–fourths of the chemical to be eliminated. The time will be 11 am of the next day. Check Use the formula to find how much of the original 16 ounces of chemical will remain after 27 hours. y = a(1 – r)t Exponential decay formula y = 16(1 – 0.05)27 Replace a with 16, r with 0.05, and t with 27. y ≈ 4.0055 Use a calculator. One–fourth of 16 is 4, so the answer seems reasonable. Three-fourths of the chemical will be eliminated in about 27 hours. This will be at 11 am of the next day. Example 2 Exponential Decay of the Form y = ae–kt. CHEMISTRY The half–life of a radioactive substance is the time it takes for half of the atoms of the substance to decay. Each element has a unique half–life. Radon–222 has a half–life of about 3.8 days, while thorium–234 has a half–life of about 24 days. Find the value of k for each element and compare their equations for decay. The equations will be of the form y = ae–kt, where t is in days. To determine the constant k for each element, let a be the initial amount of the substance. The amount y that remains after t days of the half– life is then represented by 0.5a. Use this idea to find the value of k for each element and then to write their equations. Radon–222 y = ae–kt Exponential decay formula 0.5a = ae–k(3.8) Replace y with 0.5a and t with 3.8. 0.5 = e–3.8k Divide each side by a. ln 0.5 = ln e–3.8k Property of Equality for Logarithmic Functions ln 0.5 = –3.8k Inverse Property of Exponents and Logarithms ln 0.5 –3.8 = k Divide each side by –3.8. 0.1824 ≈ k Use a calculator. Thorium–234 y = ae–kt Exponential decay formula 0.5a = ae–k(24) Replace y with 0.5a and t with 24. 0.5 = e–24k Divide each side by a. ln 0.5 = ln e–24k Property of Equality for Logarithmic Functions ln 0.5 = –24k Inverse Property of Exponents and Logarithms ln 0.5 –24 = k Divide each side by –24. 0.0289 ≈ k Use a calculator. The equations for radon–222 and thorium–234 are y = ae–0.1824t and y = ae–0.0289t, respectively. For both equations, t represents time in days. In comparing the equations, it appears that the longer the half–life, the smaller the value of k. Standardized Test EXAMPLE Example 3 Exponential Growth of the Form y = a(1 + r)t. In 1980, the value of farming land in a region of Wyoming was $150 per acre. Since then, the value has increased by exactly 0.75% per year. If the land continues to increase in value at this rate, what will the approximate value of the land per acre be in 2005? A. $610 B. $330 C. $181 D. $175 Read the Test Item You need to find the value of land 2005 – 1980 or 25 years later. Since the land is increasing in value at a fixed percent each year, use the formula y = a(1 + r)t. Solve the Test Item The initial value a is 150, the percent of increase is 0.75% or 0.0075, and the time t is 25 years. y = a(1 + r)t Exponential growth formula y = 150(1 + 0.0075)25 Replace a with 150, r with 0.0075, and t with 25. y = 150(1.0075)25 Simplify. y ≈ 180.81 Use a calculator. The answer is C. Example 4 Exponential Growth of the Form y = aekt. SAVINGS Sue invests $1000 at 5% interest compounded continuously and Norma invests $1250 at 3.5% interest compounded continuously. When interest is compounded continuously, the amount A in an account after t years is found using the formula A = Pert, where P is the amount of principal and r is the annual interest rate. In how many years will Sue’s account be greater than Norma’s account? Since Sue’s interest rate is greater than Norma’s, it seems likely that Sue’s account will eventually be greater than Norma’s. You need to write a function for Sue’s account and for Norma’s account using the formula and then write an inequality. Let S be the amount in Sue’s account and N be the amount in Norma’s account. A = Pert A = Pert S = 1000e0.05t Sue’s principal is 1000 and rate is 5% or 0.05. N = 1250e0.035t Norma’s principal is 1250 and rate is 3.5% or 0.035. You want to find t such that S > N. 1000e0.05t > 1250e0.035t S > N ln 1000e0.05t > ln 1250e0.035t Property of Inequality for Logarithms ln 1000 + ln e0.05t > ln 1250 + ln e0.035t Product Property of Logarithms ln 1000 + 0.05t > ln 1250 + 0.035t Inverse Property of Exponents and Logarithms ln 1000 + 0.015t > ln 1250 Subtract 0.035t from each side. 0.015t > ln 1250 – ln 1000 Subtract ln 1000 from each side. t > ln 1250 – ln 1000 0.015 Divide each side by 0.015. t ≥ 14.88 Use a calculator. After about 15 years, Sue’s account will be greater than Norma’s account.
3313	https://www.sciencedirect.com/topics/chemistry/hydration-enthalpy	Hydration Enthalpy - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Hydration Enthalpy In subject area:Chemistry Hydration enthalpy is defined as the energy change associated with the process of solvation when ions, such as Ca²⁺ and Zn²⁺, interact with water molecules, with variations observed as electron stabilization effects are considered. AI generated definition based on:Coordination Chemistry Reviews, 2009 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications Third Party Recommendations Featured Authors On this page Definition Chapters and Articles Related Terms Recommended Publications Third Party Recommendations Featured Authors Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Ion–solvent interactions 1999, Ions in SolutionJohn Burgess 4.3 THERMOCHEMISTRY OF ION SOLVATION The importance of ion hydration enthalpies in determining solubilities of salts was emphasised at the very beginning of this text (Introduction and Fig. 1.1). There it was shown, using the specific example of sodium chloride, that enthalpies of solution generally represent small differences between large lattice enthalpies and large ion hydration enthalpies. The main topics of this section are the dependence of ion hydration enthalpies on the nature of ions and the indications of strengths of ion-solvent interactions given by ion solvation enthalpies (cf. spectroscopic indications of strengths of ion-solvent interactions discussed in sections 4.1 and 4.2 above). However, a few sentences on the derivation of ion solvation enthalpies is required before their values are discussed. As is apparent from Fig. 1.1(a), reproduced again here as Fig. 4.6(a), it is relatively straightforward to estimate the sum of the hydration enthalpies of the pair of ions which constitute a salt. The enthalpy of solution can be measured directly in a calorimeter if the salt is reasonably soluble. If the salt is sparingly soluble, then a satisfactory estimate for its enthalpy of solution can usually be obtained from the temperature-dependence of its solubility. As the enthalpy of solution is generally the small difference between two large quantities, the approximations inherent in this van't Hoff approach are negligible in the present context. The lattice enthalpy of a salt containing monatomic ions can be calculated with reasonable precision; the lattice enthalpies of salts containing complex cations or anions are much more difficult to calculate satisfactorily, due to uncertainties in charge distribution within polyatomic ions. Sums of ion hydration enthalpies and the measured enthalpies of solution and calculated lattice enthalpies from which they are derived are shown, for the specific examples of sodium chloride (cf. Fig. 1.1(b)) and potassium hexafluororuthenate(IV), in Fig. 4.6(b) and (c). Sign in to download full-size image Fig. 4.6. Interrelation of solution, solvation, and lattice enthalpies (a) in general; (b) for sodium chloride; (c) for potassium hexafluororuthenate(IV). All enthalpies are in kJ mol−1. By this approach, an extensive set of sums of ion hydration enthalpies can be built up. Such a set of values can be shown to be self-consistent, but no amount of arithmetical manipulation allows the extraction of hydration enthalpies for individual ions. To obtain these single ion values one has to introduce an extrathermodynamic assumption. The assumption of equal values for K+ and Cl− has often been used in various contexts over many decades. In view of the significant differences in ionic radii (Table 4.6 suggests K+ = F− or Cs+ = Cl− might be better) and in the geometry of hydration (Fig. 4.7), this seems fairly unattractive here. The currently popular single ion assumption of equality of AsP h 4+ or PP h 4+ and BP h 4− is better in view of the large size and equality of radii, though Fig. 4.7 still applies. Also the very low solubility of [Ph 4 As][BPh 4] and of its phosphorus analogue precludes direct calorimetry, and lattice enthalpy calculations are complicated by the polyatomic nature of these ions (see above). In practice the best approach has proved to be that of obtaining a good estimate for the hydration enthalpy of the proton†. In essence it involves the extrapolation of enthalpy data for a series of compounds HX to the limit when X− is infinitely large and negligibly solvated. The hydration enthalpy of H+ can then be taken as the hydration enthalpy of HX. Once the value for Δ H hydr(H+) has been fixed (−1091 kJ mol−1) then values for, e.g., Δ H hydr(X−) for X = Cl, Br, I, can be obtained from enthalpies of solution of HX and ancillary thermodynamic data, then Δ H hydr(M n+) from enthalpies of solution of halides MX n, and so on. Table 4.6. Ionic radii (in Å; Shannon and Prewitt values for six coordination) Na+1.02 F−1.33 K+1.38 Cl−1.81 Rb+1.49 Br−1.96 Cs+1.70 I−2.20 Sign in to download full-size image Fig. 4.7. Geometrical relations between a solvating water molecule and a cation and an anion. Enthalpies of hydration for a range of metal ions and for a selection of anions are listed in Table 4.7. There is a general overall correlation with ionic charges and radii but with a number of significant deviations from such simple electrostatic control. The dependence of hydration enthalpy on ionic radius, at constant charge, can readily be seen in Table 4.7 by looking down the members of a Periodic Table Group such as Li+ → Cs+ or Be 2+ → Ra 2+. The dependence on charge is qualitatively obvious on looking across Table 4.7, but is better demonstrated in the manner shown in Table 4.8. In this latter table an attempt has been made to show the effect of changing the charge but keeping the ionic radius approximately constant. This is more satisfactory than simply noting the trend across the Periodic Table, for a sequence such as K+ → Ca 2+ → Sc 3+ involves a marked decrease in ionic radius as the charge increases. Table 4.7 shows that hydration enthalplies of anions follow the same pattern as those for anions. The different modes of interaction (Fig. 4.7) result in the hydration enthalpies of anions and cations of similar radius being significantly different (see, e.g., K+ and F− or Cs+ and Cl− in Table 4.7; radii in Table 4.6). Table 4.7. Hydration enthalpies (kJ mol−1) for cations and anions Li+−515 Be 2+−2487 Na+−405 Mg 2+−1922 Al 3+−4660 K+−321 Ca 2+−1592 Sc 3+−3960 Rb+−296 Sr 2+−1445 Y 3+−3620 Cs+−263 Ba 2+−1304 La 3+−3283 Ce 4+−6490 Th 4+−4220 F−−503 CN−−365 Cl−−369 NCS−−328 Br−−336 N O 3−−328 I−−298 Cl O 3−−307 Cl O 4−−244 S O 4 2−−1145 Table 4.8. Dependence of cation hydration enthalpy on charge at constant ionic radius \| Cation \| Radius (Å) \| Δ H hydr (kJ mol−1) \| --- \| Na+ \| 1.16 \| −405 \| \| Ca 2+ \| 1.14 \| −1592 \| \| Nd 3+ \| 1.14 \| −3440 \| \| Pu 3+ \| 1.14 \| −3440 \| Now it is time to consider deviations from the simple electrostatic pattern. Table 4.9 compares hydration enthalpies for some pairs of cations of identical charge and approximately equal radii. Values for B-Group and transition metal cations are markedly more negative than for A-Group ions of similar radii. Such differences are often rationalised, as indicated in Table 4.9, in terms of the Hard and Soft Acids and Bases, HSAB, approach (see section 6.2), with the extra hydration enthalpy ascribed to polarisation or covalent interaction contributions. For transition metal cations, the variation of crystal field stabilisation energies with d-electron configuration is reflected in ion hydration enthalpies (Table 4.10). For anions such as F−, there is a possibility of increased ion–solvent interaction through hydrogen-bonding in protic solvents such as water. Table 4.9. Cation hydration enthalpies for equivalent ‘hard’ and ‘soft’ cations \| ‘Hard’ \| ‘Soft’ \| --- \| \| Cation \| Radius (Å) \| Δ H hydr (kJ mol−1) \| Cation \| Radius (Å) \| Δ H hydr (kJ mol−1) \| \| K+ \| 1.33 \| −321 \| Ag+ \| 1.29 \| −475 \| \| Mg 2+ \| 0.65 \| −1922 \| Cu 2+ \| 0.69 \| −2100 \| \| Ca 2+ \| 0.99 \| −1592 \| Cd 2+ \| 0.97 \| −1806 \| \| Sr 2+ \| 1.13 \| −1445 \| Hg 2+ \| 1.10 \| −1823 \| Table 4.10. The reflection of Crystal Field stabilisation on ion hydration enthalpies for the first row d-block 2+ cations Cation Cr 2+d 4 Mn 2+d 5 Fe 2+d 6 Co 2+d 7 Ni 2+d 8 Cu 2+d 9 Zn 2+d 10 CFSE 6 Dq 0 4 Dq 8 Dq 12 Dq 6 Dq 0 Δ H hydr(kJ mol−1)−1850−1845−1920−2054−2106−2100−2044 Table 4.11 shows how ion solvation enthalpies depend on the nature of the solvent. Such information is often presented in the form of enthalpies of transfer; the enthalpy of transfer is simply the difference between the enthalpies of solvation of an ion in two solvents. Generally one of these solvents is water, to provide a common basis for comparisons. Formally the process of transfer involves taking the ion out of one solvent into the gas phase, then placing it in the second solvent (Fig. 4.8). Table 4.12 includes a selection of ion transfer enthalpies, in all cases from water. Tables 4.11 and 4.12 show that solvation enthalpies for a cation such as K+ or Ba 2+ tend to be of the same order of magnitude, though strong donor solvents such as DMSO or HMPA do give markedly more favourable solvation enthalpies. However, for Ag+ solvation by DMSO, MeCN, and ammonia is particularly favourable. Water solvates chloride effectively, while non-aqueous solvents generally are less effective;this situation contrasts with the negative enthalpies of transfer for metal cations from water into non-aqueous media shown in Table 4.12. Table 4.11. Ion solvation enthalpies (kJ mol−l) \| Empty Cell \| Water \| liq. NH 3 \| Methanol \| Acetonitrile \| Dimethyl sulphoxide \| HMPA \| --- --- --- \| \| Li+ \| −515 \| −556 \| −531 \| \| \| \| \| K+ \| −321 \| −351 \| −351 \| −347 \| −368 \| −385 \| \| Cs+ \| −263 \| −293 \| \| \| \| \| \| Ag+ \| −475 \| −577 \| −510 \| −536 \| −544 \| \| \| Ba 2+ \| −1304 \| −1405 \| −1389 \| −1326 \| −1406 \| −1460 \| \| Cl− \| −369 \| \| −361 \| −349 \| −350 \| \| Sign in to download full-size image Fig. 4.8. Enthalpy of transfer of Li+ from water to liquid ammonia, and its relation to enthalpies of solvation. Table 4.12. Enthalpies of transfer (kJ mol−1) of ions from water into various non-aqueous solvents \| Empty Cell \| Liq.NH 3 \| Methanol \| Acetonitrile \| Dimethyl sulphoxide \| HMPA \| --- --- \| K+ \| −30 \| −30 \| −26 \| −47 \| −64 \| \| Ag+ \| −102 \| −35 \| −61 \| −69 \| \| \| Ba 2+ \| −101 \| −85 \| −22 \| −102 \| −156 \| \| Cl− \| \| +8 \| +20 \| +19 \| \| The strengths of ion-solvent interactions are important in determining solubilities of salts, relative strengths of these interactions in determining differences in solubility of a given salt in various solvents. Thus, for example, potassium chloride is much less soluble in alcohols and in dimethyl sulphoxide than in water. This can be ascribed to much less favourable solvation of chloride in these non-aqueous solvents—in the case of dimethyl sulphoxide this effect dominates over the somewhat more favourable solvation of K+ in dimethyl sulphoxide than in water. Potassium iodide and sodium nitrate have similar solubilities in water and in liquid ammonia, since significantly better solvation of K+ and of Na+ by the ammonia is offset by comparably poorer solvation of the I− and NO 3− by the ammonia. But silver iodide is very much more soluble—more than 10 8 times—in liquid ammonia than in water, because ammonia interacts much more strongly than water with the Ag+ cation. Ions such as As Ph 4+ and BPh 4– are fairly well solvated by many organic solvents, so [AsPh 4][BPh 4] is freely soluble in such solvents. But these large, singly charged, hydrophobic ions are so feebly solvated by water that [AsPh 4][BPh 4] is essentially insoluble in water despite its very small lattice energy. Although enthalpies of solvation provide perhaps the simplest indicator of the strength of ion-solvent interactions, data on other thermodynamic functions are available and informative. Particularly for water and other structured, polar, hydrogen-bonding solvents, the entropy change associated with hydration (solvation) of an ion is an important quantity. Table 4.13 includes a selection of hydration entropies for simple and complex ions, and shows the role played by charge and radius. Large ions of low charge, such as Cs+, I−, ClO 4−,NR 4+,or AuBr 4−, have large positive partial molal entropies. Their introduction into solvent water results in an increase in entropy or freedom, as they act as breakers of water structure. On the other hand, small ions, especially of charge 2± or above, have negative partial molal entropies. Such ions are heavily hydrated. Here the transfer of water molecules from bulk solvent to the ion solvation shell, with consequent increase in ordering and loss of freedom, results in a nett decrease in entropy. This electrostriction of solvent under the influence of the electric field of the ion also results in a volume decrease (Fig. 4.9). Table 4.14 shows that partial molal volumes for ions in aqueous solution show a similar pattern to partial molal entropies (Table 4.13).† Small and medium sized ions of charge 2± or more have negative partial molal hydration volumes dominated by electrostriction, but for large ions, particularly of charge only 1±, the sheer size of the ion dominates over the relatively small electrostriction contribution. Table 4.13. Standard partial molal entropies (JK−1 mol−1) for hydrated ions in aqueous solution (relative to zero for the proton) Li++11 Na++59 Mg 2+−138 Al 3+−322 K++101 Ca 2+−53 Sc 3+−255 Rb++120 Sr 2+−33 Y 3+−259 Cs++133 Ba 2++10 La 3+−218 Zn 2+−110 Ga 3+−331 Ag++73 Cd 2+−76 In 3+−264 TI++126 Hg 2+−36 Ni 2+−131 Fe 3+−300 Ac 3+−181 Th 4+−423 F−−10 OH−−11 SO 4 2−+17 Cl−+55 CN−−49 CrO 4 2−+50 Br−+80 NO 3−+125 Cr 2 O 7 2−+262 I−+109 ClO 4−+ 182 NH 4++97 Fe(CN)6 4−+95 NMe 4++210 NMe 4++270 NEt 4++283 AuBr 4−+336 Sign in to download full-size image Fig. 4.9. Illustration of the overall volume decrease when into a given volume of bulk solvent water (a) is introduced an ion of moderate (b) or large (c) charge. The darker hatching indicates electrostricted water molecules in the hydration shells of the ions. Table 4.14. Partial molal hydration volumes (cm 3 mol−1) for hydrated ions (relative to zero for the proton) Li+−0.9 Na+−1.2 Mg 2+−21.2 Al 3+−42.2 K++9.0 Ca 2+−17.9 Rb++14.1 Sr 2+−18.2 Cs++21.3 Ba 2+−12.5 La 3+−39.1 Zn 2+−21.6 Ag+−0.7 Cd 2+−20.0 T1++10.6 Hg 2+−19.3 Ni 2+−24.0 Fe 3+−44 Th 4+−54 F−−1.1 OH−−4.0 SO 4 2−+14.0 Cl−+ 17.8 NCS−+35.7 Br−+24.7 NO 3−+29.0 I−+36.2 ClO 4−+44.1 [Co(NH 3)6]3++73[Cr(ox)3]3−+122 [Co(NH 3)5 C1]2++94[Fe(CN)6]3−+121 [Fe(CN)6]4−+74 In view of the similarities described above, it is hardly surprising to find that partial molal hydration entropies and volumes correlate quite closely. This is shown, for the case of cations in aqueous solution, in Fig. 4.10. Semi-empirical correlations are now being developed, in which, for example, partial molal hydration volumes for ions can be expressed as a function of their charge, radius, and hydration number. The application of these ideas to hydrated lanthanide(III) cations lent support to the hypothesis of a change in hydration number for these ions somewhere around the middle of the series. Sign in to download full-size image Fig. 4.10. Relation between partial molar volumes (V¯⊖/cm 3 mol−1) and entropies (S¯⊖/JK−1 mol−1) for aqua-metal ions. Thus a fairly consistent picture of strengths of ion-solvent interactions can be built up from thermodynamic data which, on the whole, is compatible with the indications from ultraviolet-visible and infrared–Raman spectra (sections 4.1 and 4.2). In recent years, understanding of fundamental thermodynamic aspects of ion solvation has been furthered by the determination of enthalpies and entropies for successive additions of water molecules to ions in the gas phase. Enthalpies for the typical cases of Li+ and Cl− are set out in Table 4.15. In each case the stepwise enthalpies decrease in magnitude as the number of water molecules attached to the ion increases, but this decrease is more gradual for C1− (and for other anions) than for Li+ (and for other cations). By the time six water molecules have been attached to Li+, the overall enthalpy change is almost equal to the total enthalpy of hydration, i.e. the enthalpy of transfer of Li+ from the gas phase into aqueous solution (see the discussion above). For chloride, the attachment of four water molecules is accompanied by an enthalpy change only just over half of that for transfer of Cl− from the gas phase into aqueous solution. Table 4.15. Enthalpy changes associated with the addition of water molecules to the Li+ and Cl− in the gas phase \| Empty Cell \| Δ H(kJ mol−1) \| \| Empty Cell \| Li+ \| Cl− \| \| Empty Cell \| Stepwise \| Cumulative \| Stepwise \| Cumulative \| \| Addition of first \| −142 \| \| −55 \| \| \| \| second \| −108 \| −250 \| −53 \| −108 \| \| \| third \| −88 \| −338 \| −49 \| −157 \| \| \| fourth \| −67 \| −405 \| −46 \| −203 \| \| \| fifth \| −58 \| −463 \| \| \| \| \| sixth \| −50 \| −513 \| \| \| \| \| water molecule \| \| \| \| \| \| Cf. single ion hydration enthalpy: \| \| −515 \| \| −369 \| Show more View chapterExplore book Read full chapter URL: Book 1999, Ions in SolutionJohn Burgess Chapter Theoretical and Computational Inorganic Chemistry 2010, Advances in Inorganic ChemistryRobert J. Deeth IV Effects from d Electrons The d electrons of classical Werner-type coordination complexes have a significant impact on structure and reactivity. For example, the experimental hydration enthalpies of the divalent aqua ions (black diamonds, Fig. 2) deviate substantially from any monotonic variation on crossing the series from left to right (19). Sign in to download full-size image Fig. 2. Variation of experimental hydration enthalpies (solid diamonds) from d 0 Ca 2+ through to d 10 Zn 2+. The open circles represent the values once d electron stabilization energy effects are removed (after Johnson and Nelson (19)). The “double-hump” behavior depicted in Fig. 2 is usually rationalized in terms of the ligand field stabilization energy (LFSE). The LFSE is a function of the d configuration and the magnitude of the ligand field splitting, Δ oct (Fig. 3), which can be independently derived from spectroscopic data by fitting the d–d absorption bands of these nominally octahedral species (20). If the experimental hydration enthalpies are “corrected” using the spectroscopic Δ oct values, a remarkably smooth line is obtained (open circles in Fig. 2). Sign in to download full-size image Fig. 3. Octahedral d-orbital splitting diagram from which the one-electron contribution to the LFSE can be computed. Any computational treatment of TM systems must account for the LFSE. QM methods achieve this implicitly but d-electron effects must be explicitly added to MM (4). Some effects can be modeled within conventional MM. For example, low-spin d 8 complexes are planar by virtue of the LFSE (21,22), but a planar structure can also be enforced using a “normal” out-of-plane term (22). However, the simplest general model for describing d-orbital energies is ligand field theory (LFT) (23) which was itself derived from the earlier electrostaticcrystal field theory (CFT) (24) approach. Although the properties which can be computed are limited, LFT has provided for over half a century a reasonably useful, semi-quantitative picture of metal–ligand bonding in Werner-type coordination complexes (3,25–27). In the present context, the advantage of LFT is its computational efficiency. Therefore, we added LFT to MM to give the ligand field molecular mechanics (LFMM) method (28). Since MM is intrinsically bond centered, the most convenient way to express the LFSE is based on the angular overlap model of Schaeffer and Jorgensen (29). The total ligand field potential, V LF, is assumed to be the sum of contributions from the M-L bonds each modeled by AOM parameters describing separate σ and π interactions as illustrated in Fig. 4. Sign in to download full-size image Fig. 4. Definition of AOM e parameters in terms of local M–L bonding. The σ interaction is always destabilizing (e σ>0) while π bonding in the two mutually perpendicular directions to the M–L vector (the local z axis) is shown for a π donor (e π>0). This construction provides a more physically realistic description of the M-L bond which ultimately enables LFMM to mimic far more expensive quantum chemical methods. The AOM is superior to CFT since it retains all the appropriate symmetry behavior but allows us to focus on individual ligands. The d-orbital energies are thus, in general, a function of all the ligands although in high-symmetry cases such as O h, there may be a convenient separation into σ and π interactions. For example, in AOM terms, the octahedral splitting Δ oct is given by (5) and illustrated in Fig. 5 for the particular case of a π-donor ligand for which e π is positive. Sign in to download full-size image Fig. 5. Comparison of CFT barycenter (left) versus AOM barycenter (right). (5)Δ oct=3 e σ−4 e π LFMM does have a small computational overhead. A 5×5 matrix describing the ligand field potential, V LF, must be constructed and diagonalized to extract the d-orbital energies and hence the LFSE (30). The matrix elements depend on angular factors, F, which are functions of the coordinates of the N ligands, and the AOM radial e k parameters, where k=σ, πx and πy(6). (6)=∑l N∑k s y m m F i k l F k j l e k l In addition, LFMM also requires additional parameters over conventional MM. Inevitably, therefore, there is a trade off between relatively expensive QM calculations which can be started immediately versus very much faster LFMM calculations which may require an initial period for parameter development and optimization. Show more View chapterExplore book Read full chapter URL: Book series2010, Advances in Inorganic ChemistryRobert J. Deeth Chapter Polymer Electrolyte Materials for Electrochemical Energy Devices 2018, Reference Module in Chemistry, Molecular Sciences and Chemical EngineeringLorenz Gubler Interaction With Solvent and Solutes First, we look at the interaction of an IEM with pure solvent, in our case water. The dry membrane is hygroscopic and absorbs water exothermally, releasing the hydration enthalpy. On top of that, water is absorbed endothermally driven by osmosis, viz., the tendency of the water to dilute the ionic charges in the membrane. The equilibrium water content of a membrane is determined by a number of factors: the type and density of fixed ions, the type of counterion and solvent or solvent mixture if other than water. Evidently, the polymer itself significantly influences solvent uptake, its chemistry (hydrophilic/hydrophobic constituents, flexible/rigid chains) as well as crystallinity and degree of cross-linking. Fig. 7 shows the effect of relative humidity on the water uptake of a PEM. Sign in to download full-size image Fig. 7. Schematic water sorption isotherm for a PFSA membrane, indicating the difference in water content Δ λ liq-vap between saturated water vapor (100% r.h.) and liquid water, and illustrating the underlying polymer surface configuration. ϕ w is the volume fraction of water. Adapted from Kusoglu, A.; Weber, A. Z., New Insights Into Perfluorinated Sulfonic-Acid Ionomers. Chem. Rev.2017, 117(3), 987–1104, and Kreuer, K.-D., The Role of Internal Pressure for the Hydration and Transport Properties of Ionomers and Polyelectrolytes. Solid State Ionics2013, 252, 93–101. with permission. The water uptake of an IEM greatly increases with the ionic site density, to the point that the material dissolves if it is not crosslinked. The typical water content of a swollen IEM with practical mechanical properties is 30–50 vol.%. Regarding the influence of the type of counterion, generally the smaller the hydration sphere of the ion, the lower the water uptake of the ionomer. For example, for Nafion® 117 the water uptake decreases for the series H+, Li+, Na+, K+, Rb+, Cs+ from 22 to below 10 water molecules per sulfonic acid site.1 Obviously, water uptake leads to swelling of the material, that is, to a dimensional change, which may or may not be isotropic. This can lead to internal stress if the membrane is spatially confined in an electrochemical cell and cause mechanical problems. Water uptake furthermore causes an internal hydrostatic pressure. This swelling pressure is the balancing force between the opposing tendencies of dilution of the ionic charges in the membrane and the elasticity of the polymer matrix which opposes such dilution. The difference in hydrostatic pressure between the membrane and the solution is given by the osmotic equilibrium (cf. below). Ion-exchange material in equilibrium with pure water can develop swelling pressures on the order of 100 bar.2 Next, we consider the situation where the IEM is in contact with a solution containing a dissolved salt, acid or base. For simplicity, we assume that the membrane is a pure CEM or AEM. If the solution contains different counterions, say A and B, then those will, in general, partition differently into the membrane, hence the molar ratio of A to B in the membrane will be different from that in the solution. The selectivity of the ion-exchange material for counterion A over B is a function of the interaction with the exchange site. The distribution of counterions between the membrane and the adjacent electrolyte solution is determined by the counterion exchange equilibrium, expressed by the selectivity coefficient K B A (assuming A and B are monovalent ions): (1)X A X B=c A m c B m=K B A.c A s c B s where X A and X B represent the fraction of exchange sites occupied by the corresponding ion (X A+X B=1), c the concentration (molarity), and the superscripts “m” and “s” denote the concentration in the membrane and solution, respectively. Table 3 shows examples of cation selectivity coefficients in a Nafion® membrane. If the ion-exchange involves ions of different charge, for example, n⋅A+ ↔ B n+, then the stoichiometric factor of A, here n, is placed in the exponent of the concentration/molar fraction term: (c A)n, (X A)n. For non-ideal conditions, the concentration c is replaced by the activity a=c⋅γ, where γ is the activity coefficient. Table 3. Selectivity coefficients 1 for some metal ions M over protons in Nafion® 120 at 25°C \| Ion \| K H M \| \| \| Ion \| K H M \| --- --- \| Li+ \| 0.58 \| \| \| Mg 2+ \| 2.3 \| \| Na+ \| 1.2 \| \| \| Ca 2+ \| 3.6 \| \| K+ \| 4.0 \| \| \| Ba 2+ \| 5.6 \| \| Cs+ \| 9.1 \| \| \| Zn 2+ \| 1.0 \| \| Ag+ \| 1.1 \| \| \| \| \| Ionic strength: 0.01 M for monovalent ions, and 0.1 M for divalent ions. Counter-ion: chloride. When an IEM is in contact with a solution containing ionic species, an equilibrium state between the membrane and the electrolyte solution is established, the Donnan equilibrium, which asserts that the electrochemical potential of the mobile ionic species is the same in the two phases. This leads to the situation sketched in Fig. 8. The membrane phase not only contains the fixed ions and counterions, but also coions. Coions are mobile ions that bear the same charge as the fixed ions in the IEM. Sign in to download full-size image Fig. 8. Schematic of the Donnan equilibrium between an ion exchange membrane and the adjacent electrolyte solution. The partitioning of the mobile ions between the membrane and the surrounding electrolyte leads to a difference in the electric potential between the two phases, the Donnan potential E D: (2)E D=Δ φ=φ m−φ s=−RT z i F ln c i m c i s where c i is the concentration and z i the charge of ion i, and the superscripts “m” and “s” refer to the membrane and solution phase, respectively. In case of non-ideal solutions, again the concentration has to be replaced by activity. Eq. (2) applies individually to both cations and anions. For a CEM E D<0 and for an AEM E D>0. The Donnan equilibrium leads to the exclusion of coions from the IEM (Donnan exclusion). However, the exclusion is not complete, and the uptake of coions and, for charge neutrality reasons, additional counterions increases with increasing concentration of the salt in the external solution. Fig. 9 shows the measured concentration of Na+ and Cl− in a commercial AEM exposed to NaCl solutions of different concentration. At low salt concentration the concentration of Cl−, the counterion, in the membrane corresponds to that of the fixed ion. As the solution concentration increases, the concentration of Na+, the coion, in the membrane increases, and the Donnan potential decreases correspondingly. At around 0.2–0.3 M NaCl, the electrolyte uptake of the membrane becomes significant and Donnan exclusion is said to break down. This happens approximately when the Donnan potential is lower than the thermal energy of the ions atroom temperature (kT≈25 meV) or when the concentration of salt in the solution approaches the concentration of fixed ions in the membrane. Therefore, coion exclusion is more effective for a membrane with higher fixed ion concentration. Sign in to download full-size image Fig. 9. Concentration of counterions (Cl−) and coions (Na+) in mol per volume of sorbed water in an AR103 anion exchange membrane (fixed ion concentration: 3.58 mol/L) as a function of the external NaCl concentration.3 The associated Donnan potential is calculated using Eq. (2). The partitioning of electrolyte between the membrane and solution phase entails another effect: osmotic pressure Δ π. It is caused by the difference in the activity of water a w between the two phases: (3)Δ π=p m−p s=RT V w ln a w m a w s where p is the pressure in the respective phase and V w is the partial molar volume of water (18 cm 3/mol for pure water). The uptake of water by an IEM when it is immersed in pure water is largely an osmotic effect (cf. above), since the water has the tendency to dilute the fixed charges and counterions in the membrane. However, the water content of the membrane decreases upon increase of the ionic strength in the solution (Table 4). Hence, the membrane dehydrates when in contact with a highly concentrated electrolyte, which affects transport properties and conductivity (cf. below). If osmotic effects are significant, it affects the Donnan potential and an additional term –V i/(z i⋅F)⋅Δ π has to be added to Eq. (2), where V i is the partial molar volume of the ion i. (e.g., 23.5 cm 3/mol for Cl−4). The effect is small, though, with a contribution to E D in the range of a few mV for Δ π=1000 bar. Table 4. Water and sulfuric acid uptake by Nafion® 117 in aqueous sulfuric acid solutions of different concentration, and in-plane conductivity 5 \| c(H 2 SO 4) (mol/L) \| Water uptake (m-%) \| Acid uptake (m-%) \| Conductivity (mS cm−1) \| --- --- \| \| 0 \| 35.2 \| 0 \| 91 \| \| 0.5 \| 31.6 \| 0.4 \| 95 \| \| 1.9 \| 25.1 \| 2.4 \| 112 \| \| 3.5 \| 21.3 \| 4.3 \| 99 \| \| 4.8 \| 16.9 \| 5.7 \| 73 \| \| 6.6 \| 13.4 \| 6.4 \| 50 \| \| 7.9 \| 10.2 \| 6.2 \| 29 \| \| 10.0 \| 6.7 \| 5.9 \| 11 \| Show more View chapterExplore book Read full chapter URL: Reference work2018, Reference Module in Chemistry, Molecular Sciences and Chemical EngineeringLorenz Gubler Chapter Barrier properties, antimicrobial and antifungal activities of chitin and chitosan-based IPNs, gels, blends, composites, and nanocomposites 2020, Handbook of Chitin and ChitosanKhalina Binti Abdan, ... Lee Ching Hao 6.5.1.2.2 Electrospinning technology The electrospinnability of chitosan/poly(ethylene oxide) (PEO) antimicrobial electrospun membrane increased. The hydrophobic–hydrophilic segments of the polymer form a typical surfactant micelle due to its high hydration enthalpy. Besides, insertion of micelles into a nanofiber was found to affect the modulation of the molecular structure as well as the spinnability of the solution, thus influencing fiber morphology . Another antibacterial electrospun nanofiber was studied using chitosan with liposomes immobilized releasing gentamicin. Gentamicin in the form of cream is used clinically in the treatment of infected skin cysts, ulcers, burns, infected insect bites and stings, infected lacerations and wounds. In that study, the biocompatible chitosan served as a nanofiber mesh to entrap antibiotic-loaded liposome immobilized at the polymer surface, preventing drug degradation and promoting antibacterial activity. Regardless of the ability for the electrospinning technique to prepare films with porous structure, the control of stable electrospinnability remains a challenge. The addition of 20 wt.% of nanochitin into chitosan-based composite formed cluster structures, facilitating the formation of nanofibers in the electric field and significantly reducing the amount of defects . View chapterExplore book Read full chapter URL: Book 2020, Handbook of Chitin and ChitosanKhalina Binti Abdan, ... Lee Ching Hao Chapter Introduction to Ligand Field Theory 2013, Practical Approaches to Biological Inorganic ChemistryFrank Neese Ligand Field Stabilization Energy If we disregard the subtleties that arise from interelectronic repulsion for a moment one can obtain some insight into the thermochemistry of transition metal complexes from CFT. Imagine taking a metal 2+ ion and moving it from the gas phase to aqueous solution where it will form an octahedral hexaquo transition metal complex [M(H 2 O)6]2+. Associated with the process is a net gain in energy resulting from bond formation. This is the hydration enthalpy. It is expected that the hydration enthalpies increase along the transition series due to the increasing effective nuclear charge of the transition metal ion as one moves from left to right across the periodic table. Increased effective positive charge will lead to tighter binding to the water molecules that will bind to the metal ion via the negative end of the dipole (the oxygen atom) that is associated with water. The increased effective nuclear charge arises from the fact that with each consecutive ion a proton and an electron is added. However, that additional electron in the d-shell of the transition metal ion will shield the additional positive charge of the nucleus only incompletely. Hence, the force that a given test charge in the vicinity of the transition metal nucleus experiences will increase towards the right of the transition series. This behaviour is indeed observed experimentally. However, superimposed over the trend to higher hydration enthalpies there is a peculiar ‘double bowl’ behaviour (Figure 2.6). Sign in to download full-size image FIGURE 2.6. Illustration of the ligand field stabilization energy (LSFE). The explanation for the double bowl behaviour is readily provided by CFT. The first step consists of subtracting a straight line from the observed hydration enthalpy curve in order to account for the effective nuclear charge. What remains is the double bowl behaviour. It arises from unequal occupation of the d-orbitals of the transition metal. If all d-orbitals were at the same energy a straight-line behaviour would be expected. However, relative to the centre of gravity of the d-orbital energies the t 2g orbitals are stabilised while the e g orbitals are destabilised. It is important to emphasise ‘relative to the centre of gravity’ because overall the d-orbitals are all strongly destabilised by the presence of a negatively or partially negatively charged ligand. Nevertheless, if the splitting between the d-orbitals is equated with 10Dq then the t 2g lie at −4Dq and the eg orbitals at +6Dq relative to the centre of gravity. Obviously, the first electrons to enter the d-shell will go into the t 2g. Thus, Sc 2+, Ti 2+ and V 2+ have the electronic ground state configurations (t 2g 1 e g 0), (t 2g 2 e g 0) and (t 2g 3 e g 0), respectively. Following the logic outlined above, they are associated with ligand field stabilization energies (LFSE) (relative to the centre of gravity) of −4Dq, −8Dq and −12Dq, respectively. Assuming that Hund’s rule holds, the next electron has to enter the e g which consequently leads to a reduction of the LFSE by 6Dq to +6Dq for Cr 2+. The next electron leads to a half filled d-shell at Mn 2+ in which case the LSFE vanishes. Indeed, the hydration enthalpy of Mn 2+ nicely falls onto the straight-line behaviour. The second half of the transition series mirrors the behaviour of the first half as the remaining electrons also enter the t 2g and e g shells in the same order. Hence, there is maximal LSFE for Ni 2+ and no LSFE for Zn 2+. Show more View chapterExplore book Read full chapter URL: Book 2013, Practical Approaches to Biological Inorganic ChemistryFrank Neese Chapter Nanoporous Materials from Mineral and Organic Templates 2006, NanomaterialsKiyoshi Okadaz, Kenneth J.D. MacKenzie 10.3.1 Solid acidity and applications for catalysts Since catalytic reactions proceed on the catalyst surfaces and supports, all the porous properties, including the specific surface area, pore size, and pore volume are important parameters. Pore size is especially important. In addition to these geometrical factors, the solid acidity of nanoporous materials is also an important factor for catalyst applications. Solid acidity can be generated by several mechanisms, giving the well-known Brønsted acid and Lewis acid sites . One of the mechanisms for generating solid acidity is by chemical adsorption of H+ in sites which are locally deficient in formal charge, by substitution of lower valence cations. In nanoporous materials, solid acidity is mostly introduced by substitution of Al 3+ for Si 4+ in the tetrahedral sites. Relationships between the amount of solid acidity and the pore size of various nanoporous materials are summarized in Fig. 10.16. A general trend in the solid acidity of these nanoporous materials follows the order: acid-clays < pillared-acid-clay ≤ PCHs< K-10 < MPSs < Al–Si gel <zeolites. By contrast, the pore size of these nanoporous materials increases in the following order: zeolites < PCHs < MPSs ≤ Al-Si gel ≤ K-10 ≤ pillared-acid-clay ≤ acid-clays. Thus, the trend is towards increasing solid acidity with decreasing pore size. The solid acidity in nanoporous materials is also known to change with cation exchange treatment. In the case of smectites, solid acidity is generated by the H+ formed by polarization of hydration water associated with the exchangeable cations in the interlayers . The strength of the solid acidity should therefore correlate with the hydration enthalpy of those cations. Onaka et al. reported a good correlation between the hydration enthalpy and completion time of silylation of 1-decanol with alkyltrimethylsilane, i.e., higher activity (stronger acid) results in shorter times. This is seen for a series of cation-exchanged montmorillonites which follow the order: Ni- ≤ Zn- < Cu- < H- < Al- < K-10- < Fe- < Sn-montmorillonites. Nanoporous PCHs and PLCs formed by using clay minerals as host layer compounds should show similar trends when cation exchanged. Such treatment of PCHs and PLCs should result in new types of catalysts. Sign in to download full-size image Fig. 10.16. Relationships between solid acidity and pore size of various nanoporous materials: zeolites [74,75], MPSs [76,77], PCHs , Al–Si gel , K-10 , pillared-acid-clay , and acid-clays . Since the solid acidity in nanoporous materials is generated by substitution of Al for tetrahedral SiO 4, the solid acidity is correlated with the Al/Si ratio. The relationship between the Al/Si ratio and the solid acidity of MPSs has been studied [84,85], giving data which are more similar to SiO 2–Al 2 O 3 gels than to zeolites. This may reflect a similarity in the amorphous structures of MPSs and gels. Thus, it may be more difficult to generate solid acidity in MPSs than in zeolites. The advantages of MPSs and PCHs compared with other nanoporous materials are their pore sizes, which are larger than in zeolites and more uniform in their arrangement than in gels and acid-clays, which should show a similar width of pore size distribution. These advantages of MPSs are illustrated by the synthesis of tetraarylporphyrin (TTP) from aliphatic aldehydes and pyrrole, as reported by Shinoda et al. . Since this aromatic compound is relatively large (1.8 nm), the pore size of the catalyst is an important condition for the reaction. These researchers examined the conversion yield of TTP using three FSMs with pore sizes of 2.0, 2.8, and 3.4 nm and found that the FSM with the pore size of 2.8 nm performed best. This result strongly suggests an optimum pore size for this catalytic reaction. Commercially available K-10, prepared by sulfuric acid treatment of montmorillonite, is also an effective catalyst, showing conversion yields similar to FSM, probably due to the similarity of its pore size (2.7 nm) to that of FSM (2.8 nm). However, the pores in K-10 are formed by a card-house-like aggregation of silica-rich gel particles which is considered to be less stable than the pore structure in FSM. Therefore, the conversion yield of TTP by K-10 decreased to 0% after only two cycles of use while FSM maintained its conversion yield after repeated use. It is clear that MPSs and PCHs have a good potential as catalysts for reactions of relatively large molecules such as aromatic compounds. Many of the practical applications of MPSs as catalysts have been reviewed in detail [8,47,48]. The aforementioned PCHs and MPSs have the advantage that the surfaces of their nanopores can be tailored by manipulating the surface Si-OH groups. Many studies have been reported on organic-inorganic hybrid catalysts using MPSs . Another breakthrough for MPSs was the success in crystallizing the wall matrix by introducing phenylene groups into the silica by the use of organic-inorganic compounds . The nanostructures thus synthesized were of three different types, hexagonal with 3.8 nm pores, cubic with three-dimensionally aligned mesopores, and hexagonal with both larger mesopores (6–12.4 nm) and micropores. Although these MPSs have poor thermal stability because of the organic groups included in their structure, they have many advantages of structural regularity and variety of pore structure, compared with purely inorganic MPSs. For example, these materials provide good support for proteins and enzymes in biotechnological applications. Show more View chapterExplore book Read full chapter URL: Book 2006, NanomaterialsKiyoshi Okadaz, Kenneth J.D. MacKenzie Chapter Metal-organic frameworks for recognition and sequestration of toxic anionic pollutants 2019, Metal-Organic Frameworks (MOFs) for Environmental ApplicationsAamod V. Desai, ... Sujit K. Ghosh 4.1.1 Classification of anionic pollutants Anionic pollutants can be classified on the basis of charge, size, toxicity, mineral/organic-based anions, hydration energies, HSAB principle, and geometry. Such a classification is shown in Table 4.1. Anions can be classified on the basis of charge, that is, monoanionic, dianionic, trianionic in nature. A few examples of monoanionic pollutant anions include CN−,Cl−,MnO 4−; diatomic anions include CrO 4 2−,HAsO 4 2−,SeO 4 2−; trianionic anions include PO 4 3−,AsO 4 3−. Another convention based on the different geometry of the anions can be widely classified as linear, trigonal, tetrahedral, pentagonal, octahedral, pentagonal, bipyramidal, etc. . The geometry of anions often seldom plays a vital role in the exchange, wherein the nontoxic surrogate anion with similar geometry leads to a facile exchange phenomenon, thus validating the importance of geometry. Owing to the varied size of the pollutant anions and their charge, these anions pose varied range of hydration energies. The hydration enthalpies of pollutant anions are a crucial factor in designing materials. Anions are also classified on the basis of the hard–soft acid–base principle . An anion with a high charge/size ratio is considered as hard, while anions with a low charge/size ratio are considered as soft in nature. Anions that are hard in nature, that is, having a high negative charge can attract a positive charge. Thus it becomes facile to capture harder anions selectively in the presence of the competing softer anion counterpart. The Hofmeister effect is a crucial phenomenon linked with the observed anion exchange behavior . The Hofmeister effect describes in general the order of anion exchange based on hydration energy, solvation energy, etc. In a liquid/liquid separation medium, depending on the solvent medium, it can be summarized that in solvents/sorbents having no potential hydrogen bond donor sites the feasibility of the anion exchange is directly dependent on the size/charge ratio. The hydrophobicity of the anion also determines the feasibility of any anion exchange process. Size/charge ratio is inversely proportional to the facile behavior of the anion exchange process. The equation that is involved in the overall process can be summarized as follows: Δ G p°=Δ G S°–Δ G h° wherein Δ G p is the Gibbs energy of partition; Δ G S is Gibbs energy of solvation; and Δ G h is Gibbs energy of hydration. These are very important parameters to consider the thermodynamic parameters like hydration energies of the targeted pollutant anions in order to design materials for the capture of the anions. Hydration energies of anions with a significant Z 2/r ratio pose a high enthalpy of hydration, thus sorbate materials having multiple potential hydrogen donor sites should overcome the expense of the dehydration of the sphere of hydration. View chapterExplore book Read full chapter URL: Book 2019, Metal-Organic Frameworks (MOFs) for Environmental ApplicationsAamod V. Desai, ... Sujit K. Ghosh Chapter Bioinorganic Chemistry and Homogeneous Biomimetic Inorganic Catalysis 2023, Comprehensive Inorganic Chemistry III (Third Edition)Besim Fazliji, ... Roland K.O. Sigel 2.20.4.1 Solvation content of metal ions Mg 2+ is rather small with an ionic radius of 0.72 Å,23 has a strict octahedral coordination sphere and a strong preference for oxygen ligands. In aqueous solution, this ion is thus present in its hexaaqua form with a first p K a of the coordinated water molecules of p K a=11.44±0.1 (Ref.26) and a ligand exchange rate of 6.7×10 5 s−1.82,83 While other alkaline earth metal ions used with RNA are larger and have faster ligand exchange rates, d-elements and lanthanoids employed in numerous in vitro experiments differ in many physico-chemical properties such as size, electronic properties, preference of N/O ligands and more. A comprehensive summary of the physico-chemical properties of such metal ions associated with RNA in some way is provided in Refs 15 and 16. Due to the relatively high hydration enthalpy ΔH Hydr=1858 kJmol−1.84 which needs to be overcome when coordinating to another ligand, Mg 2+ is mostly not fully dehydrated in the context of nucleic acids. Binding of Mg 2+ to nucleic acids is thus mostly a combination of innersphere (i.e., direct) and outersphere (i.e., mediated by H 2 O) coordination (Fig.6). In addition, providing the size and shape of the binding pockets allow an equilibrium between different liganding sites may also exist. Such rather dynamic systems hamper the investigation and characterization of Mg 2+binding sites in larger nucleic acid structures and consequently many unanswered questions regarding the metal ion coordination to RNA (and DNA) still exist. Sign in to download full-size image Fig.6. A Mg 2+ bound via two innersphere interactions to N7 sites of two guanines, bridging two strands, as well as displaying numerous outersphere interactions via their coordinated water molecules Mg 2+ N° 11 in the large ribosomal subunit of Haloarcula marismortui.76 This figure has been prepared with MOLMOL 168 and is adapted from Ref.85 An evaluation of the 88 identified K+/Na+ and 116 Mg 2+ binding sites within the large ribosomal subunit of Haloarcula marismortui reveals some interesting facts: (i)assuming these metal ions are the specifically bound ones, they only account to roughly 11% of the charge compensation needed by the phosphate-sugar backbone; (ii) nine ions are fully hydrated, one is completely dehydrated, but all the others show a mixture between inner- and outersphere coordination; (iii) 106 Mg 2+ are coordinated to a phosphate oxygen, 82 of which by one, and 43 by two or more innersphere contacts.85 These numbers reflect well the importance of oxygen ligands on the one hand and on the other also the crucial involvement of water-mediated contacts to place a Mg 2+ correctly. Two databases provide more information on the importance and placement of water molecules within nucleic acid structures, both in the presence and absence of metal ions. The Solvation Web Service (SwS)86 provides a statistical survey of water molecules of the first solvation sphere around nucleobase pairs, as identified from nucleic acid structures deposited in the NDB. The just established Metal Ions in Nucleic AcidS (MINAS) database 87 compiles the first and second shell coordination sphere of all metal ions in the context of nucleic acid structures deposited in the PDB. A multitude of search options allows to gain insights into the coordination spheres of metal ions and the involvement of water molecules in their binding to nucleic acids. Show more View chapterExplore book Read full chapter URL: Reference work 2023, Comprehensive Inorganic Chemistry III (Third Edition)Besim Fazliji, ... Roland K.O. Sigel Chapter Ligand Field Theory 2017, Electrons, Atoms, and Molecules in Inorganic ChemistryJoseph J. Stephanos, Anthony W. Addison 8.1 The Advantages and Disadvantages of Crystal Field Theory What are the advantages of crystal field theory, and why was there a need for another theory? • Crystal field theory was useful tool to describe the following in coordinated compounds: ○ optical spectra of transition metals ions; ○ the number of d-d transition and their relative energies; ○ some magnetic properties; ○ hydration enthalpies; and ○ spinel structures. The success arose because crystal field theory is based firmly on molecular symmetry. • However, the theory only considers d orbitals and falls to: ○ provide an adequate description of chemical bonding between neutral metal atoms and neutral or cationic ligands; ○ predict the absolute magnitude of transition energy; ○ explain the consequence of the ligand binding on the chemistry of the electrons of the central ion; ○ explain the quantitative calculation of Dq, which is differ considerably from that found experimentally; and ○ explain the charge transfer, which is between metal and ligand and more powerful than d-d transfer. Why do the quantitative calculations of Dqs differ considerably from that found experimentally? • In the point charge approximation, the radial parameter is defined as: D q=z e 2 r 4¯6 a 5=z e 2 6 a 5∫0∞R nl⁎r 4 R nl r 2∂r where r is the distance from the origin at the center ion, a is the distance of the ligand from the origin, Chapter 7, p. 427. • Therefore, the discrepancy between the calculated and found Dq reveals the variation in the metal-ligand distance, r, indicating that the metal-ligand bond does have some covalency (i.e., orbital overlapping). • In the crystal field analysis, the covalency in the metal-ligand bond has been ignored. • Ligand field theory realizes the covalency in the bond and treats Dq and other parameters as empirical parameters, which could be estimated from the electronic spectrum. • When an orbital overlap between metal and ligand is considered, the metal orbitals can possibly form σ- or π-bonding. • The term crystal field theory is kept for the extreme in which there is no mixing of the central ion and the ligand electrons, while the term ligand field theory for all non-zero degrees of mixing. • The formulation in the entire details is the same. What is the difficulty in determining the energies of the central ion terms and how can effectively approach this difficulty? • Ligand field theory admits the covalency between the metal and its neighbors. Therefore, it is not simple to write down expressions for the Hamiltonian to compute the energies, since the orbitals involved are overlapped. • On the other hand, the symmetry and group theory approach can straightforwardly be used to decide the states that result when an ion of any given electronic configuration is located into a surrounding environment of definite symmetry. • Thus, in order to gain an understanding of this theory, symmetry considerations are essential approach. • The ligand field theory cannot evaluate the absolute energies; however, experimental data can be fitted to the theory to compute semi-empirical radial parameters that manifest the interaction between ligands and metal. • These interelectronic repulsion parameters (Racah B and C) are included in a well-defined manner to provide a precise fitting of the electronic spectrum. Show more View chapterExplore book Read full chapter URL: Book 2017, Electrons, Atoms, and Molecules in Inorganic ChemistryJoseph J. Stephanos, Anthony W. Addison Chapter Metal Ion Extraction With Ionic Liquids 2020, Liquid-Phase ExtractionMark L. Dietz, Cory A. Hawkins 18.3.2 Anion exchange Formation of cationic metal ion complexes in the IL phase, driven by cation exchange of IL+, is well established as a general phenomenon. Therefore, it does not come as a surprise that certain conditions can promote the extraction of anionic complexes via anion exchange. A general equilibrium for the formation of an anionic metal complex in the IL in 1:1 exchange for an IL anion is written in Eq. (18.4), where L is an anionic extractant: (18.4)M n++n+1 L I L m−+C+A−I L⇌C+M L n+1 m−I L+A− The tendency of anions to exchange in an aqueous/organic biphasic system typically favors partitioning of the most hydrophobic anions (i.e., lowest hydration enthalpy) toward the IL phase. A general prediction for the tendency to promote anion exchange in both IL synthesis and metal ion extraction follows a Hofmeister series: SO 4 2−<Cl−<Br−<NO 3−<I−<ClO 4−<SCN−<BF 4−<PF 6−<Tf 2 N−[87, 107]. Anion exchange for metathesis of one IL to another anionic form is, therefore, influenced largely by hydration and mass action. However, for metal ion extraction, this propensity is also influenced by the free energy of anionic metal ion complex formation and solvation. This case was clearly made by Jensen et al. in the extraction of trivalent lanthanides (Ln(III)) by the fluorinated acidic extractant 2-thenoyltrifluoroacetone (Htta) into [C 4 C 1 im][Tf 2 N], where the ion-pair complex [C 4 C 1 im][Ln(tta)4], exchanged for the relatively hydrophilic Tf 2 N−, is supported by a fourth-power pH dependence and complete dehydration of the inner coordination sphere. Although this process incurs the loss of IL anion to the aqueous phase, because of the steep acid dependence, the Ln(III) and the IL are, in theory, both recoverable by treatment of the postextraction IL phase with the conjugate acid HTf 2 N. Interestingly, by changing the IL anion to nonafluoro-1-butanesulfonate, the extracted Ln(III) complexes are 1:2 and 1:3 with increasing Htta, and they are completely hydrated at low Htta concentration. In the hydrated and 1:2 complex, the dominant partitioning mechanism tends to be IX-1 . In contrast to most molecular solvents, an IL can solubilize significant quantities of acid, which will affect the extent of ion exchange. For example, a 1:1 volume contact of pure C 4 C 1 imTf 2 N with an initial aqueous HNO 3 concentration of 7.4 M was found to dissolve 1.6 M HNO 3 in the IL phase . Billard et al. found that the U-shaped curve in the extraction of U(VI) into this IL containing TBP extractant with varying nitric acid concentration could be explained by a threefold mechanistic model. The model includes cation exchange of [UO 2(TBP)2]2+ for H+ and C 4 C 1 im+, neutral extraction of UO 2(NO 3)2(TBP)2, and anion exchange of [UO 2(NO 3)3(TBP)2]−, as HNO 3 concentration increases. The role of aqueous acid and ionic species on the IL phase behavior was explored, and it was found that TBP does not extract the acid. Instead, H+ is dissolved in the IL, and another acid cation exchange mode (i.e., distinct from IX-2) was established, involving the transfer of a cationic complex for dissolved H+. This work emphasizes the importance of considering the role of all species in a system through not just partitioning experiments but also spectroscopic data and computational modeling, if necessary. Show more View chapterExplore book Read full chapter URL: Book 2020, Liquid-Phase ExtractionMark L. Dietz, Cory A. Hawkins Related terms: Solvation Montmorillonite Hydrogen Bonding Lithium Cation Ionic Liquid Gibbs Free Energy Crystal Energy Metal Ion Stabilization Energy View all Topics Recommended publications Solid State IonicsJournal Journal of Molecular LiquidsJournal Geochimica et Cosmochimica ActaJournal The Journal of Chemical ThermodynamicsJournal Browse books and journals From other publishers ChiralityJournal Natural Product ReportsJournal Journal of Heterocyclic ChemistryJournal Organic Process Research and DevelopmentJournal Browse books and journals Featured Authors Maron, L.INSA Toulouse, Toulouse, France Citations21,837 h-index75 Publications103 Norby, Truls E.Universitetet i Oslo, Oslo, Norway Citations13,638 h-index55 Publications145 Haugsrud, ReidarUniversitetet i Oslo, Oslo, Norway Citations5,342 h-index38 Publications79 Roux de Balmann, HélèneInstitut National Polytechnique de Toulouse, Toulouse, France Citations1,605 h-index22 Publications37 About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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3314	https://www.tiger-algebra.com/drill/r3-r2-r_1=0/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Other Factorizations Other Ways to Solve Step by Step Solution Reformatting the input : Changes made to your input should not affect the solution: (1): "r2" was replaced by "r^2". 1 more similar replacement(s). Step by step solution : Step 1 : Checking for a perfect cube : 1.1 r3-r2-r+1 is not a perfect cube Trying to factor by pulling out : 1.2 Factoring: r3-r2-r+1 Thoughtfully split the expression at hand into groups, each group having two terms : Group 1: -r+1 Group 2: r3-r2 Pull out from each group separately : Group 1: (-r+1) • (1) = (r-1) • (-1) Group 2: (r-1) • (r2) Add up the two groups : (r-1) • (r2-1) Which is the desired factorization Trying to factor as a Difference of Squares : 1.3 Factoring: r2-1 Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B) Proof : (A+B) • (A-B) = A2 - AB + BA - B2 = A2 - AB + AB - B2 = A2 - B2 Note : AB = BA is the commutative property of multiplication. Note : - AB + AB equals zero and is therefore eliminated from the expression. Check : 1 is the square of 1 Check : r2 is the square of r1 Factorization is : (r + 1) • (r - 1) Multiplying Exponential Expressions : 1.4 Multiply (r - 1) by (r - 1) The rule says : To multiply exponential expressions which have the same base, add up their exponents. In our case, the common base is (r-1) and the exponents are : 1 , as (r-1) is the same number as (r-1)1 and 1 , as (r-1) is the same number as (r-1)1 The product is therefore, (r-1)(1+1) = (r-1)2 Equation at the end of step 1 : Step 2 : Theory - Roots of a product : 2.1 A product of several terms equals zero. When a product of two or more terms equals zero, then at least one of the terms must be zero. We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well. Solving a Single Variable Equation : 2.2 Solve : r+1 = 0 Subtract 1 from both sides of the equation : r = -1 Solving a Single Variable Equation : 2.3 Solve : (r-1)2 = 0 (r-1) 2 represents, in effect, a product of 2 terms which is equal to zero For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means : r-1 = 0 Add 1 to both sides of the equation : r = 1 Two solutions were found : How did we do? Why learn this Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
3315	https://ssojet.com/character-encoding-decoding/ascii-in-kotlin/	ASCII in Kotlin \| Character Encoding/Decoding \| SSOJet skip to content × Product Single Sign-on Directory Sync Multi-Factor Auth Why SSOJet? Developer Overview Documentation API Reference Integrations OIDC Tester OIDC Playground SAML Tools SAML Tester JWT Validator SAML Glossary Resources Blog White Papers News CIAM 101 ROI Calculator Enterprise SSO Directory SOC 2 Assessment Tool IAM Terminology Providers and Protocols Enterprise SSO CIAM Vendors Pricing Solutions AI Companies YC Companies Startups Enterprise Ready Solutions Sign in Start 30-day free trial Sign inStart 30-day free trial Back to Character Encoding/Decoding Glossary ASCII in Kotlin Working with raw byte data or text files in Kotlin can be tricky, especially when you need to handle character encodings accurately. This guide dives into using ASCII characters within your Kotlin projects, covering how to represent, manipulate, and encode/decode ASCII strings. You'll learn practical techniques to ensure your character data is handled correctly, preventing common errors and improving the robustness of your applications. Representing ASCII Characters in Kotlin Kotlin, much like Java, handles characters using UTF-16 encoding internally. This means that standard ASCII characters are simply a subset of Unicode, represented by their corresponding Unicode code points. You can directly assign an ASCII character literal to a Char variable, or use its integer code point. For instance, to represent the character 'A': kotlin val charA: Char = 'A' // or val charAInt: Int = 65 val charAFromInt: Char = charAInt.toChar() A common misunderstanding is assuming that every Char in Kotlin is strictly an 8-bit ASCII value. In reality, Kotlin's Char type is designed to accommodate the full range of Unicode characters, which extends far beyond the 128 ASCII characters. While ASCII characters fit within this, be mindful that Char can hold values representing non-ASCII characters too. Always rely on the Char type's inherent Unicode capabilities rather than assuming byte-level ASCII representation. Converting Between ASCII Characters and Integers In Kotlin, you can easily convert between Char types and their corresponding integer ASCII (or Unicode) values. To get the numerical code point of a character, you simply access its .code property. Conversely, to convert an integer back into a character, use the .toChar() extension function. ```kotlin val charB: Char = 'B' val asciiValueB: Int = charB.code println(asciiValueB) // Output: 66 val asciiValC: Int = 67 val charC: Char = asciiValC.toChar() println(charC) // Output: C ``` A common pitfall is forgetting that .toChar() produces a Unicode character. If you provide an integer outside the standard ASCII range (0-127), you won't necessarily get a printable ASCII character. Always ensure your integer values fall within the expected ASCII bounds if that's your specific requirement. This direct conversion is handy for many text processing tasks. Working with ASCII Strings Kotlin treats strings as sequences of Char objects, making it straightforward to work with ASCII-based strings. You can iterate through a string, accessing each character and its underlying ASCII (or Unicode) value. This allows for direct manipulation and analysis of character data. For instance, to get the ASCII code for each character in a string: kotlin val asciiString = "HELLO" for (char in asciiString) { print("${char.code} ") // Output: 72 69 76 76 79 } println() A common pitfall is attempting arithmetic operations directly on a string that you intend to be an ASCII sequence, without first converting it to its Char or Int representation. Always explicitly convert characters or strings to their numerical code points when performing mathematical operations. This ensures predictable and correct results when dealing with character values. Encoding/Decoding ASCII Data with ByteArrays Kotlin's ByteArray provides straightforward methods for handling ASCII data. To explicitly encode a String into an ASCII ByteArray, use the toByteArray() method and specify Charsets.US_ASCII. This ensures each character is represented by its 7-bit ASCII value. ```kotlin val message = "Hello Kotlin!" val asciiBytes: ByteArray = message.toByteArray(Charsets.US_ASCII) // Output: [72, 101, 108, 108, 111, 32, 75, 111, 116, 108, 105, 110, 33] println("Encoded: ${asciiBytes.joinToString()}") val decodedMessage = String(asciiBytes, Charsets.US_ASCII) // Output: Hello Kotlin! println("Decoded: $decodedMessage") ``` A common pitfall is omitting the Charsets.US_ASCII parameter. 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3316	https://www.youtube.com/watch?v=tuIrKzOEHV8	Calculate Solubility from Ksp Old School Chemistry 22000 subscribers 143 likes Description 7803 views Posted: 10 Jan 2020 Calculate the solubility of a salt or ions given Ksp. 20 comments Transcript: hi we're going to talk about calculating solubility from KSP now there are a couple of ways that they could ask this question I think this is one of the challenges in juvy in KSP is that there are several different ways that they can all ask the same question so when if you read the term calculate solubility that can also be asked in three other forms it could be what are the equilibrium concentrations of ions you could also be asked what's the amount of salt that dissolves or you could be asked what are the saturated concentrations of the ions all those are really just hey calculate solubility so put this down in your notes study that and as you do your homework I'm sure that you're going to see some questions that are worded all four ways and put them on the same folder in your brain as these all mean I just need to calculate solubility how much dissolves at equilibrium what's the maximum amount that can dissolve a equilibrium and that tells me how soluble it is so here's an example problem for us it says calculate solubility which it could have said what are the equilibrium concentrations what amount of salt dissolves what are the saturated concentrations that all would be the same question kalki the solubility of calcium hydroxide in they want to units in moles per liter and grams per liter if the KSP is five point five times ten to the minus five now it's possible that they wouldn't give you the KSP they say calculate solubility and you'd have to go to a solubility table and look that up now remember when we're doing KSP there are two pieces of information there's the K value and there's the equilibrium what are the concentrations at equilibrium when they're asking for solubility they're asking for e the equilibrium concentrations which means if they're asking for e we've got to have K you have to have K so go to a table and look it up if it's not given to you alright first place I always begin is writing this equation and it's just the dissolution of assault so we were going to have calcium hydroxide and that's a solid always understood that the salts have solid and that you drop it in water and it will dissociate into a cation and anion the cation is calcium and that's a two plus charge being aqueous plus the anion is the hydroxide that's an O h minus and I look make sure this is balanced we've got two of the hydroxide so I have to put it to here when that calcium hydroxide disassociates it produces the one calcium ion and the two hydroxide ions let's go ahead and set up our ice table so we've got I see now solace we don't use we don't count pellets or liquids in our equilibrium expression because they're constant so those are dashes initially when we put this in pure water or may say no calcium and the hydroxide we're going to put a zero there that's going to be considered negligible I know that we have kW and that the that comes into equilibrium because of kW you still put zero it'll be negligible be really small compared to this now change so this is going to break apart and when this dissociates it will produce an amount of calcium and notice the coefficient 1 so 1 mole very one mole that dissociates produces 1 mole of calcium and 2 X 2 moles of the hydroxide remember you just look at the coefficient and that's the number that you put in front of X the unknown amount that ionizes that dissociates now ionizes sorry that dissociate okay I plus C gives us X 0 plus X is X 0 plus 2 X is 2x nice now let's go ahead and write the equilibrium expression so we're going to get a KSP equals products that's going to be your calcium ion raised to the first power or I could leave that blank understand to be to the first power times the concentration of hydroxide I'd be really careful what's that coefficient a two so that's going to be squared now we can plug in all of our numbers we are money tabs of five point five times ten to the minus five as the KSP that we looked up just from the a solubility table equals alright calcium is X the hydroxide is 2x and that squared so let's do our math here 2x times 2x is 4x squared times X is 4x cubed so five point five times ten to the minus five equals 4x cubed just be careful with your algebra double-check yourself every time you do this let's go ahead and solve for X so if I divide both sides by the 4 I'm going to get one point three seven five times 10 to the minus five equals x cubed okay how do I get rid of a cube we cube root so let's take the cube root of both sides and when we put that in our calculator we're going to get x equals point zero two four zero nice nice so we found X let me come back here and write this down x equals point zero two four zero now what's the unit on this what are the units that we use for equilibrium that brackets a code remember it's molarity so we actually found the molarity this is the molarity as a calcium ion that dissociates now if we wanted the molarity of the hydroxide ion what would you do just multiply it by two maybe point zero four eight molar now let's come back what was the question it wants to know the solubility of the salt of the solid so we've got to do just a little bit of stretchy AMA tree I'm going to talk it out and also write it out to prove it to you so notice here one mole of calcium site produces one mole of calcium so if I've got if I produce point zero two four zero molar because this is a one-to-one molar ratio that means that's also the molarity of the calcium hydroxide let me prove that to you though with stoichiometry I could take point zero two four zero moles of calcium ion for every 1 liter and then one mole of calcium ion came from one mole of calcium hydroxide there's one Wolfe calcium inside of the calcium hydroxide and if we do that math moles of calcium cancel out and we will end up with point four two times one divided by one you get point o4 or o2 for moles of calcium hydroxide divided by a leader is the same thing point is zero to 4 molar calcium hydroxide so I'm going to erase this we're gonna write that down that's our first piece of information right here so the calcium hydroxide solubility is going to be point zero two four molar that's how much will dissolve 0.02 4 moles will dissolve in every one liter of solution ok so that takes care of this and all you had to do was a stoichiometry so quick dirty way to do this when you find X that's going to be one mole that came from one mole of the solid when you find X all you have to do is write that down as the molarity of the salt so you know look at it think about it check the molarity so it's a one to one ratio but X is going to be the same as the molarity of the salt okay now they wanted grams per liter well molarity is mole's per liter all we have to do is change mole store grounds using molar mass so let's do that I'm going to have point zero two four moles of my calcium I everyone leader I calculated the molar mass and the molar mass of calcium hydroxide is seventy four point one ground for every one mole so that was pretty easy notice moles cancel and we multiply what unit am I left with we are left with grams right there per liter so by o24 times seventy four point one divided by one is going to give us one point seven eight grams of the calcium hydroxide will dissolve in every one liter of solution so to take that last step simply use molar mass to go from moles to grams think your way through a look at the Unison units will tell you what to do when you break apart molarity moles per liter it's easy to see oh I can get rid of moles convert that to grams just using molar mass okay so there you have it honestly here's what I think is the hardest part it's recognizing that these four phrases all mean the same thing it's calculating the solubility the amount that dissolves to get an equilibrium concentration that's why all four of those mean so wrap your brain around this put it in that same file folder in your brain making that make sense that this is where you live okay good work tick check out the KSP equilibrium playlist if you need some more help on KSP have a wonderful day thank you
3317	https://economics.stackexchange.com/questions/16214/solving-a-maximization-problem-by-substitution-when-the-constraint-is-in-implici	Skip to main content Economics Solving a maximization problem by substitution when the constraint is in implicit form Ask Question Asked Modified 8 years, 4 months ago Viewed 836 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am trying to understand how the first order conditions for an interior solution of a maximization problem were derived using the substitution method. The problem is: maxx≥0,y≥0P(a−x)+(1−P)(b−y) subject to Pf(x)+(1−P)f(y)=c where: a,b,c>0, P∈(0,1), f:[0,+∞]→[0,+∞], increasing and strictly concave over its domain. I can see how this is solved using a lagrangian to find from the first order conditions that f′(x∗)=f′(y∗). Strict concavity of f then implies x∗=y∗. But I'm at a loss as to how we can solve it by substituting the constraint in the objective function. Since f is invertible, if y did not appear in the constraint I would find x from the constraint by inverting f and substitute it in the objective function. Doing this here leads to complications that seem unnecessary for this simple problem. There has to be a simpler way that I cannot figure out: what is it? Thanks! microeconomics optimization Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications asked Apr 11, 2017 at 8:19 jloljlol 5533 bronze badges Add a comment \| 2 Answers 2 Reset to default This answer is useful 4 Save this answer. Show activity on this post. Here are two methods. First method: the substitution can be made by inverting f. Since f is strictly increasing and continuous, f−1 is well-defined. The constraint can therefore be written x=f−1(c−(1−P)f(y)P) The objective becomes maxy≥0P[a−f−1(c−(1−P)f(y)P)]+(1−P)(b−y) The derivative of this expression with respect to y equals =(1−P)f′(y)(f−1)′(c−(1−P)f(y)P)−(1−P)(1−P)f′(y)f′∘f−1(c−(1−P)f(y)P)−(1−P)=(1−P)(f′(y)f′(x)−1) And thus f′(y∗)=f′(x∗) at the optimum. Second method: since f is invertible, we can do a change in variables and define w=f(x) and z=f(y). The problem then becomes maxw≥0,z≥0P(a−f−1(w))+(1−P)(b−f−1(z)) subject to Pw+(1−P)z=c Substituing w=(c−(1−P)z)/P in the problem delivers maxz≥0P(a−f−1(c−(1−P)zP)+(1−P)(b−f−1(z)) Differentiating with respect to z yields the same solution. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Apr 11, 2017 at 10:22 answered Apr 11, 2017 at 10:12 OlivOliv 3,2721414 silver badges2525 bronze badges 1 @jlol my pleasure, I'm glad it was helpful. – Oliv Commented Apr 11, 2017 at 20:46 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. Let g be the inverse function of f defined over range of f. Notice that g is increasing and strictly convex. We can rewrite the maximization problem as: maxu≥0, v≥0s.t.P(a−g(u))+(1−P)(b−g(v))Pu+(1−P)v=c where u=f(x) and v=f(y). Solving above is equivalent to solving minu≥0, v≥0s.t.Pg(u)+(1−P)g(v)Pu+(1−P)v=c Now you may substitute v=c−Pu1−P and rewrite the problem as: min0≤u≤cPPg(u)+(1−P)g(c−Pu1−P) Differentiating with respect to u, we get the FOC as: Pg′(u)−Pg′(c−Pu1−P)=0 Since g is strictly convex, solution is: u=c−Pu1−P i.e. u=v=c. Therefore, at the optimum x=y holds. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Apr 11, 2017 at 12:12 answered Apr 11, 2017 at 10:22 AmitAmit 9,97222 gold badges2626 silver badges181181 bronze badges 1 This is an excellent answer. Very helpful, thanks! – jlol Commented Apr 11, 2017 at 19:22 Add a comment \| Start asking to get answers Find the answer to your question by asking. 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3318	https://en.wikipedia.org/wiki/Cyclotomic_polynomial	Jump to content Search Contents 1 Examples 2 Properties 2.1 Fundamental tools 2.2 Easy cases for computation 2.3 Integers appearing as coefficients 2.4 Gauss's formula 2.5 Lucas's formula 2.6 Sister Beiter conjecture 3 Cyclotomic polynomials over a finite field and over the p-adic integers 4 Polynomial values 5 Applications 5.1 Periodic recursive sequences 6 See also 7 References 8 Further reading 9 External links Cyclotomic polynomial العربية Català Čeština Deutsch Español Français 한국어 Italiano עברית Magyar 日本語 Polski Русский Svenska Українська 中文 Edit links Article Talk Read Edit View history Tools 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 Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Irreducible polynomial whose roots are nth roots of unity In mathematics, the nth cyclotomic polynomial, for any positive integer n, is the unique irreducible polynomial with integer coefficients that is a divisor of and is not a divisor of for any k < n. Its roots are all nth primitive roots of unity , where k runs over the positive integers less than n and coprime to n (and i is the imaginary unit). In other words, the nth cyclotomic polynomial is equal to It may also be defined as the monic polynomial with integer coefficients that is the minimal polynomial over the field of the rational numbers of any primitive nth-root of unity ( is an example of such a root). An important relation linking cyclotomic polynomials and primitive roots of unity is showing that is a root of if and only if it is a d th primitive root of unity for some d that divides n. Examples [edit] If n is a prime number, then If n = 2p where p is a prime number other than 2, then For n up to 30, the cyclotomic polynomials are: The case of the 105th cyclotomic polynomial is interesting because 105 is the least positive integer that is the product of three distinct odd prime numbers (3×5×7) and this polynomial is the first one that has a coefficient other than 1, 0, or −1: Properties [edit] Fundamental tools [edit] The cyclotomic polynomials are monic polynomials with integer coefficients that are irreducible over the field of the rational numbers. Except for n equal to 1 or 2, they are palindromes of even degree. The degree of , or in other words the number of nth primitive roots of unity, is , where is Euler's totient function. The fact that is an irreducible polynomial of degree in the ring is a nontrivial result due to Gauss. Depending on the chosen definition, it is either the value of the degree or the irreducibility which is a nontrivial result. The case of prime n is easier to prove than the general case, thanks to Eisenstein's criterion. A fundamental relation involving cyclotomic polynomials is which means that each n-th root of unity is a primitive d-th root of unity for a unique d dividing n. The Möbius inversion formula allows to be expressed as an explicit rational fraction: where is the Möbius function. This provides a recursive formula for the cyclotomic polynomial , which may be computed by dividing by the cyclotomic polynomials for the proper divisors d dividing n, starting from : This gives an algorithm for computing any , provided integer factorization and division of polynomials are available. Many computer algebra systems, such as SageMath, Maple, Mathematica, and PARI/GP, have a built-in function to compute the cyclotomic polynomials. Easy cases for computation [edit] As noted above, if n = p is a prime number, then If n is an odd integer greater than one, then In particular, if n = 2p is twice an odd prime, then (as noted above) If n = pm is a prime power (where p is prime), then More generally, if n = pmr with r relatively prime to p, then These formulas may be applied repeatedly to get a simple expression for any cyclotomic polynomial in terms of a cyclotomic polynomial of square free index: If q is the product of the prime divisors of n (its radical), then This allows formulas to be given for the nth cyclotomic polynomial when n has at most one odd prime factor: If p is an odd prime number, and ⁠⁠ and m are positive integers, then For other values of n, the computation of the nth cyclotomic polynomial is similarly reduced to that of where q is the product of the distinct odd prime divisors of n. To deal with this case, one has that, for p prime and not dividing n, Integers appearing as coefficients [edit] The problem of bounding the magnitude of the coefficients of the cyclotomic polynomials has been the object of a number of research papers. If n has at most two distinct odd prime factors, then Migotti showed that the coefficients of are all in the set {1, −1, 0}. The first cyclotomic polynomial for a product of three different odd prime factors is it has a coefficient −2 (see above). The converse is not true: only has coefficients in {1, −1, 0}. If n is a product of more different odd prime factors, the coefficients may increase to very high values. E.g., has coefficients running from −22 to 23; also , the smallest n with 6 different odd primes, has coefficients of magnitude up to 532. Let A(n) denote the maximum absolute value of the coefficients of . It is known that for any positive k, the number of n up to x with A(n) > nk is at least c(k)⋅x for a positive c(k) depending on k and x sufficiently large. In the opposite direction, for any function ψ(n) tending to infinity with n we have A(n) bounded above by nψ(n) for almost all n. A combination of theorems of Bateman and Vaughan states that: 10 on the one hand, for every , we have for all sufficiently large positive integers , and on the other hand, we have for infinitely many positive integers . This implies in particular that univariate polynomials (concretely for infinitely many positive integers ) can have factors (like ) whose coefficients are superpolynomially larger than the original coefficients. This is not too far from the general Landau-Mignotte bound. Gauss's formula [edit] Let n be odd, square-free, and greater than 3. Then: for certain polynomials An(z) and Bn(z) with integer coefficients, An(z) of degree φ(n)/2, and Bn(z) of degree φ(n)/2 − 2. Furthermore, An(z) is palindromic when its degree is even; if its degree is odd it is antipalindromic. Similarly, Bn(z) is palindromic unless n is composite and n ≡ 3 (mod 4), in which case it is antipalindromic. The first few cases are Lucas's formula [edit] Let n be odd, square-free and greater than 3. Then for certain polynomials Un(z) and Vn(z) with integer coefficients, Un(z) of degree φ(n)/2, and Vn(z) of degree φ(n)/2 − 1. This can also be written If n is even, square-free and greater than 2 (this forces n/2 to be odd), for Cn(z) and Dn(z) with integer coefficients, Cn(z) of degree φ(n), and Dn(z) of degree φ(n) − 1. Cn(z) and Dn(z) are both palindromic. The first few cases are: Sister Beiter conjecture [edit] The Sister Beiter conjecture is concerned with the maximal size (in absolute value) of coefficients of ternary cyclotomic polynomials where are three odd primes. Cyclotomic polynomials over a finite field and over the p-adic integers [edit] See also: Finite field § Roots of unity Over a finite field with a prime number p of elements, for any integer n that is not a multiple of p, the cyclotomic polynomial factorizes into irreducible polynomials of degree d, where is Euler's totient function and d is the multiplicative order of p modulo n. In particular, is irreducible if and only if p is a primitive root modulo n, that is, p does not divide n, and its multiplicative order modulo n is , the degree of . These results are also true over the p-adic integers, since Hensel's lemma allows lifting a factorization over the field with p elements to a factorization over the p-adic integers. Polynomial values [edit] \| \| \| This section does not cite any sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (April 2014) (Learn how and when to remove this message) \| If x takes any real value, then for every n ≥ 3 (this follows from the fact that the roots of a cyclotomic polynomial are all non-real, for n ≥ 3). For studying the values that a cyclotomic polynomial may take when x is given an integer value, it suffices to consider only the case n ≥ 3, as the cases n = 1 and n = 2 are trivial (one has and ). For n ≥ 2, one has : if n is not a prime power, : if is a prime power with k ≥ 1. The values that a cyclotomic polynomial may take for other integer values of x is strongly related with the multiplicative order modulo a prime number. More precisely, given a prime number p and an integer b coprime with p, the multiplicative order of b modulo p, is the smallest positive integer n such that p is a divisor of For b > 1, the multiplicative order of b modulo p is also the shortest period of the representation of 1/p in the numeral base b (see Unique prime; this explains the notation choice). The definition of the multiplicative order implies that, if n is the multiplicative order of b modulo p, then p is a divisor of The converse is not true, but one has the following. If n > 0 is a positive integer and b > 1 is an integer, then (see below for a proof) where k is a non-negative integer, always equal to 0 when b is even. (In fact, if n is neither 1 nor 2, then k is either 0 or 1. Besides, if n is not a power of 2, then k is always equal to 0) g is 1 or the largest odd prime factor of n. h is odd, coprime with n, and its prime factors are exactly the odd primes p such that n is the multiplicative order of b modulo p. This implies that, if p is an odd prime divisor of then either n is a divisor of p − 1 or p is a divisor of n. In the latter case, does not divide Zsigmondy's theorem implies that the only cases where b > 1 and h = 1 are It follows from above factorization that the odd prime factors of are exactly the odd primes p such that n is the multiplicative order of b modulo p. This fraction may be even only when b is odd. In this case, the multiplicative order of b modulo 2 is always 1. There are many pairs (n, b) with b > 1 such that is prime. In fact, Bunyakovsky conjecture implies that, for every n, there are infinitely many b > 1 such that is prime. See OEIS: A085398 for the list of the smallest b > 1 such that is prime (the smallest b > 1 such that is prime is about , where is Euler–Mascheroni constant, and is Euler's totient function). See also OEIS: A206864 for the list of the smallest primes of the form with n > 2 and b > 1, and, more generally, OEIS: A206942, for the smallest positive integers of this form. \| Proofs \| \| Values of If is a prime power, then If n is not a prime power, let we have and P is the product of the for k dividing n and different of 1. If p is a prime divisor of multiplicity m in n, then divide P(x), and their values at 1 are m factors equal to p of As m is the multiplicity of p in n, p cannot divide the value at 1 of the other factors of Thus there is no prime that divides If n is the multiplicative order of b modulo p, then By definition, If then p would divide another factor of and would thus divide showing that, if there would be the case, n would not be the multiplicative order of b modulo p. The other prime divisors of are divisors of n. Let p be a prime divisor of such that n is not be the multiplicative order of b modulo p. If k is the multiplicative order of b modulo p, then p divides both and The resultant of and may be written where P and Q are polynomials. Thus p divides this resultant. As k divides n, and the resultant of two polynomials divides the discriminant of any common multiple of these polynomials, p divides also the discriminant of Thus p divides n. g and h are coprime. In other words, if p is a prime common divisor of n and then n is not the multiplicative order of b modulo p. By Fermat's little theorem, the multiplicative order of b is a divisor of p − 1, and thus smaller than n. g is square-free. In other words, if p is a prime common divisor of n and then does not divide Let n = pm. It suffices to prove that does not divide S(b) for some polynomial S(x), which is a multiple of We take The multiplicative order of b modulo p divides gcd(n, p − 1), which is a divisor of m = n/p. Thus c = bm − 1 is a multiple of p. Now, As p is prime and greater than 2, all the terms but the first one are multiples of This proves that \| Applications [edit] Using , one can give an elementary proof for the infinitude of primes congruent to 1 modulo n, which is a special case of Dirichlet's theorem on arithmetic progressions. \| Proof \| \| Suppose is a finite list of primes congruent to modulo Let and consider . Let be a prime factor of (to see that decompose it into linear factors and note that 1 is the closest root of unity to ). Since we know that is a new prime not in the list. We will show that Let be the order of modulo Since we have . Thus . We will show that . Assume for contradiction that . Since we have for some . Then is a double root of Thus must be a root of the derivative so But and therefore This is a contradiction so . The order of which is , must divide . Thus \| Periodic recursive sequences [edit] The constant-coefficient linear recurrences which are periodic are precisely the power series coefficients of rational functions whose denominators are products of cyclotomic polynomials. In the theory of combinatorial generating functions, the denominator of a rational function determines a linear recurrence for its power series coefficients. For example, the Fibonacci sequence has generating function and equating coefficients on both sides of gives for . Any rational function whose denominator is a divisor of has a recursive sequence of coefficients which is periodic with period at most n. For example, has coefficients defined by the recurrence for , starting from . But , so we may write which means for , and the sequence has period 6 with initial values given by the coefficients of the numerator. See also [edit] Cyclotomic field Aurifeuillean factorization Root of unity References [edit] ^ Roman, Steven (2008), Advanced Linear Algebra, Graduate Texts in Mathematics (Third ed.), Springer, p. 465 §18, ISBN 978-0-387-72828-5 ^ Sloane, N. J. A. (ed.), "Sequence A013595", The On-Line Encyclopedia of Integer Sequences, OEIS Foundation ^ Brookfield, Gary (2016), "The coefficients of cyclotomic polynomials", Mathematics Magazine, 89 (3): 179–188, doi:10.4169/math.mag.89.3.179, JSTOR 10.4169/math.mag.89.3.179, MR 3519075 ^ Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556 ^ Cox, David A. (2012), "Exercise 12", Galois Theory (2nd ed.), John Wiley & Sons, p. 237, doi:10.1002/9781118218457, ISBN 978-1-118-07205-9. ^ Weisstein, Eric W., "Cyclotomic Polynomial", MathWorld ^ a b Sanna, Carlo (2021), "A Survey on Coefficients of Cyclotomic Polynomials", arXiv:2111.04034 [math.NT] ^ Isaacs, Martin (2009), Algebra: A Graduate Course, AMS Bookstore, p. 310, ISBN 978-0-8218-4799-2 ^ Maier, Helmut (2008), "Anatomy of integers and cyclotomic polynomials", in De Koninck, Jean-Marie; Granville, Andrew; Luca, Florian (eds.), Anatomy of integers. Based on the CRM workshop, Montreal, Canada, March 13-17, 2006, CRM Proceedings and Lecture Notes, vol. 46, Providence, RI: American Mathematical Society, pp. 89–95, ISBN 978-0-8218-4406-9, Zbl 1186.11010 ^ Gauss, DA, Articles 356-357 ^ a b Riesel, Hans (1994), Prime Numbers and Computer Methods for Factorization (2nd ed.), Boston: Birkhäuser, pp. 309–316, 436, 443, ISBN 0-8176-3743-5 ^ Beiter, Marion (April 1968), "Magnitude of the Coefficients of the Cyclotomic Polynomial ", The American Mathematical Monthly, 75 (4): 370–372, doi:10.2307/2313416, JSTOR 2313416 ^ Lidl, Rudolf; Niederreiter, Harald (2008), Finite Fields (2nd ed.), Cambridge University Press, p. 65. ^ S. Shirali. Number Theory. Orient Blackswan, 2004. p. 67. ISBN 81-7371-454-1 Further reading [edit] Gauss's book Disquisitiones Arithmeticae [Arithmetical Investigations] has been translated from Latin into French, German, and English. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes. Gauss, Carl Friedrich (1801), Disquisitiones Arithmeticae (in Latin), Leipzig: Gerh. Fleischer Gauss, Carl Friedrich (1807) , Recherches Arithmétiques (in French), translated by Poullet-Delisle, A.-C.-M., Paris: Courcier Gauss, Carl Friedrich (1889) , Carl Friedrich Gauss' Untersuchungen über höhere Arithmetik (in German), translated by Maser, H., Berlin: Springer; Reprinted 1965, New York: Chelsea, ISBN 0-8284-0191-8 Gauss, Carl Friedrich (1966) , Disquisitiones Arithmeticae, translated by Clarke, Arthur A., New Haven: Yale, doi:10.12987/9780300194258, ISBN 978-0-300-09473-2; Corrected ed. 1986, New York: Springer, doi:10.1007/978-1-4939-7560-0, ISBN 978-0-387-96254-2 Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Berlin: Springer, doi:10.1007/978-3-662-12893-0, ISBN 978-3-642-08628-1 External links [edit] Weisstein, Eric W., "Cyclotomic polynomial", MathWorld "Cyclotomic polynomials", Encyclopedia of Mathematics, EMS Press, 2001 OEIS sequence A013595 (Triangle of coefficients of cyclotomic polynomial Phi_n(x) (exponents in increasing order)) OEIS sequence A013594 (Smallest order of cyclotomic polynomial containing n or −n as a coefficient) Retrieved from " Categories: Polynomials Algebra Number theory Hidden categories: Articles with short description Short description is different from Wikidata Articles needing additional references from April 2014 All articles needing additional references CS1 Latin-language sources (la) CS1 French-language sources (fr) CS1 German-language sources (de) Cyclotomic polynomial Add topic
3319	https://pmc.ncbi.nlm.nih.gov/articles/PMC10453035/	On the Finite Complexity of Solutions in a Degenerate System of Quadratic Equations: Exact Formula - 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 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 Entropy (Basel) . 2023 Jul 25;25(8):1112. doi: 10.3390/e25081112 Search in PMC Search in PubMed View in NLM Catalog Add to search On the Finite Complexity of Solutions in a Degenerate System of Quadratic Equations: Exact Formula Olga Brezhneva Olga Brezhneva 1 Department of Mathematics, Miami University, Oxford, OH 45056, USA; brezhnoa@miamioh.edu Methodology, Validation, Formal analysis, Investigation, Resources, Writing – original draft, Supervision Find articles by Olga Brezhneva 1,†, Agnieszka Prusińska Agnieszka Prusińska 2 Faculty of Exact and Natural Sciences, Siedlce University, ul. Konarskiego 2, 08-110 Siedlce, Poland; tret@uph.edu.pl Methodology, Validation, Formal analysis, Investigation, Resources, Writing – original draft, Writing – review & editing, Project administration, Funding acquisition Find articles by Agnieszka Prusińska 2,,†, Alexey A Tret’yakov Alexey A Tret’yakov 2 Faculty of Exact and Natural Sciences, Siedlce University, ul. Konarskiego 2, 08-110 Siedlce, Poland; tret@uph.edu.pl 3 Systems Research Institute, Polish Academy of Sciences, ul. Newelska 6, 01-447 Warszawa, Poland Conceptualization, Methodology, Validation, Formal analysis, Investigation, Resources, Supervision, Funding acquisition Find articles by Alexey A Tret’yakov 2,3,† Editor: Ravi P Agarwal Author information Article notes Copyright and License information 1 Department of Mathematics, Miami University, Oxford, OH 45056, USA; brezhnoa@miamioh.edu 2 Faculty of Exact and Natural Sciences, Siedlce University, ul. Konarskiego 2, 08-110 Siedlce, Poland; tret@uph.edu.pl 3 Systems Research Institute, Polish Academy of Sciences, ul. Newelska 6, 01-447 Warszawa, Poland Correspondence: aprus@uph.edu.pl † These authors contributed equally to this work. Roles Olga Brezhneva: Methodology, Validation, Formal analysis, Investigation, Resources, Writing – original draft, Supervision Agnieszka Prusińska: Methodology, Validation, Formal analysis, Investigation, Resources, Writing – original draft, Writing – review & editing, Project administration, Funding acquisition Alexey A Tret’yakov: Conceptualization, Methodology, Validation, Formal analysis, Investigation, Resources, Supervision, Funding acquisition Ravi P Agarwal: Academic Editor Received 2023 May 23; Revised 2023 Jul 18; Accepted 2023 Jul 19; Collection date 2023 Aug. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC10453035 PMID: 37628142 Abstract The paper describes an application of the p-regularity theory to Quadratic Programming (QP) and nonlinear equations with quadratic mappings. In the first part of the paper, a special structure of the nonlinear equation and a construction of the 2-factor operator are used to obtain an exact formula for a solution to the nonlinear equation. In the second part of the paper, the QP problem is reduced to a system of linear equations using the 2-factor operator. The solution to this system represents a local minimizer of the QP problem along with its corresponding Lagrange multiplier. An explicit formula for the solution of the linear system is provided. Additionally, the paper outlines a procedure for identifying active constraints, which plays a crucial role in constructing the linear system. Keywords: quadratic programming, singular problems, p-regularity, 2-factor-operator MSC: 65H10, 90C20, 65K05, 90C30 1. Introduction Consider the nonlinear equation with the mapping F defined by: (1) where is a matrix, is a vector, and is the map defined for by: (2) where is an symmetric matrix for . We also consider the quadratic programming (QP) problem with inequality constraints: (3) where Q is an symmetric matrix, A is an matrix, , and . The paper describes an application of the p-regularity theory to nonlinear equations with the mapping F introduced in (1) and to the quadratic programming problem (3). In recent years, there has been growing interest in nonlinear problems, including quadratic and polynomial equations, as well as nonlinear optimization problems, attracting specialists from various disciplines (see, for example, refs. [1,2,3,4] and references therein). Furthermore, it was observed that nonlinear problems are closely related to singular problems, as demonstrated in . In fact, it has been discovered that essentially nonlinear problems and singular problems are locally equivalent. In this work, we aim to provide a theoretical foundation for this claim by introducing several auxiliary concepts as proposed in . Definition 1. Let V be a neighborhood of in , and let be a neighborhood of 0.A mapping , where , is considered essentially nonlinear at if there exists a perturbation of the form: such that no nondegenerate transformation of coordinates , where , satisfies , , where is the identity map in , and: Definition 2. We say that the mapping F is singular (or degenerate) at if it fails to be regular, meaning its derivative is not onto: The relationship between the notions of essential nonlinearity and singularity is established in Theorem 1, which was derived in . Theorem 1. Let V be a neighborhood of in . Suppose is and that is a solution of . Then F is essentially nonlinear at the point if and only if F is singular at the point . The work presented in primarily focuses on the construction of p-regularity and its applications in various areas of mathematics. However, it does not specifically cover quadratic nonlinear equations and quadratic programming problems. The current paper builds upon the foundation of the p-regularity theory established in but introduces novel results. The main objective of this paper is to explore the key aspects of nonlinear problems, with a particular emphasis on systems of quadratic equations and quadratic programming problems that may involve singular solutions. Specifically, we begin by considering the nonlinear equation . One of the main goals of the paper is to derive the exact formula for a solution of the nonlinear equation using the special form of the quadratic mapping F defined in (1). We demonstrate how to use a construction of a special 2-factor-operator to transform the original problem into a system of linear equations. The construction of the 2-factor-operator combines the mapping F with its first derivative . In the second part of the paper, we apply a similar approach to the quadratic programming problem (3) in order to derive explicit formulas for the solution , where represents a local minimizer of the QP problem and is the corresponding Lagrange multiplier. Namely, using the special form of the QP problem and the 2-factor-operator, we reduce the system of optimality conditions for the QP problem to a system of linear equations, with the point as its solution. The paper also describes a procedure for identifying the active constraints, which plays a vital role in constructing the linear system. Although there is literature on solutions of degenerate systems of quadratic equations, the approach presented in this paper is novel and distinct from the methods proposed by other authors. This approach can be applied to various problems and areas of mathematics where the problem involves solving a degenerate equation with a quadratic mapping F. Such nonlinear problems can arise in the numerical solutions and analysis of ordinary differential equations, partial differential equations, optimal control problems, algebraic geometry, and other fields. In the second part of the paper, we specifically focus on using the methods developed in the first sections to solve the QP problem (3). The quadratic programming problems have attracted the attention of many researchers and scientists, so there is an extensive body of literature on the topic. Some publications in this area include [6,7,8,9,10,11,12]. The outline of the paper. The main contribution and novelty of the paper are in the exact formulas for a solution of a nonlinear equation and of the quadratic programming problems, presented in Section 3 and Section 5, respectively. In Section 2, we recall the main definitions of the p-regularity theory, as presented in , including the special case of . Additionally, we introduce the p-factor method for solving singular nonlinear equations of the form and describe various versions of the 2-factor method. Section 3 presents some of the key results of the paper, focusing on the application of a modified 2-factor method to solve the nonlinear equation with the mapping F defined as where is a matrix, is a vector, and is defined by (2). In this section, we introduce multiple approaches to obtain exact formulas for a solution to the nonlinear equation , demonstrating that the proposed methods converge to a solution of the nonlinear equation in just one iteration. Section 4 focuses on an auxiliary result used in other parts of the paper. We present a theorem that describes the properties of a special mapping , which enables us to propose a procedure for determining r linearly independent vectors , , at the solution of , without needing to know the exact value of . This procedure relies on information about the system of vectors at some point x within a small neighborhood of . Section 5 presents other novel results, focusing on deriving exact formulas for a solution of quadratic programming problems. The section is divided into three parts. In Section 5.1, we consider regular quadratic programming problems and propose three approaches to solving the QP problem and obtaining a formula for its solution. These approaches are based on the construction of the 2-factor-operator. Section 5.2 addresses the issue of identifying the active constraints and proposes strategies for numerically determining the set of active constraints . These techniques are then applied in the final part, Section 5.3, to address degenerate QPs. The paper concludes with some closing remarks in Section 6. Notation. Let denote the rows of the matrix A in problem (3), and let , so that and for . The active set at any feasible point of problem (3) is the set of indices of the active constraints at , i.e., . Furthermore, denotes the null-space (kernel) of a given linear operator , and is its image space. Let be a bilinear symmetric mapping. The 2-form associated with B is the map defined by for . We also use the following notation: and . We denote by and neighborhoods of a point , where is an -neighborhood of , i.e., an open ball of radius centered at . The notation for the scalar (dot) product of vectors x and y in , used in the paper, is . We denote by the linear span of the given vectors . We also denote by the distance between a point x and a set S. 2. Elements of the -Regularity Theory We begin this section with the main definitions of the p-regularity theory, which are given in . The primary focus is on the sufficiently smooth mapping F from to , defined as: (4) where for . After presenting the general case, we focus on the specific case of . We introduce the definitions of the 2-regular mapping and the 2-factor-operator, which play a central role in the subsequent sections. 2.1. The Main Definitions and Constructions of the p-Regularity Theory Throughout this section, we consider the nonlinear equation: (5) where F is defined in Equation (4). Let represent a solution to the nonlinear Equation (5). The mapping F is called regular at if: (6) or in other words, if: where is the Jacobian matrix of the mapping F at . Conversely, the mapping F is called nonregular (irregular, degenerate) if the regularity condition (6) is not satisfied. Let the space be decomposed into the direct sum: (7) where is defined as the closure of the image of the first derivative of F evaluated at , and p is chosen as the minimum number for which Equation (7) holds. The remaining spaces are defined as follows. Let , and let be a closed complementary subspace to . Let be the projection operator onto along . Define as the closed linear span of the image of the quadratic map . More generally, we define inductively for as: where is a choice of a complementary subspace for with respect to Y, , and is the projection operator onto along with respect to Y, . Finally, we let Define the following mappings: (8) where is the projection operator onto along with respect to for . Then F can be represented as or equivalently as Definition 3. The linear operator , where , , is defined by: and is called the p-factor operator. Consider the nonlinear operator defined by: Notice that . Definition 4. The p-kernel of the operator at the point is the set defined by: where: Please note that , where: is the k-kernel of . Definition 5. A mapping F is called p-regular at along h if . Definition 6. A mapping F is called p-regular at if it is p-regular along all or . Now, we will focus on the special case of , which we are using in the paper. We denote the image of the Jacobian matrix by : and the orthogonal complementary subspace of in by . Then: We also denote an matrix of the orthogonal projection onto in by , Similarly to Equation (8), we introduce the mappings: The p-factor operator plays the central role in the p-regularity theory. We give the following definition of the p-factor-operator for . Definition 7. We define a 2–factor-operator of the mapping F at with respect to some vector , , as a linear operator from to , defined by one of the following equations: (9) (10) (11) Now we are ready to introduce another very important definition of the 2-regularity theory. Definition 8. The mapping F is called 2-regular at the point with respect to the element h if the image of a 2–factor-operator, defined by one of the Equations (9)–(11) is equal to . Definition 9. The mapping F is called 2-regular at if it is 2-regular at with respect to all the elements h from a set that is defined as: 1. for defined by (9); 2. for defined by (10); 3. for defined by (11). 2.2. The p-Factor-Method for Solving Singular Nonlinear Equations In this section, we introduce the p-factor-method for solving the singular nonlinear equation . Then we consider the special case of and describe several versions of the 2-factor-method. Consider Equation (5) in the case when mapping is singular at . In this case, the p-factor method is an iterative procedure defined by: (12) where , for , and vector h, , is chosen in such a way that the p-factor operator is nonsingular, which implies that the mapping F is p-regular at along h. The following theorem is valid for the p-factor-method (12). Theorem 2. Assume that mapping and there exists vector h, , such that the p-factor operator is nonsingular. Given a point , where is sufficiently small and is a neighborhood of , the sequence defined by Equation (12) converges quadratically to the solution of (5): (13) where is an independent constant. Now, we are ready to describe several versions of the 2-factor-method. For solving singular nonlinear Equation (5), the following iterative method, called the 2-factor-method, was proposed in : (14) where the vector h, , is chosen in such a way that matrix is invertible. The following theorem states the convergence properties of the 2-factor-method (14). Theorem 3. Given a mapping , let be a solution of Equation (5). Assume that there exists a vector such that and F is 2-regular at the point with respect to the vector h with the 2-factor-operator defined by (10). Then there is a neighborhood of in such that for any , the sequence generated by the 2-factor-method (14) converges to and: (15) where is some constant. Proof. Since is the orthoprojector onto subspace , then for the mapping: we have . Moreover, because and the mapping F is 2-regular with respect to the vector h, by Definition 8 with defined by (10), we obtain that . Hence, the matrix is invertible. Therefore, the 2-factor-method given in (14) is an application of Newton’s method to system in a sufficiently small neighborhood of . Then the statement of the theorem follows from the properties of Newton’s method (Proposition 1.4.1). □ Now, we will introduce a modified version of the 2-factor-method (14). Assume that there exists a vector such that and the matrix is invertible. Then for solving Equation (5), we can use the following modified 2-factor-method: (16) The following theorem states the convergence properties of method (16). Theorem 4. Given a mapping , let be a solution of Equation (5). Assume that there exists a vector such that , , and F is 2-regular at the point with respect to the vector h, where the 2-factor-operator is defined by (11). Then there is a neighborhood of in such that for any , the sequence generated by the 2-factor-method (16) converges to , and relation (15) holds with some . The proof is similar to one of Theorem 3. Now we introduce another version of the 2-factor-method. Assume that the following conditions hold: (17) Note that if conditions (17) are satisfied, then to solve Equation (5), we can use the following modified 2-factor-method: (18) For numerical realization of the 2-factor-method in the form (18), we only have to construct vector h satisfying conditions (17). Specifics of some problems allow us to choose vector h without any knowledge of the solution . We discuss the choice of the vector h in the following sections of the paper. The following theorem states the convergence properties of method (18). Theorem 5. Given a mapping , let be a solution of Equation (5). Assume that there exists a vector such that and conditions (17) are satisfied. Then, there is a neighborhood of in such that for any , the sequence generated by the 2-factor-method (18) converges to and relation (15) holds with some . The proof is similar to one of Theorem 3. 3. Nonlinear Equations with Quadratic Mappings: the Exact Solution Formula In this section, we consider the mapping F defined by Equation (1) as follows: where is a matrix, is a vector, and is the map defined by (2). The mapping B is twice continuously differentiable , and its derivatives are given by and for some arbitrary . Let denote a solution of the equation . We will now illustrate the application of the 2-factor method (18) for solving the nonlinear equation with the mapping F defined by (1). We will present multiple approaches to obtain an exact formula for , with the first approach being a specific case of the second approach. Additionally, we will show that for the mapping F, the method (18) converges to in just one iteration. First approach to obtain an exact formula for the solution . For the mapping F defined by (1), the assumptions (17) of Theorem 5 can be simplified to the existence of a vector h that satisfies the following conditions: (19) Under these assumptions (19), for the mapping F defined by (1) and a given point , the first iteration of the 2-factor-method (18) can be written as: which is equivalent to: Using the property , the last equation implies a one-step method for calculating and, consequently, finding the solution : (20) where the vector h satisfies conditions (19). The numerical determination of the vector h depends on the specific characteristics of the problem. Alternatively, it can be obtained using the same method as described in the third approach below, which involves transforming the initial system into a system that is completely degenerate at the point . Second approach to obtain an exact formula for the solution . Now we present an alternative approach for obtaining a formula for the solution of the equation using the same mapping F defined by (1). This second approach is applicable to a broader variety of problems compared to the first approach. Let denote the projector onto , and let denote the projector onto , which is the orthogonal complementary subspace of in . We note that and: Then for the mapping F defined in (1), Assume that there exists a vector satisfying the conditions: (21) Given the definition of , it follows that . Substituting this into (1), we obtain . Hence, the point satisfies the following identities: By adding these equations and assuming (21), we obtain the exact formula for the solution : (22) Remark 1. In the case when and, hence, , assumptions (21) become (19), and Equation (22) reduces to (20). Example 1. Consider mapping given by: (23) We can represent the mapping F in the form (1) with: The equation has a locally unique solution . In this example, and . Hence, by Remark 1, we apply Equation (20) with to obtain: as claimed. In a numerical implementation, an additional procedure is required to construct the vector h. Since the exact point is not known in advance, we only assume that a sufficiently small neighborhood of is provided to apply the procedure. Third approach to obtain an exact formula for the solution . While the first two approaches rely on knowledge of the element h, which is determined by , the third approach does not require such knowledge. Instead, all we need is for the starting point to belong to a sufficiently small neighborhood of . Specifically, we have , where is sufficiently small. Suppose that at the point , the first r vectors are linearly independent, where is defined in (4) for . Assume also that the other vectors are linear combinations of the first r vectors. Therefore, there exist coefficients such that: Let us introduce the subspace defined by: We denote the orthogonal projection on the subspace as . Then, there exist coefficients such that: In addition, introduce the notation: Then: (24) Notice that is also a solution of the equation , where is defined as: The definition of implies that is 2-regular at the point . In the case that some of the vectors , are not zero vectors, transformation (24) can be used to reduce those vectors to zero vectors. This ensures that for all . Therefore, without loss of generality, we can assume that the mapping satisfies for . An example of a mapping that satisfies these conditions is: where , , , , , and Suppose there exist vectors , , and , and indices , such that the system: is linearly independent. Then the mapping defined by: (25) has as its zero, that is . At the same time, compared to the Jacobian matrix of , the matrix: is nonsingular. We can, therefore, consider the method: (26) Theorem 6. Given a mapping , let be a solution of Equation (5). Assume that there exist vectors , , and , such that mapping defined in (25) is nonsingular at . Let , where is a neighborhood of and is sufficiently small. Then the sequence , defined by (26) is convergent to the point with the quadratic rate of convergence, that is: where is an independent constant. Using definition of mapping F given by Equation (1), mappings introduced in (4) will have the following form: where is an symmetric matrix, , and , Given an initial point , we use the iterative method (26) to obtain: Because matrix is symmetric for any index i, then for any index j, we have: Therefore, (27) Example 1 (Continuation). Consider mapping defined in (23): where In this example, is a solution of and: Therefore, mapping defined in (25) takes the form: where h is chosen in such a way that the matrix is nonsingular, and vectors are not used. For example, we can take . Then Equation (27) has the form: (28) which is a solution of in this example. The approaches described above can be modified to derive other methods for solving the equation . For example, using the equation , where , we obtain the following method: The sequence converges to under the assumption that is nonsingular. In this modification, unlike the second approach, we can construct an element h without the knowledge of the point , based on the information at an initial point . Applying the modified method to Example 1, we obtain the same formulas and results as shown in Equation (28) above. To implement this approach, it is necessary to determine the vectors , , which correspond to linearly independent vectors This can be achieved using information at a point , where is sufficiently small. If the assumption of p-regularity is satisfied, the identification of linearly independent vectors is performed using the method described in the next section. 4. Procedure for Identifying Zero Elements The procedure for identifying zero elements could be used to implement the methods described in the previous sections numerically. Let be defined as: (29) In this section, we present a theorem that describes the properties of a special mapping , which allows us to propose the method for determining r linear independent vectors , at the solution of . This procedure is based on the information about the system of vectors at some point x in a small neighborhood of . As a result, we can define the mapping with the first r components , corresponding to the linearly independent vectors , Let be 2-regular at the point . For some , where is sufficiently small, we define the following mappings: and: (30) where denotes the distance between an element x and the set S. Note that if , then . The mapping is used to determine the maximum number r of linearly independent vectors in the system using a special procedure that relies on the information about the mapping at the point . The properties of the mapping are stated in the following theorem, and the proof can be found in . Theorem 7 (Minorant theorem). Let be 2-regular at the point , and . Then there exist constants , and such that the following inequality holds for any : where function is defined in (30). In addition to the properties of the mapping given in Theorem 7, we also need the following lemma (for the proof, see ). Lemma 1. For the non-negative mappings and , let the following inequalities hold: where , , and σ are positive constants, with and . Then, there exists a sufficiently small such that one of the following conditions holds: 1. If for all , then 2. If for all , then Remark 2. Based on the assumptions of Lemma 1, there exists a sufficiently small such that if the inequality is satisfied for any , then the inequality is satisfied for all , and hence . Similarly, if the inequality is satisfied for any , then the inequality is satisfied for all , and hence . Now we are ready to introduce an iterative method that determines indices corresponding to the linearly independent vectors , Method for determining linearly independent gradients at (identifying zero elements). Using Lemma 1 and Remark 2, for a sufficiently small , , and , consider two possible cases: Case 1.. Case 2.. In Case 1, according to Remark 2, it follows that , whereas in Case 2, we have . In addition to the properties of the mapping Let be defined by (29) and be a solution of . Let x be in , where is sufficiently small. Define function using Equation (30). Step 1. Identify the smallest index in the set such that . According to Case 2 above, this implies that . Step 2. Use Step 1 to identify if the set has at least one index j such that . If it does not, the method is finished. Otherwise, identify the next smallest index in the set such that the following condition is satisfied: According to Case 2 above, it means that the vectors and are linearly independent. Step k. By this step, we have identified linearly independent vectors , …, , where . Use Step 1 to identify if the set has at least one index j such that . If it does not, the method is finished. Otherwise, identify the next smallest index such that the following condition is satisfied: The inequality implies that the vectors , are linearly independent. Repeat Step k until the method is finished. Without loss of generality, assume that the first r vectors are linearly independent and define mapping as: (31) where vectors are defined in such a way that: Namely, let: be a linear combination of the vectors , …, Coefficients are determined by solving the following system of equations: In addition, define to be a nonsingular matrix of the form: Let: Define the following vectors: These vectors allow us to transform the mapping to , where and The purpose of this transformation is to simplify the structure of the projection operators. We present a simple example to illustrate an application of the proposed method. Example 2. Let , , where: Then is a solution of . Take and consider . The Jacobian matrix of F is: It is easy to see that vectors and are linearly dependent. We can check this by applying the method introduced above. By using Equation (30), we define function , where: and α is the angle between vectors and . Note that: and hence: Using , we obtain: We are ready to apply the method described above. In Step 1, we obtain because: Hence, vector and . Then in Step 1 of the method with vector , we also verify whether the following inequality holds: Using point , we obtain: Therefore, we conclude that . Thus, in this example, the mapping defined in (31) has the form , where and . 5. Quadratic Programming Problems In this section, we consider the quadratic programming (QP) problem (3): where Q is an symmetric matrix, A is an matrix, , and . The Lagrangian for problem (3) is defined by: (32) where is the vector of Lagrange multipliers and is the i th row of the matrix A. The Karush-Kuhn-Tucker (KKT) conditions are satisfied at with some if: (33) The point at which relations (33) are satisfied is called a stationary point or a KKT point. Observe that is a solution of the following system: (34) We denote by the set of indices of the active constraints at : The following constraint qualification is used in the paper. Definition 10 (Linear independence constraint qualification). The linear independence constraint qualification (LICQ) holds at a feasible point if the row-vectors , , corresponding to the active at constraints, are linearly independent. The modified second-order sufficient conditions (MSOSC) state that there exist a Lagrange multiplier vector and a scalar such that: (35) for all satisfying: We divide the presentation in this section into three parts. We start by considering regular QP problems in Section 5.1. Then, in Section 5.2, we discuss the issue of identifying the active constraints and propose numerical strategies for determining the set . We apply these techniques to degenerate QP problems in Section 5.3. 5.1. Regular Quadratic Programming In this section, we consider regular quadratic programming (QP) problem (3). In other words, we assume that the Linear Independence Constraint Qualification (LICQ) and the Mangasarian-Fromovitz Constraint Qualification (MFCQ) conditions (35) hold. Recall that A is an matrix of coefficients representing the constraints in problem (3). Without loss of generality, assume that the first p constraints are active at , so that: Then we can rewrite the matrix A in the following form: (36) where is a matrix of coefficients corresponding to the active constraints at , and is an matrix of coefficients corresponding to the nonactive constraints at . It is important to note that we do not have prior knowledge of the set . We will discuss possible numerical realizations to approximate the set of active constraints in Section 5.2. Additionally, we introduce the following notation associated with the active constraints at the point : Similarly, In the following subsections, we will introduce three approaches to solving the QP problem (3) and provide formulas for the solution. 5.1.1. First Approach to Solving the QP Problem In this subsection, we present an approach to solving the QP problem and obtaining a formula for its solution. This approach is based on the construction of the 2-factor-operator. For our consideration below, we need the following lemma. Lemma 2. Let V be an matrix, G be a matrix, such that the columns of are linearly independent, L be an matrix, be a diagonal full rank matrix, and: (37) Then matrix Γ defined by: (38) is nonsingular. Proof. To prove the lemma, we must prove that the matrix defined by (38) has zero nullspace. Consider the following system that defines the nullspace of in the form of a vector , where , , and : (39) Since is a full-rank diagonal matrix, the third equation in the system (39) implies that . Then, using the first equation, we obtain: Consequently, ; otherwise, , which contradicts the assumption (37) of the lemma. Therefore, the first equation in (39) reduces to , and since the columns of are linearly independent, we obtain . Thus, the matrix (38) has a zero nullspace, , and therefore, is nonsingular. This concludes the proof of the lemma. □ Let , and mapping be defined in (34), so that . Introduce mappings and as: Recall that matrix is defined in (36), and introduce vector such that: where , , Define mapping as: (40) Recall that is the i th row of the matrix A and . Then: and mapping defined in (40) can be rewritten as: Introduce matrix . Then, taking into account the definition of and , we obtain: Observe that if is a solution of (34), it is also a solution of or, equivalently, (41) To obtain the formula for the solution , we rewrite the system (41) as: Assuming that LICQ and MSOSC hold and apply Lemma 2, we obtain that the matrix: is invertible and obtain the formula for : (42) 5.1.2. Second Approach to Solving the QP Problem Assume that we can estimate the set , which is in our notation . Taking into account that and that , system (34) can be reduced to the following one: (43) which can be written as: (44) Under the assumptions LICQ and MSOSC, the following matrix is invertible, and system (44) yields the formula for the solution : (45) Remark 3. System (41) reduces to system (43) by removing equations , corresponding to the nonactive constraints. Similarly, Equation (42) reduces to (45). Remark 4. We note that solutions of QP problems have the following specific property: if is a solution of the QP problem and for the vector , then the points are also solutions of the QP problem. 5.1.3. Examples In this section, we illustrate the two described approaches with examples. Namely, we consider the construction of system (41) required for the first approach. Then we illustrate using the exact formula (45) derived in the second approach. Example 2. Consider the problem: (46) The matrix A in this example is and The solution to this problem is the point with and . Hence, , , and . Moreover, By choosing and , the system (41) reduces to the linear system: Solving the system yields , as claimed. Now, let us illustrate the second approach. Specifically, using the formula (45) for the solution of problem (46) with , we obtain: as claimed. Example 3. Consider the problem: (47) The solution to this problem is the point with and . Hence, . Moreover, By choosing and , the system (41) reduces to the following linear system for problem (47): Solving yields , as claimed. To illustrate the second approach, we rewrite the exact formula (45) for the solution of problem (47) in the form: (48) as claimed. 5.1.4. Third Approach to Solving the QP Problem In this subsection, we present another approach to solving the QP problem. A formula that we obtain for the solution of the QP problem is also based on the construction of the 2-factor-operator. First, we replace the inequality constraints in the QP problem with equality constraints of the form: where . We then define the Lagrangian as follows: (49) Introduce the notation: Then the point is a solution of the following system: (50) The Jacobian matrix of the system (50) is given by: Then with , we obtain Assuming that LICQ and MSOSC hold, matrix is singular if and only if the strict complementarity condition does not hold. In other words, the set of indices of the weakly active constraints, is not empty. Let be the matrix of the orthoprojector onto , and be the matrix of the orthoprojector onto . Note that will be a projector onto the linear part of the mapping , while will be a projector onto the quadratic part of . Introduce vector such that . Then, or and is defined by: (51) Observe that , i.e., . Define H as a diagonal matrix with elements in the rows corresponding to the components of the vector , and K as a diagonal matrix with elements of the vector , so that: Then: The 2-factor-operator for the mapping is defined as: or We choose a vector according to (51) so that matrix: is nonsingular. Then can be determined using the following formula: 5.2. Identification of the Active Constraints In this section, we address the issue of identifying the active constraints and propose strategies for numerically identifying the set of active constraints . We begin by considering the mapping , where . We can also represent h as an n-vector of functions , …, , such that . Theorem 8. Let be 2-regular at the point , and let be a sufficiently small neighborhood of in . Assume that there exists a function such that and for all , we have: (52) where are independent constants. Then there exists a sufficiently small δ such that , and for any and any point , the following holds: Either , which implies that Or , which implies that Proof. The proof is similar to the one in . □ Let: where denotes the distance between a vector a and a set S. It turns out that if we take , and g is 2-regular at , then inequality (52) holds with . Theorem 8 can be used for the numerical determination of the set of active constraints in the QP problem. To apply Theorem 8, we need to define a function that satisfies the conditions of the theorem. Recall that for QP problem (3), we denote the Lagrange function defined in (32) by . Under the assumptions of LICQ and MSOSC, the following holds for and : where is sufficiently small (see, for example, ). Hence, the required function can be defined by: Then, according to Theorem 8, for every , if: then it follows that . Moreover, if we introduce the function: where: , then satisfies the estimate: for , where is a sufficiently small number. Then, for any , if: then . Here, represents the set of constraints that are weakly active, i.e, for which the associated multipliers are equal to zero, while denotes the set of constraints that are strongly active at the point , i.e., the associated Lagrange multipliers are positive. 5.3. General Case Consider the Lagrange function in the form: In this case, if is a solution of problem (3), then there exist multipliers and , not all zeros, such that , , and the point is a solution of the following system: (53) Introduce the notation: (54) We are making the following assumption for the rest of the section. Assumption A1. Assume that there exists and a sufficiently small such that for any , the following holds: Remark 5. It is easy to see that for any , where is an independent constant. As follows from Assumption 1 and Theorem 8, for those indices that satisfy the inequalty: we make a conclusion that . We can illustrate Assumption 1 with the following examples, where Assumption 1 holds. Example 4. This example illustrates a choice of the function ξ in a more general setting. Consider mapping F defined by either: or In each of the two cases, . Introduce function defined as: It follows that for any , the inequality: holds, where C is an independent constant. Example 5. Consider the problem: (55) The solution to this problem is the point , so . Moreover, the system (53) in this example is given by: (56) We also introduce the function , which can be defined using Equation (54), but in this case, we define it as . Under Assumption 1, we use the function to determine the set . We also take into account the fact that the constraints in the problem are linear and the rank of the matrix is 1. This implies that the constraints are linearly dependent. Consequently, we can eliminate, for instance, the second constraint from problem (55) and simplify system (56) to the following one: Now, by introducing: we construct the modified 2-factor-system: This system implies that the solution is . Now we will demonstrate the application of the approach described in Section 5.1.2 to problem (55). By removing the first constraint, we obtain a regular QP problem with . Additionally, in this example, Then application of Equation (45) derived in Section 5.1.2 yields: as claimed. There are various directions in which the approach proposed in this paper can be extended. The next example illustrates a degenerate QP problem, in which MSOSC does not hold at the solution. However, an approach proposed in this paper can be applied to find a solution to this problem. Moreover, the solution set is locally not unique. There are various directions in which the approach proposed in this paper can be extended. The next example illustrates a degenerate QP problem in which MSOSC does not hold at the solution. However, the approach proposed in this paper can still be applied to find a solution to this problem. It is worth noting that the solution set in this case is locally not unique. Example 6. Consider the problem: (57) The solution to this problem is the set of points . We observe that the objective function in this example is satisfied as an equality for any . Additionally, the system (53) for this example consists of one linear equation and three quadratic equations: Denote the projection of the point x onto the set by . Also, define the notation . For any point , we have the inequality: where and is sufficiently small. Consider, for example, the point . In problem (57), we replace the inequality with the equation , where . We then introduce the Lagrange function in the form of (49) as follows: If is a Lagrange multiplier corresponding to the solution , then the point is a solution of the following system: (58) The Jacobian matrix of this system is given by: This Jacobian matrix becomes singular at . To overcome this singularity, we can apply the approach described in the paper. Specifically, we notice that for . Moreover, the point is one of the solutions of the system defined in (58), corresponding to a solution of the QP problem (57). Additionally, The 2-factor-operator of with respect to the vector , which is defined similarly to the operator in Equation (11), is given by: Note that the 2-factor-operator is nonsingular and the system: has the point as its regular solution. 6. Conclusions The paper focused on applying the p-regularity theory to nonlinear equations with quadratic mappings and quadratic programming (QP) problems. The first part of the paper used the special structure of the nonlinear equation and the construction of a 2-factor operator to derive a formula for the solution of the equation. In the second part, the QP problem was reduced to a system of linear equations using a 2-factor-operator. The solution of the system is a local minimizer of the QP problem with a corresponding Lagrange multiplier. The formula for the solution of the linear system was given. The paper also described a procedure for identifying the active constraints, which was used in constructing the linear system. The paper primarily focuses on the case where the matrix is degenerate at the solution of the nonlinear equation . However, the matrix does not need to be degenerate. While we do not explicitly address the identification of degeneracy at a solution point, it is possible to determine the degeneracy of the matrix by examining the behavior of the mapping F in a small neighborhood of the solution . Specifically, a function can be defined, such that: for some natural number p and constants and . Based on the conclusion about the degeneracy of the matrix , an appropriate method can be chosen to solve the system of equations , as stated in the following theorem. Theorem 9. Let be such that , and let there exist , where is sufficiently small. Then we have the following two cases: In the first case, for all , we have: In this case, , indicating that F is not degenerate at . In the second case, there exists an index such that: In this case, , indicating that F is degenerate at . Certainly, the construction of the function is an important consideration. One approach to constructing such a function is provided in the following lemma, specifically for the case of . Lemma 3. Let and assume that either exists, or for any , there exists with . Then, there exists a sufficiently small such that the following inequality holds for all : where C is a positive constant. Based on this lemma, one can choose the function . It is worth noting that the proposed approach also covers the case where the system of equations consists of both linear and quadratic equations. Moreover, the approach can be extended to solve multilinear equations with polynomials of degree p, given by the equation: where is k-multilinear mapping for . Additionally, polynomial programming problems can be formulated as follows: where are polynomial mappings. There are various possible directions for future research work, based on the results obtained in this paper. While the focus of the current work was on obtaining exact formulas for the solutions of nonlinear equations with quadratic mappings and quadratic programming problems, it would be interesting to generalize the proposed approaches to other classes of problems, including systems of equations with both linear and quadratic mappings. Another direction would be an extension of presented methods to polynomial equations and polynomial programming problems. Another direction of future research could be focusing on numerical studies and the implementation of the methods described in the paper. Acknowledgments The authors thank the anonymous reviewers for their careful reading of our manuscript and for their insightful comments and suggestions that helped us improve the quality of the paper. Author Contributions Conceptualization, A.A.T.; methodology, O.B., A.P. and A.A.T.; validation, O.B., A.P. and A.A.T.; formal analysis, O.B., A.P. and A.A.T.; investigation, O.B., A.P. and A.A.T.; resources, O.B. and A.A.T.; writing—original draft preparation, O.B. and A.P.; supervision, O.B. and A.A.T.; project administration, A.P.; funding acquisition, A.P. and A.A.T. All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement Not applicable. Data Availability Statement No new data were created or analyzed in this study. Data sharing is not applicable to this article. Conflicts of Interest The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript. Funding Statement This research was funded by the Ministry of Education and Science, grant number 144/23/B. Footnotes Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. References 1.Barvinok A., Rudelson M. When a system of real quadratic equations has a solution. Adv. Math. 2022;403:108391. doi: 10.1016/j.aim.2022.108391. 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Conclusions Acknowledgments Author Contributions Institutional Review Board Statement Data Availability Statement Conflicts of Interest Funding Statement Footnotes References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
3320	https://www.droracle.ai/articles/291863/jock-itch-oral-meds	What are the oral medications for jock itch (tinea cruris)? Select Language▼ What are the oral medications for jock itch (tinea cruris)? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: August 29, 2025 • View editorial policy Oral Medications for Tinea Cruris (Jock Itch) Oral terbinafine is the first-line oral medication for tinea cruris (jock itch) when systemic therapy is required, due to its high cure rate, tolerability, and low cost. When to Consider Oral Therapy Oral antifungal medications are indicated for tinea cruris in the following scenarios: Extensive disease Failed topical treatment Immunocompromised patients Severe or recalcitrant infection First-Line Oral Medication Options Terbinafine Dosage: 250 mg once daily Duration: 1-2 weeks Advantages: High cure rate, well-tolerated, cost-effective 1 Fluconazole Dosage options: 50-100 mg daily for 2-3 weeks 150 mg once weekly for 2-4 weeks Advantages: Convenient weekly dosing option, good penetration into stratum corneum 2, 3 Itraconazole Dosage options: 100 mg daily for 2 weeks 200 mg daily for 7 days Advantages: Effective against most dermatophytes 2 Griseofulvin FDA-approved specifically for tinea cruris not adequately treated by topical therapy Requires longer treatment duration compared to other options Should be used after confirmation of dermatophyte infection 4 Treatment Algorithm Confirm diagnosis through microscopy (KOH preparation) or fungal culture First attempt topical therapy with terbinafine cream or butenafine cream for 1-2 weeks If topical therapy fails or case is severe/extensive: First choice: Oral terbinafine 250 mg daily for 1-2 weeks Alternative: Fluconazole 150 mg weekly for 2-4 weeks (if terbinafine contraindicated) Alternative: Itraconazole 100 mg daily for 2 weeks (if both above contraindicated) Important Considerations Medication interactions: Azoles (fluconazole, itraconazole) have more drug interactions than terbinafine Liver function: Monitor liver function tests when using oral antifungals for extended periods Compliance: Weekly fluconazole may improve compliance compared to daily medications Cost: Terbinafine is generally less expensive than itraconazole Prevention of Recurrence After successful treatment: Keep the groin area clean and dry Wear loose-fitting cotton underwear Change clothes after sweating Use antifungal powders prophylactically in susceptible individuals Treat concurrent tinea pedis if present (common source of reinfection) Cautions Oral antifungals may cause hepatotoxicity, especially with prolonged use Confirm diagnosis before initiating oral therapy, as other conditions (eczema, psoriasis, bacterial infections) can mimic tinea cruris Oral antifungals are not justified for minor infections that will respond to topical agents alone 4 Remember that while topical antifungals are first-line therapy for most cases of tinea cruris, oral medications provide a valuable option for extensive, resistant, or recurrent infections. References 1 Research Diagnosis and management of tinea infections. American family physician, 2014 2 Research Oral therapy of common superficial fungal infections of the skin. Journal of the American Academy of Dermatology, 1999 3 Research Fluconazole in the treatment of tinea corporis and tinea cruris. Dermatology (Basel, Switzerland), 1998 4 Drug Official FDA Drug Label For griseofulvin (PO) FDA, 2025 Related Questions What is the treatment for jock itch (tinea cruris)?What is the recommended treatment for jock itch (tinea cruris)?What is the best treatment for tinea cruris (jock itch)?What is the treatment for jock itch (tinea cruris)?What is the best antifungal powder for treating tinea cruris (jock itch)?What initial labs are recommended for a 14-week pregnant patient?What hazard ratio (HR) and 95% confidence interval (CI) supports the conclusion that albiglutide is noninferior in efficacy to insulin lispro for type 2 diabetes mellitus treatment?What hazard ratio (HR) and 95% confidence interval (CI) supports the conclusion that albiglutide is noninferior in efficacy to insulin lispro for type 2 diabetes mellitus treatment?Is fosnetupitant (Netupitant) as effective as fosaprepitant (Aprepitant) for preventing nausea and vomiting from highly emetogenic chemotherapy?What is the appropriate management for a patient with a cholecystostomy tube for acute cholecystitis?Is the new oral antibiotic noninferior to the standard IV treatment for community-acquired bacterial pneumonia? 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3321	https://www.skuola.net/fisica/elettricita-magnetismo/esperimenti-faraday.html	Esperimenti di Faraday Trova un tutor esperto su questo argomento Chiedi Tutor AI Invia appunti Accedi AppuntiAppunti MedieItalianoMatematicaStoriaScienzeGeografiaEducazione CivicaEducazione TecnicaEducazione ArtisticaMusicaInglese Appunti SuperioriItalianoTemi SvoltiRecensioni LibriDanteManzoniStoriaFilosofiaMatematicaFisicaVersioni LatinoVersioni Greco Appunti UniversitàAppunti per EsameAppunti per AteneoAppunti per FacoltàAppunti per Professore NotizieNotizie ScuolaInchiesteMetodo Studio ScolasticoPCTOManifestazioni e ScioperiDiritti degli StudentiBack to SchoolLibri Usati Notizie UniversitàConsigli per fuorisedeTutorial UniversitàMetodo Studio UniversitarioTesi di LaureaErasmus TrendsTech e SocialSpettacoloLove&SexPatenteViaggiQuizCuriosità Lavoro Skuola TV GuideEsami di Terza MediaGuide e notizie Terza MediaTesine di Terza Media Orientamento Superiori Prove INVALSI MaturitàGuide e Consigli MaturitàCommissioni MaturitaMaterie MaturitàPrima Prova MaturitàSeconda Prova Maturità100 giorni MaturitàTesine MaturitàCollegamenti Test d'IngressoTest MedicinaTest VeterinariaTest IngegneriaTest ArchitetturaTest Professioni SanitarieTest Scienze FormazioneTest EconomiaAltri Test d'Ingresso Orientamento UniversitarioOpen DayClassifica Università Italiane Istituti Tecnici Superiori Formazione ForumForum MedieMatematicaEsame Terza MediaOrientamento Superiori Forum SuperioriItalianoMatematicaFisicaLatinoGrecoIngleseFranceseChimicaMaturitàAltre Materie Forum UniversitàMatematicaTest di AmmissioneDiscussioni GeneraliUniversità di MilanoUniversità di RomaUniversità di NapoliUniversità di BolognaUniversità di GenovaUniversità di CataniaUniversità di Cagliari Forum GeneraleDiscussioni generaliOff-TopicI ProfessoriDiritti StudentiAmore & Co.Cinema & TvSportVideogiochi Matematicamente Ripetizioni SEGUICI SU Invia appunti Appunti Fisica Elettricità e magnetismo Esperimenti di Faraday Esperimenti di Faraday Appunto di fisica che spiega che se una corrente elettrica genera un campo magnetico, anche un campo magnetico genera una corrente. Gli esperimenti di Faraday che lo dimostrano di Langello Ominide 1 min Vota 3/5 Stampa Segnala Condividi Chiedi Tutor AI Concetti Chiave Una corrente elettrica può stabilire un campo magnetico, e viceversa un campo magnetico può indurre una corrente. Il movimento relativo tra un magnete e una spira genera una corrente indotta nel circuito, che scompare quando il movimento cessa. La direzione della corrente indotta dipende dal movimento del polo nord o sud del magnete rispetto alla spira. Un movimento più rapido tra il magnete e la spira produce una corrente indotta più intensa. Faraday ha scoperto che una variazione nel campo magnetico che attraversa una spira induce corrente e forza elettromotrice (fem). Come siamo venuti a conoscenza che una corrente elettrica stabilisce un campo magnetico, si può dimostrare che un campo magnetico può stabilire una corrente. Primo esperimento Abbiamo una spira collegata ad un amperometro o galvanometro. Se si muove un magnete vicino la spira nel circuito si instaura improvvisamente una corrente: Si genera corrente solo se la spira e il magnete sono in moto relativo. La corrente scompare se moto cessa. Un movimento più veloce produce una corrente più intensa. Avvicinando il polo nord alla spira si provoca una corrente in senso, poniamo orario, mentre allontanandolo la corrente circola in senso antiorario. Se si tratta di polo sud la situazione si inverte. La corrente che compare nella spira è detta corrente indotta e il lavoro svolto per unità di carica è detta fem indotta. La generazione di corrente e fem è detta induzione. Secondo esperimento Abbiamo due spire (una collegata ad un amperometro e l’altra collegata ad una batteria) non collegate tra loro. Se si aziona l’interruttore della batteria della seconda spira, viene a crearsi per pochi istanti una corrente indotta nell’altra. Se si spegne l’interruttore si ricrea una corrente indotta. Faraday capì che la corrente e la fem vengono indotte nel circuito quando varia la quantità di campo magnetico che attraversa la spira e quindi la densità delle linee di campo. Appunti correlati #### Esperimenti di Oersted, Faraday e Ampere #### Esperimento di Faraday Neumann #### Prima legge di Faraday #### Legge di Faraday-Neumann Recensioni 3/5 1 recensione 5 stelle 4 stelle 3 stelle 2 stelle 1 stella 0 0 1 0 0 Ti è piaciuto questo appunto? Si è verificato un errore durante l'invio della tua recensione Simonediroma 17 Settembre 2016 Mostra recensioni (x) Domande e risposte Hai bisogno di aiuto? 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3322	https://www.doubtnut.com/qna/11445774	A glass bulb contains air and mercury. What fraction of the bulb must be occupied by mercury if the volume of air in the bulb is to remain constant at all temperatures? The coefficient of linear expansion of glass is 9×10−6/K. 320 32 520 12 The correct Answer is:A Volume of the space above the mercury will remain constant if increase in volume of the vessel if equal to the increase in the volume of the mercury for any increase in temperature. Let V' be the volume of mercury and V the volume of the vessel. Then V×27×10−6×ΔT=V'×1.8×10−4ΔT or V'V=27×10−61.8×10−4=320 Topper's Solved these Questions Explore 18 Videos Explore 25 Videos Explore 23 Videos Explore 20 Videos Explore 1 Video Similar Questions A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass=9×10−6/∘C and the coefficient of volume expansion of Hg=1.8×10−4/∘C ? A glass bulb contains air and mercury. If the volume of air in bulb is to remain constant at all temperature then the fraction of the volume of the bulb that must be occupied by mercury is x20. Find out the value of 'x' (cofficient of linear epansion of glass 9×10−6K−1, coefficient of volume expansion of mercury is 1.8×10−4K−1) Knowledge Check A glass container of volume V contains some mercury. The coefficients of cubical expansion for glass and mercury are γg and γm , respectively. If the volume of air inside the flask remains constant at different temperature, then the volume of mercury in the flask is A two litre glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains the same. The volume of the mercury inside the flask is (αfor glass =9×10−6/∘C,γ for mercury 1.8×10−4/∘C) A one litre flask contains certain quantity of mercury. If the volume of air inside the flask remains the same at all temperatures then the volume of mercury in the flask is (volume expansion coefficient of mercury is 20 times that of flask) A flask of one liter volume contains some mercury . It is found that volume of air inside , the flask remains constant at all temperatures .If 'a' of glass is 9×10−6 SI units ,coefficient of real expansion of mercury is 180×10−6 SI units volume of mercury in the flask is The volume of a glass vessel is 1000 cc at 20C. What volume of mercury chould be poured into it at this temperature so that the volume of the remaining space doed not change with temperature? Coefficients of cubical expansion of mercury and glass are 1.8 xx 10^(-4) o^C(-1) respectively. A glass flask with volume 200 cm3 is filled to the brim with mercury at 20∘C. How much mercury overflows when the temperature of the system is raised to 100∘C? The coefficient of linear expansion of the glass is 0.40×10−5K−1. Cubical expansion of mercury =18×10−5K−1. In observing thereal expansion of a liquid a certain part of glass vessel is to be filled with a liquid of volume expansion γ=18×10−5 in order to exclude the effect of the change in remaining volume of the vessel during heating. What fraction of the vessel should be filled wiht the liquid? The coefficient of linear expansion of glass is =9.0×10−6. In a cylidrical glass container a solid silica is placed vertically at its bottom and remaining space is filled with mercury upto the top level of the silica. The volume of silica remains unchanged with variation in temperature. The coefficient of cubical expansion of mercury is γ and coefficient of linear expansion of glass is α. If the top surface of silica and mercury level remain at the same level wiht the variation in temperature then the ratio of volume of silica tothe volume of mercury is equal: CENGAGE PHYSICS-CALORIMETRY-Exercise 1.2 Two large holes are cut in a metal sheet. If this is heated, will thei... In the above question, will the distance between the holes increase or... A long metal rod is bent to form a ring with a small gap if this is he... Two iron spheres of the same diameter are heated to the same temperatu... A steel rod is 3.000 cm at 25^@C. A brass ring has an interior diamete... A clock with a metallic pendulum gains 5 s each day at a temperature o... The design of some physical instrument requires that there be a consta... A metal rod of 30 cm length expands by 0.075 cm when its temperature i... A brass scale is graduated at 10^@C. What is the true length of a zinc... A long horizontal glass capillary tube open at both ends contains a me... A mercury in glass thermometer has a stem of internal diameter 0.06 cm... A sphete of deamrter 7.0 cm and mass 266.5 g float in a bath of liquid... A mercury thermometer is to be made with glass tubing of internal bore... On a Celsius thermometer the distance between the readings 0^@C and 10... The height of a mercury column measured with a brass scale, which is c... A glass bulb contains air and mercury. What fraction of the bulb must ... Two rods of equal cross sections, one of copper and the other of steel... A glass vessel measures exactly 10 cm xx 10 cm xx 10 cm at 0^@C. It is... A metal ball immersed in alcohol weights W1 at 0^@C and W2 at 50^@C. T... In figure which strip (brass or steel) has higher coefficient of linea... 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3323	https://www.sciencedirect.com/science/article/pii/S1751616125002553	Skip to article My account Sign in View PDF Journal of the Mechanical Behavior of Biomedical Materials Volume 171, November 2025, 107139 The effect of presence and location of microcalcifications on atherosclerotic plaque rupture: A tissue-engineering approach Author links open overlay panel, , , , , , rights and content Under a Creative Commons license Open access Highlights ¢ We developed cap analogs with hydroxyapatite (HA) to replicate microcalcifications. ¢ HA clusters, especially when located luminally, increased local collagen dispersion. ¢ The presence of HA clusters reduced failure tensile stress and strain of the analogs. ¢ Rupture initiation location shifted toward the HA cluster sites. ¢ Rupture initiation was consistently found in high-strain regions. Abstract Rupture of the cap of an atherosclerotic plaque can trigger thrombotic cardiovascular events. It has been suggested, through computational models, that the presence and specific location of microcalcifications in the atherosclerotic cap can increase the risk of cap rupture. However, the experimental confirmation of this hypothesis is lacking. In this study, we investigated how the presence and location of microcalcifications, relative to the lumen, influence (local) mechanics and rupture behavior of atherosclerotic plaque caps. Using tissue-engineered fibrous cap analogs with hydroxyapatite (HA) clusters to mimic calcifications in human plaque caps, we replicated the microcalcification distribution observed in human carotid plaques, as identified by our histological analysis. The analogs were imaged using multiphoton microscopy with second-harmonic generation to assess local collagen fiber orientation and dispersion. Subsequently, they underwent uniaxial tensile testing to failure, during which local strain and failure characteristics were analyzed. Our results revealed that HA clusters, particularly those in the luminal region, contribute to increased local collagen fiber dispersion. Moreover, the presence of HA clusters reduced both failure tensile stress and strain in the TE cap analogs. Besides, the rupture location shifted toward the site of HA clusters. Additionally, rupture initiation was consistently found in high-strain regions, and in 86 % of the analogs, even at the highest strain location in the sample. Our findings suggest that microcalcification clusters in plaque caps may increase the cap rupture risk and relocate the rupture site. Moreover, local strain measurements can serve as an additional tool for plaque cap rupture risk assessment. Graphical abstract Keywords Atherosclerosis Plaque rupture Microcalcifications Mechanics Collagen Tissue engineering Data availability Data will be made available on request. Cited by (0) 1 : The authors Hanneke Crielaard and Imke Jansen contributed equally to the work. © 2025 The Authors. Published by Elsevier Ltd.
3324	https://cdn1.byjus.com/wp-content/uploads/2020/12/rd-sharma-dec2020-solutions-for-class-10-chapter-9.pdf	R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.1 Page No: 9.5 1. Write the first terms of each of the following sequences whose nth term are: (i) an = 3n + 2 (ii) an = (n – 2)/3 (iii) an = 3n (iv) an = (3n – 2)/ 5 (v) an = (-1)n . 2n (vi) an = n(n – 2)/2 (vii) an = n2 - n + 1 (viii) an = n2 - n + 1 (ix) an = (2n – 3)/ 6 Solutions: (i) an = 3n + 2 Given sequence whose an = 3n + 2 To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 and we get a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4) + 2 = 12 + 2 = 14 a5 = (3 × 5) + 2 = 15 + 2 = 17 ∴ the required first five terms of the sequence whose nth term, an = 3n + 2 are 5, 8, 11, 14, 17. (ii) an = (n – 2)/3 Given sequence whose On putting n = 1, 2, 3, 4, 5 then can get the first five terms ∴ the required first five terms of the sequence whose nth term, (iii) an = 3n Given sequence whose an = 3n To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above a1 = 31 = 3; a2 = 32 = 9; a3 = 27; a4 = 34 = 81; R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a5 = 35 = 243. ∴ the required first five terms of the sequence whose nth term, an = 3n are 3, 9, 27, 81, 243. (iv) an = (3n – 2)/ 5 Given sequence whose To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above And, we get ∴ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5 (v) an = (-1)n2n Given sequence whose an = (-1)n2n To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above. a1 = (-1)1.21 = (-1).2 = -2 a2 = (-1)2.22 = (-1).4 = 4 a3 = (-1)3.23 = (-1).8 = -8 a4 = (-1)4.24 = (-1).16 = 16 a5 = (-1)5.25 = (-1).32 = -32 ∴ the first five terms of the sequence are – 2, 4, - 8, 16, - 32. (vi) an = n(n – 2)/2 The given sequence is, To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5. And, we get R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ∴ the required first five terms are -1/2, 0, 3/2, 4, 15/2 (vii) an = n2 - n + 1 The given sequence whose, an = n2 - n + 1 To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5. And, we get a1 = 12 - 1 + 1 = 1 a2 = 22 - 2 + 1 = 3 a3 = 32 - 3 + 1 = 7 a4 = 42 - 4 + 1 = 13 a5 = 52 - 5 + 1 = 21 ∴ the required first five terms of the sequence are 1, 3, 7, 13, 21. (viii) an = 2n2 - 3n + 1 The given sequence whose an = 2n2 - 3n + 1 To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5. And, we get a1 = 2.12 - 3.1 + 1 = 2 - 3 + 1 = 0 a2 = 2.22 - 3.2 + 1 = 8 - 6 + 1 = 3 a3 = 2.32 - 3.3 + 1 = 18 - 9 + 1 = 10 a4 = 2.42 - 3.4 + 1 = 32 - 12 + 1 = 21 a5 = 2.52 - 3.5 + 1 = 50 - 15 + 1 = 36 ∴ the required first five terms of the sequence are 0, 3, 10, 21, 36. (ix) an = (2n – 3)/ 6 Given sequence whose, To get the first five terms of the sequence we put n = 1, 2, 3, 4, 5. And, we get R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ∴ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.2 Page No: 9.8 1. Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given, an = 5n – 7 Now putting n = 1, 2, 3, 4 we get, a1 = 5(1) – 7 = 5 – 7 = -2 a2 = 5(2) – 7 = 10 – 7 = 3 a3 = 5(3) – 7 = 15 – 7 = 8 a4 = 5(4) – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (-2) = 5 a3 – a2 = 8 – (3) = 5 a4 – a3 = 13 – (8) = 5 Since the difference between the terms is common, we can conclude that the given sequence defined by an = 5n – 7 is an A.P with common difference 5. 2. Show that the sequence defined by an = 3n2 – 5 is not an A.P. Solution: Given, an = 3n2 – 5 Now putting n = 1, 2, 3, 4 we get, a1 = 3(1)2 – 5= 3 – 5 = -2 a2 = 3(2)2 – 5 = 12 – 5 = 7 a3 = 3(3)2 – 5 = 27 – 5 = 22 a4 = 3(4)2 – 5 = 48 – 5 = 43 We can see that, a2 – a1 = 7 – (-2) = 9 a3 – a2 = 22 – 7 = 15 a4 – a3 = 43 – 22 = 21 Since the difference between the terms is not common and varying, we can conclude that the given sequence defined by an = 3n2 – 5 is not an A.P. 3. The general term of a sequence is given by an = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference. Solution: Given, an = -4n + 15 Now putting n = 1, 2, 3, 4 we get, a1 = -4(1) + 15 = -4 + 15 = 11 a2 = -4(2) + 15 = -8 + 15 = 7 a3 = -4(3) + 15 = -12 + 15 = 3 a4 = -4(4) + 15 = -16 + 15 = -1 We can see that, R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a2 – a1 = 7 – (11) = -4 a3 – a2 = 3 – 7 = -4 a4 – a3 = -1 – 3 = -4 Since the difference between the terms is common, we can conclude that the given sequence defined by an = -4n + 15 is an A.P with common difference of -4. Hence, the 15th term will be a15 = -4(15) + 15 = -60 + 15 = -45 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.3 Page No: 9.11 1. For the following arithmetic progressions write the first term a and the common difference d: (i) - 5, -1, 3, 7,… (ii) 1/5, 3/5, 5/5, 7/5,… (iii) 0.3, 0.55, 0.80, 1.05,… (iv) -1.1, – 3.1, – 5.1, – 7.1,… Solution: We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d,…. (i) – 5, –1, 3, 7,… Given arithmetic series is – 5, –1, 3, 7,... c a, a + d, a + 2d + a + 3d,…. Thus, by comparing these two we get, a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7 First term (a) = – 5 By subtracting second and first term, we get (a + d) - (a) = d -1 - (- 5) = d 4 = d ⇒ Common difference (d) = 4. (ii) 1/5, 3/5, 5/5, 7/5, ............. Given arithmetic series is 1/5, 3/5, 5/5, 7/5, ............... It is seen that, it’s of the form of 1/5, 2/5, 5/5, 7/5, ........... a, a + d, a + 2d, a + 3d, Thus, by comparing these two, we get a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5 First term (a) = 1/5 By subtracting first term from second term, we get d = (a + d)-(a) d = 3/5 - 1/5 d = 2/5 ⇒ common difference (d) = 2/5 (iii) 0.3, 0.55, 0.80, 1.05, ............ Given arithmetic series 0.3, 0.55, 0.80, 1.05, .......... It is seen that, it’s of the form of a, a + d, a + 2d, a + 3d, Thus, by comparing we get, a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05 First term (a) = 0.3. By subtracting first term from second term. We get d = (a + d) - (a) d = 0.55 - 0.3 d = 0.25 ⇒ Common difference (d) = 0.25 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions (iv) –1.1, – 3.1, – 5.1, –7.1, ........ General series is –1.1, – 3.1, – 5.1, –7.1, ........ It is seen that, it’s of the form of a, a + d, a + 2d, a + 3d, ........... Thus, by comparing these two, we get a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –71 First term (a) = –1.1 Common difference (d) = (a + d) - (a) = -3.1 – ( – 1.1) ⇒Common difference (d) = – 2 2. Write the arithmetic progression when first term a and common difference d are as follows: (i) a = 4, d = – 3 (ii) a = –1, d = 1/2 (iii) a = –1.5, d = – 0.5 Solution: We know that, if first term (a) = a and common difference = d, then the arithmetic series is: a, a + d, a + 2d, a + 3d, (i) a = 4, d = -3 Given, first term (a) = 4 Common difference (d) = -3 Then arithmetic progression is: a, a + d, a + 2d, a + 3d, ...... ⇒ 4, 4 - 3, a + 2(-3), 4 + 3(-3), ...... ⇒ 4, 1, – 2, – 5, – 8 ........ (ii) a = -1, d = 1/2 Given, first term (a) = -1 Common difference (d) = 1/2 Then arithmetic progression is: a, a + d, a + 2d, a + 3d, ⇒-1, -1 + 1/2, -1, 2½, -1 + 3½, ... ⇒-1, -1/2, 0, 1/2 (iii) a = –1.5, d = – 0.5 Given First term (a) = –1.5 Common difference (d) = – 0.5 Then arithmetic progression is; a, a + d, a + 2d, a + 3d, ...... ⇒-1.5, -1.5, -0.5, –1.5 + 2(– 0.5), –1.5 + 3(– 0.5) ⇒ – 1.5, – 2, – 2.5, – 3, ....... 3. In which of the following situations, the sequence of numbers formed will form an A.P.? (i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre. (ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder. (iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions the end of first year, second year, third year, …, and so on. Solution: (i) Given, Cost of digging a well for the first meter (c1) = Rs.150. And, the cost rises by Rs.20 for each succeeding meter Then, Cost of digging for the second meter (c2) = Rs.150 + Rs 20 = Rs 170 Cost of digging for the third meter (c3) = Rs.170 + Rs 20 = Rs 210 Hence, its clearly seen that the costs of digging a well for different lengths are 150, 170, 190, 210, .... Evidently, this series is in A∙P. With first term (a) = 150, common difference (d) = 20 (ii) Given, Let the initial volume of air in a cylinder be V liters each time 3th/4 of air in a remaining i.e 1 -1/4 First time, the air in cylinder is V. Second time, the air in cylinder is 3/4 V. Third time, the air in cylinder is (3/4)2 V. Thus, series is V, 3/4 V, (3/4)2 V,(3/4)3 V, .... Hence, the above series is not a A.P. (iii) Given, Divya deposited Rs 1000 at compound interest of 10% p.a So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100 And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210 And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331 Cleary, these amounts 1100, 1210 and 1331 are not in an A.P since the difference between them is not the same. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.4 Page No: 9.24 1. Find: (i) 10th tent of the AP 1, 4, 7, 10.... (ii) 18th term of the AP √2, 3√2, 5√2, ……. (iii) nth term of the AP 13, 8, 3, -2, .......... (iv) 10th term of the AP -40, -15, 10, 35, ............. (v) 8th term of the AP 11, 104, 91, 78, ............... (vi) 11th tenor of the AP 10.0, 10.5, 11.0, 11.2, .............. (vii) 9th term of the AP 3/4, 5/4, 7/4 + 9/4, ........... Solution: (i) Given A.P. is 1, 4, 7, 10, .......... First term (a) = 1 Common difference (d) = Second term - First term = 4 - 1 = 3. We know that, nth term in an A.P = a + (n - 1)d Then, 10th term in the A.P is 1 + (10 - 1)3 = 1 + 9x3 = 1 + 27 = 28 ∴ 10th term of A. P. is 28 (ii) Given A.P. is √2, 3√2, 5√2, ……. First term (a) = √2 Common difference = Second term – First term = 3√2 - √2 ⇒ d = 2√2 We know that, nth term in an A. P. = a + (n - 1)d Then, 18th term of A. P. = √2 + (18 - 1)2√2 = √2 + 17.2√2 = √2 (1+34) = 35√2 ∴ 18th term of A. P. is 35√2 (iii) Given A. P. is 13, 8, 3, - 2, ............ First term (a) = 13 Common difference (d) = Second term first term = 8 - 13 = – 5 We know that, nth term of an A.P. an = a +(n - 1)d = 13 + (n - 1) - 5 = 13 - 5n + 5 ∴ nth term of the A.P is an = 18 - 5n (iv) Given A. P. is - 40, -15, 10, 35, .......... First term (a) = -40 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Common difference (d) = Second term - fast term = -15 - (- 40) = 40 - 15 = 25 We know that, nth term of an A.P. an = a + (n - 1)d Then, 10th term of A. P. a10 = -40 + (10 - 1)25 = – 40 + 9.25 = – 40 + 225 = 185 ∴ 10th term of the A. P. is 185 (v) Given sequence is 117, 104, 91, 78, ............. First term (a) = 117 Common difference (d) = Second term - first term = 104 - 117 = – 13 We know that, nth term = a + (n - 1)d Then, 8th term = a + (8 - 1)d = 117 + 7(-13) = 117 - 91 = 26 ∴ 8th term of the A. P. is 26 (vi) Given A. P is 10.0, 10.5, 11.0, 11.5, First term (a) = 10.0 Common difference (d) = Second term - first term = 10.5 - 10.0 = 0.5 We know that, nth term an = a + (n - 1)d Then, 11th term a11 = 10.0 + (11 - 1)0.5 = 10.0 + 10 x 0.5 = 10.0 + 5 =15.0 ∴ 11th term of the A. P. is 15.0 (vii) Given A. P is 3/4, 5/4, 7/4, 9/4, ............ First term (a) = 3/4 Common difference (d) = Second term - first term = 5/4 - 3/4 = 2/4 We know that, nth term an = a + (n - 1)d Then, 9th term a9 = a + (9 - 1)d R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ∴ 9th term of the A. P. is 19/4. 2.(i) Which term of the AP 3, 8, 13, .... is 248? (ii) Which term of the AP 84, 80, 76, ... is 0? (iii) Which term of the AP 4. 9, 14, .... is 254? (iv) Which term of the AP 21. 42, 63, 84, ... is 420? (v) Which term of the AP 121, 117. 113, ... is its first negative term? Solution: (i) Given A.P. is 3, 8, 13, ........... First term (a) = 3 Common difference (d) = Second term - first term = 8 - 3 = 5 We know that, nth term (an) = a + (n - 1)d And, given nth term an = 248 248 = 3+(n - 1)5 248 = -2 + 5n 5n = 250 n =250/5 = 50 ∴ 50th term in the A.P is 248. (ii) Given A. P is 84, 80, 76, ............ First term (a) = 84 Common difference (d) = a2 - a = 80 - 84 = – 4 We know that, nth term (an) = a +(n - 1)d And, given nth term is 0 0 = 84 + (n - 1) - 4 84 = +4(n - 1) n - 1 = 84/4 = 21 n = 21 + 1 = 22 ∴ 22nd term in the A.P is 0. (iii) Given A. P 4, 9, 14, ............ R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions First term (a) = 4 Common difference (d) = a2 – a1 = 9 - 4 = 5 We know that, nth term (an) = a + (n - 1)d And, given nth term is 254 4 + (n - 1)5 = 254 (n - 1)∙5 = 250 n - 1 = 250/5 = 50 n = 51 ∴ 51th term in the A.P is 254. (iv) Given A. P 21, 42, 63, 84, ......... a = 21, d = a2 – a1 = 42 - 21 = 21 We know that, nth term (an) = a +(n - 1)d And, given nth term = 420 21 + (n - 1)21 = 420 (n - 1)21 = 399 n - 1 = 399/21 = 19 n = 20 ∴ 20th term is 420. (v) Given A.P is 121, 117, 113, ........... Fiat term (a) = 121 Common difference (d) = 117 - 121 = - 4 We know that, nth term an = a + (n - 1)d And, for some nth term is negative i.e., an < 0 121 + (n - 1) - 4 < 0 121 + 4 - 4n < 0 125 - 4n < 0 4n > 125 n > 125/4 n > 31.25 The integer which comes after 31.25 is 32. ∴ 32nd term in the A.P will be the first negative term. 3.(i) Is 68 a term of the A.P. 7, 10, 13,… ? (ii) Is 302 a term of the A.P. 3, 8, 13, …. ? (iii) Is -150 a term of the A.P. 11, 8, 5, 2, … ? Solutions: (i) Given, A.P. 7, 10, 13,… Here, a = 7 and d = a2 – a1 = 10 – 7 = 3 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions We know that, nth term an = a + (n - 1)d Required to check nth term an = 68 a + (n - 1)d = 68 7 + (n - 1)3 = 68 7 + 3n – 3 = 68 3n + 4 = 68 3n = 64 ⇒ n = 64/3, which is not a whole number. Therefore, 68 is not a term in the A.P. (ii) Given, A.P. 3, 8, 13,… Here, a = 3 and d = a2 – a1 = 8 – 3 = 5 We know that, nth term an = a + (n - 1)d Required to check nth term an = 302 a + (n - 1)d = 302 3 + (n - 1)5 = 302 3 + 5n – 5 = 302 5n - 2 = 302 5n = 304 ⇒ n = 304/5, which is not a whole number. Therefore, 302 is not a term in the A.P. (iii) Given, A.P. 11, 8, 5, 2, … Here, a = 11 and d = a2 – a1 = 8 – 11 = -3 We know that, nth term an = a + (n - 1)d Required to check nth term an = -150 a + (n - 1)d = -150 11 + (n - 1)(-3) = -150 11 – 3n + 3 = -150 3n = 150 + 14 3n = 164 ⇒ n = 164/3, which is not a whole number. Therefore, -150 is not a term in the A.P. 4. How many terms are there in the A.P.? (i) 7, 10, 13, ….., 43 (ii) -1, -5/6, -2/3, -1/2, … , 10/3 (iii) 7, 13, 19, …, 205 (iv) 18, 15½, 13, …., -47 Solution: (i) Given, A.P. 7, 10, 13, ….., 43 Here, a = 7 and d = a2 – a1 = 10 – 7 = 3 We know that, nth term an = a + (n - 1)d And, given nth term an = 43 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a + (n - 1)d = 43 7 + (n - 1)(3) = 43 7 + 3n – 3 = 43 3n = 43 – 4 3n = 39 ⇒ n = 13 Therefore, there are 13 terms in the given A.P. (ii) Given, A.P. -1, -5/6, -2/3, -1/2, … , 10/3 Here, a = -1 and d = a2 – a1 = -5/6 – (-1) = 1/6 We know that, nth term an = a + (n - 1)d And, given nth term an = 10/3 a + (n - 1)d = 10/3 -1 + (n - 1)(1/6) = 10/3 -1 + n/6 – 1/6 = 10/3 n/6 = 10/3 + 1 + 1/6 n/6 = (20 + 6 + 1)/6 n = (20 + 6 + 1) ⇒ n = 27 Therefore, there are 27 terms in the given A.P. (iii) Given, A.P. 7, 13, 19, …, 205 Here, a = 7 and d = a2 – a1 = 13 – 7 = 6 We know that, nth term an = a + (n - 1)d And, given nth term an = 205 a + (n - 1)d = 205 7 + (n - 1)(6) = 205 7 + 6n – 6 = 205 6n = 205 – 1 n = 204/6 ⇒ n = 34 Therefore, there are 34 terms in the given A.P. (iv) Given, A.P. 18, 15½, 13, …., -47 Here, a = 7 and d = a2 – a1 = 15½ – 18 = 5/2 We know that, nth term an = a + (n - 1)d And, given nth term an = -47 a + (n - 1)d = 43 18 + (n - 1)(-5/2) = -47 18 – 5n/2 + 5/2 = -47 36 – 5n + 5 = -94 5n = 94 + 36 + 5 5n = 135 ⇒ n = 27 Therefore, there are 27 terms in the given A.P. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 5. The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms. Solution: Given, a = 5 and d = 3 We know that, nth term an = a + (n - 1)d So, for the given A.P. an = 5 + (n - 1)3 = 3n + 2 Also given, last term = 80 ⇒ 3n + 2 = 80 3n = 78 n = 78/3 = 26 Therefore, there are 26 terms in the A.P. 6. The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term. Solution: Given, a6 = 19 and a17 = 41 We know that, nth term an = a + (n - 1)d So, a6 = a + (6-1)d ⇒ a + 5d = 19 …… (i) Similarity, a17 = a + (17 - 1)d ⇒ a + 16d = 41 …… (ii) Solving (i) and (ii), (ii) – (i) ⇒ a + 16d – (a + 5d) = 41 – 19 11d = 22 ⇒ d = 2 Using d in (i), we get a + 5(2) = 19 a = 19 – 10 = 9 Now, the 40th term is given by a40 = 9 + (40 - 1)2 = 9 + 78 = 87 Therefore the 40th term is 87. 7. If 9th term of an A.P. is zero, prove its 29th term is double the 19th term. Solution: Given, a9 = 0 We know that, nth term an = a + (n - 1)d So, a + (9 - 1)d = 0 ⇒ a + 8d = 0 ……(i) Now, R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 29th term is given by a29 = a + (29 - 1)d ⇒ a29 = a + 28d And, a29 = (a + 8d) + 20d [using (i)] ⇒ a29 = 20d ….. (ii) Similarly, 19th term is given by a19 = a + (19 - 1)d ⇒ a19 = a + 18d And, a19 = (a + 8d) + 10d [using (i)] ⇒ a19 = 10d …..(iii) On comparing (ii) and (iii), it’s clearly seen that a29 = 2(a19) Therefore, 29th term is double the 19th term. 8. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero. Solution: Given, 10 times the 10th term of an A.P. is equal to 15 times the 15th term. We know that, nth term an = a + (n - 1)d ⇒ 10(a10) = 15(a15) 10(a + (10 - 1)d) = 15(a + (15 - 1)d) 10(a + 9d) = 15(a + 14d) 10a + 90d = 15a + 210d 5a + 120d = 0 5(a + 24d) = 0 a + 24d = 0 a + (25 – 1)d = 0 ⇒ a25 = 0 Therefore, the 25th term of the A.P. is zero. 9. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term. Solution: Given, A10 = 41 and a18 = 73 We know that, nth term an = a + (n - 1)d So, a10 = a + (10 - 1)d ⇒ a + 9d = 41 …… (i) Similarity, a18 = a + (18 - 1)d ⇒ a + 17d = 73 …… (ii) Solving (i) and (ii), R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions (ii) – (i) ⇒ a + 17d – (a + 9d) = 73 – 41 8d = 32 ⇒ d = 4 Using d in (i), we get a + 9(4) = 41 a = 41 – 36 = 5 Now, the 26th term is given by a26 = 5 + (26 - 1)4 = 5 + 100 = 105 Therefore the 26th term is 105. 10. In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term. Solution: Given, 24th term is twice the 10th term. We know that, nth term an = a + (n - 1)d ⇒ a24 = 2(a10) a + (24 - 1)d = 2(a + (10 - 1)d) a + 23d = 2(a + 9d) a + 23d = 2a + 18d a = 5d …. (1) Now, the 72nd term can be expressed as a72 = a + (72 - 1)d = a + 71d = a + 5d + 66d = a + a + 66d [using (1)] = 2(a + 33d) = 2(a + (34 - 1)d) = 2(a34) ⇒ a72 = 2(a34) Hence, the 72nd term is twice the 34th term of the given A.P. 11. The 26th, 11th and the last term of an A.P. are 0, 3 and -1/5, respectively. Find the common difference and the number of terms. Solution: Given, a26 = 0 , a11 = 3 and an (last term) = -1/5 of an A.P. We know that, nth term an = a + (n - 1)d Then, a26 = a + (26 - 1)d ⇒ a + 25d = 0 …..(1) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions And, a11 = a + (11 - 1)d ⇒ a + 10d = 3 …… (2) Solving (1) and (2), (1) – (2) ⇒ a + 25d – (a + 10d) = 0 – 3 15d = -3 ⇒ d = -1/5 Using d in (1), we get a + 25(-1/5) = 0 a = 5 Now, given that the last term an = -1/5 ⇒ 5 + (n - 1)(-1/5) = -1/5 5 + -n/5 + 1/5 = -1/5 25 – n + 1 = -1 n = 27 Therefore, the A.P has 27 terms and its common difference is -1/5. 12. If the nth term of the A.P. 9, 7, 5, …. is same as the nth term of the A.P. 15, 12, 9, … find n. Solution: Given, A.P1 = 9, 7, 5, …. and A.P2 = 15, 12, 9, … And, we know that, nth term an = a + (n - 1)d For A.P1, a = 9, d = Second term – first term = 9 – 7 = -2 And, its nth term an = 9 + (n - 1)(-2) = 9 – 2n + 2 an = 11 – 2n …..(i) Similarly, for A.P2 a = 15, d = Second term – first term = 12 – 15 = -3 And, its nth term an = 15 + (n - 1)(-3) = 15 – 3n + 3 an = 18 - 3n …..(ii) According to the question, its given that nth term of the A.P1 = nth term of the A.P2 ⇒ 11 – 2n = 18 - 3n n = 7 Therefore, the 7th term of the both the A.Ps are equal. 13. Find the 12th term from the end of the following arithmetic progressions: (i) 3, 5, 7, 9, …. 201 (ii) 3,8,13, … ,253 (iii) 1, 4, 7, 10, … ,88 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Solution: In order the find the 12th term for the end of an A.P. which has n terms, its done by simply finding the ((n -12) + 1)th of the A.P And we know, nth term an = a + (n - 1)d (i) Given A.P = 3, 5, 7, 9, …. 201 Here, a = 3 and d = (5 - 3) = 2 Now, find the number of terms when the last term is known i.e, 201 an = 3 + (n - 1)2 = 201 3 + 2n – 2 = 201 2n = 200 n = 100 Hence, the A.P has 100 terms. So, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term. ⇒ a89 = 3 + (89 - 1)2 = 3 + 88(2) = 3 + 176 = 179 Therefore, the 12th term from the end of the A.P is 179. (ii) Given A.P = 3,8,13, … ,253 Here, a = 3 and d = (8 - 3) = 5 Now, find the number of terms when the last term is known i.e, 253 an = 3 + (n - 1)5 = 253 3 + 5n – 5 = 253 5n = 253 + 2 = 255 n = 255/5 n = 51 Hence, the A.P has 51 terms. So, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term. ⇒ a40 = 3 + (40 - 1)5 = 3 + 39(5) = 3 + 195 = 198 Therefore, the 12th term from the end of the A.P is 198. (iii) Given A.P = 1, 4, 7, 10, … ,88 Here, a = 1 and d = (4 - 1) = 3 Now, find the number of terms when the last term is known i.e, 88 an = 1 + (n - 1)3 = 88 1 + 3n – 3 = 88 3n = 90 n = 30 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Hence, the A.P has 30 terms. So, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term. ⇒ a89 = 1 + (19 - 1)3 = 1 + 18(3) = 1 + 54 = 55 Therefore, the 12th term from the end of the A.P is 55. 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Let’s consider the first term and the common difference of the A.P to be a and d respectively. Then, we know that an = a + (n - 1)d Given conditions, 4th term of an A.P. is three times the first Expressing this by equation we have, ⇒ a4 = 3(a) a + (4 - 1)d = 3a 3d = 2a ⇒ a = 3d/2…….(i) And, 7th term exceeds twice the third term by 1 ⇒ a7 = 2(a3) + 1 a + (7 – 1)d = 2(a + (3–1)d) + 1 a + 6d = 2a + 4d + 1 a – 2d +1 = 0 ….. (ii) Using (i) in (ii), we have 3d/2 – 2d + 1 = 0 3d – 4d + 2 = 0 d = 2 So, putting d = 2 in (i), we get a ⇒ a = 3 Therefore, the first term is 3 and the common difference is 2. 15. Find the second term and the nth term of an A.P. whose 6th term is 12 and the 8th term is 22. Solution: Given, in an A.P a6 = 12 and a8 = 22 We know that an = a + (n - 1)d So, a6 = a + (6-1)d = a + 5d = 12 …. (i) And, R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a8 = a + (8-1)d = a + 7d = 22 ……. (ii) Solving (i) and (ii), we have (ii) - (i) ⇒ a + 7d – (a + 5d) = 22 – 12 2d = 10 d = 5 Putting d in (i) we get, a + 5(5) = 12 a = 12 – 25 a = -13 Thus, for the A.P: a = -13 and d = 5 So, the nth term is given by an = a + (n-1)d an = -13 + (n-1)5 = -13 + 5n – 5 ⇒ an = 5n – 18 Hence, the second term is given by a2 = 5(2) – 18 = 10 – 18 = -8 16. How many numbers of two digit are divisible by 3? Solution: The first 2 digit number divisible by 3 is 12. And, the last 2 digit number divisible by 3 is 99. So, this forms an A.P. 12, 15, 18, 21, …. , 99 Where, a = 12 and d = 3 Finding the number of terms in this A.P ⇒ 99 = 12 + (n-1)3 99 = 12 + 3n – 3 90 = 3n n = 90/3 = 30 Therefore, there are 30 two digit numbers divisible by 3. 17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term. Solution: Given, an A.P of 60 terms And, a = 7 and a60 = 125 We know that an = a + (n - 1)d ⇒ a60 = 7 + (60 - 1)d = 125 7 + 59d = 125 59d = 118 d = 2 So, the 32nd term is given by a32 = 7 + (32 -1)2 = 7 + 62 = 69 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ⇒ a32 = 69 18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P. Solution: Given, in an A.P The sum of 4th and 8th terms of an A.P. is 24 ⇒ a4 + a8 = 24 And, we know that an = a + (n - 1)d [a + (4-1)d] + [a + (8-1)d] = 24 2a + 10d = 24 a + 5d = 12 …. (i) Also given that, the sum of the 6th and 10th terms is 34 ⇒ a6 + a10 = 34 [a + 5d] + [a + 9d] = 34 2a + 14d = 34 a + 7d = 17 …… (ii) Subtracting (i) form (ii), we have a + 7d – (a + 5d) = 17 – 12 2d = 5 d = 5/2 Using d in (i) we get, a + 5(5/2) = 12 a = 12 – 25/2 a = -1/2 Therefore, the first term is -1/2 and the common difference is 5/2. 19. The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P. Solution: Given, an A.P whose a = 5 and a100 = -292 We know that an = a + (n - 1)d a100 = 5 + 99d = -292 99d = -297 d = -3 Hence, the 50th term is a50 = a + 49d = 5 + 49(-3) = 5 – 147 = -142 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 20. Find a30 – a20 for the A.P. (i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, …… Solution: We know that an = a + (n - 1)d So, a30 – a20 = (a + 29d) – (a + 19) =10d (i) Given A.P. -9, -14, -19, -24 Here, a = -9 and d = -14 – (-9) = = -14 + 9 = -5 So, a30 – a20 = 10d = 10(-5) = -50 (ii) Given A.P. a, a+d, a+2d, a+3d, …… So, a30 – a20 = (a + 29d) – (a + 19d) =10d 21. Write the expression an – ak for the A.P. a, a+d, a+2d, ….. Hence, find the common difference of the A.P. for which (i) 11th term is 5 and 13th term is 79. (ii) a10 – a5 = 200 (iii) 20th term is 10 more than the 18th term. Solution: Given A.P. a, a+d, a+2d, ….. So, an = a + (n-1)d = a + nd –d And, ak = a + (k-1)d = a + kd – d an - ak = (a + nd – d) – (a + kd – d) = (n – k)d (i) Given 11th term is 5 and 13th term is 79, Here n = 13 and k = 11, a13 – a11 = (13 – 11)d = 2d ⇒79 – 5 = 2d d = 74/2 = 37 (ii) Given, a10 – a5 = 200 ⇒(10 - 5)d = 200 5d = 200 d = 40 (iii) Given, 20th term is 10 more than the 18th term. ⇒a20 – a18 = 10 (20 - 18)d = 10 2d = 10 d = 5 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 22. Find n if the given value of x is the nth term of the given A.P. (i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, …; x = -151 (iii) 5½, 11, 16½, 22, ….; x = 550 (iv) 1, 21/11, 31/11, 41/11, …; x = 171/11 Solution: (i) Given A.P. 25, 50, 75, 100, …… ,1000 Here, a = 25 d = 50 – 25 = 25 Last term (nth term) = 1000 We know that an = a + (n - 1)d ⇒ 1000 = 25 + (n-1)25 1000 = 25 + 25n – 25 n = 1000/25 n = 40 (ii) Given A.P. -1, -3, -5, -7, …., -151 Here, a = -1 d = -3 – (-1) = -2 Last term (nth term) = -151 We know that an = a + (n - 1)d ⇒ -151 = -1 + (n-1)(-2) -151 = -1 - 2n + 2 n = 152/2 n = 76 (iii) Given A.P. 5½, 11, 16½, 22, … , 550 Here, a = 5½ d = 11 – (5½) = 5½ = 11/2 Last term (nth term) = 550 We know that an = a + (n - 1)d ⇒ 550 = 5½ + (n-1)(11/2) 550 x 2 = 11+ 11n – 11 1100 = 11n n = 100 (iv) Given A.P. 1, 21/11, 31/11, 41/11, 171/11 Here, a = 1 d = 21/11 – 1 = 10/11 Last term (nth term) = 171/11 We know that an = a + (n - 1)d ⇒ 171/11 = 1 + (n-1)10/11 171 = 11 + 10n – 10 n = 170/10 n = 17 23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term. Solution: Given, an A.P in which, R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a8 = 1/2(a2) a11 = 1/3(a4) + 1 We know that an = a + (n - 1)d ⇒ a8 = 1/2(a2) a + 7d = 1/2(a + d) 2a + 14d = a + d a + 13d = 0 …… (i) And, a11 = 1/3(a4) + 1 a + 10d = 1/3(a + 3d) + 1 3a + 30d = a + 3d + 3 2a + 27d = 3 …… (ii) Solving (i) and (ii), by (ii) – 2x(i) ⇒ 2a + 27d – 2(a + 13d) = 3 - 0 d = 3 Putting d in (i) we get, a + 13(3) = 0 a = -39 Thus, the 15th term a15 = -39 + 14(3) = -39 + 42 = 3 24. Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12. Solution: Given, in an A.P a3 = 16 and a7 = a5 + 12 We know that an = a + (n - 1)d ⇒ a + 2d = 16…… (i) And, a + 6d = a + 4d + 12 2d = 12 ⇒ d = 6 Using d in (i), we have a + 2(6) = 16 a = 16 – 12 = 4 Hence, the A.P is 4, 10, 16, 22, ……. 25. The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P. Solution: Given, a7 = 32 and a13 = 62 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions From an - ak = (a + nd – d) – (a + kd – d) = (n – k)d a13 – a7 = (13 - 7)d = 62 – 32 = 30 6d = 30 d = 5 Now, a7 = a + (7 - 1)5 = 32 a + 30 = 32 a = 2 Hence, the A.P is 2, 7, 12, 17, …… 26. Which term of the A.P. 3, 10, 17, …. will be 84 more than its 13th term ? Solution: Given, A.P. 3, 10, 17, …. Here, a = 3 and d = 10 – 3 = 7 According the question, an = a13 + 84 Using an = a + (n - 1)d, 3 + (n - 1)7 = 3 + (13 - 1)7 + 84 3 + 7n – 7 = 3 + 84 + 84 7n = 168 + 7 n = 175/7 n = 25 Therefore, it the 25th term which is 84 more than its 13th term. 27. Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms? Solution: Let the two A.Ps be A.P1 and A.P2 For A.P1 the first term = a and the common difference = d And for A.P2 the first term = b and the common difference = d So, from the question we have a100 – b100 = 100 (a + 99d) – (b + 99d) = 100 a - b = 100 Now, the difference between their 1000th terms is, (a + 999d) – (b + 999d) = a – b = 100 Therefore, the difference between their 1000th terms is also 100. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.5 Page No: 9.30 1. Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Given, (8x + 4), (6x – 2) and (2x + 7) are in A.P. So, the common difference between the consecutive terms should be the same. (6x – 2) – (8x + 4) = (2x + 7) – (6x – 2) ⇒ 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2 ⇒ -2x – 6 = -4x + 9 ⇒ -2x + 4x = 9 + 6 ⇒ 2x = 15 Therefore, x = 15/2 2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x. Solution: Given, x + 1, 3x and 4x + 2 are in A.P. So, the common difference between the consecutive terms should be the same. 3x – x – 1 = 4x + 2 – 3x ⇒ 2x – 1 = x + 2 ⇒ 2x – x = 2 + 1 ⇒ x = 3 Therefore, x = 3 3. Show that (a – b)², (a² + b²) and (a + b)² are in A.P. Solution: If (a – b)², (a² + b²) and (a + b)² have to be in A.P. then, It should satisfy the condition, 2b = a + c [for a, b, c are in A.P] Thus, 2 (a² + b²) = (a – b)² + (a + b)² 2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab 2 (a² + b²) = 2a² + 2b² = 2 (a² + b²) LHS = RHS Hence proved. 4. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms. Solution: Let’s consider the three terms of the A.P. to be a – d, a, a + d so, the sum of three terms = 21 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ⇒ a – d + a + a + d = 21 ⇒ 3a = 21 ⇒ a = 7 And, product of the first and 3rd = 2nd term + 6 ⇒ (a – d) (a + d) = a + 6 a² – d² = a + 6 ⇒ (7 )² – d² = 7 + 6 ⇒ 49 – d² = 13 ⇒ d² = 49 – 13 = 36 ⇒ d² = (6)² ⇒ d = 6 Hence, the terms are 7 – 6, 7, 7 + 6 ⇒ 1, 7, 13 5. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers. Solution: Let the three numbers of the A.P. be a – d, a, a + d From the question, Sum of these numbers = 27 a – d + a + a + d = 27 ⇒ 3a = 27 a = 27/3 = 9 Now, product of these numbers = 648 (a - d)(a)(a + d) = 648 a(a2 – d2) = 648 a2 – 648/a = d2 92 – (648/9) = d2 93 – 648 = 9d2 729 – 648 = 9d2 81 = 9d2 d2 = 9 d = 3 or -3 Hence, the terms are 9-3, 9 and 9+3 ⇒ 6, 9, 12 or 12, 9, 6 (for d = -3) 6. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least. Solution: Let’s consider the four terms of the A.P. to be (a – 3d), (a – d), (a + d) and (a + 3d). From the question, Sum of these terms = 50 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50 ⇒ a – 3d + a – d + a + d + a – 3d= 50 ⇒ 4a = 50 ⇒ a = 50/4 = 25/2 And, also given that the greatest number = 4 x least number ⇒ a + 3d = 4 (a – 3d) ⇒ a + 3d = 4a – 12d ⇒ 4a – a = 3d + 12d ⇒3a = 15d ⇒a = 5d Using the value of a in the above equation, we have ⇒25/2 = 5d ⇒ d = 5/2 So, the terms will be: (a – 3d) = (25/2 – 3(5/2)), (a – d) = (25/2 – 5/2), (25/2 + 5/2) and (25/2 + 3(5/2)). ⇒ 5, 10, 15, 20 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Exercise 9.6 Page No: 9.50 1. Find the sum of the following arithmetic progressions: (i) 50, 46, 42, ... to 10 terms (ii) 1, 3, 5, 7, ... to 12 terms (iii) 3, 9/2, 6, 15/2, ... to 25 terms (iv) 41, 36, 31, ... to 12 terms (v) a + b, a - b, a - 3b, ... to 22 terms (vi) (x - y)2, (x2 + y2), (x + y)2, to 22 tams (viii) – 26, – 24, – 22, .... to 36 terms Solution: In an A.P if the first term = a, common difference = d, and if there are n terms. Then, sum of n terms is given by: (i) Given A.P.is 50, 46, 42 to 10 term. First term (a) = 50 Common difference (d) = 46 - 50 = – 4 nth term (n) = 10 = 5{100 - 9.4} = 5{100 - 36} = 5 × 64 ∴ S10 = 320 (ii) Given A.P is, 1, 3, 5, 7, .....to 12 terms. First term (a) = 1 Common difference (d) = 3 - 1 = 2 nth term (n) = 12 = 6 × {2 + 22} = 6.24 ∴ S12 = 144 (iii) Given A.P. is 3, 9/2, 6, 15/2, ... to 25 terms First term (a) = 3 Common difference (d) = 9/2 - 3 = 3/2 Sum of n terms Sn, given n = 25 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions (iv) Given expression is 41, 36, 31, ….. to 12 terms. First term (a) = 41 Common difference (d) = 36 - 41 = -5 Sum of n terms Sn, given n = 12 (v) a + b, a – b, a - 3b, ….. to 22 terms First term (a) = a + b Common difference (d) = a - b - a - b = -2b Sum of n terms Sn = n/2{2a(n - 1). d} Here n = 22 S22 = 22/2{2.(a + b) + (22 - 1). -2b} = 11{2(a + b) - 22b) = 11{2a - 20b} R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions = 22a - 440b ∴S22 = 22a - 440b (vi) (x - y)2,(x2 + y2), (x + y)2,... to n terms First term (a) = (x - y)2 Common difference (d) = x2 + y2 - (x - y)2 = x2 + y2 - (x2 + y2 - 2xy) = x2 + y2 - x2 + y2 + 2xy = 2xy Sum of nth terms Sn = n/2{2a + (n - 1). d} = n/2{2(x - y)2 + (n - 1). 2xy} = n{(x - y)2 + (n - 1)xy} ∴ Sn = n{(x — y)2 + (n — 1). xy) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions (viii) Given expression -26, - 24. -22, to 36 terms First term (a) = -26 Common difference (d) = -24 - (-26) = -24 + 26 = 2 Sum of n terms, Sn = n/2{2a + (n - 1)d) for n = 36 Sn = 36/2{2(-26) + (36 - 1)2} = 18[-52 + 70] = 18x18 = 324 ∴ Sn = 324 2. Find the sum to n terms of the A.P. 5, 2, –1, – 4, –7, ... Solution: Given AP is 5, 2, -1, -4, -7, ..... Here, a = 5, d = 2 - 5 = -3 We know that, Sn = n/2{2a + (n - 1)d} = n/2{2.5 + (n - 1) - 3} = n/2{10 - 3(n - 1)} = n/2{13 - 3n) ∴ Sn = n/2(13 - 3n) 3. Find the sum of n terms of an A.P. whose the terms is given by an = 5 - 6n. Solution: Given nth term of the A.P as an = 5 - 6n Put n = 1, we get a1 = 5 - 6.1 = -1 So, first term (a) = -1 Last term (an) = 5 - 6n = 1 Then, Sn = n/2(-1 + 5 - 6n) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions = n/2(4 - 6n) = n(2 - 3n) 4. Find the sum of last ten terms of the A.P. : 8, 10, 12, 14, .. , 126 Solution: Given A.P. 8, 10, 12, 14, .. , 126 Here, a = 8 , d = 10 – 8 = 2 We know that, an = a + (n - 1)d So, to find the number of terms 126 = 8 + (n - 1)2 126 = 8 + 2n - 2 2n = 120 n = 60 Next, let’s find the 51st term a51 = 8 + 50(2) = 108 So, the sum of last ten terms is the sum of a51 + a52 + a53 + ……. + a60 Here, n = 10, a = 108 and l = 126 S = 10/2 [108 + 126] = 5(234) = 1170 Hence, the sum of last ten terms of the A.P is 1170. 5. Find the sum of first 15 terms of each of the following sequences having nth term as: (i) an = 3 + 4n (ii) bn = 5 + 2n (iii) xn = 6 - n (iv) yn = 9 – 5n Solution: (i) Given an A.P. whose nth term is given by an = 3 + 4n To find the sum of the n terms of the given A.P., using the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given an, we get a = 3 + 4(1) = 3 + 4 = 7 For the last term (l), here n = 15 a15 = 3 + 4(15) = 63 So, Sn = 15(7 + 63)/2 = 15 x 35 = 525 Therefore, the sum of the 15 terms of the given A.P. is S15 = 525 (ii) Given an A.P. whose nth term is given by bn = 5 + 2n R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions To find the sum of the n terms of the given A.P., using the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given bn, we get a = 5 + 2(1) = 5 + 2 = 7 For the last term (l), here n = 15 a15 = 5 + 2(15) = 35 So, Sn = 15(7 + 35)/2 = 15 x 21 = 315 Therefore, the sum of the 15 terms of the given A.P. is S15 = 315 (iii) Given an A.P. whose nth term is given by xn = 6 - n To find the sum of the n terms of the given A.P., using the formula Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given xn, we get a = 6 – 1 = 5 For the last term (l), here n = 15 a15 = 6 – 15 = -9 So, Sn = 15(5 – 9)/2 = 15 x (-2) = -30 Therefore, the sum of the 15 terms of the given A.P. is S15 = -30 (iv) Given an A.P. whose nth term is given by yn = 9 – 5n To find the sum of the n terms of the given A.P., using the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given yn, we get a = 9 - 5(1) = 9 – 5 = 4 For the last term (l), here n = 15 a15 = 9 - 5(15) = -66 So, Sn = 15(4 - 66)/2 = 15 x (-31) = -465 Therefore, the sum of the 15 terms of the given A.P. is S15 = -465 6. Find the sum of first 20 terms the sequence whose nth term is an = An + B. Solution: Given an A.P. whose nth term is given by, an = An + B We need to find the sum of first 20 terms. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions To find the sum of the n terms of the given A.P., we use the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term, Putting n = 1 in the given an, we get a = A(1) + B = A + B For the last term (l), here n = 20 A20 = A(20) + B = 20A + B S20 = 20/2((A + B) + 20A + B) = 10[21A + 2B] = 210A + 20B Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B 7. Find the sum of first 25 terms of an A.P whose nth term is given by an = 2 - 3n. Solution: Given an A.P. whose nth term is given by an = 2 – 3n To find the sum of the n terms of the given A.P., we use the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given an, we get a = 2 - 3(1) = -1 For the last term (l), here n = 25 a25 = 2 - 3(25) = -73 So, Sn = 25(-1 - 73)/2 = 25 x (-37) = -925 Therefore, the sum of the 25 terms of the given A.P. is S25 = -925 8. Find the sum of first 25 terms of an A.P whose nth term is given by an = 7 - 3n. Solution: Given an A.P. whose nth term is given by an = 7 – 3n To find the sum of the n terms of the given A.P., we use the formula, Sn = n(a + l)/ 2 Where, a = the first term l = the last term. Putting n = 1 in the given an, we get a = 7 - 3(1) = 7 – 3 = 4 For the last term (l), here n = 25 a15 = 7 - 3(25) = -68 So, Sn = 25(4 - 68)/2 = 25 x (-32) = -800 Therefore, the sum of the 15 terms of the given A.P. is S25 = -800 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term. Solution: Given the sum of the certain number of terms of an A.P. = 116 We know that, Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So for the given A.P.(25, 22, 19,...) Here we have, the first term (a) = 25 The sum of n terms Sn = 116 Common difference of the A.P. (d) = a2 - a1 = 22 – 25 = -3 Now, substituting values in Sn ⟹ 116 = n/2[2(25) + (n − 1)(−3)] ⟹ (n/2)[50 + (−3n + 3)] = 116 ⟹ (n/2)[53 − 3n] = 116 ⟹ 53n - 3n2 = 116 x 2 Thus, we get the following quadratic equation, 3n2 - 53n + 232 = 0 By factorization method of solving, we have ⟹ 3n2 - 24n - 29n + 232 = 0 ⟹ 3n( n - 8 ) - 29 ( n - 8 ) = 0 ⟹ (3n - 29)( n - 8 ) = 0 So, 3n - 29 = 0 ⟹ n = 29/3 Also, n - 8 = 0 ⟹ n = 8 Since, n cannot be a fraction, so the number of terms is taken as 8. So, the term is: a8 = a1 + 7d = 25 + 7(-3) = 25 - 21 = 4 Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4. 10. (i) How many terms of the sequence 18, 16, 14.... should be taken so that their sum is zero. (ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40? (iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636? (iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? (v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? Solution: (i) Given AP. is 18, 16, 14, ... We know that, Sn = n/2[2a + (n − 1)d] Here, The first term (a) = 18 The sum of n terms (Sn) = 0 (given) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Common difference of the A.P. (d) = a2 - a1 = 16 - 18 = – 2 So, on substituting the values in Sn ⟹ 0 = n/2[2(18) + (n − 1)(−2)] ⟹ 0 = n/2[36 + (−2n + 2)] ⟹ 0 = n/2[38 − 2n] Further, n/2 ⟹ n = 0 Or, 38 - 2n = 0 ⟹ 2n = 38 ⟹ n = 19 Since, the number of terms cannot be zer0, hence the number of terms (n) should be 19. (ii) Given, the first term (a) = -14, Filth term (a5) = 2, Sum of terms (Sn) = 40 of the A.P. If the common difference is taken as d. Then, a5 = a + 4d ⟹ 2 = -14 + 4d ⟹ 2 + 14 = 4d ⟹ 4d = 16 ⟹ d = 4 Next, we know that Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Now, on substituting the values in Sn ⟹ 40 = n/2[2(−14) + (n − 1)(4)] ⟹ 40 = n/2[−28 + (4n − 4)] ⟹ 40 = n/2[−32 + 4n] ⟹ 40(2) = - 32n + 4n2 So, we get the following quadratic equation, 4n2 - 32n - 80 = 0 ⟹ n2 - 8n - 20 = 0 On solving by factorization method, we get n2 - 10n + 2n - 20 = 0 ⟹ n(n - 10) + 2( n - 10 ) = 0 ⟹ (n + 2)(n - 10) = 0 Either, n + 2 = 0 ⟹ n = -2 Or, n - 10 = 0 ⟹ n = 10 Since the number of terms cannot be negative. Therefore, the number of terms (n) is 10. (iii) Given AP is 9, 17, 25,... We know that, Sn = n/2[2a + (n − 1)d] R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Here we have, The first term (a) = 9 and the sum of n terms (Sn) = 636 Common difference of the A.P. (d) = a2 - a1 = 17 - 9 = 8 Substituting the values in Sn, we get ⟹ 636 = n/2[2(9) + (n − 1)(8)] ⟹ 636 = n/2[18 + (8n − 8)] ⟹ 636(2) = (n)[10 + 8n] ⟹ 1271 = 10n + 8n2 Now, we get the following quadratic equation, ⟹ 8n2 + 10n - 1272 = 0 ⟹ 4n2+ 5n - 636 = 0 On solving by factorisation method, we have ⟹ 4n2 - 48n + 53n - 636 = 0 ⟹ 4n(n - 12) + 53(n - 12) = 0 ⟹ (4n + 53)(n - 12) = 0 Either 4n + 53 = 0 ⟹ n = -53/4 Or, n - 12 = 0 ⟹ n = 12 Since, the number of terms cannot be a fraction. Therefore, the number of terms (n) is 12. (iv) Given A.P. is 63, 60, 57,... We know that, Sn = n/2[2a + (n − 1)d] Here we have, the first term (a) = 63 The sum of n terms (Sn) = 693 Common difference of the A.P. (d) = a2 - a1 = 60 - 63 = –3 On substituting the values in Sn we get ⟹ 693 = n/2[2(63) + (n − 1)(−3)] ⟹ 693 = n/2[126+(−3n + 3)] ⟹ 693 = n/2[129 − 3n] ⟹ 693(2) = 129n - 3n2 Now, we get the following quadratic equation. ⟹ 3n2 - 129n + 1386 = 0 ⟹ n2 - 43n + 462 Solving by factorisation method, we have ⟹ n2 - 22n - 21n + 462 = 0 ⟹ n(n - 22) -21(n - 22) = 0 ⟹ (n - 22) (n - 21) = 0 Either, n - 22 = 0 ⟹ n = 22 Or, n - 21 = 0 ⟹ n = 21 Now, the 22nd term will be a22 = a1 + 21d = 63 + 21( -3 ) = 63 - 63 = 0 So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions (v) Given A.P. is 27, 24, 21. . . We know that, Sn = n/2[2a + (n − 1)d] Here we have, the first term (a) = 27 The sum of n terms (Sn) = 0 Common difference of the A.P. (d) = a2 - a1 = 24 - 27 = -3 On substituting the values in Sn, we get ⟹ 0 = n/2[2(27) + (n − 1)( − 3)] ⟹ 0 = (n)[54 + (n - 1)(-3)] ⟹ 0 = (n)[54 - 3n + 3] ⟹ 0 = n [57 - 3n] Further we have, n = 0 Or, 57 - 3n = 0 ⟹ 3n = 57 ⟹ n = 19 The number of terms cannot be zero, Hence, the numbers of terms (n) is 19. 11. Find the sum of the first (i) 11 terms of the A.P. : 2, 6, 10, 14, . . . (ii) 13 terms of the A.P. : -6, 0, 6, 12, . . . (iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8. Solution: We know that the sum of terms for different arithmetic progressions is given by Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms (i) Given A.P 2, 6, 10, 14,... to 11 terms. Common difference (d) = a2 - a1 = 10 - 6 = 4 Number of terms (n) = 11 First term for the given A.P. (a) = 2 So, S11 = 11/2[2(2) + (11 − 1)4] = 11/2[2(2) + (10)4] = 11/2[4 + 40] = 11 × 22 = 242 Hence, the sum of first 11 terms for the given A.P. is 242 (ii) Given A.P. – 6, 0, 6, 12, ... to 13 terms. Common difference (d) = a2 - a1 = 6 - 0 = 6 Number of terms (n) = 13 First term (a) = -6 So, S13 = 13/2[2(− 6) + (13 –1)6] = 13/2[(−12) + (12)6] = 13/2 = 390 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Hence, the sum of first 13 terms for the given A.P. is 390 (iii) 51 terms of an AP whose a2 = 2 and a4 = 8 We know that, a2 = a + d 2 = a + d ...(2) Also, a4 = a + 3d 8 = a + 3d ... (2) Subtracting (1) from (2), we have 2d = 6 d = 3 Substituting d = 3 in (1), we get 2 = a + 3 ⟹ a = -1 Given that the number of terms (n) = 51 First term (a) = -1 So, Sn = 51/2[2(−1) + (51 − 1)(3)] = 51/2[−2 + 150] = 51/2 = 3774 Hence, the sum of first 51 terms for the A.P. is 3774. 12. Find the sum of (i) the first 15 multiples of 8 (ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6. (iii) all 3 - digit natural numbers which are divisible by 13. (iv) all 3 - digit natural numbers which are multiples of 11. Solution: We know that the sum of terms for an A.P is given by Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms (i) Given, first 15 multiples of 8. These multiples form an A.P: 8, 16, 24, …… , 120 Here, a = 8 , d = 61 – 8 = 8 and the number of terms(n) = 15 Now, finding the sum of 15 terms, we have R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions \ Hence, the sum of the first 15 multiples of 8 is 960 (ii)(a) First 40 positive integers divisible by 3. Hence, the first multiple is 3 and the 40th multiple is 120. And, these terms will form an A.P. with the common difference of 3. Here, First term (a) = 3 Number of terms (n) = 40 Common difference (d) = 3 So, the sum of 40 terms S40 = 40/2[2(3) + (40 − 1)3] = 20[6 + (39)3] = 20(6 + 117) = 20(123) = 2460 Thus, the sum of first 40 multiples of 3 is 2460. (b) First 40 positive integers divisible by 5 Hence, the first multiple is 5 and the 40th multiple is 200. And, these terms will form an A.P. with the common difference of 5. Here, First term (a) = 5 Number of terms (n) = 40 Common difference (d) = 5 So, the sum of 40 terms S40 = 40/2[2(5) + (40 − 1)5] = 20[10 + (39)5] = 20 (10 + 195) = 20 (205) = 4100 Hence, the sum of first 40 multiples of 5 is 4100. (c) First 40 positive integers divisible by 6 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Hence, the first multiple is 6 and the 40th multiple is 240. And, these terms will form an A.P. with the common difference of 6. Here, First term (a) = 6 Number of terms (n) = 40 Common difference (d) = 6 So, the sum of 40 terms S40 = 40/2[2(6) + (40 − 1)6] = 20[12 + (39)6] =20(12 + 234) = 20(246) = 4920 Hence, the sum of first 40 multiples of 6 is 4920. (iii) All 3 digit natural number which are divisible by 13. So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988. And, these terms form an A.P. with the common difference of 13. Here, first term (a) = 104 and the last term (l) = 988 Common difference (d) = 13 Finding the number of terms in the A.P. by, an = a + (n − 1)d We have, 988 = 104 + (n - 1)13 ⟹ 988 = 104 + 13n -13 ⟹ 988 = 91 + 13n ⟹ 13n = 897 ⟹ n = 69 Now, using the formula for the sum of n terms, we get S69 = 69/2[2(104) + (69 − 1)13] = 69/2[208 + 884] = 69/2 = 69(546) = 37674 Hence, the sum of all 3 digit multiples of 13 is 37674. (iv) All 3 digit natural number which are multiples of 11. So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990. And, these terms form an A.P. with the common difference of 11. Here, first term (a) = 110 and the last term (l) = 990 Common difference (d) = 11 Finding the number of terms in the A.P. by, an = a + (n − 1)d We get, 990 = 110 + (n - 1)11 ⟹ 990 = 110 + 11n -11 ⟹ 990 = 99 + 11n ⟹ 11n = 891 ⟹ n = 81 Now, using the formula for the sum of n terms, we get S81 = 81/2[2(110) + (81 − 1)11] R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions = 81/2[220 + 880] = 81/2 = 81(550) = 44550 Hence, the sum of all 3 digit multiples of 11 is 44550. 13. Find the sum: (i) 2 + 4 + 6 + . . . + 200 (ii) 3 + 11 + 19 + . . . + 803 (iii) (-5) + (-8) + (-11) + . . . + (- 230) (iv) 1 + 3 + 5 + 7 + . . . + 199 (vi) 34 + 32 + 30 + . . . + 10 (vii) 25 + 28 + 31 + . . . + 100 Solution: We know that the sum of terms for an A.P is given by Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Or Sn = n/2[a + l] Where; a = first term for the given A.P. ;l = last term for the given A.P (i) Given series. 2 + 4 + 6 + . . . + 200 which is an A.P Where, a = 2 ,d = 4 – 2 = 2 and last term (an = l) = 200 We know that, an = a + (n - 1)d So, 200 = 2 + (n - 1)2 200 = 2 + 2n – 2 n = 200/2 = 100 Now, for the sum of these 100 terms S100 = 100/2 [2 + 200] = 50(202) = 10100 Hence, the sum of terms of the given series is 10100. (ii) Given series. 3 + 11 + 19 + . . . + 803 which is an A.P Where, a = 3 ,d = 11 – 3 = 8 and last term (an = l) = 803 We know that, an = a + (n - 1)d So, 803 = 3 + (n - 1)8 803 = 3 + 8n – 8 n = 808/8 = 101 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Now, for the sum of these 101 terms S101 = 101/2 [3 + 803] = 101(806)/2 = 101 x 403 = 40703 Hence, the sum of terms of the given series is 40703. (iii) Given series (-5) + (-8) + (-11) + . . . + (- 230) which is an A.P Where, a = -5 ,d = -8 – (-5) = -3 and last term (an = l) = -230 We know that, an = a + (n - 1)d So, -230 = -5 + (n - 1)(-3) -230 = -5 - 3n + 3 3n = -2 + 230 n = 228/3 = 76 Now, for the sum of these 76 terms S76 = 76/2 [-5 + (-230)] = 38 x (-235) = -8930 Hence, the sum of terms of the given series is -8930. (iv) Given series. 1 + 3 + 5 + 7 + . . . + 199 which is an A.P Where, a = 1 ,d = 3 – 1 = 2 and last term (an = l) = 199 We know that, an = a + (n - 1)d So, 199 = 1 + (n - 1)2 199 = 1 + 2n – 2 n = 200/2 = 100 Now, for the sum of these 100 terms S100 = 100/2 [1 + 199] = 50(200) = 10000 Hence, the sum of terms of the given series is 10000. (v) Given series which is an A.P Where, a = 7, d = 10 ½ - 7 = (21 – 14)/2 = 7/2 and last term (an = l) = 84 We know that, an = a + (n - 1)d So, 84 = 7 + (n - 1)(7/2) 168 = 14 + 7n – 7 n = (168 – 7)/7 = 161/7 = 23 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Now, for the sum of these 23 terms S23 = 23/2 [7 + 84] = 23(91)/2 = 2093/2 Hence, the sum of terms of the given series is 2093/2. (vi) Given series, 34 + 32 + 30 + . . . + 10 which is an A.P Where, a = 34 ,d = 32 – 34 = -2 and last term (an = l) = 10 We know that, an = a + (n - 1)d So, 10 = 34 + (n - 1)(-2) 10 = 34 - 2n + 2 n = (36 – 10)/2 = 13 Now, for the sum of these 13 terms S13 = 13/2 [34 + 10] = 13(44)/2 = 13 x 22 = 286 Hence, the sum of terms of the given series is 286. (vii) Given series, 25 + 28 + 31 + . . . + 100 which is an A.P Where, a = 25 ,d = 28 – 25 = 3 and last term (an = l) = 100 We know that, an = a + (n - 1)d So, 100 = 25 + (n - 1)(3) 100 = 25 + 3n - 3 n = (100 – 22)/3 = 26 Now, for the sum of these 26 terms S100 = 26/2 [25 + 100] = 13(124) = 1625 Hence, the sum of terms of the given series is 1625. 14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Solution: Given, the first term of the A.P (a) = 17 The last term of the A.P (l) = 350 The common difference (d) of the A.P. = 9 Let the number of terms be n. And, we know that; l = a + (n - 1)d So, 350 = 17 + (n- 1) 9 ⟹ 350 = 17 + 9n - 9 ⟹ 350 = 8 + 9n R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions ⟹ 350 - 8 = 9n Thus we get, n = 38 Now, finding the sum of terms Sn = n/2[a + l] = 38/2(17 + 350) = 19 × 367 = 6973 Hence, the number of terms is of the A.P is 38 and their sum is 6973. 15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms. Solution: Let’s consider the first term as a and the common difference as d. Given, a3 = 7 .... (1) and, a7 = 3a3 + 2 .... (2) So, using (1) in (2), we get, a7 = 3(7) + 2 = 21 + 2 = 23 .... (3) Also, we know that an = a +(n - 1)d So, the 3th term (for n = 3), a3 = a + (3 - 1)d ⟹ 7 = a + 2d (Using 1) ⟹ a = 7 - 2d .... (4) Similarly, for the 7th term (n = 7), a7 = a + (7 - 1) d 24 = a + 6d = 23 (Using 3) a = 23 - 6d .... (5) Subtracting (4) from (5), we get, a - a = (23 - 6d) - (7 - 2d) ⟹ 0 = 23 - 6d - 7 + 2d ⟹ 0 = 16 - 4d ⟹ 4d = 16 ⟹ d = 4 Now, to find a, we substitute the value of d in (4), a =7 - 2(4) ⟹ a = 7 - 8 a = -1 Hence, for the A.P. a = -1 and d = 4 For finding the sum, we know that Sn = n/2[2a + (n − 1)d] and n = 20 (given) S20 = 20/2[2(−1) + (20 − 1)(4)] = (10)[-2 + (19)(4)] = (10)[-2 + 76] = (10) = 740 Hence, the sum of first 20 terms for the given A.P. is 740 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference. Solution: Given, The first term of the A.P (a) = 2 The last term of the A.P (l) = 50 Sum of all the terms Sn = 442 So, let the common difference of the A.P. be taken as d. The sum of all the terms is given as, 442 = (n/2)(2 + 50) ⟹ 442 = (n/2)(52) ⟹ 26n = 442 ⟹ n = 17 Now, the last term is expressed as 50 = 2 + (17 - 1)d ⟹ 50 = 2 + 16d ⟹ 16d = 48 ⟹ d = 3 Thus, the common difference of the A.P. is d = 3. 17. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms? Solution: Let us take the first term as a and the common difference as d. Given, a12 = -13 S4 = 24 Also, we know that an = a + (n - 1)d So, for the 12th term a12 = a + (12 - 1)d = -13 ⟹ a + 11d = -13 a = -13 - 11d .... (1) And, we that for sum of terms Sn = n/2[2a + (n − 1)d] Here, n = 4 S4 = 4/2[2(a) + (4 − 1)d] ⟹ 24 = (2)[2a + (3)(d)] ⟹ 24 = 4a + 6d ⟹ 4a = 24 - 6d Subtracting (1) from (2), we have R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Further simplifying for d, we get, ⟹ -19 × 2 = 19d ⟹ d = – 2 On substituting the value of d in (1), we find a a = -13 - 11(-2) a = -13 + 22 a = 9 Next, the sum of 10 term is given by S10 = 10/2[2(9) + (10 − 1)(−2)] = (5)[19 + (9)(-2)] = (5)(18 - 18) = 0 Thus, the sum of first 10 terms for the given A.P. is S10 = 0. 18. Find the sum of n terms of the series (4 - (1/n)) + (4 - (2/n)) + (4 - (3/n)) + .... Solution: Let the first term be taken as a. Given, a = 4 - 1/n So, common difference is d = (4 - (2/n)) - (4 - (1/n)) = (4n - 2 - 4n + 1)/n = -1/n By using the formula, Sn = n/2[2a + (n − 1)d] We get, = n/2[2(4−(1/n)) + (n − 1)(-1/n)] = n/2[8 - (2/n) + (-1 + (1/n))] R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions = n/2[(8n - 2 - n + 1)/n] = 1/2(7n - 1) Hence, the sum of n terms of the series is 1/2(7n - 1). 19. In an A.P., if the first term is 22, the common difference is – 4 and the sum to n terms is 64, find n. Solution: Given that, a = 22, d = – 4 and Sn = 64 Let us consider the number of terms as n. For sum of terms in an A.P, we know that Sn = n/2[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So, ⟹ Sn = n/2[2(22) + (n − 1)(−4)] ⟹ 64 = n/2[2(22) + (n − 1)(−4)] ⟹ 64(2) = n(48 - 4n) ⟹ 128 = 48n - 4n2 After rearranging the terms, we have a quadratic equation 4n2 - 48n + 128 = 0, n2 - 12n + 32 = 0 [dividing by 4 on both sides] n2 - 12n + 32 = 0 Solving by factorisation method, n2 - 8n - 4n + 32 = 0 n ( n - 8 ) - 4 ( n - 8 ) = 0 (n - 8) (n - 4) = 0 So, we get n - 8 = 0 ⟹ n = 8 Or, n - 4 = 0 ⟹ n = 4 Hence, the number of terms can be either n = 4 or 8. 20. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms? Solution: Let’s take the first term as a and the common difference to be d Given that, a5 = 30 and a12 = 65 And, we know that an = a + (n - 1)d So, a5 = a + (5 - 1)d 30 = a + 4d a = 30 - 4d .... (i) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Similarly, a12 = a + (12 - 1) d 65 = a + 11d a = 65 - 11d .... (ii) Subtracting (i) from (ii), we have a - a = (65 - 11d) - (30 - 4d) 0 = 65 - 11d - 30 + 4d 0 = 35 - 7d 7d = 35 d = 5 Putting d in (i), we get a = 30 - 4(5) a = 30 - 20 a = 10 Thus for the A.P; d = 5 and a = 10 Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P., Sn = n/2[2a + (n − 1)d] Where; a = first term of the given A.P. d = common difference of the given A.P. n = number of terms Here n = 20, so we have S20 = 20/2[2(10) + (20 − 1)(5)] = (10)[20 + (19)(5)] = (10)[20 + 95] = (10) = 1150 Hence, the sum of first 20 terms for the given A.P. is 1150 21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively. Solution: Let’s take the first term as a and the common difference as d. Given that, a2 = 14 and a3 = 18 And, we know that an = a + (n - 1)d So, a2 = a + (2 - 1)d ⟹ 14 = a + d ⟹ a = 14 - d .... (i) Similarly, a3 = a + (3 - 1)d ⟹ 18 = a + 2d ⟹ a = 18 - 2d .... (ii) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Subtracting (i) from (ii), we have a - a = (18 - 2d) - (14 - d) 0 = 18 - 2d - 14 + d 0 = 4 - d d = 4 Putting d in (i), to find a a = 14 - 4 a = 10 Thus, for the A.P. d = 4 and a = 10 Now, to find sum of terms Sn = n/2(2a + (n − 1)d) Where, a = the first term of the A.P. d = common difference of the A.P. n = number of terms So, using the formula for n = 51, ⟹ S51 = 51/2[2(10) + (51 - 1)(4)] = 51/2[20 + (40)4] = 51/2 = 51(110) = 5610 Hence, the sum of the first 51 terms of the given A.P. is 5610 22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms. Solution: Given, Sum of 7 terms of an A.P. is 49 ⟹ S7 = 49 And, sum of 17 terms of an A.P. is 289 ⟹ S17 = 289 Let the first term of the A.P be a and common difference as d. And, we know that the sum of n terms of an A.P is Sn = n/2[2a + (n − 1)d] So, S7 = 49 = 7/2[2a + (7 - 1)d] = 7/2 [2a + 6d] = 7[a + 3d] ⟹ 7a + 21d = 49 a + 3d = 7 ….. (i) Similarly, S17 = 17/2[2a + (17 - 1)d] = 17/2 [2a + 16d] = 17[a + 8d] ⟹ 17[a + 8d] = 289 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions a + 8d = 17 ….. (ii) Now, subtracting (i) from (ii), we have a + 8d – (a + 3d) = 17 – 7 5d = 10 d = 2 Putting d in (i), we find a a + 3(2) = 7 a = 7 – 6 = 1 So, for the A.P: a = 1 and d = 2 For the sum of n terms is given by, Sn = n/2[2(1) + (n − 1)(2)] = n/2[2 + 2n - 2] = n/2[2n] = n2 Therefore, the sum of n terms of the A.P is given by n2. 23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. Solution: Sum of first n terms of an A.P is given by Sn = n/2(2a + (n − 1)d) Given, First term (a) = 5, last term (an) = 45 and sum of n terms (Sn) = 400 Now, we know that an = a + (n - 1)d ⟹ 45 = 5 + (n - 1)d ⟹ 40 = nd - d ⟹ nd - d = 40 .... (1) Also, Sn = n/2(2(a) + (n − 1)d) 400 = n/2(2(5) + (n − 1)d) 800 = n (10 + nd - d) 800 = n (10 + 40) [using (1)] ⟹ n = 16 Putting n in (1), we find d nd - d = 40 16d - d = 40 15d = 40 d = 8/3 Therefore, the common difference of the given A.P. is 8/3. 24. In an A.P. the first term is 8, nth term is 33 and the sum of first n term is 123. Find n and the d, the common difference. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Solution: Given, The first term of the A.P (a) = 8 The nth term of the A.P (l) = 33 And, the sum of all the terms Sn = 123 Let the common difference of the A.P. be d. So, find the number of terms by 123 = (n/2)(8 + 33) 123 = (n/2)(41) n = (123 x 2)/ 41 n = 246/41 n = 6 Next, to find the common difference of the A.P. we know that l = a + (n - 1)d 33 = 8 + (6 - 1)d 33 = 8 + 5d 5d = 25 d = 5 Thus, the number of terms is n = 6 and the common difference of the A.P. is d = 5. 25. In an A.P. the first term is 22, nth term is -11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given, The first term of the A.P (a) = 22 The nth term of the A.P (l) = -11 And, sum of all the terms Sn = 66 Let the common difference of the A.P. be d. So, finding the number of terms by 66 = (n/2)[22 + (−11)] 66 = (n/2)[22 − 11] (66)(2) = n(11) 6 × 2 = n n = 12 Now, for finding d We know that, l = a + (n - 1)d - 11 = 22 + (12 - 1)d -11 = 22 + 11d 11d = – 33 d = – 3 Hence, the number of terms is n = 12 and the common difference d = -3 26. The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find the common difference. R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Solution: Given, First term (a) = 7, last term (an) = 49 and sum of n terms (Sn) = 420 Now, we know that an = a + (n - 1)d ⟹ 49 = 7 + (n - 1)d ⟹ 43 = nd - d ⟹ nd - d = 42 ..... (1) Next, Sn = n/2(2(7) + (n − 1)d) ⟹ 840 = n[14 + nd - d] ⟹ 840 = n[14 + 42] [using (1)] ⟹ 840 = 54n ⟹ n = 15 .... (2) So, by substituting (2) in (1), we have nd - d = 42 ⟹ 15d - d = 42 ⟹ 14d = 42 ⟹ d = 3 Therefore, the common difference of the given A.P. is 3. 27. The first and the last terms of an A.P are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference. Solution: Given, First term (a) = 5 and the last term (l) = 45 Also, Sn = 400 We know that, an = a + (n - 1)d ⟹ 45 = 5 + (n - 1)d ⟹ 40 = nd - d ⟹ nd - d = 40 ..... (1) Next, Sn = n/2(2(5) + (n − 1)d) ⟹ 400 = n[10 + nd - d] ⟹ 800 = n[10 + 40] [using (1)] ⟹ 800 = 50n ⟹ n = 16 .... (2) So, by substituting (2) in (1), we have nd - d = 40 ⟹ 16d - d = 40 ⟹ 15d = 40 ⟹ d = 8/3 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Therefore, the common difference of the given A.P. is 8/3. 28. The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1: 2. Find the first and 15th term of the A.P. Solution: Let a be the first term and d be the common difference. And we know that, sum of first n terms is: Sn = n/2(2a + (n − 1)d) Also, nth term is given by an = a + (n - 1)d From the question, we have Sq = 162 and a6 : a13 = 1 : 2 So, 2a6 = a13 ⟹ 2 [a + (6 - 1d)] = a + (13 - 1)d ⟹ 2a + 10d = a + 12d ⟹ a = 2d .... (1) And, S9 = 162 ⟹ S9 = 9/2(2a + (9 − 1)d) ⟹ 162 = 9/2(2a + 8d) ⟹ 162 × 2 = 9[4d + 8d] [from (1)] ⟹ 324 = 9 × 12d ⟹ d = 3 ⟹ a = 2(3) [from (1)] ⟹ a = 6 Hence, the first term of the A.P. is 6 For the 15th term, a15 = a + 14d = 6 + 14 × 3 = 6 + 42 Therefore, a15 = 48 29. If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term. Solution: Let’s consider a to be the first term and d be the common difference. And we know that, sum of first n terms is: Sn = n/2(2a + (n − 1)d) and nth term is given by: an = a + (n - 1)d Now, from the question we have S10 = 120 ⟹ 120 = 10/2(2a + (10 − 1)d) ⟹ 120 = 5(2a + 9d) ⟹ 24 = 2a + 9d .... (1) Also given that, a10 = 21 ⟹ 21 = a + (10 - 1)d ⟹ 21 = a + 9d .... (2) Subtracting (2) from (1), we get R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 24 - 21 = 2a + 9d - a - 9d ⟹a = 3 Now, on putting a = 3 in equation (2), we get 3 + 9d = 21 9d = 18 d = 2 Thus, we have the first term(a) = 3 and the common difference(d) = 2 Therefore, the nth term is given by an = a + (n - 1)d = 3 + (n - 1)2 = 3 + 2n -2 = 2n + 1 Hence, the nth term of the A.P is (an) = 2n + 1. 30. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. Solution: Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms Sn = n/2(2a + (n − 1)d) Given that sum of the first 7 terms of an A.P. is 63. S7 = 63 And sum of next 7 terms is 161. So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms S14 = 63 + 161 = 224 Now, having S7 = 7/2(2a + (7 − 1)d) ⟹ 63(2) = 7(2a + 6d) ⟹ 9 × 2 = 2a + 6d ⟹ 2a + 6d = 18 . . . . (1) And, S14 = 14/2(2a + (14 − 1)d) ⟹ 224 = 7(2a + 13d) ⟹ 32 = 2a + 13d .... (2) Now, subtracting (1) from (2), we get ⟹ 13d - 6d = 32 - 18 ⟹ 7d = 14 ⟹ d = 2 Using d in (1), we have 2a + 6(2) = 18 2a = 18 - 12 a = 3 Thus, from nth term ⟹ a28 = a + (28 - 1)d = 3 + 27 (2) = 3 + 54 = 57 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions Therefore, the 28th term is 57. 31. The sum of first seven terms of an A.P. is 182. If its 4th and 17th terms are in ratio 1: 5, find the A.P. Solution: Given that, S17 = 182 And, we know that the sum of first n term is: Sn = n/2(2a + (n − 1)d) So, S7 = 7/2(2a + (7 − 1)d) 182 × 2 = 7(2a + 6d) 364 = 14a + 42d 26 = a + 3d a = 26 - 3d ... (1) Also, it’s given that 4th term and 17th term are in a ratio of 1: 5. So, we have ⟹ 5(a4) = 1(a17) ⟹ 5 (a + 3d) = 1 (a + 16d) ⟹ 5a + 15d = a + 16d ⟹ 4a = d .... (2) Now, substituting (2) in (1), we get ⟹ 4 ( 26 - 3d ) = d ⟹ 104 - 12d = d ⟹ 104 = 13d ⟹ d = 8 Putting d in (2), we get ⟹ 4a = d ⟹ 4a = 8 ⟹ a = 2 Therefore, the first term is 2 and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . .. 32. The nth term of an A.P is given by (-4n + 15). Find the sum of first 20 terms of this A.P. Solution: Given, The nth term of the A.P = (-4n + 15) So, by putting n = 1 and n = 20 we can find the first ans 20th term of the A.P a = (-4(1) + 15) = 11 And, a20 = (-4(20) + 15) = -65 Now, for find the sum of 20 terms of this A.P we have the first and last term. So, using the formula Sn = n/2(a + l) S20 = 20/2(11 + (-65)) = 10(-54) R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions = -540 Therefore, the sum of first 20 terms of this A.P. is -540. 33. In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P. Solution: Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms Sn = n/2(2a + (n − 1)d) Given that sum of the first 10 terms of an A.P. is -150. S10 = -150 And the sum of next 10 terms is -550. So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms S20 = -150 + -550 = -700 Now, having S10 = 10/2(2a + (10 − 1)d) ⟹ -150 = 5(2a + 9d) ⟹ -30 = 2a + 9d ⟹ 2a + 9d = -30 . . . . (1) And, S20 = 20/2(2a + (20 − 1)d) ⟹ -700 = 10(2a + 19d) ⟹ -70 = 2a + 19d .... (2) Now, subtracting (1) from (2), we get ⟹ 19d - 9d = -70 – (-30) ⟹ 10d = -40 ⟹ d = -4 Using d in (1), we have 2a + 9(-4) = -30 2a = -30 + 36 a = 6/2 = 3 Hence, we have a = 3 and d = -4 So, the A.P is 3, -1, -5, -9, -13,….. 34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term. Solution Given, First term of the A.P is 1505 and S14 = 1505 We know that, the sum of first n terms is Sn = n/2(2a + (n − 1)d) So, S14 = 14/2(2(10) + (14 − 1)d) = 1505 7(20 + 13d) = 1505 R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions 20 + 13d = 215 13d = 215 – 20 d = 195/13 d =15 Thus, the 25th term is given by a25 = 10 + (25 -1)15 = 10 + (24)15 = 10 + 360 = 370 Therefore, the 25th term of the A.P is 370 35. In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P. Solution: Given, The first term of the A.P. (a) = 2 The last term of the A.P. (l) = 29 And, sum of all the terms (Sn) = 155 Let the common difference of the A.P. be d. So, find the number of terms by sum of terms formula Sn = n/2 (a + l) 155 = n/2(2 + 29) 155(2) = n(31) 31n = 310 n = 10 Using n for the last term, we have l = a + (n - 1)d 29 = 2 + (10 - 1)d 29 = 2 + (9)d 29 - 2 = 9d 9d = 27 d = 3 Hence, the common difference of the A.P. is d = 3 36. The first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Solution: Given, In an A.P first term (a) = 17 and the last term (l) = 350 And, the common difference (d) = 9 We know that, an = a + (n - 1)d so, R D Sharma Solutions For Class 10 Maths Chapter 9 - Arithmetic Progressions an = l = 17 + (n - 1)9 = 350 17 + 9n – 9 = 350 9n = 350 – 8 n = 342/9 n = 38 So, the sum of all the term of the A.P is given by Sn = n/2 (a + l) = 38/2(17 + 350) = 19(367) = 6973 Therefore, the sum of terms of the A.P is 6973. 37. Find the number of terms of the A.P. –12, –9, –6, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained. Solution: Given, First term, a = -12 Common difference, d = a2 - a1 = – 9 – (- 12) d = - 9 + 12 = 3 And, we know that nth term = an = a + (n - 1)d ⟹ 21 = -12 + (n - 1)3 ⟹ 21 = -12 + 3n - 3 ⟹ 21 = 3n - 15 ⟹ 36 = 3n ⟹ n = 12 Thus, the number of terms is 12. Now, if 1 is added to each of the 12 terms, the sum will increase by 12. Hence, the sum of all the terms of the A.P. so obtained is ⟹ S12 + 12 = 12/2[a + l] + 12 = 6[-12 + 21] + 12 = 6 × 9 + 12 = 66 Therefore, the sum after adding 1 to each of the terms in the A.P is 66.
3325	https://en.wikipedia.org/wiki/Post%27s_theorem	Jump to content Search Contents (Top) 1 Background 2 Post's theorem and corollaries 3 Proof of Post's theorem 3.1 Formalization of Turing machines in first-order arithmetic 3.2 Implementation example 3.3 Recursively enumerable sets 3.4 Oracle machines 3.5 Turing jump 3.6 Higher Turing jumps 4 References Post's theorem Čeština فارسی Français 日本語 Português Русский Українська Edit links Article Talk Read Edit View history Tools 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 Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Theorem in computability theory In computability theory Post's theorem, named after Emil Post, describes the connection between the arithmetical hierarchy and the Turing degrees. Background [edit] See also: Arithmetical hierarchy § Relation to Turing machines The statement of Post's theorem uses several concepts relating to definability and recursion theory. This section gives a brief overview of these concepts, which are covered in depth in their respective articles. The arithmetical hierarchy classifies certain sets of natural numbers that are definable in the language of Peano arithmetic. A formula is said to be if it is an existential statement in prenex normal form (all quantifiers at the front) with alternations between existential and universal quantifiers applied to a formula with bounded quantifiers only. Formally a formula in the language of Peano arithmetic is a formula if it is of the form where contains only bounded quantifiers and Q is if m is even and if m is odd. A set of natural numbers is said to be if it is definable by a formula, that is, if there is a formula such that each number is in if and only if holds. It is known that if a set is then it is for any , but for each m there is a set that is not . Thus the number of quantifier alternations required to define a set gives a measure of the complexity of the set. Post's theorem uses the relativized arithmetical hierarchy as well as the unrelativized hierarchy just defined. A set of natural numbers is said to be relative to a set , written , if is definable by a formula in an extended language that includes a predicate for membership in . While the arithmetical hierarchy measures definability of sets of natural numbers, Turing degrees measure the level of uncomputability of sets of natural numbers. A set is said to be Turing reducible to a set , written , if there is an oracle Turing machine that, given an oracle for , computes the characteristic function of . The Turing jump of a set is a form of the Halting problem relative to . Given a set , the Turing jump is the set of indices of oracle Turing machines that halt on input when run with oracle . It is known that every set is Turing reducible to its Turing jump, but the Turing jump of a set is never Turing reducible to the original set. Post's theorem uses finitely iterated Turing jumps. For any set of natural numbers, the notation indicates the –fold iterated Turing jump of . Thus is just , and is the Turing jump of . Post's theorem and corollaries [edit] Post's theorem establishes a close connection between the arithmetical hierarchy and the Turing degrees of the form , that is, finitely iterated Turing jumps of the empty set. (The empty set could be replaced with any other computable set without changing the truth of the theorem.) Post's theorem states: A set is if and only if is recursively enumerable by an oracle Turing machine with an oracle for , that is, if and only if is . The set is -complete for every . This means that every set is many-one reducible to . Post's theorem has many corollaries that expose additional relationships between the arithmetical hierarchy and the Turing degrees. These include: Fix a set . A set is if and only if is . This is the relativization of the first part of Post's theorem to the oracle . A set is if and only if . More generally, is if and only if . A set is defined to be arithmetical if it is for some . Post's theorem shows that, equivalently, a set is arithmetical if and only if it is Turing reducible to for some m. Proof of Post's theorem [edit] Formalization of Turing machines in first-order arithmetic [edit] The operation of a Turing machine on input can be formalized logically in first-order arithmetic. For example, we may use symbols , , and for the tape configuration, machine state and location along the tape after steps, respectively. 's transition system determines the relation between and ; their initial values (for ) are the input, the initial state and zero, respectively. The machine halts if and only if there is a number such that is the halting state. The exact relation depends on the specific implementation of the notion of Turing machine (e.g. their alphabet, allowed mode of motion along the tape, etc.) In case halts at time , the relation between and must be satisfied only for k bounded from above by . Thus there is a formula in first-order arithmetic with no unbounded quantifiers, such that halts on input at time at most if and only if is satisfied. Implementation example [edit] For example, for a prefix-free Turing machine with binary alphabet and no blank symbol, we may use the following notations: is the 1-ary symbol for the configuration of the whole tape after steps (which we may write as a number with LSB first, the value of the m-th location on the tape being its m-th least significant bit). In particular is the initial configuration of the tape, which corresponds the input to the machine. is the 1-ary symbol for the Turing machine state after steps. In particular, , the initial state of the Turing machine. is the 1-ary symbol for the Turing machine location on the tape after steps. In particular . is the transition function of the Turing machine, written as a function from a doublet (machine state, bit read by the machine) to a triplet (new machine state, bit written by the machine, +1 or -1 machine movement along the tape). is the j-th bit of a number . This can be written as a first-order arithmetic formula with no unbounded quantifiers. For a prefix-free Turing machine we may use, for input n, the initial tape configuration where cat stands for concatenation; thus is a length string of followed by and then by . The operation of the Turing machine at the first steps can thus be written as the conjunction of the initial conditions and the following formulas, quantified over for all : . Since M has a finite domain, this can be replaced by a first-order quantifier-free arithmetic formula. The exact formula obviously depends on M. . Note that at the first steps, never arrives at a location along the tape greater than . Thus the universal quantifier over j can be bounded by +1, as bits beyond this location have no relevance for the machine's operation. T halts on input at time at most if and only if is satisfied, where: This is a first-order arithmetic formula with no unbounded quantifiers, i.e. it is in . Recursively enumerable sets [edit] Let be a set that can be recursively enumerated by a Turing machine. Then there is a Turing machine that for every in , halts when given as an input. This can be formalized by the first-order arithmetical formula presented above. The members of are the numbers satisfying the following formula: This formula is in . Therefore, is in . Thus every recursively enumerable set is in . The converse is true as well: for every formula in with k existential quantifiers, we may enumerate the –tuples of natural numbers and run a Turing machine that goes through all of them until it finds the formula is satisfied. This Turing machine halts on precisely the set of natural numbers satisfying , and thus enumerates its corresponding set. Oracle machines [edit] Similarly, the operation of an oracle machine with an oracle O that halts after at most steps on input can be described by a first-order formula , except that the formula now includes: A new predicate, , giving the oracle answer. This predicate must satisfy some formula to be discussed below. An additional tape - the oracle tape - on which has to write the number m for every call O(m) to the oracle; writing on this tape can be logically formalized in a similar manner to writing on the machine's tape. Note that an oracle machine that halts after at most steps has time to write at most digits on the oracle tape. So the oracle can only be called with numbers m satisfying . If the oracle is for a decision problem, is always "Yes" or "No", which we may formalize as 0 or 1. Suppose the decision problem itself can be formalized by a first-order arithmetic formula . Then halts on after at most steps if and only if the following formula is satisfied: where is a first-order formula with no unbounded quantifiers. Turing jump [edit] If O is an oracle to the halting problem of a machine , then is the same as "there exists such that starting with input m is at the halting state after steps". Thus: where is a first-order formula that formalizes . If is a Turing machine (with no oracle), is in (i.e. it has no unbounded quantifiers). Since there is a finite number of numbers m satisfying , we may choose the same number of steps for all of them: there is a number , such that halts after steps precisely on those inputs for which it halts at all. Moving to prenex normal form, we get that the oracle machine halts on input if and only if the following formula is satisfied: (informally, there is a "maximal number of steps" such every oracle that does not halt within the first steps does not stop at all; however, for every, each oracle that halts after steps does halt). Note that we may replace both and by a single number - their maximum - without changing the truth value of . Thus we may write: For the oracle to the halting problem over Turing machines, is in and is in . Thus every set that is recursively enumerable by an oracle machine with an oracle for , is in . The converse is true as well: Suppose is a formula in with existential quantifiers followed by universal quantifiers. Equivalently, has > existential quantifiers followed by a negation of a formula in ; the latter formula can be enumerated by a Turing machine and can thus be checked immediately by an oracle for . We may thus enumerate the –tuples of natural numbers and run an oracle machine with an oracle for that goes through all of them until it finds a satisfaction for the formula. This oracle machine halts on precisely the set of natural numbers satisfying , and thus enumerates its corresponding set. Higher Turing jumps [edit] More generally, suppose every set that is recursively enumerable by an oracle machine with an oracle for is in . Then for an oracle machine with an oracle for , is in . Since is the same as for the previous Turing jump, it can be constructed (as we have just done with above) so that in . After moving to prenex formal form the new is in . By induction, every set that is recursively enumerable by an oracle machine with an oracle for , is in . The other direction can be proven by induction as well: Suppose every formula in can be enumerated by an oracle machine with an oracle for . Now Suppose is a formula in with existential quantifiers followed by universal quantifiers etc. Equivalently, has > existential quantifiers followed by a negation of a formula in ; the latter formula can be enumerated by an oracle machine with an oracle for and can thus be checked immediately by an oracle for . We may thus enumerate the –tuples of natural numbers and run an oracle machine with an oracle for that goes through all of them until it finds a satisfaction for the formula. This oracle machine halts on precisely the set of natural numbers satisfying , and thus enumerates its corresponding set. References [edit] Rogers, H. The Theory of Recursive Functions and Effective Computability, MIT Press. ISBN 0-262-68052-1; ISBN 0-07-053522-1 Soare, R. Recursively enumerable sets and degrees. Perspectives in Mathematical Logic. Springer-Verlag, Berlin, 1987. ISBN 3-540-15299-7 Retrieved from " Categories: Theorems in the foundations of mathematics Computability theory Mathematical logic hierarchies Hidden categories: Articles with short description Short description matches Wikidata Post's theorem Add topic
3326	https://internetbookofemergencymedicine.com/wp-content/uploads/2020/10/pediatric-shock-pdf.pdf	Emergency Department Management of Pediatric Shock Jenny Mendelson, MDa,b, INTRODUCTION Shock is a state of acute energy failure stemming from a decrease in adenosine triphosphate production and subsequent failure to meet the acute metabolic demands of the body. More simply put, it is a state of inadequate oxygen supply to meet the body’s cellular demands. Hypoxemia or decreased perfusion results in decreased ox-ygen delivery to the tissues, causing a shift from more efficient aerobic pathways to anaerobic metabolism, resulting in the production of lactic acid. As oxygen deprivation persists, cellular hypoxia leads to the disruption of critical biochemical processes, eventually resulting in cell membrane ion pump dysfunction, intracellular edema, inad-equate regulation of intracellular pH, and cell death. Disclosures: None. a Pediatrics, Division of Pediatric Critical Care Medicine, University of Arizona College of Med-icine, Banner-University Medical Center, 1501 North Campbell Avenue, PO Box 245073, Tucson, AZ 85724-5073, USA; b Emergency Medicine, University of Arizona College of Medicine, Banner-University Medical Center, 1501 North Campbell Avenue, Tucson, AZ 85724-5073, USA Pediatrics, Division of Pediatric Critical Care Medicine, University of Arizona College of Med-icine, Banner-University Medical Center, 1501 North Campbell Avenue, PO Box 245073, Tucson, AZ 85724-5073. E-mail address: jmendelson@peds.arizona.edu KEYWORDS Pediatric Children Shock Hypotension Sepsis Vasopressors Emergency department KEY POINTS Clinical history and physical examination findings are crucial for the early recognition and classification of shock in the pediatric patient. Hypotension is a late and ominous finding in the pediatric patient in shock. Rapid fluid resuscitation is the first line of treatment in most forms of shock. Three 20 mL/kg isotonic crystalloid boluses should be given within the first 20 to 60 minutes after shock is identified. Epinephrine is usually the preferred vasopressor in pediatric shock and should be started peripherally if central access is not present. Emerg Med Clin N Am - (2018) -–- emed.theclinics.com 0733-8627/18/ª 2018 Elsevier Inc. All rights reserved. Oxygen delivery to the tissues is determined by cardiac output and arterial oxygen content. Cardiac output depends on heart rate and stroke volume. Stroke volume is determined by preload (the amount of filling of the ventricle at end-diastole), after-load (the force against which the ventricle must work to eject blood during systole, which is greatly affected by systemic vascular resistance [SVR]); contractility (the force generated by the ventricle during systole), and lusitropy (the degree of myocardial relaxation during diastole). In children, compared with adults, cardiac output is more dependent on heart rate than stroke volume owing to myocardial immaturity, which limits the ability to increase contractility. Arterial oxygen content depends on hemoglobin concentration, arterial oxygen saturation, and the arterial partial pressure of oxygen, with most oxygen being carried on hemoglobin and a small portion delivered as dissolved O2.1 Under normal conditions of increased oxygen demand, such as exercise, oxygen delivery must increase by redistribution of blood flow. Similarly, in pathologic in-stances of increased oxygen demand or decreased oxygen delivery (shock), initial compensatory mechanisms occur to preserve tissue perfusion. In compensated shock, vital organ function is maintained and blood pressure remains normal. In un-compensated shock, hypotension develops and organ and cellular function deterio-rate. Left untreated, uncompensated shock progresses to irreversible shock, characterized by irreversible organ failure, cardiovascular collapse, cardiac arrest, and death. Pediatric shock results in a significant amount of morbidity and mortality world-wide. Sepsis and hypovolemia owing to infectious gastroenteritis are leading causes of child mortality worldwide, with an estimated 3 to 5 billion cases of acute gastro-enteritis and nearly 2 million deaths occurring each year in children under 5 years of age, with 98% of those deaths occurring developing countries.2 In developed countries like the United States, shock is also a common occurrence in the emer-gency department (ED). These children have a higher mortality rate compared with patients not in shock (11.4% vs 2.6%). The presence of shock is also associated with worse outcomes in a variety of emergency conditions, including traumatic brain injury and cardiac arrest.3,4 CLASSIFICATIONS OF SHOCK Several classifications of shock exist (Table 1). Rapid identification of the etiology may help to guide specific therapies. Table 1 Categories of shock Category Hemodynamics Causes Hypovolemic YPreload, [SVR, YCO Gastrointestinal loses, renal loses, hemorrhage, third spacing, burns Distributive YPreload, YYSVR, Y[CO Sepsis, anaphylaxis, neurogenic shock Cardiogenic [Preload, [SVR, YCO Congenital heart disease, arrhythmia, cardiomyopathy, myocarditis, severe anemia Obstructive Y[Preload, [SVR, YCO Pulmonary embolus, pericardial tamponade, tension pneumothorax, certain congenital heart lesions Abbreviations: CO, cardiac output; SVR, systemic vascular resistance. Mendelson 2 Hypovolemic Shock Hypovolemia is the most common cause of shock in children5 and is a leading cause of child mortality worldwide. Hypovolemic shock occurs owing to inappropriately low intravascular blood volume (either owing to intravascular volume loss or hemorrhage), leading to decreased cardiac output. Additionally, hemorrhagic shock decreases oxygen-carrying capacity secondary to direct loss of available hemoglobin. Intravascular volume loss can occur owing to gastrointestinal, renal, skin (ie, burns), or interstitial (ie, third spacing) losses. Hypovolemia can develop rapidly! Children with gastroenteritis can lose a significant percentage of their circulating volume within a few hours. Even if there is ongoing vomiting or diarrhea, it is usually preferable to attempt oral rehydration if dehydration is mild to moderate. Several studies including large metaanalyses have shown oral rehydration to be highly successful (<5% failure rate) and resulting in shorter ED stays and fewer adverse events compared with intra-venous (IV) hydration.6 If a patient shows signs of decreased end-organ function, how-ever, forego attempts at oral rehydration and proceed to IV resuscitation. Capillary leak syndrome owing to sepsis, burns, or other systemic inflammatory diseases can result in profound intravascular volume loss in patients that may otherwise seem to be edematous and volume overloaded. Hemorrhage may occur from traumatic or nontraumatic bleeding. Hemorrhagic shock can be further broken down into stages of severity based on percent volume loss and physical examination findings (Table 2). In an infant/toddler in shock with un-clear etiology, consider occult hemorrhage owing to nonaccidental trauma. Distributive Shock In distributive shock, normal peripheral vascular tone becomes inappropriately relaxed. In this state, vasodilation results in effective hypovolemia, although a net fluid loss may not have actually occurred. Common causes of distributive shock include sepsis, anaphylaxis, neurologic injury (ie, spinal shock), or drug-related causes. In Table 2 Classification of pediatric hemorrhagic shock by clinical signs Class I Very Mild Blood Loss (<15%) Class II Mild Blood Loss (15%–30%) Class III Moderate Blood Loss (30%–40%) Class IV Severe Blood Loss (>40%) HR Normal to mildly increased Tachycardic Tachycardic Severely tachycardic Pulse quality Normal Peripheral pulses decreased Peripheral pulses decreased Central pulses decreased Respiratory rate Normal Tachypneic Tachypneic Severely tachypneic Mental status Normal/ slightly anxious Anxious/irritable Irritable/confused Confused/lethargic/ obtunded Urine output Normal Decreased Decreased Anuric Skin Warm/pink Cool/mottled Cool/mottled/pallor Cold/pallor/cyanotic Adapted from American College of Surgeons. Advanced trauma life support (ATLS) study guide. Chicago (IL): American College of Surgeon; 2012; with permission. Management of Pediatric Shock 3 sepsis, massive inflammatory response along with nitric oxide and cytokine release lead to peripheral vasodilation. In anaphylaxis, mast cell degranulation leads to vaso-dilatory cytokine release. In spinal shock, injury to the cranial portion of the spinal cord disrupts the sympathetic chain of the autonomic nervous system, resulting in unop-posed parasympathetic vasodilation. Spinal shock, unlike most types of shock, often presents with bradycardia owing to unopposed vagal effects. Sepsis is a common clinical syndrome that complicates severe infection and is char-acterized by immune dysregulation, systemic inflammation, microcirculatory derange-ments, and end-organ dysfunction. Sepsis is 10 times more common in children under 1 year than in older children and adolescents.7 Pediatric sepsis is commonly encoun-tered in the ED, and is a major cause of morbidity, mortality, and health care use costs worldwide. The American College of Critical Care Medicine (ACCM) defines septic shock as a clinical diagnosis made when children have suspected infection manifested by hypothermia or hyperthermia and clinical signs of inadequate tissue perfusion including any of the following: decreased/altered mental status, abnormal capillary refill time (CRT) or pulse characteristic, or decreased urine output (<1 mL/kg/h). Hypoten-sion is not required for the clinical diagnosis of septic shock.8 Septic shock can present in one of two ways: cold shock or warm shock. Cold shock is characterized by high SVR resulting in cool/cold extremities, delayed CRT (<2 seconds), diminished peripheral pulses or differential between peripheral and cen-tral pulses, and narrow pulse pressure. Warm shock is characterized by low SVR, with warm/dry extremities with brisk (“flash”) CRT, tachycardia, and bounding pulses with a wide pulse pressure. Cardiogenic Shock Cardiogenic shock can result from a variety of conditions that impair cardiac output. In children, cardiac failure is most commonly due to congenital heart disease, cardiomy-opathies, or myocarditis. Additionally, arrhythmias can result in decreased cardiac output and shock. Categories of heart failure and cardiogenic shock can be classified according to the presence/absence of 2 traits: venous congestion (owing to increased filling pressures) and hypoperfusion (owing to decreased cardiac output or myocardial contractility). This concept is summarized in Box 1. The presence of venous congestion is consid-ered “wet” and the absence is described as “dry,” and hypoperfusion is “cold” and normal perfusion is “warm.” The wet patient may have findings including edema, hepatomegaly, ascites, jugular venous distension, S3 gallop, or crackles on lung auscultation owing to pulmonary edema. The cold patient may have cool extremities, weak pulses with narrow pulse pressure, delayed CRT, altered mental status, or hypotension.9,10 Obstructive Shock Obstructive shock occurs when either pulmonary or systemic blood flow is impaired, resulting in impaired cardiac output. Causes of obstruction to heart function may be intracardiac or extracardiac and may be congenital or acquired. Examples include obstructive congenital heart lesions, cardiac tamponade, tension pneumothorax, massive pulmonary embolism, severe pulmonary hypertension, and hypertrophic car-diomyopathy. Obstructive shock in infants occurs when congenital lesions interfere with the outflow of blood from the heart, requiring the systemic output to be supplied by the pulmonary artery system via the ductus arteriosus. When the ductus closes within the first few days or weeks after birth, these infants present with severe shock. Obstructive shock generally requires prompt recognition with medical management Mendelson 4 (eg, initiation of prostaglandin therapy for a ductal-dependent lesion) and/or proce-dural management (eg, pericardiocentesis for tamponade or tube thoracostomy for tension pneumothorax). RECOGNITION Clinical history is important in children presenting to the ED in shock, and it may help to classify the etiology of shock and help direct therapies. Attention should be paid to past medical history and medication use (especially immunosuppression or steroid use). All infants under 3 months of age presenting in shock should be considered sep-tic until proven otherwise. A history of fever or trauma may be particularly elucidative; however, often the history of a child in shock is often nonspecific with symptoms such as lethargy, fussiness, poor feeding, or decreased urine output. Children are usually able to compensate for shock with tachycardia and increased SVR to maintain cardiac output and critical organ perfusion. Tachycardia is the most common presenting physical examination finding in pediatric shock. Persistent tachy-cardia in a calm, afebrile child should be concerning to the emergency provider and should prompt further investigation. For normal heart rates and blood pressure by age, see Table 3.11 Increased SVR manifests as delayed CRT and diminished periph-eral pulses. A recent metaanalysis showed that children with prolonged CRT have a 4-fold greater risk of dying compared with children with normal CRT. They found CRT to be highly specific, but not sensitive, for mortality.12 Another study showed the com-bination of prolonged CRT and hypotension has a staggering mortality rate of 26.9%.13 When compensatory mechanisms fail, hypotension occurs. Guidelines define hypo-tension as a systolic blood pressure of less than the 5th percentile for age.5 In addition Box 1 Hemodynamic profiles in pediatric heart failure, classified by presence of hypoperfusion and/ or venous congestion (increased filling pressures) Warm and dry: Normal perfusion and no congestion. Well-compensated but may have significant cardiac dysfunction. Cold and dry: Poor perfusion without venous congestion. Decompensating. Sick appearing. Increased peripheral vascular resistance. May have oliguria and altered mental status. Warm and wet: Normal perfusion with venous congestion. Still partially compensated. May benefit from diuretics or inodilators. Cold and wet: Poor perfusion with venous congestion. The sickest group of all. Usually requires inotropes. May require mechanical support. Management of Pediatric Shock 5 to hypotension, a child with decompensated shock will present with signs of inade-quate end-organ perfusion, including depressed mental status, decreased urine output, metabolic acidosis, tachypnea, weak central pulses, and worsening peripheral perfusion. These signs of hypoperfusion are highly specific for the development of or-gan dysfunction, even in the absence of hypotension.14 ULTRASOUND EXAMINATION Another useful tool increasingly used for the assessment of children in shock is point-of-care ultrasound (POCUS) examination.15 The focused assessment with sonography for trauma (FAST) examination is used routinely in both pediatric and adult trauma to identify hemoperitoneum, hemopericardium, and hemothorax (plus pneumothorax in the extended e-FAST). One small study found that, when combined with increased liver transaminases of greater than 100 IU/L, the specificity of the FAST examination was 98%, suggesting a negative FAST and transaminases of less than 100 IU/L have a low likelihood of significant intraabdominal injury and should prompt patient observa-tion instead of abdominal computed tomography scanning.16 Although standard mea-surements of the inferior vena cava and aorta are not established in children (as they are in adults), in the evaluation of pediatric hypovolemic shock, both the ratio of the aorta to inferior vena cava and the dynamic assessment of inferior vena cava collapsibility have been studied and both metrics may correlate with hydration status.17–19 In the adult emergency medicine/critical care literature, several POCUS algorithms for the assessment of shock exist, and may dramatically affect treatment decisions and improve survival.20–24 Evidence for the use of POCUS for the assessment of pe-diatric cardiac function, volume status, and shock management has lagged behind that for adult patients, yet the concepts remain similar. Clearly, further studies are needed in this area. TREATMENT Because shock is a problem of inadequate oxygen delivery, every child in shock should be given supplemental oxygen. Place the child on continuous cardiorespiratory and pulse oximetry monitors and obtain peripheral IV access as soon as possible. Check blood sugar and correct hypoglycemia if present. Hypocalcemia (ionized calcium <1.1 mmol/L) may contribute to cardiac dysfunction and should also be corrected. Fluid Resuscitation Isotonic crystalloid solutions are the fluid of choice for resuscitation of a child in shock. Crystalloids are preferred because of their safety, effectiveness, low cost, and wide Table 3 Pediatric heart rate ranges and hypotensive systolic blood pressure levels by age Age HR (bpm) Hypotensive SBP (mm Hg) <1 mo 110–180 <60 1–12 mo 100–170 <70 1–2 y 85–150 <70 1 (2 age in y) 3–5 y 70–140 <70 1 (2 age in y) 6–10 y 60–110 <70 1 (2 age in y) >10 50–100 <90 Mendelson 6 availability.25 In less common circumstances, such as in resource-limited settings in developing countries with a high incidence of malaria, anemia, and malnutrition, take caution with IV fluid resuscitation and consider use of colloid (5% albumin) or early transfusion for suspected anemia.26,27 Treat signs of shock with a fluid bolus of 20 mL/kg, even if blood pressure is normal, and give additional boluses if systemic perfusion fails to improve. In neonates or children with suspected cardiogenic shock, use 10 mL/kg boluses and reassess the patient frequently for signs of volume over-load, including hepatomegaly, S3 gallop, or pulmonary rales/crackles. Volume resus-citation in hypovolemia and sepsis commonly requires 40 to 60 mL/kg, but may require as much as 200 mL/kg. POCUS may be useful to help determine if shock is still volume responsive. It is generally accepted that children remaining in shock after 60 mL/kg of IV fluid should be started on vasopressors. Fluid administration in shock should be as rapid as possible. In infants and children with smaller gauge IVs, a “push/pull” method should be used. Push/pull uses a 3-way stopcock to manually draw a large syringe of fluid from the IV bag (pull) and then rapidly deliver it to the patient (push), and then repeat this process until the full volume is delivered. In 1 trial, fluid administration rates were equivalent in children us-ing a pressure bag versus push/pull system, and both were faster than gravity or an IV infusion pump. Investigators have shown that 20 mL/kg of fluid can be delivered in 5 minutes or less via pressure bag or push methods.28 Although placement of a central venous line (CVL) is common in resuscitation in adults, this is unnecessary in children, at least in the initial stages. For the manage-ment of a child in shock, the goal should be placement of PIVs of the largest bore possible. If the child is in extremis and without access, intraosseous access should be placed without delay. Vasoactive Medications When shock remains refractory to fluid resuscitation, vasoactive infusions should be initiated (Box 2). Although infusion of vasoactive medications through a CVL is preferred, placement may be difficult in children. If the child is in fluid-refractory shock, Box 2 Usual dosing ranges for vasoactive medications Inotropes Epinephrine 0.05 to 1.00 (or more) mg/kg/min Dopamine: 5 to 20 mg/kg/min Dobutamine 5 to 20 mg/kg/min Vasopressors Norepinephrine 0.05 to 0.50 (or more) mg/kg/min Dopamine 10 to 20 mg/kg/min Vasopressin 0.0005 to 0.0100 U/kg/min Inotropes increase cardiac contractility. Vasopressors cause vasoconstriction, increasing sys-temic vascular resistance. Some medications fit into both categories. Start at the low end of the range and titrate rapidly until shock reversal is achieved. If administering via peripheral an intravenous line, dilute the solution (usually 10 the usual central concentration). Addi-tional “driver” fluid (3–5 mL/h of saline) may be needed if the infusion rate is very low (<1 mL/h). Management of Pediatric Shock 7 start vasopressors through whatever line is available (peripheral IV access, intraoss-eous access, or a CVL). A recent small study of peripheral vasoactive medication use in children in a pediatric intensive care unit found IV infiltration and extravasation to occur in only 2% of patients, with none requiring medical or surgical intervention.29 Another larger study in adults found similarly low rates of complications with the pe-ripheral administration of vasoactive medications, including norepinephrine, dopa-mine, and phenylephrine.30 If a peripheral IV is used for vasopressor administration, the medication solution should be diluted and the IV site should be assessed frequently for problems. The use of peripheral vasopressors may be particularly rele-vant in children requiring transport to a higher level of care. Transport should not be delayed for CVL placement.31 The choice of which vasoactive medication to use depends on the clinical picture. Dopamine has long been the initial medication of choice in pediatric shock; however, several recent studies in both adults and children have challenged this dogma.32–34 In adults, dopamine is associated with increased mortality and occurrence of arrhyth-mias compared with norepinephrine.32 Two recent pediatric studies randomized epinephrine versus dopamine use in septic shock. One showed children receiving epinephrine versus dopamine for fluid-refractory septic shock had a lower mortality rate (7% vs 20%). Both groups used peripheral IVs for the initiation of vasoactive med-ications until central lines could be placed.33 The other study did not show a difference in mortality, but children in the epinephrine group had faster resolution of shock and less organ dysfunction than those receiving dopamine.34 In response to this study and other data, the newest ACCM guidelines recommend epinephrine as first-line treatment for cold fluid-refractory shock, with dopamine use (5–10 mg/kg/min) reserved for when epinephrine is unavailable.8 Warm shock is seen much less commonly in children than adults (for whom warm shock predominates). For children in warm shock, norepinephrine is recommended as first-line therapy. Dopamine may be used if norepinephrine is unavailable, and generally requires higher doses than in cold shock (10–20 mg/kg/min). In adults with septic shock, vasopressin levels are frequently low and this finding is thought to contribute to vasodilation. This state has not been found consistently in children, and trials of vasopressin for shock have failed to show benefit.35 However, vaso-pressin remains available as an adjunctive therapy for refractory vasodilatory shock not responsive to norepinephrine. Intubation Airway management and ventilatory support is often necessary in children in shock. Often underrecognized, intubation of a child in shock may be indicated for hemody-namic instability alone. A significant portion of a child’s oxygen consumption (up to 40%) goes into the work of breathing. Support with mechanical ventilation can reduce this oxygen consumption and divert critical cardiac output to vital organs. Care must be taken to fluid resuscitate (and sometimes even start on peripheral vasoactive medication) as best as possible before intubation because initiation of positive-pressure ventilation will decrease venous return and exacerbate hypoten-sion. In a child with decreased cardiac function, the increased intrathoracic pressure associated with mechanical ventilation will afterload reduce the left ventricle and improve cardiac output. Consider the etiology of shock when choosing intubation medications. Although etomidate has been shown to facilitate endotracheal intubation in infants and chil-dren with minimal hemodynamic effect, it is not recommended for use in patients with suspected sepsis owing to its adrenal-suppressive effects. In children and Mendelson 8 adults with septic shock, use of etomidate is associated with increased mortal-ity.36,37 Ketamine has a favorable hemodynamic profile, but without the adrenal suppression, and is the recommended choice for children in septic shock. For shock without sepsis, such as in trauma, the choice of either medication is reasonable. Antibiotics When sepsis is suspected, administer broad-spectrum antibiotics within the first hour of presentation. In a study examining adult patients with sepsis, each hour of delay in antibiotic administration was associated with a mean decrease in survival of 7.6%.38 If possible, obtain cultures to identify the source of infection before antibiotic delivery. Antibiotics should not be delayed if there is difficulty obtaining specimens. Factors such as local antibiotic resistance patterns, recent antibiotic use, existing immunosup-pression, drug allergies, and suspected source of infection may influence what antibiotic is chosen. Steroids If shock persists despite escalating vasoactive medication doses (catecholamine-resistant shock), consider the adjunctive use of stress-dose corticosteroids. Pa-tients with known or suspected adrenal insufficiency (ie, steroid use within the last 6 months, known pituitary or adrenal abnormalities, or sepsis with purpura ful-minans) should receive stress-dose hydrocortisone as soon as possible after shock is identified. Evidence for the use of steroids in pediatric shock is limited and demonstrates conflicting results. In 1 study, children with sepsis who received corticosteroids had no improvement in mortality, days of vasoactive infusion, or hospital duration of stay.39 A recent metaanalysis showed no difference in mor-tality rates between those who did and did not receive steroids.40 For catecholamine-resistant shock, hydrocortisone dosing of 50 to 100 mg/m2/d or 2 to 4 mg/kg/d is generally used, although some investigators advocate for doses as high as 50 mg/kg/d in refractory shock.8 Ideally, a baseline cortisol level should be drawn before hydrocortisone dosing. RESUSCITATION ENDPOINTS Reversal of shock depends on the reestablishment of sufficient oxygen delivery to the body. The Surviving Sepsis Campaign identifies these therapeutic endpoints for resuscitation of pediatric shock: restoration of a CRT of less than 2 seconds, normal blood pressure for age, normal pulses, warm extremities, normal urine output, and normal mental status. Goal-Directed Therapy In addition to clinical resuscitation endpoints, the Surviving Sepsis Campaign and ACCM recommend that resuscitation of children in septic shock should target a mixed venous saturation (SvO2) of 70% or greater, a perfusion pressure (mean arte-rial pressure – central venous pressure) of 55 1 1.5 age in years, and cardiac index between 3.3 and 6.0 L/min/m2. Low cardiac output is associated with increased mor-tality in children with septic shock; a cardiac index between 3.3 and 6.0 is associated with the best outcomes in pediatric septic shock patients compared with patients without shock for whom a cardiac index above 2.0 L/min/m2 is sufficient. Cardiac output measurement can be measured invasively or noninvasively with a variety of devices. Additionally, to maximize oxygen and glucose delivery to help reverse Management of Pediatric Shock 9 Hemorrhagic shock Obstructive shock Neurogenic shock Hypovolemic shock Septic shock Anaphylacc shock Cardiogenic shock Signs of shock ↑HR AMS ↓UOP ↑CRT ↓BP Historical/physical Examinaon Clues Classiﬁcaon of shock Unique treatments Bleeding? Abnormal heart/lung examinaon? Abnormal neurologic Fluid-loss? ↑↓ Temperature? Immunocompromise? Exposure to allergen? Wheeze? Hives? Airway obstruction? Other: Nontraumac tension PTX Massive Pulmonary Embolism Adrenal Insuﬃciency Toxic Ingeson Hypothyroidism RBCs, blood products, massive transfusion protocol, surgical control Pericardiocentesis Chest tube or needle decompression Vasopressors steroids, surgical stabilizaon Cultures, anbiocs Epinephrine IM infusion, anhistamine, steroids PALS PGE infusion, Epinephrine, cardiology/CT surgery consult Inotrope aerload reducon diuresis anarrhythmic a examination ? examination / bradycardia ? with/without with/without with/without with/without with/without A Signs of shock: ↑HR, AMS, ↓UOP, ↑CRT •Manage ABCs, •Apply Supplemental O2 •Obtain vascular access Unresponsive to 60 mL/kg? Fluid-refractory shock Catecolamine-resistant shock Fluid-resuscitate Goal 60 mL/kg given in ﬁrst 30 min Consider 10 mL/kg boluses in neonates Consider history/physical examinaon features suggesve of specific etiology and treat accordingly as needed Warm? Cold? Norepinephrine Epinephrine Still in Shock? Bolus 20 mL/kg Reassess Consider adrenal suﬃciency Goal-directed therapy Transfuse to Hgb >10 g/ dL Check SvO2, perfusion pressure, CI Check lactate Add/titrate: Vasopressor (warm shock, low BP, low SVR) Inotrope (cold shock, low BP, high SVR) Vasodilator (cold shock, normal BP, High SVR) Draw corsol • Give hydrocorsone Sll in shock? ECMO B Fig. 1. Algorithmic approach to the pediatric shock patient. (A) Recognition/classification of pediatric shock. (B) Treatment of pediatric shock. For all patients, (1) Manage ABC’s. (2) Apply supplemental O2 & obtain vascular access. (3) Use history/physical exam 1/- POCUS to classify shock and guide treatment. (4) Frequently reassess response to treatment. (5) Clinical goals 5 normalization of heart rate, mental status, perfusion, blood pressure, urine output. Signs of shock: [HR 5 tachycardia; YUOP 5 decreased urine output; [CRT 5 delayed capillary refill time; YBP 5 hypotension; ABCs, airway, breathing, circulation; AMS, altered mental status; BP, blood pressure; CHD, congenital heart disease; CI, cardiac index; CT, computed 10 shock, the ACCM recommends transfusion to a hemoglobin concentration of greater than 10 g/dL and the administration of maintenance fluids containing D10 (D10 normal saline or D10 ½ normal saline).8,41 In the Surviving Sepsis Campaign’s nonpediatric recommendations, a lactate concentration 4 mmol/L or greater is iden-tified as a key marker of tissue hypoperfusion, and normalization of lactate is a key resuscitation goal. Several pediatric studies have shown that increased lactate levels and failure to clear lactate correlate with mortality and organ dysfunction.42,43 Lactate clearance, however, was notably excluded from the pediatric guidelines as a resuscitation endpoint based on the observation that many children in shock have normal lactate levels as well as the fact that lactate may be increased for many reasons other than cellular hypoxia.8 Targeted resuscitation has its foundation in the classic Rivers’ early goal-directed therapy (EGDT) trial, which showed a significant mortality benefit when specific resus-citation goals were used in the ED management of adults with septic shock.44 A pe-diatric trial of EGDT found significant mortality reduction and decreased organ dysfunction when resuscitation was titrated using SvO2 goals.45 However, EGDT (particularly the requirement for invasive CVP measurement and continuous SvO2 monitoring) has lost some support after 3 recent large methodologically robust trials in adults with septic shock comparing EGDT with usual care showed no benefit in either mortality or secondary clinical and economic outcomes.46 Resuscitation to specific EGDT goals may eventually go by the wayside in pediatric algorithms, but for now the ACCM continues to advocate for the titration of therapies to SVO2, perfusion pressure, and cardiac index goals. In the initial ED management, if invasive monitoring is not used, then usual care must mean vigilant, attentive care. Early recognition with prompt delivery of IV fluids and antibiotics and frequent reas-sessment is critical. Consider trending lactate levels and using noninvasive methods such as POCUS to assess the adequacy of resuscitation. PUTTING IT ALL TOGETHER When a child presents to the ED with tachycardia and signs/symptoms of shock, the most immediate concern should be stabilization of the airway, breathing, and circula-tion, followed by a rapid assessment of historical clues, physical examination findings, and laboratory studies that may aid classification and help to guide treatment. Refer to Fig. 1 for an algorithmic approach to pediatric shock management. Some types of shock require specific therapies. Most shock requires some degree of fluid resuscita-tion, but be cautious if there is concern for a cardiogenic etiology. If shock remains re-fractory to fluids, add inotropes and/or vasopressors. If catecholamine-resistant shock occurs, advanced hemodynamic monitoring may be required to help to titrate therapies. Consider hydrocortisone supplementation. At multiple points along the way, POCUS may assist diagnosis and help to guide therapies including assessment of preload, fluid responsiveness, and cardiac function. At each step, reassess for response to treatment. = tomography; ECMO, extracorporeal membrane oxygenation; Hgb, hemoglobin concentra-tion; IM, intramuscular; PALS, pediatric advanced life support guidelines; PGE, prostaglandin E infusion; POCUS, point of care ultrasound examination (used to help diagnose reasons for shock and assess volume responsiveness and cardiac function); PTX, tension pneumo-thorax; RBCs, red blood cells; SvO2, venous oxygen saturation; SVR, systemic vascular resistance. 11 Management of Pediatric Shock SUMMARY Shock is an unstable pathophysiologic state of inadequate tissue perfusion that must be identified and treated promptly. Failure to recognize and reverse shock can have catastrophic results. In the ED, initial therapies should be titrated to normalize vital signs and physical examination abnormalities. 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Limited echocardiography-guided ther-apy in sub-acute shock is associated with change in management and improved outcomes. J Crit Care 2014;29(5):700–5. 25. Medeiros DN, Ferranti JF, Delgado AF, et al. Colloids for the initial management of severe sepsis and septic shock in pediatric patients: a systematic review. Pediatr Emerg Care 2015;31(11):e11–16. 26. Maitland K, Kiguli S, Opoka RO, et al. Mortality after fluid bolus in African children with severe infection. N Engl J Med 2011;364:2483–95. 27. de Caen AR, Berg MD, Chameides L, et al. Part 12: pediatric advanced life sup-port: 2015 American Heart Association guidelines update for cardiopulmonary resuscitation and advanced emergency cardiovascular care. Circulation 2015; 132:S526–42. 28. Stoner MJ, Goodman DG, Cohen DM, et al. Rapid fluid resuscitation in pediatrics: testing the American College of Critical Care Medicine guideline. Ann Emerg Med 2007;50(5):601–7. 29. Patregnani JT, Sochet AA, Klugman D. Short-term peripheral vasoactive infusions in pediatrics: where is the harm? Pediatr Crit Care Med 2017;18(8):e378–81. 30. Cardenas-Garcia J, Schaub KF, Belchikov YG, et al. Safety of peripheral intrave-nous administration of vasoactive medication. J Hosp Med 2015;10(9):581–5. 31. Turner DA, Kleinman ME. The use of vasoactive agents via peripheral intravenous access during the transport of critically ill infants and children. Pediatr Emerg Care 2010;26(8):563–6. Management of Pediatric Shock 13 32. De Backer D, Biston P, Devriendt J, et al, SOAP II Investigators. Comparison of dopamine and norepinephrine in the treatment of shock. N Engl J Med 2010; 362:779–89. 33. Ventura AM, Shieh HH, Bousso A, et al. Double-blind prospective randomized controlled trial of dopamine versus epinephrine as first-line vasoactive drugs in pediatric septic shock. Crit Care Med 2015;43:2292–302. 34. Ramaswamy KN, Singhi S, Jayashree M, et al. Double-blind randomized clinical trial comparing dopamine and epinephrine in pediatric fluid-refractory hypoten-sive septic shock. Pediatr Crit Care Med 2016;17:e502–512. 35. Choong K, Bohn D, Fraser DD, et al, Canadian Critical Care Trials Group. Vaso-pressin in pediatric vasodilatory shock: a multicenter randomized controlled trial. Am J Respir Crit Care Med 2009;180:632–9. 36. den Brinker M, Hokken-Koelega AC, Hazelzet JA, et al. One single dose of eto-midate negatively influences adrenocortical performance for at least 24h in chil-dren with meningococcal sepsis. Intensive Care Med 2008;24(1):163–8. 37. Sprung CL, Annane D, Keh D, et al. Hydrocortisone therapy for patients with sep-tic shock. N Engl J Med 2008;258(2):111–24. 38. Kumar A, Roberts D, Wood KE, et al. Duration of hypotension before initiation of effective antimicrobial therapy is the critical determinant of survival in human sep-tic shock. Crit Care Med 2006;34:1589–96. 39. Zimmerman JJ, Williams MD. Adjunctive corticosteroid therapy in pediatric se-vere sepsis: observations from the RESOLVE study. Pediatr Crit Care Med 2011;12:2–8. 40. Menon K, McNally D, Choong K, et al. A systematic review and meta-analysis on the effect of steroids in pediatric shock. Pediatr Crit Care Med 2013;14(5): 474–80. 41. Dellinger RP, Levy MM, Rhodes A, et al. Surviving sepsis campaign: International guidelines for management of severe sepsis and septic shock: 2012. Crit Care Med 2013;41:580–637. 42. Choudhary R, Sitaraman S, Choudhary A. Lactate clearance as a predictor of outcome in pediatric septic shock. J Emerg Trauma Shock 2017;10(2):55–9. 43. Scott HF, Brou L, Deakyn SJ, et al. 30-day mortality in clinically suspected sepsis in children. JAMA Pediatr 2017;171(3):249–55. 44. Rivers E, Nguyen B, Havstad S, et al. Early goal-directed therapy in the treatment of severe sepsis and septic shock. N Engl J Med 2001;345:1368–77. 45. deOliveira CF, deOliveira DS, Gottschald AF, et al. ACCM/PALS haemodynamic support guidelines for paediatric septic shock: an outcomes comparison with and without monitoring central venous oxygen saturation. Intensive Care Med 2008;34:1065–75. 46. The PRISM Investigators. Early, goal-directed therapy for septic shock—a patient-level meta-analysis. N Engl J Med 2017;376:2223–34. Mendelson 14
3327	https://www.youtube.com/playlist?list=PLSQl0a2vh4HDL79RlGmKk77ODVXikS2ez	Linear inequalities \| Algebra I \| Khan Academy - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Linear inequalities \| Algebra I \| Khan Academy by Khan Academy • Playlist•26 videos•45,159 views Play all PLAY ALL Linear inequalities \| Algebra I \| Khan Academy 26 videos 45,159 views Last updated on Jul 9, 2018 Save playlist Shuffle play Share Show more Khan Academy Khan Academy Subscribe Play all Linear inequalities \| Algebra I \| Khan Academy by Khan Academy Playlist•26 videos•45,159 views Play all 1 4:31 4:31 Now playing Compound inequalities \| Linear inequalities \| Algebra I \| Khan Academy Khan Academy Khan Academy • 453K views • 14 years ago • Fundraiser 2 6:07 6:07 Now playing Reasoning through inequality expressions \| Linear inequalities \| Algebra I \| Khan Academy Khan Academy Khan Academy • 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3328	https://courses.lumenlearning.com/ccbcmd-math/chapter/using-cramers-rule-to-solve-a-system-of-three-equations-in-three-variables/	Using Cramer’s Rule to Solve a System of Three Equations in Three Variables \| Applied Algebra and Trigonometry Skip to main content Applied Algebra and Trigonometry Chapter 2.5: Solving Systems with Cramer’s Rule Search for: Using Cramer’s Rule to Solve a System of Three Equations in Three Variables Evaluating the Determinant of a 3 × 3 Matrix Finding the determinant of a 2×2 matrix is straightforward, but finding the determinant of a 3×3 matrix is more complicated. One method is to augment the 3×3 matrix with a repetition of the first two columns, giving a 3×5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example. Find the determinant of the 3×3 matrix. A=⎡⎢⎣a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3⎤⎥⎦A=[a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3] Augment A A with the first two columns. d e t(A)=\|a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3\|a 1 a 2 a 3 b 1 b 2 b 3\|d e t(A)=\|a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3\|a 1 a 2 a 3 b 1 b 2 b 3\| From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal. Figure 2 The algebra is as follows: \|A\|=a 1 b 2 c 3+b 1 c 2 a 3+c 1 a 2 b 3−a 3 b 2 c 1−b 3 c 2 a 1−c 3 a 2 b 1\|A\|=a 1 b 2 c 3+b 1 c 2 a 3+c 1 a 2 b 3−a 3 b 2 c 1−b 3 c 2 a 1−c 3 a 2 b 1 Example 3: Finding the Determinant of a 3 × 3 Matrix Find the determinant of the 3 × 3 matrix given A=⎡⎢⎣0 2 1 3−1 1 4 0 1⎤⎥⎦A=[0 2 1 3−1 1 4 0 1] Solution Augment the matrix with the first two columns and then follow the formula. Thus, \|A\|=\|0 2 1 3−1 1 4 0 1\|0 3 4 2−1 0\|=0(−1)(1)+2(1)(4)+1(3)(0)−4(−1)(1)−0(1)(0)−1(3)(2)=0+8+0+4−0−6=6\|A\|=\|0 2 1 3−1 1 4 0 1\|0 3 4 2−1 0\|=0(−1)(1)+2(1)(4)+1(3)(0)−4(−1)(1)−0(1)(0)−1(3)(2)=0+8+0+4−0−6=6 Try It 2 Find the determinant of the 3 × 3 matrix. d e t(A)=\|1−3 7 1 1 1 1−2 3\|d e t(A)=\|1−3 7 1 1 1 1−2 3\| Solution Q & A Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2×2 2×2 and 3×3 3×3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramer’s Rule to Solve a System of Three Equations in Three Variables Now that we can find the determinant of a 3 × 3 matrix, we can apply Cramer’s Rule to solve a system of three equations in three variables. Cramer’s Rule is straightforward, following a pattern consistent with Cramer’s Rule for 2 × 2 matrices. As the order of the matrix increases to 3 × 3, however, there are many more calculations required. When we calculate the determinant to be zero, Cramer’s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system. Consider a 3 × 3 system of equations. Figure 3 x=D x D,y=D y D,z=D z D,D≠0 x=D x D,y=D y D,z=D z D,D≠0 where Figure 4 If we are writing the determinant D x D x, we replace the x x column with the constant column. If we are writing the determinant D y D y, we replace the y y column with the constant column. If we are writing the determinant D z D z, we replace the z z column with the constant column. Always check the answer. Example 4: Solving a 3 × 3 System Using Cramer’s Rule Find the solution to the given 3 × 3 system using Cramer’s Rule. x+y−z=6 3 x−2 y+z=−5 x+3 y−2 z=14 x+y−z=6 3 x−2 y+z=−5 x+3 y−2 z=14 Solution Use Cramer’s Rule. D=\|1 1−1 3−2 1 1 3−2\|,D x=\|6 1−1−5−2 1 14 3−2\|,D y=\|1 6−1 3−5 1 1 14−2\|,D z=\|1 1 6 3−2−5 1 3 14\|D=\|1 1−1 3−2 1 1 3−2\|,D x=\|6 1−1−5−2 1 14 3−2\|,D y=\|1 6−1 3−5 1 1 14−2\|,D z=\|1 1 6 3−2−5 1 3 14\| Then, x=D x D=−3−3=1 y=D y D=−9−3=3 z=D z D=6−3=−2 x=D x D=−3−3=1 y=D y D=−9−3=3 z=D z D=6−3=−2 The solution is (1,3,−2)(1,3,−2). Try It 3 Use Cramer’s Rule to solve the 3 × 3 matrix. x−3 y+7 z=13 x+y+z=1 x−2 y+3 z=4 x−3 y+7 z=13 x+y+z=1 x−2 y+3 z=4 Solution Example 5: Using Cramer’s Rule to Solve an Inconsistent System Solve the system of equations using Cramer’s Rule. 3 x−2 y=4(1)6 x−4 y=0(2)3 x−2 y=4(1)6 x−4 y=0(2) Solution We begin by finding the determinants D,D x,and D y D,D x,and D y. D=\|3−2 6−4\|=3(−4)−6(−2)=0 D=\|3−2 6−4\|=3(−4)−6(−2)=0 We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables. Multiply equation (1) by −2−2. Add the result to equation (2)(2). −6 x+4 y=−8 6 x−4 y=0 ___ 0=8−6 x+4 y=−8 6 x−4 y=0 _____ 0=8 We obtain the equation 0=−8 0=−8, which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. Figure 5 Example 6: Use Cramer’s Rule to Solve a Dependent System Solve the system with an infinite number of solutions. x−2 y+3 z=0(1)3 x+y−2 z=0(2)2 x−4 y+6 z=0(3)x−2 y+3 z=0(1)3 x+y−2 z=0(2)2 x−4 y+6 z=0(3) Solution Let’s find the determinant first. Set up a matrix augmented by the first two columns. \|1−2 3 3 1−2 2−4 6\|1−2 3 1 2−4\|\|1−2 3 3 1−2 2−4 6\|1−2 3 1 2−4\| Then, 1(1)(6)+(−2)(−2)(2)+3(3)(−4)−2(1)(3)−(−4)(−2)(1)−6(3)(−2)=0 1(1)(6)+(−2)(−2)(2)+3(3)(−4)−2(1)(3)−(−4)(−2)(1)−6(3)(−2)=0 As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out. Multiply equation (1) by −2−2 and add the result to equation (3): −2 x+4 y−6 x=0 2 x−4 y+6 z=0 0=0−2 x+4 y−6 x=0 2 x−4 y+6 z=0 0=0 Obtaining an answer of 0=0 0=0, a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. Figure 6 Candela Citations CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution PreviousNext
3329	https://pubmed.ncbi.nlm.nih.gov/19887978/	Contemporary management of intracranial complications of otitis media - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content Service Alert: Planned Maintenance beginning July 25th Most services will be unavailable for 24+ hours starting 9 PM EDT. Learn more about the maintenance. An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Contemporary management of intracranial complications of otitis media George B Wanna1,Latif M Dharamsi,Jonathan R Moss,Marc L Bennett,Reid C Thompson,David S Haynes Affiliations Expand Affiliation 1 Department of Otolaryngology-Head and Neck Surgery, Division of Otology and Neurotology, Vanderbilt University Medical Center, Nashville, Tennessee 37232, USA. george.wanna@vanderbilt.edu PMID: 19887978 DOI: 10.1097/MAO.0b013e3181c2a0a8 Item in Clipboard Contemporary management of intracranial complications of otitis media George B Wanna et al. Otol Neurotol.2010 Jan. Show details Display options Display options Format Otol Neurotol Actions Search in PubMed Search in NLM Catalog Add to Search . 2010 Jan;31(1):111-7. doi: 10.1097/MAO.0b013e3181c2a0a8. Authors George B Wanna1,Latif M Dharamsi,Jonathan R Moss,Marc L Bennett,Reid C Thompson,David S Haynes Affiliation 1 Department of Otolaryngology-Head and Neck Surgery, Division of Otology and Neurotology, Vanderbilt University Medical Center, Nashville, Tennessee 37232, USA. george.wanna@vanderbilt.edu PMID: 19887978 DOI: 10.1097/MAO.0b013e3181c2a0a8 Item in Clipboard Full text links Cite Display options Display options Format Abstract Objectives: Intracranial complications as a result of otogenic infections occur even in the antibiotic era. Meningitis is the most common reported intracranial complication, followed by brain abscess and lateral sinus thrombosis. The purpose of this study is to review our experience and management of these serious complications. Materials and methods: A retrospective chart review was performed at a tertiary referral medical center for the period from 1998 to 2007. Charts with acute or chronic otitis media as primary diagnosis were reviewed, and intracranial complications secondary to either were included in the study. Age, sex, clinical presentation, radiographic findings, management, and outcome were studied. Patients with meningitis or petrous apicitis were not included in the study. Results: Ten cases reviewed had intracranial complications. Five patients had brain abscesses, 1 patient had a subdural empyema, and 4 patients had lateral sinus thrombosis. All patients received broad-spectrum intravenous antibiotics for 6 weeks. Mastoidectomy was performed in all patients, but not all patients were treated with direct drainage of the intracranial abscess, especially if clinical and serial radiographic response was favorable. Conclusion: Otogenic intracranial complications can be fatal if not managed appropriately. Broad-spectrum intravenous antibiotics for 6 weeks is usually sufficient treatment. Management of the intracranial disease takes precedence, but direct drainage of the abscess may not be necessary if a patient's symptoms, neurologic status, and radiographic findings progress favorably. A high index of suspicion should be maintained on all patients presenting with symptoms not typically seen with routine otitis media. PubMed Disclaimer Similar articles [Pathogenesis and methods of treatment of otogenic brain abscess].Derić D, Arsović N, Dordević V.Derić D, et al.Med Pregl. 1998 Jan-Feb;51(1-2):51-5.Med Pregl. 1998.PMID: 9531775 Croatian. Otogenic intracranial complications. a 7-year retrospective review.Hafidh MA, Keogh I, Walsh RM, Walsh M, Rawluk D.Hafidh MA, et al.Am J Otolaryngol. 2006 Nov-Dec;27(6):390-5. doi: 10.1016/j.amjoto.2006.03.004.Am J Otolaryngol. 2006.PMID: 17084222 Acute complications of otitis media in adults.Leskinen K, Jero J.Leskinen K, et al.Clin Otolaryngol. 2005 Dec;30(6):511-6. doi: 10.1111/j.1749-4486.2005.01085.x.Clin Otolaryngol. 2005.PMID: 16402975 Intracranial complications of otitis media: 15 years of experience in 33 patients.Penido Nde O, Borin A, Iha LC, Suguri VM, Onishi E, Fukuda Y, Cruz OL.Penido Nde O, et al.Otolaryngol Head Neck Surg. 2005 Jan;132(1):37-42. doi: 10.1016/j.otohns.2004.08.007.Otolaryngol Head Neck Surg. 2005.PMID: 15632907 Review. Suppurative intracranial complications of upper respiratory tract infections.Yogev R.Yogev R.Pediatr Infect Dis J. 1987 Mar;6(3):324-7. doi: 10.1097/00006454-198703000-00047.Pediatr Infect Dis J. 1987.PMID: 2883627 Review. See all similar articles Cited by Nontypeable Haemophilus influenzae Otitis Media: Mastoiditis and Meningitis Complicated with Central Venous Thrombosis in an Immunocompetent Child.Gönüllü E, Özkan N, Soysal A, Acıoğlu E, Tavil EB, Ötgün SN, Karaböcüoğlu M.Gönüllü E, et al.Case Rep Infect Dis. 2021 Mar 13;2021:8845200. doi: 10.1155/2021/8845200. eCollection 2021.Case Rep Infect Dis. 2021.PMID: 34007496 Free PMC article. Subdural empyema-a rare complication of chronic otitis media: a case report.Fekadu ET, Daniel N, Mengistu ST, Fekadu GT.Fekadu ET, et al.J Med Case Rep. 2024 Aug 3;18(1):351. doi: 10.1186/s13256-024-04671-4.J Med Case Rep. 2024.PMID: 39095925 Free PMC article. Neurological Complications of Acute and Chronic Otitis Media.Hutz MJ, Moore DM, Hotaling AJ.Hutz MJ, et al.Curr Neurol Neurosci Rep. 2018 Feb 14;18(3):11. doi: 10.1007/s11910-018-0817-7.Curr Neurol Neurosci Rep. 2018.PMID: 29445883 Review. Subdural empyema in children.Hendaus MA.Hendaus MA.Glob J Health Sci. 2013 Aug 14;5(6):54-9. doi: 10.5539/gjhs.v5n6p54.Glob J Health Sci. 2013.PMID: 24171874 Free PMC article.Review. [Clinical analysis of 11 cases of otogenic intracranial complications treated by multidisciplinary collaboration].Song Z, Liu W, Wang N, Fu Y, Li Z, Wang C, Sun Y.Song Z, et al.Lin Chuang Er Bi Yan Hou Tou Jing Wai Ke Za Zhi. 2023 Oct;37(10):819-824;828. doi: 10.13201/j.issn.2096-7993.2023.10.011.Lin Chuang Er Bi Yan Hou Tou Jing Wai Ke Za Zhi. 2023.PMID: 37828887 Free PMC article.Chinese. 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3330	https://commons.wikimedia.org/wiki/File:Guide_for_the_use_of_the_International_System_of_Units_(SI)_(IA_guideforuseofint811thom).pdf	Contents File:Guide for the use of the International System of Units (SI) (IA guideforuseofint811thom).pdf Captions Captions Summary \| Guide for the use of the International System of Units (SI) (Wikidata search (Cirrus search) Wikidata query (SPARQL) Create new Wikidata item based on this file) \| \| \| --- \| Author \| Thompson, E. Ambler Taylor, Barry N. \| image of artwork listed in title parameter on this page \| \| Title \| Guide for the use of the International System of Units (SI) \| \| Volume \| NIST special publication 811 \| \| Description \| Subjects: \| \| Language \| English \| \| Publication date \| 2008 \| \| Current location \| IA Collections: NISTSpecialPublications; NISTresearchlibrary; fedlink; americana \| \| Accession number \| guideforuseofint811thom \| \| Notes \| No copyright page found. \| \| Source \| Internet Archive identifier: guideforuseofint811thom \| Licensing \| \| \| Public domainPublic domainfalsefalse \| \| \| \| \| --- \| Public domain \| This work is in the public domain in the United States because it is a work prepared by an officer or employee of the United States Government as part of that person’s official duties under the terms of Title 17, Chapter 1, Section 105 of the US Code. Note: This only applies to original works of the Federal Government and not to the work of any individual U.S. state, territory, commonwealth, county, municipality, or any other subdivision. This template also does not apply to postage stamp designs published by the United States Postal Service since 1978. (See § 313.6(C)(1) of Compendium of U.S. Copyright Office Practices). It also does not apply to certain US coins; see The US Mint Terms of Use. \| \| \| This file has been identified as being free of known restrictions under copyright law, including all related and neighboring rights. \| \| \| Commons Public Domain Mark 1.0falsefalse File history Click on a date/time to view the file as it appeared at that time. \| \| Date/Time \| Thumbnail \| Dimensions \| User \| Comment \| --- --- --- \| \| current \| 16:37, 21 July 2020 \| Thumbnail for version as of 16:37, 21 July 2020 \| 1,429 × 1,893, 90 pages (4.79 MB) \| Fæ (talk \| contribs) \| FEDLINK - United States Federal Collection guideforuseofint811thom (User talk:Fæ/IA books#Fork8) (batch 1993-2020 #17480) \| You cannot overwrite this file. File usage on Commons The following page uses this file: Metadata This file contains additional information such as Exif metadata which may have been added by the digital camera, scanner, or software program used to create or digitize it. If the file has been modified from its original state, some details such as the timestamp may not fully reflect those of the original file. The timestamp is only as accurate as the clock in the camera, and it may be completely wrong. \| \| \| --- \| \| Short title \| Guide for the use of the International System of Units (SI) \| \| Author \| Thompson, E. Ambler \| \| Software used \| Digitized by the Internet Archive \| \| Conversion program \| Recoded by LuraDocument PDF v2.65 \| \| Encrypted \| no \| \| Page size \| 686 x 909 pts 662 x 853 pts 658 x 871 pts 649 x 865 pts 661 x 871 pts 659 x 864 pts 661 x 877 pts 659 x 865 pts 657 x 878 pts 657 x 864 pts 659 x 878 pts 658 x 865 pts 658 x 877 pts 655 x 864 pts 662 x 871 pts 656 x 865 pts 662 x 878 pts 661 x 865 pts 658 x 878 pts 656 x 864 pts 658 x 864 pts 654 x 864 pts 653 x 871 pts 660 x 864 pts 654 x 871 pts 657 x 866 pts 659 x 871 pts 654 x 865 pts 655 x 877 pts 659 x 877 pts 655 x 865 pts 655 x 866 pts 657 x 871 pts 652 x 864 pts 647 x 865 pts 650 x 865 pts 650 x 866 pts 653 x 865 pts 648 x 865 pts 653 x 866 pts \| \| Version of PDF format \| 1.5 \| Structured data Items portrayed in this file depicts copyright status public domain media type application/pdf
3331	https://stackoverflow.com/questions/48306941/efficient-magnitude-calculation-of-3d-vector	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Efficient Magnitude Calculation of 3D Vector Ask Question Asked Modified 2 years, 1 month ago Viewed 4k times 3 I'm working on an inertial measurement project using a 3-axis accelerometer and an Arduino. I want the Arduino to take in the x, y, and z g-values and spit out the magnitude. Since the \|a\| = sqrt(x^2 + y^2 + z^2) is computationally expensive, I wanted to investigate whether there was an alternative algorithm that could be used to speed it up (I'm willing to sacrifice a little accuracy). I read about the Alpha-max, Beta-min method, but that appears to only work for 2D vectors. Is there anything similar for 3D vectors? EDIT: Program language is C++ c++ algorithm vector magnitude Share Improve this question edited Jan 17, 2018 at 18:21 Justinas Marozas 2,71211 gold badge2121 silver badges4040 bronze badges asked Jan 17, 2018 at 17:38 RaddyRaddy 3922 silver badges77 bronze badges 7 It would be helpful if you'd tag a programming language you're using as performance implications will differ based on that. If you're using Python numpy is usually a performant option for calculations. This answer shows an example of how to use it to calculate vector magnitude. Justinas Marozas – Justinas Marozas 2018-01-17 17:43:24 +00:00 Commented Jan 17, 2018 at 17:43 Its being written in C++, added that to the original post. Thanks! Raddy – Raddy 2018-01-17 17:53:45 +00:00 Commented Jan 17, 2018 at 17:53 Related on math.se. Nelfeal – Nelfeal 2018-01-17 18:11:08 +00:00 Commented Jan 17, 2018 at 18:11 A sqrt calculation on a microcontroller can be expensive, but how fast do you need it to be? How much precision are you willing to trade off? Is there anything you know about your inputs that can be leveraged (e.g., are they floats or ints and what range will they be in?) Adrian McCarthy – Adrian McCarthy 2018-01-17 19:45:49 +00:00 Commented Jan 17, 2018 at 19:45 I'm shooting for 100Hz sampling. I haven't had a chance to see if that's possible given the other tasks it is has to do, but since I'm still somewhat in the planning stages, I thought it would be best to start optimizing efficiency now. Raddy – Raddy 2018-01-22 15:37:44 +00:00 Commented Jan 22, 2018 at 15:37 \| Show 2 more comments 3 Answers 3 Reset to default 2 If you have a fast way of calculating two-dimensional magnitude, then perhaps the three-dimensional magnitude can be restructured in those terms. The three-dimensional magnitude can be derived from the Pythagorean theorem. \|a\| = sqrt(sqrt(x^2 + y^2)^2 + z^2) = sqrt(x^2 + y^2 + z^2) Share Improve this answer edited Jan 17, 2018 at 18:38 MAhipal Singh 4,79511 gold badge4444 silver badges5858 bronze badges answered Jan 17, 2018 at 18:02 MattHMattH 13199 bronze badges 1 Comment Raddy Raddy This is a good idea, I'll try this and see if it's faster. Thanks! 0 Normalizing Spatial Vectors without Square Root If you use what this guy posted to calculate the unit vector, you can then divide the x of the original vector by the calculated unit vector's x. Share Improve this answer answered May 13, 2020 at 15:16 August van CasterenAugust van Casteren 1 Comments 0 There is also std::hypot, which computes the length of a 2D vector (since C++11) or 3D vector (since C++17). For in-between versions of C++, you can compute the length of a 3D vector using the 2D version of the function as std::hypot(std::hypot(x, y), z). Hypot is more robust against over- and underflow (especially during squaring of the individual components) compared to computing the formula manually. It might or might not be faster, depending on your standard library and hardware. Share Improve this answer answered Aug 23, 2023 at 5:50 EphEph 26711 silver badge66 bronze badges Comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions c++ algorithm vector magnitude See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Linked How do you get the magnitude of a vector in Numpy? Normalizing Spatial Vectors without Square Root Related Calculate direction vector 0 How do I calculate the magnitude of a 2 vector whose components are very large? Efficient way to reduce a vectors magnitude by a specific length? 1 three dimensional vectors in C++ 3 c++ fastest loop scan over 3D vector 0 computing angle between two 3d vectors - how to implement 3 How do I efficiently determine the scale factor of two parallel vectors? 2 Cheapest way to find Vector magnitude from a given point and angle 0 3 dimensional vector in c++ 1 How to get angle between two vectors in 3D Hot Network Questions Passengers on a flight vote on the destination, "It's democracy!" Mishearing Monica’s line in Friends: “beacon that only dogs…” — is there a “then”? Exchange a file in a zip file quickly Numbers Interpreted in Smallest Valid Base Another way to draw RegionDifference of a cylinder and Cuboid is the following argument valid? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? 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3332	https://blogs.millersville.edu/adecaria/files/2021/11/esci341_lesson06_processes.pdf	ESCI 341 – Atmospheric Thermodynamics Lesson 6 – Thermodynamic Processes References: An Introduction to Atmospheric Thermodynamics, Tsonis Introduction to Theoretical Meteorology, Hess Physical Chemistry (4th edition), Levine Thermodynamics and an Introduction to Thermostatistics, Callen ISOTHERMAL PROCESSES If a process is isothermal (dT = 0) then for an ideal gas the first law becomes α pd dq = For an ideal gas we can substitute for p from the ideal gas law to get α α d T R dq ′ = which integrates to i f T R q α α ln ′ = . We can also use the enthalpy form of the ideal gas law, which for an isothermal process becomes dq dp α = − . When this is integrated we get ln f i p q R T p ′ = − . ISOCHORIC PROCESSES If a process is isochoric (constant volume) then the first law for an ideal gas becomes dT c dq v = . This can be integrated to get (assuming cv is constant) ( ) i f v T T c q − = . ISOBARIC PROCESSES For an isobaric process, dp = 0. Therefore the first law for an ideal gas becomes dT c dq p = which integrates to 2 ( ) i f p T T c q − = . ADIABATIC PROCESSES An adiabatic process is one in which there is no heat transfer (dq = 0). The two forms of the first law of thermodynamics for an adiabatic process in an ideal gas are . dp dT c pd dT c p v α α = − = If we start with the first form of the first law for an ideal gas (the one involving cv) and substitute for pressure from the ideal gas law, we get 0 = ′ + α α d R T dT cv . ο Integrating this gives . ln ln const R T cv = ′ + α which can also be written as . const T v c R = ′ α (1) ο We’ve previously shown that p v c c R′ − = . Therefore, we can write Eqn. (1) as ( ) . const T v v p c c c = − α and defining the ratio p v c c γ ≡ we get . 1 const T = − γ α (2) ο Using the ideal gas law, this equation can also be written as . const p = γ α (3) or ( ) . 1 const p T = − γ γ (4) Equations (2), (3), and (4) are known as the Poisson relations (note that the constant on the right-hand-side is not necessarily the same in each equation. ( ) . . . 1 1 const p T const p const T = = = − − γ γ γ γ α α Poisson relations 3 The Poisson relations relate T, p, and α α α α in ideal gases undergoing quasi-static, adiabatic processes. If you know the initial values of two of these variables, and one of their final values, you can find the other two final values by using these relations. It is important to realize that Poisson’s relations are only valid for ideal gases undergoing quasi-static adiabatic processes! It is inappropriate to use them for nonadiabatic processes. POTENTIAL TEMPERATURE Potential temperature (denoted as θ θ θ θ ) is defined as the temperature an air parcel would have if it were moved dry-adiabatically to a reference pressure, p0, of 1000 mb. From the Poisson relation for T and p [Eqn. (4)] we get (see Exercise 11) p d c R p p T         = 0 θ . If an air parcel undergoes an adiabatic process its potential temperature is conserved. WORK IN AN ADIABATIC PROCESS For an adiabatic process the change in internal energy is solely due to work done on or by the system, du = dw. ο Note that this is true for any system (not just ideal gasses) and regardless of whether the adiabatic process is quasi-static or not. For an ideal gas, du = cvdT = dw. ADIABATIC FREE EXPANSION If an ideal gas is allowed to adiabatically freely expand, unopposed, its temperature will not change. To see why, recall that for an ideal gas undergoing adiabatic expansion v c dT dw = . In a free expansion there is no work done, so there is no change in temperature. But what about the expression 4 dw pdα = − ? The gas expanded, so specific volume changed, so shouldn’t there be work accomplished? KEY POINT: Remember that the expression dw pdα = − only applies to quasi-static processes. A free expansion is not quasi-static, so we can’t calculate work using this expression. EXERCISES 1. Show that for an isothermal process for an ideal gas i f p p T R q ln ′ − = . 2. For an isothermal process for an ideal gas, show that the work done by the system is i f T R w α α ln ′ − = or i f p p T R w ln ′ = . 3. a. For an isobaric process show that ( ) ( ) f i i f p p T T c u α α − + − = ∆ . b. Is this true for all gasses, or only ideal gasses? 4. Starting with cpdT = dq +α α α αdp, derive the Poisson relation ( ) . 1 const p T = − γ γ 5. A 1.5-kg parcel of dry air is at a temperature of 15° ° ° °C and a pressure of 1013 mb. 5 a. How many moles of air are in the parcel? (The molecular weight of air is 28.96 g/mol) b. What is the volume of the parcel? c. What is the specific volume of the parcel? d. If 50 KJ of heat are added to the parcel while its volume is held constant, what is the new temperature of the parcel? (The specific heat of air at constant volume is 717 J-kg− − − −1-K− − − −1). 6. An parcel of dry air is at a temperature of 15° ° ° °C and a pressure of 1013 mb. Heat is added to the parcel to cause it to expand. It expands at constant pressure to 1.5 times its original volume. a. What is the new temperature of the parcel? b. How much work (per unit mass) was done by the parcel during this expansion? c. What was the change in specific internal energy of the air parcel? d. What was the amount of heat per unit mass that was added to the air parcel? 7. An air parcel is at a temperature of 15° ° ° °C and a pressure of 1013 mb. Heat is added to the parcel to cause it to expand. It expands at constant temperature until its volume is 1.5 time it original volume. a. What is the new pressure of the air parcel? 6 b. How much heat per unit mass was added to the air parcel? c. How much work per unit mass was done in expanding the air parcel? d. What was the change in specific internal energy of the air parcel? 8. A dry air parcel at an initial temperature of 20° ° ° °C and a pressure of 950 mb is forced to rise adiabatically up a mountain slope. The top of the mountain is at a pressure of 720 mb. a. What is the temperature of the air parcel when it reaches the top of the mountain? b. What is the work done by the air parcel? 9. A cylinder filled with helium (a monatomic ideal gas) has a volume of 1.8x106 cm3, a pressure of 1.2x105 mb, and a temperature of 300K. The cylinder is contained in an evacuated room with a volume of 16 m3. The cylinder ruptures and helium fills the room. a. What is the pressure in the room after the cylinder ruptures? b. What is the temperature in the room after the cylinder ruptures? c. What is the work done by the expanding helium? 10. A parcel of dry air is initially at a pressure of 900 mb and a temperature of 15° ° ° °C. It rises to the 400 mb level. a. What amount of heat (per mass) must be exchanged with its surroundings if the temperature is to remain constant at 15° ° ° °C during the ascent? Will the heat be gained or lost by the parcel? 7 b. If the parcel first ascends adiabatically to 400 mb, and then heat is added to it to raise its temperature back to 15° ° ° °C, how much heat must be added? Is this the same amount of heat as the previous question? If not, why not? 11. Using the Poisson relation ( ) . 1 const p T = − γ γ show that ( ) p d c R p p T 0 = θ . 12. For a non-ideal gas, will an adiabatic free expansion result in a temperature change? Explain.
3333	http://columbus.cps.edu/uploads/8/3/5/5/83552742/fact_families.pdf	Printable Worksheets @ www.mathworksheets4kids.com Name : 10 2 8 1) = = = = 2 8 10 10 + + – – 8 2 8 2 17 14 3 2) = = = = 14 3 17 17 + + – – 3 14 3 14 9 4 5 = = = = 4 5 9 9 + + – – 5 4 5 4 4) 3) 8 7 1 = = = = 7 1 8 8 + + – – 1 7 1 7 5) 12 10 2 6) = = = = 10 2 12 12 + + – – 2 10 2 10 Each triangle contains the numbers in a fact family. Use the numbers to complete the facts. Sheet 1 Fact Family - Addition & Subtraction 15 9 6 = = = = 9 6 15 15 + + – – 6 9 9 6 Printable Worksheets @ www.mathworksheets4kids.com Name : Answer key 10 2 8 1) = = = = 2 8 10 10 + + – – 8 2 8 2 10 10 2 8 17 14 3 2) = = = = 14 3 17 17 + + – – 3 14 3 14 17 17 14 3 9 4 5 = = = = 4 5 9 9 + + – – 5 4 5 4 9 9 4 5 4) 3) 8 7 1 = = = = 7 1 8 8 + + – – 1 7 1 7 8 8 7 1 5) 12 10 2 6) = = = = 10 2 12 12 + + – – 2 10 2 10 12 12 10 2 Each triangle contains the numbers in a fact family. Use the numbers to complete the facts. Sheet 1 Fact Family - Addition & Subtraction 15 9 6 = = = = 9 6 15 15 + + – – 6 9 9 6 15 15 6 9
3334	https://www.youtube.com/watch?v=eRN8HIGH5NQ	What are Odd Functions? \| Functions and Relations, Function Symmetry, Even and Odd Functions Wrath of Math 290000 subscribers 21 likes Description 986 views Posted: 20 Jul 2019 What is an odd function? Odd functions, like even functions, have a special type of symmetry. The graph of an odd function has rotational symmetry about the origin. So, if we rotate the graph of an odd function 180 degrees about the origin, it will appear as though we haven't rotated it at all! It's pretty neat, and we go into further detail in the full video lesson, including the definition of an odd function, and how to show that a function is odd! In order for a function to be odd, every element of its domain must satisfy the equation f(-x) = - f(x). If this is the case, the function is odd! My lesson on even functions: SOLUTIONS TO PRACTICE PROBLEMS: f(x) = x^5 - x^2 f(-x) = (-x)^5 - (-x)^2 = (-x)^5 - x^2 -f(x) = -x^5 + x^2 Thus, f(-x) is not equal to -f(x), so f is not odd. g(x) = x^3 + 3x g(-x) = (-x)^3 - 3x = (-x)^3 - 3x -g(x) = -(x^3) - 3x = (-x)^3 - 3x Thus, g(-x) = -g(x), so g is odd. I hope you find this video helpful, and be sure to ask any questions down in the comments! The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work. Vallow Bandcamp: Vallow Spotify: Vallow SoundCloud: +WRATH OF MATH+ ◆ Support Wrath of Math on Patreon: Follow Wrath of Math on... ● Instagram: ● Facebook: ● Twitter: My Music Channel: 5 comments Transcript: hey everyone welcome back to wrath of math in today's video we'll be answering the question what is an odd function just yesterday I posted a lesson on even function so be sure to check that out if you're interested otherwise let's just jump into today's lesson just like even functions odd functions have a special type of symmetry so here is a graph of an odd function what symmetry does this function have it's a little tougher to pick out than with the even functions but odd functions have rotational symmetry about the origin so if we were to rotate this function 180 degrees about the origin it would look like we hadn't rotated it at all and that is true of all odd functions let's try to see this rotational symmetry in action I've made this beautiful orange copy of our graph let's go ahead and rotate it 180 degrees about the origin I'll do the best I can this is a little difficult because my fingers get in the way of the screen but we're rotating this function 180 degrees about the origin and what happens when we complete this rotation well we're just about right back where we started make that a little bigger you can see it looks like we never moved the function at all so hopefully that helps you get to see exactly what we mean by a 180 degree rotation about the origin and that is the special type of symmetry that odd functions have and just like with even functions knowing a function is odd can be useful because if we know it's odd and we know its behavior on one side of the y axis then we also know its behavior on the other side of the y axis and I'll bet you've seen a function like this before let's go back to the function that we began with this beautiful odd function so we've talked a bit about the symmetry now what exactly is the definition of an odd function how can we demonstrate mathematically that a function is odd well in order for a function to be odd every element in its domain must satisfy a particular and I'll bet with a bit of thinking you might be able to figure out what that equation is I'll write down most of it here it's an equation like this F of negative x the function evaluated at the value negative x is equal to negative something let's go ahead and represent this part of the equation on the graph for starters let's just write down what this function is in case you're not familiar with it this is a graph of we'll call it G of X to avoid confusion G of X is equal to X cubed so remember our equation F of negative x is equal to negative something this function evaluated at let's say negative 2 is equal to negative 8 this point here is negative 2 negative 8 so again F of negative x is equal to negative something or the opposite of something this function which is an odd function evaluated at negative 2 is equal to negative 8 negative 8 is the opposite of positive 8 and what is positive a equal to with relation to this function 8 is the value of this function at x equals positive 2 and notice of course that 2 is the opposite of negative 2 so this whole equation is f of negative x is equal to negative f of positive x so if we evaluate an odd function at some value negative x that will be equal to the negative of the function evaluated at positive x it's pretty neat and that's really all there is to it if every element in the domain of a function satisfies this fairly simple equation then the function is odd and you might be convinced by now that this function is odd but how could we demonstrate that for sure to see if our function G is odd we want to find out what G of negative x is and what negative G of X is and if they're equal then the function is odd so we can start with either one of these let's just start by looking at G of negative X define G of negative x we just have to evaluate this function at negative X so we'll substitute negative x in for X thus G of negative x is equal to negative X cubed here we've got negative 1 times X all raised to the power of 3 so we can rewrite this as negative 1 whoops negative 1 to the power of 3 multiplied by X to the power of 3 a negative number raised to an odd power is always negative so we know that we'll have negative 1 as a factor here and then X cubed we can leave that as it is so this is equal to negative x cubed so we have evaluated G of negative X what about negative G of X well we know that G of X is equal to X cubed so to find negative G of X we just multiply by negative 1 negative G of X is equal to negative x cubed and would you look at that this is the same thing that we got up here for G of negative X thus we have demonstrated that G of negative x which is equal to negative 1 times X cubed is equal to negative G of X which is also equal to negative 1 times X cubed therefore we know that G is an odd function so that is how we can demonstrate that a function is odd and if it had turned out that these two values were not equal then we would know that it is not an odd function and before we go you know I always like to leave you with some examples so here are two examples to try on your own tell me which one of these functions is odd and which one is not in the comments as always I'll leave the solution in the description so I hope this video helped you understand what functions are as well as how to determine if a function is odd let me know in the comments if you have any questions need anything clarified or have any other video requests thank you very much for watching I'll see you next time and be sure to subscribe for the swankiest math lessons on the and a big thanks to valo who upon my request kindly gave me permission to use his music in my math lessons linked to his music in the description [Music]
3335	https://captaincalculator.com/math/root/fifth-root-calculator/	Fifth Root Calculator Skip to content Menu Mathexpand child menu Combinationexpand child menu Combinations (nCr) Permutation (nPr) Numberexpand child menu Absolute Value (Modulus) Factorial (n!) 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The fifth root of a number is the number that would have to be multiplied by itself 5 times to get the original number. For example, the fifth root of 243 is 3 as 3 x 3 x 3 x 3 x 3 is 243. The fifth root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,024. Formula – How to calculate the fifth root Even for perfect fifth root numbers, the fifth root can be difficult to calculate by hand. The most basic techniques involve trial and error. Example: Find the fifth root of 32,768 1) Try a number – 5 : 5 x 5 x 5 x 5 x 5 = 3,125 (too low) 2) Try another number that is more than 5 – 6 : 6 x 6 x 6 x 6 x 6= 7,776(too low) 3) Try a number that is more than 6 – 10 – 10 x 10 x 10 x 10 x 10 = 100,000 (too high) 4) Try a number in between 6 and 10 – 8 – 8 x 8 x 8 x 8 x 8 = 32,768 (answer) A fifth root can also be calculated as the same number to the exponent 1/5. 5√number = number 1/5 Table – Fifth Root (5√) of 1 – 100 The Root 5 of...\| Find the 5th root of... \| \| The root 5 \| --- \| 5√ 1 \| = \| 1.0000000000 \| \| 5√ 2 \| = \| 1.1486983550 \| \| 5√ 3 \| = \| 1.2457309396 \| \| 5√ 4 \| = \| 1.3195079108 \| \| 5√ 5 \| = \| 1.3797296615 \| \| 5√ 6 \| = \| 1.4309690811 \| \| 5√ 7 \| = \| 1.4757731616 \| \| 5√ 8 \| = \| 1.5157165665 \| \| 5√ 9 \| = \| 1.5518455739 \| \| 5√ 10 \| = \| 1.5848931925 \| \| 5√ 11 \| = \| 1.6153942662 \| \| 5√ 12 \| = \| 1.6437518295 \| \| 5√ 13 \| = \| 1.6702776523 \| \| 5√ 14 \| = \| 1.6952182031 \| \| 5√ 15 \| = \| 1.7187719276 \| \| 5√ 16 \| = \| 1.7411011266 \| \| 5√ 17 \| = \| 1.7623403478 \| \| 5√ 18 \| = \| 1.7826024580 \| \| 5√ 19 \| = \| 1.8019831273 \| \| 5√ 20 \| = \| 1.8205642030 \| \| 5√ 21 \| = \| 1.8384162873 \| \| 5√ 22 \| = \| 1.8556007363 \| \| 5√ 23 \| = \| 1.8721712306 \| \| 5√ 24 \| = \| 1.8881750226 \| \| 5√ 25 \| = \| 1.9036539387 \| More Root Calculators nth Root Square Root Cubed Root 4th Root 5th Root … more root calculators More Math Calculators Exponents Percentages Number Roots Statistics … more math calculators SPONSORED Discover more Calculator math Math Mathematics calculators Calculators Financial calculator Activewear Financial Calculators Mathexpand child menu Combinationexpand child menu Combinations (nCr) Permutation (nPr) Numberexpand child menu Absolute Value (Modulus) Factorial (n!) 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3336	https://www.jstage.jst.go.jp/article/katejournal/28/0/28_4/_pdf	【理論的研究論文】 A Consideration of Corpus Use in English Teaching and Learning in Japan: Referencing a Learner Corpus and a Corpus of Native Speakers HATA Kazuki Newcastle University Abstract The aim of this paper is to examine the use of a learner corpus and a corpus of native speakers in foreign or second language (L2) teaching and learning in Japan. It is revealed that learner corpora can function as a reference to identify and clarify difficult materials for a particular learner group, which is significant step for better language learning (Meunier, 2002, p. 125; Nesselhauf, 2004, p. 125-126). In this paper, two corpora are employed in language learning; the Japanese English as a Foreign Language Learner Corpus (JEFLL corpus) and the Corpus of Contemporary American English (COCA). These corpora represent the language produced by native speakers and non-native speakers with a particular language background, respectively American first language (L1) speakers of English and Japanese learners of English. It is arguable that L2 teaching and learning with these corpora will allow learners to compare language patterns between native and non-native speakers of English. This paper proposes and discusses possible activities using the corpora. These concrete examples are provided to facilitate an understanding of the difficulties faced by Japanese learners of English in classroom practice; and additionally, demonstrate the efficacy of applying corpora to scaffold learning and pedagogic practice. Keyword: Corpus, Data-Driven Learning (DDL), Inductive Approach, Notice-the-Gap Introduction Learner corpora are electronic collections which contain language samples produced by L2 learners. While native corpora are most frequently encountered in Corpus Linguistics, it is arguable that learner corpora can be a critical source to aid language learners because “[l]earner corpora tend to contain a much higher error rate than native corpora” (Granger 2009, p. 23). This is important because learner corpora can be used to identify and emphasise difficult materials particularly for a certain group; and evidences the typical difficulties for learners of a certain language (Nesselhauf, 2004, p. 125-126). Similarly, Meunier (2002) argues that (a) a learner corpus can work as a further reference revealing problematic language forms for learners and (b) it allows teachers to take into account learner’s L1 (p. 125). Moreover, referencing actual data in learner corpora is considered an efficient way to clarify difficult materials for specific learner groups. It is authentic to the target group and reveals features that may be unclear or unsurprisingly non-evident in native corpora (Granger, 1998, p. 7; Granger & Tribble, 1998, p. 201). One approach towards language learning where corpora are involved is ‘data-driven learning (DDL)’. Johns (1991) argues that language learning should be driven by authentic language data because a language learner will also be a researcher who applies this data to L2 learning. In fact, it has been reported that DDL makes L2 learners aware of linguistic differences between their inter-language and the target language (see Johns, 1994; Meunier, 2002, p. 134; O’Keeffe, McCarthy & Carter, 2007). In the field of L2 teaching and learning, it is considered likely that raising learner’s awareness of these gaps is key to successful learning; in other words, language acquisition can be advanced by noticing the gap between a learners’ own language and the target language (Ellis, 1993, p. 99; Joyce & Burns, 1999, p. 48). Regarding the significance of facilitating notice-the-gap tasks with corpora, Tribble and Jones (1990) propose that learners could be motivated by having access to language in use, which is not formulated for pedagogical purposes (p. 35-36). Additionally, as concordance data from the corpora tend to be substantial, the output is frequently divided into some parts for dyadic or group work to help learners cope with the messy data in the concordance (p. 45). Therefore, it can be said that DDL with corpora will not only raise learner’s awareness but also facilitate interaction in the classroom. This paper discusses the utilisation of two types of corpus in English teaching and learning in Japan. Even though many applications of learner corpora have been examined, there are “very few studies which relate the findings from learner corpora to actual classroom practice” (Tono, 2003, p. 806). Consequently, there is significance in discussing the application of learner and native corpora for L2 teaching and learning; and concomitant to this, providing exemplars which situate the theory firmly within language learning. For the purposes of this paper, two corpora are employed: the Japanese EFL Learner Corpus (JEFLL corpus) and the Corpus of Contemporary American English (COCA). Utilising these corpora allows comparative focus on the gaps between the productions of L2 learners and those of native speakers. Corpora employed 2.1 Japanese EFL Leaner Corpus (JEFLL corpus) This paper assumes the use of the JEFLL corpus employed as a learner corpus for L2 learning and teaching. This corpus is a collection of written language by Japanese native speakers learning English as L2. More than 10,000 Japanese junior high school students (from elementary to intermediate level) have been involved and approximately 700,000 words were collected from the writing tasks in class. This corpus also has sub-corpora, in other words searchable results accessible from different settings: genre or topic of their writing, the learner’s year, as well as school category, e.g., public or private school. This feature is arguably useful across multifarious research contexts, for instance, work focussing on specific learner’s level. However, it should be noted that these sub-corpora are not employed in this paper because the total amount of information available is not sufficient for my investigation. In other words, it is difficult to compare sub-corpora with for example a major corpus like the International Corpus of Learner English (see Granger, 2003) because using sub-corpora can often lead to a lack of data. Consequently, this paper assumes the usage of the JEFLL corpus without any filters for the concordance. Significantly, it is arguable that the JEFLL corpus can be more easily employed by Japanese learners of English than corpora situated in other countries and contexts due to the following vantage points. Firstly, this corpus is totally free of charge for research and pedagogical purpose. Some corpora cannot be used without licence or permission; and this has financial implications which are obviated through use of the JEFLL corpus. Secondly, learners can select the operating language through which the corpus is manipulated, Japanese or English. Lastly, a great deal of the content and instruction are written in Japanese. Thus, Japanese learners will find the data more accessible and understand how to use this corpus more effectively. 2.2 Corpus of Contemporary American English (COCA) The COCA is the first large and genre-balanced corpus of any language, which enables researchers, teachers and learners to examine the recent language usage (Davies, 2010, p. 447). This corpus contains over 450 million words of American English, reflecting up-to-date usage. Due to this large amount of authentic data, it is highly possible to state that this corpus is a reliable source of information which can be referenced in language learning. Consequently, in this paper, this corpus is employed as a reference corpus in L2 learning. As for language variations in genre, this paper takes this into account by only making use of academic contexts. It is thought that this will reduce the occurrence of inaccurate and infrequent examples; in explanation, even a native corpus has some tokens showing ungrammatical or unacceptable usage in some contexts. For example, eleven tokens of “discuss about” are presented in the COCA. However, the verb “discuss” is not commonly used with the preposition “about” . On the other hand, only two examples of “discuss about” are confirmed in the same corpus when the academic contexts are set. It is therefore rational to suggest that focusing on academic contexts will control for this variable when employing the COCA. The design of the interface on the COCA also helps to facilitate the comprehension of the Keyword in Context (KWIC) concordance by learners and teachers; and which is commonly employed to output results (Barnbrook, 1996, p. 67). In the COCA, words are coded by different colours. Due to this word coding, learners can focus on a target category of words without reading whole texts and identifying words by themselves. As a result, it is possible for teachers to let learners focus on a particular word category within a colour category to understand language structure. Thus, this word coding highlighted by colour is highly helpful for learners especially at beginner or intermediate level, since they might have difficulties in understanding authentic language data (Oghigian & Chujo, 2010, p. 203). Examples of corpus use This paper proposes that L2 learners can focus on frequently produced common errors made by Japanese learners of English by comparing the language productions of native speakers and L2 learners. Significantly, the concordance can be used as an actual reference offering a more inductive approach to L2 learning (Meunier, 2002, p. 130). In other words, learners can likely find solutions and gain an understanding of confusing rules in use by referencing the concordance without explicit teaching. For example, an activity with two corpora can demonstrate the word category following particular prepositions; and this has been reported a confusing issue for non-native English speakers especially those at elementary or intermediate level (Hultfors, 1986, p. 108-111). In fact, Figure 1 actually shows that Japanese learners of English used three word categories: noun, verb (basic form) and gerund after the phrase “looking forward to” . On the other hand, Figure 2 suggests that native speakers do not commonly use the basic verb forms after that phrase. This comparative result can help learners understand that Japanese L2 learners tend to be confused when distinguishing between “to” as a preposition or to-infinitive; the former is mainly followed by a noun phrase including gerunds but the latter is followed by the basic form of verb. When L2 learners compare the language data, it is likely to scaffold their understanding and facilitate comprehension of the differences between sentences produced native speakers and L2 learners. Figure 1 . The concordance of “looking forward to” in the JEFLL corpus. Figure 2 . The KWIC concordance of “looking forward to” in the COCA. The language learning with corpora can also help learners focus on the complicated structures of English. For example, the phenomenon called ‘tough-movement’ can be introduced through comparing the data seen in both corpora. As seen in Figure 3, the JEFLL corpus shows that learners tend to use “it” as a formal subject in most cases with an adjective “difficult” followed by the ‘to-infinitive’ structure. Conversely, there are two examples in which a noun subject precedes a verb phrase. Those sentences arise from a phenomenon called ‘tough-movement’; an object of the infinitive clause can be moved out of the position of object and moved into the position of subject. If this phenomenon is a completely new issue for L2 learners, comparing different sentence patterns seen on the concordance can be an effective approach to expand their grammatical knowledge through activities. In addition, the concordance found in the L1 corpora should be referenced if possible, in order to double-check the acceptability of the ‘tough-movement’ structure (see Figure 4). Figure 3 . The concordance of “is difficult to” in the JEFLL corpus. Figure 4 . The KWIC concordance of “is difficult to” in the COCA. Significantly, these activities utilising authentic data can let L2 learners examine not only complicated grammatical features but also unexpected language patterns. As discussed above, ‘tough-movement’ can be seen in the productions made by L2 learners and those by L1 speakers. In the examples above, it is unsurprising if learners assume that the ‘it + is + adjective (adj) + to-infinitive’ structure can always be paraphrased into the ‘tough-movement’ structure. However, learners can come to realise that this paraphrasing might not be always acceptable by analysing the data of target adjectives in native corpora. Figure 5 shows the total number of tokens accounting for these two sentence patterns in cases with four adjectives 2. This figure might suggest that native speakers use the ‘tough-movement’ structure with three adjectives; “easy” , “difficult” and “impossible” in some cases. On the other hand, significantly few examples of the ‘tough-movement’ structure with “possible” are found in the COCA. This result seems to imply that the ‘it + is + adj + to-infinitive’ structure would be more appropriate with an adjective “possible” . As discussed, the concordance can help learners and teachers analyse these unexpected patterns of language. Therefore, it is conceivable that the JEFLL corpus and the COCA can support L2 teaching and learning, helping learners understand patterns of target words in use. Phrase it + is + adj + to tough-movement Total is easy to 80 20 100 is difficult to 69 31 100 is possible to 100 0 100 is impossible to 90 10 100 Figure 5. A total number of tokens in the KWIC concordance of the COCA. Furthermore, the concordances can offer language learners the opportunity to examine the different language usage or patterns between spoken and written language. Figure 6 shows that language might be expressed differently depending on the context: spoken and written. Reppen and Simpson-Vlach (2010) suggest that it is very difficult or nearly impossible to see this kind of data by only utilising textbooks or dictionaries designed for pedagogical objective, as they are lacking in clues to reveal patterns of words in use (p. 103). Corpora on the other hand contain this data and can facilitate L2 learning. Therefore, it is suggested that activities conducted with corpora can cover the differences in language usage depending on the different context, e.g., spoken vs. written language or academic vs. non-academic. Genre it + is + adj + to tough-movement Total ACADEMIC 80 20 100 SPOKEN 53 47 100 Figure 6. A total number of tokens of “easy” in the KWIC concordance of the COCA. Classroom activities with corpora As discussed above, the basic aim of DDL with corpora is to teach language inductively and let L2 learners discover the actual usage through DDL activities. According to Horváth (2001), learners have to find new knowledge about language without explicit language teaching or instruction for better L2 learning (p. 57). From this viewpoint, activities in the language classroom should allow learners to have access to enough data of language in order to let them learn inductively. As such, I argue that concordances taken from corpora can be effectively applied to classroom activities because they are comprised of sufficient language data. There are arguably two major ways to introduce the concordances for classroom activities: paper-based or computer-based activities. 4.1 Paper-based activity The first way of making the concordance available in the classroom is through a paper-based activity. Teachers can make an activity sheet which contains language materials from the concordance and distribute them to learners. The great merit of paper-based materials lies in the fact that the teacher can edit or change the appearance of the data when appropriate; and this is because learners do not see the raw results directly on the screen. In fact, it is argued that successful activities with corpora for language learning might strongly depend on appropriate mediation between corpora and users (Braun, 2005). Hence, it is an essential task for teachers to edit or mediate the data of interest; and make them suitable for learners depending on their proficiency level. In addition, it is possible to conduct paper-based activities when facilities available are insufficient, for example, in institutions when there isn’t a computer for each learner in the class. Moreover, this method obviates the need to provide digital training to learners but still allows for utilisation of a corpus. In line with the DDL objectives, a possible paper-based activity with the JEFLL corpus and the COCA can be designed to make learners aware of the different patterns in language productions between native speakers and L2 learners; and this can obviate any potential challenges that occur due to technological limitations, e.g., lack of facilities or the difficulty of data manipulation (see below in detail). For example, Figure 1 and 2 presented above can be used and described on the activity sheet, encouraging them to notice the gap of language usage by analysing and discussing the question “which category of words comes after looking forward to ?” with other learners if possible. 4.2 Computer-based activity Instead of paper-based activities, learners and teachers can actually use corpora on the computer for L2 learning in the class. A distinct advantage of a computer-based classroom is that teachers can easily change or add materials related to the topic depending on learner preference and immediate needs. For example, if some learners raise an unexpected question or problem, teachers can adjust the activities by introducing new data or conducting another activity. Hence, it is credible that computer-based activities can be more flexible than paper-based. Computer-based activities allow learners to notice the gap in language productions by virtue of advantageous real-time learning. Due to the flexibility of a corpus and the concordancing software, learners can broaden a search to deliver new results by applying different search settings within the classroom. For example, learners can deal with the exceptional usage of particular adjectives by analysing the KWIC concordance in the COCA and counting tokens of target words or phrases, followed by a further search according to their needs. Here, two adjectives “hard” and “fun” are analysed as an example task (see Figure 7). Phrase it + is + adj + to tough-movement Total is hard to 67 33 100 is fun to 10 6 16 Figure 7 . A total number of tokens of “hard” and “fun” in the KWIC concordance of the COCA. From these results, learners can examine whether it is possible to make use of a particular adjective they want to employ within the ‘tough-movement’ structure. They will also see that there are few cases of “fun” in any of these cases. This result implies that native speakers tend not to use this word as adjective with both structures in daily interaction. As seen, the benefit of this activity means that learners can analyse and discover different patterns of particular words by themselves. In this approach, it is important that every learner has access to a computer, so technological limitations should always be taken into account before employing this type of DDL. Firstly, teachers and learners should have enough equipment and be connected to the internet to facilitate access to the information held in the corpora. Contemporarily, a large proportion of freely available corpora have datasets that are only accessible when users have an internet connection. Indeed, this is in an issue for many schools, as Oghigian and Chujo (2010) point out that not all schools possess sufficient equipment to undertake computer-based language learning with learners (p. 210). Secondly, the complexity of successfully implementing computer-based activities should not be overlooked; every learner needs pre-class preparations in order to successfully engage with the materials and equipment during class. If this is not considered, learners might use corpora without sufficient understanding and this may generate findings that bear little resemblance to true reality of the interactional aspect under consideration. Consequently, it is argued that teachers must be aware of the limitations of computer-based activities and consider targeted solutions for their learners. Limitation Previous discussion has revealed that learners can have access to authentic language by referencing the corpus data. Unfortunately, DDL with authentic materials will not be suitable in every context and for every learner. Meunier (2002) notes that the use of corpora for DDL should not be “dogma” despite the recommendation that it is extremely useful in clarifying grammatical features through analysis of authentic language data (p. 129). As for the limitation, the studies have reported two important points for consideration. The first limitation may arise as a result of the still-inadequate data found in the concordance. Unfortunately, it is possible for authentic interaction to lose important background context when they inputted and stored in a corpora; the data in both the JEFLL corpus and the COCA is unimodal, in other words texts without audio and video information. Certainly, with regard to this issue, Kaltenböck and Mehlmauer-Larcher (2005) suggest that learners can only engage with the external data of a piece of language and no longer see the co-contextual language. Additional to this, challenges resulting from cultural background should also not be overlooked. Because background information other than language will not be seen on the concordance of most corpora, there is a possibility that L2 learners might not guess some of the meaning from the context. If the context is unusual, or some words are very specific and commonly used only by a particular culture, it may be difficult for non-native learners to fully comprehend the results; and as such the importance of paraphrasing and mediation is highlighted when employing DDL (O’Keeffe et al., 2007, p. 26). The second limitation is results from employing DDL for L2 learning. Regarding this point, Meunier (2002) proposes three particular weaknesses. Firstly, DDL activity is a time consuming activity for learners to adopt. Secondly, learners have different preferences in regard to their style for language learning; inductive vs. deductive. Finally, DDL activities might make learners confused and frustrated because there is vagueness in the data. Arguably, this point is critical when considering employing DDL with L2 learners who are at a beginner or intermediate level when they engage with the target language. For example, the following sentences taken from the concordance of the COCA may be difficult for lower level learners to comprehend: (5.1) It is possible to do the right things in the right order to make this… (5.2) the origin of IKOV is difficult to determine… (5.3) yet I do think it, you know, is possible to create something on… These sentences can be used in order to examine two sentence patterns with particular adjectives. The sentence (5.1) might be simple for L2 learners because a formal subject “it” is followed by “is + adj” . For the same reason, it will also be easy to analyse the sentence (5.2) if they have already learnt the ‘tough-movement’ structure. Contrary to these two sentences, however, the sentence (5.3) will be troublesome since there is a discourse marker “you know” (see Carter & McCarthy, 2006), which is unlikely to be frequently encountered in a text book. If learners do not know the function of these words, then the sentence is open to misinterpretation. Although these markers might be effectively learnt with authentic data utilising a concordance, this type of language will be ignored here since activities presented in this study do not focus on the function of these markers. Conversely, some studies have counter-argued that benefits of language learning with corpora outweigh the posited limitations. Yoon and Hirvela (2004), for instance, conclude that L2 learners are very positive toward using corpora for L2 learning. Interestingly, they do not report serious problem in learning with corpora. In addition, teachers can deal with the difficulty of materials by supplementing the DDL with paper-based activities which will allow teachers to mediate data dependent on the needs of the L2 learners (see Braun, 2005). Consequently, these studies might imply the possibility of language learning with two types of corpus. However, it is still important to keep limitations to a manageable level. In order to reduce those unwelcome outcomes, some researchers have suggested two precautions that should be considered in DDL activities. Firstly, teachers should carefully select or mediate language materials for learners (O’Keeffe et al., 2007, p. 27). This mediation will likely avoid problems derived from a lack of proficiency level or cultural differences. Secondly, instruction by teachers should be clear and adequate; and additionally should take place before a DDL activity. Comprehensive instruction is mandatory since many learners might be confused when they first engage with a KWIC concordance (Tribble & Jones, 1990, p. 36). Therefore, successful activities with corpora depend to what extent teachers can make activities appropriate for their learners. Conclusion So far, this paper has discussed the possibility of utilising two corpora, i.e., the JEFLL corpus and the COCA, for English teaching and learning. In conclusion, activities with these corpora can be recommended for the following reasons: 1) Learner corpora allow teachers and learners to focus on the typical errors produced by a particular learner group. 2) Learners can analyse the actual usage of words in English by comparing the data in the JEFLL corpus and those in the COCA without explicit grammar teaching. 3) The data in the COCA can help learners expand the grammatical knowledge inductively, dependent on their demands without explicit instruction. 4) DDL activities, which might be difficult for individual learning, can be facilitated with pair or group work in the classroom. On one hand, teachers and learners should know the limitations of activities with corpora. Here, it should be highlighted that DDL activities with the concordance can facilitate L2 learning, provided that teachers successfully control or adjust learning materials and activities properly for learners. From this viewpoint, it is suggested that potential limitations should be taken into account before implementation. The main concerns are summarised as follows: 1) It should be understood that the data in the JEFLL corpus and the COCA cannot be used to focus on the language beyond what is actually seen on the concordance. 2) Teachers should take into account the different factors with regard to learners, e.g., learning style, requiring time to adapt DDL and how or to what extent they are frustrated with these activities. 3) If computer-based activities are employed, both teachers and learners have to be given sufficient equipment to use corpora in the classroom. Further studies should be conducted, firstly because it is important to know whether DDL employing the JEFLL corpus and the COCA actually interests Japanese native speakers who have learnt English as L2. Secondly, examination of the extent to which the concordance helps these learners understand the authentic materials stored in these corpora should be a focus for future research attention. It is therefore suggested that researchers conduct a study recruiting Japanese native speakers learning English as L2 in order to analyse the positive and negative effect of using two types of corpus for classroom activities. If the outcomes are positive, DDL with these corpora can be supported as an effective way of language learning for Japanese learners of English. Notes Inductive language learning is an approach which contrasts with deductive learning. This approach is based on learner discovery of the pertinent grammatical rules by engaging with the target language without explicit instruction by teachers (see Richards & Schmidt, 2002, p. 146). 2. The result of the KWIC concordance can be varied if the result is sorted by setting the number of items to be reported, displaying either 100, 200, 500 or 1,000 tokens which are selected from total number of tokens available in the corpus. This paper set ‘100 hits’ per target word or phrase; and all figures were captured in November, 2013. References Barnbrook, G. (1996). Language and computers: A practical introduction to the computer analysis of language .Edinburgh: Edinburgh University Press. Braun, S. (2005). From pedagogically relevant corpora to authentic language learning contents. ReCALL, 17 (1), 47-64. Carter, R. & McCarthy, M. (2006). Cambridge grammar of English: A comprehensive guide: Spoken and written English grammar and usage . Cambridge: Cambridge University Press. Davies, M. (2010). The Corpus of Contemporary American English as the first reliable monitor corpus of English. Literary and Linguistic Computing, 25 (4), 447-464. Ellis, R. (1993). The Structural syllabus and second language acquisition. TESOL Quarterly, 27 (1), 92-114. Granger, S. (1998). The computer learner corpus: a versatile new source of data for SLA research. In S. Granger (Ed.), Learner English on computer (pp.3-18). London: Longman. Granger, S. (2003). The International Corpus of Learner English: A new resource for foreign language learning and teaching and second language acquisition research. TESOL Quarterly, 37 (3), 538-546. Granger, S. (2009). The contribution of learner corpora to second language acquisition and foreign language teaching. In K. Aijmer (Ed.) Corpora and language teaching (pp.13-32). Philadelphia: John Benjamins. Granger, S. & Tribble, C. (1998). Learner corpus data in the foreign language classroom: form-focused instruction and data-driven learning. In S. Granger (Ed.), Learner English on computer (pp.199-209). London: Longman. Horváth J. (2001). Advanced writing in English as a foreign language: A corpus-based study of processes and products . Pécs: Lingua Franca Csoport. Hultfors, P. (1986). Reactions to non-native English: Native English-speakers' assessments of errors in the use of English made by non-native users of the language . Stockholm: Almqvist & Wiksell International. Johns, T. (1991). Should you be persuaded – two sample of data-driven learning materials. In T. Johns & P. King (Eds.), Classroom concordancing (pp.1-16). Birmingham: University of Birmingham. Johns, T. (1994). From printout to handout: Grammar and vocabulary teaching in the context of data-driven learning. In T. Odlin (Ed.), Perspectives on pedagogical grammar (pp.293-313). Cambridge: Cambridge University Press. Joyce, H. & Burns, A. (1999). Focus on grammar . Sydney: Macquarie University. Kaltenb öck, G. & Mehlmauer-Larcher, B. (2005). ‘Computer corpora and the language classroom: On the potential and limitations of computer corpora in language teaching.’ ReCALL, 17 (1) , 65-84. Meunier, F. (2002). The pedagogical value of native and learner corpora in EFL grammar teaching. In S. Granger, J. Hung & S. Petch-Tyson (Eds.), Computer learner corpora, second language acquisition and foreign language teaching (pp.119-142). Philadelphia: John Benjamins. Nesselhauf, N. (2004). Learner corpora and their potential for language teaching. In J. Sinclair (Ed.), How to use corpora in language teaching (pp.125-152). Philadelphia: John Benjamins. Oghigian, K & Chujo, K. (2010). An effective way to use corpus exercises to learn grammar basics in English. Language Education in Asia, 1 (1) , 200-214. O’Keeffe, A., McCarthy, M. & Carter, R. (2007). From corpus to classroom . Cambridge: Cambridge University Press. Reppen, R. & Simpson-Vlach, R. (2010). Corpus linguistics. In N. Schmitt (Ed.), An introduction to applied linguistics (pp.89-105). 2 nd ed, London: Hodder Education. Richards, J. C. & Schmidt, R. (2002). Longman dictionary of language teaching and applied linguistics . 3 rd ed, New York: Longman. Tono, Y. (2003). Learner corpora: design, development and applications. In Proceedings of the 2003 Corpus Linguistics Conference (pp.800-809). Tribble. C. & Jones, G. (1990). Concordances in the classroom: A resource book for teachers . Harlow: Longman. Yoon, H. & Hirvela, A. (2004). ESL student attitudes toward corpus use in L2 writing. Journal of Second Language Writing, 13, 257-283. Appendix: Corpora employed Japanese EFL (English as a Foreign Language) Learner Corpus (Shogakukan Corpus Network): approximately 700,000 words. Available at: [21/11/2013]. Corpus of Contemporary American English (Brigham Young University, 1990-2012): 450 million words. Available at: < [21/11/2013]. Acknowledgment I would like to express my deepest appreciation to my supervisor Dawn Knight for the useful comments and suggestions with regard to this article.
3337	https://mathworld.wolfram.com/Integral.html	Integral Download Wolfram Notebook The term "integral" can refer to a number of different concepts in mathematics. The most common meaning is the the fundamenetal object of calculus corresponding to summing infinitesimal pieces to find the content of a continuous region. Other uses of "integral" include values that always take on integer values (e.g., integral embedding, integral graph), mathematical objects for which integers form basic examples (e.g., integral domain), and particular values of an equation (e.g., integral curve), In calculus, an integral is a mathematical object that can be interpreted as an area or a generalization of area. Integrals, together with derivatives, are the fundamental objects of calculus. Other words for integral include antiderivative and primitive. The process of computing an integral is called integration (a more archaic term for integration is quadrature), and the approximate computation of an integral is termed numerical integration. The Riemann integral is the simplest integral definition and the only one usually encountered in physics and elementary calculus. In fact, according to Jeffreys and Jeffreys (1988, p. 29), "it appears that cases where these methods [i.e., generalizations of the Riemann integral] are applicable and Riemann's [definition of the integral] is not are too rare in physics to repay the extra difficulty." The Riemann integral of the function over from to is written \| \| \| --- \| \| \| (1) \| Note that if , the integral is written simply \| \| \| --- \| \| \| (2) \| as opposed to . Every definition of an integral is based on a particular measure. For instance, the Riemann integral is based on Jordan measure, and the Lebesgue integral is based on Lebesgue measure. Moreover, depending on the context, any of a variety of other integral notations may be used. For example, the Lebesgue integral of an integrable function over a set which is measurable with respect to a measure is often written \| \| \| --- \| \| \| (3) \| In the event that the set in () is an interval , the "subscript-superscript" notation from (2) is usually adopted. Another generalization of the Riemann integral is the Stieltjes integral, where the integrand function defined on a closed interval can be integrated against a real-valued bounded function defined on , the result of which has the form \| \| \| --- \| \| \| (4) \| or equivalently \| \| \| --- \| \| \| (5) \| Yet another scenario in which the notation may change comes about in the study of differential geometry, throughout which the integrand is considered a more general differential k-form and can be integrated on a set using either of the equivalent notations \| \| \| --- \| \| \| (6) \| where is the above-mentioned Lebesgue measure. Worth noting is that the notation on the left-hand side of equation () is similar to that in expression () above. There are two classes of (Riemann) integrals: definite integrals such as (5), which have upper and lower limits, and indefinite integrals, such as \| \| \| --- \| \| \| (7) \| which are written without limits. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals, since if is the indefinite integral for , then \| \| \| --- \| \| \| (8) \| What's more, the first fundamental theorem of calculus can be rewritten more generally in terms of differential forms (as in () above) to say that the integral of a differential form over the boundary of some orientable manifold is equal to the exterior derivative of over the interior of , i.e., \| \| \| --- \| \| \| (9) \| Written in this form, the first fundamental theorem of calculus is known as Stokes' Theorem. Since the derivative of a constant is zero, indefinite integrals are defined only up to an arbitrary constant of integration , i.e., \| \| \| --- \| \| \| (10) \| Wolfram Research maintains a web site that can find the indefinite integral of many common (and not so common) functions. Differentiating integrals leads to some useful and powerful identities. For instance, if is continuous, then \| \| \| --- \| \| \| (11) \| which is the first fundamental theorem of calculus. Other derivative-integral identities include \| \| \| --- \| \| \| (12) \| the Leibniz integral rule \| \| \| --- \| \| \| (13) \| (Kaplan 1992, p. 275), its generalization \| \| \| --- \| \| \| (14) \| (Kaplan 1992, p. 258), and \| \| \| --- \| \| \| (15) \| as can be seen by applying (14) on the left side of (15) and using partial integration. Other integral identities include \| \| \| --- \| \| \| (16) \| \| \| \| --- \| \| \| (17) \| \| \| \| \| \| --- --- \| \| \| \| \| (18) \| \| \| \| \| (19) \| and the amusing integral identity \| \| \| --- \| \| \| (20) \| where is any function and \| \| \| --- \| \| \| (21) \| as long as and is real (Glasser 1983). Integrals with rational exponents can often be solved by making the substitution , where is the least common multiple of the denominator of the exponents. See also A-Integrable, Abelian Integral, Calculus, Chebyshev-Gauss Quadrature, Chebyshev Quadrature, Darboux Integral, Definite Integral, Denjoy Integral, Derivative, Differential Geometry, Differential k-Form, Double Exponential Integration, Double Integral, Euler Integral, Form Integration, Fundamental Theorem of Gaussian Quadrature, Gauss-Jacobi Mechanical Quadrature, Gaussian Quadrature, Haar Integral, Hermite-Gauss Quadrature, Indefinite Integral, Integral Calculus, Integration, Jacobi-Gauss Quadrature, Laguerre-Gauss Quadrature, Lebesgue Integral, Lebesgue-Stieltjes Integral, Legendre-Gauss Quadrature, Leibniz Integral Rule, Lobatto Quadrature, Multiple Integral, Nested Function, Newton-Cotes Formulas, Numerical Integration, Perron Integral, Quadrature, Radau Quadrature, Recursive Monotone Stable Quadrature, Repeated Integral, Romberg Integration, Riemann Integral, Singular Integral, Stieltjes Integral, Stokes' Theorem, Triple Integral Explore this topic in the MathWorld classroom Portions of this entry contributed by Christopher Stover Explore with Wolfram\|Alpha More things to try: integral integral e^(2x) integral e^(-2x) References Beyer, W. H. "Integrals." CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 233-296, 1987.Boros, G. and Moll, V. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. Cambridge, England: Cambridge University Press, 2004.Bronstein, M. Symbolic Integration I: Transcendental Functions. New York: Springer-Verlag, 1996.Dubuque, W. G. "Re: Integrals done free on the Web." math-fun@cs.arizona.edu posting, Sept. 24, 1996.Glasser, M. L. "A Remarkable Property of Definite Integrals." Math. Comput. 40, 561-563, 1983.Gordon, R. A. The Integrals of Lebesgue, Denjoy, Perron, and Henstock. Providence, RI: Amer. Math. Soc., 1994.Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, 2000.Jeffreys, H. and Jeffreys, B. S. Methods of Mathematical Physics, 3rd ed. Cambridge, England: Cambridge University Press, p. 29, 1988.Kaplan, W. Advanced Calculus, 4th ed. Reading, MA: Addison-Wesley, 1992.Piessens, R.; de Doncker, E.; Uberhuber, C. W.; and Kahaner, D. K. QUADPACK: A Subroutine Package for Automatic Integration. New York: Springer-Verlag, 1983.Ritt, J. F. Integration in Finite Terms: Liouville's Theory of Elementary Methods. New York: Columbia University Press, p. 37, 1948.Shanks, D. Solved and Unsolved Problems in Number Theory, 4th ed. New York: Chelsea, p. 145, 1993.Wolfram Research. "The Integrator." Referenced on Wolfram\|Alpha Integral Cite this as: Stover, Christopher and Weisstein, Eric W. "Integral." From MathWorld--A Wolfram Resource. Subject classifications Find out if you already have access to Wolfram tech through your organization
3338	https://pmc.ncbi.nlm.nih.gov/articles/PMC10649463/	Possible Roles of Transition Metal Cations in the Formation of Interstellar Benzene via Catalytic Acetylene Cyclotrimerization - 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 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 Molecules . 2023 Nov 6;28(21):7454. doi: 10.3390/molecules28217454 Search in PMC Search in PubMed View in NLM Catalog Add to search Possible Roles of Transition Metal Cations in the Formation of Interstellar Benzene via Catalytic Acetylene Cyclotrimerization Tatsuhiro Murakami Tatsuhiro Murakami 1 Department of Chemistry, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan; n.matsumoto.503@ms.saitama-u.ac.jp (N.M.); fuji@chem.saitama-u.ac.jp (T.F.) 2 Department of Materials & Life Sciences, Faculty of Science & Technology, Sophia University, 7-1 Kioicho, Chiyoda-ku, Tokyo 102-8554, Japan Conceptualization, Methodology, Validation, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization, Supervision, Project administration Find articles by Tatsuhiro Murakami 1,2,, Naoki Matsumoto Naoki Matsumoto 1 Department of Chemistry, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan; n.matsumoto.503@ms.saitama-u.ac.jp (N.M.); fuji@chem.saitama-u.ac.jp (T.F.) Writing – review & editing Find articles by Naoki Matsumoto 1, Takashi Fujihara Takashi Fujihara 1 Department of Chemistry, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan; n.matsumoto.503@ms.saitama-u.ac.jp (N.M.); fuji@chem.saitama-u.ac.jp (T.F.) 3 Comprehensive Analysis Center for Science, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan Writing – review & editing Find articles by Takashi Fujihara 1,3, Toshiyuki Takayanagi Toshiyuki Takayanagi 1 Department of Chemistry, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan; n.matsumoto.503@ms.saitama-u.ac.jp (N.M.); fuji@chem.saitama-u.ac.jp (T.F.) Conceptualization, Methodology, Validation, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization, Supervision, Project administration, Funding acquisition Find articles by Toshiyuki Takayanagi 1, Editor: Felice Grandinetti Author information Article notes Copyright and License information 1 Department of Chemistry, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan; n.matsumoto.503@ms.saitama-u.ac.jp (N.M.); fuji@chem.saitama-u.ac.jp (T.F.) 2 Department of Materials & Life Sciences, Faculty of Science & Technology, Sophia University, 7-1 Kioicho, Chiyoda-ku, Tokyo 102-8554, Japan 3 Comprehensive Analysis Center for Science, Saitama University, Shimo-Okubo 255, Sakura-ku, Saitama City 338-8570, Japan Correspondence: murakamit@mail.saitama-u.ac.jp (T.M.); tako@mail.saitama-u.ac.jp (T.T.); Tel.: +81-48-858-9113 (T.M. & T.T.) Roles Tatsuhiro Murakami: Conceptualization, Methodology, Validation, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization, Supervision, Project administration Naoki Matsumoto: Writing – review & editing Takashi Fujihara: Writing – review & editing Toshiyuki Takayanagi: Conceptualization, Methodology, Validation, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization, Supervision, Project administration, Funding acquisition Felice Grandinetti: Academic Editor Received 2023 Oct 12; Revised 2023 Oct 28; Accepted 2023 Nov 3; Collection date 2023 Nov. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC10649463 PMID: 37959873 Abstract Polycyclic aromatic hydrocarbons (PAHs) are ubiquitous interstellar molecules. However, the formation mechanisms of PAHs and even the simplest cyclic aromatic hydrocarbon, benzene, are not yet fully understood. Recently, we reported the statistical and dynamical properties in the reaction mechanism of Fe+-catalyzed acetylene cyclotrimerization, whereby three acetylene molecules are directly converted to benzene. In this study, we extended our previous work and explored the possible role of the complex of other 3d transition metal cations, TM+ (TM = Sc, Ti, Mn, Co, and Ni), as a catalyst in acetylene cyclotrimerization. Potential energy profiles for bare TM+-catalyst (TM = Sc and Ti), for TM+NC−-catalyst (TM = Sc, Ti, Mn, Co, and Ni), and for TM+-(H 2 O)8-catalyst (TM = Sc and Ti) systems were obtained using quantum chemistry calculations, including the density functional theory levels. The calculation results show that the scandium and titanium cations act as efficient catalysts in acetylene cyclotrimerization and that reactants, which contain an isolated acetylene and (C 2 H 2)2 bound to a bare (ligated) TM cation (TM = Sc and Ti), can be converted into a benzene–metal–cation product complex without an entrance barrier. We found that the number of electrons in the 3d orbitals of the transition metal cation significantly contributes to the catalytic efficiency in the acetylene cyclotrimerization process. On-the-fly Born–Oppenheimer molecular dynamics (BOMD) simulations of the Ti+-NC− and Ti+-(H 2 O)8 complexes were also performed to comprehensively understand the nuclear dynamics of the reactions. The computational results suggest that interstellar benzene can be produced via acetylene cyclotrimerization reactions catalyzed by transition metal cation complexes. Keywords: astrochemistry, interstellar medium, density functional theory, polycyclic aromatic hydrocarbons, transition metal catalysis, cyclotrimerization 1. Introduction Polycyclic aromatic hydrocarbons (PAHs) play an important role in regulating the physical and chemical conditions of interstellar medium. Moreover, they are possible sources of unidentified infrared emission bands (UIBs) and diffuse interstellar bands (DIBs) [1,2,3,4,5,6,7]. Exploring the mechanisms of PAH formation may clarify their abundance and presence in space. Understanding the mechanism underlying the formation of benzene, the simplest building block of PAHs with an aromatic ring structure, from small precursor molecules under interstellar conditions is also important [1,2,3]. However, it is generally difficult to understand the formation mechanism of interstellar benzene directly from astrochemical observations. Several scenarios have been previously proposed. For example, Woods et al. proposed a stepwise ion–molecule reaction mechanism involving the reactions C 4 H 3+ + C 2 H 2 → c-C 6 H 5+ + hν and c-C 6 H 5+ + H 2 → c-C 6 H 7+ + hn, followed by dissociative electron attachment: c-C 6 H 7+ + e− → C 6 H 6 + H. The validity of this mechanism was examined by comparing the benzene column density observed in CRL 618 with calculations using the chemical evolution model. Kaiser et al. studied the neutral C 2 H + C 4 H 6 (1,3-butadiene) → C 6 H 6 + H reaction in the gas phase using sophisticated crossed molecular beam experiments. This neutral bimolecular reaction is barrier-less and highly exothermic, leading to the presumption that it can proceed efficiently in cold molecular clouds. Alternatively, El-Shall et al. [10,11] suggested that the benzene radical cation is formed via the ionization and subsequent polymerization of large acetylene clusters, (C 2 H 2)n, and that these processes may be responsible for the formation of benzene and other PAHs in space. Zhao et al. experimentally showed that the catalytic conversion reactions of acetylene on the surface of a solid SiC grain efficiently form benzene and PAHs and emphasized the importance of the dangling bonds of the SiC solid in the catalytic p-bond breaking of acetylene. The direct synthesis of benzene from three acetylene molecules is generally referred to as [2 + 2 + 2] acetylene cyclotrimerization. Due to the large activation barrier associated with the cleavage of π-bonds, [13,14] non-catalytic cyclotrimerization is very slow even at high temperatures . The formation mechanism of metal-catalyzed acetylene cyclotrimerization, therefore, was extensively studied in the field of organic chemistry [16,17]. Acetylene complexes with transition metal ions as catalysts were studiously investigated by Duncan and coworkers [18,19,20,21,22,23,24,25]. They observed the infrared spectra of acetylene complexes with several transition metal ions, including iron, nickel, [19,20] copper, silver, gold, zinc, and vanadium . The spectroscopic data provided evidence of cyclotrimerization of the V+(C 2 H 2)3 complexes to form V+(benzene) via metallacycle intermediates . Moreover, Duncan and coworkers showed that there is a large activation barrier for the cyclotrimerization reaction . The activation barrier becomes the key limitation in the cyclotrimerization process under the interstellar condition. Through density functional theory (DFT) calculations of the quantum chemical reaction paths, we previously demonstrated that the conversion of the (C 2 H 2)3-Fe+-L complex to the C 6 H 6-Fe+-L complex proceeds via transition states, with relatively lower barriers compared to those in the cyclotrimerization of the complexes with the bare iron ion [26,27,28]. It was found that L = (H 2 O)n (n = 8, 10, 12, and 18) clusters as the ligand reduced the number of transition states [26,28]. We also studied the cyclotrimerization of (three) acetylene units catalyzed by Fe+ with several ligands, namely, F−, Cl−, OH−, SH−, NC−, NH 3, H 2 O, cyclopentadienyl anion (Cp−), CN−, benzene (Bz), and PH 3 . The Fe+-catalyzed cyclotrimerization using three of these ligands (CN−, Bz, and PH 3) had more than two additional transition state barriers, whereas the former eight ligand complexes formed benzene through one transition state barrier. Moreover, in dynamics simulations, 44 trajectories out of 50 generated benzene forms within 1 ps with the NC− ligand . From these results, we tentatively speculate that the TM+-(H 2 O)n and TM+-NC− compounds may act as an effective catalysis for forming benzene via acetylene cyclotrimerization. Iron is the sixth most abundant element in astrophysical environments . However, the role of transition metals is not yet understood from an astrochemical viewpoint [30,31]. Herein, we extended our previous work on Fe+ to earlier 3d transition metal cations, namely Sc+ and Ti+, and later ones, which are Mn+, Co+, and Ni+. Notably, atomic metals and their cations, including Ti and Mn, have been previously detected in the circumstellar envelope of the asymptotic giant branch carbon star IRC+10216 . Furthermore, TiO and TiO 2 have been detected in the gas phase through rotational transitions at submillimeter wavelengths toward the red supergiant VY CMa . The interstellar spectra of ζ Ophiuchi have been observed, where the lines of cobalt and nickel ions were detected [34,35]. Scandium has been studied as a key element of Am star due to the deficiency of Sc . Nevertheless, weak lines of scandium were also detected in Am star . Although the Sc+-ligand complex might not be effective for cyclotrimerization in the interstellar medium due to the Sc deficiency, investigating the reaction mechanism using scandium, which is the earliest in the 3d transition metal series, may be important for understanding the cyclotrimerization mechanisms involving 3d transition metals from a theoretical viewpoint. In this study, using quantum chemistry calculations and reaction dynamics simulations, we investigate the catalytic efficiency of compounds composed of the aforementioned transition metals in acetylene cyclotrimerization reactions leading to benzene-metal complexes. The results are discussed in Section 2. In the first place, the potential energy profiles of the bare Sc+ and Ti+ catalyst systems without any ligand for three acetylene cyclotrimerization were investigated and are discussed in Section 2.1. Early transition metals such as scandium and titanium, which have more unoccupied d orbitals than Mn+, Co+, and Ni+, are the focus in the subsection. We discuss the barrier-reduction effects of bare Sc and Ti ions. Next, the reaction mechanisms for the complexes containing TM+NC− compounds (TM = Sc and Ti) are discussed and compared with the bare TM catalyst system cases in Section 2.2. In the previous study, the gas-phase FeCN and MgCN/MgNC molecule was detected by observing the rotational spectra toward the envelope of IRC+10216 [38,39,40,41]. It was also expected that the NC− ligand would act as an effective catalysis from our previous theoretical study . We also calculated the potential energy profiles for the acetylene complexes with the TM+NC− ligands (TM = Mn, Co, and Ni) and compared the barrier height with those of ScNC and TiNC catalyst systems in the same subsection. Thirdly, the potential energy profiles of the Sc+ and Ti+ catalysts with the (H 2 O)8 cluster as a ligand are discussed in Section 2.3. Solid water (ice) is known to play a very important role in interstellar surface reactions . As in our previous theoretical study, the (H 2 O)8 cluster was utilized. The previous study showed that the cubic structure with a highly symmetric D 2 d character is the most stable for the (H 2 O)8 cluster. Thus, the effects of the various low-lying structures can be neglected because of the complicated hydrogen-bonding motif. Note that the (H 2 O)8 cluster with bare transition metal ions might not retain the cubic structure because of the large exothermic energy due to the relatively strong attractive interaction between the ice cluster and the bare metal ions. However, if complex transition metal systems such as TM+-NC− and TM+-H 2 O catalysts are adsorbed onto the ice cluster, the exothermic interaction energies could be reduced, indicating that the (H 2 O)8 cluster and larger clusters can possibly maintain the cubic structure without forming a solvated metal ion structure. In the study of a catalysis reaction on interstellar ice, the complex structures of MgCN with 17H 2 O and 24H 2 O were investigated, as well as the structure of the ice cluster . Therefore, the (H 2 O)8 cluster was been selected as the ligands in this study. In the last part of Section 2, the results of reaction dynamics simulations in terms of the acetylene cyclotrimerization for Ti+NC− and Ti+(H 2 O)8 catalyst systems are discussed. The potential energy profiles through the intrinsic reaction coordinate (IRC) using quantum chemistry calculations effectively provide statistical information such as the properties of stational points to understand the reaction mechanism. The reaction dynamics, however, do not often follow the IRC [44,45]. Non-IRC dynamics play a significant role in many organic chemistry reactions, including catalytic reactions. Our previous study in terms of Fe+-catalyzed acetylene cyclotrimerization presented the fact that many trajectories exhibited non-IRC dynamics . The performing of molecular dynamics simulations could be of critical importance to understand the dynamic properties of the chemical reactions. Motivated by this, we performed the molecular dynamics simulations in this study. Finally, a summary and future directions of this study are presented in Section 3. The computational details of the quantum chemistry calculations and molecular dynamics simulations are presented in the last part of this paper. 2. Results and Discussion 2.1. Bare Sc+ and Ti+ Catalyst Systems Figure 1a,b shows the potential energy profiles (obtained from the B3LYP-D3(BJ)/def2-TZVP calculations) of the acetylene cyclotrimerization reactions for the conversion of the reactant (C 2 H 2 + (C 2 H 2)2-TM+) to the product complex (PC) (C 6 H 6-TM+) catalyzed by the TM+ (TM = Sc and Ti) complexes. The horizontal axis represents the reaction path length measured within the mass-weighted coordinate system, and the origin of the corresponding axis is taken to be the first intermediate complex, which is primarily produced by the association of the (C 2 H 2)2-TM+ reactant with an additional C 2 H 2 molecule. Zero energy is defined as the energy level of the C 2 H 2 + (C 2 H 2)2 TM+ reactants, and no zero-point energy correction was included (the harmonic zero-point energy values for all stationary points found in this study are reported in the Supplementary Materials). Figure 1a,b clearly shows that both reactions from the reactants to PCs are totally exothermic. Further details such as the molecular structures at the stationary points and the reaction and the reaction paths are discussed later. The potential energies at each stationary point from the reactants to products (benzene + TM+) via the intermediate (INT1) and transition (TS1) states and their vertical excitation energies are shown in Table 1. The spin state i of the reactants corresponds to the superscript of i{1 C 2 H 2, i Sc+(C 2 H 2) (i = 1, 3, and 5)} and i{1 C 2 H 2, i Ti+(C 2 H 2) (i 2, 4, and 6)}. The spin states of the products are represented as i{1 benzene, i Sc+(C 2 H 2) (i 1, 3, and 5)} and i{1 benzene, i Ti+ (i = 2, 4, and 6)}, respectively. From the reactants to PCs, the potential energies of the lowest spin state were found to be consistently lower than that of the higher spin state. For example, in the case of Sc+, the spin state of the electronic ground state of the isolated Sc+ cation is known to be a triplet; however, the spin states of the (C 2 H 2)3 Sc+ in the electronic ground state are singlets. Similarly for Ti+, the potential energy surface of the doublet was lower than that of the quartet and sextet. The Sc- and Ti-catalyzed product fragments have triplet and quartet electronic ground states, as expected. The highest spin states for both systems do not seem to contribute to the cyclotrimerization because their potential energies through the overall reaction pathway relatively represent large values. The quintet and sextet energies of the products are extremely high because the electron configuration of isolated Sc+ is [Ar]3d 1 4s 1 and that of Ti+ is [Ar]3d 2 4s 1, and excitation from the inner shell is required. Herein, an affordable pathway to obtain the final product fragments should be emphasized. Under low-density conditions, the intramolecular redistribution of vibrational energies derived from the exothermic energy could play an important role in releasing the transition metal cation. Furthermore, one of the most reasonable pathways to obtain the final product fragments is the dissociative electron attachment: (benzene-TM+) + e− → benzene + TM [46,47]. For example, the reaction involving a spin transition, (1 benzene-1 Ti) → 1 benzene + 3 Ti, is only a 0.5 kcal/mol endothermic reaction (see Table S1). The vibrational continuum and resonance states of the neutral product complex involved in the electron attachment provide relatively large vibrational energies to the complex. The metal–molecule complex therefore enables the dissociation to the benzene and neutral metal fragments. Alternative pathways are photodissociation by irradiation of the cosmic rays and collisional dissociation by encountering other systems in the dense clouds [46,47]. Figure 1. Open in a new tab Potential energy profiles for the acetylene cyclotrimerization reactions catalyzed by (a) Sc+ (singlet spin) and (b) Ti+ (doublet spin) compounds as a function of the reaction path length obtained at the B3LYP-D3(BJ)/def2-TZVP DFT-D3 level. The zero energy is defined as the energy level of the C 2 H 2 + TM+-(C 2 H 2)2 reactants (TM = Sc and Ti). Table 1. Vertical excitation energies (in kcal/mol) at the stationary points for Sc+(C 6 H 6) and Ti+(C 6 H 6) calculated using B3LYP-D3(BJ)/def2-TZVP. \| \| Reactant \| INT1 \| TS1 \| PC \| Product \| :---: :---: :---: \| \| Sc+-C 6 H 6 \| \| \| \| \| \| \| Quintet \| 130.5 \| 80.3 \| 79.5 \| −24.9 \| 549.9 \| \| Triplet \| 36.6 \| 5.3 \| 6.8 \| −118.7 \| −79.5 \| \| Singlet \| 0.0 \| −59.2 \| −56.8 \| −133.0 \| −41.8 \| \| Ti+-C 6 H 6 \| \| \| \| \| \| \| Sextet \| 151.3 \| 81.4 \| 79.2 \| −12.5 \| 626.1 \| \| Quartet \| 40.6 \| 2.8 \| 5.0 \| −90.5 \| −74.3 \| \| Doublet \| 0.0 \| −55.3 \| −47.9 \| −126.9 \| −55.3 \| Open in a new tab Figure 2a,c shows the optimized geometries and their Kohn–Sham molecular orbitals (MOs) for the reactant fragments TM+(C 2 H 2)2 (TM = 1 Sc and 2 Ti). We also optimized 3 Sc+(C 2 H 2)2 and 4 Ti+(C 2 H 2)2, which are shown in Figure 2b,d, with their MOs. It is known that the formation of a metallacycle between a transition metal and acetylene (alkyne) is affected by the electron donor–acceptor interactions between the d orbitals of the metal and π-orbitals of the acetylene . Two π-orbitals, which are parallel (π∥) and perpendicular (π⊥) to the TMCC plane, respectively, interact with the unoccupied d-orbitals of the metal. Moreover, the d-donor orbital interacts with the π∥ orbital, which is known as π-back-donation. This metal–olefin bonding belongs to the Dewar–Chatt–Duncanson model . In this study, the number of electrons in the d-donor orbital could be two and three for Sc+ and Ti+, respectively, from the electron configurations of isolated Sc+ and Ti+. As shown in Figure 2a, the 24 α and 25 α orbitals of the singlet state, which are the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO), respectively, contribute to π-back-donation. Thus, one acetylene could only be influenced by the d-π∥ interaction. The metallacycle is formed by d-π∥ and d-π⊥ interactions only on the other side. The MOs correlated with the d-π∥ and d-π⊥ interactions are shown in Figure S1 in the Supplementary Materials. The C 2 H 2 moiety on the right side of the Sc cation was influenced by the back-donation from 24 α orbital. The C-C bond was elongated to 1.33 Å and the C-C-H bond angle was distorted to 131.5° from the equilibrium structure of the isolated singlet C 2 H 2 (the C-C bond = 1.20 Å and the C-C-H angle = 180.0° ). The coordinated structure was quite similar to the triplet C 2 H 2 (the C-C bond = 1.32 Å and the C-C-H angle = 128.3° ). It is represented that the triplet configuration of acetylene plays quite a significant role in the π-donation and π-back-donation process between acetylene and the 3d metal. The result is quite consistent with the chemisorption of the acetylene on the copper surface . On the other hand, the C 2 H 2 part on the left side had the singlet configuration, which indicates that the occupied π-orbital of the singlet C 2 H 2 donates directly to the unoccupied metal d-orbital. Moreover, the Sc-C bond distance (2.66 Å) was larger than the right side one (2.02 Å). The π-back-donation is significantly correlated with the 3d-metal-acetylene coordination. From these acetylene structures, the nature of the bond formation between the Sc cation and acetylene based on the donor and acceptor processes of π-electrons follows a spin-uncoupling mechanism, [50,51,52] which is the coordination process whereby unpaired electron configurations for both metal and unsaturated hydrocarbons interact with each other. In contrast with the singlet Sc+(C 2 H 2)2, the singly occupied highest and second highest molecular orbitals (SOMO and SOMO-1) of triplet Sc+(C 2 H 2)2 are the 25 α and 24 α orbitals, as shown in Figure 2b. The two half-occupied orbitals symmetrically contribute to π-back-donation and then the molecular structure became C 2v. Both C-C bond lengths (1.24 Å) are longer than the isolated singlet configuration and shorter than the triplet one. In comparison with the singlet Sc+(C 2 H 2)2, the Sc-C bond distances are longer than the length of the C 2 H 2 moiety under the strong influence of back-donation and shorter than the bond distance of the acetylene part, which the back-donation does not affect. It is suggested that π-back-donation plays an important role in the spin-uncoupling process and the acetylene conformation change. Note that the potential energy of the triplet Sc+(C 2 H 2)2 was 9.7 kcal/mol, which is higher than the one of the reactant fragment. Figure 2c,d shows the doublet (0.0 kcal/mol) and quartet (16.0 kcal/mol) Ti+(C 2 H 2)2 equilibrium structures and the Kohn–Sham molecular orbitals at their minima. The 25 α (SOMO) and 24 α (HOMO) were assigned as the d-π∥ orbitals for the doublet state, whereas the quartet-state compound has three back-donation orbitals, namely the 26 α (SOMO), 25 α (SOMO-1), and 24 α (SOMO-2) orbitals. The acetylene configurations of both doublet and quartet Ti+(C 2 H 2)2 compounds are influenced by the π-back-donation from the occupied 3d-orbitals of Ti cation, which is similar to the case of Sc-catalyst systems. The Ti-C bond distances are relatively shorter than the related Sc-C lengths. Especially, the Ti-C bond distance (2.28 Å) on the left side of Ti shown in Figure 2c was extremely shorter than the one (2.66 Å) of Sc shown in Figure 2a because the 25 α (SOMO) of doublet Ti+(C 2 H 2)2 contributed to the Ti-acetylene-coordination, while the related orbital character for singlet Sc+(C 2 H 2)2 belonged to the 25 α (LUMO). The back-donation from metal cations is therefore of critical importance for the coordination. These results are quite consistent with the relation between the metal-ligand distance and back-donation in the Dewar–Chatt–Duncanson model . Figure 2. Open in a new tab Optimized geometries (upper panel) and molecular orbitals (lower panel) contributed to π-back-donation for Sc+(C 2 H 2)2 of (a) singlet and (b) triplet states, (b) Ti+(C 2 H 2)2 of (c) doublet and (d) quartet (red). The potential energies for the reactants are set as 0 kcal/mol. As shown in Figure 1a,b, the TM+C 4 H 4 five-membered metallacycle structures (INT1) were formed without potential barriers, which was confirmed through the geometry optimization (See Figure S2 in the Supplementary Materials). It was predicted that the occupied d-donor orbitals were insufficient to maintain the forms of the three acetylenes for the 1 Sc+- and 2 Ti+-catalyst systems. Figure 3 shows molecular structures at optimization steps 10, 20, and 52 for the 2 Ti+-catalyst system (See Figure S2b) and the crucial molecular orbitals. In 31 α (HOMO) at step 10, one d-donor orbital is shared with C1-C3 π and C2-C4 π orbitals of each acetylene, where the C1-C2 length is 2.03 Å. The 29 α (HOMO-2) at step 10 is due to the π-donation to the unoccupied metal d-orbital from the occupied C1-C3 π and C2-C4 π orbitals. The Ti-C1 and Ti-C2 coordination interactions are correlated with 33 α (LUMO). During the optimization process to step 20, the C1-C2 bond (1.52 Å) was generated with the C1-C2 σ-orbital in 29 α (HOMO-2). This MO was stabilized (31 α → 29 α) and the MO included a node between C1 and C2 was destabilized (29 α → 30 α) from steps 10 to 20. The Ti-C1 (Ti-C2) bond was elongated from 2.18 (2.27) Å to 2.28 (2.31) Å with the 35 α (LUMO+2) which does not interact between the 3d and C1-C3 (C2-C4) π orbitals. Eventually, a five-membered metallacycle intermediate of the doublet Ti+(C 4 H 4)C 2 H 2 was generated in step 52. This metallacycle formation from the reactants is similar to the titanium metallacyclobutane formation from an ethylene and titanium methylidene, which is the barrierless reaction . The π-back-donation does not contribute to the reaction. As mentioned above, the coordination conformations between the metal and unsaturated hydrocarbon are critically influenced by the π-back-donation based on the spin-uncoupling mechanism. In the cases of the 2 Ti+-catalyst as well as 1 Sc+-catalyst systems, the d-donor electrons contributing to the π-back-donation are not enough to keep three acetylenes separately. It is suggested that a small number of occupied d-donor orbitals induce the barrierless formation of the five-membered metallacycle. Figure 3. Open in a new tab Molecular structures and molecular orbitals at optimization steps 10, 20, and 52 for 2 Ti+-catalyst system. The entire potential energy minimization profiles are shown in Figure S2b in the Supplementary Materials. Figure 1a,b shows that TM+-benzene (PC) is produced through a relatively low transition state (TS1) from the 3/5 dimetallacycle (INT1). The IRC profiles are found to be similar to the pathway to the vanadium cation ([Ar]3d 3 4s 1) case . The barrier heights of TS1 are 2.4 and 7.4 kcal/mol for the Sc+ and Ti+ ions, respectively. The π-back-donation is not correlated with the Sc+C 2 H 2 coordination bond, whereas the formation of Ti+C 2 H 2 is influenced by the d-π∥ interaction involving one SOMO. A weaker Sc+C 2 H 2 coordination could relatively reduce this barrier. Linares and coworker presented that the metal–ligand distance becomes more lenient in the absence of back-donation in the Dewar–Chatt–Duncanson model . Our work quite agrees with the previous study. As mentioned above, it is found that this three acetylene cyclotrimerization with the TM+ catalysis is a totally exothermic reaction. 2.2. TM+NC− Catalyst System Figure 4 shows the potential energy profiles (obtained from B3LYP-D3(BJ)/def2-TZVP calculations) of the acetylene cyclotrimerization reactions (from the C 2 H 2 + (C 2 H 2)2-TM-NC reactant to the C 6 H 6-TM-NC product complex) catalyzed by the TM+NC− (TM = Sc and Ti) compounds. The lowest electronic states in the overall reaction pathways to the PCs for ScNC and TiNC catalyst systems are the singlet and doublet state, respectively (See Table S2), and the benchmark results are shown in Tables S3 and S4 in the Supplementary Materials. As shown in Figure 4a, the two acetylene molecules of the (C 2 H 2)2-ScNC reactant complex spontaneously dimerized to form a Sc-C 4 H 4-type five-membered metallacycle structure without an entrance barrier. This was confirmed by the geometry optimization (see Figure S2c). A second transition state (TS2) was also observed in the reaction pathway from the Sc+C 6 H 6 seven-membered metallacycle intermediate (INT2) to the benzene product complex (PC), although the corresponding barrier height measured for this intermediate was extremely small (0.1 kcal/mol). This suggests that the seven-membered metallacycle intermediate (INT2) is located in a shallow potential energy well. Another interesting result was observed for the TiNCs, as shown in Figure 4b. In this case, the Ti-C 4 H 4 five-membered metalacyclic structure was not observed in the initial (C 2 H 2)2-TiNC reactant; however, there was no barrier between the C 2 H 2 + (C 2 H 2)2-TiNC reactants and the C 2 H 2-TiNC-C 4 H 4 intermediate complex. Thus, the association of the third acetylene group with (C 2 H 2)2-TiNC to form the C 2 H 2-TiNC-C 4 H 4 intermediate complex (INT1) is barrierless. This behavior was confirmed by geometric optimization (see Figure S2d), minimum energy path calculations (see Figure S3a), and BOMD calculations (see below). The INT1 connects to the seven-membered metallacycle intermediate (INT2) via a transition state (TS1), as shown in Figure 4b, while there is no second intermediate in the bare Ti+ system, as shown in Figure 1b. One possible reason for the existence of INT2 is that the interactions between the d orbital of the titanium cation and the occupied nonbonding two π and σ orbitals of NC− may stabilize the potential energy of the ring-opening C 6 H 6 compound. Further details are out of scope in this study because the TS2 barrier height was extremely small (0.6 kcal/mol), which indicates that the INT2 is located in a very shallow potential energy well. Eventually, another transition state (TS2) connects the INT2 to the TiNC-benzene PC. The overall reaction process including these two transition states is found to be highly exothermic. Figure 4. Open in a new tab Potential energy profiles for the acetylene cyclotrimerization reactions catalyzed by (a) Sc+NC− (singlet spin) and (b) Ti+NC− (doublet spin) compounds as a function of the reaction path length obtained at the B3LYP-D3(BJ)/def2-TZVP DFT-D3 level. Zero energy is defined as the energy level of the C 2 H 2 + TM+NC−-(C 2 H 2)2 reactants (TM = Sc and Ti). Figure 5 shows the difference in the barrier height of the bare metal versus the TMNC catalysts. The potential energy of the intermediate complex (INT1) was set to 0 kcal/mol. The zero point on the horizontal axis representing the reaction path length was set at the TS1 structure. In the case of scandium, the NC−-ligated complex has a slightly higher energy (0.5 kcal/mol) than that of the bare metal compounds shown in Figure 5a. In contrast to the 1 Sc+ complexes, the barrier height at TS1 with bare titanium (7.4 kcal/mol) was significantly decreased by the NC− ligand. The potential energy of the TiNC compound in TS1 was 2.2 kcal/mol. Figure 6 shows the imaginary frequencies at the TS1. While the frequencies for the Sc+ (215 i cm−1) and ScNC (266 i cm−1) catalyst systems are similar, the frequency of the TiNC system (72 i cm−1) is quite a bit smaller than the one of the Ti+ system (269 i cm−1), which indicates that the potential curvature along the reaction coordinate around TS1 for the TiNC system is quite lenient compared with the curvature for Ti+ catalyst system. This behavior leads to a reduction in barrier height. Figure 5. Open in a new tab Relative energies as a function of the reaction path length for (a) scandium and (b) titanium compounds. The red and black lines represent the potential energies for the bare TM+ catalyst and TMNC catalyst systems, respectively (TM = Sc and Ti). Zero energy is defined as the energy level of each intermediate complex, INT1. Figure 6. Open in a new tab Imaginary frequencies, their vibrational vectors, and important CC distances at TS1 with their structures for Sc+, ScNC, Ti+, and TiNC catalyst systems. The potential energy profiles calculated by the B3LYP-D3(BJ)/def2-SVPP level for the MnNC, CoNC, and NiNC catalyst systems are shown in Figure 7. In contrast with the Sc+ and Ti+ systems, the Mn+, Co+, and Ni+ systems have a stable structure corresponding to the TM-NC-(C 2 H 2)3 structure, in which the three acetylene molecules are coordinated to the transition metal cation center. In addition, there is a substantial barrier between the TM and NC–(C 2 H 2)3 structure and the C 2 H 2-TM-NC-C 4 H 4 intermediate complex, as shown in Figure 7 In particular, for Co+ and Ni+, the energy levels of these barriers are higher than the energy levels of the reactant. The results suggest that the catalytic formation of benzene from the three acetylene molecules is energetically favored. However, one can easily understand that the ScNC and TiNC compounds can act as efficient catalysts in the cyclotrimerization of acetylene because there is no substantial barrier along the reaction pathways. The catalytic efficiency of the Sc and Ti compounds in the acetylene cyclotrimerization process can be qualitatively understood by considering the number of electrons in the 3d orbitals of the transition metal cation. Three unoccupied 3d orbitals of the metal cation are required to interact strongly with three π-orbitals of the three acetylene molecules, which can form three 3d-π bonding orbitals. Note that each acetylene molecule has two π-orbitals, which may account for the absence of a large barrier in the cyclotrimerization pathways from acetylene to benzene for the ScNC and TiNC catalysts. Therefore, we investigated the catalytic efficiencies of the Sc+(H 2 O)n and Ti+(H 2 O)n clusters in acetylene cyclotrimerization. Figure 7. Open in a new tab Potential energy profiles for (a) Mn+NC− (quintet spin), (b) Co+NC− (triplet spin), and (c) Ni+NC− (doublet spin) obtained at the B3LYP-D3(BJ)/def2-SVPP DFT-D3 level. 2.3. TM+(H 2 O)8 Catalyst System As a tentative model catalyst system, we employed the TM+(H 2 O)8-ligand to form a cubic cluster structure throughout the reaction-path calculations in this study. The potential energies were calculated at the B3LYP-D3(BJ)/def2-SVPP level. Figure 8a shows the reaction pathways for acetylene cyclotrimerization catalyzed by the Sc+(H 2 O)8 complex. The spin state of the lowest potential energy surface was a singlet, similar to those of bare Sc+ and ScNC. The scandium cation was preferentially bound to the O atom of water with no dangling OH bonds. This is similar to the binding in the Fe+ system . Moreover, clearly the Sc-C 4 H 4 five-membered metallacycle structure was already formed in the reactant complex. This is similar to the structure of ScNC. There is only one transition state in the reaction pathway from the intermediate complex to the benzene complex, the barrier height of which is 3.9 kcal/mol, as measured from the intermediate energy level, and is slightly higher than that for the ScNC system (2.3 kcal/mol of the def2-SVPP level, see Table S3). From this figure, it is qualitatively deduced that the association of the third acetylene group with the C 4 H 4-Sc+(H 2 O)8 complex spontaneously produces the benzene-Sc+(H 2 O)8 complex. Figure 8b shows similar results for the Ti+(H 2 O)8 complexes. The barrierless reaction was confirmed by MEP calculation (see Figure S3). The reaction pathway is similar to that in the case of TiNC. Figure 8. Open in a new tab Potential energy profiles for (a) Sc+(H 2 O)8 (singlet) and (b) Ti+(H 2 O)8 (doublet). OH bond lengths (in Å) in water directly bound to the transition metal cation are shown. 2.4. Molecular Dynamics Simulations Although the static reaction path calculations presented in Figure 4b and Figure 8b show that TiNC and Ti+(H 2 O)8 can serve as efficient catalysts for benzene formation via acetylene trimerization, it would be interesting to understand these reaction processes from a dynamical perspective. To this end, BOMD calculations were performed. The BOMD calculations were initiated with the optimized structures of both the TiNC-(C 2 H 2)2 (or Ti+(H 2 O)8-(C 2 H 2)2) reactant complex and C 2 H 2, where the third acetylene molecule is initially located in such a manner that the distance between Ti and the midpoint of C≡C is 4–5 Å, with a random orientation. Initially, no kinetic energy was applied to the atoms in the system for qualitative simulation of the reaction at very low temperatures under interstellar conditions. However, these atoms can obtain kinetic energy during the trajectory evolution owing to the large attractive force (~50 kcal/mol, see Figure 4b and Figure 8b) between C 2 H 2 and the Ti+ moiety in the complex. These initial conditions are physically unrealistic; however, such computational conditions are reasonable for understanding the reaction pathways of benzene formation from a dynamic perspective. Six and five BOMD trajectories were integrated for the TiNC and Ti+(H 2 O)8 cases, respectively. Figure 9a shows a typical BOMD trajectory (trajectory #1) for the TiNC complex, where the potential energy is plotted as a function of the simulation time, along with a few selected structures. In the early stage of this trajectory (t< 600 fs), the third acetylene molecule approaches the Ti+ moiety in the complex because of attractive forces, as previously mentioned. At t ≈ 700 fs, the two carbon atoms in the two acetylenes approach each other, forming a CC bond, thereby generating a five-membered metallacycle intermediate with a C 2 H 2-TiNC-C 4 H 4 structure. This intermediate structure was maintained up to t = 1500 fs. At t ≈ 1500 fs, a new CC bond is formed between C 2 H 2 and the C 4 H 4 moiety, producing a seven-membered metallacycle intermediate with a TiNC-C 6 H 6 structure. The lifetime of this intermediate is not long (~300 fs), and the benzene structure is readily formed at t ≈ 1900 fs in this trajectory. These real-time nuclear dynamic behaviors are qualitatively consistent with the potential energy profile of the reaction path presented in Figure 4b. The results for the other four BOMD trajectories are shown in Figure 9b. In trajectories of the #2–#5 cases, benzene was finally formed within a simulation time of 2.5 ps, although the lifetime of the intermediate complex differed slightly depending on the nature of the BOMD trajectories. We calculated another trajectory (Trajectory #6 in Figure 9b) in which the third acetylene molecule was initially placed on the other side of the Ti moiety of the TiNC–(C 2 H 2)2 complex (see the inset structure in Figure 9b). Interestingly, the third acetylene molecule does not approach Ti+ because of the presence of the other two C 2 H 2 molecules. In this case, the trajectory consistently bounced dynamically between the third acetylene and the two acetylenes bound to Ti+ during the simulation. This suggests that the initial position of the third acetylene molecule plays an important role in the dynamics of benzene formation. Figure S4 shows the initial configuration of the TiNC–(C 2 H 2)2 complex with the isolated acetylene for all trajectories. At t 0 fs, the acetylene was located below the CCCC plane of the two acetylene molecules bound to Ti+ and there was no obstacle between the isolated acetylene and Ti+ in Trajectories #1–#5. As mentioned above, the bond formation between Ti+ and acetylene in Trajectory #6 was prevented by the two acetylenes. The benzene formation movie is presented as Video S1 in the Supplementary Materials. Figure 9. Open in a new tab (a) Potential energy profile of the benzene formation trajectory for the (C 2 H 2)3-TiNC system plotted as a function of simulation time. (b) Potential energy profiles of five trajectories, where four trajectories show benzene formation while one shows non-reactive behavior (see main text). Figure 10a shows a typical benzene-forming BOMD trajectory (Trajectory #1) calculated for the Ti+(H 2 O)8 complex, where the potential energy is plotted as a function of time, along with selected pictures of the geometry. In this trajectory, the C 2 H 2-Ti-C 4 H 4 five-membered metallacycle intermediate was formed in the early stages (after t ≈ 200 fs). This intermediate structure was maintained up to t ≈ 500 fs and a CC bond was formed between the C 2 H 2 and the C 4 H 4 moieties in this intermediate complex to generate a TiNC-C 6 H 6 seven-membered metallacycle structure at t ≈ 850 fs, followed by benzene formation. Thus, a benzene complex is formed within a short period (1 ps) in this trajectory. Interestingly, proton transfer from the water molecule directly attached to the Ti+ cation to another water molecule occurs at t ≈ 580 fs. Thus, after this time, the structure of the Ti+(H 2 O)8 complex approximately assumes the form of Ti+OH−·(H 2 O)5·(H 5 O 2+) or Ti+OH−·(H 2 O)6·(H 3 O+), where the H 5 O 2+ and H 3 O+ structures are generally called Zundel and Eigen ions, respectively . This is very interesting because the calculation of the reaction path presented in Figure 8b does not indicate any proton transfer along the reaction pathway. In this regard, the trajectory presented in Figure 10a shows a non-IRC dynamics feature [44,45]. The occurrence of proton transfer suggests that the OH bond in water, which is directly bound to Ti+, is somewhat weak in the complex. This behavior is consistent with the fact that the corresponding OH bond lengths for all the stationary points on the potential energy surface were slightly longer (1.05–1.08 Å, see Figure 8b) than those for an isolated neutral water molecule (~0.96 Å). Thus, energy transfer can also occur from the (C 2 H 2)3-Ti+ moiety to the water clusters through proton transfer during the dynamic process, although the amount of energy transferred is not very large because of the light nature of protons. This can be understood from evaluation of the water cluster dynamics after formation of the benzene complex. Although the hydrogen-bonded structure was deformed to a certain degree at this stage, water evaporation was not observed within the simulation time. Thus, the nuclear dynamics are in contrast to those of a simple TiNC complex. Benzene formation movies from the two viewpoint angles are presented as Videos S2 and S3 in the Supplementary Materials. Figure 10. Open in a new tab (a) Potential energy profile of the benzene formation trajectory for the (C 2 H 2)3-Ti+(H 2 O)8 system plotted as a function of simulation time. (b) Potential energy profiles of four trajectories, where one trajectory shows benzene formation while three trajectories show the five-membered metallacycle intermediate formation. The results obtained with the other four BOMD trajectories are shown in Figure 10b. Three of these trajectories led to the formation of the benzene complex, although the formation time depended on the nature of the trajectory. However, the benzene complex was formed later (2–3 ps) for these trajectories compared to benzene formation in Trajectory #1 (Figure 10a). Thus, the process of generating the benzene complex (within 1 ps), as shown in Figure 10a, was remarkably fast. The slow production of the benzene complex presented in Figure 10b contrasts with the trajectories calculated for the TiNC complex; however, the results are consistent with the potential energy profile along the reaction path presented in Figure 9b, where the barrier height from the C 2 H 2-Ti-C 4 H 4 five-membered metallacycle intermediate is 5.2 kcal/mol. This barrier height was slightly larger than that of the TiNC complex (2.0 kcal/mol, see Table S3). Moreover, the number of nuclear degrees of freedom for the Ti+(H 2 O)8 system is higher than that for the TiNC system, which may affect the relaxation of the kinetic behavior. Thus, it is reasonable to assume that the lifetimes of the five- or seven-membered metallacycle intermediates in the Ti+(H 2 O)8 system were longer than those in the TiNC system. To estimate the average benzene formation time, several BOMD trajectories with longer simulation times were integrated to obtain statistically meaningful outcomes. However, these calculations are beyond the scope of this study. Trajectory #3 does not lead to benzene formation up to t = 3.5 ps but leads to a seven-membered metallacycle intermediate structure. Interestingly, the geometric structure at t = 3–3.5 ps was nearly planar (Figure 10b). Trajectory #3 also shows non-IRC dynamics because such a planar structure cannot be observed along the IRC pathway presented in Figure 8b. Finally, proton transfer in a water molecule directly bound to Ti+ was observed in all trajectories. A detailed analysis indicated rearrangement of the hydrogen bonding in the water molecule directly bound to Ti+. Initially, this water molecule is hydrogen bonded to two other water molecules by donating two hydrogen atoms (see Figure 8b). During trajectory evolution, the structure of the cluster is rearranged such that the corresponding water is hydrogen bonded to one water molecule. 3. Conclusions and Future Directions In this study, we first performed reaction path analyses using quantum chemistry calculations of the acetylene cyclotrimerization reaction catalyzed by transition metal cation complexes (Sc+, Ti+, Mn+, Co+, and Ni+) with NC− and (H 2 O)8 ligands, as well as bare Sc+- and Ti+-catalyst systems to understand the possible roles of such processes in interstellar benzene formation. The results indicate that Sc+ and Ti+ complexes act as efficient catalysts for cyclotrimerization because they do not have an entrance barrier. The number of d-donor orbitals plays an important role in determining the height of entrance barriers. The NC− ligand reduced the barrier for formation of the Ti+ compound. On-the-fly BOMD trajectories were subsequently calculated for TiNC and Ti+(H 2 O)8 to understand the mechanism of benzene formation from a dynamic perspective. Each trajectory calculation was started from the C 2 H 2 + TiNC-(C 2 H 2)2 or C 2 H 2 + Ti+(H 2 O)8-(C 2 H 2)2 reactants, where the initial distance between C 2 H 2 and Ti+ was set as 4–5 Å. No initial kinetic energy was applied to the atoms in the reaction system to qualitatively understand the overall reaction pathway and the reaction under low-temperature interstellar conditions. BOMD calculations revealed benzene formation via a sequential C-C bond formation in both complexes. Thus, this computational study suggests that interstellar benzene can be formed via acetylene cyclotrimerization catalyzed by a Ti+ complex with a solid water (ice) cluster and NC−. It is important to describe the astrophysical implications of the computational study. The most important outcome of the present quantum chemistry calculations is that the benzene formation process starting from C 2 H 2 + (C 2 H 2)2-Ti+-L and C 2 H 2 + (C 2 H 2)2-Sc+-L is barrierless. Nevertheless, it is unlikely that complexes of such metal cyanides with acetylene molecules exist in low-density spaces because of the lack of a third body for absorbing the excess energy. In contrast with isolated ligands, it is likely that the solid surface, such as water and silicate, acts as a ligand for the metal cations . Once a metal cation attaches to the metal center, this could lead to efficient cyclotrimerization. Note that this scenario works only for low-temperature solid water conditions because the metal cation-attached water surface structure is energetically metastable. It should be emphasized that the metal cation can be internally solvated by water molecules owing to the strong ion–water attractive interaction. In fact, many previous experimental studies on large metal cation–(H 2 O)n complexes have shown that the metal cation is surrounded by several water molecules . Further studies should be performed to further evaluate the quantitative astrophysical implications of the present barrierless cyclotrimerization processes on the dust surface. In the near future, we will extend the present computational study to include HCN/HNC instead of acetylene in the catalytic cyclotrimerization processes as HCN and its isomer, HNC, are isoelectronic with C 2 H 2 and have been detected in interstellar media . We intend to discuss the possible roles of transition metal compounds in the formation of pyridine, pyrazine, and pyrimidine via catalytic cyclotrimerization processes, where these compounds play an important role as building blocks in interstellar PAH formation, including heteroatoms. Moreover, CO is abundant in the interstellar medium. We hope to study and discuss PAH formation using CO as the ligand. 4. Computational Procedure Following our previous benchmark quantum chemistry calculations for open-shell transition metal atom systems, [26,27,28] we carried out an unrestricted version of the B3LYP DFT functional with def2-SVPP and def2-TZVP basis sets. To account for the dispersion effects in the B3LYP functional, Grimme’s D3 dispersion corrections (DFT-D3) and the Becke–Johnson (BJ) damping function were employed throughout the present calculations . The potential energies at the stationary points were obtained using the Gaussian09 program package . For reliable computational outcomes, that is, smooth potential energy surfaces for the transition metal compounds with open-shell electronic structures, we applied the “stable = opt” option implemented in the Gaussian09 code at every step of geometry optimization. When this option is selected, the self-consistent field (SCF) solution is forced to converge to the most stable SCF solution without internal instability. All stationary points and their associated IRC were obtained from global reaction route mapping (GRRM) calculations, which automatically explored the potential minima and transition states without human intuition [61,62,63]. The IRC as well as the potential energies at the stationary points calculated by this DFT-D3 level qualitatively agreed with the results obtained by higher-level computational methods. The benchmarked results are represented in the Supplementary Materials. We also performed Born–Oppenheimer molecular dynamics (BOMD) calculations implemented in the Gaussian09 code, where energy gradients were employed to solve the classical mechanical equations of motion. BOMD calculations were performed for TiNC(C 2 H 2)2 and Ti+(H 2 O)8(C 2 H 2)2 using isolated acetylene systems. Initially, the isolated acetylene was placed approximately 4 Å away from the titanium ion at a constant velocity. The momenta of the TiNC(C 2 H 2)2 and Ti+(H 2 O)8(C 2 H 2)2 complexes were set as 0. The cation–molecule interactions can affect the activation of nuclear motion. Supplementary Materials The following supporting information can be downloaded at: Figure S1: Molecular orbitals correlated with the π-donation and π-back-donation; Figure S2: Potential energy minimization profiles; Figure S3: Minimum energy paths (MEPs) for the doublet TiNC(C 2 H 2)2 with isolated acetylene and for doublet Ti+(H 2 O)8(C 2 H 2)2 with isolated acetylene; Figure S4: Initial configurations of the TiNC(C 2 H 2)3 from molecular dynamics; Table S1: Potential energies for neutral catalyst systems; Table S2: Vertical excitation energies at the stationary points; Table S3: Potential energy data by the benchmark calculations; Table S4: Relative energies calculated by 2-state averaged DF-CASPT2/cc-pVTZ; Video S1: Benzene formation movie by Trajectory #1 (up to 2 ps) for the TiNC(C 2 H 2)3 system; Video S2: Benzene formation movie by Trajectory #1 (up to 1 ps) for the Ti+(H 2 O)8(C 2 H 2)3 system; Video S3: Movie of Trajectory #1 for the Ti+(H 2 O)8(C 2 H 2)3 system from another angle. Cartesian coordinates for all stationary points (in Angstrom) and the potential energies (in Hartree) and zero-point energies (in Hartree) [64,65,66,67]. Click here for additional data file. (7.3MB, zip) Author Contributions Conceptualization, T.M. and T.T.; Methodology, T.M. and T.T.; validation T.M. and T.T.; investigation, T.M. and T.T.; data curation, T.M. and T.T.; writing—original draft preparation, T.M. and T.T.; writing—review and editing, T.M., N.M., T.F. and T.T.; visualization, T.M. and T.T.; supervision, T.M. and T.T.; project administration, T.M. and T.T.; funding acquisition, T.T. All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement Data are contained within the article. Conflicts of Interest The authors declare no conflict of interest. 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3339	https://math.mit.edu/research/highschool/primes/materials/2013/Oh.pdf	Towards Generalizing Thrackles to Arbitrary Graphs Jin-Woo Bryan Oh Abstract In the 1950s, John Conway came up with the notion of thrackles, graphs with embeddings in which no edge crosses itself, but every pair of distinct edges intersects each other exactly once. He conjectured that \|E(G)\| ≤\|V(G)\| for any thrackle G, a question unsolved to this day. In this paper, we discuss some of the known properties of thrackles and contribute a few new ones. Only a few sparse graphs can be thrackles, and so it is of interest to ﬁnd an analogous notion that applies to denser graphs as well. In this paper we introduce a generalized version of thrackles called near-thrackles, and prove some of their properties. We also discuss a large number of conjectures about them which seem very obvious but nonetheless are hard to prove. In the ﬁnal section, we introduce thrackleability, a number between 0 and 1 that turns out to be an accurate measure of how far away a graph is from being a thrackle. 1 Introduction In 1952, John Conway came up with the notion of thrackles and formulated his famous Thrackle Conjecture, which remains one of the most difﬁcult open problems in combinatorics. One of the reasons why this is of interest is the apparent simplicity of these objects, and the beautiful known mathematics surrounding them. Another reason is the fact that — its seeming simplicity notwithstanding — it appears to be an extremely challenging problem, since it combines argu-ments from graph theory with arguments from topology and geometry. Throughout this paper, we will only consider graphs without loops and multiple edges. Conway conjectured that a thrackleable graph, a graph in which no edge crosses itself but every pair of edges intersects each other precisely once, has at most as many edges as vertices. The conjecture is still open, but better and better bounds have been proven over the years. It is known that \|E(G)\| ≤c\|V(G)\| for some constant c, which implies that thrackles are sparse graphs. The constant c has also gotten closer and closer to 1 over the years. Lov´ asz, Pach and Szegedy proved the result for c ≈3, and Cairns and Nikolayevsky proved it for c ≈1.5. Today the best-known bound is c ≈1.428, due to Fulek and Pach . The conjecture is known to be true for some subclasses of thrackles, such as for linear thrackles, which are graphs that have thrackle drawings that use only straight lines. This paper is arranged as follows. In Section 2, we will brieﬂy set up the preliminaries and notation that we will use throughout the rest of the paper. Then in Section 3, we will brieﬂy discuss some basic properties of thrackles and formulate a new theorem related to their chromatic number. 1 Theorem 1.1. A thrackle G has chromatic number at most 3. In Section 4 we will switch gears and talk about the problem of generalizing thrackles. Thrackles are interesting objects, but the problem is that they only apply to sparse graphs, since the number of edges is only linear in the number of vertices. To ﬁx this, we wish to ﬁnd some embedding for any graph G that is in some sense “near” a thrackle drawing of G. This is the motivation behind the deﬁnition of near-thrackle drawings. A near-thrackle drawing will be an embedding of a graph that will ﬁrst maximize the number of pairs of edges that intersect once, and then maximize the number of pairs of edges that do not cross, and then the ones that cross twice, then thrice, and so on. In particular, if G is a thrackle, its near-thrackle drawing should just be a thrackle drawing of G. In Section 5, we will consider the theory of near-thrackle drawings in the case when we only allow straight lines, forming linear near-thrackles. We will ﬁnally turn to asymptotics in Section 6, and then discuss the notion of thrackleability and pose some of the interesting questions it raises. 2 Preliminaries and Notation We start with some basic deﬁnitions. Deﬁnition 2.1. A thrackle drawing is a graph embedding where no edge crosses itself, but every pair of distinct edges intersects each other exactly once; this point of intersection is allowed to be a common endpoint, but cannot be tangential between the two edges. A thrackle is a graph that admits a thrackle drawing. For instance, the following graphs (the cycle graphs C3, C5, C6 and the path graph P4) are all thrackles because of the way we have drawn them. However, note that the 4-cycle C4 is not a thrackle. v3 v2 v4 v1 If v1v2v3v4v1 had been a thrackle, we would be able to start from v1 and trace out the cycle and get to v4 without violating the thrackle property. But then note that v4v1 would have to 2 intersect v2v3, and then we would end up in the small triangle as in the ﬁgure above, and there would be no way to escape and connect back to v1 without creating superﬂuous crossings. In fact, it is known (and we will prove shortly) that Cn for n ≥3 is a thrackle except when n = 4. Note that it is obvious that if G is a thrackle, then so is any subgraph G′ of G, since we can just start with a thrackle drawing of G and then delete edges and vertices from that same drawing to get a thrackle drawing of G′ ⊆G. 3 Properties of Thrackles 3.1 Known Results The following results are classic and can be found readily in literature. (See, for instance, [4, 5, 6]) Theorem 3.1. If G is a linear thrackle (constrained to be drawn only using straight lines), then \|E(G)\| ≤\|V(G)\|. The proof of this theorem is due to Pach and Sterling . We now prove the result mentioned earlier about almost all cycles being thrackles. It is clear that C3 is a thrackle. We can also easily verify that C6 also has a thrackle drawing. The following claim, due to Wehner , then gives us a two-step inductive proof that every Ck is a thrackle for k ∈N{1,2,4}. Claim 3.1. If Cn is a thrackle, then Cn+2 is also a thrackle. Proof. If we replace any path of length 3 to a path of length 5 as follows, each pair of edges will still intersect exactly once, and so we are left with a thrackle. In fact, Lov´ asz, Pach and Szegedy proved the following structural property of thrackles. We omit the proof since it is too technical, though in essence it just involves comparing parities in two different ways. Claim 3.2. A thrackle cannot contain two vertex-disjoint odd cycles. Let us restate an easy claim mentioned once before as a proposition, since it makes it easier to motivate an approach to Conway’s conjecture. The proof is trivial. Proposition 3.1. If G is a thrackle, then any subgraph G′ of G is also a thrackle. Corollary 3.1.1. If G has a C4-subgraph, G is not a thrackle. Using these tools, one can carefully prove the following claim. The details are left as an exercise for the enterprising reader. 3 Claim 3.3. If Conway’s Conjecture is false, then a minimal counterexample will be topologi-cally homeomorphic to one of the following three shapes, as shown in . Furthermore, Rubinstein (unpublished) showed that if any one of these counterexamples exists, then so do the other two. 3.2 Chromatic Numbers of Thrackles In this subsection, we will prove Theorem 1.1. Recall that the theorem states that every thrackle is 3-colorable. Note that this is not a very obvious result in spite of the fact that thrackles do not have C4-subgraphs and hence K4-subgraphs. We might expect that if a graph does not contain small cycles, then it looks locally like a tree, and is therefore 2-colorable. However, this is false, since graphs of arbitrarily large girth and chromatic number are known, a deep result due to Erd˝ os . Proof of Theorem 1.1. We use induction on the number of vertices n. Clearly for n ∈{1,2,3}, the result is trivial. Suppose now for n ≥4 we have some thrackle G on n vertices. First we claim that G has a vertex of degree at most 2. If not, then all vertices in G have degree at least 3, and so summing the degrees we get 2\|E(G)\| = ∑v∈V(G) deg(v) ≥3\|V(G)\|, so that \|E(G)\| ≥1.5\|V(G)\|, which contradicts the known bound of \|E(G)\| ≤1.428\|V(G)\| < 1.5\|V(G)\| due to Fulek and Pach . So assume v ∈V(G) has degree at most 2. v G′ G Consider the graph G′ = G{v}. Clearly G′ is a subgraph of G and is therefore a thrackle. By the induction hypothesis, G′ is 3-colorable. Now when we add v back in to form G from G′, we can extend a proper 3-coloring of G′ to a proper 3-coloring of G, since the neighbors of v in G use up at most two of our three available colors. Hence G is 3-colorable as well, completing the proof. 4 4 Near-Thrackle Drawings 4.1 Deﬁnition and examples Deﬁnition 4.1. For any graph G, a near-thrackle drawing of G is an embedding of G in the plane satisfying the following conditions: • First, out of all the embeddings of G, choose only the ones that maximize the number of pairs of edges that cross exactly once. • Then, out of these embeddings of G, choose only the ones that maximize the number of pairs of edges that do not cross. • Iterate the process by maximizing the number of pairs of edges that cross 2,3,4,... times. Note that if G is a thrackle, this algorithm stops after the ﬁrst step. We have the following conjecture that seems true based on an extensive search. Conjecture 4.1. For any input graph G, the algorithm to determine a near-thrackle drawing (Deﬁnition 4.1) stops after the ﬁrst two steps. That is, for every graph G, there exists a (not necessarily unique) drawing that maximizes the 1-crossings and only uses 0- and 1-crossings. The reason we expect this conjecture to be true is the following intuition. Suppose for a graph G, a near-thrackle drawing has m1 pairs of edges that cross once, m0 pairs of edges that do not cross, and m≥2 pairs that cross twice or more, for m≥2 ≥1. Then, the total number of pairs of edges of G, which is constant for any given G, is m1 + m0 + m≥2. We already know there exists an embedding of G. In that embedding, suppose there are m′ 1 pairs of edges that cross once and m′ 0 pairs that do not cross. Then, m′ 1 +m′ 0 = m1 +m0 +m≥2 > m1 +m0. Since we started by maximizing m1, we have m1 ≥m′ 1, so that m′ 0 < m0, which appears to be an extremely hard example to construct, though not entirely impossible. An interesting question would be to ﬁgure out what such a drawing would look like, if it exists, and what, if anything, it would tell us about the underlying topology. If this conjecture were false, let us see what kind of conﬁgurations we might expect in a counterexample. Clearly, if m≥2 ≥1, then one of the above conﬁgurations must appear in our graph drawing (since there must be a pair of edges intersecting more than once). However, if the conﬁguration above appears in our drawing with nothing in the bounded region, then we can get a better drawing just by “pulling” the two edges apart, like a type II Reidemeister move. Therefore, the bounded region must contain a vertex of degree at least 2. This fact is straightforward to prove. We can prove a small number of other statements about the bounded region, but not enough to get a counterexample or a contradiction. The following is an example of a graph G and a near-thrackle drawing. In this near-thrackle drawing, it is easy to check that there are 27 pairs of edges crossing each other exactly once, and one pair of edges which do not cross, namely a1a2 and ac. 5 c a a1 a2 c1 b1 b Figure 1: An example of a near-thrackle drawing of a graph G Claim 4.1. Figure 1 does indeed depict a near-thrackle drawing of the graph G. Proof. It sufﬁces to prove that G is not thrackleable. This can be shown by arguing that the wedge of two triangles (which is a subgraph of G) is not a thrackle. Since a triangle has a unique thrackle embedding, we can without loss of generality ﬁx one of them. Now consider what happens when we add the two edges adjacent to the “wedge” vertex. It is easy to see that there are essentially four ways to add this pair of edges and still maintain the thrackle property. In one of them, the two degree-1 vertices lie inside the original triangle; in another, they both lie outside; and in two other ways they lie in different regions. In these two drawings, we have to connect the two degree-1 vertices by an edge, which should intersect each side of the original triangle exactly once (to maintain the thrackle property). But clearly, by parity counting, this edge will then end up on a different region away from where we want it to connect. The two remaining cases are the following. 6 We can no longer use a parity argument, but by explicitly starting from one of the degree-1 vertices, we can argue topologically (using the Jordan curve theorem implicitly) that we cannot complete the drawing in any way. We leave it to the reader to verify the details. This implies that G is not thrackleable, and we are done. Finding a near-thrackle drawing of K4 is similar. D C B A If a graph contains C4, it cannot be a thrackle. The usual (non-planar) way of drawing K4 turns out to be a near-thrackle drawing. In this particular drawing, we can see 13 pairs of 1-crossing and 2 pairs of 0-crossing edges. Since we know that K4 is not a thrackle and hence there cannot be 15 pairs of 1-crossing edges, our goal at best would be to ﬁnd 14 1-crossing pairs of edges with either a 0-crossing or an n-crossing (for some n ≥2). In the drawing above, the two 0-crossing pairs of edges are AB and CD, and BC and DA, so let’s try to make 14 pairs of 1-crossing edges by crossing AB and CD. D C B A However, when we cross AB and CD, it is unavoidable to cross one of already crossed lines such as AC, BD, AD, BC, CD. As a result, it gives a strictly worse embedding than the standard drawing of K4 itself. So this standard drawing is the near-thrackle embedding. This is not a rigorous proof by any means, but we leave it to the reader to check that it can be made into one easily. Finding a near-thrackle embedding of K5 turns out to be much harder. So far, this seems to be the best embedding. 7 Here 37 pairs of edges intersect once, 6 pairs do not intersect, and 2 pairs of edges intersect twice. If this embedding is a near-thrackle drawing, Conjecture 4.1 ends up false. It can be proven with some additional work that there can be at most 39 pairs of edges intersecting exactly once in any drawing of K5. 4.2 Strong and weak deletion conjectures Since near-thrackle drawings are generalized forms of thrackles, we expect them to have some characteristics of thrackles. It seemed at ﬁrst glance that the so-called strong deletion conjec-ture would be true. This was due to the following: Suppose we have a near-thrackle drawing of a graph G. Pick any e ∈E(G), and delete e from that drawing. Then this drawing would be expected to be a near-thrackle drawing of G{e}. However, this conjecture turns out to be false, as evidenced by the following counterexample. Note that this is a counterexample to both the strong deletion conjecture for edges, as well as to the one for vertices, which states that given a near-thrackle drawing, we can delete any vertex to obtain a near-thrackle drawing for the corresponding subgraph. In the following ﬁgure, there are two ways to delete an edge, and since all edges in K4 are equivalent, the resulting subgraph should be the same. Out of 10 pairs of edges, the graph on the left would have 9 pairs of 1-crossing and 1 pair of 0-crossing edges. However, the graph on the right would have 8 pairs of 1-crossing and 2 pairs of 0-crossing edges. Thus the strong edge deletion conjecture is false. To disprove the strong vertex deletion conjecture, we can use Figure 1. Deleting vertex b leaves a non-thrackle drawing of a thrackleable graph, and hence the conjecture is false. 8 c a a1 a2 c1 b1 b The disproof follows because the graph above has a thrackle embedding. This leads to the so-called weak deletion conjectures, which has been found true for all studied examples so far. Conjecture 4.2 (Weak deletion for edges). Suppose we have a near-thrackle drawing of a graph G. Then there exists some e ∈E(G) such that deleting e from this drawing yields a near-thrackle drawing of G{e}. Conjecture 4.3 (Weak deletion for vertices). Suppose we have a near-thrackle drawing of a graph G. Then there exists some v ∈V(G) such that deleting v from this drawing yields a near-thrackle drawing of G{v}. Note that the conjecture for edges does not imply the one for vertices in case of weak deletion. 5 Linear Near-Thrackles 5.1 Deﬁnition and examples Deﬁnition 5.1. For any graph G, a linear near-thrackle drawing of G is a near-thrackle em-bedding of G subject to the constraint that all edges must be drawn as straight lines. Note that any two non-overlapping straight lines can intersect at most once. So the edges of a linear near-thrackle drawing intersect each other at most once. So unlike general near-thrackle drawings, it is clear that the algorithm to determine a linear near-thrackle drawing (Deﬁnition 5.1) actually does stop after the second step. An example would be Figure 2. The following theorem follows because of the well-known fact that the number of crossings is maximized when points are placed in a convex position. For a convex n-gon, the number of interior crossings is n 4 , since any interior crossing is formed by two diagonals which are deﬁned uniquely by the quadrilateral formed by choosing four of the n vertices. Theorem 5.1. A linear near-thrackle drawing of Kn is obtained by taking the n vertices in convex position, and then drawing all possible edges between them. 9 Figure 2: An example of a linear near-thrackle. In general, this seems to be the unique linear near-thrackle drawing of Kn. A nice conse-quence of the theorem about complete graphs above is that linear near-thrackle drawings of Kn have n 4 +n n−1 2 = n(n−1)(n−2)(n+9) 24 pairs of edges that cross exactly once, and the remaining pairs do not cross at all. The same result seems to be true for complete bipartite graphs as well. It is veriﬁable for K2,3 and K3,3. Conjecture 5.1. A linear near-thrackle drawing of Km,n is obtained by taking m+n vertices in convex position, and then deﬁning m contiguous ones as one side of the partition, the n others as the other side of the partition, and drawing all possible edges between them. If the conjecture is true, a similar expression can be obtained for Km,n. It can be easily checked that in this case, linear near-thrackle drawings of Km,n have m n 2 + n m 2 + n 2 m 2 pairs of edges that cross exactly once. 5.2 Strong and weak deletion conjectures for linear near-thrackles We can formulate the deletion conjectures exactly as for general near-thrackles. Note that these are different questions, given the additional constraints in drawing linear near-thrackles. However, since the near-thrackle drawing and the linear near-thrackle drawing are the same for K4, the strong edge deletion conjecture for linear near-thrackle also turns out to be false, by the same counterexample. For strong vertex deletion conjecture for linear near-thrackle, we will use Figure 2 to prove that the strong vertex deletion conjecture is false. To prove that, we will delete vertex a1 from Figure 2. 10 b c a b1 a2 c1 a1 To ﬁnish off the argument, note that the graph above has a strictly better drawing by shifting the vertex a2, so this cannot be a near-thrackle drawing. b c a b1 a2 c1 We once again wrap up this section by stating the weak deletion conjecture, this time for linear near-thrackles. Conjecture 5.2 (Weak deletion for edges, linear case). Suppose we have a linear near-thrackle drawing of a graph G. Then there exists some e ∈E(G) such that deleting e from this drawing yields a linear near-thrackle drawing of G{e}. Conjecture 5.3 (Weak deletion for vertices, linear case). Suppose we have a linear near-thrackle drawing of a graph G. Then there exists some v ∈V(G) such that deleting v from this drawing yields a linear near-thrackle drawing of G{v}. It is worth observing that the conjectures for the general case do not, in fact, imply the ones for the linear case. We leave it to the reader to verify this straightforward fact. 6 Thrackleability Because of the fact that, given a graph G, there is no easy way to determine m1, the number of pairs of edges from E(G), that cross each other exactly once in a near-thrackle drawing of G (or in other words, the maximum number of pairs of edges that cross each other exactly once over all drawings of G on the plane), we can consider for any graph the ratio m1/ \|E(G)\| 2 and use it as a measure of how “close” G is to being a thrackle, in some sense. This yields the following deﬁnition. 11 Deﬁnition 6.1. The thrackleability ϕ(G) of a graph G with at least two edges is deﬁned as the quantity m1 \|E(G)\| 2 , where m1 is the number of pairs of edges in E(G) that cross each other exactly once in any near-thrackle drawing of G. Note that ϕ(G) = 1 if and only if G is a thrackle. Furthermore, note that ϕ(G) ̸= 0 for any G, since any graph with two edges has a drawing in which m1 ≥1. It is easy to compute the thrackleability of the graphs we have used throughout this paper as examples. An interesting question is what happens in the asymptotics limit. Stated precisely, if F is a family of graphs of increasing size, then what are the families F = (G1,G2,...) for which ϕ(Gn) converges to a ﬁnite value as n →∞? If this happens, we denote this value by ϕ(F). For an example, suppose F = (G1,G2,...) is the family of triangle-wedges, where Gn is the wedge of n triangles. What can we say about ϕ(F)? 1 3 2 n Proposition 6.1. If F is the family of triangle-wedges, then ϕ(F) exists and is bounded above by 8/9. Proof. We know by our earlier arguments that G2 is not a thrackle. So in every copy of G2 in Gn for n ≥2, there is a pair of edges that do not cross each other exactly once. In particular, note that for any two triangles in Gn, we have a unique copy of G2, and hence a pair of edges corresponding only to that copy of G2 that do not intersect exactly once. So, there are at least n 2 pairs of edges in Gn that do not intersect each other, and so ϕ(Gn) ≤1− n 2 3n 2 . In the limit n →∞, therefore, we get ϕ(F) ≤1−lim n→∞ n 2 3n 2 = 1−n2 9n2 = 8 9. We can even ask the same questions for linear near-thrackles. Deﬁnition 6.2. The linear thrackleability ϕ+(G) to be m′ 1/ \|E(G)\| 2 , where m′ 1 is the number of pairs of edges that intersect each other exactly once in a linear near-thrackle drawing of G. 12 As before, 0 < ϕ+(G) ≤1 for any G, with ϕ+(G) = 1 if and only if G is a linear thrackle, for instance, any odd cycle. We can ask the same bound-related questions about ϕ+(G). For instance, we have the following proposition. Proposition 6.2. Let G be the family of complete graphs {Kn}n∈N. Then, ϕ+(G ) = 1/3, where ϕ+ for a family of graphs is deﬁned as for ϕ. Proof. From the discussions following Theorem 5.1, we get ϕ+(G ) = lim n→∞ n(n−1)(n−2)(n+9) 24 (n 2) 2 = 8 24 = 1 3. Proposition 6.3. Let G be the family of complete graphs {Kn}n∈N, and G \e be the family {Kn \ e}n∈N, the family of complete graphs with precisely one edge deleted from each. Then, ϕ+(G \e) = 1/3, where ϕ+ for a family of graphs is deﬁned as the obvious restriction of ϕ to linear drawings. Proof. From the discussions following Theorem 5.1, we get ϕ+(G ) ≤lim n→∞ n(n−1)(n−2)(n+9) 24 (n 2) 2 = 8 24 = 1 3. ϕ+(e) ≥2(n−2). ϕ+(G \e) ≤lim n→∞ n(n−1)(n−2)(n+9) 24 −2(n−2) (n 2)−1 2 = 8 24 = 1 3. Proposition 6.4. Let G be the family of complete graphs {Kn}n∈N, and G \te be the family of complete graphs{Kn}n∈N with t disjoint edges deleted where t ≤⌊n⌋ 2 Then, ϕ+(G \te) = 1/3, where ϕ+ for a family of graphs is deﬁned as for ϕ. Proof. Starting from any drawing of Kn, any edge we delete gets read of at least 2(n −2) crossings, corresponding to its degree in the rest of the graph. So after deleting t disjoint edges, we have destroyed at least 2t(n −2) 1-crossings. Since the number of 1-crossings in a linear near-thrackle drawing of Kn is known from Theorem 5.1, we get an upperbound to ϕ+(G \te) to be ϕ+(G \te) ≤lim n→∞ n(n−1)(n−2)(n+9) 24 −2t(n−2) (n 2)−1 2 = lim n→∞ n(n−1)(n−2)(n+9) 24 −2 ⌊n⌋ 2 (n−2) (n 2)−1 2 = 8 24 = 1 3. The fact that this bound can be attained can be seen by drawing a usual linear near-thrackle drawing of Kn, picking t disjoint edges along with the outer boundary of this drawing and deleting them. 13 Conjecture 6.1. Let B be the family of complete bipartite graphs {Km,n }m,n∈N. Then, ϕ+(B) = 1/2, where ϕ+ for a family of graphs is deﬁned as for ϕ. The reason why we expect this to be true is the following. If the Conjecture 5.1 is true, we get ϕ+(B) = lim m,n→∞ m n 2 +n m 2 + n 2 m 2 m+n 2 = 1 2. This raises an obvious question. How close to 0 can ϕ(G) or ϕ+(G) get? Are there speciﬁc families of graphs for which asymptotically ϕ or ϕ+ gets arbitrarily close to 0? If not, what is a good lower bound for ϕ or ϕ+? Can we get ϕ(G) to be at most 1/2? 7 Acknowledgement I would like to thank Rik Sengupta, an incredible mentor who helped me with the Thrackle Conjecture. Also, I want to credit Andrey Grinshpun who provided helpful suggestions to prove a theorem, and Dr. DouglassWoodall who kindly shared his research paper on thrackles with me. Furthermore, I would like to extend my gratitude to the PRIMES-USA program staff, namely Dr. Ben Elias, PRIMES-USA’s director, Dr. Tanya Khovanova, who provided me with great feedback, Dr. Slava Gerovitch, PRIMES director, and Dr. Jacob Fox, who originally inspired me to solve this problem. 14 References G. Cairns and Y. Nikolayevsky, Bounds for generalized thrackles, Discrete and Computa-tional Geometry, 23:2 (2000), 191-206. P. Erd˝ os, Graph Theory and Probability, Canadian Journal of Mathematics 11 (1959), 34-38. R. Fulek and J. Pach, A computational approach to Conway’s thrackle conjecture, Com-putational Geometry 44:6-7 (2011), 345-355. L. Lov´ asz, J. Pach and M. Szegedy, On Conway’s thrackle conjecture, Discrete and Com-putational Geometry 18:4 (1997), 369-376. J. Pach and E. Sterling, Conway’s conjecture for monotone thrackles, American Mathe-matical Monthly 118:6 (2011), 544-548. S. Wehner, On the thrackle problem, ac-cessed 9/25/2013.
3340	https://www.learnleansigma.com/guides/normality-test/	Published Time: 2023-09-10T13:40:21+00:00 Guide: Normality Test » Learn Lean Sigma Get 28 Days of Templates and Training Content Get Free Stuff! We respect your privacy. Unsubscribe at anytime. 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With more than ten years of experience applying his skills across various industries, Daniel specializes in optimizing processes and improving efficiency. His approach combines practical experience with a deep understanding of business fundamentals to drive meaningful change. View Templates View Guides A Normality Test is a statistical procedure that helps you determine if a given set of data follows a normal distribution or not. This is an important aspect of statistical analysis, quality control, and even machine learning models. Why? Many statistical techniques, such as t-tests, ANOVAs, and regression models, assume that the underlying data is normally distributed. If this assumption is violated, the results may be unreliable, leading to inaccurate conclusions and misguided decisions. The concept of a “normal distribution” might sound complex, but it’s essentially the famous “bell curve” that many of us learned about in school. In a normal distribution, the majority of the data points cluster around the mean, and the frequencies taper off symmetrically towards both ends, forming a bell-like shape. As illustrated above, the first graph represents a normal distribution, where the data is symmetrically distributed around the mean. The second graph, on the other hand, represents a non-normal (exponential in this case) distribution. You can see that the majority of data points are skewed towards one side. Understanding whether your data follows a normal distribution is crucial in fields like manufacturing, logistics, and especially in methodologies like Lean Six Sigma, where data-driven decision-making is key. This guide aims to provide you with a comprehensive understanding of Normality Tests, from the theory behind them to practical ways of performing these tests using various software tools. Now that you have a basic understanding of what a Normality Test is and why it’s important, let’s delve into the different methods for testing normality. Why Normality Matters Understanding the distribution of your data is not just an academic exercise; it has direct implications for how you interpret data and make decisions in various settings. Below are some key areas where the concept of normality plays a critical role: Statistical Assumptions Many statistical tests, such as t-tests, ANOVAs, and linear regression models, are based on the assumption that the data is normally distributed. If the data doesn’t meet this criterion, these tests can produce misleading results, which in turn may lead to incorrect conclusions. Quality Control in Lean Six Sigma In methodologies like Lean Six Sigma, ensuring that processes are stable and predictable is crucial. Understanding the distribution of data related to these processes can help you identify variations, anomalies, or trends that need to be addressed. For example, if a manufacturing process is assumed to be normally distributed, you can set control limits and identify outliers more reliably. Predictive Modeling Machine learning and predictive analytics models also often assume that the errors are normally distributed. Knowing the distribution of your data can help you choose the right model or make necessary adjustments to improve the model’s performance. Graphical Example: Control Charts Control charts are a staple in quality control and Lean Six Sigma projects. These charts often assume that the process data follows a normal distribution. Below is a simple control chart illustrating how data points are distributed around the control limits, with the assumption of normality. You might also be interested in our Contol Chart Tool By understanding the normality of your process data, you can set these control limits with a higher degree of confidence. This makes your quality control efforts more effective and reliable. The control chart is a practical example of why understanding data normality is essential, particularly in methodologies like Lean Six Sigma where data-driven decision-making is vital. Common Methods for Testing Normality Understanding the distribution of your dataset is a cornerstone of statistical analysis and is particularly important in methodologies like Lean Six Sigma. Testing for normality can be broadly categorized into two methods: Parametric Tests and Graphical Methods. Let’s delve into each in more detail: Parametric Tests Parametric tests are statistical tests that make certain assumptions about the parameters of the population distribution from which the samples are drawn. Here are the most commonly used parametric tests for checking normality: Shapiro-Wilk Test When to Use: This test is most accurate when used on small sample sizes (n < 50). How It Works: The test calculates a W statistic that represents a ratio: the squared sum of the differences between the observed and expected values of a normally distributed variable, divided by the sample variance. A W statistic close to 1 indicates that the data is normally distributed. Limitations: The test is sensitive to sample size. As the sample size increases, the test may show that the data is not normal even if the deviation from normality is trivial. Kolmogorov-Smirnov Test When to Use: This test is better suited for larger sample sizes. How It Works: It compares the empirical distribution function of the sample with the distribution expected if the sample were drawn from a normal population. The maximum difference between these two distributions is the D statistic. Limitations: It is less powerful for identifying deviations from normality at the tails of the distribution. Anderson-Darling Test When to Use: This test is a modified version of the Kolmogorov-Smirnov test and is used when more weight needs to be given to the tails. How It Works: It squares the differences between observed and expected values and gives more weight to the tails of the distribution. Limitations: Like the Shapiro-Wilk test, it is sensitive to sample size. Lilliefors Test When to Use: This is an adaptation of the Kolmogorov-Smirnov test for small sample sizes. How It Works: It operates similarly to the Kolmogorov-Smirnov test but corrects for the bias caused by the estimation of parameters from the sample data itself. Limitations: It’s less commonly used and not as powerful as the Shapiro-Wilk test for very small sample sizes. Graphical Methods Graphical methods provide a visual approach to understanding the distribution of your data. Here are some commonly used graphical methods: QQ-Plot (Quantile-Quantile Plot) A plot of the quantiles of the sample data against the quantiles of a standard normal distribution. A 45-degree line is often added as a reference. If the data points fall along this line, it suggests that the sample data is normally distributed. P-P Plot (Probability-Probability Plot) Similar to a QQ-Plot but plots the cumulative probabilities of the sample data against a standard normal distribution. Useful when you are interested in the fit of different types of distributions, not just the normal. Histogram A bar graph that shows the frequency of data points in different ranges. If the data is normally distributed, the histogram will resemble a bell curve. Box Plot Provides a visual representation of the data’s spread, skewness, and potential outliers. A symmetric box indicates normality, while skewness or outliers suggest non-normality. These are the most common methods for testing normality, each with its own advantages and limitations. Choosing the right method depends on your specific needs, the size of your dataset, and the importance of the tails in your analysis. Want exclusive templates, tools and guides? Join our email list below and for the next 28 days, we will send you exclusive tools, templates and guides unavailable on the website. We developed a short and simple 28-day program designed to develop your ability to implement Lean and Six Sigma methods daily. Step-by-Step Guide to Performing a Normality Test After understanding the importance of normality and the methods available for testing it, the next step is to actually perform these tests. This section will guide you through conducting normality tests using popular software tools such as Minitab, SPSS, and R. Using Software Tools Software tools offer a convenient and efficient way to perform normality tests, particularly when dealing with large datasets. Below we’ll walk you through how to use each of these tools for this purpose. Minitab Introduction to Minitab as a Statistical Software Minitab is a widely-used statistical software package that offers a range of data analysis capabilities. It is particularly popular in industries like manufacturing and services where Lean Six Sigma methodologies are employed. Steps for Conducting a Normality Test in Minitab Load Your Data: Import your dataset into Minitab. Navigate to the Test: Go to Stat>Basic Statistics>Normality Test. Select Variables: Choose the variable(s) you want to test for normality. Run the Test: Click OK to run the test. Minitab will generate an output with the results. Interpretation of Minitab Output P-value: A P-value less than 0.05 generally indicates that the data is not normally distributed. Test Statistic: Look for the W statistic in the case of a Shapiro-Wilk test. Graphical Output: Minitab also provides QQ-Plots and histograms for visual inspection. SPSS How to Use SPSS for Normality Tests SPSS is another comprehensive statistical software package used in various fields such as social sciences, healthcare, and market research. Load Data: Import your dataset into SPSS. Go to Test Option: Navigate to Analyze>Descriptive Statistics>Explore. Select Variables: Add the variables you want to test. Run: Click OK to run the test and review the output for the Shapiro-Wilk or Kolmogorov-Smirnov statistics and P-values. R Using R for Normality Tests R is a free and open-source software environment that is highly extensible and offers numerous packages for statistical analysis. Load Data: Use functions like read.csv() to load your data into R. Perform Test: Use functions such as shapiro.test() for the Shapiro-Wilk test or ks.test() for the Kolmogorov-Smirnov test. Interpret Output: A P-value less than 0.05 typically indicates non-normality. Using Python Python, with its rich ecosystem of data science libraries, offers a powerful environment for conducting normality tests. Below, we’ll explore two examples: the Shapiro-Wilk test and generating a QQ-Plot. Example 1: Shapiro-Wilk Test Python Code Snippet for Performing Shapiro-Wilk Test You can use the scipy.stats library to perform the Shapiro-Wilk test. First, you’ll need to import the library and then apply the shapiro() function to your dataset. Here’s how you can do it: ```python from scipy import stats Sample data data = [your_data_here] Perform Shapiro-Wilk test shapiro_result = stats.shapiro(data) Output the result print("Shapiro-Wilk Statistic:", shapiro_result) print("P-value:", shapiro_result) ``` Copy Interpretation of Results The output will consist of two values: Shapiro-Wilk Statistic: A value close to 1 indicates that the data is normally distributed. P-value: A P-value less than 0.05 generally indicates that the data is not normally distributed. A P-value greater than or equal to 0.05 suggests that the data is normally distributed. Example 2: QQ-Plot Python Code Snippet for Generating a QQ-Plot You can use the statsmodels library to generate a QQ-Plot. The qqplot() function is used for this purpose. Here’s a sample code snippet: ```python import statsmodels.api as sm import matplotlib.pyplot as plt Your data here data = [your_data_here] Create QQ-Plot fig, ax = plt.subplots(figsize=(10, 6)) sm.qqplot(data, line='45', ax=ax) plt.title('QQ-Plot') plt.show() ``` Copy Graphical Interpretation Points Along the 45-Degree Line: If the points fall along this line, it suggests that the data is normally distributed. Points Deviating from the Line: If the points significantly deviate from the 45-degree line, especially at the tails, then the data is not normally distributed. By using Python, you can easily perform normality tests and visualize the distribution of your dataset. The examples above provide you with the code snippets and interpretation guidelines to get you started. Advanced Topics Once you’re familiar with the basics of normality testing, you may encounter situations that require a more nuanced approach. This section delves into advanced topics such as dealing with non-normal data, data transformation techniques, non-parametric test alternatives, and the impact of sample size on test power. Dealing with Non-Normal Data Not all data sets are normally distributed, and that’s okay. The question is, what do you do when your data is not normal? Check the Importance: First, consider how crucial the normality assumption is for your specific analysis or project. In some cases, slight deviations from normality may not significantly impact your results. Use Robust Methods: Some statistical methods are robust to deviations from normality. These methods can often be used as a direct replacement for their non-robust counterparts. Data Transformation Techniques If normality is essential for your analysis, you might consider transforming your data to fit a normal distribution better. Common transformation techniques include: Log Transformation: Useful for reducing right skewness. Square Root Transformation: Effective for count data. Box-Cox Transformation: A more generalized form that encompasses many other types of transformations. ```python Example using Python for Box-Cox Transformation from scipy import stats Perform the transformation transformed_data, lambda_value = stats.boxcox(original_data) ``` Copy Non-Parametric Tests as Alternatives Non-parametric tests don’t assume any specific distribution and can be a useful alternative when dealing with non-normal data. Examples include: Mann-Whitney U Test: An alternative to the independent samples t-test. Wilcoxon Signed-Rank Test: An alternative to the paired samples t-test. Kruskal-Wallis Test: An alternative to the one-way ANOVA. Power and Sample Size How Sample Size Affects the Power of a Normality Test Small Sample Sizes: Normality tests are generally less reliable with small sample sizes. They may not detect non-normality even when it exists. Large Sample Sizes: On the other hand, with large sample sizes, the tests can detect even trivial deviations from normality, which might not be practically significant. Understanding the relationship between sample size and test power can help you make more informed decisions when planning your data collection and analysis strategies. Conclusion This comprehensive guide aimed to serve as a one-stop resource on the extensive topic of understanding and conducting normality tests, a cornerstone in statistical analyses and continuous improvement methodologies like Lean Six Sigma. Starting with the foundational principles, the guide navigated through various methods to test for normality, including parametric tests like Shapiro-Wilk and graphical methods such as QQ-Plots. Special attention was given to the practical application of these tests using popular software tools like Minitab, SPSS, R, and Python, providing step-by-step procedures and code snippets. The guide also ventured into advanced topics, offering insights into handling non-normal data through transformations and non-parametric tests. The power dynamics influenced by sample size, often overlooked, were highlighted to ensure a more nuanced understanding. Real-world case studies from industries like automotive, FMCG, and logistics were included to bridge the gap between theory and practice. Whether you’re a seasoned professional or a beginner in the realm of data analysis and continuous improvement, this guide aspires to equip you with the essential skills and knowledge to perform normality tests confidently and interpret their results effectively. Your journey towards mastering this critical aspect of data analysis begins here. References Das, K.R. and Imon, A.H.M.R., 2016. A brief review of tests for normality.American Journal of Theoretical and Applied Statistics,5(1), pp.5-12. Yazici, B. and Yolacan, S., 2007. A comparison of various tests of normality.Journal of Statistical Computation and Simulation,77(2), pp.175-183. Jarque, C.M. and Bera, A.K., 1987. A test for normality of observations and regression residuals.International Statistical Review/Revue Internationale de Statistique, pp.163-172. Q: What is the importance of performing a normality test? A: Normality tests are essential for determining whether your data follows a normal distribution, a foundational assumption in many statistical analyses. If your data is not normally distributed, using techniques that assume normality may lead to incorrect or misleading results. Normality tests help you validate this assumption before proceeding with further analyses. Q: Can I still perform statistical analyses if my data is not normally distributed? A: Yes, you can. There are non-parametric tests designed to analyze non-normal data. These tests do not assume any specific distribution and are often used as alternatives to their parametric counterparts. Additionally, you can transform your data to make it more normal-like and then apply parametric tests. Q: Are normality tests reliable for small sample sizes? A: Normality tests are generally less reliable for small sample sizes. With fewer data points, it’s difficult to accurately determine the distribution of the dataset. Therefore, caution should be exercised when interpreting the results of a normality test on small samples. Q: What's the difference between a QQ-Plot and a P-P Plot? A: Both QQ-Plots and P-P Plots are graphical methods for assessing the distribution of a dataset. A QQ-Plot compares the quantiles of the sample data against a theoretical distribution, while a P-P Plot compares the cumulative probabilities. QQ-Plots are more sensitive to deviations in the tails, whereas P-P Plots focus on deviations across all data points. Q: Can normality tests be automated in software like Minitab or Python? A: Yes, software tools like Minitab and Python libraries offer functionalities to automate normality testing. In Minitab, you can use macros to run the test on multiple datasets, while in Python, you can use loops and functions to perform the tests programmatically. This is particularly useful when dealing with large datasets or running repetitive analyses. Author #### Daniel Croft Daniel Croft-Bednarski is a Continuous Improvement Manager with a passion for Lean Six Sigma and continuous improvement. With years of experience in developing operational excellence, Daniel specializes in simplifying complex concepts and engaging teams to drive impactful changes. He shares his expertise through LearnLeanSigma.com, offering tools, guides, and insights to help others implement Lean methods effectively. Daniel is committed to cultivating a culture of improvement, across the industry, through practical resources, innovative strategies, and a hands-on approach to leadership. View Posts Free Lean Six Sigma Templates Improve your Lean Six Sigma projects with our free templates. Explore Templates Free Lean Six Sigma Templates Improve your Lean Six Sigma projects with our free templates. Explore Templates Menu Blog Calculators About Usage Policy Privacy Policy Cookie Policy (EU) Accessibility statement Focus Topics What is Lean Six Sigma? 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3341	https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.03%3A_Electron_Configurations-_How_Electrons_Occupy_Orbitals	Skip to main content 8.3: Electron Configurations- How Electrons Occupy Orbitals Last updated : Aug 14, 2020 Save as PDF 8.2: The Development of the Periodic Table 8.4: Electron Configurations, Valence Electrons, and the Periodic Table Page ID : 37944 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Derive the predicted ground-state electron configurations of atoms Identify and explain exceptions to predicted electron configurations for atoms and ions Relate electron configurations to element classifications in the periodic table Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom. Orbital Energies and Atomic Structure The energy of atomic orbitals increases as the principal quantum number, n, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of l differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure 8.3.1 depicts how these two trends in increasing energy relate. The 1s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2s and then 2p, 3s, and 3p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3d orbital is higher in energy than the 4s orbital. Such overlaps continue to occur frequently as we move up the chart. Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5p orbitals fill immediately after the 4d, and immediately before the 6s. The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n, increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f. Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have +Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1s through 3p), the increase in energy due to n is more significant than the increase due to l; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order. The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure 8.3.2): The number of the principal quantum shell, n, The letter that designates the orbital type (the subshell, l), and A superscript number that designates the number of electrons in that particular subshell. For example, the notation 2p4 (read "two–p–four") indicates four electrons in a p subshell (l = 1) with a principal quantum number (n) of 2. The notation 3d8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., l = 2) of the principal shell for which n = 3. The Aufbau Principle To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure 8.3.3), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure 8.3.3 illustrates the traditional way to remember the filling order for atomic orbitals. Since the arrangement of the periodic table is based on the electron configurations, Figure 8.3.4 provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3d orbitals. We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure 8.3.3 or 8.3.4, we would expect to find the electron in the 1s orbital. By convention, the ms=+12 value is usually filled first. The electron configuration and the orbital diagram are: Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron (n = 1, l = 0, ml = 0, ms=+12). The second electron also goes into the 1s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, ms=−12. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are: The n = 1 shell is completely filled in a helium atom. The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure 8.3.3 or 8.3.4). Thus, the electron configuration and orbital diagram of lithium are: An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital. An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals (ml = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling. Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals have identical n, l, and ms quantum numbers and differ in their ml quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are: Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are: The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s22s22p63s1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called valence electrons, and those occupying the inner shell orbitals are called core electrons ( Figure \PageIndex5\PageIndex5). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s22s22p6) and our abbreviated or condensed configuration is [Ne]3s1. Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells. Li:[He]2s1Na:[Ne]3s1 The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s2 configuration, is analogous to its family member beryllium, [He]2s2. Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s23p1, is analogous to its family member boron, [He]2s22p1. The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure 8.3.6 shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements. When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3d level but is, instead, added to the 4s level (Figure 8.3.3 or 8.3.4). As discussed previously, the 3d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4s, which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4s1. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4s subshell and calcium has an electron configuration of [Ar]4s2. This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium. Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [d orbitals], there are 2l + 1 = 5 values of ml, meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons (l = 3, 2l + 1 = 7 ml values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons. Example 8.3.1: Quantum Numbers and Electron Configurations What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added? Solution The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons: The last electron added is a 3p electron. Therefore, n = 3 and, for a p-type orbital, l = 1. The ml value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these ml values is correct. For unpaired electrons, convention assigns the value of +12 for the spin quantum number; thus, ms=+12. Exercise 8.3.1 Identify the atoms from the electron configurations given: [Ar]4s23d5 [Kr]5s24d105p6 Answer a : Mn Answer b : Xe The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure 8.3.3 or 8.3.4. For instance, the electron configurations of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling. In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4s into the 3d orbital to gain the extra stability of a half-filled 3d subshell (in Cr) or a filled 3d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5s24d3. Experimentally, we observe that its ground-state electron configuration is actually [Kr]5s14d4. We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5s orbital are larger than the gap in energy between the 5s and 4d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells. Electron Configurations and the Periodic Table As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure 8.3.6), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements. Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react. It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure 8.3.6, which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure 8.3.6 show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom. Main group elements (sometimes called representative elements) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure 8.3.6. This category includes all the nonmetallic elements, as well as many metals and the intermediate semimetallic elements. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar]4s23d104p1, which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons. Transition elements or transition metals. These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and (n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure 8.3.6) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure 8.3.6), and we will adopt this usage in this textbook. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure 8.3.6. The valence shells of the inner transition elements consist of the (n – 2)f, the (n – 1)d, and the ns subshells. There are two inner transition series: The lanthanide series: lanthanide (La) through lutetium (Lu) The actinide series: actinide (Ac) through lawrencium (Lr) Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons. Electron Configurations of Ions We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the (n – 1)d or (n – 2)f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle. Example 8.3.2: Predicting Electron Configurations of Ions What is the electron configuration and orbital diagram of: Na+ P3– Al2+ Fe2+ Sm3+ Solution First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable. Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals. Na: 1s22s22p63s1. Sodium cation loses one electron, so Na+: 1s22s22p63s1 = Na+: 1s22s22p6. P: 1s22s22p63s23p3. Phosphorus trianion gains three electrons, so P3−: 1s22s22p63s23p6. Al: 1s22s22p63s23p1. Aluminum dication loses two electrons Al2+: 1s22s22p63s23p1 = Al2+: 1s22s22p63s1. Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 = 1s22s22p63s23p63d6. Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6. Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm3+: 1s22s22p63s23p64s23d104p65s24d105p66s24f6 = 1s22s22p63s23p64s23d104p65s24d105p64f5. Exercise 8.3.2 Which ion with a +2 charge has the electron configuration 1s22s22p63s23p63d104s24p64d5? Which ion with a +3 charge has this configuration? Answer a : Tc2+ Answer b : Ru3+ Summary The relative energy of the subshells determine the order in which atomic orbitals are filled (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals). Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements (s and p orbitals), transition elements (d orbitals), and inner transition elements (f orbitals). Glossary Aufbau principle : procedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time core electron : electron in an atom that occupies the orbitals of the inner shells electron configuration : electronic structure of an atom in its ground state given as a listing of the orbitals occupied by the electrons Hund’s rule : every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin orbital diagram : pictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow valence electrons : electrons in the outermost or valence shell (highest value of n) of a ground-state atom; determine how an element reacts valence shell : outermost shell of electrons in a ground-state atom; for main group elements, the orbitals with the highest n level (s and p subshells) are in the valence shell, while for transition metals, the highest energy s and d subshells make up the valence shell and for inner transition elements, the highest s, d, and f subshells are included 8.2: The Development of the Periodic Table 8.4: Electron Configurations, Valence Electrons, and the Periodic Table
3342	https://www.youtube.com/watch?v=AUHe0emnAPo	sigma(1, infinity) 2^n / (1 + 3^n) Determine whether the series converges or diverges. MSolved Tutoring 66100 subscribers 78 likes Description 30288 views Posted: 12 Feb 2017 sigma(1, infinity) 2^n / (1 + 3^n) Determine whether the series converges or diverges. 9 comments Transcript:
3343	https://www.youtube.com/watch?v=HIjFpRxX_Hg	How to deal with any recursive sequence. Michael Penn 326000 subscribers 988 likes Description 19680 views Posted: 21 Nov 2021 Suggest a problem: Please Subscribe: Patreon: Merch: Personal Website: Randolph College Math: Randolph College Math and Science on Facebook: Research Gate profile: Google Scholar profile: If you are going to use an ad-blocker, considering using brave and tipping me BAT! Buy textbooks here and help me out: Buy an amazon gift card and help me out: Books I like: Sacred Mathematics: Japanese Temple Geometry: Electricity and Magnetism for Mathematicians: Abstract Algebra: Judson(online): Judson(print): Dummit and Foote: Gallian: Artin: Differential Forms: Bachman: Number Theory: Crisman(online): Strayer: Andrews: Analysis: Abbot: How to think about Analysis: Calculus: OpenStax(online): OpenStax Vol 1: OpenStax Vol 2: OpenStax Vol 3: My Filming Equipment: Camera: Lense: Audio Recorder: Microphones: Lights: White Chalk: Color Chalk: 86 comments Transcript: Introduction [Music] in a previous video i used a trick related to the roots of a certain quadratic polynomial and the closed form of a recursively defined sequence and i asked if there was anyone out there that wanted to see that trick proven and a couple of you guys said yes and so this is that video where we go over that trick okay so let's recall what we did back then so if we've got this recursively defined sequence a n plus 2 equals capital a a sub n plus 1 plus capital b a sub n then we know the closed form from this for this sequence a n is of the form capital r r 1 to the n plus capital s r 2 to the n where r 1 and r 2 are the distinct roots of the following polynomial which is known as the characteristic polynomial related to this recursive sequence so we've got x squared equals ax minus b equals 0. you might say well what about the seeds and so this allows us to define the n plus second term in terms of the n plus first term and the nth term well what about the zeroth term in the first term or the first term and the second term depending on how we define this well it turns out that that will not influence these roots at all that will influence these numbers here capital r and capital s which are constants so that's something that you would take care of afterwards looking at this polynomial and finding the roots will give you a rough idea of the closed form so i think if you were to look at a standard proof of this fact you would start with an educated guess that the sequence has this following closed form and then show any sequence with this closed form obeys this recursion and that's a reasonable way to do this because obeying this recursion has some sort of uniqueness property but that's not the way we're going to prove this and that's because to a fault i like to do things constructively and that's exactly what we're going to do here we're going to use a generating function okay so let's start off by defining f of x to be the sum as n goes from 0 up to infinity of a sub n x to the n so in other words that's our generating function for our recursively defined sequence okay now we'll take out the first term and the second term or maybe the zeroth term in the first term depending on how you're counting so we'll have a zero plus a one x plus now left over we'll have the sum as n goes from two up to infinity of a sub n x to the n now we can reindex this stuff right here so that the n's are replaced with n plus 2's and that'll be nice because then we'll have a sub n plus 2 and we can apply our defining recursion relation okay so let's see what we get we'll have a 0 plus a 1 x plus the sum now n will go from zero up to infinity because if n plus two is equal to two n is equal to zero and then we'll have a sub n plus two x to the n plus two now next up like i alluded to we will apply our recursion so i'll replace this a n plus 2 with capital a a n plus 1 plus capital b b a n and then i'll pull that apart into two sums so that's going to give us a naught plus a1 x plus ax times the sum as n goes from zero up to infinity of a n plus one x to the n plus 1 plus b x squared and then the sum as n goes from 0 up to infinity of a and x to the n so i did a little bit of simplification in the middle there notice here i have a times a n plus 1 times x to the n plus 2. i took our capital a out and one of the x's out leaving us with x to the n plus 1 and then the n plus first term of our sequence we really want this index here to match with this exponent then i did something similar over here but now we can notice that these objects that we've just created look a lot like our our starting generating function in fact this one that i am squaring in blue is exactly our starting generating function and this one is almost our starting generating function maybe we could re-index this and see exactly how it differs from f of x so let's maybe take n and replace it with n minus 1. so we'll do that so we'll have a and x to the n and then notice we will start at n equals 1 now okay so let's write this down we have a naught plus a 1 x plus a times x times the sum as n goes from 1 up to infinity of a and x to the n and then we have plus bx squared times f of x like that now this guy right here is our original generating function but it's missing the n equals zero term but we can easily fix that let's add in the zeroth term here but if we add in the zeroth term we also have to subtract the zeroth term but the zeroth term is just a zero times x to the zero or just a zero but let's recall that this sum was multiplied by a x that means the correction term that we include also must be multiplied by ax so now let's take all of this and rewrite it a little bit before we move on to the next board so this bit right here will be equal to a times x times f of x minus a zero so now look up here we've got f of x equals a naught plus a one x plus a times x times f of x minus a zero plus b bx squared f of x so that's a nice equation that we can use to solve for f of x but let's maybe bring the results of this board to the top and we'll move on to the next the last calculation we did gave us the following equation that we can solve for f of x which recall was our generating function for our recursively defined sequence now we can move some things around and easily solve for f of x so let's see we've got f of x is equal to a naught plus a one minus a a naught times x over let's see it'll be one minus ax minus b x squared so i've combined a couple steps but we get this from moving this term over this term over factoring in f of x out of the whole left hand side and then dividing and now look what we've got we've got a linear polynomial in the numerator we have a quadratic polynomial in the denominator but it's not quite the quadratic polynomial that we have over here squared in blue but it's very related to that polynomial which is over there squared in blue and it's related via the following limit which i won't prove but i'll let you guys prove it's pretty easy to check and that is if x squared minus ax minus b factors as x minus r1 times x minus r2 which we're assuming that that happens given that r1 and r2 are roots of this quadratic polynomial then the one that we have encountered factors kind of similarly so 1 minus ax minus bx squared factors like 1 minus r1x 1 minus r2x great so like i said i'll let you guys check that that's not too hard to do but now we will apply this result to our rational function version of our generating function okay so let's see we've got a naught plus a one minus a a naught x over one minus r one x times one minus r two times x but now we can do a partial fraction decomposition here and that's possible because we have factored the denominator into distinct linear terms so in fact this factorization will look like capital r over 1 minus r 1 x plus capital s over 1 minus r 2 x and here r and s those are going to depend on our first term of our recursion this capital a as well as a0 and a1 so i'll say a a0 and a1 like that okay nice but now we can re-expand this like geometric series and that will give us our sum as n goes from zero to infinity of r times r one to the n plus s times r two to the n all times x to the n so really i expanded those as two geometric series and then smash them together but now from here let's bring down our definition of f of x which was the sum as n goes from 0 up to infinity of a sub n x to the n and we see just by comparing coefficients of x to the n on both sides we have built the closed form of our recursively defined sequence and it is it is exactly what we would like it to be but all of this only works if r1 and r2 are distinct roots so what happens if they are not distinct roots well luckily we can start from this step right here with a slightly different limit and finish that off pretty quickly so we're done looking at Repeated Roots the case when we have distinct roots for the characteristic polynomial of our recursively defined sequence now we want to see what happens when we have repeated roots so in other words r1 and r2 are equal and they're equal to something which we will denote by little r or lowercase r so the following limit follows from the limit that we had on the last board just for the case when you have non-distinct roots and that is if x squared minus ax minus b factors like x minus r squared then 1 minus ax minus bx squared factors is 1 minus rx squared that means we can take the rational function version of our generating function and define it like or and write it like this so we've got a linear polynomial in the numerator and then this 1 minus r times x quantity squared in the denominator but again v is some partial fraction decomposition we can pull this apart into two pieces and now the pieces will be r over 1 minus rx plus s over 1 minus rx squared so let's notice that if we were to put these back together we would multiply this capital r by 1 minus rx it would become a linear polynomial we would add them back together and we would have a linear polynomial in the denom in the numerator so that gives you some idea of how this capital r and s are related to a not a1 and capital a okay so from here we'll expand each of these kind of similar to what we did before but we're in a little bit of trouble because this is not exactly a geometric series but luckily it's the derivative of the geometric series so in fact this is equal to well let's see if we can get it right s over r times the derivative with respect to x of 1 minus r x so that means that we can take each of these expand them keeping in mind that we have to take the derivative of the second one so this is going to give us r times the sum as n goes from zero up to infinity of r and x to the n plus s over little r and then we'll have the derivative of the sum as n goes from zero to infinity of r to the n x to the n but that derivative is pretty easy to take just term by term and that will give us the sum as n goes from zero up to infinity of n times r n times x to the n minus 1. finally we can play some tricks that are pretty similar to what we did before to push these two together and what we will end up with is the sum as n goes from zero to infinity of r plus n times s times r to the n x to the n where this capital r and s have been renamed appropriately from our original r and s to absorb any of the constants that were there but now rewriting f of x as its original definition so the sum as n goes from 0 to infinity of a sub n x to the n and then extracting the coefficient of x to the n from both sides we see that we've found a closed form in this case where we do not have distinct roots okay so let's finish this video off with a quick example okay so let's Twostep Recursion finish this off with one example so let's say we've got the following recursively defined sequence so the zeroth term a0 is four the first term a1 is one and then our two-step recursion is given as follows so a n plus 2 is equal to 3 a n plus 1 minus 2 a n so what we've got over here gives us a nice polynomial which we should look at and that will allow us to quickly write down a rough draft of the closed form that we can maybe tweak until we've got the exact closed form okay so let's jump into it so we'll look at maybe this thing that i'm calling the characteristic polynomial of the recursively defined sequence so in this case it will be x squared minus 3x plus 2 equals 0. we need to find the roots of that that's because capital a here like over here is equal to 3 and capital b is equal to negative 2. all right but now let's notice that this thing factors pretty nicely this is x minus 2 times x minus 1 equals 0. okay so that means a n is equal to capital r times 2 to the n plus capital s times 1 to the n because 2 and 1 are the roots to that polynomial well obviously 1 to the n is always n so that's not super interesting but now we can apply these initial conditions to find values of r and s so let's notice that on the one hand we are given that a 0 is 4 but on the other hand we know a0 has the form r times 2 to the 0 which is 1 plus s times 1 to the 0 which is also 1. so we've got this equation 4 is equal to r plus s and then likewise we're given that a1 is equal to 1 but on the other hand it's equal to 2r plus s just by plugging 1 into this rough draft of our closed form that gives us a system of equations for r and s let's see what we can do with it maybe we'll take this second equation and subtract the first equation so that'll have the nice simplifying effect of getting rid of the s so we'll have 2r minus r is just r and then we'll have 1 minus 4 is negative 3. so that means this number right here must be equal to negative 3. but now we can plug this value of r into either of these two equations in order to get a value for s and you'll see that s is equal to seven so that finally gives us our closed form a n is equal to minus three times 2 to the n plus 7 and that's a good place to stop
3344	https://curriculum.illustrativemathematics.org/MS/students/1/5/4/index.html	Illustrative Mathematics - Students \| IM Demo Skip to main content Professional LearningContact Us For full sampling or purchase, contact an IM Certified Partner:Imagine LearningKendall HuntKiddom Grade 6 Middle SchoolGrade 6Grade 7Grade 8 Unit 5 Grade 6Unit 1Unit 2Unit 3Unit 4Unit 5Unit 6Unit 7Unit 8Unit 9 Lesson 123456789101112131415 Lesson 4 Adding and Subtracting Decimals with Many Non-Zero Digits Let’s practice adding and subtracting decimals. LessonPractice 4.1: The Cost of a Photo Print Here are three ways to write a subtraction calculation. What do you notice? What do you wonder? Description:Three set-ups for the subtraction calculation 5 subtract 0 point 1 7. In the leftmost calculation, 5 is on top with the subtract 0 point 1 7 beneath, and the 5 and 0 line up vertically. In the center calculation, 5 is on top with the subtract 0 point 1 7 beneath, and the 5 and 1 line up vertically. In the rightmost calculation, 5 is on top with the subtract 0 point 1 7 beneath and the 5 and 7 line up vertically. Clare bought a photo for 17 cents and paid with a $5 bill. Look at the previous question. Which way of writing the numbers could Clare use to find the change she should receive? Be prepared to explain how you know. Find the amount of change that Clare should receive. Show your reasoning, and be prepared to explain how you calculate the difference of 0.17 and 5. 4.2: Decimals All Around Find the value of each expression. Show your reasoning. (11.3 - 9.5) (318.8 - 94.63) (0.02 - 0.0116) Discuss with a partner: Which method or methods did you use in the previous question? Why? In what ways were your methods effective? Was there an expression for which your methods did not work as well as expected? Lin’s grandmother ordered needles that were 0.3125 inches long to administer her medication, but the pharmacist sent her needles that were 0.6875 inches long. How much longer were these needles than the ones she ordered? Show your reasoning. There is 0.162 liter of water in a 1-liter bottle. How much more water should be put in the bottle so it contains exactly 1 liter? Show your reasoning. One micrometer is 1 millionth of a meter. A red blood cell is about 7.5 micrometers in diameter. A coarse grain of sand is about 70 micrometers in diameter. Find the difference between the two diameters in meters. Show your reasoning. 4.3: Missing Numbers Write the missing digits in each calculation so that the value of each sum or difference is correct. Be prepared to explain your reasoning. Description:2 decimal addition problems. Problem 1, 404 thousandth + blank, blank, blank, blank = 1, blank, blank, blank. Problem 2, 9 and 8765 ten thousandths +blank, blank, blank, blank, blank, blank = 1, 0,blank, blank, blank, blank. Description:Calculations with missing digits. Third calculation, 0 point 7 minus an unknown quantity with four digits is equal to 0 point 0 1 2. Fourth calculation, 7 minus an unknown quantity with five digits is equal to 3 point 4 5 6 7. Fifth calculation, 70 minus an unknown quantity with six digits is equal to 0 point 0 0 8 9. Are you ready for more? In a cryptarithmetic puzzle, the digits 0-9 are represented using the first 10 letters of the alphabet. Use your understanding of decimal addition to determine which digits go with the letters A, B, C, D, E, F, G, H, I, and J. How many possibilities can you find? Summary Base-ten diagrams work best for representing subtraction of numbers with few non-zero digits, such as (0.16 - 0.09). For numbers with many non-zero digits, such as (0.25103 - 0.04671), it would take a long time to draw the base-ten diagram. With vertical calculations, we can find this difference efficiently. Thinking about base-ten diagrams can help us make sense of this calculation. Description:A setup for the subtraction calculation 0 point 2 5 1 0 3 subtract 0 point 0 4 6 7 1 results in 0 point 1 0 4 3 2. The number 0 point 2 5 1 0 3 is on top with the subtract 0 point 0 4 6 7 1 beneath, and the 0 from the first number lines up vertically with the 0 from the second number, the 2 from the first number lines up vertically with the 0 from the second, the 5 from the first number lines up vertically with the 4 from the second, and so on. The 1 in the thousandths place of the first number is unbundled to make ten groups of ten thousandths. The five in the hundredths place has 1 unbundled to make 4 hundredths and 10 thousandths. The thousandth in 0.25103 is unbundled (or decomposed) to make 10 ten-thousandths so that we can subtract 7 ten-thousandths. Similarly, one of the hundredths in 0.25103 is unbundled (or decomposed) to make 10 thousandths. Video Summary Video VLS G6U5V1 Adding and Subtracting Decimals (Lessons 2–4) available at About IM In the News Curriculum Grades K-5 Grades 6-8 Grades 9-12 Professional Learning Standards and Tasks Jobs Privacy Policy Facebook Twitter IM Blog Contact Us 855-741-6284 What is IM Certified™? IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Spanish translation of the "B" assessments are copyright 2020 byIllustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.
3345	https://pmc.ncbi.nlm.nih.gov/articles/PMC11186013/	Mixed Metal Amide-Hydride Solid Solutions for Potential Energy Storage Applications - 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 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 Inorg Chem . 2024 May 30;63(24):11233–11241. doi: 10.1021/acs.inorgchem.4c01016 Search in PMC Search in PubMed View in NLM Catalog Add to search Mixed Metal Amide-Hydride Solid Solutions for Potential Energy Storage Applications Thi Thu Le Thi Thu Le †Institute of Hydrogen Technology, Helmholtz-Zentrum hereon GmbH, Max-Planck-Straße 1, Geesthacht D-21502, Germany Find articles by Thi Thu Le †,, Simone Bordignon Simone Bordignon ‡Department of Chemistry, University of Torino, V. P. Giuria 7, Torino I-10125, Italy Find articles by Simone Bordignon ‡, Michele R Chierotti Michele R Chierotti ‡Department of Chemistry, University of Torino, V. P. Giuria 7, Torino I-10125, Italy Find articles by Michele R Chierotti ‡, Yuanyuan Shang Yuanyuan Shang †Institute of Hydrogen Technology, Helmholtz-Zentrum hereon GmbH, Max-Planck-Straße 1, Geesthacht D-21502, Germany Find articles by Yuanyuan Shang †, Alexander Schökel Alexander Schökel §Deutsches Elektronen-Synchrotron DESY, Notkestraße 85, Hamburg D-22607, Germany Find articles by Alexander Schökel §, Thomas Klassen Thomas Klassen †Institute of Hydrogen Technology, Helmholtz-Zentrum hereon GmbH, Max-Planck-Straße 1, Geesthacht D-21502, Germany ∥Helmut Schmidt University, Holstenhofweg 85, Hamburg D-22043, Germany Find articles by Thomas Klassen †,∥, Claudio Pistidda Claudio Pistidda †Institute of Hydrogen Technology, Helmholtz-Zentrum hereon GmbH, Max-Planck-Straße 1, Geesthacht D-21502, Germany Find articles by Claudio Pistidda †, Author information Article notes Copyright and License information †Institute of Hydrogen Technology, Helmholtz-Zentrum hereon GmbH, Max-Planck-Straße 1, Geesthacht D-21502, Germany ‡Department of Chemistry, University of Torino, V. P. Giuria 7, Torino I-10125, Italy §Deutsches Elektronen-Synchrotron DESY, Notkestraße 85, Hamburg D-22607, Germany ∥Helmut Schmidt University, Holstenhofweg 85, Hamburg D-22043, Germany Email: thi.le@hereon.de. Email: claudio.pistidda@hereon.de. Received 2024 Mar 12; Accepted 2024 May 16; Revised 2024 May 13; Collection date 2024 Jun 17. © 2024 The Authors. Published by American Chemical Society Permits the broadest form of re-use including for commercial purposes, provided that author attribution and integrity are maintained ( PMC Copyright notice PMCID: PMC11186013 PMID: 38815249 Abstract Mixed solid solutions have played an important role in improving the kinetics and performance of hydrogen storage materials, as reported for the Li–Mg–N–H, K–Mg–N–H, and Rb–Mg–N–H systems. Besides, the formation of a homogeneous solid solution, mostly due to partial ionic substitution, is known to be an effective approach to improve the ionic conductivity of a material, which is an important property in electrochemical applications. We have reported a series of solid solutions based on mixed amide-hydride materials of the Group 1 elements, e.g., K(NH 2)x H 1–x, Rb(NH 2)x H 1–x, and Cs(NH 2)x H 1–x, via the exchange of NH 2–/H– anions with the change of the lattice cell of the solid solution. Extending the research in this direction, we study the M–N–H solid solution in the MNH 2–MH systems (M = K, Rb, Cs, and their combinations), i.e., KNH 2–RbH, RbNH 2–KH, RbNH 2–CsH, and CsNH 2–RbH via ex situ/in situ XRD, IR, and 1 H 2D solid-state NMR. The results obtained confirm the formation of mixed metal amide-hydride solid solutions associated with an exchange between both anionic (NH 2– and H–) and cationic species (K+, Rb+, and Cs+). With this study, we aim to create an accessible library of M–N–H solid solutions for further studies as additives for hydrogen storage materials or ionic conductors. Short abstract Structural and chemical similarities between alkali metal (K, Rb, and Cs) amide and hydride compounds lead to the formation of mixed metal solid solutions that can be further investigated for hydrogen storage and ionic conduction. 1. Introduction Metal amide-hydride materials have been extensively investigated in recent years for use in energy storage applications and specifically for storage applications in hydrogen technology (e.g., LiNH 2-LiH,1−6 Mg(NH 2)2-2LiH,7−14 Mg(NH 2)2-KH,15 and Mg(NH 2)2-RbH)16) and as solid electrolytes17−19 for all solid-state batteries. In hydrogen storage applications, amide-hydride systems prove promising candidates especially due to their high hydrogen storage capacity and tunable thermodynamics, which allows hydrogen desorption/absorption to occur at temperatures below 150 °C.20 However, the sluggish desorption/absorption kinetics of these amide-hydride systems limit their potential employment in commercial applications. In such cases, metal hydrides-based additives such as KH, RbH, and CsH can effectively improve the hydrogen sorption kinetics and alter the thermodynamics of amide-hydride systems, in particular of Mg(NH 2)2-2LiH.21−25 These metal hydrides promote the formation of intermediate phases/solid solutions, which reduce the kinetic barrier of the amide/hydride interfacial reaction and lower the operating temperature of the Li–Mg–N–H system. Other solid solutions, such as the Li–N–H ones (e.g., Li 2+x(NH)1–x N x H x and Li 2+x NH), were synthesized to modify the thermal stability and ammonia reactivity of the LiNH-LiH system,6 while Ca 2 NH–Ca 2 N 2 solid solutions were studied as catalysts to promote low-temperature ammonia synthesis.26 In general, solid solutions are not only important for hydrogen storage but also for the ionic conduction of complex metal hydrides. The formation of a homogeneous solid solution by ion mixing is one of the most common ways to improve the ionic conductivity of a material,27 most likely due to the structural perturbations or modifications that facilitate the migration of ions, as observed in the LiBH 4–LiX (X = Cl–, Br–, and I–)28,29 and NaBH 4–NaI.30 Our previous studies reported the formation of mixed amide-hydride solid solutions, such as K(NH 2)x H 1–x, Rb(NH 2)x H 1–x, and Cs(NH 2)x H 1–x, via the exchange of anionic species (NH 2– in amides and H– in hydrides) for the KNH 2–KH,31 RbNH 2–RbH,16 and CsNH 2–CsH32 systems using both experimental determinations and theoretical calculations. The relative amides (i.e., KNH 2, RbNH 2, and CsNH 2) undergo a structural transition from monoclinic space group (s.g.) P 2 1/m (for both KNH 2 and RbNH 2) and tetragonal s.g. P 4/nmm (for CsNH 2) to cubic crystal structures with the same s.g. Fm 3̅m within the temperature range of 50–80 °C. On the contrary, the corresponding hydrides, namely, KH, RbH, and CsH, have similar cubic crystal structures (s.g. Fm 3̅m), which remain unchanged upon temperature variation. Furthermore, the dissolution of NH 2– into the KH (or RbH and CsH) hydride structure leads to the structural expansion of the M(NH 2)x H 1–x solid solutions (M = K, Rb, and Cs). This change in the lattice structure could lead to significant changes in the material’s functional properties, such as ionic conductivity, similarly to what observed in ionic conducting systems like LiBH 4–LiX (X = Cl–, Br–, and I–)28,29 and NaBH 4–NaI.30 In particular, KNH 2 exhibits an ionic conductivity of 3.56 × 10–4 S cm–1 at 150 °C, which can be further enhanced by introducing structural disorder.18 These findings suggest a high potential for the application of these amide-hydride solid solutions (with modified structures) in the energy storage field. In this work, we report the formation of solid solutions in the MNH 2–MH system, namely, KNH 2–RbH, RbNH 2–KH, CsNH 2–RbH, and RbNH 2–CsH, via the exchange of anion and cation species. It is well-known that several factors determine the limits of solubility. These are expressed as a series of rules, the so-called William Hume–Rothery Rules: (i) atomic size factor (extensive substitutional solid solution occurs only if the relative difference between the atomic radii of the two species is less than 15%); (ii) crystal structures of the two elements must be identical; (iii) two species (the solute and the solvent atoms) should typically have the same valence; and (iv) electronegativity difference was close to 0. In the present study, there are clear structural analogies among these MNH 2 amides and between them and the MH hydrides. Moreover, comparing the relative difference between the size of K+ (r ion = 1.52 Å), Rb+ (r ion = 1.66 Å), and Cs+ (r ion= 1.81 Å) cations and between the size of NH 2– (r ion = 1.73 Å) and H– (r ion = 1.53 Å) anions, the radii difference for any of their combinations is less than 15%, which, according to the Hume–Rothery rules, allows for the formation of M(NH 2)x H 1–x solid solutions, based on the substitution of both anion and cation species. The proposed diagram for ion substitutions in these systems is illustrated in Figure1. As mentioned above, ionic hydrides (KH, RbH, and CsH) have been shown to be effective additives to improve the kinetic and thermodynamic properties of the most promising amide-hydride hydrogen storage system, Mg(NH 2)2-2LiH. Therefore, the use of such mixed metal amide-hydride as additives in the hydrogen storage systems might be possible. Moreover, the formation of solid solutions displaying crystal lattice disorder (due to the different ionic radius of exchanged cations and anions) could create channels for ion diffusion and therefore increase the ionic conduction within the solid lattice, highlighting the potential use of these alkali metal mixed amide-hydride solid solutions for further studies on the ionic conductivity or as dopants/catalysts for hydrogen storage materials. Figure 1. Open in a new tab Schematic diagram of the amide-hydride reaction mechanism. 2. Experimental Method 2.1. Materials Preparation Potassium hydride (KH) in paraffin was purchased commercially (35.8 wt % KH purity, Sigma-Aldrich). The synthesis of potassium amide (KNH 2), rubidium amide (RbNH 2), rubidium hydride (RbH), cesium hydride (CsH), and cesium amide (CsNH 2) followed the same procedures, as reported in refs (16 and 31). KNH 2 was synthesized by repeated ball-milling of the potassium hydride in a Pulverisette planetary mill at 500 rpm with a ball-to-powder ratio (BPR) of 20:1 under 7 bar of NH 3 for 5 h. This process was repeated four times for a total milling time of 20 h. Between each repetition, the high-pressure vessel was evacuated and refilled with fresh NH 3. RbH was synthesized by ball-milling metallic rubidium (Rb 99.8%, Alfa Aesar) in 50 bar of H 2 at 500 rpm for 13 h with a BPR of 60:1, followed by further annealing at 180 °C for 5 h under 70 bar of H 2. RbNH 2 was synthesized by heat treatment of metallic Rb in 7 bar of NH 3 at 250 °C for 13 h. The mixtures x KNH 2 + (1 – x)RbH and x RbNH 2 + (1 – x)KH with x = {0, 0.5, 0.7, 1} were prepared by hand-grinding using an agate mortar followed by heat treatment at 270 °C for 3 h. To avoid oxidation of the materials, all samples were prepared under a protective atmosphere in a continuous Ar-filled glovebox (MBraun, Germany) with an oxygen and humidity concentration of less than 1 ppm. 2.2. Materials Characterization Ex situ powder X-ray diffraction (XRD) experiments were performed using a D8 Discover diffractometer (Bruker AXS GmbH, Karlsruhe, Germany) equipped with a Cu Kα beam (λ = 1.54184 Å) and 2D VANTEC detector. The diffractograms were acquired in the 2θ range from 10° to 90°, in nine steps with an exposure time of 400 s per step. A small amount of the sample was placed on a flat commercial sample holder and sealed with an airtight poly(methyl methacrylate) lid to prevent oxidation. In situ synchrotron powder X-ray diffraction (in situ SR-PXD) measurements were performed at the Powder Diffraction and Total Scattering Beamline (P02.1) of Petra III (Desy Hamburg, Germany)33 using a monochromatic X-ray beam (λ = 0.20734 Å). The diffraction patterns were collected by a Varex4343CT detector with an array of 2880 × 2880 pixels and a pixel size of 150 μm × 150 μm with an exposure time of 10 s per pattern. Samples were loaded into sapphire capillaries under a purified Ar atmosphere and then mounted on an in-house developed in situ cell, in which operating temperatures and pressures can be controlled.34,35 The measurements for all mixtures were carried out under 1 bar of Ar, with the sample heated up from room temperature (RT) to 270 °C at a heating rate of 5–10 °C/min, held isothermally at 270 °C for 30 min, and then cooled down to RT. A small difference in the measurement conditions of the starting materials (KNH 2, RbH, RbNH 2, and KH) is that they were kept isothermally at 270 °C for 10 min instead of 30 min. The 2D diffraction images were integrated into 1D diffractograms using the Fit2d software, and the quantitative analyses were performed using the Rietveld refinement method with the Material Analysis Using Diffraction software (MAUD).36 Structural information on known phases was obtained from the International Crystal Structure Database (ICSD) using the ICSD-Desktop software. The pure and mixed samples were characterized using the Fourier transform infrared spectroscopy (Cary 630 FT-IR spectrometer, Agilent Technologies Deutschland GmbH, Waldbronn, Germany). The FT-IR spectrometer was placed in an Ar-circulated glovebox with oxygen and moisture concentrations below 5 ppm. The background was calibrated for each measurement; a small amount of material was placed on the diamond ATR top plate, and the FT-IR spectrum was acquired at RT in a full frequency range of 4000–650 cm–1 with a spectral resolution of 4 cm–1 and a number of scans of 300. Solid-state nuclear magnetic resonance (SSNMR) experiments were run on a Jeol ECZR 600 instrument, operating at a frequency of 600.13 MHz for 1 H and equipped with a 3.2 mm probe. Rotors were packed inside a glovebox to prevent sample decomposition. The 1 H MAS spectra were acquired at probe temperature at a spinning speed of 20 kHz (4 scans; optimized relaxation delays equal to 200 or 280 s, corresponding to 5·T 1 for quantitative measurements). The 2D 1 H double-quantum (DQ) MAS experiments were performed at probe temperature at a spinning speed of 20 kHz with the back-to-back (BABA)-xy16 recoupling pulse sequence with excitation time durations of eight rotor periods (1 H 90° = 2.2 μs; 4 scans; t 1 increments = 64; relaxation delay = 50 or 72 s, corresponding to 1.27·T 1). The 1 H chemical shift scale was calibrated with adamantane (1 H signal at 1.87 ppm with respect to primary standard tetramethylsilane) as an external standard. 3. Results and Discussion 3.1. KNH 2–RbH System Figure2 shows the room temperature XRD patterns of the annealed x KNH 2 + (1 – x)RbH samples, where x = {0, 0.5, 0.7, 1}. Note that in the XRD pattern of the compositions featuring x = {0.5, 0.7}, the reflections of KNH 2 (monoclinic, s.g. P 2 1/m, labeled ) are not visible. This indicates that KNH 2 reacted with RbH, accompanied by the formation of a single cubic phase (, s.g. Fm 3̅m). In addition, the metallic K (labeled ϕ) and an unknown phase (labeled ?) are also present. This may indicate a partial decomposition of reactants under given conditions. In addition, for the x = 0.5 composition, the peaks appear broadened, particularly at the high Q-values, suggesting an overlap of multiple phases due to incomplete reactions between the amide and hydride. To clarify these, in situ SR-PXD measurements were carried out. Figure 2. Open in a new tab RT XRD of the annealed x KNH 2 + (1 – x)RbH samples, with x = {0, 0.5, 0.7, 1}. = Rb–K–N–H solid solution (Fm 3̅m), = KNH 2 (P 2 1/m), = RbH (Fm 3̅m), ϕ = K metallic (Im 3̅m), and ? = unknown phase. In order to assess the formation of the single cubic phase seen in Figure2, as well as the microstructural evolution of the investigated material under temperature and pressure variations, in situ SR-PXD experiments were carried out on the initial and mixed samples. Figure3 shows the in situ SR-PXD data of the x KNH 2 + (1 – x)RbH samples, where x = {0, 0.5, 0.7, 1}. As can be seen in Figure3, the in situ SR-PXD data of the KNH 2 starting material show two phase transitions: the first one corresponds to the phase transformation of KNH 2 from a monoclinic (s.g. P 2 1/m) to a tetragonal structure (s.g. P 4/nmm, ⧫) at about 55 °C and the second one is the phase conversion of KNH 2 from the tetragonal (s.g. P 4/nmm) to the cubic (s.g. Fm 3̅m, ) structure at about 75 °C, while RbH (s.g. Fm 3̅m, ) does not undergo structural changes during the heating and cooling periods. These observations agree with those reported in references and literatures.16,31 For mixed samples with x = {0.5, 0.7}, their in situ SR-PXD data appear similar, and both differ from the in situ SR-PXD data of the starting samples (KNH 2 and RbH). At the beginning of the heating period, both x = 0.5 and x = 0.7 mixtures show the existence of monoclinic KNH 2 (), cubic RbH (). Besides, tiny peaks of metallic K (ϕ) and unknown phase (?) are also detected. Trying to understand the reason for the presence of metallic K and the unknown phase found for the mixed samples, the gas evolution for the x = 0.7 composition during the mixing under inert atmosphere of KNH 2 and RbH was monitored by mass spectrometer (MS). As demonstrated in ESI-Figure S1, the investigation’s findings indicated signs of hydrogen and ammonia release during mixing. The presence of metallic K and an unknown phase, along with the gas evolution found by MS, indicates a partial decomposition of reactants. However, according to the Rietveld refinement, the cell parameter of the RbH-like structure for the x = 0.7 composition (ESI-Figure S2), a, is 6.0405617 Å, thus larger than the RbH cell parameter (a = 6.0363336 Å) of the pure sample (ESI-Figure S3). This increase corresponds to a volume expansion of RbH of 0.21%. This supports the assumption that the solid solution formation by the cation and anion substitution, which underlies the interaction between KNH 2 and RbH, already occurred during grinding. As the temperature increases, phase transitions are found for KNH 2 from the monoclinic () to the tetragonal (⧫) and then to the cubic () structure, as predicted, while the RbH peaks () remain unchanged, like what was observed for the pure KNH 2 and RbH samples. At T ∼ 160 °C, solid solution formation begins, corresponding to the disappearance of the cubic KNH 2 diffraction peaks and the shift of the cubic RbH peaks toward the lower Q values, indicating a gradual transformation from the RbH to the solid solution structure. This process continues during the heating period and is completed at 270 °C, with the observation of a complete single cubic solid solution phase (s.g. Fm 3̅m, ). This phase () is stable on cooling and at RT after the measurement. No Bragg peaks of initial materials (i.e., monoclinic KNH 2 and cubic RbH) are detected, suggesting the stability of the solid solution. In addition, small peaks of elemental K (ϕ) and unknown phase (?) are back at RT after the measurement for both x = 0.5 and x = 0.7 compositions, like what was observed in the ex situ XRD (see Figure2). Figure 3. Open in a new tab In situ SR-PXD data of the (a) KNH 2, (b) 0.5KNH 2 + 0.5RbH, (c) 0.7KNH 2 + 0.3RbH, and (d) RbH. = Rb–K–N–H solid solution (Fm 3̅m), = KNH 2 (P 2 1/m), ⧫ = KNH 2 (P 4/nmm), = KNH 2 (Fm 3̅m), = RbH (Fm 3̅m), ϕ = K (Im 3̅m), and ? = unknown phase. 3.2. RbNH 2–KH System Similarly, the room temperature XRD diffractograms of the x RbNH 2 + (1 – x)KH samples after being annealing at 270 °C are shown in Figure4. The data show that for the compositions x = {0.5, 0.7}, the Bragg peaks of the RbNH 2 (ψ) and KH (δ) phases are not detected, but instead, a single cubic phase (, s.g. Fm 3̅m) is observed, indicating the dissolution of RbNH 2 and KH within the structure. Small peaks related to elemental K (ϕ) and an unknown phase (?) are also observed, similarly to the KNH 2–RbH system (Section 3.1). It should be noted that peak broadening is also observed for the x = 0.5 composition. This will be discussed in detail in the Rietveld refinements of XRD data, which are displayed in Figures6 and 7. Figure 4. Open in a new tab RT XRD data of the x RbNH 2 + (1 – x)KH samples after annealing at 270 °C, with x = {0, 0.5, 0.7, 1}. ψ = RbNH 2 (P 2 1/m), δ = KH (Fm 3̅m), = Rb–K–N–H solid solution (Fm 3̅m), ϕ = K (Im 3̅m), and ? = unknown phase. Figure 6. Open in a new tab Rietveld refinements of the in situ SR-PXD data collected at 270 °C for x KNH 2 + (1 – x)RbH samples, where x = {0, 0.5, 0.7, 1}. Figure 7. Open in a new tab Rietveld refinements of the in situ SR-PXD data collected at 270 °C for x RbNH 2 + (1 – x)KH samples, where x = {0, 0.5, 0.7, 1}. As before, the phase evolution under real conditions was monitored with the conduction of in situ SR-PXD measurements. As shown in Figure5, the in situ SR-PXD data of the x RbNH 2 + (1 – x)KH samples, where x = {0, 0.5, 0.7, 1}, were also collected. The in situ SR-PXD data of starting material RbNH 2 show a phase transition of RbNH 2 from a monoclinic (s.g. P 2 1/m, labeled ψ) to a cubic structure (s.g. Fm 3̅m, labeled φ) at around 68 °C, while the RT-cubic structure of KH (s.g. Fm 3̅m, labeled δ) does not undergo a phase transition under the given experimental conditions, similar to the RbH. These results agree with those reported in refs (16 and 31). For compositions with x = {0.5, 0.7}, their in situ SR-PXD data show similarities, i.e., both show the transformation of RbNH 2 from the monoclinic (ψ) to the cubic (φ) structure at about 65 °C during heating, as expected. Subsequently, the mutual dissolution of RbNH 2 and KH takes place, accompanied by the substitution process of cubic RbNH 2 (φ) and cubic KH (δ) peaks observed starting from a temperature of about 150 °C and further increased during the isothermal period. As a result, the completion of solubility is achieved and the formation of a single cubic solid solution is observed for both compositions with x = {0.5, 0.7}, similar to the KNH 2–RbH system. In addition, metallic K (ϕ) and an unknown phase (?) are observed at room temperature, similarly to the KNH 2–RbH system. Figure 5. Open in a new tab In situ SR-PXD data of the (a) RbNH 2, (b) 0.5RbNH 2 + 0.5KH, (c) 0.7RbNH 2 + 0.3KH, and (d) KH. ψ = RbNH 2 (P 2 1/m), φ = RbNH 2 (Fm 3̅m), δ = KH (Fm 3̅m), = Rb–K–N–H solid solution (Fm 3̅m), ϕ = K (Im 3̅m), and ? = unknown phase. To elucidate phase compositions in the mixed samples, Rietveld refinement of in situ SR-PXD data was performed (Rietveld refinement details can be found in ESI-Figure S4). Figure6 and Table1 present the Rietveld refinements of diffraction data collected at 270 °C and cell parameters of phases within the KNH 2–RbH system. For the x = {0.5, 0.7} compositions, the results confirmed the formation of a single cubic solid solution this these mixtures, also considering the partial desorption of ammonia and hydrogen during heating. Additionally, unreacted RbH is also observed for the x = 0.5 sample, indicating a not complete solubility of amide and hydride at this temperature. Similarly, for the RbNH 2–KH system, Rietveld refinement of XRD data is performed, as depicted in Figure7 and Table2. For the x = 0.7 sample, a single cubic solid solution of amide and hydride is formed, while for the x = 0.5 sample, an overlap of two solid solution phases corresponding to broader peaks observed in the XRD diffraction data is detected. Additionally, unreacted KH is also observed. Tables1 and 2 present the phase compositions and lattice parameters of phases in both systems. The existence of cubic polymorphs of KNH 2 and RbNH 2 at temperature higher than 75 °C and their structural similarities with RbH and KH (e.g., cation and anion radii: r K+ = 1.52 Å, r Rb+ = 1.66 Å, r NH2-= 1.73 Å, r H– = 1.53 Å and cell parameters at 270 °C: KNH 2 (a = 6.19766 Å), RbNH 2 (a = 6.10687 Å), KH (a = 5.74913 Å), and RbH (a = 6.10687 Å)) pave the path to the formation of mixed solid solutions. It is observed that for the KNH 2–RbH system, the cubic KNH 2 likely dissolves into the cubic RbH structure, as seen by the disappearance of the KNH 2 diffraction peaks and the change in the RbH peaks (in situ XRD data in Figure3). In contrast, the cubic KH dissolves into the cubic RbNH 2, detected by the shift of RbNH 2 peaks at T ∼ 270 °C, when the solubility takes place (in situ XRD data in Figure5). Besides, the diffraction data of the two systems (i.e., KNH 2–RbH (Figure6) and RbNH 2–KH (Figure7)) also show that the diffraction peaks are shifted, and their relative intensity varies gradually with increasing the amount of amide (i.e., x). This is indicative of the expected ionic substitution and change of the unit cell (see Tables1 and 2). Table 1. Lattice Parameters of the Solid Solution in the KNH 2–RbH System Obtained from the Rietveld Refinement of the SR-PXD Patterns Acquired at 270 °C. \| system 1 \| phase compositions \| unit cell (a/Å) \| :---: \| KNH 2 \| KNH 2 \| 6.19766 \| \| 0.7KNH 2 + 0.3RbH \| Rb 0.8 K 0.2(NH 2)0.4 H 0.6 (100 wt %) \| 6.17582 \| \| 0.5KNH 2 + 0.5RbH \| Rb 0.9 K 0.1(NH 2)0.44 H 0.56(78.9 wt%) \| 6.17899 \| \| \| RbH-like cubic (21.1 wt %) \| 6.10120 \| \| RbH \| RbH \| 6.10687 \| Open in a new tab Table 2. Lattice Parameters of the Solid Solution in the RbNH 2–KH System Obtained from the Rietveld Refinement of the SR-PXD Patterns Acquired at 270 °C. \| system 2 \| phase compositions \| unit cell (a/Å) \| :---: \| RbNH 2 \| RbNH 2 \| 6.48491 \| \| 0.7RbNH 2 + 0.3KH \| Rb 0.77 K 0.23(NH 2)0.68 H 0.32(97.4 wt%) \| 6.38822 \| \| \| KH-like cubic (2.6 wt %) \| 5.78233 \| \| 0.5RbNH 2 + 0.5KH \| Rb 0.94 K 0.06(NH 2)0.91 H 0.09 (48.2 wt %) \| 6.32481 \| \| \| K 0.99(Rb 0.01)NH 2 (48 wt %) \| 6.23548 \| \| \| KH-like cubic (3.8 wt %) \| 5.77403 \| \| KH \| KH \| 5.74913 \| Open in a new tab The vibrational frequencies of the NH 2 group in the pure amides and mixed amide-hydride samples were characterized by FT-IR. As shown in ESI-Figure S5a for the KNH 2–RbH system, the stretching vibrational modes of the N–H bonds in KNH 2 are observed at 3253 and 3207 cm–1. These signals slightly shift to lower frequencies in the mixed amide-hydride samples, probably due to volume expansion, which increases the N–H bond length, and the ionic substitution (K with Rb). In addition, the magnitude of these bands is weaker in the mixed samples with a reduced amount of amide compared to the original KNH 2, indicating the decrease of the dipole moment caused by the reduction of the negative charge on the nitrogen atom induced by a substituent such as hydrogen. This observation is similar to what has been observed in the Cs–N–H system32 and for C–H bonds.37,38 Likewise, a similar observation on the vibrational modes is also found for the RbNH 2–KH system, where the intensity of the signals of the stretching modes decreases because of the low amount of amide in the samples (ESI-Figure S5b). It is noteworthy that although partial desorption of ammonia and hydrogen is observed, the formation of mixed metal amide-hydride solid solution is evident. Further confirmation of the solid solution formation for the KNH 2–RbH and RbNH 2–KH systems was achieved by solid-state NMR measurements. Figure8a shows the 1 H MAS spectra of the 0.5KNH 2 + 0.5RbH and 0.5RbNH 2 + 0.5KH systems. In both spectra, similar chemical shifts for the hydride (6.2–6.3 ppm) and amide (−2.9 ppm) anions are observed, indicating the similarity of the chemical environment between these two mixed samples. Important note that, the presence of imide is not notied, thi Figure 8. Open in a new tab (a) 1 H (600.13 MHz) MAS SSNMR spectra of the 0.5KNH 2 + 0.5RbH and 0.5RbNH 2 + 0.5KH compositions, acquired at probe temperature at a spinning speed of 20 kHz; 2D 1 H (600.13 MHz) DQ MAS SSNMR spectra of the (b) 0.5KNH 2 + 0.5RbH and (c) 0.5RbNH 2 + 0.5KH systems, recorded at a spinning speed of 20 kHz at probe temperature. Red lines highlight the DQ correlation between the amide and hydride signals in both systems. A direct assessment of the solid solution formation was provided by the 2D 1 H DQ MAS spectra, as shown in Figure8b,c. For the 0.5KNH 2 + 0.5RbH sample (Figure8b), the DQ correlation between the amide (δ SQ = −2.9 ppm) and hydride (δ SQ = 6.2 ppm) signals is observed at δ DQ = 3.3 ppm. This implies that the protons of KNH 2 and RbH are in close spatial proximity to each other, i.e., within 3.5 Å, which implies that they belong to a homogeneous phase.39,40 A similar correlation is also observed for the 0.5KNH 2 + 0.5RbH sample, where a DQ correlation between the amide (δ SQ = −2.9 ppm) and the hydride (δ SQ = 6.3 ppm) signals is found at δ DQ = 3.4 ppm (Figure8c), also indicating the intimate spatial proximity of the RbNH 2 and KH protons. The results of the in situ/ex situ XRD and SSNMR measurements provide adequate evidence to confirm the formation of solid solutions for both the KNH 2–RbH and RbNH 2–KH systems. 3.3. RbNH 2–CsH and CsNH 2–KH Systems Extending the investigation in this line of research, we observed that solid solutions were also formed in the RbNH 2–CsH and CsNH 2–RbH systems. Figure9 shows the phase evolution in these mixed amide-hydride composites in the temperature range from RT to 240 °C. As shown in Figure9a for the in situ SR-PXD data of the RbNH 2–CsH composite, a phase transformation of RbNH 2 occurs, from a monoclinic (s.g. P 2 1/m) to a cubic structure (s.g. Fm 3̅m), while the cubic CsH structure (s.g. Fm 3̅m) does not undergo any phase transition. These observations are similar to those reported earlier in this work and in ref (32). It is noted that at T> 65 °C, both RbNH 2 and CsH have similar crystal structures (s.g. Fm 3̅m), which might facilitate the ion mobility and consequently the formation of the solid solution. At the temperature of 240 °C, the disappearance of the cubic RbNH 2 phase and the change of the cubic CsH peaks are observed, indicating the mutual solubility of the RbNH 2 and CsH to form a solid solution with the same space group. This single cubic solid solution is stable and remains unchanged during cooling. Similarly, for the CsNH 2–RbH composite, the in situ SR-PXD data seen in Figure9b show the phase transition of CsNH 2 from a tetragonal structure (s.g. P 4/nmm) to cubic ones (s.g. Pm 3̅m and Fm 3̅m), comparable to what was reported in ref (32), while the cubic RbH structure (s.g. Fm 3̅m) remains stable. A single cubic phase (s.g. Fm 3̅m) is observed at the isothermal temperature of 240 °C, indicating the mutual solubility of RbH and CsNH 2. This CsNH 2–RbH cubic solid solution is stable during cooling. The results derived from the in situ SR-PXD data of RbNH 2–CsH and CsNH 2–RbH are comparable to those obtained for the KNH 2–RbH and RbNH 2–KH systems. Therefore, it is reasonable to assume that the formation of a solid solution was similarly achieved for these systems. Figure 9. Open in a new tab In situ SR-PXD data of the RbNH 2–CsH (a) and CsNH 2–RbH (b) composites. Ψ = RbNH 2 (P 2 1/m), φ = RbNH 2 (Fm 3̅m), σ = CsH (Fm 3̅m), = RbNH 2–CsH solid solution (Fm 3̅m). ο = CsNH 2 (P 4/nmm), ⧫ = CsNH 2 (Pm 3̅m), • = CsNH 2 (Fm 3̅m), = RbH (Fm 3̅m), and = CsNH 2–RbH solid solution (Fm 3̅m). 4. Conclusions In this work, the formation of solid solutions in the amide-hydride system of alkali metals (K, Rb, and Cs), namely, KNH 2–RbH, RbNH 2–KH, RbNH 2–CsH, and CsNH 2–RbH, has been elucidated by various techniques (i.e., 1 H 1D and 2D SSNMR, FT-IR, and ex situ/in situ XRD), leading to the following conclusive points: Due to the structural and physicochemical similarities between MNH 2 amide and MH hydride, the mixed metal solid solutions are formed by both cation and anion exchange. The formed MNH 2-MH solid solutions such as RbNH 2–KH, KNH 2–RbH, RbNH 2–CsH, etc., could replace ionic hydrides (KH, RbH, and CsH) as additives for hydrogen storage systems. Follow-up research will be conducted to investigate the hydrogen storage properties of the Mg(NH 2)2-2LiH system in the presence of mixed alkali amide-hydride solid solutions. It is worth noting that the formation of mixed metal amide-hydride solid solutions with an expanded lattice could lead to disorder of the crystal structure (due to the size difference between cations and anions substituted) and possibly facilitate the ion transport. This aspect will be further investigated in a future work. The findings of this work, in conjunction with our earlier studies on the amide-hydride solid solutions, serve as the foundation for additional research aimed at expanding this area of study and identifying analogies with other systems. Acknowledgments This research is funded by dtec. bw—Digitalization and Technology Research Center of Bundeswehr which we gratefully acknowledge. Supporting Information Available The Supporting Information is available free of charge at MS data, Rietveld refinement data, and FT-IR spectra (PDF) Author Contributions T.-T.L.: conceptualization, methodology, formal analysis, investigation, validation, and writing-original draft preparation; S.B. and M.R.C.: resources and investigation; Y.S.: investigation; A.S.: resources; T.K.: supervision and funding acquisition; C.P.: supervision; and all authors contributed to revising the manuscript. The authors declare no competing financial interest. Supplementary Material ic4c01016_si_001.pdf (667.2KB, pdf) References Chen P.; Xiong Z.; Luo J.; Lin J.; Tan K. L. Interaction of hydrogen with metal nitrides and imides. Nature 2002, 420 (6913), 302–304. 10.1038/nature01210. [DOI] [PubMed] [Google Scholar] Ichikawa T.; Isobe S.; Hanada N.; Fujii H. 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3346	https://en.wikipedia.org/wiki/Centered_square_number	Jump to content Centered square number العربية Deutsch Emiliàn e rumagnòl Esperanto Français 한국어 Italiano Magyar 日本語 Romnă Русский Suomi Svenska தமிழ் Українська 中文 Edit links From Wikipedia, the free encyclopedia Centered figurate number that gives the number of dots in a square with a dot in the center \| \| \| --- \| \| \| This article includes a list of references, related reading, or external links, but its sources remain unclear because it lacks inline citations. Please help improve this article by introducing more precise citations. (January 2016) (Learn how and when to remove this message) \| In elementary number theory, a centered square number is a centered figurate number that gives the number of dots in a square with a dot in the center and all other dots surrounding the center dot in successive square layers. That is, each centered square number equals the number of dots within a given city block distance of the center dot on a regular square lattice. While centered square numbers, like figurate numbers in general, have few if any direct practical applications, they are sometimes studied in recreational mathematics for their elegant geometric and arithmetic properties. The figures for the first four centered square numbers are shown below: : \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Each centered square number is the sum of successive squares. Example: as shown in the following figure of Floyd's triangle, 25 is a centered square number, and is the sum of the square 16 (yellow rhombus formed by shearing a square) and of the next smaller square, 9 (sum of two blue triangles): Relationships with other figurate numbers [edit] Let Ck,n generally represent the nth centered k-gonal number. The nth centered square number is given by the formula: That is, the nth centered square number is the sum of the nth and the (n − 1)th square numbers. The following pattern demonstrates this formula: : \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| The formula can also be expressed as: That is, the nth centered square number is half of the nth odd square number plus 1, as illustrated below: : \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Like all centered polygonal numbers, centered square numbers can also be expressed in terms of triangular numbers: where is the nth triangular number. This can be easily seen by removing the center dot and dividing the rest of the figure into four triangles, as below: : \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| The difference between two consecutive octahedral numbers is a centered square number (Conway and Guy, p.50). Another way the centered square numbers can be expressed is: where Yet another way the centered square numbers can be expressed is in terms of the centered triangular numbers: where List of centered square numbers [edit] The first centered square numbers (C4,n < 4500) are: : 1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, … (sequence A001844 in the OEIS). Properties [edit] All centered square numbers are odd, and in base 10 one can notice the one's digit follows the pattern 1-5-3-5-1. All centered square numbers and their divisors have a remainder of 1 when divided by 4. Hence all centered square numbers and their divisors end with digit 1 or 5 in base 6, 8, and 12. Every centered square number except 1 is the hypotenuse of a Pythagorean triple (3-4-5, 5-12-13, 7-24-25, ...). This is exactly the sequence of Pythagorean triples where the two longest sides differ by 1. (Example: 52 + 122 = 132.) This is a consequence of (2n − 1)2 + (2n2 − 2n)2 = (2n2 − 2n + 1)2. Generating function [edit] The generating function that gives the centered square numbers is: References [edit] Alfred, U. (1962), "n and n + 1 consecutive integers with equal sums of squares", Mathematics Magazine, 35 (3): 155–164, doi:10.1080/0025570X.1962.11975326, JSTOR 2688938, MR 1571197. Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001. Beiler, A. H. (1964), Recreations in the Theory of Numbers, New York: Dover, p. 125. Conway, John H.; Guy, Richard K. (1996), The Book of Numbers, New York: Copernicus, pp. 41–42, ISBN 0-387-97993-X, MR 1411676. \| v t e Figurate numbers \| \| --- \| \| 2-dimensional \| \| \| \| --- \| \| centered \| Centered triangular numbers Centered square numbers Centered pentagonal numbers Centered hexagonal numbers Centered heptagonal numbers Centered octagonal numbers Centered nonagonal numbers Centered decagonal numbers Star numbers \| \| non-centered \| Triangular numbers Square numbers Pentagonal numbers Hexagonal numbers Heptagonal numbers Octagonal numbers Nonagonal numbers Decagonal numbers Dodecagonal numbers \| \| \| 3-dimensional \| \| \| \| --- \| \| centered \| Centered tetrahedral numbers Centered cube numbers Centered octahedral numbers Centered dodecahedral numbers Centered icosahedral numbers \| \| non-centered \| Cube numbers Octahedral numbers Dodecahedral numbers Icosahedral numbers Stella octangula numbers \| \| pyramidal \| Tetrahedral numbers Square pyramidal numbers \| \| \| 4-dimensional \| \| \| \| --- \| \| non-centered \| Pentatope numbers Squared triangular numbers Tesseractic numbers \| \| \| Higher dimensional \| \| \| \| --- \| \| non-centered \| 5-hypercube numbers 6-hypercube numbers 7-hypercube numbers 8-hypercube numbers \| \| \| v t e Classes of natural numbers \| \| --- \| \| \| Powers and related numbers \| \| --- \| \| Achilles Power of 2 Power of 3 Power of 10 Square Cube Fourth power Fifth power Sixth power Seventh power Eighth power Perfect power Powerful Prime power \| \| \| \| \| \| Of the form a × 2b ± 1 \| \| --- \| \| Cullen Double Mersenne Fermat Mersenne Proth Thabit Woodall \| \| \| \| \| \| Other polynomial numbers \| \| --- \| \| Hilbert Idoneal Leyland Loeschian Lucky numbers of Euler \| \| \| \| \| \| Recursively defined numbers \| \| --- \| \| Fibonacci Jacobsthal Leonardo Lucas Narayana Padovan Pell Perrin \| \| \| \| \| \| Possessing a specific set of other numbers \| \| --- \| \| Amenable Congruent Knödel Riesel Sierpiński \| \| \| \| \| \| Expressible via specific sums \| \| --- \| \| Nonhypotenuse Polite Practical Primary pseudoperfect Ulam Wolstenholme \| \| \| \| \| \| Figurate numbers \| \| --- \| \| \| \| \| \| \| \| \| --- --- --- \| \| 2-dimensional \| \| \| \| --- \| \| centered \| Centered triangular Centered square Centered pentagonal Centered hexagonal Centered heptagonal Centered octagonal Centered nonagonal Centered decagonal Star \| \| non-centered \| Triangular Square Square triangular Pentagonal Hexagonal Heptagonal Octagonal Nonagonal Decagonal Dodecagonal \| \| \| 3-dimensional \| \| \| \| --- \| \| centered \| Centered tetrahedral Centered cube Centered octahedral Centered dodecahedral Centered icosahedral \| \| non-centered \| Tetrahedral Cubic Octahedral Dodecahedral Icosahedral Stella octangula \| \| pyramidal \| Square pyramidal \| \| \| 4-dimensional \| \| \| \| --- \| \| non-centered \| Pentatope Squared triangular Tesseractic \| \| \| \| \| \| \| \| Combinatorial numbers \| \| --- \| \| Bell Cake Catalan Dedekind Delannoy Euler Eulerian Fuss–Catalan Lah Lazy caterer's sequence Lobb Motzkin Narayana Ordered Bell Schröder Schröder–Hipparchus Stirling first Stirling second Telephone number Wedderburn–Etherington \| \| \| \| \| \| Primes \| \| --- \| \| Wieferich Wall–Sun–Sun Wolstenholme prime Wilson \| \| \| \| \| \| Pseudoprimes \| \| --- \| \| Carmichael number Catalan pseudoprime Elliptic pseudoprime Euler pseudoprime Euler–Jacobi pseudoprime Fermat pseudoprime Frobenius pseudoprime Lucas pseudoprime Lucas–Carmichael number Perrin pseudoprime Somer–Lucas pseudoprime Strong pseudoprime \| \| \| \| \| \| Arithmetic functions and dynamics \| \| --- \| \| \| \| \| --- \| \| Divisor functions \| Abundant Almost perfect Arithmetic Betrothed Colossally abundant Deficient Descartes Hemiperfect Highly abundant Highly composite Hyperperfect Multiply perfect Perfect Practical Primitive abundant Quasiperfect Refactorable Semiperfect Sublime Superabundant Superior highly composite Superperfect \| \| Prime omega functions \| Almost prime Semiprime \| \| Euler's totient function \| Highly cototient Highly totient Noncototient Nontotient Perfect totient Sparsely totient \| \| Aliquot sequences \| Amicable Perfect Sociable Untouchable \| \| Primorial \| Euclid Fortunate \| \| \| \| \| \| \| Other prime factor or divisor related numbers \| \| --- \| \| Blum Cyclic Erdős–Nicolas Erdős–Woods Friendly Giuga Harmonic divisor Jordan–Pólya Lucas–Carmichael Pronic Regular Rough Smooth Sphenic Størmer Super-Poulet \| \| \| \| \| \| Numeral system-dependent numbers \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Arithmetic functions and dynamics \| Persistence + Additive + Multiplicative \| \| \| --- \| \| Digit sum \| Digit sum Digital root Self Sum-product \| \| Digit product \| Multiplicative digital root Sum-product \| \| Coding-related \| Meertens \| \| Other \| Dudeney Factorion Kaprekar Kaprekar's constant Keith Lychrel Narcissistic Perfect digit-to-digit invariant Perfect digital invariant + Happy \| \| \| P-adic numbers-related \| Automorphic + Trimorphic \| \| Digit-composition related \| Palindromic Pandigital Repdigit Repunit Self-descriptive Smarandache–Wellin Undulating \| \| Digit-permutation related \| Cyclic Digit-reassembly Parasitic Primeval Transposable \| \| Divisor-related \| Equidigital Extravagant Frugal Harshad Polydivisible Smith Vampire \| \| Other \| \| \| \| \| \| \| \| Binary numbers \| \| --- \| \| Evil Odious Pernicious \| \| \| \| \| \| Generated via a sieve \| \| --- \| \| Lucky Prime \| \| \| \| \| \| Sorting related \| \| --- \| \| Pancake number Sorting number \| \| \| \| \| \| Natural language related \| \| --- \| \| Aronson's sequence Ban \| \| \| \| \| \| Graphemics related \| \| --- \| \| \| \| \| \| \| Mathematics portal \| \| Retrieved from " Categories: Figurate numbers Quadrilaterals Hidden categories: Articles with short description Short description is different from 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3347	https://math.stackexchange.com/questions/4407752/proving-that-sqrtx21x0-for-all-x	algebra precalculus - Proving that $\sqrt{x^2+1}+x>0$ for all $x$ - 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 Proving that √x 2+1+x>0 x 2+1−−−−−√+x>0 for all x x Ask Question Asked 3 years, 6 months ago Modified3 years ago Viewed 193 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Recently while dealing with inverse hyperbolic functions, I came across the expression sinh−1 x=ln(x+√x 2+1) sinh−1 x=ln(x+x 2+1−−−−−√) We know that f(x)=sinh−1 x f(x)=sinh−1 x is defined for all real values of x x since the range for sinh x sinh x is all real numbers. But I would like to prove this by proving that ln(x+√x 2+1)ln(x+x 2+1−−−−−√) is defined for all real values. For this, clearly, x+√x 2+1>0 x+x 2+1−−−−−√>0 but how do you prove this? I can see that for x>0,x>0,√x 2+1>√x 2=x>0 x 2+1−−−−−√>x 2−−√=x>0 ⟹f(x)=x+√x 2+1>0[∵x>0] But what about when x<0? I understand that √x 2+1 will always be greater than 0 but how do you deal with the +x? Another idea I have is to consider \|x−√x 2+1\| and use the fact that 0<\|x−√x 2+1\|<1 to show that √x 2+1 and x cannot be "too far apart" and since √x 2+1>1, x+√x 2+1>0 for all real x. But this method doesn't seem very robust, so I was wondering if there was a more convincing argument and less roundabout way of doing this. PS –I may be missing some simple algebra, but am only in high school, so would appreciate it if any answers are not too convoluted. algebra-precalculus inequality proof-explanation radicals hyperbolic-functions Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 19, 2022 at 11:40 Peter 86.8k 17 17 gold badges 87 87 silver badges 238 238 bronze badges asked Mar 19, 2022 at 11:34 devam_04devam_04 828 6 6 silver badges 15 15 bronze badges 2 10 √x 2+1>√x 2=\|x\|≥−x.Kavi Rama Murthy –Kavi Rama Murthy 2022-03-19 11:36:52 +00:00 Commented Mar 19, 2022 at 11:36 @KaviRamaMurthy Thank you so much! This is exactly what I was missing devam_04 –devam_04 2022-03-20 00:34:08 +00:00 Commented Mar 20, 2022 at 0:34 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. (√x 2+1+x)(√x 2+1−x)=1 One of the factors is positive because either x or −x is positive, so the other factor is positive as well. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 19, 2022 at 12:21 Empy2Empy2 52.4k 1 1 gold badge 48 48 silver badges 99 99 bronze badges 1 1 For benefit of OP - note how this follows the (familiar?) pattern (a+b)(a−b)=a 2−b 2 where a and b may be expressions involving square roots. When you first see such things they can be confusing or seem mysterious, but once understood they can come in handy.Mark Bennet –Mark Bennet 2022-03-19 12:26:19 +00:00 Commented Mar 19, 2022 at 12:26 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Note that the inequality is obvious when x is zero or positive. It remains for us to check that x is a negative number. For any negative x, \|x\|=−x. Hence, √x 2+1>√x 2=\|x\|=−x. We then have √x 2+1>−x√x 2+1+x>0. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 19, 2022 at 12:30 NightyNighty 2,228 2 2 gold badges 31 31 silver badges 52 52 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. WLOG x=cot 2 y,0<2 y<π using Principal values √x 2+1+x=csc 2 y+cot 2 y=⋯=cot y>0 as 0<y<π 2 and cos y≠0 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 19, 2022 at 12:10 lab bhattacharjeelab bhattacharjee 279k 20 20 gold badges 213 213 silver badges 337 337 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. 1)x≥0; the inequality is satisfied. 2)x<0; Rewrite as √x 2+1−√x 2= 1√x 2+1+√x 2>0, and we are done. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 19, 2022 at 15:23 Peter SzilasPeter Szilas 21.2k 2 2 gold badges 20 20 silver badges 31 31 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. x = 0; the inequality is satisfied 1 > 0 x < 0; √x 2+1−x>0; x > 0; √x 2+1+x>0; Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 25, 2022 at 23:53 VEILANCEVEILANCE 3 2 2 bronze badges 0 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 inequality proof-explanation radicals hyperbolic-functions See similar questions with these tags. 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3348	https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_10?srsltid=AfmBOordjY51zyC7R29wvwqJ8X8g_KwI5xDKcCzse264rYsvwnitRop0	Art of Problem Solving 2021 Fall AMC 12A 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. 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 2021 Fall AMC 12A Problems/Problem 10 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2021 Fall AMC 12A Problems/Problem 10 The following problem is from both the 2021 Fall AMC 10A #12 and 2021 Fall AMC 12A #10, so both problems redirect to this page. Contents [hide] 1 Problem 2 Solution 1 (Modular Arithmetic) 3 Solution 2 (Powers of 9) 4 Remark 5 Video Solution 6 Video Solution 7 Video Solution 8 Video Solution by WhyMath 9 See Also Problem The base-nine representation of the number is What is the remainder when is divided by Solution 1 (Modular Arithmetic) Recall that We expand by the definition of bases: ~Aidensharp ~Kante314 ~MRENTHUSIASM ~anabel.disher Solution 2 (Powers of 9) We need to first convert into a regular base- number: Now, consider how the last digit of changes with changes of the power of Note that if is odd, then On the other hand, if is even, then Therefore, we have Note that for the odd case, may simplify the process further, as given by Solution 1. ~Wilhelm Z Remark By the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit modulo , negate it, and add it to the next digit, and repeat the process until you reach the units digit. ~oinava Video Solution ~Education, the Study of Everything Video Solution Video Solution Video Solution by WhyMath ~savannahsolver See Also 2021 Fall AMC 12A (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•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions 2021 Fall AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 11Followed by Problem 13 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 10 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.
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Caret Package in R Python Introduction to Python Setup Python environment for ML Decorators in Python Generators in Python Iterators in Python Python Module Object Oriented Programming (OOPS) in Python List Comprehension Requests in Python Python Collections Python Logging Plots Matplotlib Tutorial Matplotlib Histogram Bar Plot in Python Python Boxplot Waterfall Plot in Python Top 50 matplotlib Visualizations Matplotlib Tutorial Matplotlib Pyplot Python Scatter Plot Matplotlib Subplots Data Wrangling 101 NumPy Exercises for Data Analysis (Python) 101 Pandas Exercises for Data Analysis 101 Pandas Exercises for Data Analysis Dask Modin Numpy Tutorial data.table in R 101 Python datatable Exercises (pydatatable) 101 R data.table Exercises Advanced Python Conda create environment and everything you need to know to manage conda virtual environment Python @Property Explained pdb – How to use Python debugger Python JSON – Guide cProfile – How to profile your python code Python Yield Lambda Function in Python What does Python Global Interpreter Lock Install opencv python Install pip mac Scrapy vs. Beautiful Soup Add Python to PATH PySpark Introduction to Pyspark Power of Pyspark Install PySpark on Windows Install PySpark on MAC Install PySpark on Linux What is Sparksession Read and Write files using PySpark Pyspark Show Run SQL Queries with PySpark PySpark Pandas API Select columns in PySpark dataframe PySpark withColumn() Pyspark Drop Columns PySpark Rename Columns PySpark Filter vs Where PySpark orderBy() and sort() PySpark GroupBy() PySpark Pivot PySpark Joins PySpark Union PySpark Connect to MySQL PySpark Connect to PostgreSQL PySpark Connect to SQL Serve PySpark Connect to Redshift PySpark Connect to Snowflake PySpark Linear Regression PySpark Logistic Regression PySpark Decision Tree PySpark Ridge Regression PySpark Lasso Regression PySpark Random Forest PySpark Gradient Boosting model PySpark Mllib K-Means Clustering PySpark Statistics Mean PySpark Statistics Median PySpark Statistics Mode PySpark Statistics Standard Deviation PySpark Statistics Variance PySpark Statistics Deciles and Quartiles PySpark Correlation PySpark Chi-Square Test PySpark Variable type Identification PySpark Outlier Detection and Treatment PySpark Missing Data Imputation PySpark Variance Inflation Factor (VIF) PySpark StringIndexer PySpark OneHot Encoding PySpark Exercises – 101 PySpark Exercises for Data Analysis Others Deployment Population Stability Index (PSI) Deploy ML model in AWS Ec2 Julia Julia – Programming Language Linear Regression in Julia Logistic Regression in Julia For-Loop in Julia While-loop in Julia Function in Julia DataFrames in Julia Linux ls command in Linux – Mastering the “ls” command in Linux mkdir command in Linux – A comprehensive guide for mkdir command cd command in linux – Mastering the ‘cd’ command in Linux cat command in Linux – Mastering the ‘cat’ command in Linux Linux Commands List with Examples Machine Learning Deep Learning TensorFlow vs PyTorch How to use tf.function to speed up Python code in Tensorflow How to implement Linear Regression in TensorFlow NLP Complete Guide to Natural Language Processing (NLP) Text Summarization Approaches for NLP 101 NLP Exercises (using modern libraries) Gensim Tutorial LDA in Python Topic Modeling with Gensim (Python) Lemmatization Approaches with Examples in Python Topic modeling visualization Cosine Similarity spaCy Tutorial Training Custom NER models in SpaCy to auto-detect named entities Building chatbot with Rasa and spaCy SpaCy Text Classification Algorithms K-Means Clustering Algorithm from Scratch Simulated Annealing Algorithm Explained from Scratch How Naive Bayes Algorithm Works? Feature selection using FRUFS and VevestaX Principal Component Analysis Gradient Boosting Feature Selection – Ten Effective Techniques with Examples Projects Evaluation Metrics for Classification Models Deploy ML model in AWS Ec2 Portfolio Optimization with Python using Efficient Frontier Bias Variance Tradeoff Specific Topics Logistic Regression Complete Introduction to Linear Regression in R Caret Package Brier Score Time Series Granger Causality Test Augmented Dickey Fuller Test (ADF Test) KPSS Test for Stationarity ARIMA Model Time Series Analysis in Python Vector Autoregression (VAR) Prob and Stats Probability Introduction to Probability Odds and Odds Ratios Independent and Dependent Events Mutually Exclusive Events Joint Probability Conditional Probability Bayes’ Theorem Expected Value Probability frequency distribution Discrete Frequency Distributions Continuous Frequency Distributions Partial Correlation Chi-Square Test – Theory & Math Gentle Introduction to Markov Chain What is P-Value? 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SQL for Data Science – Window Functions 10. Data Pre-processing and EDA 11. Linear regression and regularisation 12. Classification: Logistic Regression 13. Imbalanced Classification 14. Supervised ML Algorithms 15. Ensemble Learning 16. ML Deployment in AWS EC2 17. Deploy in AWS Lamda 18. Deploy in AWS Sagemaker 19. PySpark for Data Science – I: Fundamentals 20. PySpark for Data Science – II: Statistics for Big Data 21. Introduction to Time Series Analaysis 22. Time Series Analysis – I (Beginners) 23. Time Series Analysis – II (Intermediate ) 24. Time Series Forecasting Part 1: Statistical Models 25. Time Series Forecasting Part 2: ARIMA modeling and Tests 26. Time Series Forecasting Part 3: Vector Auto Regression 27. Time Series Analysis – III: Singular Spectrum Analysis 28. Feature Engineering for Time Series Project: I 29. Feature Engineering for Time Series Projects: II 31. Estimating customer lifetime value for business 32. Microsoft malware detection project 33. Credit card fraud detection 34. Restaurant Visitor Forecasting 35. Optimizing Marketing Budget Spend with Marketing Mix Modeling 36. Predict Rating given Amazon Product review using NLP 37. Foundations of Deep Learning in Python 38. Foundations of Deep Learning: Part 2 39. Applied Deep Learning with PyTorch 40. Detecting defects in Steel sheet with Computer vision 41. Project Text Generation using Language models with LSTM 42. Project Classifying Sentiment of reviews using BERT NLP 43. Spacy for NLP 44. Base R Programming 45. Dplyr for Data Wrangling 46. Wrangling Data with Data Table 47. GGPlot2 Visualization for Data Analysis 48. Statistical foundation for ML in R 49. Regression Model in R 50. Caret Package in R Python Introduction to Python Setup Python environment for ML Decorators in Python Generators in Python Iterators in Python Python Module Object Oriented Programming (OOPS) in Python List Comprehension Requests in Python Python Collections Python Logging Plots Matplotlib Tutorial Matplotlib Histogram Bar Plot in Python Python Boxplot Waterfall Plot in Python Top 50 matplotlib Visualizations Matplotlib Tutorial Matplotlib Pyplot Python Scatter Plot Matplotlib Subplots Data Wrangling 101 NumPy Exercises for Data Analysis (Python) 101 Pandas Exercises for Data Analysis 101 Pandas Exercises for Data Analysis Dask Modin Numpy Tutorial data.table in R 101 Python datatable Exercises (pydatatable) 101 R data.table Exercises Advanced Python Conda create environment and everything you need to know to manage conda virtual environment Python @Property Explained pdb – How to use Python debugger Python JSON – Guide cProfile – How to profile your python code Python Yield Lambda Function in Python What does Python Global Interpreter Lock Install opencv python Install pip mac Scrapy vs. Beautiful Soup Add Python to PATH PySpark Introduction to Pyspark Power of Pyspark Install PySpark on Windows Install PySpark on MAC Install PySpark on Linux What is Sparksession Read and Write files using PySpark Pyspark Show Run SQL Queries with PySpark PySpark Pandas API Select columns in PySpark dataframe PySpark withColumn() Pyspark Drop Columns PySpark Rename Columns PySpark Filter vs Where PySpark orderBy() and sort() PySpark GroupBy() PySpark Pivot PySpark Joins PySpark Union PySpark Connect to MySQL PySpark Connect to PostgreSQL PySpark Connect to SQL Serve PySpark Connect to Redshift PySpark Connect to Snowflake PySpark Linear Regression PySpark Logistic Regression PySpark Decision Tree PySpark Ridge Regression PySpark Lasso Regression PySpark Random Forest PySpark Gradient Boosting model PySpark Mllib K-Means Clustering PySpark Statistics Mean PySpark Statistics Median PySpark Statistics Mode PySpark Statistics Standard Deviation PySpark Statistics Variance PySpark Statistics Deciles and Quartiles PySpark Correlation PySpark Chi-Square Test PySpark Variable type Identification PySpark Outlier Detection and Treatment PySpark Missing Data Imputation PySpark Variance Inflation Factor (VIF) PySpark StringIndexer PySpark OneHot Encoding PySpark Exercises – 101 PySpark Exercises for Data Analysis Others Deployment Population Stability Index (PSI) Deploy ML model in AWS Ec2 Julia Julia – Programming Language Linear Regression in Julia Logistic Regression in Julia For-Loop in Julia While-loop in Julia Function in Julia DataFrames in Julia Linux ls command in Linux – Mastering the “ls” command in Linux mkdir command in Linux – A comprehensive guide for mkdir command cd command in linux – Mastering the ‘cd’ command in Linux cat command in Linux – Mastering the ‘cat’ command in Linux Linux Commands List with Examples Machine Learning Deep Learning TensorFlow vs PyTorch How to use tf.function to speed up Python code in Tensorflow How to implement Linear Regression in TensorFlow NLP Complete Guide to Natural Language Processing (NLP) Text Summarization Approaches for NLP 101 NLP Exercises (using modern libraries) Gensim Tutorial LDA in Python Topic Modeling with Gensim (Python) Lemmatization Approaches with Examples in Python Topic modeling visualization Cosine Similarity spaCy Tutorial Training Custom NER models in SpaCy to auto-detect named entities Building chatbot with Rasa and spaCy SpaCy Text Classification Algorithms K-Means Clustering Algorithm from Scratch Simulated Annealing Algorithm Explained from Scratch How Naive Bayes Algorithm Works? Feature selection using FRUFS and VevestaX Principal Component Analysis Gradient Boosting Feature Selection – Ten Effective Techniques with Examples Projects Evaluation Metrics for Classification Models Deploy ML model in AWS Ec2 Portfolio Optimization with Python using Efficient Frontier Bias Variance Tradeoff Specific Topics Logistic Regression Complete Introduction to Linear Regression in R Caret Package Brier Score Time Series Granger Causality Test Augmented Dickey Fuller Test (ADF Test) KPSS Test for Stationarity ARIMA Model Time Series Analysis in Python Vector Autoregression (VAR) Prob and Stats Probability Introduction to Probability Odds and Odds Ratios Independent and Dependent Events Mutually Exclusive Events Joint Probability Conditional Probability Bayes’ Theorem Expected Value Probability frequency distribution Discrete Frequency Distributions Continuous Frequency Distributions Partial Correlation Chi-Square Test – Theory & Math Gentle Introduction to Markov Chain What is P-Value? How to implement common statistical significance tests and find the p value? Mahalanobis Distance T Test (Students T Test) Confidence Interval in Statistics Standard Error in Statistics One Sample T Test Descriptive and inferential statistics Types of data in statistics Measures of central tendency Quantiles and Percentiles Measures of dispersion Skewness and kurtosis Central Limit Theroem Law of large numbers Standard Error Sampling and sampling distributions Correlation SQL SQL Tutorial – The Introduction SQL Subquery (advanced) SQL Window Functions (advanced) SQL Window Functions Exercises – Set 1 SQL Window Functions Exercises – Set 2 Intro to SQL SQL Select SQL Select Distinct SQL Where SQL Order by SQL Insert Into SQL AND, OR, and NOT SQL Null Values SQL Update SQL DELETE SQL SELECT TOP SQL MIN and MAX Functions SQL Count(), Avg(), Sum() SQL LIKE SQL Wildcards SQL IN SQL BETWEEN SQL Aliases SQL Joins SQL Inner Join SQL Left Join SQL Right Join SQL Full Join SQL Self Join SQL UNION SQL GROUP BY SQL HAVING SQL EXISTS SQL ANY, ALL Operators How to transpose columns to rows in SQL? How to select only rows with max value on a column? SQL Select Into SQL Insert Into Select SQL Case SQL Null Functions SQL Comments SQL Operators SQL Create Table SQL Drop Table SQL Primary Key SQL Foreign Key Sort multiple columns in SQL and in different directions? Count the number of work days between two dates? Compute maximum of multiple columns, aks row wise max? GROUP BY clause on multiple columns in SQL? Linear Algebra 01. Introduction to Linear Algebra 02. Types of Tensors 03. Scalars 04. Vectors 05. Vectors Linear Algebra 06. Matrix Types 07. Matrix Operations 08. Orthogonal and Ortrhonormal Matrix 09. Eigenvectors and Eigenvalues 10. Affine Transformation 11. Singular Value Decomposition (SVD) 12. System of Equations 13. Linear Regression Algorithm 14. Principal Component Analysis Maximum A Posteriori (MAP) Estimation – Clearly Explained Join thousands of students who advanced their careers with MachineLearningPlus. Go from Beginner to Data Science (AI/ML/Gen AI) Expert through a structured pathway of 9 core specializations and build industry grade projects. View Course June 2, 2025 Selva Prabhakaran Maximum A Posteriori (MAP) estimation is a Bayesian method for finding the most likely parameter values given observed data and prior knowledge. Unlike maximum likelihood estimation which only considers the data, MAP combines what we observe with what we already know (or believe) about the parameters. Ever wondered how your smartphone’s autocorrect gets better over time? Or how recommendation systems seem to know your preferences even with limited data? That’s the principle of MAP estimation at work – a smart way to make predictions by combining new evidence with existing knowledge. Think of MAP as your wise friend who gives advice. They don’t just look at today’s events (the data) but also consider everything they know about you from the past (prior knowledge) to give you the best guidance. 1. The Intuition Behind MAP Estimation Let me start with a simple story that will make everything clear. Imagine you’re a detective investigating a crime. You have two pieces of information: Evidence from the crime scene (this is your data/likelihood) What you know about the suspect’s history (this is your prior knowledge) A good detective doesn’t just look at today’s evidence. They combine it with background knowledge to reach the most reasonable conclusion. That’s exactly what MAP estimation does in machine learning. MAP estimation was formally introduced in the context of Bayesian statistics, with significant contributions from Thomas Bayes (18th century) and later formalized by Pierre-Simon Laplace. The modern computational approach to MAP estimation was developed alongside the advancement of Bayesian methods in the mid-20th century. 2. The Mathematical Foundation Here’s the core idea in one equation. Don’t let the math scare you – I’ll break it down: MAP estimate = argmax P(θ\|data) = argmax P(data\|θ) × P(θ) Let me translate this into plain English: P(data\|θ), aka Likelihood: “How likely is our data given a specific parameter value?” P(θ), aka Prior: “What did we believe about θ before seeing any data?” P(θ\|data), aka Posterior: Your updated belief after seeing the data, which is, “What’s the probability of our parameter θ given the data we observed?” MAP estimation is based on Bayes’ theorem, which you can think of as a way to update your beliefs when you get new information: Posterior ∝ Likelihood × Prior MAP finds the parameter values that maximize this posterior probability. The original concept was formalized by Thomas Bayes in the 18th century and later developed into modern MAP estimation by statisticians in the 20th century. Let’s expand on the idea a little bit more. 3. Core Idea behind MAP Estimation: (I’ll try to Keep It Simple, but skip and move to the code if this is too much) MAP estimation is a Bayesian approach to estimate an unknown parameter (θ) by finding the value that maximizes the posterior probability distribution. Posterior Probability (P(θ\|X)): The probability of the parameter θ given the observed data X. Computed using Bayes’ Theorem: P(\theta\|X) = \frac{P(X\|\theta) \cdot P(\theta)}{P(X)} Likelihood (P(X\|θ)): Probability of data (X) given θ (from your model). Prior (P(θ)): Initial belief about θ (e.g., in coin toss, θ is P(head) is near 0.5). Evidence (P(X)): Probability of data (is constant, independent of θ). Goal of MAP: Find the θ that maximizes the posterior probability. \theta_{\text{MAP}} = \underset{\theta}{\arg\max} \, P(\theta\|X) Since $(P(X)$ in the denominator is constant, because the observed data does not change, we ignore it and maximize the numerator: Now since the numerator contains multiplication and since the probability values are between 0 and 1, multiplying priors and likelihoods are going to give very small fractions. So, instead we take ‘log’ so that we will be able to add the values instead. Logarithm Trick: Convert products to sums (easier for optimization): \theta_{\text{MAP}} = \underset{\theta}{\arg\max} \, \left[ \log P(X\|\theta) + \log P(\theta) \right] $\log P(X\|\theta)$: Log-likelihood (measures data fit). $\log P(\theta)$: Log-prior (regularizes using prior knowledge). Optimize: Solve using calculus (e.g., set derivative to zero) or numerical methods. That’s the theory, but that doesn’t do much if we are not able to relate to a practical example. So, let’s see one and then get to the code implementations. 4. A Practical Explanation: The Cookie Jar Example Imagine you have 2 cookie jars: Jar A: 70% chocolate chips, 30% raisins Jar B: 30% chocolate chips, 70% raisins You randomly pick a jar (you don’t know which) and draw 3 cookies: Chocolate, Raisin, Raisin What jar did it come from? Your Prior Belief: You think both jars are equally likely to be picked. → P(Jar A) = 50%, P(Jar B) = 50% 2. The Data (Likelihood): Probability of drawing one chocolate and two raisins from Jar A: 0.7 (choc) × 0.3 (raisin) × 0.3 (raisin) = 0.063 Probability from Jar B: 0.3 × 0.7 × 0.7 = 0.147 MAP Combines Both: Multiply prior belief × data probability: Jar A: 0.5 × 0.063 = 0.0315 Jar B: 0.5 × 0.147 = 0.0735 MAP picks Jar B because 0.0735 > 0.0315! → It’s the most probable jar given your data + prior belief. Why Not Just Use Data Alone? (MLE vs. MAP) MLE (Maximum Likelihood):Only uses data. In our example, MLE would pick Jar B too (because 0.147 > 0.063). But if your prior changes… Suppose you know Jar A is used 90% of the time: P(Jar A) = 0.9, P(Jar B) = 0.1 Now: Jar A: 0.9 × 0.063 = 0.0567 Jar B: 0.1 × 0.147 = 0.0147 MAP picks Jar A! → Your strong prior belief overruled the data. In one sentence: MAP is your “best guess” after combining new evidence with what you already believed. 5. Setting Up Your Environment Before we dive into the code, let’s get everything ready. I assume you’re comfortable with Python and have VS Code set up. ```python !pip install numpy scipy matplotlib seaborn pandas scikit-learn jupyter ``` ```python import numpy as np import matplotlib.pyplot as plt import scipy.stats as stats from scipy.optimize import minimize_scalar import seaborn as sns Set up plotting style plt.style.use('seaborn-v0_8') sns.set_palette("husl") ``` Let’s also create some helper functions we’ll use throughout: ```python def plot_distribution(x, y, title, xlabel, ylabel): """Helper function to create clean plots""" plt.figure(figsize=(10, 6)) plt.plot(x, y, linewidth=2) plt.title(title, fontsize=14, fontweight='bold') plt.xlabel(xlabel, fontsize=12) plt.ylabel(ylabel, fontsize=12) plt.grid(True, alpha=0.3) plt.show() def normalize_distribution(values): """Normalize values to form a probability distribution""" return values / np.trapz(values) ``` 4. A Real-World Example: Estimating Coin Bias Let’s start with something concrete. Imagine you find a coin and want to know if it’s fair. You flip it 10 times and get 7 heads. Question: What’s your best estimate of the coin’s bias (probability of heads)? Let me show you three different approaches and see how they compare: ```python Our data: 10 flips, 7 heads n_flips = 10 n_heads = 7 print(f"Observed data: {n_heads} heads out of {n_flips} flips") print(f"That's {n_heads/n_flips:.1%} heads") ``` Observed data: 7 heads out of 10 flips That's 70.0% heads 4.1 Maximum Likelihood Estimation (MLE) First, let’s see what Maximum Likelihood Estimation would tell us: ```python MLE simply uses the observed frequency mle_estimate = n_heads / n_flips print(f"MLE estimate: {mle_estimate:.3f} (or {mle_estimate:.1%})") ``MLE estimate: 0.700 (or 70.0%)` MLE says: “Based purely on the data, the coin has a 70% chance of heads.” 4.2 MAP Estimation with a Fair Coin Prior Now let’s use MAP estimation. We’ll start with a prior belief that most coins are roughly fair: ```python Define our prior belief (Beta distribution) Beta(2,2) is centered at 0.5 but allows some uncertainty prior_alpha = 2 prior_beta = 2 After observing data, posterior is Beta(prior_alpha + heads, prior_beta + tails) posterior_alpha = prior_alpha + n_heads posterior_beta = prior_beta + (n_flips - n_heads) MAP estimate is the mode of the posterior Beta distribution map_estimate = (posterior_alpha - 1) / (posterior_alpha + posterior_beta - 2) print(f"MAP estimate: {map_estimate:.3f} (or {map_estimate:.1%})") print(f"Posterior parameters: α={posterior_alpha}, β={posterior_beta}") ``` MAP estimate: 0.667 (or 66.7%) Posterior parameters: α=9, β=5 4.3 Visualizing the Difference Let’s see this graphically to understand what’s happening: ```python Create range of possible bias values theta_range = np.linspace(0, 1, 1000) Calculate likelihood for each possible bias likelihood = stats.binom.pmf(n_heads, n_flips, theta_range) Calculate prior for each possible bias prior = stats.beta.pdf(theta_range, prior_alpha, prior_beta) Calculate posterior (likelihood × prior, normalized) posterior = stats.beta.pdf(theta_range, posterior_alpha, posterior_beta) Create the plot fig, axes = plt.subplots(2, 2, figsize=(15, 10)) fig.suptitle('MAP vs MLE: Coin Bias Estimation', fontsize=16, fontweight='bold') Plot 1: Prior belief axes[0,0].plot(theta_range, prior, 'g-', linewidth=2, label='Prior') axes[0,0].set_title('Prior Belief: Coins are Usually Fair') axes[0,0].set_xlabel('Bias (P(Heads))') axes[0,0].set_ylabel('Probability Density') axes[0,0].axvline(0.5, color='g', linestyle='--', alpha=0.7, label='Prior Peak') axes[0,0].legend() axes[0,0].grid(True, alpha=0.3) Plot 2: Likelihood axes[0,1].plot(theta_range, likelihood, 'b-', linewidth=2, label='Likelihood') axes[0,1].set_title('Likelihood: What the Data Tells Us') axes[0,1].set_xlabel('Bias (P(Heads))') axes[0,1].set_ylabel('Likelihood') axes[0,1].axvline(mle_estimate, color='b', linestyle='--', alpha=0.7, label=f'MLE = {mle_estimate:.2f}') axes[0,1].legend() axes[0,1].grid(True, alpha=0.3) Plot 3: Posterior axes[1,0].plot(theta_range, posterior, 'r-', linewidth=2, label='Posterior') axes[1,0].set_title('Posterior: Combining Prior and Likelihood') axes[1,0].set_xlabel('Bias (P(Heads))') axes[1,0].set_ylabel('Probability Density') axes[1,0].axvline(map_estimate, color='r', linestyle='--', alpha=0.7, label=f'MAP = {map_estimate:.2f}') axes[1,0].legend() axes[1,0].grid(True, alpha=0.3) Plot 4: Comparison axes[1,1].plot(theta_range, prior, 'g-', linewidth=2, label='Prior', alpha=0.7) axes[1,1].plot(theta_range, likelihood/max(likelihood) max(prior), 'b-', linewidth=2, label='Likelihood (scaled)', alpha=0.7) axes[1,1].plot(theta_range, posterior, 'r-', linewidth=2, label='Posterior') axes[1,1].axvline(0.5, color='g', linestyle='--', alpha=0.5) axes[1,1].axvline(mle_estimate, color='b', linestyle='--', alpha=0.5) axes[1,1].axvline(map_estimate, color='r', linestyle='--', alpha=0.7, linewidth=2) axes[1,1].set_title('All Together: The MAP Compromise') axes[1,1].set_xlabel('Bias (P(Heads))') axes[1,1].set_ylabel('Probability Density') axes[1,1].legend() axes[1,1].grid(True, alpha=0.3) plt.tight_layout() plt.show() print(f"\nSummary:") print(f"Prior belief: Fair coin (50% heads)") print(f"MLE estimate: {mle_estimate:.1%} (purely data-driven)") print(f"MAP estimate: {map_estimate:.1%} (balanced view)") ``` Summary: Prior belief: Fair coin (50% heads) MLE estimate: 70.0% (purely data-driven) MAP estimate: 66.7% (balanced view) Notice how MAP gives us a more conservative estimate? It’s pulling the purely data-driven MLE estimate back toward our prior belief that coins are usually fair. 5. The Power of Priors: Strong vs Weak Beliefs The posterior probability also depends on how strongly we believe in the priors. Let’s see how different prior beliefs affect our MAP estimates: ```python Let's try different prior beliefs priors = { 'Weak prior (uniform)': (1, 1), # No strong belief 'Fair coin belief': (2, 2), # Mild belief in fairness 'Strong fair belief': (10, 10), # Very strong belief in fairness 'Biased coin belief': (1, 3) # Belief that coins are usually biased toward tails } results = {} for name, (alpha, beta) in priors.items(): # Calculate MAP estimate post_alpha = alpha + n_heads post_beta = beta + (n_flips - n_heads) map_est = (post_alpha - 1) / (post_alpha + post_beta - 2) results[name] = { 'prior': (alpha, beta), 'posterior': (post_alpha, post_beta), 'map_estimate': map_est } print(f"{name}:") print(f" Prior: Beta({alpha}, {beta})") print(f" MAP estimate: {map_est:.3f} ({map_est:.1%})") print() print(f"MLE estimate (for comparison): {mle_estimate:.3f} ({mle_estimate:.1%})") ``` ``` Weak prior (uniform): Prior: Beta(1, 1) MAP estimate: 0.700 (70.0%) Fair coin belief: Prior: Beta(2, 2) MAP estimate: 0.667 (66.7%) Strong fair belief: Prior: Beta(10, 10) MAP estimate: 0.571 (57.1%) Biased coin belief: Prior: Beta(1, 3) MAP estimate: 0.583 (58.3%) MLE estimate (for comparison): 0.700 (70.0%) ``` See how strong priors resist being changed by the data? This is both a strength (prevents overfitting to small datasets) and a potential weakness (can ignore important new evidence). 6. MAP in Linear Regression with Ridge Regularization Now let’s see MAP in action for a more complex problem: linear regression. It turns out that Ridge regression is actually MAP estimation in disguise! Fore more details, check notes here. ```python Generate some sample data np.random.seed(42) n_samples = 50 n_features = 5 True coefficients (what we're trying to estimate) true_coefficients = np.array([2, -1.5, 0.8, -0.3, 1.2]) Generate features X = np.random.randn(n_samples, n_features) y = X @ true_coefficients + 0.3 np.random.randn(n_samples) # Add some noise print(f"True coefficients: {true_coefficients}") print(f"Generated {n_samples} samples with {n_features} features") ``` True coefficients: [ 2. -1.5 0.8 -0.3 1.2] Generated 50 samples with 5 features 6.1 Regular Linear Regression (MLE) First, let’s fit a regular linear regression model: ```python from sklearn.linear_model import LinearRegression Fit regular linear regression (this is MLE) mle_model = LinearRegression() mle_model.fit(X, y) print(f"MLE coefficients: {mle_model.coef_}") print(f"True coefficients: {true_coefficients}") print(f"MLE error: {np.mean((mle_model.coef_ - true_coefficients)2):.4f}") ``` MLE coefficients: [ 1.83533532 -1.5667978 0.83609877 -0.32856023 1.14017299] True coefficients: [ 2. -1.5 0.8 -0.3 1.2] MLE error: 0.0075 6.2 Ridge Regression (MAP with Gaussian Prior) Now let’s use Ridge regression, which is MAP estimation with a Gaussian prior on the coefficients: ```python from sklearn.linear_model import Ridge Fit Ridge regression (this is MAP with Gaussian prior) regularization_strength = 1.0 map_model = Ridge(alpha=regularization_strength) map_model.fit(X, y) print(f"MAP coefficients: {map_model.coef_}") print(f"True coefficients: {true_coefficients}") print(f"MAP error: {np.mean((map_model.coef_ - true_coefficients)2):.4f}") ``` MAP coefficients: [ 1.79932082 -1.53701643 0.8093644 -0.3294865 1.11706068] True coefficients: [ 2. -1.5 0.8 -0.3 1.2] MAP error: 0.0099 6.3 Understanding the Connection Let me show you why Ridge regression is MAP estimation: ```python def manual_map_regression(X, y, prior_variance=1.0): """ Manual implementation of MAP regression to show the connection """ # MAP estimate for linear regression with Gaussian prior # θ_MAP = (X^T X + λI)^(-1) X^T y # where λ = σ²/σ₀² (noise variance / prior variance) lambda_reg = 1.0 / prior_variance # Regularization parameter # Calculate MAP estimate XtX = X.T @ X identity = np.eye(X.shape) map_coefficients = np.linalg.inv(XtX + lambda_reg identity) @ X.T @ y return map_coefficients Calculate using our manual implementation manual_map_coef = manual_map_regression(X, y, prior_variance=1.0) print("Comparison of implementations:") print(f"Sklearn Ridge: {map_model.coef_}") print(f"Manual MAP: {manual_map_coef}") print(f"Difference: {np.mean((map_model.coef_ - manual_map_coef)2):.6f}") ``` Comparison of implementations: Sklearn Ridge: [ 1.79932082 -1.53701643 0.8093644 -0.3294865 1.11706068] Manual MAP: [ 1.79970476 -1.53673869 0.80945523 -0.32985766 1.11685307] Difference: 0.000000 The regularization parameter in Ridge regression controls how strongly we believe the coefficients should be close to zero (our prior). 7. MAP vs MLE vs Bayesian: The Full Picture Let’s compare all three approaches side by side: ```python def compare_estimation_methods(n_samples_list=[5, 20, 100, 500]): """Compare MLE, MAP, and full Bayesian approaches with different sample sizes""" results = {'n_samples': [], 'mle_error': [], 'map_error': [], 'true_coef': true_coefficients} for n in n_samples_list: # Generate data with different sample sizes X_temp = np.random.randn(n, n_features) y_temp = X_temp @ true_coefficients + 0.3 np.random.randn(n) # MLE (regular regression) mle_temp = LinearRegression().fit(X_temp, y_temp) mle_error = np.mean((mle_temp.coef_ - true_coefficients)2) # MAP (Ridge regression) map_temp = Ridge(alpha=1.0).fit(X_temp, y_temp) map_error = np.mean((map_temp.coef_ - true_coefficients)2) results['n_samples'].append(n) results['mle_error'].append(mle_error) results['map_error'].append(map_error) print(f"n={n:3d} \| MLE error: {mle_error:.4f} \| MAP error: {map_error:.4f}") return results Run the comparison np.random.seed(42) comparison_results = compare_estimation_methods() ``` n= 5 \| MLE error: 0.5091 \| MAP error: 0.7225 n= 20 \| MLE error: 0.0029 \| MAP error: 0.0027 n=100 \| MLE error: 0.0008 \| MAP error: 0.0015 n=500 \| MLE error: 0.0002 \| MAP error: 0.0002 Plot the results to see the pattern: ```python plt.figure(figsize=(12, 6)) plt.subplot(1, 2, 1) plt.plot(comparison_results['n_samples'], comparison_results['mle_error'], 'o-', label='MLE', linewidth=2, markersize=6) plt.plot(comparison_results['n_samples'], comparison_results['map_error'], 's-', label='MAP', linewidth=2, markersize=6) plt.xlabel('Number of Samples') plt.ylabel('Mean Squared Error') plt.title('MLE vs MAP: Performance with Different Sample Sizes') plt.legend() plt.grid(True, alpha=0.3) plt.yscale('log') plt.subplot(1, 2, 2) Show the bias-variance tradeoff concept sample_sizes = np.array(comparison_results['n_samples']) mle_errors = np.array(comparison_results['mle_error']) map_errors = np.array(comparison_results['map_error']) plt.bar(np.arange(len(sample_sizes)) - 0.2, mle_errors, 0.4, label='MLE', alpha=0.7) plt.bar(np.arange(len(sample_sizes)) + 0.2, map_errors, 0.4, label='MAP', alpha=0.7) plt.xticks(range(len(sample_sizes)), sample_sizes) plt.xlabel('Number of Samples') plt.ylabel('Mean Squared Error') plt.title('Error Comparison by Sample Size') plt.legend() plt.grid(True, alpha=0.3) plt.tight_layout() plt.show() ``` When to use which estimation method: Small dataset, strong prior knowledge → MAP Small dataset, weak prior knowledge → Full Bayesian (if computational resources allow) Large dataset, any prior knowledge → MLE or MAP (similar results) Need point estimates quickly → MAP Need uncertainty quantification → Full Bayesian Regularization is important → MAP (Ridge, Lasso) Interpretability is key → MAP with appropriate priors 9. Real-World Applications Let me show you where MAP estimation shines in practice: 9.1 Text Classification with Naive Bayes ```python Simulate text classification problem def naive_bayes_example(): """ MAP estimation in Naive Bayes classification """ # Simulate word counts for spam/ham classification np.random.seed(42) # Prior probabilities (what we know before seeing any emails) prior_spam = 0.3 # 30% of emails are spam prior_ham = 0.7 # 70% of emails are legitimate # Likelihoods (how often words appear in spam vs ham) word_probs = { 'free': {'spam': 0.8, 'ham': 0.1}, 'money': {'spam': 0.7, 'ham': 0.05}, 'meeting': {'spam': 0.1, 'ham': 0.6}, 'urgent': {'spam': 0.6, 'ham': 0.2} } # New email: "urgent free money" email_words = ['urgent', 'free', 'money'] # Calculate MAP classification spam_posterior = prior_spam ham_posterior = prior_ham for word in email_words: spam_posterior = word_probs[word]['spam'] ham_posterior = word_probs[word]['ham'] # Normalize total = spam_posterior + ham_posterior spam_prob = spam_posterior / total ham_prob = ham_posterior / total print(f"Email words: {email_words}") print(f"P(Spam\|email) = {spam_prob:.3f}") print(f"P(Ham\|email) = {ham_prob:.3f}") print(f"MAP classification: {'SPAM' if spam_prob > ham_prob else 'HAM'}") naive_bayes_example() ``` Email words: ['urgent', 'free', 'money'] P(Spam\|email) = 0.993 P(Ham\|email) = 0.007 MAP classification: SPAM 9.2 Image Denoising ```python Simulate MAP estimation for image denoising def image_denoising_demo(): """ Simplified example of MAP estimation in image denoising """ np.random.seed(42) # Create a simple "image" (1D signal for simplicity) true_signal = np.sin(np.linspace(0, 4np.pi, 50)) noise = 0.5 np.random.randn(50) noisy_signal = true_signal + noise # MAP estimation with smoothness prior # We believe neighboring pixels should have similar values def map_denoise(noisy_data, smoothness_weight=1.0): """Simple MAP denoising with smoothness prior""" n = len(noisy_data) result = noisy_data.copy() # Iterative approach (simplified) for _ in range(10): new_result = result.copy() for i in range(1, n-1): # Data term: stay close to observations data_term = noisy_data[i] # Prior term: stay close to neighbors (smoothness) smoothness_term = (result[i-1] + result[i+1]) / 2 # Combine with weights new_result[i] = (data_term + smoothness_weight smoothness_term) / (1 + smoothness_weight) result = new_result return result # Apply MAP denoising denoised = map_denoise(noisy_signal, smoothness_weight=2.0) # Plot results plt.figure(figsize=(12, 4)) plt.subplot(1, 3, 1) plt.plot(true_signal, 'g-', label='True signal', linewidth=2) plt.title('Original Clean Signal') plt.legend() plt.grid(True, alpha=0.3) plt.subplot(1, 3, 2) plt.plot(noisy_signal, 'r-', alpha=0.7, label='Noisy signal') plt.plot(true_signal, 'g-', label='True signal', linewidth=2) plt.title('Noisy Observation') plt.legend() plt.grid(True, alpha=0.3) plt.subplot(1, 3, 3) plt.plot(denoised, 'b-', label='MAP denoised', linewidth=2) plt.plot(true_signal, 'g-', label='True signal', linewidth=2) plt.title('MAP Denoised Result') plt.legend() plt.grid(True, alpha=0.3) plt.tight_layout() plt.show() # Calculate improvement noisy_error = np.mean((noisy_signal - true_signal)2) denoised_error = np.mean((denoised - true_signal)2) print(f"Noisy signal error: {noisy_error:.4f}") print(f"MAP denoised error: {denoised_error:.4f}") print(f"Improvement: {((noisy_error - denoised_error) / noisy_error) 100:.1f}%") image_denoising_demo() ``` Noisy signal error: 0.2263 MAP denoised error: 0.0796 Improvement: 64.8% More Articles Statistics ### Understanding Confidence Intervals: A spelled out guide to clarify misconceptions Jun 06, 2025 Statistics ### Ridge Regression as MAP Estimation – Supporting notes Jun 02, 2025 Statistics ### Maximum A Posteriori (MAP) Estimation – Clearly Explained Statistics ### F Statistic Formula – Explained Aug 04, 2024 Statistics ### Correlation – Connecting the Dots, the Role of Correlation in Data Analysis Sep 23, 2023 Statistics ### Hypothesis Testing – A Deep Dive into Hypothesis Testing, The Backbone of Statistical Inference Sep 21, 2023 Similar Articles Complete Introduction to Linear Regression in R Selva Prabhakaran 12/03/2017 7 Comments Read More » How to implement common statistical significance tests and find the p value? 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3350	https://smw.ch/index.php/smw/article/view/3549	Skip to main navigation menu Skip to main content Skip to site footer Peripheral lymphadenopathy of unknown origin in adults: a diagnostic approach emphasizing the malignancy hypothesis Ivana Hanzalova+− Maurice Matter+− Fulltext PDF Fulltext HTML Cite this as: : Swiss Med Wkly. 2024;154:3549 Published : 31.07.2024 Summary The term lymphadenopathy refers to an abnormality in size, consistency or morphological aspect of one or several lymph nodes. Although lymphadenopathies are commonly observed in everyday clinical practice, the difficulty of differentiating benign and malignant disease may delay therapeutic approaches. The present review aims to update diagnostic algorithms in different clinical situations based on the currently available literature. A literature review was performed to assess current knowledge of and to update the diagnostic approach. A short clinical vignette was used as an example of a typical clinical presentation. This case of metastatic lymphadenopathy with incomplete patient history demonstrates how misleading such lymphadenopathy may be, leading to a delayed diagnosis and even a fatal outcome. Any lymphadenopathy persisting for more than 2 weeks should be considered suspicious and deserves further investigation. Precise clinical examination, meticulous history-taking and a search for associated symptomatology are still cornerstones for diagnosing the origin of the condition. The next diagnostic step depends on the anatomical region and the specific patient’s situation. Imaging starts with ultrasound, while computed tomography (CT) and magnetic resonance imaging (MRI) allow assessment of the surrounding structures. If the diagnosis remains uncertain, tissue sampling and histological analyses should be performed. Except for head and neck loco-regional lymphadenopathy, there are no methodical guidelines for persistent lymphadenopathy. The present review clarifies several confusing and complex situations. The accuracy of fine needle aspiration cytology could be increased by using core needle biopsy with immunocytologic and flow cytometric methods. Notably, except in the head and neck area, open biopsy remains the best option when lymphoma is suspected or when inconclusive results of previous fine needle aspiration cytology or core needle biopsy are obtained. The incidence of malignant lymphadenopathy varies with its location and the various diagnostic strategies. In metastatic lymphadenopathy of unknown primary origin, European Society for Medical Oncology (ESMO) guidelines and modern methods like next-generation sequencing (NGS) may help to manage such complex cases. References Gaddey HL, Riegel AM. Unexplained Lymphadenopathy: evaluation and differential diagnosis . 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3351	https://sedl.org/secc/common_core_videos/grade.php?action=view&id=694	Common Core Videos for Grade 6, Mathematics 6.NS.7d SEDL merged with the American Institutes for Research (AIR) in 2015. This archived website contains the work of SEDL legacy projects and rich resources from the past 50 years. << Return to AIR.org SECC portal ARCHIVE Previous Work October 2005 to September 2019 These resources were published under a previous SECC contract; therefore, information contained therein may have changed and is not updated. Follow Us Mathematics Grade: 6, Mathematics 6.NS.7d Part 1: View Transcript 6.NS.7d Transcript This is Common Core State Standards support video for mathematics; this is standard 6.NS.7d. This standard states: understand ordering and absolute value of rational numbers; distinguish comparisons of absolute value from statements about order. Let’s look at the introductory statement for this standard; understand ordering and absolute value of rational numbers. Now, the idea of ordering of numbers is covered in earlier standards. For example, fractions, ordering them, is addressed back in the fourth grade in standard 4.NF.2. So, this is where students learn how to distinguish between, say, which is larger, three fourths or two thirds. Going on to some other types of rational numbers, ordering decimals is covered in Grades 4 and 5. In Grade 4, it’s in standard 4.NF.7, and in Grade 5, it’s in standard 5.NBT.3b, where, again, students learn to order decimals such as 7.6 compared to 7.34. And then here, in Grade 6, the notion of integers is introduced. So, students learn in previous standards, in this grade, to order integers in standard 6.NS.6c and standard 6.NS.7a, where again, they learn to distinguish, for example, that negative 12 will be less than negative 4. Let’s look at the idea of absolute value. This actually occurs right before this standard in standard 6.NS.7c, and it states: understand the absolute value of a rational number as its distance from zero on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. So, basically, what this is saying is that absolute value simply refers to the distance from zero regardless of the direction, and there are real-world contexts for this. So, here we have two numbers, negative 4 and 4. Both are the same distance from zero. But we have a little bit of a problem because, obviously, this is not a true statement. Negative 4 is not equal to positive 4. So, mathematically we need some type of symbolism, something to adjust this, and the symbol for absolute value would be two bars, something like this that again, indicates that we’re talking about the distance of 4. So, then, that would actually make this a true statement because we are talking only about the distance from zero, and both of these numbers are four away from zero, just in different directions. Let’s look at a real-life context. Let’s say I’m at home, and I drive 8 miles to work. So, let’s let home be zero. So, if we are talking about location, I’m starting at zero. I’m starting at home, and then I’m going to drive 8 miles to work. But then, I’m going to drive 8 miles in the opposite direction and go back home. So, I’m back where I started. I’m back at zero. I’m back home. Now, that’s a little bit different situation as opposed to distance traveled. This would be like, okay, the odometer on my car. Well, if I drive 8 miles to work and then 8 miles back, my odometer isn’t going to register that I didn’t travel. There’s no more distance, no more mileage on my car. It would be nice if that was the case, but it isn’t. So, what happens here is that, okay, I started home. I go 8 miles to work, but then, when I go back home, I’m still traveling 8 miles even though it’s in a different direction. So, here, I have to indicate that I want just the actual distance traveled. And so, my odometer is going to say that I traveled 16 miles, not zero miles. Now let’s look at the specific standard here, d: distinguish comparisons of absolute value from statements about order. Now, when we’re dealing with positive rational numbers, there really isn’t any confusion between the two ideas, absolute value versus order, because they will be one and the same. So, for example, let’s say we’re talking about a distance of 5. . . 1, 2, 3, 4, 5. So, a distance of 5 and 5 being compared to say, zero, there’s no confusion. Where it does get a little bit confusing for students would be when we’re talking about absolute value and order for negative rational numbers. And the best way to really get a handle on what this is saying is to just look at a few real-life contexts. Okay, in a real-life context, debt implies negative numbers. So, let’s say, on one credit card, I owe $400, and on another credit card I owe $1,000. Now, mathematically, negative 1,000 is less than negative 400. But, here’s where it gets a little bit sticky. A real- life statement might be something like a debt of $1,000 is greater than a debt of $400. Now, here’s where again, there might be a little bit of confusion because we’re saying greater than, even though back here, the comparison between those two numbers is less than, that again, negative 1,000 is less than negative 400. But what you’re really dealing with here in this statement is absolute value because we’re talking about the distance from zero. A debt of $1,000 is going to put me further away from zero than a debt of $400 is. So, again, students need to be aware of the distinction here, that, again, negative 1,000 isn’t greater than negative 400, but the distance from zero is. Let’s take a horizontal number line, and let’s make it vertical and apply a temperature context. Now, colder implies a lower temperature. So, for example, here, negative 40 degrees is less than negative 20 degrees. But a real-life statement might be that 40 below zero is more cold than 20 below zero; again, the confusion with this idea of more cold, even though negative 40 is less than negative 20. But again, what we’re dealing with here is absolute value. We’re talking about the distance from zero where, if the temperature is minus 40 degrees, I am, in fact, further away from zero than I would be at negative 20 degrees. Let’s take one more context, again, a vertical number line application where we’re talking about depth. Now, depth implies the distance from the surface of the water in some contexts. So, in this scenario, a depth of negative 4,000 is less than negative 2,000. But then, when we talk about this in a real-life kind of way, we say that 4,000 feet below sea level is a greater depth than 2,000 feet below sea level; again, this idea of greater. But again, we are dealing with absolute value even though absolute value is never actually stated in the sentence. But, again, what this is really saying is that negative, a depth of 4,000 is further away from zero, from the surface, than a depth of 2,000. So, again, these are just a few examples of the care that needs to be taken when you’re dealing with comparisons from an absolute value perspective versus talking about comparing the two numbers numerically. Again, just be more careful with negative rational numbers because that is where it might get a little bit confusing. Students understand that in one case, you’re dealing with absolute value; in the other case, you’re making a comparison between the two numbers. Related Standards referred to in video 4.NF.2 Extend understanding of fraction equivalence and ordering. Numbers and Operations - Fractions: Domain Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. (Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.) 5.NBT.3b Understand the place value system. Numbers and Operations Base Ten: Domain Read, write, and compare decimals to thousandths. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. 6.NS.7a Apply and extend previous understandings of numbers to the system of rational numbers. The Number System: Domain Understand ordering and absolute value of rational numbers. Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right. 6.NS.7c Apply and extend previous understandings of numbers to the system of rational numbers. The Number System: Domain Understand ordering and absolute value of rational numbers. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write \|–30\| = 30 to describe the size of the debt in dollars. The Number System Apply and extend previous understandings of numbers to the system of rational numbers. Understand ordering and absolute value of rational numbers. Distinguish comparisons of absolute value from statements about order. Related Resources The Problem with Math is English Mathematics School Improvement and Accountability Success Stories Follow on Twitter Terms of use (PDF file) The contents of this site were developed under grant number S283B120009 from the U.S. Department of Education. However, the contents do not necessarily represent the policy of the Department of Education, and you should not assume endorsement by the federal government. Contact \| Terms of Use \| Privacy Policy Copyright ©2025 American Institutes for Research®. All rights reserved. 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3352	https://en.wikipedia.org/wiki/Aluminium	Jump to content Search Contents 1 Physical characteristics 1.1 Isotopes 1.2 Electron shell 1.3 Bulk 2 Chemistry 2.1 Inorganic compounds 2.2 Organoaluminium compounds and related hydrides 3 Natural occurrence 3.1 Space 3.2 Earth 4 History 5 Etymology 6 Naming and spelling history 6.1 Early proposals (1808–1812) 6.2 19th-century spelling and usage 6.3 20th-century standardization and regional usage 6.4 Other proposed names 7 Production and refinement 7.1 Bayer process 7.2 Hall–Héroult process 7.3 Recycling 8 Applications 8.1 Metal 8.2 Compounds 9 Biology 9.1 Toxicity 9.2 Effects 9.3 Exposure routes 9.4 Treatment 10 Environmental effects 11 See also 12 Notes 13 References 14 Bibliography 15 Further reading 16 External links Aluminium Afrikaans Alemannisch አማርኛ अंगिका العربية Aragonés Արեւմտահայերէն Armãneashti অসমীয়া Asturianu Avañe'ẽ Azərbaycanca تۆرکجه Basa Bali বাংলা 閩南語 / Bn-lm-gí Башҡортса Беларуская Беларуская (тарашкевіца) भोजपुरी Bikol Central Български བོད་ཡིག Bosanski Brezhoneg Буряад Català Чӑвашла Cebuano Čeština ChiTumbuka Corsu Cymraeg Dansk الدارجة Davvisámegiella Deutsch Diné bizaad Eesti Ελληνικά Эрзянь Español Esperanto Estremeñu Euskara فارسی Fiji Hindi Føroyskt Français Frysk Furlan Gaeilge Gaelg Gàidhlig Galego Gĩkũyũ ગુજરાતી 客家語 / Hak-k-ngî Хальмг 한국어 Hawaiʻi Հայերեն हिन्दी Hrvatski Ido Ilokano Bahasa Indonesia Interlingua Interlingue Ирон IsiZulu Íslenska Italiano עברית Jawa Kabɩyɛ ಕನ್ನಡ Kapampangan ქართული Қазақша Kernowek Ikinyarwanda Kiswahili Коми Kreyòl ayisyen Kriyòl gwiyannen Kurdî Кыргызча Кырык мары Ladin ລາວ Latina Latviešu Lëtzebuergesch Лезги Lietuvių Ligure Limburgs Livvinkarjala La .lojban. 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Chemical element with atomic number 13 (Al) Aluminium, 13Al \| Aluminium \| \| Pronunciation \| aluminium: /ˌæljəˈmɪniəm/ ⓘ(AL-yə-MIN-ee-əm) aluminum: /əˈluːmənəm/ ⓘ1 \| \| Alternative name \| Aluminum (U.S., Canada) \| \| Appearance \| Silvery gray metallic \| \| Standard atomic weight Ar°(Al) \| \| 26.9815384±0.0000003 26.982±0.001 (abridged) \| \| \| \| Aluminium in the periodic table \| \| \| \| \| \| --- \| Hydrogen \| \| Helium \| \| Lithium \| Beryllium \| \| Boron \| Carbon \| Nitrogen \| Oxygen \| Fluorine \| Neon \| \| Sodium \| Magnesium \| \| Aluminium \| Silicon \| Phosphorus \| Sulfur \| Chlorine \| Argon \| \| Potassium \| Calcium \| \| Scandium \| Titanium \| Vanadium \| Chromium \| Manganese \| Iron \| Cobalt \| Nickel \| Copper \| Zinc \| Gallium \| Germanium \| Arsenic \| Selenium \| Bromine \| Krypton \| \| Rubidium \| Strontium \| \| \| Yttrium \| Zirconium \| Niobium \| Molybdenum \| Technetium \| Ruthenium \| Rhodium \| Palladium \| Silver \| Cadmium \| Indium \| Tin \| Antimony \| Tellurium \| Iodine \| Xenon \| \| Caesium \| Barium \| Lanthanum \| Cerium \| Praseodymium \| Neodymium \| Promethium \| Samarium \| Europium \| Gadolinium \| Terbium \| Dysprosium \| Holmium \| Erbium \| Thulium \| Ytterbium \| Lutetium \| Hafnium \| Tantalum \| Tungsten \| Rhenium \| Osmium \| Iridium \| Platinum \| Gold \| Mercury (element) \| Thallium \| Lead \| Bismuth \| Polonium \| Astatine \| Radon \| \| Francium \| Radium \| Actinium \| Thorium \| Protactinium \| Uranium \| Neptunium \| Plutonium \| Americium \| Curium \| Berkelium \| Californium \| Einsteinium \| Fermium \| Mendelevium \| Nobelium \| Lawrencium \| Rutherfordium \| Dubnium \| Seaborgium \| Bohrium \| Hassium \| Meitnerium \| Darmstadtium \| Roentgenium \| Copernicium \| Nihonium \| Flerovium \| Moscovium \| Livermorium \| Tennessine \| Oganesson \| B↑Al↓Ga magnesium ← aluminium → silicon \| \| Atomic number (Z) \| 13 \| \| Group \| group 13 (boron group) \| \| Period \| period 3 \| \| Block \| p-block \| \| Electron configuration \| [Ne] 3s2 3p1 \| \| Electrons per shell \| 2, 8, 3 \| \| Physical properties \| \| Phase at STP \| solid \| \| Melting point \| 933.47 K (660.32 °C, 1220.58 °F) \| \| Boiling point \| 2743 K (2470 °C, 4478 °F) \| \| Density (at 20 °C) \| 2.699 g/cm3 \| \| when liquid (at m.p.) \| 2.375 g/cm3 \| \| Heat of fusion \| 10.71 kJ/mol \| \| Heat of vaporization \| 284 kJ/mol \| \| Molar heat capacity \| 24.20 J/(mol·K) \| \| Vapor pressure \| P (Pa) \| 1 \| 10 \| 100 \| 1 k \| 10 k \| 100 k \| --- --- --- \| at T (K) \| 1482 \| 1632 \| 1817 \| 2054 \| 2364 \| 2790 \| \| \| Atomic properties \| \| Oxidation states \| common: +3 −2, −1, 0, +1, +2 \| \| Electronegativity \| Pauling scale: 1.61 \| \| Ionization energies \| 1st: 577.5 kJ/mol 2nd: 1816.7 kJ/mol 3rd: 2744.8 kJ/mol (more) \| \| Atomic radius \| empirical: 143 pm \| \| Covalent radius \| 121±4 pm \| \| Van der Waals radius \| 184 pm \| \| Spectral lines of aluminium \| \| Other properties \| \| Natural occurrence \| primordial \| \| Crystal structure \| face-centered cubic (fcc) (cF4) \| \| Lattice constant \| a = 404.93 pm (at 20 °C) \| \| Thermal expansion \| 22.87×10−6/K (at 20 °C) \| \| Thermal conductivity \| 237 W/(m⋅K) \| \| Electrical resistivity \| 26.5 nΩ⋅m (at 20 °C) \| \| Magnetic ordering \| paramagnetic \| \| Molar magnetic susceptibility \| +16.5×10−6 cm3/mol \| \| Young's modulus \| 70 GPa \| \| Shear modulus \| 26 GPa \| \| Bulk modulus \| 76 GPa \| \| Speed of sound thin rod \| (rolled) 5000 m/s (at r.t.) \| \| Poisson ratio \| 0.35 \| \| Mohs hardness \| 2.75 \| \| Vickers hardness \| 160–350 MPa \| \| Brinell hardness \| 160–550 MPa \| \| CAS Number \| 7429-90-5 \| \| History \| \| Naming \| from alumine, obsolete name for alumina \| \| Prediction \| Antoine Lavoisier (1782) \| \| Discovery \| Hans Christian Ørsted (1824) \| \| Named by \| Humphry Davy (1812[a]) \| \| Isotopes of aluminium v e \| \| \| \| \| Main isotopes \| Decay \| --- \| \| Isotope \| abundance \| half-life (t1/2) \| mode \| product \| \| 26Al \| trace \| 7.17×105 y \| β+ \| 26Mg \| \| 27Al \| 100% \| stable \| \| \| \| \| Category: Aluminium view talk edit \| references \| Aluminium (the Commonwealth and preferred IUPAC name) or aluminum in North American English is a chemical element; it has symbol Al and atomic number 13. It has a density lower than other common metals, about one-third that of steel. Aluminium has a great affinity towards oxygen, forming a protective layer of oxide on the surface when exposed to air. It visually resembles silver, both in its color and in its great ability to reflect light. It is soft, nonmagnetic, and ductile. It has one stable isotope, 27Al, which is highly abundant, making aluminium the 12th-most abundant element in the universe. The radioactivity of 26Al leads to it being used in radiometric dating. Chemically, aluminium is a post-transition metal in the boron group; as is common for the group, aluminium forms compounds primarily in the +3 oxidation state. The aluminium cation Al3+ is small and highly charged; as such, it has more polarizing power, and bonds formed by aluminium have a more covalent character. The strong affinity of aluminium for oxygen leads to the common occurrence of its oxides in nature. Aluminium is found on Earth primarily in rocks in the crust, where it is the third-most abundant element, after oxygen and silicon, rather than in the mantle, and virtually never as the free metal. It is obtained industrially by mining bauxite, a sedimentary rock rich in aluminium minerals. The discovery of aluminium was announced in 1825 by Danish physicist Hans Christian Ørsted. The first industrial production of aluminium was initiated by French chemist Henri Étienne Sainte-Claire Deville in 1856. Aluminium became much more available to the public with the Hall–Héroult process developed independently by French engineer Paul Héroult and American engineer Charles Martin Hall in 1886, and the mass production of aluminium led to its extensive use in industry and everyday life. In 1954, aluminium became the most produced non-ferrous metal, surpassing copper. In the 21st century, most aluminium was consumed in transportation, engineering, construction, and packaging in the United States, Western Europe, and Japan. The standard atomic weight of aluminium is low in comparison with many other metals,[b] giving it the low density responsible for many of its uses. Despite its prevalence in the environment, no living organism is known to metabolize aluminium salts, but aluminium is well tolerated by plants and animals. Because of the abundance of these salts, the potential for a biological role for them is of interest, and studies are ongoing. Physical characteristics [edit] Isotopes [edit] Main article: Isotopes of aluminium Aluminium has one stable isotope, 27Al, which comprises virtually all of the naturally-occurring element. This is common for elements with an odd atomic number.[c] It is therefore a mononuclidic element for standard atomic weight, which is determined completely by that isotope. Aluminium is useful in nuclear magnetic resonance (NMR), as its single stable isotope (though quadrupolar) has a high NMR sensitivity. All other isotopes of aluminium are radioactive. The most stable of these is 26Al, with a half-life of 717,000 years: while it was present along with stable 27Al in the interstellar medium from which the Solar System formed (believe to have been produced by stellar nucleosynthesis also), no detectable amount could have survived the time since the formation of the planet. However, minute traces of 26Al are still produced from decay of argon in the atmosphere induced by ionizing radiation of cosmic rays. The ratio of 26Al to 10Be has been used for radiodating of geological processes over 105 to 106 year time scales, in particular transport, deposition, sediment storage, burial times, and erosion. Most meteorite scientists believe that the energy released by the decay of 26Al was responsible for the melting and differentiation of some asteroids after their formation 4.55 billion years ago. The other known isotopes of aluminium, with mass numbers ranging from 20 to 43, all have half-lives less than 7 minutes, as do the four detected metastable states. Electron shell [edit] An aluminium atom has 13 electrons, arranged in an electron configuration of [Ne] 3s2 3p1, with three electrons beyond a stable noble gas configuration. Accordingly, the combined first three ionization energies of aluminium are far lower than the fourth ionization energy alone. Such an electron configuration is shared with the other well-characterized members of its group, boron, gallium, indium, and thallium; it is also expected for nihonium. Aluminium can surrender its three outermost electrons in many chemical reactions (see below). The electronegativity of aluminium is 1.61 (Pauling scale). A free aluminium atom has a radius of 143 pm. With the three outermost electrons removed, the radius shrinks to 39 pm for a 4-coordinated atom or 53.5 pm for a 6-coordinated atom. At standard temperature and pressure, aluminium atoms (when not affected by atoms of other elements) form a face-centered cubic crystal system bound by metallic bonding provided by atoms' outermost electrons; hence aluminium (at these conditions) is a metal. This crystal system is shared by many other metals, such as lead and copper; the size of a unit cell of aluminium is comparable to that of those other metals. The system, however, is not shared by the other members of its group: boron has ionization energies too high to allow metallization, thallium has a hexagonal close-packed structure, and gallium and indium have unusual structures that are not close-packed like those of aluminium and thallium. The few electrons that are available for metallic bonding in aluminium are a probable cause for it being soft with a low melting point and low electrical resistivity. Bulk [edit] Aluminium metal has an appearance ranging from silvery white to dull gray depending on its surface roughness.[d] Aluminium mirrors provides high reflectivity for light in the ultraviolet, visible (on par with silver), and the far infrared region. Aluminium is also good at reflecting solar radiation, although prolonged exposure to sunlight in air can deteriorate the reflectivity of the metal; this may be prevented if aluminium is anodized, which adds a protective layer of oxide on the surface. The density of aluminium is 2.70 g/cm3, about 1/3 that of steel, much lower than other commonly encountered metals, making aluminium parts easily identifiable through their lightness. Aluminium's low density compared to most other metals arises from the fact that its unit cell size is relatively large in proportion to the number of nucleons. The only lighter metals are the metals of groups 1 and 2, which apart from beryllium and magnesium are too reactive for structural use (and beryllium is very toxic). Aluminium is not as strong or stiff as steel, but the low density makes up for this in the aerospace industry and for many other applications where light weight and relatively high strength are crucial. Pure aluminium is quite soft and lacking in strength. In most applications various aluminium alloys are used instead because of their higher strength and hardness. The yield strength of pure aluminium is 7–11 MPa, while aluminium alloys have yield strengths ranging from 200 MPa to 600 MPa. Aluminium is ductile, with a percent elongation of 50–70%, and malleable allowing it to be easily drawn and extruded. It is also easily machined and cast. Aluminium is an excellent thermal and electrical conductor, and the amount of aluminum required to match the same amperage in copper weighs only half as much. Aluminium is capable of superconductivity, with a superconducting critical temperature of 1.2 kelvin and a critical magnetic field of about 100 gauss (10 milliteslas). It is paramagnetic and thus essentially unaffected by static magnetic fields. The high electrical conductivity, however, means that it is strongly affected by alternating magnetic fields through the induction of eddy currents. Chemistry [edit] Main article: Compounds of aluminium Aluminium combines characteristics of pre- and post-transition metals. Since it has few available electrons for metallic bonding, like the heavier group 13 elements, it has the characteristic physical properties of a post-transition metal, with longer-than-expected interatomic distances. Furthermore, as Al3+ is a small and highly charged cation, it is strongly polarizing and bonding in aluminium compounds tends towards covalency; this behavior is similar to that of beryllium (Be2+), and the two display an example of a diagonal relationship. The underlying core under aluminium's valence shell is that of the preceding noble gas, whereas those of the heavier group 13 elements gallium, indium, thallium, and nihonium also include a filled d-subshell and in some cases a filled f-subshell. Hence, the inner electrons of aluminium shield the valence electrons almost completely, unlike those of the heavier group 13 elements. As such, aluminium is the most electropositive metal in its group, and its hydroxide is in fact more basic than that of gallium.[e] Aluminium also bears minor similarities to the boron (a metalloid) in the same group: AlX3 compounds are valence isoelectronic to BX3 compounds (they have the same valence electronic structure), and both behave as Lewis acids and readily form adducts. Additionally, one of the main motifs of boron chemistry is regular icosahedral structures, and aluminium forms an important part of many icosahedral quasicrystal alloys, including the Al–Zn–Mg class. Aluminium has a high chemical affinity to oxygen, which renders it suitable for use as a reducing agent in the thermite reaction. A fine powder of aluminium reacts explosively on contact with liquid oxygen; under normal conditions, however, aluminium forms a thin oxide layer (~5 nm at room temperature) that protects the metal from further corrosion by oxygen, water, or dilute acid, a process termed passivation. Aluminium is not attacked by oxidizing acids because of its passivation. This allows aluminium to be used to store reagents such as nitric acid, concentrated sulfuric acid, and some organic acids. In hot concentrated hydrochloric acid, aluminium reacts with water with evolution of hydrogen, and in aqueous sodium hydroxide or potassium hydroxide at room temperature to form aluminates—protective passivation under these conditions is negligible. Aqua regia also dissolves aluminium. Aluminium is corroded by dissolved chlorides, such as common sodium chloride. The oxide layer on aluminium is also destroyed by contact with mercury due to amalgamation or with salts of some electropositive metals. As such, the strongest aluminium alloys are less corrosion-resistant due to galvanic reactions with alloyed copper, and aluminium's corrosion resistance is greatly reduced by aqueous salts, particularly in the presence of dissimilar metals. Aluminium reacts with most nonmetals upon heating, forming compounds such as aluminium nitride (AlN), aluminium sulfide (Al2S3), and the aluminium halides (AlX3). It also forms a wide range of intermetallic compounds involving metals from every group on the periodic table. Inorganic compounds [edit] The vast majority of compounds, including all aluminium-containing minerals and all commercially significant aluminium compounds, feature aluminium in the oxidation state 3+. The coordination number of such compounds varies, but generally Al3+ is either six- or four-coordinate. Almost all compounds of aluminium(III) are colorless. In aqueous solution, Al3+ exists as the hexaaqua cation [Al(H2O)6]3+, which has an approximate Ka of 10−5. Such solutions are acidic as this cation can act as a proton donor and progressively hydrolyze until a precipitate of aluminium hydroxide, Al(OH)3, forms. This is useful for clarification of water, as the precipitate nucleates on suspended particles in the water, hence removing them. Increasing the pH even further leads to the hydroxide dissolving again as aluminate, [Al(H2O)2(OH)4]−, is formed. Aluminium hydroxide forms both salts and aluminates and dissolves in acid and alkali, as well as on fusion with acidic and basic oxides. This behavior of Al(OH)3 is termed amphoterism and is characteristic of weakly basic cations that form insoluble hydroxides and whose hydrated species can also donate their protons. One effect of this is that aluminium salts with weak acids are hydrolyzed in water to the aquated hydroxide and the corresponding nonmetal hydride: for example, aluminium sulfide yields hydrogen sulfide. However, some salts like aluminium carbonate exist in aqueous solution but are unstable as such; and only incomplete hydrolysis takes place for salts with strong acids, such as the halides, nitrate, and sulfate. For similar reasons, anhydrous aluminium salts cannot be made by heating their "hydrates": hydrated aluminium chloride is in fact not AlCl3·6H2O but [Al(H2O)6]Cl3, and the Al–O bonds are so strong that heating is not sufficient to break them and form Al–Cl bonds. This reaction is observed instead: : 2[Al(H2O)6]Cl3 heat→ Al2O3 + 6 HCl + 9 H2O All four trihalides are well known. Unlike the structures of the three heavier trihalides, aluminium fluoride (AlF3) features six-coordinate aluminium, which explains its involatility and insolubility as well as high heat of formation. Each aluminium atom is surrounded by six fluorine atoms in a distorted octahedral arrangement, with each fluorine atom being shared between the corners of two octahedra. Such {AlF6} units also exist in complex fluorides such as cryolite, Na3AlF6.[f] AlF3 melts at 1,290 °C (2,354 °F) and is made by reaction of aluminium oxide with hydrogen fluoride gas at 700 °C (1,300 °F). With heavier halides, the coordination numbers are lower. The other trihalides are dimeric or polymeric with tetrahedral four-coordinate aluminium centers.[g] Aluminium trichloride (AlCl3) has a layered polymeric structure below its melting point of 192.4 °C (378 °F) but transforms on melting to Al2Cl6 dimers. At higher temperatures those increasingly dissociate into trigonal planar AlCl3 monomers similar to the structure of BCl3. Aluminium tribromide and aluminium triiodide form Al2X6 dimers in all three phases and hence do not show such significant changes of properties upon phase change. These materials are prepared by treating aluminium with the halogen. The aluminium trihalides form many addition compounds or complexes; their Lewis acidic nature makes them useful as catalysts for the Friedel–Crafts reactions. Aluminium trichloride has major industrial uses involving this reaction, such as in the manufacture of anthraquinones and styrene; it is also often used as the precursor for many other aluminium compounds and as a reagent for converting nonmetal fluorides into the corresponding chlorides (a transhalogenation reaction). Aluminium forms one stable oxide with the chemical formula Al2O3, commonly called alumina. It can be found in nature in the mineral corundum, α-alumina; there is also a γ-alumina phase. Its crystalline form, corundum, is very hard (Mohs hardness 9), has a high melting point of 2,045 °C (3,713 °F), has very low volatility, is chemically inert, and a good electrical insulator, it is often used in abrasives (such as sandpaper), as a refractory material, and in ceramics, as well as being the starting material for the electrolytic production of aluminium. Sapphire and ruby are impure corundum contaminated with trace amounts of other metals. The two main oxide-hydroxides, AlO(OH), are boehmite and diaspore. There are three main trihydroxides: bayerite, gibbsite, and nordstrandite, which differ in their crystalline structure (polymorphs). Many other intermediate and related structures are also known. Most of these Al-O-OH systems are produced from ores by a variety of wet processes using acid and base. Heating the hydroxides leads to formation of corundum. These materials are of central importance to the production of aluminium and are themselves extremely useful. Some mixed oxide phases are also very useful, such as spinel (MgAl2O4), Na-β-alumina (NaAl11O17), and tricalcium aluminate (Ca3Al2O6, an important mineral phase in Portland cement). The only stable chalcogenides under normal conditions are aluminium sulfide (Al2S3), selenide (Al2Se3), and telluride (Al2Te3). All three are prepared by direct reaction of their elements at about 1,000 °C (1,800 °F) and quickly hydrolyze completely in water to yield aluminium hydroxide and the respective hydrogen chalcogenide. As aluminium is a small atom relative to these chalcogens, these have four-coordinate tetrahedral aluminium with various polymorphs having structures related to wurtzite, with two-thirds of the possible metal sites occupied either in an orderly (α) or random (β) fashion; the sulfide also has a γ form related to γ-alumina, and an unusual high-temperature hexagonal form where half the aluminium atoms have tetrahedral four-coordination and the other half have trigonal bipyramidal five-coordination. Four pnictides – aluminium nitride (AlN), aluminium phosphide (AlP), aluminium arsenide (AlAs), and aluminium antimonide (AlSb) – are known. They are all III-V semiconductors isoelectronic to silicon and germanium, all of which but AlN have the zinc blende structure. All four can be made by high-temperature (and possibly high-pressure) direct reaction of their component elements. Aluminium alloys well with most other metals (with the exception of most alkali metals and group 13 metals) and over 150 intermetallics with other metals are known. Preparation involves heating fixed metals together in certain proportion, followed by gradual cooling and annealing. Bonding in them is predominantly metallic and the crystal structure primarily depends on efficiency of packing. There are few compounds with lower oxidation states. A few aluminium(I) compounds exist: AlF, AlCl, AlBr, and AlI exist in the gaseous phase when the respective trihalide is heated with aluminium, and at cryogenic temperatures. A stable derivative of aluminium monoiodide is the cyclic adduct formed with triethylamine, Al4I4(NEt3)4. Al2O and Al2S also exist but are very unstable. Very simple aluminium(II) compounds are invoked or observed in the reactions of Al metal with oxidants. For example, aluminium monoxide, AlO, has been detected in the gas phase after explosion and in stellar absorption spectra. More thoroughly investigated are compounds of the formula R4Al2 which contain an Al–Al bond and where R is a large organic ligand. Organoaluminium compounds and related hydrides [edit] Main article: Organoaluminium chemistry A variety of compounds of empirical formula AlR3 and AlR1.5Cl1.5 exist. The aluminium trialkyls and triaryls are reactive, volatile, and colorless liquids or low-melting solids. They catch fire spontaneously in air and react with water, thus necessitating precautions when handling them. They often form dimers, unlike their boron analogues, but this tendency diminishes for branched-chain alkyls (e.g. Pri, Bui, Me3CCH2); for example, triisobutylaluminium exists as an equilibrium mixture of the monomer and dimer. These dimers, such as trimethylaluminium (Al2Me6), usually feature tetrahedral Al centers formed by dimerization with some alkyl group bridging between both aluminium atoms. They are hard acids and react readily with ligands, forming adducts. In industry, they are mostly used in alkene insertion reactions, as discovered by Karl Ziegler, most importantly in "growth reactions" that form long-chain unbranched primary alkenes and alcohols, and in the low-pressure polymerization of ethene and propene. There are also some heterocyclic and cluster organoaluminium compounds involving Al–N bonds. The industrially most important aluminium hydride is lithium aluminium hydride (LiAlH4), which is used as a reducing agent in organic chemistry. It can be produced from lithium hydride and aluminium trichloride. The simplest hydride, aluminium hydride or alane, is not as important. It is a polymer with the formula (AlH3)n, in contrast to the corresponding boron hydride that is a dimer with the formula (BH3)2. Natural occurrence [edit] See also: List of countries by bauxite production Space [edit] Aluminium's per-particle abundance in the Solar System is 3.15 ppm (parts per million).[h] It is the twelfth most abundant of all elements and third most abundant among the elements that have odd atomic numbers, after hydrogen and nitrogen. The only stable isotope of aluminium, 27Al, is the eighteenth most abundant nucleus in the universe. It is created almost entirely after fusion of carbon in massive stars that will later become Type II supernovas: this fusion creates 26Mg, which upon capturing free protons and neutrons, becomes aluminium. Some smaller quantities of 27Al are created in hydrogen burning shells of evolved stars, where 26Mg can capture free protons. Essentially all aluminium now in existence is 27Al. 26Al was present in the early Solar System with abundance of 0.005% relative to 27Al but its half-life of 728,000 years is too short for any original nuclei to survive; 26Al is therefore extinct. Unlike for 27Al, hydrogen burning is the primary source of 26Al, with the nuclide emerging after a nucleus of 25Mg catches a free proton. However, the trace quantities of 26Al that do exist are the most common gamma ray emitter in the interstellar gas; if the original 26Al were still present, gamma ray maps of the Milky Way would be brighter. Earth [edit] Overall, the Earth is about 1.59% aluminium by mass (seventh in abundance by mass). Aluminium occurs in greater proportion in the Earth's crust than in the universe at large. This is because aluminium easily forms the oxide and becomes bound into rocks and stays in the Earth's crust, while less reactive metals sink to the core. In the Earth's crust, aluminium is the most abundant metallic element (8.23% by mass) and the third most abundant of all elements (after oxygen and silicon). A large number of silicates in the Earth's crust contain aluminium. In contrast, the Earth's mantle is only 2.38% aluminium by mass. Aluminium also occurs in seawater at a concentration of 0.41 μg/kg. Because of its strong affinity for oxygen, aluminium is almost never found in the elemental state; instead it is found in oxides or silicates. Feldspars, the most common group of minerals in the Earth's crust, are aluminosilicates. Aluminium also occurs in the minerals beryl, cryolite, garnet, spinel, and turquoise. Impurities in alumina yield gemstones: for example, chromium yields ruby and iron yields sapphire. Native aluminium metal is extremely rare and can only be found as a minor phase in low oxygen fugacity environments, such as the interiors of certain volcanoes. Native aluminium has been reported in cold seeps in the northeastern continental slope of the South China Sea. It is possible that these deposits resulted from bacterial reduction of tetrahydroxoaluminate Al(OH)4−. Although aluminium is a common and widespread element, not all aluminium minerals are economically viable sources of the metal. Almost all metallic aluminium is produced from the ore bauxite (AlOx(OH)3–2x). Bauxite occurs as a weathering product of low iron and silica bedrock in tropical climatic conditions. In 2017, most bauxite was mined in Australia, China, Guinea, and India. History [edit] Main article: History of aluminium The history of aluminium has been shaped by usage of alum. The first written record of alum, made by Greek historian Herodotus, dates back to the 5th century BCE. The ancients are known to have used alum as a dyeing mordant and for city defense as a fire-resistant coating for wood. After the Crusades, alum, an indispensable good in the European fabric industry, was a subject of international commerce; it was imported to Europe from the eastern Mediterranean until the mid-15th century. The nature of alum remained unknown. Around 1530, Swiss physician Paracelsus suggested alum was a salt of an earth of alum. In 1595, German doctor and chemist Andreas Libavius experimentally confirmed this. In 1722, German chemist Friedrich Hoffmann announced his belief that the base of alum was a distinct earth. In 1754, German chemist Andreas Sigismund Marggraf synthesized alumina by boiling clay in sulfuric acid and subsequently adding potash. Attempts to produce aluminium date back to 1760. The first successful attempt, however, was completed in 1824 by Danish physicist and chemist Hans Christian Ørsted. He reacted anhydrous aluminium chloride with potassium amalgam, yielding a lump of metal looking similar to tin. He presented his results and demonstrated a sample of the new metal in 1825. In 1827, German chemist Friedrich Wöhler repeated Ørsted's experiments but did not identify any aluminium. (The reason for this inconsistency was only discovered in 1921.) He conducted a similar experiment in the same year by mixing anhydrous aluminium chloride with potassium (the Wöhler process) and produced a powder of aluminium. In 1845, he was able to produce small pieces of the metal and described some physical properties of this metal. For many years thereafter, Wöhler was credited as the discoverer of aluminium. As Wöhler's method could not yield great quantities of aluminium, the metal remained rare; its cost exceeded that of gold. The first industrial production of aluminium was established in 1856 by French chemist Henri Etienne Sainte-Claire Deville and companions. Deville had discovered that aluminium trichloride could be reduced by sodium, which was more convenient and less expensive than potassium, which Wöhler had used. Even then, aluminium was still not of great purity and produced aluminium differed in properties by sample. Because of its electricity-conducting capacity, aluminium was used as the cap of the Washington Monument, completed in 1885, the tallest building in the world at the time. The non-corroding metal cap was intended to serve as a lightning rod peak. The first industrial large-scale production method was independently developed in 1886 by French engineer Paul Héroult and American engineer Charles Martin Hall; it is now known as the Hall–Héroult process. The Hall–Héroult process converts alumina into metal. Austrian chemist Carl Joseph Bayer discovered a way of purifying bauxite to yield alumina, now known as the Bayer process, in 1889. Modern production of aluminium is based on the Bayer and Hall–Héroult processes. As large-scale production caused aluminium prices to drop, the metal became widely used in jewelry, eyeglass frames, optical instruments, tableware, and foil, and other everyday items in the 1890s and early 20th century. Aluminium's ability to form hard yet light alloys with other metals provided the metal with many uses at the time. During World War I, major governments demanded large shipments of aluminium for light strong airframes; during World War II, demand by major governments for aviation was even higher. From the early 20th century to 1980, the aluminium industry was characterized by cartelization, as aluminium firms colluded to keep prices high and stable. The first aluminium cartel, the Aluminium Association, was founded in 1901 by the Pittsburgh Reduction Company (renamed Alcoa in 1907) and Aluminium Industrie AG. The British Aluminium Company, Produits Chimiques d’Alais et de la Camargue, and Société Electro-Métallurgique de Froges also joined the cartel. By the mid-20th century, aluminium had become a part of everyday life and an essential component of housewares. In 1954, production of aluminium surpassed that of copper,[i] historically second in production only to iron, making it the most produced non-ferrous metal. During the mid-20th century, aluminium emerged as a civil engineering material, with building applications in both basic construction and interior finish work, and increasingly being used in military engineering, for both airplanes and armored vehicle engines. Earth's first artificial satellite, launched in 1957, consisted of two separate aluminium semi-spheres joined and all subsequent space vehicles have used aluminium to some extent. The aluminium can was invented in 1956 and employed as a storage for drinks in 1958. Throughout the 20th century, the production of aluminium rose rapidly: while the world production of aluminium in 1900 was 6,800 metric tons, the annual production first exceeded 100,000 metric tons in 1916; 1,000,000 tons in 1941; 10,000,000 tons in 1971. In the 1970s, the increased demand for aluminium made it an exchange commodity; it entered the London Metal Exchange, the oldest industrial metal exchange in the world, in 1978. The output continued to grow: the annual production of aluminium exceeded 50,000,000 metric tons in 2013. The real price for aluminium declined from $14,000 per metric ton in 1900 to $2,340 in 1948 (in 1998 United States dollars). Extraction and processing costs were lowered over technological progress and the scale of the economies. However, the need to exploit lower-grade poorer quality deposits and the use of fast increasing input costs (above all, energy) increased the net cost of aluminium; the real price began to grow in the 1970s with the rise of energy cost. Production moved from the industrialized countries to countries where production was cheaper. Production costs in the late 20th century changed because of advances in technology, lower energy prices, exchange rates of the United States dollar, and alumina prices. The BRIC countries' combined share in primary production and primary consumption grew substantially in the first decade of the 21st century. China is accumulating an especially large share of the world's production thanks to an abundance of resources, cheap energy, and governmental stimuli; it also increased its consumption share from 2% in 1972 to 40% in 2010. In the United States, Western Europe, and Japan, most aluminium was consumed in transportation, engineering, construction, and packaging. In 2021, prices for industrial metals such as aluminium have soared to near-record levels as energy shortages in China drive up costs for electricity. Etymology [edit] The names aluminium and aluminum are derived from the word alumine, an obsolete term for alumina,[j] the primary naturally occurring oxide of aluminium. Alumine was borrowed from French, which in turn derived it from alumen, the classical Latin name for alum, the mineral from which it was collected. The Latin word alumen stems from the Proto-Indo-European root alu- meaning "bitter" or "beer". The English name alum does not come directly from Latin, whereas alumine/alumina comes from the Latin word alumen (on declension, alumen changes to alumin-). Naming and spelling history [edit] Early proposals (1808–1812) [edit] British chemist Humphry Davy, who performed a number of experiments aimed to isolate the metal, is credited as the person who named the element. The first name proposed for the metal to be isolated from alum was alumium, which Davy suggested in an 1808 article on his electrochemical research, published in Philosophical Transactions of the Royal Society. It appeared that the name was created from the English word alum and the Latin suffix -ium; but it was customary then to give elements names originating in Latin, so this name was not adopted universally. The name alumium was criticized by contemporary chemists from France, Germany, and Sweden, who insisted the metal should be named for the oxide, alumina, from which it would be isolated. One example was Essai sur la Nomenclature chimique (July 1811), written in French by a Swedish chemist, Jöns Jacob Berzelius, in which the name aluminium is given to the element that would be synthesized from alum.[k] (Another article in the same journal issue also refers to the metal whose oxide is the basis of sapphire, i.e. the same metal, as to aluminium.) A January 1811 summary of one of Davy's lectures at the Royal Society mentioned the name aluminium as a possibility. In 1812, Davy published his chemistry text Elements of Chemical Philosophy in which he used the spelling aluminum. 19th-century spelling and usage [edit] In 1812, British scientist Thomas Young wrote an anonymous review of Davy's book, in which he proposed the name aluminium instead of aluminum, which he thought had a "less classical sound". This name persisted: although the -um spelling was occasionally used in Britain, the American scientific language used -ium from the start. The French have used the spelling aluminium from the start. However, in England and Germany Davy's spelling aluminum was initially used; until Wöhler published his account of the Wöhler process in 1827 in which he used the spelling aluminium[l], which caused that spelling's largely wholesale adoption in England and Germany, with the exception of a small number of what Richards characterized as "patriotic" English chemists that were "averse to foreign innovations" who occasionally still used aluminum. Most scientists throughout the world used -ium in the 19th century; and it was entrenched in several other European languages, such as French, German, and Dutch.[m] In 1828, an American lexicographer, Noah Webster, entered only the aluminum spelling in his American Dictionary of the English Language. In the 1830s, the -um spelling gained usage in the United States; by the 1860s, it had become the more common spelling there outside science. In 1892, Hall used the -um spelling in his advertising handbill for his new electrolytic method of producing the metal, despite his constant use of the -ium spelling in all the patents he filed between 1886 and 1903. It is unknown whether this spelling was introduced by mistake or intentionally, but Hall preferred aluminum since its introduction because it resembled platinum, the name of a prestigious metal. By 1890, both spellings had been common in the United States, the -ium spelling being slightly more common; by 1895, the situation had reversed; by 1900, aluminum had become twice as common as aluminium; in the next decade, the -um spelling dominated American usage. 20th-century standardization and regional usage [edit] In 1925, the American Chemical Society adopted this spelling. The International Union of Pure and Applied Chemistry (IUPAC) adopted aluminium as the standard international name for the element in 1990. In 1993, they recognized aluminum as an acceptable variant; the most recent 2005 edition of the IUPAC nomenclature of inorganic chemistry also acknowledges this spelling. IUPAC official publications use the -ium spelling as primary, and they list both where it is appropriate.[n] Both spellings have coexisted since. Their usage is currently regional: aluminum dominates in the United States and Canada; aluminium is prevalent in the rest of the English-speaking world. Other proposed names [edit] German physicist Ludwig Wilhelm Gilbert had proposed Thonerde-metall, after the German "Thonerde"[o] for alumina, in his Annalen der Physik but that name never caught on at all even in Germany. American chemist Joseph W. Richards[p] in 1891 found just one occurrence of argillium in Swedish, from the French "argille"[q] for clay. Production and refinement [edit] See also: List of countries by primary aluminium production World's largest producing countries of aluminium, 2024 \| Country \| Output(thousand tons) \| \| China \| 43,000 \| \| India \| 4,200 \| \| Russia \| 3,800 \| \| Canada \| 3,300 \| \| United Arab Emirates \| 2,700 \| \| Bahrain \| 1,600 \| \| Australia \| 1,500 \| \| Norway \| 1,300 \| \| Brazil \| 1,100 \| \| Malaysia \| 870 \| \| Iceland \| 780 \| \| United States \| 670 \| \| Other countries \| 6,800 \| \| Total \| 72,000 \| The production of aluminium starts with the extraction of bauxite rock from the ground. The bauxite is processed and transformed using the Bayer process into alumina, which is then processed using the Hall–Héroult process, resulting in the final aluminium. Aluminium production is highly energy-consuming, and so the producers tend to locate smelters in places where electric power is both plentiful and inexpensive. Production of one kilogram of aluminium requires 7 kilograms of oil energy equivalent, as compared to 1.5 kilograms for steel and 2 kilograms for plastic. As of 2024, the world's largest producers of aluminium were China, India, Russia, Canada, and the United Arab Emirates, while China is by far the top producer of aluminium with a world share of over 55%. According to the International Resource Panel's Metal Stocks in Society report, the global per capita stock of aluminium in use in society (i.e. in cars, buildings, electronics, etc.) is 80 kg (180 lb). Much of this is in more-developed countries (350–500 kg (770–1,100 lb) per capita) rather than less-developed countries (35 kg (77 lb) per capita). Bayer process [edit] Main article: Bayer process See also: List of countries by bauxite production Bauxite is converted to alumina by the Bayer process. Bauxite is blended for uniform composition and then is ground fine. The resulting slurry is mixed with a hot solution of sodium hydroxide; the mixture is then treated in a digester vessel at a pressure well above atmospheric, dissolving the aluminium hydroxide in bauxite while converting impurities into relatively insoluble compounds: Al(OH)3 + Na+ + OH− → Na+ + [Al(OH)4]− After this reaction, the slurry is at a temperature above its atmospheric boiling point. It is cooled by removing steam as pressure is reduced. The bauxite residue is separated from the solution and discarded. The solution, free of solids, is seeded with small crystals of aluminium hydroxide; this causes decomposition of the [Al(OH)4]− ions to aluminium hydroxide. After about half of aluminium has precipitated, the mixture is sent to classifiers. Small crystals of aluminium hydroxide are collected to serve as seeding agents; coarse particles are converted to alumina by heating; the excess solution is removed by evaporation, (if needed) purified, and recycled. Hall–Héroult process [edit] Main articles: Hall–Héroult process and Aluminium smelting See also: List of countries by aluminium oxide production The conversion of alumina to aluminium is achieved by the Hall–Héroult process. In this energy-intensive process, a solution of alumina in a molten (940 and 970 °C (1,720 and 1,780 °F)) mixture of cryolite (Na3AlF6) with calcium fluoride is electrolyzed to produce metallic aluminium. The liquid aluminium sinks to the bottom of the solution and is tapped off, and usually cast into large blocks called aluminium billets for further processing. Anodes of the electrolysis cell are made of carbon—the most resistant material against fluoride corrosion—and either bake at the process or are prebaked. The former, also called Söderberg anodes, are less power-efficient and fumes released during baking are costly to collect, which is why they are being replaced by prebaked anodes even though they save the power, energy, and labor to prebake the cathodes. Carbon for anodes should be preferably pure so that neither aluminium nor the electrolyte is contaminated with ash. Despite carbon's resistivity against corrosion, it is still consumed at a rate of 0.4–0.5 kg per each kilogram of produced aluminium. Cathodes are made of anthracite; high purity for them is not required because impurities leach only very slowly. The cathode is consumed at a rate of 0.02–0.04 kg per each kilogram of produced aluminium. A cell is usually terminated after 2–6 years following a failure of the cathode. The Hall–Heroult process produces aluminium with a purity of above 99%. Further purification can be done by the Hoopes process. This process involves the electrolysis of molten aluminium with a sodium, barium, and aluminium fluoride electrolyte. The resulting aluminium has a purity of 99.99%. Electric power represents about 20 to 40% of the cost of producing aluminium, depending on the location of the smelter. Aluminium production consumes roughly 5% of electricity generated in the United States. Because of this, alternatives to the Hall–Héroult process have been researched, but none has turned out to be economically feasible. Recycling [edit] Main article: Aluminium recycling Recovery of the metal through recycling has become an important task of the aluminium industry. Recycling was a low-profile activity until the late 1960s, when the growing use of aluminium beverage cans brought it to public awareness. Recycling involves melting the scrap, a process that requires only 5% of the energy used to produce aluminium from ore, though a significant part (up to 15% of the input material) is lost as dross (ash-like oxide). An aluminium stack melter produces significantly less dross, with values reported below 1%. White dross from primary aluminium production and from secondary recycling operations still contains useful quantities of aluminium that can be extracted industrially. The process produces aluminium billets, together with a highly complex waste material. This waste is difficult to manage. It reacts with water, releasing a mixture of gases including, among others, acetylene, hydrogen sulfide and significant amounts of ammonia. Despite these difficulties, the waste is used as a filler in asphalt and concrete. Its potential for hydrogen production has also been considered and researched. Applications [edit] Metal [edit] See also: Aluminium alloy The global production of aluminium in 2016 was 58.8 million metric tons.[outdated statistic] It exceeded that of any other metal except iron (1,231 million metric tons). Aluminium is almost always alloyed, which markedly improves its mechanical properties, especially when tempered. For example, the common aluminium foils and beverage cans are alloys of 92% to 99% aluminium. The main alloying agents for both wrought and cast aluminium are copper, zinc, magnesium, manganese, and silicon (e.g., duralumin) with the levels of other metals in a few percent by weight. The major uses for aluminium are in: Transportation (automobiles, aircraft, trucks, railway cars, marine vessels, bicycles, spacecraft, etc.). Aluminium is used because of its low density; Packaging (cans, foil, frame, etc.). Aluminium is used because it is non-toxic (see below), non-adsorptive, and splinter-proof; Building and construction (windows, doors, siding, building wire, sheathing, roofing, etc.). Since steel is cheaper, aluminium is used when lightness, corrosion resistance, or engineering features are important; Electricity-related uses (conductor alloys, motors, and generators, transformers, capacitors, etc.). Aluminium is used because it is relatively cheap, highly conductive, has adequate mechanical strength and low density, and resists corrosion; A wide range of household items, from cooking utensils to furniture. Low density, good appearance, ease of fabrication, and durability are the key factors of aluminium usage; Machinery and equipment (processing equipment, pipes, tools). Aluminium is used because of its corrosion resistance, non-pyrophoricity, and mechanical strength. Compounds [edit] The great majority (about 90%) of aluminium oxide is converted to metallic aluminium. Being a very hard material (Mohs hardness 9), alumina is widely used as an abrasive; being extraordinarily chemically inert, it is useful in highly reactive environments such as high pressure sodium lamps. Aluminium oxide is commonly used as a catalyst for industrial processes; e.g. the Claus process to convert hydrogen sulfide to sulfur in refineries and to alkylate amines. Many industrial catalysts are supported by alumina, meaning that the expensive catalyst material is dispersed over a surface of the inert alumina. Another principal use is as a drying agent or absorbent. Several sulfates of aluminium have industrial and commercial application. Aluminium sulfate (in its hydrate form) is produced on the annual scale of several millions of metric tons. About two-thirds is consumed in water treatment. The next major application is in the manufacture of paper. It is also used as a mordant in dyeing, in pickling seeds, deodorizing of mineral oils, in leather tanning, and in production of other aluminium compounds. Two kinds of alum, ammonium alum and potassium alum, were formerly used as mordants and in leather tanning, but their use has significantly declined following availability of high-purity aluminium sulfate. Anhydrous aluminium chloride is used as a catalyst in chemical and petrochemical industries, the dyeing industry, and in synthesis of various inorganic and organic compounds. Aluminium hydroxychlorides are used in purifying water, in the paper industry, and as antiperspirants. Sodium aluminate is used in treating water and as an accelerator of solidification of cement. Many aluminium compounds have niche applications, for example: Aluminium acetate in solution is used as an astringent. Aluminium phosphate is used in the manufacture of glass, ceramic, pulp and paper products, cosmetics, paints, varnishes, and in dental cement. Aluminium hydroxide is used as an antacid, and mordant; it is used also in water purification, the manufacture of glass and ceramics, and in the waterproofing of fabrics. Lithium aluminium hydride is a powerful reducing agent used in organic chemistry. Organoaluminiums are used as Lewis acids and co-catalysts. Methylaluminoxane is a co-catalyst for Ziegler–Natta olefin polymerization to produce vinyl polymers such as polyethene. Aqueous aluminium ions (such as aqueous aluminium sulfate) are used to treat against fish parasites such as Gyrodactylus salaris. In many vaccines, certain aluminium salts serve as an immune adjuvant (immune response booster) to allow the protein in the vaccine to achieve sufficient potency as an immune stimulant. Until 2004, most of the adjuvants used in vaccines were aluminium-adjuvanted. Biology [edit] Despite its widespread occurrence in the Earth's crust, aluminium has no known function in biology. At pH 6–9 (relevant for most natural waters), aluminium precipitates out of water as the hydroxide and is hence not available; most elements behaving this way have no biological role or are toxic. Aluminium sulfate has an LD50 of 6207 mg/kg (oral, mouse), which corresponds to 435 grams (about one pound) for a 70 kg (150 lb) mouse. Toxicity [edit] Aluminium is classified as a non-carcinogen by the United States Department of Health and Human Services.[r] A review published in 1988 said that there was little evidence that normal exposure to aluminium presents a risk to healthy adult, and a 2014 multi-element toxicology review was unable to find deleterious effects of aluminium consumed in amounts not greater than 40 mg/day per kg of body mass. Most aluminium consumed will leave the body in feces, and any that enters the bloodstream will be excreted via urine. Effects [edit] Aluminium, although rarely, can cause vitamin D-resistant osteomalacia, erythropoietin-resistant microcytic anemia, and central nervous system alterations. People with kidney insufficiency are especially at a risk. Chronic ingestion of hydrated aluminium silicates (for excess gastric acidity control) may result in aluminium binding to intestinal contents and increased elimination of other metals, such as iron or zinc; sufficiently high doses (>50 g/day) can cause anemia. During the 1988 Camelford water pollution incident, people in Camelford had their drinking water contaminated with aluminium sulfate for several weeks. A final report into the incident in 2013 concluded it was unlikely that this had caused long-term health problems. Aluminium has been suspected of being a possible cause of Alzheimer's disease, but research into this for over 40 years has found, as of 2018[update], no good evidence of causal effect. Aluminium increases estrogen-related gene expression in human breast cancer cells cultured in the laboratory. In very high doses, aluminium is associated with altered function of the blood–brain barrier. A small percentage of people have contact allergies to aluminium and experience itchy red rashes, headache, muscle pain, joint pain, poor memory, insomnia, depression, asthma, irritable bowel syndrome, or other symptoms upon contact with products containing aluminium. Exposure to powdered aluminium or aluminium welding fumes can cause pulmonary fibrosis. Fine aluminium powder can ignite or explode, posing another workplace hazard. Exposure routes [edit] Food is the main source of aluminium. Drinking water contains more aluminium than solid food; however, aluminium in food may be absorbed more than aluminium from water. Major sources of human oral exposure to aluminium include food (due to its use in food additives, food and beverage packaging, and cooking utensils), drinking water (due to its use in municipal water treatment), and aluminium-containing medications (particularly antacid/antiulcer and buffered aspirin formulations). Dietary exposure in Europeans averages to 0.2–1.5 mg/kg/week but can be as high as 2.3 mg/kg/week. Higher exposure levels of aluminium are mostly limited to plumbers, masons, electrical workers, machinists, and surgeons. Consumption of antacids, antiperspirants, vaccines, and cosmetics provide possible routes of exposure. Consumption of acidic foods or liquids with aluminium enhances aluminium absorption, and maltol has been shown to increase the accumulation of aluminium in nerve and bone tissues. Treatment [edit] In case of suspected sudden intake of a large amount of aluminium, the only treatment is deferoxamine mesylate which may be given to help eliminate aluminium from the body by chelation therapy. However, this should be applied with caution as this reduces not only aluminium body levels, but also those of other metals such as copper or iron. Environmental effects [edit] High levels of aluminium occur near mining sites; small amounts of aluminium are released to the environment at coal-fired power plants or incinerators. Aluminium in the air is washed out by the rain or normally settles down but small particles of aluminium remain in the air for a long time. Acidic precipitation is the main natural factor to mobilize aluminium from natural sources and the main reason for the environmental effects of aluminium; however, the main factor of presence of aluminium in salt and freshwater are the industrial processes that also release aluminium into air. In water, aluminium acts as a toxiс agent on gill-breathing animals such as fish when the water is acidic, in which aluminium may precipitate on gills, which causes loss of plasma- and hemolymph ions leading to osmoregulatory failure. Organic complexes of aluminium may be easily absorbed and interfere with metabolism in mammals and birds, even though this rarely happens in practice. Aluminium is primary among the factors that reduce plant growth on acidic soils. Although it is generally harmless to plant growth in pH-neutral soils, in acid soils the concentration of toxic Al3+ cations increases and disturbs root growth and function. Wheat has developed a tolerance to aluminium, releasing organic compounds that bind to harmful aluminium cations. Sorghum is believed to have the same tolerance mechanism. Aluminium production possesses its own challenges to the environment on each step of the production process. The major challenge is the emission of greenhouse gases. These gases result from electrical consumption of the smelters and the byproducts of processing. The most potent of these gases are perfluorocarbons, namely CF4 and C2F6, from the smelting process. Biodegradation of metallic aluminium is extremely rare; most aluminium-corroding organisms do not directly attack or consume the aluminium, but instead produce corrosive wastes. The fungus Geotrichum candidum can consume the aluminium in compact discs. The bacterium Pseudomonas aeruginosa and the fungus Cladosporium resinae are commonly detected in aircraft fuel tanks that use kerosene-based fuels (not avgas), and laboratory cultures can degrade aluminium. See also [edit] Chemistry portal Aluminium granules Aluminium joining Aluminium–air battery Aluminized steel, for corrosion resistance and other properties Aluminized screen, for display devices Aluminized cloth, to reflect heat Aluminized mylar, to reflect heat Panel edge staining Quantum clock Notes [edit] ^ Davy's 1812 written usage of the word aluminum was predated by other authors' usage of aluminium. However, Davy is often mentioned as the person who named the element; he was the first to coin a name for aluminium: he used alumium in 1808. Other authors did not accept that name, choosing aluminium instead. See below for more details. ^ Most other metals have greater standard atomic weights: for instance, that of iron is 55.845; copper 63.546; lead 207.2. which has consequences for the element's properties (see below) ^ No elements with odd atomic number have more than two stable isotopes, while even-numbered elements from oxygen to lead (atomic numbers 8 to 82) all have more than two. See Even and odd atomic nuclei for more details. ^ The two sides of aluminium foil differ in their luster: one is shiny and the other is dull. The difference is due to the small mechanical damage on the surface of dull side arising from the technological process of aluminium foil manufacturing. Both sides reflect similar amounts of visible light, but the shiny side reflects a far greater share of visible light specularly whereas the dull side almost exclusively diffuses light. Both sides of aluminium foil serve as good reflectors (approximately 86%) of visible light and an excellent reflector (as much as 97%) of medium and far infrared radiation. ^ In fact, aluminium's electropositive behavior, high affinity for oxygen, and highly negative standard electrode potential are all better aligned with those of scandium, yttrium, lanthanum, and actinium, which like aluminium have three valence electrons outside a noble gas core; this series shows continuous trends whereas those of group 13 is broken by the first added d-subshell in gallium and the resulting d-block contraction and the first added f-subshell in thallium and the resulting lanthanide contraction. ^ These should not be considered as [AlF6]3− complex anions as the Al–F bonds are not significantly different in type from the other M–F bonds. ^ Such differences in coordination between the fluorides and heavier halides are not unusual, occurring in SnIV and BiIII, for example; even bigger differences occur between CO2 and SiO2. ^ Abundances in the source are listed relative to silicon rather than in per-particle notation. The sum of all elements per 106 parts of silicon is 2.6682×1010 parts; aluminium comprises 8.410×104 parts. ^ Compare annual statistics of aluminium and copper production by USGS. ^ The spelling alumine comes from French, whereas the spelling alumina comes from Latin. ^ Davy discovered several other elements, including those he named sodium and potassium, after the English words soda and potash. Berzelius referred to them as to natrium and kalium. Berzelius's suggestion was expanded in 1814 with his proposed system of one or two-letter chemical symbols, which are used up to the present day; sodium and potassium have the symbols Na and K, respectively, after their Latin names. ^ Wöhler had previously used aluminium in 1824, when translating a paper by Jöns Jacob Berzelius from Swedish. ^ Some European languages, like Spanish or Italian, use a different suffix from the Latin -um/-ium to form a name of a metal, some, like Polish or Czech, have a different base for the name of the element, and some, like Russian or Greek, do not use the Latin script altogether. ^ For instance, see the November–December 2013 issue of Chemistry International: in a table of (some) elements, the element is listed as "aluminium (aluminum)". ^ a historic spelling, nowadays spelled "Tonerde" ^ founder and later president of the Electrochemical Society ^ nowadays spelled "argile" ^ While aluminium per se is not carcinogenic, Söderberg aluminium production is, as is noted by the International Agency for Research on Cancer, likely due to exposure to polycyclic aromatic hydrocarbons. References [edit] ^ "aluminum". Oxford English Dictionary (Online ed.). Oxford University Press. (Subscription or participating institution membership required.) ^ "Standard Atomic Weights: Aluminium". CIAAW. 2017. ^ a b Prohaska, Thomas; Irrgeher, Johanna; Benefield, Jacqueline; Böhlke, John K.; Chesson, Lesley A.; Coplen, Tyler B.; Ding, Tiping; Dunn, Philip J. H.; Gröning, Manfred; Holden, Norman E.; Meijer, Harro A. J. (May 4, 2022). "Standard atomic weights of the elements 2021 (IUPAC Technical Report)". Pure and Applied Chemistry. doi:10.1515/pac-2019-0603. ISSN 1365-3075. ^ Zhang, Yiming; Evans, Julian R. G.; Yang, Shoufeng (2011). "Corrected Values for Boiling Points and Enthalpies of Vaporization of Elements in Handbooks". J. Chem. Eng. Data. 56 (2): 328–337. doi:10.1021/je1011086. ^ a b c Arblaster, John W. (2018). Selected Values of the Crystallographic Properties of Elements. Materials Park, Ohio: ASM International. ISBN 978-1-62708-155-9. ^ Al(−2) has been observed in Sr14[Al4]2[Ge]3, see Wemdorff, Marco; Röhr, Caroline (2007). "Sr14[Al4]2[Ge]3: Eine Zintl-Phase mit isolierten [Ge]4–- und [Al4]8–-Anionen / Sr14[Al4]2[Ge]3: A Zintl Phase with Isolated [Ge]4–- and [Al4]8– Anions". Zeitschrift für Naturforschung B (in German). 62 (10): 1227. doi:10.1515/znb-2007-1001. S2CID 94972243. ^ Al(–1) has been reported in Na5Al5; see Haopeng Wang; Xinxing Zhang; Yeon Jae Ko; Andrej Grubisic; Xiang Li; Gerd Ganteför; Hansgeorg Schnöckel; Bryan W. Eichhorn; Mal-Soon Lee; P. Jena; Anil K. Kandalam; Boggavarapu Kiran; Kit H. Bowen (2014). "Aluminum Zintl anion moieties within sodium aluminum clusters". The Journal of Chemical Physics. 140 (5). Bibcode:2014JChPh.140e4301W. doi:10.1063/1.4862989. ^ Unstable carbonyl of Al(0) has been detected in reaction of Al2(CH3)6 with carbon monoxide; see Sanchez, Ramiro; Arrington, Caleb; Arrington Jr., C. A. (December 1, 1989). 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External links [edit] Aluminium at Wikipedia's sister projects Definitions from Wiktionary Media from Commons Texts from Wikisource Resources from Wikiversity Aluminium at The Periodic Table of Videos (University of Nottingham) Toxicological Profile for Aluminum (PDF) (September 2008) – 357-page report from the United States Department of Health and Human Services, Public Health Service, Agency for Toxic Substances and Disease Registry Aluminum entry (last reviewed 30 October 2019) in the NIOSH Pocket Guide to Chemical Hazards published by the CDC's National Institute for Occupational Safety and Health Current and historical prices (1998–present) for aluminum futures on the global commodities market The short film Aluminum is available for free viewing and download at the Internet Archive. \| v t e Aluminium compounds \| \| Al(I) \| AlBr AlCl AlF AlI Al2O AlOH \| \| \| --- \| \| Organoaluminium(I) compounds \| Al(C5(CH3)5) \| \| \| Al(II) \| AlB2 AlB12 AlO \| \| Al(III) \| AlAs Al(BH4)3 AlBr3 Al(CN)3 AlCl3 NaAlCl4 AlF3 AlH3 AlI3 AlN Al(NO3)3 Al2(CO3)3 Al(OH)3 Al(OH)2OAc Al(OH)(OAc)2 Al(OAc)3 Al2SO4(OAc)4 AlP AlPO4 AlSb Al(C5H7O2)3 Al(MnO4)3 Al2(MoO4)3 Al2O3 Al2S3 Al2(SO4)3 Al2Se3 Al2Te3 Al2SiO5 AlAsO4 Al4C3 AlOHO Al(OH)2CO2C17H5 NaAlH2(OC2H4OCH3)2 LiAlH2(OC2H4OCH3)2 K2Al2B2O7 K3AlF6 Al(C3H5O3)3 C36H69AlO6 \| \| \| --- \| \| Alums \| (NH4)Al(SO4)2 KAl(SO4)2 NaAl(SO4)2 \| \| Organoaluminium(III) compounds \| (Al(CH3)3)2 poly-AlCH3O (Al(C2H5)3)2 Al(CH2CH(CH3)2)3 Al(C2H5)2Cl Al(C2H5)2CN Al(CH2CH(CH3)2)2H Al(C2H5)2Cl2C2H5Cl Ti(C5H5)2CH2ClAl(CH3)2 \| \| \| v t e Periodic table \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| \| 1 \| 2 \| \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| 17 \| 18 \| \| 1 \| H \| \| He \| \| 2 \| Li \| Be \| \| B \| C \| N \| O \| F \| Ne \| \| 3 \| Na \| Mg \| \| Al \| Si \| P \| S \| Cl \| Ar \| \| 4 \| K \| Ca \| \| Sc \| Ti \| V \| Cr \| Mn \| Fe \| Co \| Ni \| Cu \| Zn \| Ga \| Ge \| As \| Se \| Br \| Kr \| \| 5 \| Rb \| Sr \| \| Y \| Zr \| Nb \| Mo \| Tc \| Ru \| Rh \| Pd \| Ag \| Cd \| In \| Sn \| Sb \| Te \| I \| Xe \| \| 6 \| Cs \| Ba \| La \| Ce \| Pr \| Nd \| Pm \| Sm \| Eu \| Gd \| Tb \| Dy \| Ho \| Er \| Tm \| Yb \| Lu \| Hf \| Ta \| W \| Re \| Os \| Ir \| Pt \| Au \| Hg \| Tl \| Pb \| Bi \| Po \| At \| Rn \| \| 7 \| Fr \| Ra \| Ac \| Th \| Pa \| U \| Np \| Pu \| Am \| Cm \| Bk \| Cf \| Es \| Fm \| Md \| No \| Lr \| Rf \| Db \| Sg \| Bh \| Hs \| Mt \| Ds \| Rg \| Cn \| Nh \| Fl \| Mc \| Lv \| Ts \| Og \| \| \| \| \| \| \| \| --- --- \| \| s-block \| f-block \| d-block \| p-block \| \| \| v t e Aluminium alloys \| \| Introduction \| Aluminium Aluminum powder Aluminium alloys History of aluminium \| \| Al 1000 series (pure) \| 1050 1060 1070 1100 1145 1199 1200 1230 1350 1370 1420 1421 1424 1430 1440 1441 1445 1450 1460 1461 1464 1469 \| \| Al-Cu 2000 series \| 2004 2011 2014 2017 2020 2024 2025 2029 2036 2048 2055 2080 2090 2091 2094 2095 2097 2098 2099 2124 2195 2196 2197 2198 2218 2219 2224&2324 2297 2319 2397 2519 2524 2618 \| \| Al-Mn 3000 series \| 3003 3004 3005 3102 3102&3303 3105 3203 \| \| Al-Si 4000 series \| 4006 4007 4015 4032 4043 4047 4543 \| \| Al-Mg 5000 series \| 5005&5657 5010 5019 5024 5026 5050 5052&5652 5056 5059 5083 5086 5154&5254 5182 5252 5356 5454 5456 5457 5557 5754 \| \| Al-Mg-Si 6000 series \| 6005 (6005A) 6009 6010 6013 6022 6060 6061 6063 6065 6066 6070 6081 6082 6101 6105 6113 6151 6162 6201 6205 6262 6351 6463 6951 \| \| Al-Zn 7000 series \| 7005 7010 7022 7034 7039 7046 7050 7055 7065 7068 7072 7075 7079 7085 7090 7091 7093 7116 7129 7150 7178 7255 7475 \| \| 8000 series (misc.) \| 8006 8009 8011 8014 8019 8025 8030 8090 8091 8093 8176 \| \| Named alloys \| Aluminium–lithium alloys AlBeMet Alclad Alnico AlSiC Alumel Aluminium granules Alusil Birmabright Devarda's alloy Duralumin Hiduminium (aka R.R. alloys) Hydronalium Italma Lo-Ex Magnalium Magnox 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3353	http://physics.bu.edu/~duffy/semester2/c27_malus.html	theta= phi = alpha = J2S._Canvas2D (emwave4.EMWave) "EMWaveApp"[x] Polarizers: polarization by selective absorption Another way to polarize light is by selectively absorbing light with electric field vectors pointing in a particular direction. Certain materials, known as dichroic materials, do this, absorbing light polarized one way but not absorbing light polarized perpendicular to that direction. If the material is thick enough to absorb all the light polarized in one direction, the light emerging from the material will be linearly polarized. Polarizers (such as the lenses of polarizing sunglasses) are made from this kind of material. If unpolarized light passes through a polarizer, the intensity of the transmitted light will be 1/2 of what it was coming in. If linearly polarized light passes through a polarizer, the intensity of the light transmitted is given by Malus' law: I 1 = I o cos 2(θ) where θ is the angle between the polarization direction of the light and the transmission axis of the polarizer. EMWaveApp ready
3354	https://www.geeksforgeeks.org/physics/problems-based-on-coulombs-law/	Problems Based on Coulomb's Law Last Updated : 03 Jun, 2025 Suggest changes 6 Likes Every physical occurrence in the universe of physics includes some type of attraction or repulsion, which causes the universe to exist in a unique way. The environment remains in a well-equipped and well-balanced state due to the attractions and repulsions between particles. When these attractions and repulsions are manipulated to disturb or vary, the outcomes in observations can be remarkable. Consider an atom's electrons: if the attraction between protons in the nucleus and electrons in the shells is disrupted, the atom can be destroyed. Physicists have long been fascinated by the quantitative side of physics since it aids in the understanding of concepts and the development of new theories and ideas. In 1785, a French physicist named Charles Augustin de Coulomb coined a measurable mathematical relationship between two electrically charged bodies. Coulomb's law, often known as Coulomb's inverse-square law, is an equation that helps to determine the extent of repulsion or attraction force between two charged particles. Coulomb's Law Coulomb's Law describes the force that exists between two point charges. The term point charge refers to the fact that the size of linearly charged objects is relatively small in comparison to the distance between them in physics. As a result, we treat them as point charges since calculating the force of attraction/repulsion between them is simple. Therefore, Coulomb's Law can be stated as: The force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, according to Coulomb's law. It acts on the line that connects the two charges that are called point charges. In general, the statement has two charges, q1and q2. The attraction/repulsion force between the charges is ‘F', and the distance between them is ‘r'. Then mathematically Coulomb's law is given as: The electrostatic force, F is directly proportional to the product of the magnitude of the charges in contact i.e. F ∝ q1q2 The electrostatic force, F is inversely proportional to the square of the distance between the two charges in contact i.e. F ∝ 1/r2 Let's combine the above to relation as , F ∝ q1q2 / r2 Now, introduce a new proportionality constant 'k' as, F = k q1q2 / r2 where 'k' is proportionality constant and equals 1/4πε0, here ε0 is called epsilon not which indicates the permittivity of a vacuum. The calculated value of k is 9 × 109 Nm2/ C2. Like charges repel one other, while unlike charges attract each other, according to Coulomb. This means that charges of the same sign will repel each other, while charges of opposite signs will attract each other. Coulomb’s Law in Vector Form There are two sorts of physical quantities: Scalars (with just one magnitude) and Vectors (with many magnitudes) (those quantities with magnitude and direction). Because it has both magnitude and direction, force is a vector quantity. In the form of vectors, Coulomb's law can be rewritten. Remember that the vector “F” is denoted as F, and the vector “r” is denoted as r, and so on. Let's say there are two charges, q1 and q2, with r1 and r2 as their location vectors. Because both charges are of the same sign, a repulsive force will exist between them. Let F12 be the force on the q1 charge due to q2, and F21 be the force on the q2 charge due to the q1 charge. The r21 vector is the matching vector from q1 to q2. F21=4πϵ∘1r212q1q2r^21F21=4πϵ∘1r213q1q2r21 When using Coulomb's Law to calculate the force between two point charges, keep the following points in mind. Because both charges are inherently opposite, the vector form of the equation is unaffected by their signs. Owing to the change in position vector, the repulsive force F21, which is the force on charge q1 due to q2, and another repulsive force F21, which is the force on charge q2 due to q1, have opposite signs. F12 = – F21 This is due to the fact that the position vector for force F12 is r12, while the position vector for force F21 is r21. r21 = r2 – r1 r12 = r1 – r2 Because the signs of r21 and r12 are opposed, they produce forces with opposing signs. This demonstrates that Coulomb's Law is compatible with Newton's Third Law, which states that every action has an equal and opposite reaction. When two charges are present in a vacuum, Coulomb's Law determines the force between them. This is due to the fact that in a vacuum, charges are unaffected by other matter or particles. Limitations of Coulomb’s Law Coulomb's Law is based on a set of assumptions and cannot be employed in the same way as other universal formulas. The following points are covered by the law: If the charges are static, we can utilise the formula (in rest position) When dealing with charges that have a normal and smooth shape, the formula is simple to employ; but, when dealing with charges that have irregular shapes, it becomes too complicated. Only when the solvent molecules between the particles are significantly larger than both charges is the formula valid. Problems Based on Coulomb's Law Problem 1: What will be the electrostatic force between the two-point charges of charges +2μC and +4μC repel each other with a force of 20N when a charge of −6μC is added to each of them? Given that, The first charge, q1 is +2μC. The second charge, q2 is +4μC. The third charge, q3 is -6μC. The electrostatic force in the first case, F is 20 N. Therefore, F = k q1q2 / r2 or 20 N = k (+2μC × +4μC) / r2 ...... (1) Now, when a third charge q3 is introduces to q1 and q2 then, charges on both q1 and q2 changes as: q1' = (2 - 6) μC = -4 μC q2' = (4 - 6) μC = -2 μC Then, the electrostatic force in this new case is: F' = k q'1q'2 / r2 = k (-4 μC × -2 μC) / r2 ...... (2) Now, in order to obtain F', lets divide equation (2) by (1) as, F' / 20 N = [k (-4 μC × -2 μC) / r2] / [20 N = k (+2μC × +4μC) / r2 ] F' = +20 N Hence, the electrostatic force when third charge is introduced is +20 N. Problem 2: Determine the electrostatic force between the two charges of magnitude 2 C and -1 C separated by a distance 1m in air. Solution: Given that, The first charge, q1 is +2 C. The second charge, q2 is -1C. The distance between the two charges, r is 1 m. The formula to calculate electrostatic force between the charges is: F = (k q1q2 )/ r2 Substitute the given values in the above expression as, F = (9 × 109 Nm2/ C2)(+2 C)(-1 C) / (1 m)2 = -18 × 109 N Problem 3: The distance between the two electrons in contact is equal to 1Å. Determine the Coulomb force between them. Solution: The charge on an electron, q is -1.6 × 10-19 C. The distance between the two charges, r is 1 Å. The formula to calculate electrostatic force between the two electrons is: F = k (q2 / r2) Substitute the given values in the above expression as, F = (9 × 109 Nm2/ C2) [(-1.6 × 10-19 C)2 / (1 Å)2] = 2.3 × 10−8 N Problem 4: When held apart at a certain distance, two spherical conductors B and C with similar radii and carrying equal charges repel each other with a force F. A third spherical conductor, with the same radius as B but no charge, is brought into contact with B, then with C, and ultimately removed from both. What is the new repulsion force between B and C? Solution: For the given case, Initially the electrostatic force on the conductors is defined as: F = k (q2 / r2) ......(1) But when a third spherical conductor comes in contact alternately with B and C then removed, so charges on B and C are Q / 2 and 3Q / 4 respectively. Therefore, the New force becomes as: F' = k [Q / 2) (3Q / 4)/ r2] ......(2) Comparing equation (1) and (2), we get: F' = 3/8F Problem 5: Consider a system of two charges of magnitude 2 × 10-7 C and 4.5 × 10-7 C which is acted upon by a force of 0.1 N. What is the distance between the two charges? Solution: Given that, The first charge, q1 is 2 × 10-7 C. The second charge, q2 is 4.5 × 10-7 C. The force acted upon them, F is 0.1 N. The formula to calculate electrostatic force between the charges is: F = k q1q2 / r2 Substitute the given values in the above expression as, 0.1 N = (9 × 109 Nm2/ C2)(2 × 10-7 C)(4.5 × 10-7 C) / (r)2 r = 0.09 m Hence, the distance between the two charges, r is 0.9 m. Problem 6: Determine the magnitude of the two identical charges, when the electrostatic force between these two identical charges is 1000 N and are separated by a distance of 0.1 m. Solution: Given that, The distance between the two charges, r is 0.1 m. The force acted upon them, F is 1000 N. The formula to calculate electrostatic force between the charges is: F = k q2 / r2 where q is the charge. Rearrange the above formula for q as, q2 = Fr2 / k Substitute the given values in the above expression as, q2 = (1000 N) (0.1 m)2 / (9 × 109 Nm2/ C2) q = 0.33 × 10-5 C Hence, the magnitude of the charge is 0.33 × 10-5 C. Problem 7: Consider two opposite charges with the same magnitude, placed at a distance from each other such that the force of F N acts between these two charges. If 60% of the charge from one is transferred to another. Determine how much the value of force changes in this case. Solution: Initially, the electrostatic force between the two charges is given by, F = k q2 / r2 ......(1) Now, when the charge is transferred, the electrostatic force becomes, F' = k q1q2 / r2 ......(2) The transferred charge is, 60 % of q = 60 / 100 × q = 3 / 5 q Therefore, charge q1 = q - 3 / 5 q = 2 / 5 q And the charge q2 = q + 3 / 5 q = 8 / 5 q Thus, the net force between these charges is, F' = k q1q2 / r2 = k (2 / 5 q) (8 / 5 q) / r2 = 16 / 25 F Problem 8: A specific charge Q is split into two components, q, and Q-q. What is the relation between Q and q if the two portions are separated by r and have the greatest Coulomb repulsion? Solution: Given that, the given charge Q is divided into charges Q-q and q separated by a distance r. We know, F = k q1q2 / r2 i.e. F = k q (Q-q) / r2 Now, to maximize this force lets take: d F / d q = 0 which implies, (Q-q) - q = 0 2 q =Q or q = Q / 2 A amanarora3dec Improve Article Tags : School Learning Physics Class 12 Physics-Class-12 Explore Physics Tutorial 15+ min read Mechanics Rest and Motion 10 min readForce 11 min readWhat is Pressure? 7 min readFriction 8 min readInertia Meaning 8 min readNewton's Laws of Motion \| Formula, Examples and Questions 9 min readUniversal Law of Gravitation 15 min readWhat is Gravity? 12 min readLaw of Conservation of Energy 11 min readFree Body Diagram 10 min readInclined Plane 10 min readWork Done 12 min readConservative Forces - Definition, Formula, Examples 7 min readEnergy 10 min readFrame of Reference 6 min read Kinematics Kinematics \| Definition, Formula, Derivation, Problems 10 min readWhat is Motion? 9 min readDistance and Displacement 5 min readSpeed and Velocity 13 min readAcceleration 9 min readWhat is Momentum Equation? 6 min readEquations of Motion: Derivations and Examples 11 min readUniform Circular Motion 9 min readProjectile Motion 15+ min readRelative Motion 10 min read Rotational Mechanics Concepts of Rotational Motion 10 min readAngular Motion 7 min readAngular Frequency 10 min readRotational Kinetic Energy 7 min readTorque 10 min readAngular Momentum 10 min readCentre of Mass 15 min readCentre of Gravity 8 min readRadius of Gyration 11 min readMoment of Inertia 15+ min read Fluid Mechanics Mechanical Properties of Fluids 11 min readWhat is Viscosity? 10 min readBuoyant Force 13 min readArchimedes Principle 12 min readPascal's Law 10 min readReynolds Number 6 min readStreamline Flow 7 min readLaminar and Turbulent Flow 9 min readBernoulli's Principle 14 min readPoiseuilles Law Formula 4 min readStoke's Law 11 min read Solid Mechanics What is Stress? 9 min readStress and Strain 12 min readStress-Strain Curve 11 min readElasticity and Plasticity 9 min readModulus of Elasticity 12 min readModulus of Rigidity 11 min readYoung's Modulus 8 min readBulk Modulus Formula 7 min readShear Modulus and Bulk Modulus 7 min readPoisson's Ratio 9 min readStress, Strain and Elastic Potential Energy 9 min read Thermodynamics Basics Concepts of Thermodynamics 12 min readZeroth Law of Thermodynamics 7 min readFirst Law of Thermodynamics 8 min readSecond Law of Thermodynamics 10 min readThermodynamic Cycles 15 min readThermodynamic State Variables and Equation of State 5 min readEnthalpy: Definition, Formula and Reactions 12 min readState Functions 7 min readCarnot Engine 5 min readHeat Engine - Definition, Working, PV Diagram, Efficiency, Types 14 min read Wave and Oscillation Introduction to Waves - Definition, Types, Properties 11 min readWave Motion 12 min readOscillation 8 min readOscillatory Motion Formula 3 min readAmplitude Formula 6 min readWhat is Frequency? 9 min readAmplitude, Time Period and Frequency of a Vibration 5 min readEnergy of a Wave Formula 7 min readSimple Harmonic Motion 15+ min readDisplacement in Simple Harmonic Motion 10 min read Sound Production and Propagation of Sound 6 min readWhat are the Characteristics of Sound Waves? 7 min readSpeed of Sound 12 min readReflection of Sound 9 min readRefraction of Sound 5 min readHow do we hear? 7 min readAudible and Inaudible Sounds 10 min readExplain the Working and Application of SONAR 8 min readNoise Pollution 8 min readDoppler Effect - Definition, Formula, Examples 7 min readDoppler Shift Formula 3 min read Electrostatics Electrostatics 13 min readElectric Charge 8 min readCoulomb's Law 9 min readElectric Dipole 11 min readDipole Moment 6 min readElectrostatic Potential 12 min readElectric Potential Energy 15+ min readPotential due to an Electric Dipole 7 min readEquipotential Surfaces 9 min readCapacitor and Capacitance 11 min read
3355	https://en.wikipedia.org/wiki/Polynomial_evaluation	Jump to content Search Contents (Top) 1 Background 2 General methods 2.1 Horner's rule 2.1.1 Multivariate 2.1.2 Estrin's scheme 2.2 Evaluation with preprocessing 2.2.1 Example 2.3 Multipoint evaluation 2.4 Dynamic evaluation 3 Specific polynomials 3.1 Evaluation of powers 3.2 Polynomial families 3.3 Hard polynomials 4 Matrix polynomials 5 See also 6 References Polynomial evaluation Русский Edit links Article Talk Read Edit View history Tools 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 Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Algorithms for polynomial evaluation In mathematics and computer science, polynomial evaluation refers to computation of the value of a polynomial when its indeterminates are substituted for some values. In other words, evaluating the polynomial at consists of computing See also Polynomial ring § Polynomial evaluation For evaluating the univariate polynomial the most naive method would use multiplications to compute , use multiplications to compute and so on for a total of multiplications and additions. Using better methods, such as Horner's rule, this can be reduced to multiplications and additions. If some preprocessing is allowed, even more savings are possible. Background [edit] This problem arises frequently in practice. In computational geometry, polynomials are used to compute function approximations using Taylor polynomials. In cryptography and hash tables, polynomials are used to compute k-independent hashing. In the former case, polynomials are evaluated using floating-point arithmetic, which is not exact. Thus different schemes for the evaluation will, in general, give slightly different answers. In the latter case, the polynomials are usually evaluated in a finite field, in which case the answers are always exact. General methods [edit] Horner's rule [edit] See also: Horner's method Horner's method evaluates a polynomial using repeated bracketing: This method reduces the number of multiplications and additions to just Horner's method is so common that a computer instruction "multiply–accumulate operation" has been added to many computer processors, which allow doing the addition and multiplication operations in one combined step. Multivariate [edit] If the polynomial is multivariate, Horner's rule can be applied recursively over some ordering of the variables. E.g. can be written as An efficient version of this approach was described by Carnicer and Gasca. Estrin's scheme [edit] See also: Estrin's scheme While it's not possible to do less computation than Horner's rule (without preprocessing), on modern computers the order of evaluation can matter a lot for the computational efficiency. A method known as Estrin's scheme computes a (single variate) polynomial in a tree like pattern: Combined by Exponentiation by squaring, this allows parallelizing the computation. Evaluation with preprocessing [edit] Arbitrary polynomials can be evaluated with fewer operations than Horner's rule requires if we first "preprocess" the coefficients . An example was first given by Motzkin who noted that can be written as where the values are computed in advance, based on . Motzkin's method uses just 3 multiplications compared to Horner's 4. The values for each can be easily computed by expanding and equating the coefficients: Example [edit] To compute the Taylor expansion , we can upscale by a factor 24, apply the above steps, and scale back down. That gives us the three multiplication computation Improving over the equivalent Horner form (that is ) by 1 multiplication. Some general methods include the Knuth–Eve algorithm and the Rabin–Winograd algorithm. Multipoint evaluation [edit] Evaluation of a degree-n polynomial at multiple points can be done with multiplications by using Horner's method times. Using the above preprocessing approach, this can be reduced by a factor of two; that is, to multiplications. However, it is possible to do better and reduce the time requirement to just . The idea is to define two polynomials that are zero in respectively the first and second half of the points: and . We then compute and using the Polynomial remainder theorem, which can be done in time using a fast Fourier transform. This means and by construction, where and are polynomials of degree at most . Because of how and were defined, we have Thus to compute on all of the , it suffices to compute the smaller polynomials and on each half of the points. This gives us a divide-and-conquer algorithm with , which implies by the master theorem. In the case where the points in which we wish to evaluate the polynomials have some structure, simpler methods exist. For example, Knuth section 4.6.4 gives a method for tabulating polynomial values of the type Dynamic evaluation [edit] In the case where are not known in advance, Kedlaya and Umans gave a data structure for evaluating polynomials over a finite field of size in time per evaluation after some initial preprocessing. This was shown by Larsen to be essentially optimal. The idea is to transform of degree into a multivariate polynomial , such that and the individual degrees of is at most . Since this is over , the largest value can take (over ) is . Using the Chinese remainder theorem, it suffices to evaluate modulo different primes with a product at least . Each prime can be taken to be roughly , and the number of primes needed, , is roughly the same. Doing this process recursively, we can get the primes as small as . That means we can compute and store on all the possible values in time and space. If we take , we get , so the time/space requirement is just Kedlaya and Umans further show how to combine this preprocessing with fast (FFT) multipoint evaluation. This allows optimal algorithms for many important algebraic problems, such as polynomial modular composition. Specific polynomials [edit] While general polynomials require operations to evaluate, some polynomials can be computed much faster. For example, the polynomial can be computed using just one multiplication and one addition since Evaluation of powers [edit] Main articles: Exponentiation by squaring and Addition-chain exponentiation A particularly interesting type of polynomial is powers like . Such polynomials can always be computed in operations. Suppose, for example, that we need to compute ; we could simply start with and multiply by to get . We can then multiply that by itself to get and so on to get and in just four multiplications. Other powers like can similarly be computed efficiently by first computing by 2 multiplications and then multiplying by . The most efficient way to compute a given power is provided by addition-chain exponentiation. However, this requires designing a specific algorithm for each exponent, and the computation needed for designing these algorithms are difficult (NP-complete), so exponentiation by squaring is generally preferred for effective computations. Polynomial families [edit] Often polynomials show up in a different form than the well known . For polynomials in Chebyshev form we can use Clenshaw algorithm. For polynomials in Bézier form we can use De Casteljau's algorithm, and for B-splines there is De Boor's algorithm. Hard polynomials [edit] The fact that some polynomials can be computed significantly faster than "general polynomials" suggests the question: Can we give an example of a simple polynomial that cannot be computed in time much smaller than its degree? Volker Strassen has shown that the polynomial cannot be evaluated with less than multiplications and additions. At least this bound holds if only operations of those types are allowed, giving rise to a so-called "polynomial chain of length ". The polynomial given by Strassen has very large coefficients, but by probabilistic methods, one can show there must exist even polynomials with coefficients just 0's and 1's such that the evaluation requires at least multiplications. For other simple polynomials, the complexity is unknown. The polynomial is conjectured to not be computable in time for any . This is supported by the fact that, if it can be computed fast, then integer factorization can be computed in polynomial time, breaking the RSA cryptosystem. Matrix polynomials [edit] Sometimes the computational cost of scalar multiplications (like ) is less than the computational cost of "non scalar" multiplications (like ). The typical example of this is matrices. If is an matrix, a scalar multiplication takes about arithmetic operations, while computing takes about (or using fast matrix multiplication). Matrix polynomials are used, for example, for computing matrix exponentials. Paterson and Stockmeyer showed how to compute a degree polynomial using only non scalar multiplications and scalar multiplications. Thus a matrix polynomial of degree n can be evaluated in time, where ⁠⁠ is the time needed for multiplying two ⁠⁠ matices. If this is where ⁠⁠ or ⁠⁠ deprending whether usual or fast matrix multiplication is used. This is to be compared to the usual Horner method, which gives ⁠⁠ or ⁠⁠ respectively. This method works as follows: For a polynomial let k be the least integer not smaller than The powers are computed with matrix multiplications, and are then computed by repeated multiplication by Now, : , where for i ≥ n. This requires just more non-scalar multiplications. The direct application of this method uses non-scalar multiplications, but combining it with Evaluation with preprocessing, Paterson and Stockmeyer show you can reduce this to . Methods based on matrix polynomial multiplications and additions have been proposed allowing to save nonscalar matrix multiplications with respect to the Paterson-Stockmeyer method.[clarification needed] See also [edit] Estrin's scheme to facilitate parallelization on modern computer architectures Arithmetic circuit complexity theory studies the computational complexity of evaluating different polynomials. References [edit] ^ Carnicer, J.; Gasca, M. (1990). "Evaluation of Multivariate Polynomials and Their Derivatives". Mathematics of Computation. 54 (189): 231–243. doi:10.2307/2008692. JSTOR 2008692. ^ Motzkin, T. S. (1955). "Evaluation of polynomials and evaluation of rational functions". Bulletin of the American Mathematical Society. 61 (163): 10. ^ Rabin, Michael O.; Winograd, Shmuel (July 1972). "Fast evaluation of polynomials by rational preparation". Communications on Pure and Applied Mathematics. 25 (4): 433–458. doi:10.1002/cpa.3160250405. ^ Von Zur Gathen, Joachim; Jürgen, Gerhard (2013). Modern computer algebra. Cambridge University Press. Chapter 10. ISBN 9781139856065. ^ Knuth, Donald (2005). Art of Computer Programming. Vol. 2: Seminumerical Algorithms. Addison-Wesley. ISBN 9780201853926. ^ Kedlaya, Kiran S.; Umans, Christopher (2011). "Fast Polynomial Factorization and Modular Composition". SIAM Journal on Computing. 40 (6): 1767–1802. doi:10.1137/08073408x. hdl:1721.1/71792. S2CID 412751. ^ Larsen, K. G. (2012). "Higher Cell Probe Lower Bounds for Evaluating Polynomials". 2012 IEEE 53rd Annual Symposium on Foundations of Computer Science. Vol. 53. IEEE. pp. 293–301. doi:10.1109/FOCS.2012.21. ISBN 978-0-7695-4874-6. S2CID 7906483. ^ Downey, Peter; Leong, Benton; Sethi, Ravi (1981). "Computing Sequences with Addition Chains". SIAM Journal on Computing. 10 (3): 638–646. doi:10.1137/0210047. Retrieved 27 January 2024. ^ Strassen, Volker (1974). "Polynomials with Rational Coefficients Which are Hard to Compute". SIAM Journal on Computing. 3 (2): 128–149. doi:10.1137/0203010. ^ Schnorr, C. P. (1979), "On the additive complexity of polynomials and some new lower bounds", Theoretical Computer Science, Lecture Notes in Computer Science, vol. 67, Springer, pp. 286–297, doi:10.1007/3-540-09118-1_30, ISBN 978-3-540-09118-9 ^ Chen, Xi, Neeraj Kayal, and Avi Wigderson. Partial derivatives in arithmetic complexity and beyond. Now Publishers Inc, 2011. ^ Paterson, Michael S.; Stockmeyer, Larry J. (1973). "On the Number of Nonscalar Multiplications Necessary to Evaluate Polynomials". SIAM Journal on Computing. 2 (1): 60–66. doi:10.1137/0202007. ^ Fasi, Massimiliano (1 August 2019). "Optimality of the Paterson–Stockmeyer method for evaluating matrix polynomials and rational matrix functions" (PDF). Linear Algebra and Its Applications. 574: 185. doi:10.1016/j.laa.2019.04.001. ISSN 0024-3795. Retrieved from " Category: Polynomials Hidden categories: CS1: long volume value Articles with short description Short description matches Wikidata Wikipedia articles needing clarification from May 2025 Polynomial evaluation Add topic
3356	https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23?srsltid=AfmBOopeftAqC0UB21g7MtkrH5nliJ6rCW4AVklKfbwGOfShjRWtTBFd	Art of Problem Solving 2002 AMC 12B Problems/Problem 23 - 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 2002 AMC 12B Problems/Problem 23 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2002 AMC 12B Problems/Problem 23 Contents 1 Problem 2 Solution 2.1 Solution 1: Pythagorean Theorem 2.2 Solution 2: Law of Cosines 2.3 Solution 3: Stewart's Theorem 2.4 Solution 4: Pappus's Median Theorem 2.5 Video Solution by TheBeautyofMath 3 See also Problem In , we have and . Side and the median from to have the same length. What is ? Solution Solution 1: Pythagorean Theorem Let be the foot of the altitude from to extended past . Let and . Using the Pythagorean Theorem, we obtain the equations Subtracting equation from and , we get Then, subtracting from and rearranging, we get , so ~greenturtle 11/28/2017 Solution 2: Law of Cosines Let be the foot of the median from to , and we let . Then by the Law of Cosines on , we have Since , we can add these two equations and get Hence and . Solution 3: Stewart's Theorem From Stewart's Theorem, we have Simplifying, we get - awu2014 Solution 4: Pappus's Median Theorem There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have , and you draw a median from point to side (label this as ), then: . Note that is the length of side , is the length of side , and is length of side . Let . Then . Now, we can plug into the formula given above: , , , and . After some simple algebra, we find . Then, . -Flames Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with . ~Puck_0 aka Apollonius' Theorem - Orion 2010 Video Solution by TheBeautyofMath ~IceMatrix See also 2002 AMC 12B (Problems • Answer Key • Resources) Preceded by Problem 22Followed by Problem 24 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Introductory Geometry Problems 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.
3357	https://www.youtube.com/watch?v=f8svAm237xM	More on why SSA is not a postulate \| Congruence \| Geometry \| Khan Academy Khan Academy 9090000 subscribers 397 likes Description 182125 views Posted: 26 Sep 2011 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: SSA special cases including RSH Watch the next lesson: Missed the previous lesson? Geometry on Khan Academy: We are surrounded by space. And that space contains lots of things. And these things have shapes. In geometry we are concerned with the nature of these shapes, how we define them, and what they teach us about the world at large--from math to architecture to biology to astronomy (and everything in between). Learning geometry is about more than just taking your medicine ("It's good for you!"), it's at the core of everything that exists--including you. Having said all that, some of the specific topics we'll cover include angles, intersecting lines, right triangles, perimeter, area, volume, circles, triangles, quadrilaterals, analytic geometry, and geometric constructions. Wow. That's a lot. To summarize: it's difficult to imagine any area of math that is more widely used than geometry. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Geometry channel: Subscribe to Khan Academy: 41 comments Transcript: Several videos ago, I very quickly went through why side-side-angle is not a valid postulate. And what I want to do in this video is explore it a little bit more. And it's not called angle-side-side for obvious reasons, because then the acronym would make people giggle in geometry class. And I guess we don't want people giggling while they're doing mathematics. So let's just think about a triangle here. So let's say I have a triangle. Let me draw it. Let's have a triangle that looks something like this. If I have a triangle that looks something-- I have trouble drawing straight triangles. So let's say the triangle looks something like that. And let's say that we've found another triangle that has a congruent side, a side that is congruent to this side right over here. I guess any side on a triangle is next to the other two sides. Next to that is a side that is congruent to this side right over here. And then that side is one of the sides of an angle. So it forms one of the parts of an angle right over here. And that other triangle has a congruent angle right over here. So this is the angle that that first side is not a part of. Only that second side is part of this angle. So this is side-side-angle. Or you could call it angle-side-side and giggle a little bit about it. Now, how do we know that this doesn't by itself show that this is congruent? Well, we'd have to show that this could actually imply two different triangles. And to think about that, let's say we know that the angle-- we know that this other triangle has that same yellow angle there, which means that the blue side is going to have to look something like that, just the way we drew it over here. This side down here, I'll make it a green side. This green side down here we know nothing about. We never said that this is congruent to anything. If we knew, then we could use side-side-side. We only know that this side is congruent and this side is congruent, and this angle is congruent. So this green side, and I'll draw it as a dotted line, it could be of any length. We don't know what the length is of that green side. Now we have this magenta side. We have another side that is congruent here. So this thing could pivot over here. We know nothing about this angle so it could form any angle. But it does have to get to this other side. So one possibility is that maybe the triangles are congruent. So maybe this side does go down just like that, in which case, we actually would have congruent triangles. But the kind of aha moment here, or the reason why SSA isn't possible, is that this side, could also come down like this. There's two ways to get down to this base, if you want to call it that way. It can come out that way or it could kind of come in this way. And so that's why SSA by itself with no other information is ambiguous. It does not give you enough information to say that those triangles are definitely the same. Now there are special cases. So in this situation right over here, our angle, the angle in our SSA, our angle was acute. This is an acute angle right over here. And when you have an acute angle as one of the sides of your triangle, the other sides of the triangle, you could still have an obtuse angle. Remember, acute means less than 90 degrees, obtuse means greater than 90 degrees. So you could still have an obtuse angle. So that's why this is an option. So one option is that you have two other acute angles. So this could be acute. This is also acute, also acute, also acute. But then you have the option where this is even more acute, even narrower, and then this becomes an obtuse angle. And that's only possible if you don't-- you can't have two obtuse angles in the same triangle. You can't have two things that have larger than 90-degree measure in the same triangle. And so that's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse-- and that is the A in the SSA. So you have the angle. And I were to say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, then it's not so ambiguous. Because we could try to draw that. So let's draw that same congruent obtuse angle. We know nothing about this side down here because we haven't said that that's necessarily congruent. So that could be of any length. We do know that this triangle is going to have the same length for this side of the angle. So it looks like this. And then we know that this side-- let me do that in orange. We know that this side is also going to be the same length. We haven't told you anything about this angle right over here. So this side could pivot over here. We can kind of rotate it over there. But there's only one way, now, that this orange side can reach this green side. Now the only way is this way over here. And we were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one. And so then it constrains what the triangle can become. So I don't want to make you say, in general, SSA, you do not want to use it as a postulate. I just wanted to make it clear that there is the special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. And then finally, there's a circumstance that this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle. So where you have the A in SSA is a right angle. So if you had it like this. If you have a right angle and you have some base of unknown length but you fix this length right over here-- if you know that this is fixed because you're saying it's congruent to some other triangle, and if you know that the next length is fixed-- and if you think about it, this next side is going to be the side opposite the right angle. It's going to have to be the hypotenuse of the right angle. Then you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you could do it is to bring it down over here. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. And here, they wrote the angle first. You could view this as angle-side-side. And they were able to do it because now they can write "right angle," and so it doesn't form that embarrassing acronym. And this would also be a little bit common sense. Because if you know two sides of a right triangle-- and we haven't gone into depth on this in the geometry playlist, but you might already be familiar with it-- by Pythagorean theorem, you can always figure out the third side. So if you have this information about any triangle, you can always figure out the third side. And then you can use side-side-side. So I just wanted to show you this little special case. But in general, the important thing is that you can't just use SSA unless you have more information.
3358	https://sdbidoon.com/document/digital-electonics-and-microprocessor-27-04-2020-to-28-04-2020.pdf	ELECTRICAL 2ND YEAR SUBJECT- DIGITAL ELECTONICS AND MICROPROCESSOR DATE -27/04/2020 TO 28-04-2020 TIME 9:40-10:30 5 KARNAUGH MAP MINIMIZATION A Karnaugh map provides a systematic method for simplifying Boolean expressions and, if properly used, will produce the simplest SOP or POS expression possible, known as the minimum expression. As you have seen, the effectiveness of algebraic simplification depends on your familiarity with all the laws, rules, and theorems of Boolean algebra and on your ability to apply them. The Karnaugh map, on the other hand, provides a "cookbook" method for simplification. A Karnaugh map is similar to a truth table because it presents all of the possible values of input variables and the resulting output for each value. Instead of being organized into columns and rows like a truth table, the Karnaugh map is an array of cells in which each cell represents a binary value of the input variables. The cells are arranged in a way so that simplification of a given expression is simply a matter of properly grouping the cells. Karnaugh maps can be used for expressions with two, three, four. and five variables. Another method, called the Quine-McClusky method can be used for higher numbers of variables. The number of cells in a Karnaugh map is equal to the total number of possible input variable combinations as is the number of rows in a truth table. For three variables, the number of cells is 23 = 8. For four variables, the number of cells is 24 = 16. The 3-Variable Karnaugh Map The 3-variable Karnaugh map is an array of eight cells. as shown in Fig.(5- 1)(a). In this case, A, B, and C are used for the variables although other letters could be used. Binary values of A and B are along the left side (notice the sequence) and the values of C are across the top. The value of a given cell is the binary values of A and B at the left in the same row combined with the value of C at the top in the same column. For example, the cell in the upper left corner has a binary value of 000 and the cell in the lower right corner has a binary value of 101. Fig.(5-1)( b) shows the standard product terms that are represented by each cell in the Karnaugh map. (a) (b) Fig.(5-1) A 3-variable Karnaugh map showing product terms. The 4-Variable Karnaugh Map The 4-variable Karnaugh map is an array of sixteen cells, as shown in Fig.(5-2)(a). Binary values of A and B are along the left side and the values of C and D are across the top. The value of a given cell is the binary values of A and B at the left in the same row combined with the binary values of C and D at the top in the same column. For example, the cell in the upper right corner has a binary value of 0010 and the cell in the lower right corner has a binary value of 1010. Fig.(5-2)(b) shows the standard product terms that are represented by each cell in the 4-variable Karnaugh map. (a) (b) Fig.(5-2) A 4-variable Karnaugh map. Cell Adjacency The cells in a Karnaugh map are arranged so that there is only a single- variable change between adjacent cells. Adjacency is defined by a single- variable change. In the 3-variable map the 010 cell is adjacent to the 000 cell, the 011 cell, and the 110 cell. The 010 cell is not adjacent to the 001 cell, the 111 cell, the 100 cell, or the 101 cell. Fig.(5-3) Adjacent cells on a Karnaugh map are those that differ by only one variable. Arrows point between adjacent cells. KARNAUGH MAP SOP MINIMIZATION For an SOP expression in standard form, a 1 is placed on the Karnaugh map for each product term in the expression. Each 1 is placed in a cell corresponding to the value of a product term. For example, for the product term ABC, a 1 goes in the 10l cell on a 3-variable map. Example Map the following standard SOP expression on a Karnaugh map: see Fig.(5-4). Example Map the following standard SOP expression on a Karnaugh map: See Fig.(5-5). Fig.(5-4) Fig.(5-5) Example Map the following SOP expression on a Karnaugh map: Solution The SOP expression is obviously not in standard form because each product term does not have three variables. The first term is missing two variables, the second term is missing one variable, and the third term is standard. First expand the terms numerically as follows: Example Map the following SOP expression on a Karnaugh map: Solution The SOP expression is obviously not in standard form because each product term does not have four variables. Map each of the resulting binary values by placing a 1 in the appropriate cell of the 4- variable Karnaugh map. Karnaugh Map Simplification of SOP Expressions Grouping the 1s, you can group 1s on the Karnaugh map according to the following rules by enclosing those adjacent cells containing 1s. The goal is to maximize the size of the groups and to minimize the number of groups.  A group must contain either 1, 2, 4, 8, or 16 cells, which are all powers of two. In the case of a 3-variable map, 23 = 8 cells is the maximum group.  Each cell in a group must be adjacent to one or more cells in that same group.  Always include the largest possible number of 1s in a group in accordance with rule 1.  Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s. Example: Group the 1s in each of the Karnaugh maps in Fig.(5-6). Fig.(5-6) Solution: The groupings are shown in Fig.(5-7). In some cases, there may be more than one way to group the 1s to form maximum groupings. Fig.(5-7) Determine the minimum product term for each group. a. For a 3-variable map: (1) A l-cell group yields a 3-variable product term (2) A 2-cell group yields a 2-variable product term (3) A 4-cell group yields a 1-variable term (4) An 8-cell group yields a value of 1 for the expression b. For a 4-variable map: (1) A 1-cell group yields a 4-variable product term (2) A 2-cell group yields a 3-variable product term (3) A 4-cell group yields a 2-variable product term (4) An 8-cell group yields a 1-variable term (5) A 16-cell group yields a value of 1 for the expression Example: Determine the product terms for each of the Karnaugh maps in Fig.(5-7) and write the resulting minimum SOP expression. Fig.(5-8) Solution: The resulting minimum product term for each group is shown in Fig.(5-8). The minimum SOP expressions for each of the Karnaugh maps in the figure are: (a) AB+BC+ABC (C) AB + AC + ABD (b) B + A C + AC (d) D + ABC + BC Example: Use a Karnaugh map to minimize the following standard SOP expression: ABC + ABC + ABC + ABC + ABC Example: Use a Karnaugh map to minimize the following SOP expression: "Don't Care" Conditions Sometimes a situation arises in which some input variable combinations are not allowed. For example, recall that in the BCD code there are six invalid combinations: 1010, 1011, 1100, 1101, 1110, and 1111. Since these unallowed states will never occur in an application involving the BCD code, they can be treated as "don't care" terms with respect to their effect on the output. That is, for these "don't care" terms either a 1 or a 0 may be assigned to the output: it really does not matter since they will never occur. The "don't care" terms can be used to advantage on the Karnaugh map. Fig.(5-9) shows that for each "don't care" term, an X is placed in the cell. When grouping the 1 s, the Xs can be treated as 1s to make a larger grouping or as 0s if they cannot be used to advantage. The larger a group, the simpler the resulting term will be. The truth table in Fig.(5-9)(a) describes a logic function that has a 1 output only when the BCD code for 7,8, or 9 is present on the inputs. If the "don't cares" are used as 1s, the resulting expression for the function is A + BCD, as indicated in part (b). If the "don't cares" are not used as 1s, the resulting expression is ABC + ABCD: so you can see the advantage of using "don't care" terms to get the simplest expression. Fig.(5-9) KARNAUGH MAP POS MINIMIZATION In this section, we will focus on POS expressions. The approaches are much the same except that with POS expressions, 0s representing the standard sum terms are placed on the Karnaugh map instead of 1s. For a POS expression in standard form, a 0 is placed on the Karnaugh map for each sum term in the expression. Each 0 is placed in a cell corresponding to the value of a sum term. For example, for the sum term A + B + C, a 0 goes in the 0 1 0 cell on a 3-variable map. When a POS expression is completely mapped, there will be a number of 0s on the Karnaugh map equal to the number of sum terms in the standard POS expression. The cells that do not have a 0 are the cells for which the expression is 1. Usually, when working with POS expressions, the 1s are left off. The following steps and the illustration in Fig.(5-10) show the mapping process. Step 1. Determine the binary value of each sum term in the standard POS expression. This is the binary value that makes the term equal to 0. Step 2. As each sum term is evaluated, place a 0 on the Karnaugh map in the corresponding cell. Fig.(5-10) Example of mapping a standard POS expression. Example: Map the following standard POS expression on a Karnaugh map: Solution: Karnaugh Map Simplification of POS Expressions The process for minimizing a POS expression is basically the same as for an SOP expression except that you group 0s to produce minimum sum terms instead of grouping 1s to produce minimum product terms. The rules for grouping the 0s are the same as those for grouping the 1s that you learned before. Example: Use a Karnaugh map to minimize the following standard POS expression: Also, derive the equivalent SOP expression. Solution: Example: Use a Karnaugh map to minimize the following POS expression: Example: Using a Karnaugh map, convert the following standard POS expression into a minimum POS expression, a standard SOP expression, and a minimum SOP expression. FIVE-VARIABLE KARNAUGH MAPS Boolean functions with five variables can be simplified using a 32-cell Karnaugh map. Actually, two 4-variable maps (16 cells each) are used to construct a 5-variable map. You already know the cell adjacencies within each of the 4- variable maps and how to form groups of cells containing 1s to simplify an SOP expression. All you need to learn for five variables is the cell adjacencies between the two 4-variable maps and how to group those adjacent 1s. A Karnaugh map for five variables (ABCDE) can be constructed using two 4-variable maps with which you are already familiar. Each map contains 16 cells with all combinations of variables B, C, D, and E. One map is for A = 0 and the other is for A = 1, as shown in Fig.(5-11). Cell Adjacencies You already know how to determine adjacent cells within the 4-variable map. The best way to visualize cell adjacencies between the two 16-cel1 maps is to imagine that the A = 0 map is placed on top of the A = 1 map. Each cell in the A = 0 map is adjacent to the cell directly below it in the A = 1 map, see Fig.(5-12). Fig.(5-11) Fig.(5-12) The simplified SOP expression yields x = DE + BCE + ABD + BC DE Example: Use a Karnaugh map to minimize the following standard SOP 5-variable expression:
3359	https://en.wikipedia.org/wiki/Proof_of_impossibility	Jump to content Proof of impossibility العربية Español Français Українська Edit links From Wikipedia, the free encyclopedia Category of mathematical proof \| \| \| Part of a series on \| \| Mathematics \| \| History Index \| \| \| \| \| Areas Number theory Geometry Algebra Calculus and Analysis Discrete mathematics Logic Set theory Probability Statistics and Decision theory \| \| Relationship with sciences Physics Chemistry Geosciences Computation Biology Linguistics Economics Philosophy Education \| \| \| Mathematics Portal \| \| v t e \| In mathematics, an impossibility theorem is a theorem that demonstrates a problem or general set of problems cannot be solved. These are also known as proofs of impossibility, negative proofs, or negative results. Impossibility theorems often resolve decades or centuries of work spent looking for a solution by proving there is no solution. Proving that something is impossible is usually much harder than the opposite task, as it is often necessary to develop a proof that works in general, rather than to just show a particular example. Impossibility theorems are usually expressible as negative existential propositions or universal propositions in logic. The irrationality of the square root of 2 is one of the oldest proofs of impossibility. It shows that it is impossible to express the square root of 2 as a ratio of two integers. Another consequential proof of impossibility was Ferdinand von Lindemann's proof in 1882, which showed that the problem of squaring the circle cannot be solved because the number π is transcendental (i.e., non-algebraic), and that only a subset of the algebraic numbers can be constructed by compass and straightedge. Two other classical problems—trisecting the general angle and doubling the cube—were also proved impossible in the 19th century, and all of these problems gave rise to research into more complicated mathematical structures. Some of the most important proofs of impossibility found in the 20th century were those related to undecidability, which showed that there are problems that cannot be solved in general by any algorithm, with one of the more prominent ones being the halting problem. Gödel's incompleteness theorems were other examples that uncovered fundamental limitations in the provability of formal systems. In computational complexity theory, techniques like relativization (the addition of an oracle) allow for "weak" proofs of impossibility, in that proofs techniques that are not affected by relativization cannot resolve the P versus NP problem. Another technique is the proof of completeness for a complexity class, which provides evidence for the difficulty of problems by showing them to be just as hard to solve as any other problem in the class. In particular, a complete problem is intractable if one of the problems in its class is. Proof techniques [edit] See also: Types of proof Contradiction [edit] One of the widely used types of impossibility proof is proof by contradiction. In this type of proof, it is shown that if a proposition, such as a solution to a particular class of equations, is assumed to hold, then via deduction two mutually contradictory things can be shown to hold, such as a number being both even and odd or both negative and positive. Since the contradiction stems from the original assumption, this means that the assumed premise must be impossible. In contrast, a non-constructive proof of an impossibility claim would proceed by showing it is logically contradictory for all possible counterexamples to be invalid: at least one of the items on a list of possible counterexamples must actually be a valid counterexample to the impossibility conjecture. For example, a conjecture that it is impossible for an irrational power raised to an irrational power to be rational was disproved, by showing that one of two possible counterexamples must be a valid counterexample, without showing which one it is. By descent [edit] Main article: Proof by infinite descent Another type of proof by contradiction is proof by descent, which proceeds first by assuming that something is possible, such as a positive integer solution to a class of equations, and that therefore there must be a smallest solution (by the Well-ordering principle). From the alleged smallest solution, it is then shown that a smaller solution can be found, contradicting the premise that the former solution was the smallest one possible—thereby showing that the original premise that a solution exists must be false. Counterexample [edit] The obvious way to disprove an impossibility conjecture is by providing a single counterexample. For example, Euler proposed that at least n different nth powers were necessary to sum to yet another nth power. The conjecture was disproved in 1966, with a counterexample involving a count of only four different 5th powers summing to another fifth power: : 275 + 845 + 1105 + 1335 = 1445. Proof by counterexample is a form of constructive proof, in that an object disproving the claim is exhibited. Economics [edit] Arrow's theorem: Rational ranked-choice voting [edit] In social choice theory, Arrow's impossibility theorem shows that it is impossible to devise a ranked-choice voting system that is both non-dictatorial and satisfies a basic requirement for rational behavior called independence of irrelevant alternatives. Gibbard's theorem: Non-dictatorial strategyproof games [edit] Gibbard's theorem shows that any strategyproof game form (i.e. one with a dominant strategy) with more than two outcomes is dictatorial. The Gibbard–Satterthwaite theorem is a special case showing that no deterministic voting system can be fully invulnerable to strategic voting in all circumstances, regardless of how others vote. Revelation principle: Non-honest solutions [edit] The revelation principle can be seen as an impossibility theorem showing the "opposite" of Gibbard's theorem, in a colloquial sense: any game or voting system can be made resistant to strategy by incorporating the strategy into the mechanism. Thus, it is impossible to design a mechanism with a solution that is better than can be obtained by a truthful mechanism. Geometry [edit] Expressing mth roots rationally [edit] The proof by Pythagoras about 500 BCE has had a profound effect on mathematics. It shows that the square root of 2 cannot be expressed as the ratio of two integers. The proof bifurcated "the numbers" into two non-overlapping collections—the rational numbers and the irrational numbers. There is a famous passage in Plato's Theaetetus in which it is stated that Theodorus (Plato's teacher) proved the irrationality of taking all the separate cases up to the root of 17 square feet ... . A more general proof shows that the mth root of an integer N is irrational, unless N is the mth power of an integer n. That is, it is impossible to express the mth root of an integer N as the ratio a⁄b of two integers a and b, that share no common prime factor, except in cases in which b = 1. Euclidean constructions [edit] Greek geometry was based on the use of the compass and a straightedge (though the straightedge is not strictly necessary). The compass allows a geometer to construct points equidistant from each other, which in Euclidean space are equivalent to implicitly calculations of square roots. Four famous questions asked how to construct: a pair of lines trisecting a given angle; a cube with a volume twice the volume of a given cube; a square equal in area to that of a given circle; an equilateral polygon with an arbitrary number of sides. For more than 2,000 years unsuccessful attempts were made to solve these problems; at last, in the 19th century it was proved that the desired constructions are mathematically impossible without admitting additional tools other than a compass. All of these are problems in Euclidean construction, and Euclidean constructions can be done only if they involve only Euclidean numbers (by definition of the latter). Irrational numbers can be Euclidean. A good example is the square root of 2 (an irrational number). It is simply the length of the hypotenuse of a right triangle with legs both one unit in length, and it can be constructed with a straightedge and a compass. But it was proved centuries after Euclid that Euclidean numbers cannot involve any operations other than addition, subtraction, multiplication, division, and the extraction of square roots. Both trisecting the general angle and doubling the cube require taking cube roots, which are not constructible numbers. is not a Euclidean number ... and therefore it is impossible to construct, by Euclidean methods a length equal to the circumference of a circle of unit diameter Because was proved in 1882 to be a transcendental number, it is not a Euclidean number; Hence the construction of a length from a unit circle is impossible. Constructing an equilateral n-gon [edit] The Gauss-Wantzel theorem showed in 1837 that constructing an equilateral n-gon is impossible for most values of n. Deducing Euclid's parallel postulate [edit] Main article: Non-Euclidean geometry The parallel postulate from Euclid's Elements is equivalent to the statement that given a straight line and a point not on that line, only one parallel to the line may be drawn through that point. Unlike the other postulates, it was seen as less self-evident. Nagel and Newman argue that this may be because the postulate concerns "infinitely remote" regions of space; in particular, parallel lines are defined as not meeting even "at infinity", in contrast to asymptotes. This perceived lack of self-evidence led to the question of whether it might be proven from the other Euclidean axioms and postulates. It was only in the nineteenth century that the impossibility of deducing the parallel postulate from the others was demonstrated in the works of Gauss, Bolyai, Lobachevsky, and Riemann. These works showed that the parallel postulate can moreover be replaced by alternatives, leading to non-Euclidean geometries. Nagel and Newman consider the question raised by the parallel postulate to be "...perhaps the most significant development in its long-range effects upon subsequent mathematical history". In particular, they consider its outcome to be "of the greatest intellectual importance," as it showed that "a proof can be given of the impossibility of proving certain propositions [in this case, the parallel postulate] within a given system [in this case, Euclid's first four postulates]." Number theory [edit] Impossibility of Fermat triples [edit] Main article: Fermat's Last Theorem Fermat's Last Theorem was conjectured by Pierre de Fermat in the 1600s, states the impossibility of finding solutions in positive integers for the equation with . Fermat himself gave a proof for the n = 4 case using his technique of infinite descent, and other special cases were subsequently proved, but the general case was not proven until 1994 by Andrew Wiles. Integer solutions of Diophantine equations: Hilbert's tenth problem [edit] Main articles: Matiyasevich's theorem and Hilbert's problems The question "Does any arbitrary Diophantine equation have an integer solution?" is undecidable. That is, it is impossible to answer the question for all cases. Franzén introduces Hilbert's tenth problem and the MRDP theorem (Matiyasevich-Robinson-Davis-Putnam theorem) which states that "no algorithm exists which can decide whether or not a Diophantine equation has any solution at all". MRDP uses the undecidability proof of Turing: "... the set of solvable Diophantine equations is an example of a computably enumerable but not decidable set, and the set of unsolvable Diophantine equations is not computably enumerable". Decidability [edit] Richard's paradox [edit] Main article: Richard's paradox This profound paradox presented by Jules Richard in 1905 informed the work of Kurt Gödel and Alan Turing. A succinct definition is found in Principia Mathematica: Richard's paradox ... is as follows. Consider all decimals that can be defined by means of a finite number of words [“words” are symbols; boldface added for emphasis]; let E be the class of such decimals. Then E has [an infinite number of] terms; hence its members can be ordered as the 1st, 2nd, 3rd, ... Let X be a number defined as follows [Whitehead & Russell now employ the Cantor diagonal method]. If the n-th figure in the n-th decimal is p, let the n-th figure in X be p + 1 (or 0, if p = 9). Then X is different from all the members of E, since, whatever finite value n may have, the n-th figure in X is different from the n-th figure in the n-th of the decimals composing E, and therefore X is different from the n-th decimal. Nevertheless we have defined X in a finite number of words [i.e. this very definition of “word” above.] and therefore X ought to be a member of E. Thus X both is and is not a member of E. — Principia Mathematica, 2nd edition 1927, p. 61 Kurt Gödel considered his proof to be “an analogy” of Richard's paradox, which he called "Richard's antinomy" (see below). Alan Turing constructed this paradox with a machine and proved that this machine could not answer a simple question: will this machine be able to determine if any machine (including itself) will become trapped in an unproductive ‘infinite loop’ (i.e. it fails to continue its computation of the diagonal number). Complete and consistent axiomatic system [edit] Main article: Gödel's incompleteness theorems To quote Nagel and Newman (p. 68), "Gödel's paper is difficult. Forty-six preliminary definitions, together with several important preliminary theorems, must be mastered before the main results are reached". In fact, Nagel and Newman required a 67-page introduction to their exposition of the proof. But if the reader feels strong enough to tackle the paper, Martin Davis observes that "This remarkable paper is not only an intellectual landmark but is written with a clarity and vigor that makes it a pleasure to read" (Davis in Undecidable, p. 4). Gödel proved, in his own words: : "It is reasonable... to make the conjecture that ...[the] axioms [from Principia Mathematica and Peano] are ... sufficient to decide all mathematical questions which can be formally expressed in the given systems. In what follows it will be shown that this is not the case, but rather that ... there exist relatively simple problems of the theory of ordinary whole numbers which cannot be decided on the basis of the axioms" (Gödel in Undecidable, p. 4). Gödel compared his proof to "Richard's antinomy" (an "antinomy" is a contradiction or a paradox; for more see Richard's paradox): : "The analogy of this result with Richard's antinomy is immediately evident; there is also a close relationship with the Liar Paradox (Gödel's footnote 14: Every epistemological antinomy can be used for a similar proof of undecidability) ... Thus, we have a proposition before us which asserts its own unprovability . (His footnote 15: Contrary to appearances, such a proposition is not circular, for, to begin with, it asserts the unprovability of a quite definite formula)". Proof of halting [edit] Main article: Turing's proof The Entscheidungsproblem, the decision problem, was first answered by Church in April 1935 and preceded Turing by over a year, as Turing's paper was received for publication in May 1936. Turing's proof is made difficult by number of definitions required and its subtle nature. See Turing machine and Turing's proof for details. Turing's first proof (of three) follows the schema of Richard's paradox: Turing's computing machine is an algorithm represented by a string of seven letters in a "computing machine". Its "computation" is to test all computing machines (including itself) for "circles", and form a diagonal number from the computations of the non-circular or "successful" computing machines. It does this, starting in sequence from 1, by converting the numbers (base 8) into strings of seven letters to test. When it arrives at its own number, it creates its own letter-string. It decides it is the letter-string of a successful machine, but when it tries to do this machine's (its own) computation it locks in a circle and can't continue. Thus, we have arrived at Richard's paradox. (If you are bewildered see Turing's proof for more). A number of similar undecidability proofs appeared soon before and after Turing's proof: April 1935: Proof of Alonzo Church ("An Unsolvable Problem of Elementary Number Theory"). His proof was to "...propose a definition of effective calculability ... and to show, by means of an example, that not every problem of this class is solvable" (Undecidable p. 90)) 1946: Post correspondence problem (cf Hopcroft and Ullman p. 193ff, p. 407 for the reference) April 1947: Proof of Emil Post (Recursive Unsolvability of a Problem of Thue) (Undecidable p. 293). This has since become known as "The Word problem of Thue" or "Thue's Word Problem" (Axel Thue proposed this problem in a paper of 1914 (cf References to Post's paper in Undecidable, p. 303)). Rice's theorem: a generalized formulation of Turing's second theorem (cf Hopcroft and Ullman p. 185ff) Greibach's theorem: undecidability in language theory (cf Hopcroft and Ullman p. 205ff and reference on p. 401 ibid: Greibach "The undecidability of the ambiguity problem for minimal lineal grammars," Information and Control 6:2, 117–125, also reference on p. 402 ibid: Greibach "A note on undecidable properties of formal languages", Math Systems Theory 2:1, 1–6.) Penrose tiling questions. Information theory [edit] Compression of random strings [edit] Main article: Chaitin's incompleteness theorem For an exposition suitable for non-specialists, see Beltrami p. 108ff. Also see Franzen Chapter 8 pp. 137–148, and Davis pp. 263–266. Franzén's discussion is significantly more complicated than Beltrami's and delves into Ω—Gregory Chaitin's so-called "halting probability". Davis's older treatment approaches the question from a Turing machine viewpoint. Chaitin has written a number of books about his endeavors and the subsequent philosophic and mathematical fallout from them. A string is called (algorithmically) random if it cannot be produced from any shorter computer program. While most strings are random, no particular one can be proved so, except for finitely many short ones: : "A paraphrase of Chaitin's result is that there can be no formal proof that a sufficiently long string is random..." Beltrami observes that "Chaitin's proof is related to a paradox posed by Oxford librarian G. Berry early in the twentieth century that asks for 'the smallest positive integer that cannot be defined by an English sentence with fewer than 1000 characters.' Evidently, the shortest definition of this number must have at least 1000 characters. However, the sentence within quotation marks, which is itself a definition of the alleged number is less than 1000 characters in length!" Natural sciences [edit] Main article: No-go theorem In natural science, impossibility theorems are derived as mathematical results proven within well-established scientific theories. The basis for this strong acceptance is a combination of extensive evidence of something not occurring, combined with an underlying theory, very successful in making predictions, whose assumptions lead logically to the conclusion that something is impossible. Two examples of widely accepted impossibilities in physics are perpetual motion machines, which violate the law of conservation of energy, and exceeding the speed of light, which violates the implications of special relativity. Another is the uncertainty principle of quantum mechanics, which asserts the impossibility of simultaneously knowing both the position and the momentum of a particle. There is also Bell's theorem: no physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics. While an impossibility assertion in natural science can never be absolutely proved, it could be refuted by the observation of a single counterexample. Such a counterexample would require that the assumptions underlying the theory that implied the impossibility be re-examined. See also [edit] List of unsolved problems in mathematics – Solutions of these problems are still being searched for. In contrast, the above problems are known to have no solution. Paradoxes of set theory Notes and references [edit] ^ Pudlák, pp. 255–256. ^ Weisstein, Eric W. "Circle Squaring". mathworld.wolfram.com. Retrieved 2019-12-13. ^ Raatikainen, Panu (2018), "Gödel's Incompleteness Theorems", in Zalta, Edward N. (ed.), The Stanford Encyclopedia of Philosophy (Fall 2018 ed.), Metaphysics Research Lab, Stanford University, retrieved 2019-12-13 ^ Baker, Theodore; Gill, John; Solovay, Robert (1975). "Relativizations of the P=?NP Question". SIAM Journal on Computing. 4 (4): 431–442. doi:10.1137/0204037. Retrieved 2022-12-11. ^ More generally, proof by infinite descent is applicable to any well-ordered set. ^ Hardy and Wright, p. 42 ^ Hardy and Wright, p. 40 ^ Nagel and Newman p. 8 ^ Hardy and Wright p. 159 ^ Hardy and Wright p. 176 ^ Hardy and Wright p. 159 referenced by E. Hecke. (1923). Vorlesungen über die Theorie der algebraischen Zahlen. Leipzig: Akademische Verlagsgesellschaft ^ Jump up to: a b Nagel and Newman, p. 9 ^ Nagel and Newman, p. 10 ^ Franzén p.71 ^ Nagel, Ernest; Newman, James R. (1958). Gödel's proof. Lulu.com. pp. 60 ff. ISBN 0-359-07926-1. OCLC 1057623639. {{cite book}}: ISBN / Date incompatibility (help) ^ Principia Mathematica, 2nd edition 1927, p. 61, 64 in Principia Mathematica online, Vol.1 at University of Michigan Historical Math Collection ^ Jump up to: a b Gödel in Undecidable, p. 9 ^ Also received for publication in 1936 (in October, later than Turing) was a short paper by Emil Post that discussed the reduction of an algorithm to a simple machine-like "method" very similar to Turing's computing machine model (see Post–Turing machine for details). ^ Jump up to: a b c John E. Hopcroft, Jeffrey D. Ullman (1979). Introduction to Automata Theory, Languages, and Computation. Addison-Wesley. ISBN 0-201-02988-X. ^ "...there can be no machine E which ... will determine whether M [an arbitrary machine] ever prints a given symbol (0 say)" (Undecidable p. 134). Turing makes an odd assertion at the end of this proof that sounds remarkably like Rice's Theorem: : "...each of these "general process" problems can be expressed as a problem concerning a general process for determining whether a given integer n has a property G(n)... and this is equivalent to computing a number whose nth figure is 1 if G(n) is true and 0 if it is false" (Undecidable p 134). Unfortunately he doesn't clarify the point further, and the reader is left confused. 21. ^ Beltrami p. 109 22. ^ Beltrami, p. 108 Bibliography [edit] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Clarendon Press, Oxford England, 1979, reprinted 2000 with General Index (first edition: 1938). The proofs that e and pi are transcendental are not trivial, but a mathematically adept reader will be able to wade through them. Alfred North Whitehead and Bertrand Russell, Principia Mathematica to 56, Cambridge at the University Press, 1962, reprint of 2nd edition 1927, first edition 1913. Chap. 2.I. "The Vicious-Circle Principle" p. 37ff, and Chap. 2.VIII. "The Contradictions" p. 60ff. Turing, A.M. (1936), "On Computable Numbers, with an Application to the Entscheidungsproblem", Proceedings of the London Mathematical Society, 2, vol. 42, no. 1 (published 1937), pp. 230–65, doi:10.1112/plms/s2-42.1.230, S2CID 73712 (and Turing, A.M. (1938), "On Computable Numbers, with an Application to the Entscheidungsproblem: A correction", Proceedings of the London Mathematical Society, 2, vol. 43, no. 6 (published 1937), pp. 544–6, doi:10.1112/plms/s2-43.6.544). online version This is the epochal paper where Turing defines Turing machines and shows that it (as well as the Entscheidungsproblem) is unsolvable. Martin Davis, The Undecidable, Basic Papers on Undecidable Propositions, Unsolvable Problems And Computable Functions, Raven Press, New York, 1965. Turing's paper is #3 in this volume. Papers include those by Godel, Church, Rosser, Kleene, and Post. Martin Davis's chapter "What is a Computation" in Lynn Arthur Steen's Mathematics Today, 1978, Vintage Books Edition, New York, 1980. His chapter describes Turing machines in the terms of the simpler Post–Turing machine, then proceeds onward with descriptions of Turing's first proof and Chaitin's contributions. Andrew Hodges, Alan Turing: The Enigma, Simon and Schuster, New York. Cf Chapter "The Spirit of Truth" for a history leading to, and a discussion of, his proof. Hans Reichenbach, Elements of Symbolic Logic, Dover Publications Inc., New York, 1947. A reference often cited by other authors. Ernest Nagel and James Newman, Gödel's Proof, New York University Press, 1958. Edward Beltrami, What is Random? Chance and Order in Mathematics and Life, Springer-Verlag New York, Inc., 1999. Torkel Franzén, Godel's Theorem, An Incomplete Guide to Its Use and Abuse, A.K. Peters, Wellesley Mass, 2005. A recent take on Gödel's Theorems and the abuses thereof. Not so simple a read as the author believes it is. Franzén's (blurry) discussion of Turing's 3rd proof is useful because of his attempts to clarify terminology. Offers discussions of Freeman Dyson's, Stephen Hawking's, Roger Penrose's and Gregory Chaitin's arguments (among others) that use Gödel's theorems, and useful criticism of some philosophic and metaphysical Gödel-inspired dreck that he's found on the web. Pavel Pudlák, Logical Foundations of Mathematics and Computational Complexity. A Gentle Introduction, Springer 2013. (See Chapter 4 "Proofs of impossibility".) \| v t e Mathematical logic \| \| --- \| \| General \| Axiom + list Cardinality First-order logic Formal proof Formal semantics Foundations of mathematics Information theory Lemma Logical consequence Model Theorem Theory Type theory \| \| Theorems (list) and paradoxes \| Gödel's completeness and incompleteness theorems Tarski's undefinability Banach–Tarski paradox Cantor's theorem, paradox and diagonal argument Compactness Halting problem Lindström's Löwenheim–Skolem Russell's paradox \| \| Logics \| \| \| \| --- \| \| Traditional \| Classical logic Logical truth Tautology Proposition Inference Logical equivalence Consistency + Equiconsistency Argument Soundness Validity Syllogism Square of opposition Venn diagram \| \| Propositional \| Boolean algebra Boolean functions Logical connectives Propositional calculus Propositional formula Truth tables Many-valued logic + 3 + finite + ∞ \| \| Predicate \| First-order + list Second-order + Monadic Higher-order Fixed-point Free Quantifiers Predicate Monadic predicate calculus \| \| \| Set theory \| \| \| \| --- \| \| Set + hereditary Class (Ur-)Element Ordinal number Extensionality Forcing Relation + equivalence + partition Set operations: + intersection + union + complement + Cartesian product + power set + identities \| \| \| Types of sets \| Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe + constructible + Grothendieck + Von Neumann \| \| Maps and cardinality \| Function/Map + domain + codomain + image In/Sur/Bi-jection Schröder–Bernstein theorem Isomorphism Gödel numbering Enumeration Large cardinal + inaccessible Aleph number Operation + binary \| \| Set theories \| Zermelo–Fraenkel + axiom of choice + continuum hypothesis General Kripke–Platek Morse–Kelley Naive New Foundations Tarski–Grothendieck Von Neumann–Bernays–Gödel Ackermann Constructive \| \| \| Formal systems (list), language and syntax \| \| \| \| --- \| \| Alphabet Arity Automata Axiom schema Expression + ground Extension + by definition + conservative Relation Formation rule Grammar Formula + atomic + closed + ground + open Free/bound variable Language Metalanguage Logical connective + ¬ + ∨ + ∧ + → + ↔ + = Predicate + functional + variable + propositional variable Proof Quantifier + ∃ + ! + ∀ + rank Sentence + atomic + spectrum Signature String Substitution Symbol + function + logical/constant + non-logical + variable Term Theory + list \| \| \| Example axiomatic systems (list) \| of arithmetic: + Peano + second-order + elementary function + primitive recursive + Robinson + Skolem of the real numbers + Tarski's axiomatization of Boolean algebras + canonical + minimal axioms of geometry: + Euclidean: - Elements - Hilbert's - Tarski's + non-Euclidean Principia Mathematica \| \| \| Proof theory \| Formal proof Natural deduction Logical consequence Rule of inference Sequent calculus Theorem Systems + axiomatic + deductive + Hilbert - list Complete theory Independence (from ZFC) Proof of impossibility Ordinal analysis Reverse mathematics Self-verifying theories \| \| Model theory \| Interpretation + function + of models Model + equivalence + finite + saturated + spectrum + submodel Non-standard model + of arithmetic Diagram + elementary Categorical theory Model complete theory Satisfiability Semantics of logic Strength Theories of truth + semantic + Tarski's + Kripke's T-schema Transfer principle Truth predicate Truth value Type Ultraproduct Validity \| \| Computability theory \| Church encoding Church–Turing thesis Computably enumerable Computable function Computable set Decision problem + decidable + undecidable + P + NP + P versus NP problem Kolmogorov complexity Lambda calculus Primitive recursive function Recursion Recursive set Turing machine Type theory \| \| Related \| Abstract logic Algebraic logic Automated theorem proving Category theory Concrete/Abstract category Category of sets History of logic History of mathematical logic + timeline Logicism Mathematical object Philosophy of mathematics Supertask \| \| Mathematics portal \| \| Retrieved from " Categories: Mathematical logic Mathematical proofs Methods of proof Possibility Hidden categories: CS1 errors: ISBN date Articles with short description Short description is different from Wikidata Pages using sidebar with the child parameter Broad-concept articles
3360	https://mathematica.stackexchange.com/questions/195410/clarification-on-how-total-can-be-used-on-multi-dimensional-array	list manipulation - Clarification on how Total[] can be used on multi-dimensional array - Mathematica Stack Exchange Join Mathematica 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 Clarification on how Total[] can be used on multi-dimensional array Ask Question Asked 6 years, 5 months ago Modified6 years, 5 months ago Viewed 368 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I have a rank 5 tensor that ultimately I want to modify so that for the first 3 dimensions, each element is summed together. The result will be a rank 2 tensor whose elements are the summed totals from the other 3 dimensions. That was pretty abstract and difficult for me to explain.. But I believe I can achieve the result that I want by using the function Total[]. I see that I can use Total to sum along a single dimension, or multiple dimensions. I'm just a bit confused about exactly how this works. Can I simply use: ```mathematica MyArray (This is the rank 5 tensor) Table[MyArray, {1, 2, 3}]; (This is supposed to sum up all elements along indices 1,2,3 , \ leaving behind a rank 2 tensor.) ``` Are there any issues with this? Should I instead call Total three separate times? list-manipulation equation-solving summation packed-arrays Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Apr 17, 2019 at 16:42 LooseyGooseLooseyGoose 51 2 2 bronze badges 7 1 Have you tried this: c = ConstantArray[1, {3, 3, 3, 3, 3}]; Total[c, {1, 3}] ?Arnoud Buzing –Arnoud Buzing 2019-04-17 16:47:53 +00:00 Commented Apr 17, 2019 at 16:47 1 Or Total[c, 3].Henrik Schumacher –Henrik Schumacher 2019-04-17 16:56:50 +00:00 Commented Apr 17, 2019 at 16:56 So the only difference between this and what I wrote is that you called the indices {1,3} and I called the indices {1,2,3}. Is that correct? Can you clarify why you made this change?LooseyGoose –LooseyGoose 2019-04-17 16:57:50 +00:00 Commented Apr 17, 2019 at 16:57 1 Excellent, I think that is what I want to do. Thanks for the help arnould and henrik!LooseyGoose –LooseyGoose 2019-04-17 17:21:44 +00:00 Commented Apr 17, 2019 at 17:21 1 @LooseyGoose You're welcome!Henrik Schumacher –Henrik Schumacher 2019-04-17 17:38:04 +00:00 Commented Apr 17, 2019 at 17:38 \|Show 2 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. A good way to test things is to create an array where all the dimensions are different, and then check the dimensions of the result. So: mathematica array = ConstantArray[1, {2, 3, 4, 5, 6}]; Total[array, 3] //Dimensions {5, 6} This shows that Total summed over the first 3 dimensions. Another example: mathematica Total[array, {2, 4}] //Dimensions {2, 6} showing that the 3 middle dimensions have been summed over. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Apr 17, 2019 at 18:05 Carl WollCarl Woll 133k 6 6 gold badges 253 253 silver badges 366 366 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Mathematica 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. Use MathJax to format equations. MathJax reference. 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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3361	https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.06%3A_The_Quantum_Harmonic_Oscillator	7.6: The Quantum Harmonic Oscillator - Physics LibreTexts Skip to main content Table of Contents menu search Searchbuild_circle Toolbarfact_check Homeworkcancel 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 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: Quantum Mechanics University Physics III - Optics and Modern Physics (OpenStax) { } { "7.01:_Prelude_to_Quantum_Mechanics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Wavefunctions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_The_Heisenberg_Uncertainty_Principle" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_The_Schrdinger_Equation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_The_Quantum_Particle_in_a_Box" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.06:_The_Quantum_Harmonic_Oscillator" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.07:_Quantum_Tunneling_of_Particles_through_Potential_Barriers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.0A:_7.A:_Quantum_Mechanics_(Answers)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.0E:_7.E:_Quantum_Mechanics_(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.0S:_7.S:_Quantum_Mechanics_(Summary)" : "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:_The_Nature_of_Light" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Geometric_Optics_and_Image_Formation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Interference" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Diffraction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:__Relativity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Photons_and_Matter_Waves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Quantum_Mechanics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Atomic_Structure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Condensed_Matter_Physics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:__Nuclear_Physics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Particle_Physics_and_Cosmology" : "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" } Sun, 16 Mar 2025 21:22:09 GMT 7.6: The Quantum Harmonic Oscillator 4531 4531 Delmar Larsen { } Anonymous Anonymous 2 false false [ "article:topic", "authorname:openstax", "Quantum Harmonic Oscillator", "harmonic oscillator", "license:ccby", "showtoc:no", "program:openstax", "licenseversion:40", "source@ ] [ "article:topic", "authorname:openstax", "Quantum Harmonic Oscillator", "harmonic oscillator", "license:ccby", "showtoc:no", "program:openstax", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article 3. 1. 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Username Password Sign in Sign in Sign in Forgot password Contents Home Bookshelves University Physics University Physics (OpenStax) University Physics III - Optics and Modern Physics (OpenStax) 7: Quantum Mechanics 7.6: The Quantum Harmonic Oscillator Expand/collapse global location University Physics III - Optics and Modern Physics (OpenStax) Front Matter 1: The Nature of Light 2: Geometric Optics and Image Formation 3: Interference 4: Diffraction 5: Relativity 6: Photons and Matter Waves 7: Quantum Mechanics 7.1: Prelude to Quantum Mechanics 7.2: Wave functions 7.3: The Heisenberg Uncertainty Principle 7.4: The Schrӧdinger Equation 7.5: The Quantum Particle in a Box 7.6: The Quantum Harmonic Oscillator 7.7: Quantum Tunneling of Particles through Potential Barriers 7.A: Quantum Mechanics (Answers) 7.E: Quantum Mechanics (Exercises) 7.S: Quantum Mechanics (Summary) 8: Atomic Structure 9: Condensed Matter Physics 10: Nuclear Physics 11: Particle Physics and Cosmology Back Matter 7.6: The Quantum Harmonic Oscillator Last updated : Mar 16, 2025 Save as PDF 7.5: The Quantum Particle in a Box 7.7: Quantum Tunneling of Particles through Potential Barriers Page ID : 4531 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents Learning Objectives The Classic Harmonic Oscillator The Quantum Harmonic Oscillator Example 7.6.17.6.1: Classical Region of Harmonic Oscillations Solution Significance Example 7.6.27.6.2: Vibrational Energies of the Hydrogen Chloride Molecule Solution Significance Exercise 7.6.17.6.1 Exercise 7.6.27.6.2 Learning Objectives By the end of this section, you will be able to: Describe the model of the quantum harmonic oscillator Identify differences between the classical and quantum models of the harmonic oscillator Explain physical situations where the classical and the quantum models coincide Oscillations are found throughout nature, in such things as electromagnetic waves, vibrating molecules, and the gentle back-and-forth sway of a tree branch. In previous chapters, we used Newtonian mechanics to study macroscopic oscillations, such as a block on a spring and a simple pendulum. In this chapter, we begin to study oscillating systems using quantum mechanics. We begin with a review of the classic harmonic oscillator. The Classic Harmonic Oscillator A simple harmonic oscillator is a particle or system that undergoes harmonic motion about an equilibrium position, such as an object with mass vibrating on a spring. In this section, we consider oscillations in one-dimension only. Suppose a mass moves back-and-forth along the xx-direction about the equilibrium position, x=0x=0. In classical mechanics, the particle moves in response to a linear restoring force given by Fx=−kxFx=−kx, where xx is the displacement of the particle from its equilibrium position. The motion takes place between two turning points, x±Ax±A, where A denotes the amplitude of the motion. The position of the object varies periodically in time with angular frequency ω=k/m−−−−√ω=k/m, which depends on the mass m of the oscillator and on the force constant kk of the net force, and can be written as x(t)=Acos(ωt+ϕ).(7.6.1)(7.6.1)x(t)=Acos⁡(ωt+ϕ). The total energy EE of an oscillator is the sum of its kinetic energy K=mu2/2K=mu2/2 and the elastic potential energy of the force U(x)=kx2/2U(x)=kx2/2, E=12mu2+12kx2.(7.6.2)(7.6.2)E=12mu2+12kx2. At turning points x=±Ax=±A, the speed of the oscillator is zero; therefore, at these points, the energy of oscillation is solely in the form of potential energy E=kA2/2E=kA2/2. The plot of the potential energy U(x)U(x) of the oscillator versus its position xx is a parabola (Figure 7.6.17.6.1). The potential-energy function is a quadratic function of xx, measured with respect to the equilibrium position. On the same graph, we also plot the total energy EE of the oscillator, as a horizontal line that intercepts the parabola at x=±Ax=±A. Then the kinetic energy KK is represented as the vertical distance between the line of total energy and the potential energy parabola. Figure 7.6.17.6.1: The potential energy well of a classical harmonic oscillator: The motion is confined between turning points at x=−Ax=−A and at x=+Ax=+A. The energy of oscillations is E=kA2/2E=kA2/2. In this plot, the motion of a classical oscillator is confined to the region where its kinetic energy is nonnegative, which is what the energy relation Equation 7.6.27.6.2 says. Physically, it means that a classical oscillator can never be found beyond its turning points, and its energy depends only on how far the turning points are from its equilibrium position. The energy of a classical oscillator changes in a continuous way. The lowest energy that a classical oscillator may have is zero, which corresponds to a situation where an object is at rest at its equilibrium position. The zero-energy state of a classical oscillator simply means no oscillations and no motion at all (a classical particle sitting at the bottom of the potential well in Figure 7.6.17.6.1). When an object oscillates, no matter how big or small its energy may be, it spends the longest time near the turning points, because this is where it slows down and reverses its direction of motion. Therefore, the probability of finding a classical oscillator between the turning points is highest near the turning points and lowest at the equilibrium position. (Note that this is not a statement of preference of the object to go to lower energy. It is a statement about how quickly the object moves through various regions.) The Quantum Harmonic Oscillator One problem with this classical formulation is that it is not general. We cannot use it, for example, to describe vibrations of diatomic molecules, where quantum effects are important. A first step toward a quantum formulation is to use the classical expression k=mω2k=mω2 to limit mention of a “spring” constant between the atoms. In this way the potential energy function can be written in a more general form, U(x)=12mω2x2.(7.6.3)(7.6.3)U(x)=12mω2x2. Combining this expression with the time-independent Schrӧdinger equation gives −ℏ2md2ψ(x)dx2+12mω2x2ψ(x)=Eψ(x).(7.6.4)(7.6.4)−ℏ2md2ψ(x)dx2+12mω2x2ψ(x)=Eψ(x). To solve Equation 7.6.47.6.4, that is, to find the allowed energies EE and their corresponding wavefunctions ψ(x)ψ(x) - we require the wavefunctions to be symmetric about x=0x=0 (the bottom of the potential well) and to be normalizable. These conditions ensure that the probability density \|ψ(x)\|2\|ψ(x)\|2 must be finite when integrated over the entire range of x from −∞−∞ to +∞+∞. How to solve Equation 7.6.47.6.4 is the subject of a more advanced course in quantum mechanics; here, we simply cite the results. The allowed energies are En=(n+12)ℏω=2n+12ℏω(7.6.5)(7.6.6)(7.6.5)En=(n+12)ℏω(7.6.6)=2n+12ℏω with n=0,1,2,3,...n=0,1,2,3,... The wavefunctions that correspond to these energies (the stationary states or states of definite energy) are ψn(x)=Nne−β2x2/2Hn(βx),n=0,1,2,3,...(7.6.7)(7.6.7)ψn(x)=Nne−β2x2/2Hn(βx),n=0,1,2,3,... where β=mω/ℏ−−−−−√β=mω/ℏ, NnNn is the normalization constant, and Hn(y)Hn(y) is a polynomial of degree nn called a Hermite polynomial. The first four Hermite polynomials are H0(y)=1H0(y)=1 H1(y)=2yH1(y)=2y H2(y)=4y2−2H2(y)=4y2−2 H3(y)=8y3−12y.H3(y)=8y3−12y. A few sample wavefunctions are given in Figure 7.6.27.6.2. As the value of the principal number increases, the solutions alternate between even functions and odd functions about x=0x=0. Figure 7.6.27.6.2: The first five wavefunctions of the quantum harmonic oscillator. The classical limits of the oscillator’s motion are indicated by vertical lines, corresponding to the classical turning points at x=±Ax=±A of a classical particle with the same energy as the energy of a quantum oscillator in the state indicated in the figure. Example 7.6.17.6.1: Classical Region of Harmonic Oscillations Find the amplitude AA of oscillations for a classical oscillator with energy equal to the energy of a quantum oscillator in the quantum state nn. Strategy To determine the amplitude AA, we set the classical energy E=kx2/2=mω2A2/2E=kx2/2=mω2A2/2 equal to EnEn given by Equation 7.6.67.6.6. Solution We obtain EnAn=mω2A2n/2=2mω2En−−−−−−−√=2mω22n+12ℏω−−−−−−−−−−−−√=(2n+1)ℏmω−−−−−−−−−−√.En=mω2An2/2An=2mω2En=2mω22n+12ℏω=(2n+1)ℏmω. Significance As the quantum number n increases, the energy of the oscillator and therefore the amplitude of oscillation increases (for a fixed natural angular frequency. For large n, the amplitude is approximately proportional to the square root of the quantum number . Several interesting features appear in this solution. Unlike a classical oscillator, the measured energies of a quantum oscillator can have only energy values given by Equation 7.6.67.6.6. Moreover, unlike the case for a quantum particle in a box, the allowable energy levels are evenly spaced, ΔE=En+1−En=2(n+1)+12ℏω−2n+12ℏω=ℏω=hf.(7.6.8)(7.6.9)(7.6.10)(7.6.8)ΔE=En+1−En(7.6.9)=2(n+1)+12ℏω−2n+12ℏω(7.6.10)=ℏω=hf. When a particle bound to such a system makes a transition from a higher-energy state to a lower-energy state, the smallest-energy quantum carried by the emitted photon is necessarily hfhf. Similarly, when the particle makes a transition from a lower-energy state to a higher-energy state, the smallest-energy quantum that can be absorbed by the particle is hfhf. A quantum oscillator can absorb or emit energy only in multiples of this smallest-energy quantum. This is consistent with Planck’s hypothesis for the energy exchanges between radiation and the cavity walls in the blackbody radiation problem. Example 7.6.27.6.2: Vibrational Energies of the Hydrogen Chloride Molecule The HClHCl diatomic molecule consists of one chlorine atom and one hydrogen atom. Because the chlorine atom is 35 times more massive than the hydrogen atom, the vibrations of the HClHCl molecule can be quite well approximated by assuming that the Cl atom is motionless and the H atom performs harmonic oscillations due to an elastic molecular force modeled by Hooke’s law. The infrared vibrational spectrum measured for hydrogen chloride has the lowest-frequency line centered at f=8.88×1013Hzf=8.88×1013Hz. What is the spacing between the vibrational energies of this molecule? What is the force constant k of the atomic bond in the HCl molecule? Strategy The lowest-frequency line corresponds to the emission of lowest-frequency photons. These photons are emitted when the molecule makes a transition between two adjacent vibrational energy levels . Assuming that energy levels are equally spaced, we use Equation 7.6.107.6.10 to estimate the spacing. The molecule is well approximated by treating the Cl atom as being infinitely heavy and the H atom as the mass mm that performs the oscillations. Treating this molecular system as a classical oscillator, the force constant is found from the classical relation k=mω2k=mω2. Solution The energy spacing is ΔE=hf=(4.14×10−15eV⋅s)(8.88×1013Hz)=0.368eV.ΔE=hf=(4.14×10−15eV⋅s)(8.88×1013Hz)=0.368eV. The force constant is k=mω2=m (2πf)2=(1.67×10 −27kg)(2π×8.88×10 13Hz)2=520N/m.k=mω2=m (2πf)2=(1.67×10 −27kg)(2π×8.88×10 13Hz)2=520N/m. Significance The force between atoms in an HCl molecule is surprisingly strong. The typical energy released in energy transitions between vibrational levels is in the infrared range. As we will see later, transitions in between vibrational energy levels of a diatomic molecule often accompany transitions between rotational energy levels . Exercise 7.6.17.6.1 The vibrational frequency of the hydrogen iodide HI diatomic molecule is 6.69×10 13Hz6.69×10 13Hz. What is the force constant of the molecular bond between the hydrogen and the iodine atoms? What is the energy of the emitted photon when this molecule makes a transition between adjacent vibrational energy levels ? Answer a : 295 N/m Answer b : 0.277 eV The quantum oscillator differs from the classic oscillator in three ways: First, the ground state of a quantum oscillator is E0=ℏω/2E0=ℏω/2, not zero. In the classical view, the lowest energy is zero. The nonexistence of a zero-energy state is common for all quantum-mechanical systems because of omnipresent fluctuations that are a consequence of the Heisenberg uncertainty principle . If a quantum particle sat motionless at the bottom of the potential well, its momentum as well as its position would have to be simultaneously exact, which would violate the Heisenberg uncertainty principle . Therefore, the lowest-energy state must be characterized by uncertainties in momentum and in position, so the ground state of a quantum particle must lie above the bottom of the potential well. - Second, a particle in a quantum harmonic oscillator potential can be found with nonzero probability outside the interval −A≤x≤+A−A≤x≤+A. In a classic formulation of the problem, the particle would not have any energy to be in this region. The probability of finding a ground-state quantum particle in the classically forbidden region is about 16%. - Third, the probability density distributions \|ψn(x)\|2\|ψn(x)\|2 for a quantum oscillator in the ground low-energy state, ψ0(x)ψ0(x), is largest at the middle of the well (x=0)(x=0). For the particle to be found with greatest probability at the center of the well, we expect that the particle spends the most time there as it oscillates. This is opposite to the behavior of a classical oscillator, in which the particle spends most of its time moving with relative small speeds near the turning points. Exercise 7.6.27.6.2 Find the expectation value of the position for a particle in the ground state of a harmonic oscillator using symmetry. Answer b : ⟨x⟩=0⟨x⟩=0 Quantum probability density distributions change in character for excited states, becoming more like the classical distribution when the quantum number gets higher. We observe this change already for the first excited state of a quantum oscillator because the distribution \|ψ1(x)\|2\|ψ1(x)\|2 peaks up around the turning points and vanishes at the equilibrium position, as seen in Figure 7.6.27.6.2. In accordance with Bohr’s correspondence principle , in the limit of high quantum numbers, the quantum description of a harmonic oscillator converges to the classical description, which is illustrated in Figure 7.6.37.6.3. The classical probability density distribution corresponding to the quantum energy of the n=12n=12 state is a reasonably good approximation of the quantum probability distribution for a quantum oscillator in this excited state. This agreement becomes increasingly better for highly excited states. Figure 7.6.17.6.1: The probability density distribution for finding the quantum harmonic oscillator in its n=12n=12 quantum state. The dashed curve shows the probability density distribution of a classical oscillator with the same energy. This page titled 7.6: The Quantum Harmonic Oscillator is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7.5: The Quantum Particle in a Box 7.7: Quantum Tunneling of Particles through Potential Barriers Was this article helpful? Yes No Recommended articles 4.6: The Quantum Harmonic OscillatorThe quantum harmonic oscillator is a model built in analogy with the model of a classical harmonic oscillator. It models the behavior of many physical... 6.6: The Quantum Harmonic OscillatorThe quantum harmonic oscillator is a model built in analogy with the model of a classical harmonic oscillator. 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3362	https://pmc.ncbi.nlm.nih.gov/articles/PMC6327898/	Are serial hematocrit measurements sensitive enough to predict intra-abdominal injuries in blunt abdominal trama? - 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 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 Open Access Emerg Med . 2019 Jan 7;11:9–13. doi: 10.2147/OAEM.S180398 Search in PMC Search in PubMed View in NLM Catalog Add to search Are serial hematocrit measurements sensitive enough to predict intra-abdominal injuries in blunt abdominal trama? Reza Mosaddegh Reza Mosaddegh 1 Emergency Medicine Management Research Center, Iran University of Medical Sciences, Tehran, Iran, dr.nsi_noohi@yahoo.com Find articles by Reza Mosaddegh 1, Neda Ashayeri Neda Ashayeri 2 Department of Pediatric Hematology and Oncology, Ali Asghar Children’s Hospital, Iran University of Medical Sciences, Tehran, Iran Find articles by Neda Ashayeri 2, Mahdi Rezai Mahdi Rezai 1 Emergency Medicine Management Research Center, Iran University of Medical Sciences, Tehran, Iran, dr.nsi_noohi@yahoo.com Find articles by Mahdi Rezai 1, Gholamreza Masoumi Gholamreza Masoumi 3 Trauma and Injury Research Center, Iran University of Medical Sciences, Tehran, Iran Find articles by Gholamreza Masoumi 3, Samira Vaziri Samira Vaziri 1 Emergency Medicine Management Research Center, Iran University of Medical Sciences, Tehran, Iran, dr.nsi_noohi@yahoo.com Find articles by Samira Vaziri 1, Fatemeh Mohammadi Fatemeh Mohammadi 4 Research and Development Center of Firoozgar Hospital, Iran University of Medical Sciences, Tehran, Iran Find articles by Fatemeh Mohammadi 4, Hamed Givzadeh Hamed Givzadeh 5 Orthopedic Research Center, Guilan University of Medical Sciences, Rasht, Iran Find articles by Hamed Givzadeh 5, Nasrin Noohi Nasrin Noohi 1 Emergency Medicine Management Research Center, Iran University of Medical Sciences, Tehran, Iran, dr.nsi_noohi@yahoo.com Find articles by Nasrin Noohi 1,✉ Author information Article notes Copyright and License information 1 Emergency Medicine Management Research Center, Iran University of Medical Sciences, Tehran, Iran, dr.nsi_noohi@yahoo.com 2 Department of Pediatric Hematology and Oncology, Ali Asghar Children’s Hospital, Iran University of Medical Sciences, Tehran, Iran 3 Trauma and Injury Research Center, Iran University of Medical Sciences, Tehran, Iran 4 Research and Development Center of Firoozgar Hospital, Iran University of Medical Sciences, Tehran, Iran 5 Orthopedic Research Center, Guilan University of Medical Sciences, Rasht, Iran ✉ Correspondence: Nasrin Noohi, Emergency Medicine Management Research Center, Iran University of Medical Sciences, Firoozgar Hospital, Valadi St., Valiasr St., Tehran, Iran, Tel +98 912 494 0802, Fax +98 218 894 2622, Email dr.nsi_noohi@yahoo.com Collection date 2019. © 2019 Mosaddegh et al. This work is published and licensed by Dove Medical Press Limited The full terms of this license are available at and incorporate the Creative Commons Attribution – Non Commercial (unported, v3.0) License ( By accessing the work you hereby accept the Terms. Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed. PMC Copyright notice PMCID: PMC6327898 PMID: 30662287 Abstract Objective Routine serial hematocrit measurements are a component of the trauma evaluation for patients without serious injury identified on initial evaluation. We sought to determine whether serial hematocrit testing was useful in predicting the probable injuries in blunt abdominal trauma. Materials and method We performed a prospective study of trauma patients admitted in our observation unit over a 12-month period. Patients routinely underwent serial hematocrit testing in 6-hour intervals (two hematocrit levels). We compared trauma patients with a hematocrit drop of 5 and 10 points or more to those without a significant hematocrit drop. Results Five hundred forty-two isolated blunt abdominal trauma patients were admitted to observation unit, and 468 patients (86.35%) had serial hematocrit during their 6-hour stay. Of these patients, 36.11% had a hematocrit drop of 5 or more and 12.61% a drop of 10 or more. Of patients with the hematocrit drop >10, 50.8% have had diagnostic manifestations of intra-abdominal injury in both ultrasonographic and computed tomography scanning (P<0.001). There was no significant correlation between hematocrit drop >5 and positive imaging. Conclusion Although serial hematocrit testing may be useful in specific situations, routine use of serial hematocrit testing in trauma patients at a level I trauma center’s observation unit did not significantly aid in the prediction of occult injuries. Keywords: Blunt abdominal trauma, serial hematocrit, ultrasonography, computed tomography Introduction Trauma is the first leading cause of mortality all over the world and most of the victims are among people aged 14–44 years.1 In 2015, WHO reported that accidents were responsible for 8.1% of deaths in Iran.2 Also, surveys have shown that abdominoplevic trauma is a main cause of morbidity and mortality in trauma patients.3 Patients are referred to emergency department because abdominal trauma may have vague symptoms and signs, and estimating of the severity of injury through abdominal examination is difficult and is not reliably possible.4–6 The benefit of serial hematocrit measurement for diagnosis of intra-abdominal injuries has been assessed in some studies and it can be used as a screening tool in trauma patients.7,8 Zehtabchi et al found that a change in hematocrit of >5 points in 4 hours predicted serious injury.7 Another study that was performed in patients with penetrating trauma stated that a hematocrit change of 6.5 points at 15 minutes predicts significant injury.8 The aim of the present study was to evaluate whether serial hematocrit testing could precisely predict intra-abdominal injuries in blunt abdominal trauma patients. Materials and methods Design and ethics This prospective study was performed on patients admitted to emergency department of Firoozgar University Hospital from May 1, 2014 to April 31, 2016. Firoozgar Hospital is a teaching hospital located at downtown of Tehran, Iran, as a level I trauma center. The ethics committee of Iran University of Medical Sciences approved the study. Participants We enrolled patients referred to emergency department with blunt trauma after obtaining written informed consent form from all patients. This study was conducted in accordance with the Declaration of Helsinki. All types of trauma such as motor vehicle collision, falling from height, and fighting were included the study. Exclusion criteria were patient death before completion of evaluations, penetrating abdominal trauma, necessity for blood transfusion in the first 6 hours of admission, abnormal electrocardiogram, abnormal head computed tomography (CT) scan, Glasgow Coma Scale score <14, injuries resulting in immediate transferring to operation room before 6 hours. Process and assessment In our emergency department the trauma patients are managed according to advanced Trauma Life Support instructions and algorithms. The primary and secondary surveys were done by emergency medicine residents supervised by emergency medicine attendings. The focused assessment with sonography for trauma (FAST exam) and chest radiography were done as an adjunct to primary survey. General surgeon, neurosurgeon, and orthopedic surgeon visited the patients if needed. In our emergency department, the trauma patients are allocated to subacute and acute area according to severity of trauma and observed by emergency medicine residents. Other ancillary tests such as supplementary radiographies and laboratory tests are pursued in those areas. As part of the observation, especially surgeons, checked hematocrit levels every 6 hours to aid in the detection of occult hemorrhagic injuries that may not have been identified in the initial trauma assessment by FAST exam and CT scanning. Physicians might order spiral abdominopelvic CT scans with intravenous contrast for some trauma patients who had indication according to trauma guidelines or their judgments. Post graduate year 3 emergency medicine residents reviewed all diagnostic and therapeutic processes of admitted trauma patients. They recorded patient characteristics, injuries, hematocrit levels, length of stay, results of FAST exam and spiral abdominopelvic CT scanning. Also the adverse outcomes including instability of vital signs, advanced airway management (intubation), death, and whether these patients were discharged from the hospital or admitted to an inpatient ward were also recorded. Significant intra-abdominal injury was defined as any injury detected during observation period that required operative repair or hospitalization in an inpatient ward. Patients were considered to have a hematocrit drop of 5 points or more if any of the hematocrit levels after the initial hematocrit level (which was drawn on patient’s ED admission) was 5 points or more lower. We used a hematocrit decrease of 5 points or more based on the previous study that established this threshold as a predictor of intra-abdominal injury.7 We also considered a hematocrit drop of 10 or more to assess whether the diagnostic values changed or not. We also followed up patients to identify outcome 30 days after trauma. The hematocrit drop and its relationship with the results of ultrasound and CT scans were assessed to determine the diagnostic values (sensitivity and specificity) of these drops. Statistics All statistical analyses were performed using SPSS 16.0 (SPSS Inc. Chicago, IL, USA). Quantitative descriptive data were reported using mean with 95% CI. To determine the diagnostic value of hematocrit drop in trauma patients, indicators of sensitivity, specificity, positive predictive value (PPV), negative predictive value (NPV), and positive and negative likelihood ratio were calculated. The level of significance was considered <5%. Results During the study period, 617 patients were admitted to the observation unit, and 524 patients (85%) had at least two hematocrits results between their initial evaluation and their observation unit stays. The observation unit trauma protocol called for hematocrit levels to be drawn every 6 hours. All the patients observed in this unit suffered from isolated sustained blunt trauma. The most common specific mechanism was motor vehicle accident (331/524, 63.2%). Of the patients, 79.2% were male, and average age was 32.3 years (median age, 24 years; Table 1). Table 1. Data of patients \| Characteristics \| Results, n (%) \| :--- \| \| Age (years), mean±SD \| 32.27±16.39 \| \| Sex (male) \| 415 (79.2) \| \| Injury mechanisms \| \| Motor vehicle accident \| 331 (63.2) \| \| Falling down and others \| 193 (36.8) \| Open in a new tab All patients had received primary and secondary trauma surveys, including FAST exam and CT scanning per the trauma team discretion. All the trauma patients had a FAST Exam (initially and serial according to the severity of injury and order of attending physician), and of these, 92 patients (17.6%) had positive FAST (free fluid in abdomen or pelvis cavity). One hundred seventy-five patients (32.29%) had a CT scan of the abdomen and pelvis performed before admission to the observation unit and of these 90 patients (17.2%) had a positive scanning and showed intra-abdominal or pelvis injuries or hematoma (Table 2). Table 2. Comparison of patients with positive FAST and abdominal CT scanning and hematocrit change of 5 and 10 or more \| Characteristics \| Results \| :--- \| \| FAST exam (positive) \| 92 (17.6) \| \| Abdominal CT scan (positive) \| 90 (17.2) \| \| Hematocrit drop (≥5) \| 86 (16.41) \| \| Hematocrit drop (≥10) \| 38 (7.25) \| \| Final outcome (within 30 days of the observation unit stay) \| \| Alive \| 515 (98.3) \| \| Dead \| 9 (1.7) \| Open in a new tab Note: Values are in n (%). Abbreviations: CT, computed tomography; FAST, focused assessment with sonography for trauma. The average changes in hematocrit were a decrease of 3.64% (median, 3% decrease). Of the 524 patients, 86 (16.41%) had a hematocrit drop of 5 points or more during their observation unit stay. The hematocrit drop of 10 points or more was seen in 38 patients (7.25%). Comparison between patients with a significant hematocrit drop (≥5% and ≥10%) and those without are listed in Table 2. Thirty-three (38.4%) of patients with a hematocrit drop of 5% or more had an abnormal FAST exam vs 59 (13.5%) of patients without a significant hematocrit drop (P=0.324; OR =1.27; 95% CI, 0.792–2.039). In patients with a hematocrit drop of 10 points or more, 30 (78.9%) patients had an abnormal FAST vs 62 (12.7%) of patients without a significant hematocrit drop (P<0.001; OR =6.72; 95% CI, 3.779–11.964). Thirty-three (38.4%) of patients with a hematocrit drop of 5% or more had abnormal CT scans vs 57 (13.0%) of patients without a significant hematocrit drop (P=0.263; OR =1.325; 95% CI, 0.824–2.130). In patients with a hematocrit drop of 10% or more, 30 (78.9%) patients had abnormal CT scans vs 60 (12.3%) patients without a significant hematocrit drop (P<0.001; OR =6.983; 95% CI, 3.918–12.446; Table 3). Table 3. Relationship of patients with hematocrit change of 5 and 10 or more vs those without change with positive FAST and abdominal CT scanning \| Characteristics \| N (%) \| P-value \| 95% CI \| OR \| :--- :--- \| Hematocrit change of 5 or more \| \| FAST exam (positive) \| 33 (20.0) \| 0.325 \| 0.792–2.039 \| 1.271 \| \| Abdominal CT scan (positive) \| 33 (20.0) \| 0.263 \| 0.824–2.130 \| 1.325 \| \| Hematocrit change of 10 or more \| \| FAST exam (positive) \| 30 (50.8) \| <0.001 \| 3.779–11.964 \| 6.724 \| \| Abdominal CT scan (positive) \| 415 (50.8) \| <0.001 \| 3.918–12.446 \| 6.983 \| Open in a new tab Abbreviations: CT, computed tomography; FAST, focused assessment with sonography for trauma. In our study, trauma patients who had intra-abdominal injuries seen in CT scan images were admitted to intensive care units (ICUs) or sent for laparotomy. Thus, we can conclude that an Hct drop of 10% or more was associated with ICU care or laparotomy (P<0.001). This study showed that a hematocrit drop of 5% or more has 35.87% sensitivity, 69.45% specificity, 20% PPV, and 83.56% NPV for abnormal FAST. The sensitivity and specificity for abnormal abdominopelvic CT scanning were 36.67% and 69.58%, respectively, and PPV and NPV for abnormal abdominopelvic CT scanning were 20.0% and 84.12%, respectively (Table 4). Table 4. Predictive values of hematocrit drop of 5 and 10 or more in positive FAST and abdominal CT scanning \| Characteristics \| Sensitivity \| Specificity \| PPV \| NPV \| LR+ \| LR− \| :--- :--- :--- \| Hematocrit change of 5 or more \| \| Positive FAST \| 35.87 \| 69.45 \| 20.0 \| 83.56 \| 1.16 \| 0.93 \| \| Positive abdominal CT scanning \| 36.67 \| 69.58 \| 20.0 \| 84.12 \| 1.18 \| 0.91 \| \| Hematocrit change of 10 or more \| \| Positive FAST \| 32.61 \| 93.29 \| 50.85 \| 86.67 \| 4.71 \| 0.72 \| \| Positive abdominal CT scanning \| 33.33 \| 93.32 \| 50.85 \| 87.10 \| 4.76 \| 0.72 \| Open in a new tab Abbreviations: CT, computed tomography; FAST, focused assessment with sonography for trauma; NPV, negative predictive value; PPV, positive predictive value. According to the results of this study, the hematocrit drop of 10% or more had 32.61% sensitivity, 93.29% specificity, 50.85% PPV, and 86.67% NPV for abnormal FAST exam, whereas, the sensitivity, specificity, PPV, and NPV for abnormal abdominopelvic CT scanning were 33.33%, 93.32%, 50.85%, and 87.10%, respectively (Table 4). We used analysis of ROC curves to test the diagnostic performance of drop in hematocrit in predicting abnormal FAST examination and CT scanning in the study subjects. In this analysis, the area under the ROC curve was significantly different (P<0.001) from the unity line. Patient records were reviewed for 30 days after the observation unit visit. Only nine of these cases (1.7%) have died during stay in observation unit or the admitting service within 30-day follow-up. Eight of them (89%) suffered from intra-abdominal injuries (free fluid in FAST exam and CT scan images) and admitted to ICUs or operation room. We compared the hematocrit drop after 6 hours in survivors and nonsurvivors and found that there was no significant differences between the two groups (−3% vs −3.64%, P=0.642; Table 2). Discussion Currently, physical examination and imaging studies such as extended FAST and CT in trauma patients apply to determine the severity of injuries and blood loss, but in certain cases the diagnosis can be challenging. Recently, bedside ultrasound in the initial assessment of trauma patients plays a significant role,9–11 and it’s sensitivity and specificity for the presence of intra-abdominal fluid in patients suffering from trauma is acceptable (75%–93.8% and 97%–100%, respectively).3,6 However, serial ultrasound examination may be needed in some situations11,12 and of course ultrasound is an operator-dependent modality. Serial hematocrit measurements are one of current screening tests as part of the trauma assessment. The aim of serial hematocrit testing is to detect occult, and may be serious intra-abdominal injuries. However, dilution of blood due to infusion of intravenous fluids may decrease the diagnostic value of dropped hematocrit (DHct). Thus, we designed this study to test the diagnostic performance of DHct in a population of trauma patients. The operating characteristics of an efficient screening test require a high sensitivity and a low likelihood ratio of a negative test. These qualities will allow a negative screening test to strongly rule out patients for the disease in question. In the present study, there was no significant difference between the normal and abnormal FAST examinations and CT scanning for DHct-6 h >5 points. The sensitivity of DHct-6 h >5 points in predicting normal FAST examination and CT scanning was low. Meanwhile there was a significant difference between the positive and negative FAST examinations and CT scanning for DHct-6 h >10%. Using a cutoff for DHct-6 h of 10% could not improve the sensitivity. Likelihood ratios using DHct-6 h of >10% had a higher +LR and a lower –LR in detecting positive FAST examination and CT scanning. Previous studies have assessed DHct within 1 hour after blood loss.8,13–14 Paradis et al8 reported a sensitivity of 20% and 27% for DHct 15 and 30 minutes after ED arrival, respectively. Kass et al14 reported a sensitivity of >90% for DHct, 30 minutes following phlebotomy of 500 mL in healthy individuals. We chose 6-hour intervals to measure DHct. According to our analysis, DHct-6 h >5 points, and DHct-6 h >10 points showed low sensitivity in predicting abnormal FAST examination and CT scanning. While the DHct proved to have a relatively low sensitivity, their specificity especially in DHct-6 h >10 points was quite high. These findings indicate that DHct could be a powerful test for ruling in abnormal FAST examination and CT scanning which could be indicative of major injury. A similarly high specificity was also found by Paradis et al.7 In summary, we showed that a drop in Hct greater than 5 and especially 10 points after 6 hours has a high specificity (>93%) and a high likelihood ratio for a positive test (4.7). A trauma patient who drops his/her Hct more than 5 or 10 points in 6 hours, compared with the one without a similar drop, has much higher odds of having major injury. The low sensitivity and likelihood ratio of a negative test for the drop in hematocrit greater than 5 or 10 points indicates that these tests cannot be used to reliably rule out abnormal FAST examination and CT scanning which could be indicative of major injury. Trauma patients with DHct <5 or 10 points have approximately the same odds of having positive FAST examination and CT scanning. There are several limitations to our study. The study analyzed a special section of trauma patients and our data represent only a portion of our hospital’s total trauma admissions. We excluded the severely injured patients with multiple organ involvement or unstable patients who received blood transfusion in order to more accurately study the effect of hemorrhage on hematocrit measurement. This exclusion, although inevitable, resulted in the elimination of a significant number of patients with major injury. Similarly, only the data of patients who stayed in the ED for 6 hours were analyzed; patients who were transferred to other units, such as the operating room, were not included in this analysis. Conclusion At a level I trauma center, in which patients receive a trauma evaluation and diagnostic testing before admission to the observation unit, the use of routine serial hematocrit testing as a screening tool during observation rarely provides additional diagnostic information. But in hospitals where emergency physicians or surgeons do not have access to CT scan machine, monitoring Hct changes can be helpful. We conclude that a drop in Hct of more than 10% during the first 6 hours is suggestive of positive imaging relative to major trauma (high specificity and +LR). However, the drop of Hct less than these degrees cannot be used to rule out abnormal imaging or major injury (low sensitivity and −LR). Acknowledgments The authors wish to thank all staff of Emergency Medicine Management Research Center for their kind help in all steps of this study. This study was supported by Iran University of Medical Sciences, Tehran, Iran. Footnotes Disclosure The authors report no conflicts of interest in this work. References 1.Center for Disease Control and Prevention . National Vital Statistics System. Atlanta, GA: National Center for Health Statistics; 2004. [Google Scholar] 2.WHO . World Health Statistics 2015. Geneva: World Health Organization; 2015. 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[DOI] [PubMed] [Google Scholar] Articles from Open Access Emergency Medicine : OAEM are provided here courtesy of Dove Press ACTIONS View on publisher site PDF (127.1 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Materials and methods Results Discussion Conclusion Acknowledgments Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
3363	https://www.ncbi.nlm.nih.gov/books/NBK2608/	Idiopathic Generalised Epilepsies - The Epilepsies - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. Panayiotopoulos CP. The Epilepsies: Seizures, Syndromes and Management. Oxfordshire (UK): Bladon Medical Publishing; 2005. The Epilepsies: Seizures, Syndromes and Management. Show details Panayiotopoulos CP. Oxfordshire (UK): Bladon Medical Publishing; 2005. Contents Search term < PrevNext > Chapter 10 Idiopathic Generalised Epilepsies Idiopathic generalised epilepsies (IGEs) constitute one-third of all epilepsies.1–6 They are genetically determined and affect otherwise normal people of both sexes and all races. IGEs manifest with typical absences, myoclonic jerks and generalised tonic clonic seizures (GTCS), alone or in varying combinations and severity. Absence status epilepticus (ASE) is common. Most syndromes of IGE start in childhood or adolescence, but some have an adult onset. They are usually life long, though a few are age related. The EEG is the most sensitive test in the diagnosis and confirmation of IGE. EEG shows generalised discharges of spikes, polyspikes or spike/polyspike-wave either ictally or interictally. These discharges are often precipitated by hyperventilation, sleep deprivation and intermittent photic stimulation. Inconspicuous clinical manifestations become apparent on video EEG and with breath counting during hyperventilation. The EEG is unlikely to be normal in untreated patients. In suspected cases with normal routine awake EEG, an EEG during sleep and awakening should be obtained. Molecular genetic analyses have led to important breakthroughs in the identification of candidate genes and loci;7,8 genetic heterogeneity is common.8–11 Genetic mutations found in γ-aminobutyric acid (GABAA) receptor subunits strongly implicate the GABA A receptor in IGEs.12 Treatment of IGEs is demanding for two main reasons. Firstly, anti-epileptic drugs (AEDs) beneficial in focal epilepsies may be deleterious in IGEs.4,13 Secondly, efficacy of AEDs differs even within IGE seizures. This is because the generation of absences, for example, is due to a predominance of inhibitory activity, in contrast to generalised convulsive seizures in which an excess of excitatory activity is present.13 Most IGEs respond well to appropriate AEDs, but treatment is often life long. The fact that nearly 50% of patients with IGE are currently taking “ill-advised AED” medication14 is a grave problem that needs to be addressed. Go to: Seizures of Idiopathic Generalised Epilepsies The syndromes of IGEs manifest with three main types of seizures alone or in combination. These are: Typical absence seizures Myoclonic seizures Generalised tonic clonic seizures ILAE Definition of Idiopathic Generalised Epilepsies The ILAE Commission1 defined IGE as follows: “Idiopathic generalised epilepsies are forms of generalised epilepsies in which all seizures are initially generalised (absences, myoclonic jerks and generalised tonic clonic seizures), with an EEG expression that is a generalised bilateral, synchronous, symmetrical discharge (such as is described in the seizure classification of the corresponding type). The patient usually has a normal interictal state, without neurological or neuroradiologic signs. In general, interictal EEGs show normal background activity and generalised discharges, such as spikes, polyspike spike-waves, and polyspike-waves □ 3 Hz. The discharges are increased by slow sleep. The various syndromes of idiopathic generalised epilepsies differ mainly in age of onset. No aetiology can be found other than a genetic predisposition towards these disorders.”1 Typical Absence Seizures Typical absences (previously known as petit mal) are brief (lasting seconds) generalised epileptic seizures of abrupt onset and abrupt termination (Table 10.1). They have two essential components: Table 10.1 Clinical and EEG manifestations of typical absence seizures a clinical component manifesting with impairment of consciousness (absence) an EEG component manifesting with generalised spike-slow wave discharges of 3–4 Hz (> 2.5 Hz).1–4,15,16 The absence seizures are fundamentally different and pharmacologically unique compared with any other type of seizure, which also makes their treatment different.4,16,17 The clinical and EEG manifestations of typical absences are extensive and syndrome-related.1–4,15,16,18 Clinical manifestations: Impairment of consciousness may be severe, moderate, mild or inconspicuous (and special cognitive testing may be required to detect it). It is often associated with other concomitant symptoms, such as myoclonia, automatisms and autonomic disturbances. Myoclonia may be rhythmic or random, mild or severe, regional (mouth or eyes) or widespread (head, limbs and trunk). Typical absences are predominantly spontaneous, though they are precipitated by hyperventilation in around 90% of untreated patients. Other specific modes of precipitation include photic, pattern, video games and thinking (reflex absences). Ictal EEG: The ictal EEG consists of generalised discharges with repetitive and rhythmic 3–4 Hz single or multiple spike-slow wave complexes (Figure 10.1). Figure 10.1 Examples from video EEG recorded generalised discharges of 3 Hz spikes or multiple spikes-waves of typical absence seizures. These seven patients had different syndromes of idiopathic and symptomatic absence epilepsies. Note (more...) These generalised spike-wave discharges (GSWD) may be brief (sometimes less than 3 s) or long (□ 30 s), and continuous or fragmented. The intradischarge frequency of the spike-wave may be relatively constant or may vary. Typical absence seizures in IGE syndromes. Typical absences are severe in childhood absence epilepsy (CAE) and juvenile absence epilepsy (JAE), but mild or inconspicuous in other syndromes, such as juvenile myoclonic epilepsy (JME). They may occur alone or in combination with other types of generalised seizures. IGE with absences may remit with age or be lifelong. Typical ASE occurs in approximately one-third of patients who suffer from typical absence seizures.19 Clinical Manifestations of Typical Absence Seizures The clinical manifestations of typical absence seizures vary significantly between patients.3,4,15,18,20–26 Impairment of consciousness may be the only clinical symptom, but it is often combined with other manifestations (Table 10.1). Typical absences are categorised as: simple absences with impairment of consciousness only complex absences when impairment of consciousness combines with other ictal motor manifestations. Complex absences are far commoner than simple absences in children. Simple absences are commoner in adults. The same patient may have both simple and complex absences. Absence with Impairment of Consciousness Only1 The classical27 and ILAE1 descriptions refer to absence seizures with severe impairment of consciousness as in CAE and JAE. Patient note “Transient loss of consciousness without conspicuous convulsions. A patient stops for a moment whatever he or she is doing, very often turns pale, may drop whatever is in the hand.....There may be a slight stoop forward, or a slight quivering of the eyelids...The attack usually lasts only a few seconds. The return of the consciousness may be sudden and the patient after the momentary lapse, may be in just the same state as before the attack, may even continue a sentence or action which was commenced before it came on, and suspended during the occurrence.” W.R.Gowers (1885) 27 The hallmark of severe absence seizures is a sudden onset and interruption of ongoing activities, often with a blank stare. If the patient is speaking, speech is slowed or interrupted; if walking, he/she stands transfixed. Usually the patient will be unresponsive when spoken to. Attacks are often aborted by auditory or sensory stimulation. In less severe absences, the patient may not stop his/her activities, though reaction time and speech may slow down. In their mildest form, absences may be inconspicuous to the patient and imperceptible to the observer (phantom absences), as disclosed by video EEG recordings showing errors and delays during breath counting or other cognitive tests during hyperventilation. Absence with Clonic Components1 During the absence, as described above, clonic motor manifestations, rhythmic or arrhythmic and singular or repetitive, are particularly frequent at the onset. They may be continuous. They may also occur at any other stage of the seizure. The most common manifestations are clonic jerking of the eyelids, eyebrows and eyeballs, together or independently, as well as random or repetitive eye closures. Fast flickering of the eyelids is probably the most common ictal clinical manifestation, and may occur during brief GSWD without discernible impairment of consciousness. Myoclonias at the corner of the mouth and jerking of the jaw are less common. Myoclonic jerks of the head, body and limbs may be singular or rhythmical and repetitive, and they may be mild or violent. In some patients with absence seizures, single myoclonic jerks of the head and less often of the limbs may occur during the progression of ictus; in my opinion, these are indicative of a bad prognosis and may constitute an epileptic syndrome, but this needs further documentation.15,16 Absence with Atonic Components Diminution of muscle tone is usual when absences are severe. This manifests with drooping of the head and, occasionally, slumping of the trunk, dropping of the arms and relaxation of the grip. Rarely, tone is sufficiently diminished to cause falls. Absence with Tonic Components Tonic seizures alone do not occur in IGEs. However, tonic muscular contractions are common concomitant manifestations during typical absence seizures. They mainly affect facial and neck muscles symmetrically or asymmetrically. The eyes and head may be drawn backwards (retropulsion) or to one side, and the trunk may arch. Tonic manifestations are prominent in myoclonic absence seizures. Absence with Automatisms Automatisms are common in typical absences when consciousness is sufficiently impaired, and they are more likely to occur 4–6 s after the onset of GSWD. They do not occur in mild absence seizures irrespective of duration as for example in ASE. Automatisms of typical absence seizures are simple and void of behavioural changes (see definitions in Chapter 12). They vary in location and character from seizure to seizure. Perioral automatisms, such as lip licking, smacking, swallowing or ‘mute’ speech movements, are the most common. Scratching, fumbling with clothes and other limb automatisms are also common. Automatisms can be evoked; passive movements, postural repositioning or other stimuli can change their pattern and distribution.24 Absence with Autonomic Components Autonomic components consist of pallor and, less frequently, flushing, sweating, dilatation of the pupils and incontinence of urine. During absence seizures, pronounced changes in cerebral oxygenation occur with a decrease in oxygenated and an increase in deoxygenated haemoglobin. Oxygenation changes start several seconds after the EEG-defined onset of the absence and outlast the clinically defined event by 20–30 s.28 Absences with Focal Motor Components, Hallucinations and Other Manifestations of Neocortical or Limbic Symptomatology During a typical absence seizure, patients frequently manifest with concomitant focal motor components (tonic or clonic) imitating focal motor seizures. Hallucinations and other manifestations such as concurrent epigastric sensations29 may occur; these are particularly more apparent during ASE.16 Patient note I was in that state of confused mind. The surroundings were vertical and flat and I lost depth perception. People around me appeared as be wearing wigs in pastel shades.30 Electroencephalography The ictal EEG is characteristic with regular and symmetrical 3–4 Hz GSWD (Figure 10.1). The intradischarge spike-wave frequency varies from onset to termination (Figure 10.2). It is usually faster and unstable in the opening phase (first second), becomes more regular and stable in the initial phase (first 3 s), and slows down towards the terminal phase (last 3 s).24 The intradischarge relationship between spike or multiple spike and slow wave frequently varies. The GSWD is often of higher amplitude in the anterior regions. A generalised discharge with an onset or a higher amplitude in the posterior regions may indicate a bad prognosis.31 Figure 10.2 Girl, Aged 7 Years, with Childhood Absence Epilepsy. The seizure occurred during hyperventilation with breath counting. She stopped counting (black arrow), opened her eyes (open arrow) and became unresponsive. Marked perioral (more...) Duration of the discharges commonly varies from 3 to 30 s (Figure 10.1). The background interictal EEG is usually normal, though some paroxysmal activity (such as spikes or spike–wave complexes) may occur. Focal abnormalities or other asymmetries are common.32,33 Sleep EEG patterns are normal. GSWD are more likely to increase, but decrease during sleep. The discharges are often shorter and usually devoid of discernible clinical manifestations in sleep, even in those patients who have numerous clinical seizures with motor manifestations during the alert state. Important Note Though the EEGGSWD of typical absences is defined as symmetrical and synchronous, this is rarely the case at its onset. Commonly, the discharge starts with single or multiple spikes-slow waves that are asymmetrical and usually have a regional onset, mainly frontal (Figures 10.1 and 10.2). Often (but not always), there is an alternating side emphasis. Unilateral onset of the GSWD may be confused with secondary bilateral synchrony. The end of the discharge may be abrupt or consists of brief rhythmic or irregular slow waves (Figure 10.2). Sometimes more focal or fragmentary spikes occur, representing a ‘forme fruste’ of GSWD. These are more often recorded from the anterior regions, but other locations are also common. Genetics IGEs with typical absences are genetically determined, as indicated by the high incidence of similar disorders among families. However, the precise mode of inheritance and the genes involved remain largely unknown.7 Currently, various chromosomal loci have been identified for IGEs, as detailed in the description of individual syndromes. Furthermore, there is now evidence available to suggest that mutations in genes encoding GABA receptors34–36 or brain expressed voltage-dependent calcium channels37 may underlie CAE. Genetic heterogeneity of the GSWD phenotype in animal models of absences favours a similar, and probably much wider, genetic heterogeneity in humans.7,13,17 Pathophysiology of Absence Seizures The pathophysiological mechanisms of absence seizures have been studied in various animal models with GSWD associated with behavioural arrest.7,13,17,38,39 It appears that the GSWD are generated and sustained by highly synchronised abnormal oscillatory rhythms in thalamocortical networks that mainly involve neocortical pyramidal cells, the reticular thalamic nucleus and the relay nuclei of the thalamus. Neither the cortex nor the thalamus alone can sustain these discharges, indicating that both structures are involved in their generation. The involvement of thalamus as the generator of GSWD is documented by the fact that: stimulation of the medial thalamus induces a cortical GSWD without leading to self-sustained activity thalamic neurons can intrinsically generate action potentials in both a tonic and a burst firing mode7,13,17,40 magnetic resonance spectroscopy (MRS) and neuroimaging in vivo provides evidence of progressive thalamic neuronal dysfunction in patients with IGE.41,42 The relative importance of the cortex in the initiation and synchronisation of GSWD is mainly documented by the finding that, following thalamectomy, instigation of GSWD persists though the thalamus is required to maintain rhythmicity once the GSWD is established. More recently, in a rat model of absence, Meeren et al.43 showed that, during GSWD, cortical and thalamic interactions lag behind an initial burst of activity in the perioral region of the primary somatosensory cortex (S1po) during the first 500 ms of GSWD activity. These findings suggest that, in this animal model, a cortical focus within S1po is the dominant factor in initiating the paroxysmal oscillation within the corticothalamic loops, and that the large scale synchronisation is mediated by an extremely fast intracortical spread of seizure activity.43 This is also supported by experiments whereby microinfusion of ethosuximide into S1po produces an immediate cessation of GSWD.44 The basic intrinsic neuronal mechanisms involve low threshold T-type calcium currents elicited by activating the low threshold calcium channels. These channels are present in high densities in thalamic neurons and trigger regenerative burst firing that drive normal and pathological thalamocortical rhythms, including the GSWD of absence seizures. Ethosuximide exerts its anti-absence effect by either reducing thalamic low threshold calcium currents, probably by a direct channel blocking action that is voltage dependent,45 or through a potent inhibitory effect in the perioral region of the primary somatosensory cortex.13,44 Clinical note It is likely that the generation of absence seizures is due to a predominance of inhibitory activity, in contrast to generalised or focal convulsive seizures in which an excess of excitatory activity is present.13 Both inhibitory and excitatory neurotransmissions are involved in the genesis and control of absence seizures. This may be the result of excessive cortical excitability due to an imbalance between inhibition and excitation, or excessive thalamic oscillations due to abnormal intrinsic neuronal properties under the control of inhibitory GABAergic mechanisms. GABAB receptors play the most prominent role by eliciting long-standing hyperpolarisation required to drive low threshold calcium channels for the initiation of sustained burst firing. Typical absences are aggravated by GABA B agonists, such as baclofen, and suppressed by GABA B antagonists. GABAergic drugs (e.g. vigabatrin, tiagabine) are pro-absence substances; they interfere with the degradation of, and the re-uptake of, GABA.4,13 The only exception to GABAergic activation inhibiting absences is the reticular thalamic nucleus, which has exclusively GABA A receptors; it functions as a pacemaker to synchronise thalamocortical oscillations.46,47 Enhanced activation of GABA A receptors in this nucleus decreases the pacemaking capacity of these cells, thereby decreasing the likelihood of generating absence seizures. Functional imaging using positron emission tomography (PET) demonstrates normal cerebral glucose metabolism and benzodiazepine receptor density in absence epilepsies, with diffuse hypermetabolism during 3 Hz GSWD.48,49 There is no evidence of any overall interictal abnormality of opioid receptors in IGE, but typical absences have been found to displace 11C-diprenorphine from the association areas of the neocortex. In contrast, binding of 11C-flumazenil to central benzodiazepine receptors has been shown to be unaffected by serial absences.49 Ictal single photon emission computed tomography (SPECT) shows an overall increase in cerebral blood flow50,51 and may be useful in detecting cases of frontal or other secondarily generalised absences.52 Interictally, relative hypoperfusion occurs in the frontal lobes, and may involve neighbouring parietal and temporal regions.51 Ictally, there is relative hyperperfusion in the same brain regions that are hypoperfused in the baseline study.51 Microdysgenesis and other cerebral structural changes were reported in some patients with CAE and JAE at autopsy53 and MRI54 studies. These results were not replicated in a more recent, blinded study.55 Microdysgenesis may be inconceivable for a benign, age-dependent and age-limited epileptic syndrome, such as CAE, though the current ion channel hypothesis for the pathogenesis of IGE does not preclude microscopic or ultramicroscopic abnormalities. Furthermore, recent quantitative MRI, PET and MRS studies have challenged the belief that IGEs are not associated with tissue pathology.42 Diagnosing Absences and Differential Diagnosis Clinical note The brief duration of absence seizures with abrupt onset and abrupt termination of ictal symptoms, daily frequency and nearly invariable provocation by hyperventilation makes the diagnosis easy.4,16,18 The differential diagnosis of typical absence seizures with severe impairment of consciousness in children is relatively straightforward. The absences may be missed if mild or void of myoclonic components. Automatisms, such as lip smacking or licking, swallowing, fumbling or aimless walking, are common and should not be taken as evidence of complex partial (focal) seizures, which require entirely different management. The EEG or, ideally, video EEG can confirm the diagnosis of typical absence seizures in more than 90% of untreated patients, mainly during hyperventilation.16 If not, the diagnosis of absences should be questioned. In practical terms, a child with suspected typical absences should be asked to overbreathe for 3 minutes, counting his or her breaths while standing with hands extended in front. Hyperventilation will provoke an absence in more than 90% of those with typical absences. This procedure should preferably be videotaped to document the clinical manifestations. It may reveal features favouring a specific epileptic syndrome and, therefore, may determine the long-term prognosis and management. Video EEG documentation may be particularly useful if absences prove resistant to treatment, if other seizures develop, or for future genetic counselling. Focal spike abnormalities and asymmetrical onset of the ictal GSWD are common and may be a cause of misdiagnosis, particularly in resistant cases.15 If video EEG is not available, documentation of absences using a camcorder or modern digital means of recording is recommended. The differentiation of typical from atypical absence seizures is shown in Table 7.4. Briefly, atypical absences differ from typical absences in the following ways: Atypical absences occur only in the context of mainly severe symptomatic or cryptogenic epilepsies of children with learning difficulties, who also suffer from frequent seizures of other types, such as atonic, tonic and myoclonic seizures. In atypical absences, onset and termination is not as abrupt as in typical absences, and changes in tone are more pronounced. The ictal EEG of atypical absence has slow (< 2.5 Hz) GSWD. These are heterogeneous, often asymmetrical, and may include irregular spike–wave complexes and other paroxysmal activity. Background interictal EEG is usually abnormal. The differentiation of typical absences from complex focal seizures, detailed in Table 10.2, may be more difficult when the motor components of the absence are unilateral and in adults in whom absences are often misdiagnosed as temporal lobe seizures.56–58 Absences occur in 10% of adult patients with epileptic seizures.56–58 Table 10.2 Differential diagnosis of typical absences from complex focal seizures Misconceptions Petit mal is seen almost exclusively in children, more rarely in adolescents and is a real curiosity in adults and the elderly.59 Contrary to the dominant view above, typical absence seizures occur in approximately 10% of adults with epilepsy.56,57 The alarming problem is that these are underdiagnosed or misinterpreted as focal seizures. Myoclonic Jerks Myoclonic jerks are shock-like, irregular and often arrhythmic, clonic-twitching movements that are singular or repetitive.21,60,61 They are of variable amplitude and force, ranging from mild and inconspicuous to sufficiently violent to make the patient fall on the ground, drop or throw things, or kick. Commonly, the same patients experience mild and violent jerks. Myoclonic jerks predominantly affect the eyelids, facial and neck muscles, the upper limbs more than the lower limbs and the body. Myoclonic jerks of IGE mainly occur on awakening. Precipitating factors are sleep deprivation, fatigue, excitement or distress and often photic stimulation. The patient is fully aware of myoclonic jerks unless they occur during impairment of consciousness in absence seizures. The location and extent of myoclonic jerks varies between IGE syndromes. Polyspikes are the EEG accompaniment of myoclonic jerks (Figures 10.2, 10.3 and 10.5). Figure 10.3 Samples from the video EEG of a woman with JME on carbamazepine. Violent myoclonic jerks occur with generalised multiple spike discharges (see also Figure 10.14 of the same patient). Figure 10.5 Samples from a Video EEG of a 6-Year-Old Normal Boy with Doose Syndrome. The background activity was normal, but there were frequent (at least every 10 s) generalised discharges of high-voltage spike/polyspike–wave complexes at (more...) Diagnosing Myoclonic Jerks Elicitation of the characteristic history of myoclonic jerks is something of an art. It is often necessary to physically demonstrate mild myoclonic jerks of the fingers and hands, and to inquire about morning clumsiness and tremors.62 Questions like ‘do you spill your morning tea?’ and ‘do you drop things in the morning?’, together with a simultaneous demonstration of how myoclonic jerks produce this effect, may be answered positively by patients who denied experiencing myoclonic jerks on direct questioning. Further elaboration is required to confirm that clumsiness was due to genuine myoclonic jerks. If the patient reports normal hypnagogic jactitations, it is reassuring that the concept of myoclonic jerks has been understood. Diagnostic yield may be improved by emphasising the close relationship between jerks and fatigue, alcohol and sleep deprivation. Some patients do not report their jerks, erroneously assuming that this is a self-inflicted normal phenomenon related to excess of alcohol and lack of sleep. Generalised Tonic Clonic Seizures in IGEs In IGEs, GTCS are primary (primarily) in the sense that they are generalised from onset without preceding auras or objective ictal focal symptoms, though they are often heralded by a series of myoclonic jerks or absences. This contrasts with secondarily GTCS of focal epilepsies that are often preceded by an aura or motor-sensory focal symptoms. Overall, primarily GTCS occur on awakening (17–53% of patients), diffusely while awake (23–36%) or during sleep (27–44%), or randomly (13–26%).63 The proportion of these patients who also have other generalised seizures, such as jerks or absences, is undetermined. Differential Diagnosis of Primarily from Secondarily Generalised Tonic Clonic Seizures GTCS are dramatic in their presentation, which is the main reason for referral for medical consultation. This firstly demands careful exclusion of syncopal and other non-epileptic events. Once an unequivocal diagnosis of genuine epileptic GTCS has been established, the main differential diagnosis is between primarily and secondarily GTCS. GTCS whether primarily (IGEs) or secondarily (focal epilepsies) are identical in their clinical presentation. Their differentiation, which is of immense clinical importance regarding overall management and AED treatment, is often easy (Table 10.3) based on: Table 10.3 Differentiation of primarily versus secondarily GTCS clinical history regarding other types of coexisting seizures, precipitating factors, circadian distribution and family history EEG manifestations brain imaging. SPECT studies have shown activation of selective frontal, parietal and temporal networks after both spontaneous and induced primarily or secondarily GTCS, but thalamic cerebral blood flow is only increased after primarily GTCS.42 Status Epilepticus in Idiopathic Generalised Epilepsies IGEs manifest with all types of generalised status epilepticus. ASE is probably the most common of all and the most likely to escape diagnosis or be misdiagnosed as focal status epilepticus or non-epileptic confusion, psychogenic or behavioural disorder.1,15,19,64–66 The new ILAE diagnostic scheme5 considers status epilepticus as a “continuous seizure” with two main categories (Table 1.2): Generalised status epilepticus Focal status epilepticus The subcategories of generalised status epilepticus are relevant to this chapter: Generalised tonic-clonic status epilepticus Clonic status epilepticus Absence status epilepticus Tonic status epilepticus Myoclonic status epilepticus Definition of Status Epilepticus There is no satisfactory definition of status epilepticus.64 Its definition is mainly influenced by the convulsive status epilepticus, because of its high morbidity, mortality and the need for early detection and early treatment. The World Health Organization defines status epilepticus as “a condition characterised by epileptic seizures that are sufficiently prolonged or repeated at sufficiently brief intervals so as to produce an unvarying and enduring epileptic condition”.67 By consensus, “sufficient length of time” was defined as being more than 30 minutes’ duration68 although more recent opinions argue for shorter periods of 5–10 minutes in defining convulsive status epilepticus.69 These short periods may not be applicable in other than convulsive types of status epilepticus. The recent ILAE glossary makes no significant contribution in defining status epilepticus as “A seizure that shows no clinical signs of arresting after a duration encompassing the great majority of seizures of that type in most patients or recurrent seizures without interictal resumption of baseline central nervous system function”.70 Also, etymologically, status epilepticus is not a ‘continuous seizure’ as the ILAE Task Force proposed;5 it is a prolonged, enduring or rapidly repeated seizure which may also be ‘discontinuous’ and often stops without medical intervention. Shorvons’ operational definition is more appealing “status epilepticus is condition in which epileptic activity persists for 30 minutes or more, causing a wide spectrum of clinical symptoms, and with a highly variable pathophysiological, anatomical and aetiological basis”.64 This definition implies that status is not simply a prolonged seizure or rapid repetition of seizures (in fact the word ‘seizure’ is no longer retained), but a condition (or group of conditions) in its own right with distinctive pathophysiological features. Classification of Absence Status Epilepticus Absence status epilepticus is divided into: typical ASE occurring in patients with IGEs who also have absence seizures atypical ASE occurring in patients with symptomatic epilepsies and epileptic encephalopathies (Figure 7.7) de novo or situation related absence status epilepticus occurring mainly in adults without a history of previous epileptic seizures commonly as the result of benzodiazepine or other drug discontinuation. 73,74 The most well-documented example is diazepine withdrawal. De novo ASE is often misdiagnosed as a psychotic state or dementia.66,73,74 Typical Absence Status Epilepticus in IGE Typical (idiopathic) absence status epilepticus in IGEs is defined as a prolonged (> 30 minutes), generalised non-convulsive seizure of impairment of the content of consciousness (absence) and EEG generalised spike/polyspike-wave discharges.19,65,66,75,76 It should be emphasised that typical ASE, like absence seizures, is of many types with impairment of cognition as a shared common symptom. Impairment of consciousness may be mild or severe. It may occur alone (Figure 10.18) or more frequently be associated with other symptoms such as those listed in Table 10.1 for complex absence seizures. Motor manifestations such as myoclonic jerks, eyelid or perioral myoclonia, may predominate and be syndrome related (Figures 10.6 and 13.7). Figure 10.18 Video EEG of a Highly Intelligent Woman, Aged 79 Years, with Phantom Absences. ASE and GTCS (case 1 in ref ). She had her first overt seizure at the age of 30 years. She was moderately confused for 12 hours prior to a GTCS. (more...) Figure 10.6 Samples of a Video EEG of a Boy in Myoclonic-Atonic Status Epilepticus. Top: The EEG showed electrical status epilepticus during wakefulness and sleep. This consisted of nearly continuous generalised discharges of spikes (or double (more...) Accordingly, typical (idiopathic) absence status epilepticus may be subdivided to: typical absence status epilepticus with impairment of consciousness only (Figure 10.18) myoclonic-atonic status epilepticus (Figure 10.6) myoclonic-absence status epilepticus (Figure 10.14) perioral myoclonic status epilepticus (Figure 10.16) eyelid myoclonic status epilepticus (Figure 13.8). Figure 10.14 Myoclonic-absence status epilepticus in JME. Video EEG of a woman with classical JME, but on carbamazepine at the time of this recording (see also Figure 10.3 from the same patient). The patient was mildly confused (more...) Figure 10.16 Diagnostic Errors in PMA. Top: Video EEG recording of an 18-year-old woman with perioral myoclonia with absences (case 6 in ref ). She was referred because of ‘focal motor seizures and secondarily (more...) The ictal EEG is characteristic with usually regular and symmetrical 3 Hz (range 1–4 Hz) GSWD, which is continuous or repetitive (Figures 10.6, 10.14, 10.16, 10.18 and 13.7). With the possible exception of CAE, all IGEs with typical absences may manifest with typical ASE, either as a spontaneous expression of their natural course or provoked by external factors or inappropriate treatment manoeuvres. Impairment of Consciousness, Memory and Higher Cognitive Functions The cardinal symptom shared by all cases of typical ASE is altered content of consciousness in a usually fully alert patient. Memory and higher cognitive intellectual functions, such as abstract thinking, computation and personal awareness, are the main areas of disturbance, which varies from very mild to very severe with intermediate states of severity occurring more often. Mild disturbance is experienced as a state of slow reaction, behaviour and mental functioning: Patient note My mind slows down and I am able to understand, but it takes longer to formulate answers. I become slow, but can communicate verbally with others. My behaviour slows down and I muddle up words. It’s like being in a trance and missing pieces of conversation. Moderate and severe impairment of consciousness manifest with varying degrees of confusion, global disorientation and inappropriate behaviour: Patient note Confused, cannot recognise people other than close relatives, disorientated in time and place, very quiet. Disturbed, vague, uncooperative, confused. Markedly confused, goes into a dreamy state, able to formulate some single word answers to simple questions, puts trousers on over pyjamas. Confused, makes coffee twice, fades away mentally and physically, disoriented in time and place. Usually, the patient is alert, attentive and cooperative. Verbal functioning is relatively well preserved, but is often slow with stereotypic and usually monosyllabic or monolectic answers. Movement and coordination are intact. Complete unresponsive is rare. Behavioural Abnormalities and Experiential Phenomena Though the most common behavioural changes refer to daily activities disturbed by the impairment of consciousness, some patients become depressed, agitated and, occasionally, hostile and aggressive. Experiential and sensational phenomena are more common than is usually appreciated and may include: Patient note Sensation of viewing the world through a different medium and a feeling of not being in the same world as everyone else. Uncontrollable rush of thoughts. A feeling of fear of losing control of my mind. A feeling of closeness. A funny feeling that I can not elaborate. A strange feeling of not being myself. Edgy, worry and uncomfortable. My character changes completely, I become extremely snappy, have a severe headache. Weird. Simple gestural and ambulatory automatisms, and automatic behavioural and fugue-like states may occur in the 20% of patients who also have severe impairment of consciousness: Patient note Replies yes to any question and fumbles with his clothes. Myoclonic Jerks in Absence Status Epilepticus Segmental myoclonic jerks, usually involving the eyelids or perioral region and less often the limbs, frequently occur during typical ASE, and vary in degree and severity. They are most likely to occur in syndromes that manifest with similar myoclonic phenomena during brief absences (see descriptions in the relevant sections of individual IGE syndromes). GTCS Associated with Typical Absence Status Epilepticus ASE ending with a GTCS is probably the rule irrespective of the syndrome. However, in one-third of patients, ASE always ends with GTCS. In the remaining two-thirds, it may also terminate spontaneously without GTCS. It is exceptional for GTCS to precede or be interspersed with ASE. It is also exceptional for more than one GTCS to occur following ASE. Duration and Frequency of Typical Absence Status Epilepticus ASE usually lasts for an average of 3–4 hours, rarely as little as half an hour, often exceeds 6–10 hours and occasionally lasts for 2–10 days. Frequency also varies from once in a lifetime to an average of 10–20 episodes/year, or be consistently catamenial. The duration and frequency depend on treatment strategies and syndromic classification. Postictal State Amnesia of the event is exceptional. Commonly patients are aware of what happens during the ASE and some are able to write down their experiences even when in status. Other patients have a patchy recollection of events and usually miss the last part prior to the GTCS. After a GTCS, the patient feels tired, has a headache and is confused for a varying duration of time. Age at Onset and Sex Mean age at onset of ASE is 29 years, with a range of 9–56 years. It is rare for ASE in IGE to start before the first decade of life. Other types of seizures, such as absences, myoclonic jerks and GTCS, may predate the first ASE by many years. ASE rarely is the first overt type of seizure.77 Precipitating Factors Inappropriate use of AEDs, such as tiagabine,78,79 vigabatrin,66 carbamazepine and phenytoin,80 as well as discontinuation of proper anti-absence medication are the commonest precipitants of ASE. Sleep deprivation, stress and excess alcohol consumption, alone or usually in combination are common precipitating factors. Some women may have consistent catamenial precipitation.81–83 Differential Diagnosis A confusional state lasting for hours and ending with GTCS creates significant diagnostic difficulties regarding its nature and cause. Causes to consider are intoxication, and psychogenic, metabolic or systemic diseases. If these are excluded, the differential diagnosis is between focal (complex, partial) or ASE. Idiopathic ASE is commonly unrecognised or misdiagnosed. It is surprising how often physicians are deceived by the general good appearance, alertness and cooperation of the patient.66 Basic testing of memory and higher cognitive functions, essential for diagnosis, are rarely carried out. It is important to remember that more than half of patients are aware of the situation when entering or during ASE, which is of great practical significance with regard to termination of the ASE and prevention of the impending GTCS by self-administered appropriate medication. Idiopathic ASE is easy to diagnose on the basis of proper identification of the IGE syndrome. Differentiation of Generalised ASE from Complex Focal Status Epilepticus Complex focal status epilepticus (including limbic status epilepticus) is rarer than generalised ASE. Patients frequently have recurring complex focal seizures with incomplete recovery between attacks, or a continuous ‘epileptic twilight state’ with cycling between unresponsiveness and partial responsiveness. The ictal EEG reveals recurrent epileptiform patterns consistent with those encountered in isolated complex focal seizures. The interictal EEG usually shows a unilateral or bilateral cortical focus. In complex focal status epilepticus, conscious levels fluctuate during the attack and patients experience postictal confusion and amnesia of the episode. Although automatisms may occur in both forms of status epilepticus, they are more complex and prolonged in complex focal status epilepticus. The commonest reason for misdiagnosis between the two conditions is because absences are not recognised or misdiagnosed as complex focal seizures (Table 10.2). A previous or new EEG invariably shows generalised discharges in IGE. It may be normal or show specific focal spikes in focal epilepsies, mainly temporal lobe epilepsy. Ictal EEG with GSWD is diagnostic of IGE. Coexisting focal abnormalities should not be interpreted as evidence of focal epilepsy. The Differentiation of Typical (Idiopathic) ASE from Atypical (Usually Symptomatic or Probably Symptomatic) ASE Is Also Easy The major distinguishing feature of atypical ASE is that it occurs mainly in children with symptomatic or cryptogenic generalised epilepsies, who also have a plethora of other types of frequent seizures, such as atypical absences, tonic and atonic seizures, myoclonic jerks and GTCS. Most of these patients also have moderate or severe learning and physical handicaps. In addition, the interictal EEG is often very abnormal with slow background activity and frequent, brief or long runs of slow generalised spike–wave complexes, paroxysmal fast activity and paroxysms of polyspikes. It is often difficult to define the boundaries, onset and termination of atypical ASE, because these children frequently have alterations of behaviour and alertness as well as long interictal slow GSWD. Atypical ASE is clinically characterised by fluctuating impairment of consciousness often with other ictal symptoms, such as repeated series of tonic or atonic seizures and segmental or generalised jerks. The ictal EEG pattern is of slow (< 2.5 Hz) GSWD. Both the clinical patterns and the EEG abnormalities are more variable than of the typical ASE. Additional discriminating features of atypical ASE are: gradual onset and offset level of consciousness and other coexisting types of seizures tend to fluctuate sometimes for weeks, with little distinction between ictal and interictal phases initiation or termination with a GTCS is exceptional incontinence is common. Go to: Epileptic Syndromes of Idiopathic Generalised Epilepsies The recognised syndromes of IGEs are shown in Table 1.5 (1989 ILAE classification)1 and Table 1.6 (new ILAE classification scheme).5 Listed according to the age at onset, these are: Benign myoclonic epilepsy in infancy (Chapter 6) Generalised epilepsy with febrile seizures plus (Chapter 6) Epilepsy with myoclonic absences Epilepsy with myoclonic-astatic seizures Childhood absence epilepsy Idiopathic generalised epilepsies with variable phenotypes – Juvenile absence epilepsy – Juvenile myoclonic epilepsy – Epilepsy with generalised tonic-clonic seizures only Other possible syndromes of IGE for consideration, which are not yet recognised by the ILAE Committees, are:1,5 IGE with absences of early childhood Perioral myoclonia with absences Idiopathic generalised epilepsy with phantom absences Jeavons syndrome (eyelid myoclonia with absences) Benign adult familial myoclonic epilepsy Autosomal dominant cortical myoclonus and epilepsy Considerations on the Classification of Idiopathic Generalised Epilepsies: The classification of IGE is probably one of the most significant and debated issues. There are two schools of thought, with diversely opposing views:2 (a). IGE is one disease, (b). IGE comprises a large group of many distinct syndromes. The evidence so far is not conclusive in favour of one or the other, and any new classification should not take sides unreasonably. In practical terms, the view that ‘IGE is one disease’ would, overall, be an easy clinical diagnostic approach, but it would discourage the diagnostic precision required for genetic studies, prognosis and management decisions. The view that ‘IGE comprises a large group of many distinct syndromes’ would be more demanding diagnostically and occasionally require exhaustive clinical and video EEG data. However, this is often the price that we, as physicians, have to pay in pursuing an accurate diagnosis, which is is the golden rule in medicine. This view also satisfies: (a). “maximum practical application to differential diagnosis,”84 which is the main reason for reorganising the classification of epileptic syndromes in the forthcoming revisions; and (b). takes advantage of “significant advances in our understanding”85 of IGEs, which constitute one-third of cases of ‘epilepsy’. Similarly, there is no justification for the unification of ‘IGEs with onset in adolescence’ in a single syndrome as has been recently proposed.86 The major conceptual problem with this proposition is that it takes ‘onset in adolescence’ as the most significant almost defining factor, which is at variance with the definition of a syndrome.1 Further, the same IGE syndromes may start in childhood, adolescence and occasionally adult life.1 On the surface, syndromes of IGE may look alike if their clinical EEG manifestations are not properly analysed and synthesised. For example, JME and JAE both manifest with absences, myoclonic jerks and GTCS. However, severe absences are the main and most disturbing type of seizure in JAE, and myoclonic jerks may not occur or be randomly distributed.16 Conversely, myoclonic jerks on awakening is the defining symptom of JME; absences are mild and occur in only one-third of patients. The clinical-EEG features of typical absence seizures that may be syndrome-related are well described in video EEG studies.16,18 Unifying all typical absence seizures as a single type is of no benefit to any cause. Animal genetic studies have documented numerous syndromes of IGE17 and this is likely to be the case in humans.2 Recognised Syndromes of IGEs in the ILAE Classification of 1989 versus the Newly Proposed Diagnostic Scheme The new ILAE diagnostic scheme5 has some significant differences in relation to the ILAE classification of 19891 regarding IGE of childhood and adolescence. These are: (1). The syndromes of JAE, JME and IGE with GTCS only are considered as phenotypical variants of IGE of adolescence (2). A new syndrome of ‘IGE with GTCS only’ has been proposed to replace ‘epilepsy with GTCS on awakening’ (3). ‘Epilepsy with myoclonic-astatic seizures’ and ‘epilepsy with myoclonic absences’ are included among idiopathic generalised epilepsies; these were previously categorised as symptomatic or cryptogenic generalised epilepsies. (4). ‘Generalised epilepsy with febrile seizures plus’ is proposed as a new syndrome in development Epilepsy with Myoclonic Absences Clinical note Epilepsy with myoclonic absences (MAE) is a rare syndrome of childhood, which demands scrupulous exclusion of other forms of symptomatic or probably symptomatic cases manifesting with the same seizure (myoclonic absences).173–177 Demographic Data Age at onset varies from the first months of life to early teens with a median of 7 years. Boys (69%) predominate. MAE is a very rare disorder with an approximate prevalence of 0.5–1% among selected patients with epileptic disorders.173,177,178 I have seen only three cases (two of which were idiopathic MAE) over a period of 15 years out of nearly 200 patients with video EEG-recorded typical absence seizures.4,24,177 Clinical Manifestations Myoclonic Absences The myoclonic absences are the hallmark of MAE.173,178 They consist of impairment of consciousness, which varies from mild to severe and rhythmic myoclonic jerks, mainly of the shoulders, arms and legs with a concomitant tonic contraction. Eyelid twitching is practically absent, but perioral myoclonias are frequent. The jerks and the tonic contraction may be unilateral or asymmetrical, and head/body deviation may be a constant feature in some patients. The tonic contraction mainly affects the shoulder and deltoid muscles, and may cause elevation of the arms. Some patients maintain awareness of the jerks. The duration of the absences varies from 8 to 60 s. Myoclonic absences occur many times a day. Other Types of Seizure Other types of seizure also occur in two-thirds of patients.173,174,179 These are infrequent GTCS and atonic seizures, which may precede or occur concurrently with the myoclonic absences. ASE is rare. Precipitating Factors Hyperventilation is the main precipitating factor.173,178 Photosensitivity is uncommon (14%).178 Non-photosensitive myoclonic absences precipitated by eye-closure or eye-opening have been described.173,178 Seizures induced by emotionally gratifying stimuli, such as cheek-kissing or after viewing pleasant or funny events have been reported in a child with inverted chromosomal 15 duplications.24 Neurological and Mental State Neurological examination is usually normal, but nearly half of patients (45%) have impaired cognitive functioning prior to the onset of absences.175,178 Considerations on the Classification of Epilepsy with Myoclonic Absences Myoclonic absences (the seizures) may feature either in normal or children with neurocognitive impairment.16,176 The 1989 ILAE Commission, discounting the idiopathic form, classified ‘epilepsy with myoclonic absences’ among ‘cryptogenic/symptomatic’ generalised epilepsies, that is in the same group of disorders as the Lennox-Gastaut syndrome and EM-AS.1 “The syndrome of epilepsy with myoclonic absences is clinically characterised by absences accompanied by severe bilateral rhythmical clonic jerks, often associated with a tonic contraction. On the EEG, these clinical features are always accompanied by bilateral, synchronous, and symmetrical discharge of rhythmical spike–waves at 3 Hz, similar to childhood absence. Seizures occur many times a day. Awareness of the jerks may be maintained. Associated seizures are rare. Age of onset is ~7 years, and there is a male preponderance. Prognosis is less favourable than in pyknolepsy owing to resistance to therapy of the seizures, mental deterioration, and possible evolution to other types of epilepsy such as Lennox-Gastaut syndrome or JME.”1 Contrary to this, the new ILAE diagnostic scheme considers only the idiopathic form (Table 1.7),5 which probably represents around one-third of the whole spectrum of epileptic disorders manifesting with myoclonic absences. The others are symptomatic or probably symptomatic cases.1 Aetiology Only one-third of patients with myoclonic absences are idiopathic cases and only these belong to this syndrome of MAE. The other two-thirds with myoclonic absences (the seizures, not the syndrome) are due to symptomatic causes including chromosomal abnormalities, such as trisomy 12p, Angelman syndrome, inverted chromosomal 15 duplications179,180 and malformations of brain development.176,177,181 Diagnostic Procedures By definition, in idiopathic MAE, all tests but the EEG should be normal. Brain MRI and chromosomal testing179 are needed to detect symptomatic cases. Electroencephalography Background EEG is usually normal at onset, but may deteriorate later or be abnormal in symptomatic cases. Interictal EEG shows brief generalised, focal or multifocal spike and slow waves in 50% of cases.173,177 Ictal EEG shows 3 Hz GSWD, even in those with unilateral or asymmetrical clinical manifestations. Polygraphic studies have revealed that each myoclonic jerk coincides with the spike component of the discharge.173,177 Differential Diagnosis The differential diagnosis of MAE from other types of syndromes with absences is easy because of the characteristic type of myoclonic absences. The difficulty is between idiopathic and symptomatic/cryptogenic cases that manifest with the same seizure type (myoclonic absences). Symptomatic patients often have an abnormal neurological state, abnormal background EEG and abnormal brain MRI. Chromosomal abnormalities are common.179 Additionally, absences with rhythmic myoclonic jerking, but less than 2.5 Hz spike/polyspike–wave complexes and other characteristics of atypical absences may occur in epileptic encephalopathies173,178,182 and include some of the cases with chromosomal abnormalities.179 Prognosis Myoclonic absences remit after an average of 5 years in one-third of patients.176,177 In the remaining (perhaps symptomatic) patients, absences continue into adult life together with other types of seizures, such as GTCS and atonic seizures, or they develop features of other types of epilepsy, such as Lennox-Gastaut syndrome or JME. Nearly half of the children with MAE have impaired cognitive functioning prior to the onset of absences, but these are probably symptomatic cases. However, half of those who were normal prior to the onset of absences develop cognitive and behavioural impairment. This may mean that the EEG discharges have a deteriorating effect on cognition unless eliminated early with treatment. Management The aim is to stop myoclonic absence seizures as early as possible. Early control of absences may prevent subsequent cognitive deterioration and secure normal development.178 Myoclonic absences are often resistant to treatment. Treatment frequently requires high doses of valproate often combined with ethosuximide or lamotrigine.177 Interestingly, Tassinari et al.173 had a few cases that responded only to phenobarbitone alone or in combination with anti-absence drugs. Newer drugs, such as levetiracetam,183 or old drugs, such as clonazepam and acetazolamide, may be tried in resistant cases. Baclofen a GABAB agonist used for the treatment of spasticity in neurologically impaired patients is contraindicated because of significant provocation of absence seizures. See also AEDs contraidicated in IGEs (page 337). Figure 10.4 Samples from a video EEG showing idiopathic and symptomatic myoclonic absence seizures. The EEG GSWD are similar with no apparent differentiating features or asymmetry in the symptomatic patient. Top: Video EEG of a boy (more...) Epilepsy with Myoclonic-Astatic Seizures Doose Syndrome Clinical note Epilepsy with myoclonic-astatic seizures (EM-AS) or Doose syndrome87–97 is considered as an IGE in the new ILAE diagnostic sceme.5 Diagnosis of this syndrome requires careful application of inclusion and exclusion criteria. Its characteristic symptom, myoclonic-astatic seizures, is shared by many other childhood syndromes and particularly epileptic encephalopathies. Demographic Data Onset occurs between 7 months and 6 years, and peaks at 2–4 years. Two-thirds are boys. EM-AS accounts for about 1–2% of all childhood epilepsies. Clinical Manifestations Doose syndrome is characterised by myoclonic-astatic seizures that often occur together with atonic, myoclonic and absence seizures; myoclonic-astatic status epilepticus is common. Children are normal prior to the onset of seizures. In two-thirds of children, febrile and afebrile generalised tonic clonic seizures appear first, several months prior to the onset of myoclonic-astatic seizures. Considerations on the Classification of EM-AS The 1989 ILAE Commission classifies EM-AS among ‘cryptogenic/symptomatic’ generalised epilepsies, that is in the same group of disorders as Lennox-Gastaut syndrome.1 It is defined as follows: “Manifestations of myoclonic-astatic seizures begin between the ages of 7 months and 6 years (mostly between the ages of 2 and 5 years), with (except if seizures begin in the first year) twice as many boys affected. There is frequently hereditary predisposition and usually a normal developmental background. The seizures are myoclonic, astatic, myoclonic-astatic, absence with clonic and tonic components, and tonic-clonic. Status frequently occurs. Tonic seizures develop late in the course of unfavorable cases. The EEG, initially often normal except for 4–7 Hz rhythms, may have irregular fast spike-wave or polyspike wave. Course and outcome are variable.”1 Contrary to this, the ILAE Task Force considers EM-AS as IGE,5 a view which is similar to that of Doose (Table 10.4):98 “Myoclonic-astatic epilepsy belongs to the epilepsies with primarily generalized seizures and thus stands in one line with absence epilepsies, JME, as well as the infantile and juvenile idiopathic epilepsy with generalized tonic clonic seizures. Like these types of epilepsy, myoclonic-astatic epilepsy is polygenically determined with little nongenetic variability. The disease is characterized by the following criteria: genetic predisposition (high incidence of seizures and/or genetic EEG patterns in relatives); mostly normal development and no neurological deficits before onset; primarily generalized myoclonic, astatic or myoclonic-astatic seizures, short absences and mostly generalized tonic clonic seizures; no tonic seizures or tonic drop attacks during daytime (except for some rare cases with a most unfavourable course); generalized EEG patterns (spikes and waves, photosensitivity, 4–7 Hz rhythms), no multifocal EEG-abnormalities (but often pseudofoci).”98 The problem may reflect lack of specific diagnostic criteria and undefined boundaries of certain epileptic syndromes and particularly epileptic encephalopathies, which may manifest with myoclonic-astatic seizures. This particularly refers to Dravet, Lennox-Gastaut syndrome and atypical benign epilepsy of childhood. Cases of benign and severe myoclonic epilepsy in infants may have been included in EM-AS.89 Other myoclonic epilepsies with brief seizures reported as intermediate cases between EM-AS and Lennox-Gastaut syndrome probably prove this point.99 However, it is generally accepted that some children with myoclonic-astatic seizures are otherwise normal with no discernible causes other than a strong genetic epileptic background and probably represent the genuine ‘Doose syndrome’ of ‘idiopathic epilepsy with myoclonic-astatic seizures’. This point is exemplified in the study of Kaminska et al.,100 who found evidence that EM-AS is distinct from Lennox-Gastaut syndrome, and the distinction appears from the first year of the disorder. A further exciting development is that myoclonic-astatic seizures frequently occur in patients with ‘generalised epilepsy with febrile seizures plus’101, a syndrome that also has strong genetic links with Dravet syndrome. Table 10.4 Diagnostic criteria for Doose syndrome (idiopathic EM-AS) Myoclonic-astatic (in fact myoclonic-atonic) seizures are the defining symptoms (100% of cases).89 These manifest with symmetrical myoclonic jerks immediately followed by loss of muscle tone (post-myoclonic atonia) (Figure 10.5). Atonic seizures of sudden, brief and severe loss of postural tone may involve the whole body or only the head. Attacks are brief, lasting 1–4 s and frequent. Generalised loss of postural tone causes a lightning-like fall. The patient collapses on the floor irresistibly. In brief and milder attacks, there is only head nodding or bending of the knees. Myoclonic jerks may precede or less often intersperse with the atonic manifestations (Figures 10.5 and 10.6). Brief absence seizures happen in more than 50% of cases. These often occur together with myoclonic jerks, facial myoclonias and atonic manifestations. Atonic and absence seizures may occur frequently, sometimes many times a day in the active period of the disease. Absence seizures alone are exceptional. Tonic seizures are an exclusion criterion. Myoclonic-atonic status epilepticus lasting for hours or even days (Figure 10.6) is common affecting one-third of patients. It manifests with varying degrees of usually severe cognitive impairment or cloudiness of consciousness interspersed with repetitive myoclonic and atonic fits. Facial myoclonus of eyelids and mouth may be continuous together with irregular jerks of the limbs and atonic seizures of head nodding or falls. Myoclonic-atonic status epilepticus may occur several times during a period of 1–2 years. Aetiology Doose syndrome may be genetically determined in a multifactorial polygenic fashion with variable penetrance.87–89 One-third of patients have familial seizure disorders and mainly IGEs.87–89 Of significant interest are the clinical and molecular studies in ‘generalised epilepsy with febrile seizures plus’ in which myoclonic-atonic seizures are common in some families.101 ‘Generalised epilepsy with febrile seizures plus’ also has strong genetic links with Dravet syndrome. Diagnostic Procedures By definition, all tests other than the EEG are normal. Electroencephalography Interictal EEG may be normal at the stage of febrile or afebrile GTCS. Rhythmic theta activity in the parasagittal regions may be the only significant abnormality. Subsequently, when myoclonic-atonic seizures appear, there are frequent clusters of 2–3 Hz GSWD interrupted by high amplitude slow waves in cases with predominant atonic or myoclonic-atonic seizures. In children with predominantly myoclonic seizures, paroxysms of irregular spikes or polyspike–wave complexes prevail. The ictal EEG of myoclonic and atonic seizures manifests with discharges of irregular spike–wave or polyspike–wave complexes at a frequency of 2.5–3 Hz or more (Figures 10.5 and 10.6). Atonia is usually concurrent with the slow wave of a single or multiple spike–wave complex and the intensity of the atonia is proportional to the amplitude of the slow wave. Drop attacks are associated with diffuse EMG paucity indicating their true atonic nature.91 The myoclonus of Doose syndrome appears to be a primary generalised epileptic phenomenon, which differs from that of Lennox-Gastaut syndrome.102 In myoclonic-atonic status epilepticus, the EEG shows continuous or discontinuous and repetitive 2–3 Hz spike-wave complexes (Figure 10.6). Differential Diagnosis Differentiation of EM-AS is mainly between: benign myoclonic epilepsy in infancy Dravet syndrome Lennox-Gastaut syndrome late onset West syndrome. In general, children with the Doose syndrome are normal prior to the development of seizures, have a strong family history of IGE, and the background EEG and brain imaging are normal. Progressive myoclonic epilepsies, such as myoclonic epilepsy with ragged-red fibres, Lafora and Unverricht-Lundborg disease, may initially imitate Doose syndrome. However, the associated relevant neurological abnormalities and, sometimes, the relentless progression and deterioration will establish the diagnosis. Atypical benign partial epilepsy of childhood may also imitate Doose syndrome, because of repeated falls, absences and diffuse slow spike–wave activity mainly in the sleep EEG.31,103–105 The main differentiating point is that these children also have nocturnal focal seizures similar to the Rolandic seizures (RS) that are often the presenting seizure symptom. Also, the EEG shows centrotemporal and other functional spikes in various locations. Atypical evolutions of RS106,107 and Panayiotopoulos syndrome108,109 may have clinico-EEG features similar to those of Doose syndrome, but they are preceded by typical presentations of these syndromes (see Chapter 9). A similar, but reversible, clinico-EEG condition may be induced by carbamazepine,31 oxcarbazepine110 and lamotrigine111 in a few children with RS. This possibility should be considered in children with RS and dramatic deterioration after treatment with these AEDs. Children with ‘epilepsy with continuous spike–wave complexes during slow-wave sleep’ may also have drop attacks due to atypical absences or negative epileptic myoclonus. Non-epileptic myoclonus of many neurological disorders rarely raises a diagnostic problem with Doose syndrome, unless it is a symptom of a degenerative disease with associated epileptic features.90 Clinical note Diagnostic tips The diagnosis of Doose syndrome is probably safe if myoclonic-atonic seizures start in a previously normal child with pre-existing febrile or afebrile GTCS and familial seizure disorders. Differential diagnostic problems from Lennox-Gastaut syndrome probably reflect ill-defined inclusion and exclusion criteria. Prognosis The prognosis is unclear probably because of different selection criteria. Half of the patients may achieve a seizure-free state and normal or near normal development.95 Myoclonic-atonic seizures remit within 1–3 years from onset despite initial resistance to treatment, but GTCS or clonic seizures tend to continue.95 These patients who have a good prognosis may correspond to the genuine Doose syndrome of the idiopathic form of EM-AS. Spontaneous remission with normal development has been observed in a few untreated cases, but these may belong to benign myoclonic epilepsy in infancy. The others, probably belonging to symptomatic or probably symptomatic cases or other syndromes, may continue with seizures, severe impairment of cognitive functions and behavioural abnormalities. Ataxia, poor motor function, dysarthria and poor language development may emerge. Management Drug therapy is dictated by seizure type. Valproate, which is effective in myoclonic jerks, atonic seizures and absences, is the most efficacious. Add-on small doses of lamotrigine have a beneficial pharmacodynamic interaction with valproate. Topiramate reduces the frequency of atonic seizures.112 Levetiracetam may be an effective substitute AED for valproate. In resistant cases, ketogenic diet, followed by adrenocorticotropin hormone (ACTH) and ethosuximide, have been found to be highly beneficial.95 Benzodiazepines, acetazolamide, sulthiame and even bromides are also used. Carbamazepine, phenytoin, and vigabatrin are contraindicated. In myoclonic-atonic status epilepticus, intravenous benzodiazepines are often efficacious, but rarely may precipitate tonic status. Childhood Absence Epilepsy Patient note “…a disease with an explosive onset between the ages of 4 and 12 years, of frequent short, very slight, monotonous minor epileptiform seizures of uniform severity, which recur almost daily for weeks, months, or years, are uninfluenced by anti-epileptic remedies, do not impede normal and psychical development, and ultimately cease spontaneously never to return. At most, the eyeballs may roll upwards, the lids may flicker, and the arms may be raised by a feeble tonic spasm. Clonic movements, however slight, obvious vasomotor disturbances, palpitations, and lassitude or confusion after the attacks are equivocal symptoms strongly suggestive of oncoming grave epilepsy, and for the present they should be considered as foreign to the more favourable disease.” 113 Patient note W. J. Adie (1924) 113 defining CAE as an epileptic syndrome Clinical note CAE is the prototype IGE of typical absence seizures.7,15,16,18,114 It is genetically determined, age related and affects otherwise normal children. Demographic Data Onset is between 4 and 10 years of age, with a peak at 5–6 years.15,114–117 The onset of typical absence seizures in CAE before 4 years of age 118–121 and after 10 years of age118–121 is uncertain or at least exceptional. Comments and Debate on the ILAE Definition of CAE The ILAE Commission of 1989 1 largely defined CAE by age at onset and frequency of absences: “Childhood absence epilepsy (pyknolepsy) occurs in children of school age (peak manifestation age 6–7 years), with a strong genetic predisposition in otherwise normal children. It appears more frequently in girls than in boys. It is characterised by very frequent (several to many per day) absences. The EEG reveals bilateral, synchronous symmetrical spike–waves, usually 3 Hz, on a normal background activity. During adolescence, GTCS often develop. Otherwise, absences may remit or more rarely, persist as the only seizure type.”1 The new ILAE diagnostic scheme also classifies CAE as IGE.5 The ILAE definition is a very broad and requires revision. Otherwise, any type of frequent absence seizures occurring in childhood would be erroneously equated with CAE. Because of this ambiguity, the epidemiology, genetics, age at onset, clinical manifestations, other types of seizures, long-term prognosis and treatment of CAE reviewed in this chapter may not accurately reflect the syndrome of CAE. It is also because of this ambiguity that some authors: (a.) have divided patients with childhood onset absence seizures into ‘subsyndromes’ including those who remit, those who persist into adolescence and develop GTCS, and those who develop both GTCS and myoclonic seizures during adolescence;138 and (b.) consider that patients with CAE ‘evolve’ into JAE or JME.139,140 The inclusion and exclusion criteria of Table 10.5 proposed by Loiseau and Panayiotopoulos114 for CAE should not be taken as an extreme position. They do not differ significantly from the ILAE (1989)1 criteria of CAE with: age at onset in childhood very frequent (several to many per day) absences presumably with severe impairment of consciousness ictal EEG with bilateral, synchronous and symmetrical 3 Hz GSWD, on a normal background activity (that presumably excludes fragmented, asymmetrical and asynchronous 3–5 Hz GSWD with intradischarge variations) GTCS accepted only if they develop later in adolescence. Also, the Commission (1989)1 by accepting ‘epilepsy with myoclonic absences’ as a separate syndrome differentiates myoclonic absences from typical absences of CAE. It is along this line that Loiseau and Panayiotopoulos114 also considered eyelid myoclonia (which is a predominantly myoclonic and less of an absence syndrome) as an exclusion criterion. Whether, perioral myoclonia or single violent jerks during the ictus of an absence seizure is an exclusion criterion may be debatable. However, their presence indicates a worse prognosis.15,16,141 The same applies to multiple spikes (more than three spikes/wave), which also indicate a bad prognosis,138,142 and coexistent myoclonic jerks or GTCS.24 Further, the Commission (1989),1 by accepting ‘typical absence seizures consistently provoked by specific stimuli’ as a specific type of reflex seizures, indicates that these may be a separate group from CAE.16,114,136,143 Table 10.5 Criteria for childhood absence epilepsy Age at Onset of Absences Does Not Determine Syndromic Classification of CAE We have studied 39 adults with IGE with typical absences starting before 10 years of age.144 All were older than 18 years (31.5±10.5; range 18–56) at the last follow-up and all had EEG (15 with video-EEG) recorded typical absence seizures. Typical absences had onset at 6.2±1.9 years (range 2–9) that persisted into adulthhood in 28 (71.8%) cases. GTCS occurred in 87.2% (onset 13±7.2 years; range 2–36). Myoclonic jerks occurred in 38.5% (onset 2.6±4.1 years; range 7–18). Women (82%) and photosensitivity (56.4%) markedly predominated. Eight patients have EMA, 5 JAE, 4 perioral myoclonia with absences, 3 JME, 3 absences with single myoclonic jerk, 1 CAE, 3 predominantly reflex absences. Twelve patients (8 with photosensitivity) could not be classified. Only 6 patients were free from all type of seizures (1 CAE, 2 JAE, 2 unclassified IGE with reflex absences). 144 Synonyms Pyknolepsy. Many European clinicians use the word ‘pyknolepsy’ (from the Greek words pyknos = dense and epilepsy) either to define a high daily frequency of absences or as a synonym to CAE. Two-thirds of patients are girls115,116,122,123 despite some studies indicating that boys and girls are equally affected.124,125 The prevalence is about 10%126–128 and the annual incidence is about 7/100,000129–131 in children with epileptic seizures who are less than 15–16 years of age. Recruitment bias explains the wide range in the reported incidence (1.9–8/100,000 of children < 16 years) and prevalence (2–37% of children with epileptic disorders) of CAE.124,129,130,132–137 Clinical Manifestations Typical Absence Seizures CAE manifests with the most characteristic and classical example of typical absence seizures, which are characterised by: short duration abrupt onset and abrupt termination severe impairment of consciousness high daily frequency. Probably any other types of seizure are incompatible with this diagnosis. Mild impairment of consciousness in untreated patients is an exclusion criterion.114 Absences are severe and frequent, with from ten to hundreds per day, for which reason CAE is also known as pyknolepsy.113 They are of abrupt onset and abrupt termination (Figures 10.2, 10.7 and 10.8). Their duration varies from 4–20 s, though most last around 10 s. Clinically, the hallmark of the absence is abrupt, brief and severe impairment of consciousness with unresponsiveness and interruption of the ongoing voluntary activity, which is not restored during the ictus. The eyes open spontaneously, overbreathing, and speech and other voluntary activity stop within the first 3 s from the onset of the discharge. Simple automatisms occur in two-thirds of seizures, but are not stereotyped. The eyes stare or move slowly. Mild myoclonic elements of the eyes, eyebrows and eyelids may feature in CAE, but are usually mild and occur at the onset of the GSWD. However, more severe and sustained myoclonic jerks of the facial muscles may indicate other IGEs with absences. Figure 10.7 Video EEG of an 8-year-old boy with classical CAE. He was referred for EEG because of “blanks and day dreaming”; the diagnosis had been overlooked despite frequent absences for 2 years, which also interfered (more...) Figure 10.8 Girl Aged 6 Years with Untreated Childhood Absence Epilepsy. Left: Long runs of bilateral, rhythmic and fusiform, high amplitude 3 Hz slow waves appear in the posterior regions (between arrows). Right: Onset of a typical absence seizure. (more...) The attack ends as abruptly as it commenced with sudden resumption of the pre-absence activity as if it was not interrupted. Typical absence seizures are nearly invariably provoked by hyperventilation. Other Types of Seizures Clinical note Seizures other than typical absences are not compatible with CAE. The only exceptions are (a). febrile seizures that may precede the onset of CAE and (b). solitary or infrequent GTCS that occur long after the onset of absence seizures and usually in adolescence after absences have remitted. GTCS or myoclonic jerks preceding the onset of typical absences or concomitant with the stage of active absence seizures do not occur in CAE.15,114,143,145,146 However, about 10% of patients may later develop a few, solitary or infrequent GTCS in adolescence or adult life.145,146 This contrasts with the fact that focal seizures, myoclonic jerks, GTCS and other more bizarre fits have been described with CAE but these are probably other epileptic syndromes starting with absences in childhood.134 Status epilepticus, whether convulsive or non-convulsive, is incompatible with CAE. If this occurs, the diagnosis of CAE should be seriously challenged. Also, though absence status epilepticus may occur in 5–16% of patients with typical absence seizures starting before the age of 10 years,117,140,147 this is probably incompatible with CAE.15,19,77 Atonic falls do not occur in CAE.136 Seizure-Precipitating Factors Typical absence seizures occur spontaneously, but they are also influenced by various other factors, mainly hyperventilation (Figures 10.2, and 10.6). Hyperventilation is the most potent precipitating factor that induces absences in more than 90% of the trials,148,149 ranging from 75%150 or 80%151 to 100%.143 A diagnosis of CAE should be seriously questioned in an untreated child who does not have an attack on hyperventilation. Other precipitating or facilitating factors are emotional (anger, sorrow, fear, surprise, embarrassment), intellectual (lack of interest, release of attention, mealtimes for some children and school-time for others) and metabolic (hypoglycaemia).114 Typical absence seizures generally do not occur when the child is busy and stimulated by physical or mental activity, or has sustained attention.136 Emotional or conflicting situations, such as reading difficulties, may provoke absences.152 In a given patient, typical absence seizures are often triggered by the same factor. There are no other precipitating factors in CAE, which is at variance with nearly all other forms of IGE. In particular, though photosensitivity is accepted according to the ILAE Commission’s definition,1 most other authors consider clinical photosensitivity as a significant exclusion criterion.15,136,143 Mild EEG photosensitivity or facilitation (not consistent provocation) of photoparoxysmal responses and absences may occur. Aetiology Although CAE is genetically determined, the precise mode of inheritance and the genes involved remain largely unidentified.7 In monozygotic twins, 84% had 3 Hz GSWD and only 75 % of pairs had clinical absence seizures. These events occurred 16 times less often in dizygotic twins.115 Currently, various chromosomal loci have been identified in families with absences of childhood onset (not necessarily equated with CAE). Linkage to chromosome 1 was found in families with absences starting in childhood and the later development of myoclonic jerks and GTCS, as in JME.9 Linkage analysis in five generations of a family in which affected patients had childhood absences and GTCS provided evidence of a locus on chromosome 8q24.9,138 The candidate region for this locus, designated ECA 1, has been refined, but a gene remains to be identified. According to the criteria proposed in this chapter, neither of these groups is CAE. There are also reports implicating chromosome 5q31.1 and 19p13.2 (see ref 7 for an excellent recent review). Furthermore, there is now evidence available to suggest that mutations in genes encoding GABA receptors34–36 or brain-expressed voltage-dependent calcium channels37 may underlie CAE. Feucht et al.34 found a significant association between a polymorphism in GABA A receptor gamma 3 subunit in chromosome 15q11 in 50 families with CAE. Marini et al.36 found GABA A receptor gamma 2 subunit gene mutations on chromosome 5 in a large family with CAE and febrile seizures (including febrile seizures plus and other seizure phenotypes). This gene mutation segregated with febrile seizures and CAE, and also occurred in individuals with the other phenotypes. The clinical and molecular data suggested that the GABA A receptor subunit mutation alone could account for the febrile seizure phenotype, but an interaction of this gene with another gene or genes was required for the childhood absence phenotype in this family. Linkage analysis for a putative second gene contributing to the childhood absence phenotype suggested possible loci on chromosomes 10, 13, 14 and 15.36 Chen et al.37 found 68 variations, including 12 missense mutations in the calcium channel CACNA1H gene in CAE patients. The identified missense mutations occurred in the highly conserved residues of the T-type calcium channel gene.37 However, another study of 33 nuclear families, each with two or more individuals with CAE each provided conclusive evidence that the genes encoding GABA A and GABA B receptors, voltage-dependent calcium channels and the ECA1 region on chromosome 8q do not account independently for the childhood absence trait in a majority of the families.153 Acquired factors may play a facilitating role. Diagnostic Procedures In typical cases nothing but an EEG is needed. Interictal EEG The interictal EEG in CAE has normal background activity, with frequent rhythmic posterior delta activity (Figure 10.8). Benign functional spikes (mainly centrotemporal and less frequently occipital or frontal) may be seen and do not alter the prognosis. Posterior rhythmic slow activity is considered a characteristic finding in CAE. This consists of long runs of bilateral rhythmic and fusiform, high amplitude slow waves at 3 Hz in the distribution regions of alpha rhythm. It is bilateral, often of higher amplitude on the right, but occasionally may be lateralised to one side or fluctuate from side to side in emphasis. It occurs in less than half of patients at a peak age of 6–10 years. Patients with this EEG finding tend to have longer absence seizures, a better prognosis and less often develop GTCS.154,155 Sleep EEG has not been studied systematically in pure forms of CAE. Generally, in IGE, GSWD are more likely to increase, but a reduction is also observed during sleep. The discharges are shorter and usually devoid of discernible clinical manifestations, even in those patients who have numerous clinical seizures during the alert state. However, clinical absence seizures that may awake the patient have been recorded during sleep. Sleep EEG patterns are normal. Photic stimulation: Clinical photosensitivity or consistent provocation of typical absences by IPS is an exclusion criterion for CAE. However, IPS may act as a facilitatory factor in that these children may have more typical absences during IPS than in the resting EEG, but these are far less common than during hyperventilation. Ictal EEG Ictal EEG consists of high amplitude 3 Hz GSWD which are of higher voltage in the anterior regions. They are rhythmic at around 3 Hz (2.5–4 Hz) with a gradual and regular slowing down of the frequency by 0.5–1 Hz from the initial to the terminal phase of the discharge. The opening phase of the discharge, 1–2 s from the onset, is usually fast and unreliable for these measurements. There are no marked variations in the relationship of the spike to the slow wave, no fluctuations in the intradischarge frequency and certainly no fragmentations of the ictal discharges (Figure 10.7). Clinical note Diagnostic tips Typical absence seizures of CAE are easy to diagnose and reproduce by hyperventilation. Any child with sudden, brief and frequent cessation of physical and mental activity should be tested clinically for absences. This is easily performed with the hyperventilation test. Ask the child to overbreathe for 3 minutes while counting his/her breaths and holding his/her hands in front. This will evoke an absence in more than 90% of children with childhood absence epilepsy. It is essential that a video EEG is performed prior to initiating treatment for appropriate confirmation and documentation of the clinical EEG characteristics of the absences. If this is not possible, the clinical manifestations should be documented with home camcorders or modern digital recorders. Differential Diagnosis CAE should be the easiest type of epileptic syndrome to diagnose, because seizures have an abrupt onset and termination, high daily frequency and are nearly invariably provoked by hyperventilation. In practical terms, a child with suspected typical absences should be asked to overbreathe for 3 minutes while standing, counting his/her breaths and with the hands extended in front. This will provoke an absence in as many as 90% of those affected. Important note Clarification CAE is not synonymous with any type of absence seizures starting in childhood. Therefore, other epilepsy syndromes with absence seizures that may be life-long and have a worse prognosis should be meticulously differentiated from CAE. Diagnosis should improve with heightened awareness and video EEG studies. Exclusion criteria for CAE are as important as inclusion criteria (Table 10.5). Automatisms have no significance in the diagnosis. They should not be taken as evidence of complex focal seizures (see Table 10.2), which require entirely different management. Prognosis Patient note “For myself I shall be well satisfied if I have made it appear probable to you that there does exist a form of epilepsy in children which is distinguishable by its clinical features and in which the prognosis is always good.” Adie (1924)113 Most of the studies on the prognosis of CAE are based on the seizure/symptom itself and not on the relevant syndrome. Absences and therefore their prognosis are syndrome related.15,24,141 Studies on the prognosis of typical absences with onset in childhood (which are not necessarily CAE) 115–117,145,147,151,154,156–169 give an overall rate of remission from 30–80%. Thus, 20–70 % of patients continue having typical absences in adult life and of these 90% will develop other generalised seizures such as GTCS and myoclonic jerks. Absences become less frequent and less severe with age. The diagnostic confusion is further indicated by the finding that some cases develop motor or versive seizures of frontal lobe epilepsies. If CAE is defined by age at onset alone, half of the patients develop GTCS later.168 The factors indicating a worse prognosis in these studies were: mental retardation, neurological and EEG background abnormalities (which by definition are against IGE) GTCS or other type of seizures preceding or coinciding with the onset of typical absences (an exclusion criterion for CAE) photosensitivity (exclusion criterion) onset in late childhood or adolescence (are these JAE? or JME?) or very early childhood (are these MAE? or EMA?). prominent ictal myoclonic components (are these absences with perioral or single myoclonic jerks? poor response to treatment. Note on the Prognosis of Childhood Absence Epilepsy In my opinion, based on prospective studies of children fulfilling the strict criteria of CAE (see Table 10.5) and retrospective evaluation of adults with absences15,56,57 the long-term prognosis of CAE is excellent. This implies complete remission of absences 2–6 years after onset and no more than 3% of patients developing infrequent GTCS in adult life. There is indirect, but not definite, evidence for this. In our series of 85 adult patients (> 16 years of age with mean age 32 years) with absences and syndromes of IGEs, none had CAE, though in 37 patients absences had started in childhood, before the age of 10 years.57 Most patients had JME (30 patients), but others had JAE (10 patients), IGE with phantom absences (12 patients), eyelid myoclonia with absences (11 patients), perioral myoclonia with absences (PMA; 7 patients), photosensitive absence syndromes (4 patients), absences with single myoclonic jerks (1 patient), and 10 patients were unclassified. Half of these patients would have been classified as CAE continuing in adult life, if the only criteria for diagnosis were age at onset and frequency of absences. This optimistic view of mine is also supported by the results of long-term studies in which the remission rate for absences starting in childhood increased with the application of exclusion criteria.114,145,154 Loiseau and associates145,146 studied 53 patients over the age of 20 years at last follow-up. Inclusion criteria were age at onset (3–10 years) of daily and EEG-recorded typical absences as a presenting symptom of normal children, with no history of preceding seizures other than febrile convulsions. Patients with EEG multiple or irregular spike–wave complexes and/or photosensitivity were excluded. Absences persisted in five children (< 10%) and were the only type of seizure in two of them. GTCS occurred in 14 patients (26%), but were isolated or rare in 11 cases. GTCS were more common among patients with onset of absences at 9–10 years and without posterior delta rhythms. Response to treatment varied. Control was achieved in 12 patients within weeks, but in most cases absences persisted for years. In this scholarly study, the prognosis could probably have been improved if video EEG had been used to exclude patients who perhaps had other types of IGE with frequent absences from childhood. In other words, patients who did not remit may have had other epileptic syndromes, particularly All the above studies involve long-term follow-up of patients with onset of absences in childhood or adolescence. Most were based on clinico-EEG documented ‘absences’, probably with severe impairment of consciousness, without reference to syndromic classification and without the help of video-EEG recordings. Myoclonic jerks or GTCS occurring independently of absences, severity of impairment of consciousness during the absence ictus, syndromic criteria of epilepsies and video-EEG documentation have not been utilised in these studies. By applying strict diagnostic criteria, an excellent prognosis may be anticipated for CAE. At a time when no anti-absence drugs existed, Adie113 concluded that even if absence seizures in CAE (he called it pyknolepsy) persisted for a long time, they ultimately ceased, never to return. Adie considered only the pure form of CAE with severe pyknoleptic absences while with the advent of EEG this group was broaden to include any type absences starting in childhood. A good prognosis is consistent with recent findings that absences of CAE, even if they persist for several years, they finally disappear with age in more than 90% of cases.18,154 In a Swedish population-based study, a 91% remission rate was found when patients with absence epilepsy had only absence seizures.162 Remission occurs before the age of 12 years. Less than 10% of patients develop infrequent or solitary GTCS in adolescence or adult life. It is exceptional for patients to continue having absence seizures when adults. Poor social adjustment has been reported in one-third of patients.122,147,154,160 This may be due to frequent absence seizures, particularly if they were not treated early at their onset, the attitudes of school mates and parents, or medication. Management Monotherapy either with valproate or ethosuximide controls absences in 80% of patients.4 Another option is lamotrigine monotherapy, though this is less effective and only nearly half of patients become seizure free.4,170,171 If monotherapy fails or unacceptable adverse reactions appear, replacement of one drug with the other is the alternative. Adding small doses of lamotrigine to valproate may be the best combination in resistant cases. There are anecdotal reports that children may not respond to syrup of valproate despite adequate levels, but seizures stop if this is replaced with tablets of valproate. It is also anecdotal experience that, once seizure cessation has been achieved, valproate may be safely reduced to more moderate doses without relapses.4 Contraindicated drugs, which make seizures worse and may induce status epilepticus are: carbamazepine, gabapentin, oxcarbazepine, phenytoin, phenobarbitone, vigabatrin and tiagabine. Withdrawing anti-epileptic medication: In the pure form of CAE, drug therapy can be gradually withdrawn (within 3–6 months) after 2–3 years free of seizures. Clinical note Note of practical significance Clinical note In evaluating the efficacy of therapy or anticipating AED withdrawal, it should be remembered that: the true frequency of typical absence seizures is difficult to assess without prolonged video EEG monitoring, because the clinical manifestations can be easily missed. parents’ assessment is often an underestimate.4,172 Juvenile Absence Epilepsy Clinical note JAE is an IGE syndrome1,5,15 mainly manifesting with severe typical absence seizures; nearly all patients (80%) also suffer from GTCS and one-fifth from sporadic myoclonic jerks.15,16,24,184,185 Demographic Data Usual age at onset is 9–13 years (70% of patients), but can range from 5 to 20 years.15,185 Myoclonic jerks and GTCS usually begin 1–10 years after the onset of absences. Rarely, GTCS may precede the onset of absences.15,184 Both sexes are equally affected. The exact prevalence of JAE is unknown because of variable criteria. In adults over 20 years of age, the prevalence of JAE may be around 2–3% of all epilepsies and around 8–10% of IGEs.57,186 Clinical Manifestations JAE manifests with severe typical absences.15,184,185 Nearly all patients also develop GTCS and one-fifth also suffer from mild myoclonic jerks. Typical Absence Seizures Frequent and severe typical absences are the defining seizure type of JAE.15,184,185 The seizures are similar to those of CAE, though they may be milder. The usual frequency of absences is approximately 1–10/day, but this may be much higher for some patients. The hallmark of the absence is abrupt, brief and severe impairment of consciousness with total or partial unresponsiveness. Mild or inconspicuous impairment of consciousness is not compatible with JAE. The ongoing voluntary activity usually stops at onset, but may be partly restored during the ictus. Automatisms are frequent, usually occurring 6–10 s after the onset of the EEG discharge (Figure 10.9). In JAE, mild myoclonic elements of the eyelids are common during the absence. However, more severe and sustained myoclonic jerks of the facial muscles may indicate other IGE with absences. Severe eyelid or perioral myoclonus, rhythmic limb jerking and single or arrhythmic myoclonic jerks of the head, trunk or limbs during the absence ictus are probably incompatible with JAE. Figure 10.9 Video EEG of a 26-Year-Old Normal Woman. She started having frequent typical absence seizures (tens of seizures each day) with severe impairment of consciousness at the age of 11 years, but medical attendance was sought (more...) Considerations on Classification The 1989 ILAE classification broadly defined JAE as follows:1 “The absences of JAE are the same as in pyknolepsy, but absences with retropulsive movements are less common. Manifestation occurs around puberty. Seizure frequency is lower than in pyknolepsy, with absences occurring less frequently than every day, mostly sporadically. Association with GTCS is frequent, and GTCS precede the absence manifestations more often than in childhood absence epilepsy, often occurring on awakening. Not infrequently, the patients also have myoclonic seizures. Sex distribution is equal. The spike-waves are often >3 Hz. Response to therapy is excellent.” Age at onset (around puberty) and frequency of seizures (less frequent than CAE) are insufficient criteria for the categorisation of any syndrome.15 Thus, epidemiology, genetics, age at onset, clinical manifestations, other types of seizure, long-term prognosis and treatment may not accurately reflect the syndrome of JAE. Recently, JAE has been redefined based on a cluster of clinical and EEG manifestations studied in video EEG recordings (Table 10.6).15,24 The ILAE Task Force has not yet reached definite conclusions regarding the definition of JAE, though there is a tendency to consider JAE as part of a broader syndrome of IGE in adolescence.5,86 Table 10.6 Main inclusion and exclusion criteria for juvenile absence epilepsy (JAE) Age at Onset of Absences Does Not Determine Syndromic Classification of JAE We have studied 71 adults with onset of typical absences after the age of 10 years (median 13 years). All were over 18 years of age and all had experienced typical absences, verified by EEG or video EEG. Two-thirds were women (43 patients). Mean age at last follow-up was 36 years. In 65 patients (92%), absences continued during adulthood. All but two patients had GTCS with a mean age at onset of 19 years. A total of 33 patients (47%) also had myoclonic jerks with a mean age at onset of 16 years. One-third of patients (26 patients were clinically or EEG photosensitive. In terms of epileptic syndromes, 21 had JME, 13 phantom absences with GTCS, 11 JAE, 5 eyelid myoclonia with absences, 3 perioral myoclonia with absences, 2 purely photosensitive IGE, 2 GTCS on awakening and 1 absences with single myoclonic jerks; 13 patients could not be classified. Patients with briefer, milder and later onset absence seizures had a worse prognosis. Duration of the absences varies from 4–30 s, but it is usually long (about 16 s). Generalised Tonic Clonic Seizures GTCS are probably unavoidable in untreated patients. They occur in 80% of patients, mainly after awakening, though nocturnal or diurnal GTCS may also be experienced.15,57,168,184,185,187 GTCS are usually infrequent, but may also become severe and intractable. Myoclonic Jerks Myoclonic jerks occur in 15–25% of patients and are infrequent, mild and of random distribution. They usually occur in the afternoon when the patient is more tired than in the morning after awakening.15,66 Absence Status Epilepticus ASE is truly generalised non-convulsive (without any type of jerks) and occurs in one-fifth of patients.19,66 Seizure-Precipitating Factors Mental and psychological arousal is the main precipitating factor for absences. Conversely, sleep deprivation, fatigue, alcohol, excitement and lights, either alone or more usually in combination, are the main precipitating factors for GTCS. Some authors have reported that 8% of JAE patients suffer from photosensitivity clinically or on the EEG.184 However, clinical photosensitivity that is consistent with provocation of seizures (absences, GTCS or jerks) may be incompatible with JAE. These patients may have another IGE syndrome.15 EEG photosensitivity (i.e. facilitation of absences by IPS) may not be uncommon. Aetiology JAE is determined by genetic factors, but mode of transmission and its relation to other forms of IGE, particularly CAE and JME, has not yet been established. A single Mendelian mode appears to be unlikely. There is an increased incidence of epileptic disorders in families of patients with JAE and there are reports of monozygotic twins with JAE.24,185,188 A proband with JAE was found in 3 of 37 families selected, because at least three members were affected by IGE in one or more generations.189 However, only one sibling also had JAE, while other members mainly had GTCS.189 In various reports JAE has been linked to chromosome 8,190 21,191 1810 and probably 5.10 Heterogeneity may be common. Autopsy192 and MRI studies54 found microdysgenesis and other cerebral changes in patients with JAE. Diagnostic Procedures All tests apart from the EEG are normal. Electroencephalography In untreated patients, absences are easily elicited by hyperventilation (Figure 10.9), if not the diagnosis of JAE should be questioned. The interictal EEG is normal or with mild abnormalities only. Focal epileptiform abnormalities and abortive asymmetrical bursts of spike/multiple spikes are common. The ictal EEG shows 3–4 Hz GSWD. The frequency at the initial phase of the discharge is usually fast (3–5 Hz). There is a gradual and smooth decline in frequency from the initial to the terminal phase. The discharge is regular, with well-formed spikes and polyspikes, which retain a constant relation with the slow waves (Figure 10.9). Differential Diagnosis In general, and particularly in adults, absences are often misdiagnosed as complex focal seizures, though they are easy to differentiate (Table 10.2).56,57 The differentiation of JAE from other IGE with absences may be more difficult without appropriate video EEG evaluation.4,15,24 In children, it is often difficult to distinguish between CAE and JAE, because their features often overlap and manifestations are similar. In JAE, absences start later, usually they are less frequent and impairment of cognition is less severe.24 Automatisms are equally prominent in both. Limb myoclonic jerks (not during the absences) and/or GTCS in the presence of severe absences indicate JAE. JAE is distinctly different from the Jeavons syndrome of very brief seizures marked with rapid eyelid myoclonia, PMA with rhythmic perioral myoclonia during the absence, and MAE with rhythmic myoclonic jerks during the absence. In adolescents, the differential diagnosis between JAE and JME should not be difficult. Severe absences are the major problem in JAE while myoclonic jerks are the main seizure type in JME (Table 10.7). Absences in JME are mild and often inconspicuous. Table 10.7 Main differences between JME and JAE Prognosis JAE is a lifelong disorder, though seizures can be controlled in 70–80% of patients. However, there is a tendency for the absences to become less severe, in terms of impairment of cognition, duration and frequency, with age and particularly after the fourth decade of life.168,184GTCS are usually infrequent and are often precipitated by sleep deprivation, fatigue and alcohol consumption. Myoclonic jerks, if present, are not troublesome to the patient. However, one-fifth of patients may have frequent and sometimes intractable absences and GTCS, and this figure may be higher if appropriate treatment is not initiated in the early stages of JAE. Management In JAE, the consensus is that because of the frequent combination of absences and GTCS, the drug of choice is valproate, which controls all seizures in 70–80% of patients.4 Lamotrigine, though less effective than valproate, probably controls absences and GTCS in half of patients and may be a monotherapy option, particularly in women of childbearing age.193,194 If seizure control with valproate monotherapy is inadequate, add-on treatment with lamotrigine or ethosuximide (if absences persist) may control the situation. If patients are unwilling to receive treatment with valproate, combining lamotrigine with levetiracetam may prove a potent efficacious option based on their different seizure efficacy and mode of action. In one study of 5 patients with JAE all became seizure free with levetiracetam alone or as add-on therapy.183 Control of absences is usually associated with good control of GTCS (90% of cases). Patients should be warned about the factors precipitating GTCS. Treatment may be lifelong, because attempts to withdraw medication nearly invariably leads to relapses even after many years free of seizures. Juvenile Myoclonic Epilepsy Janz Syndrome Clinical note Juvenile myoclonic epilepsy (JME) is one of the most important syndromes of IGE and is genetically determined.1,60,195–201 Demographic Data The triad of absences, jerks and GTCS shows a characteristic age-related onset (Figure 10.10). Absences, when a feature, begin between the ages of 5 and 16 years. Myoclonic jerks follow between 1 and 9 years later, usually around the age of 14–15 years. GTCS usually appear a few months after, occasionally earlier, the myoclonic jerks. Exceptionally, JME may start or become clinically identifiable in adult life as ‘adult myoclonic epilepsy’.202 Both sexes are equally affected. The reported prevalence of JME in hospital-based clinics has increased since the syndrome was first described, from 2.7%,201 5.7%,203,204 8.7%,205 to 10.2%.60 In community-based studies,206,207 the reported prevalence is lower and probably reflects underdiagnosis, which may improve with heightened medical awareness. Figure 10.10 Age at onset of absences, myoclonic jerks and GTCS in 66 consecutive patients with JME. Modified from Panayiotopoulos et al with the permission of the Editor of Epilepsia Clinical Manifestations Juvenile myoclonic epilepsy is characterised by the triad of: Myoclonic jerks on awakening (all patients) Generalised tonic clonic seizures (> 90% of patients) Typical absences (about one-third of patients). Seizures have an age-related onset. Myoclonic status epilepticus is common. Patient note Lots of blanks and jerks; then I had a grand mal...I usually have fits when rushing after getting up; usually does not happen later in the day198 Classification and Definition of Juvenile Myoclonic Epilepsy The 1989 ILAE Commission classified JME as a distinct syndrome of IGE and defined it as follows:1 “Juvenile myoclonic epilepsy (impulsive petit mal) appears around puberty and is characterised by seizures with bilateral, single or repetitive, arrhythmic, irregular myoclonic jerks, predominantly in the arms. Jerks may cause some patients to fall suddenly. No disturbance of consciousness is noticeable. The disorder may be inherited, and sex distribution is equal. Often, there are GTCS and, less often, infrequent absences. The seizures usually occur shortly after awakening and are often precipitated by sleep deprivation. Interictal and ictal EEG have rapid, generalized, often irregular spike–waves and polyspike–waves; there is no close phase correlation between EEG spikes and jerks. Frequently, the patients are photosensitive. Response to appropriate drugs is good.” The ILAE Task Force has not yet reached definite conclusions regarding the definition of JME though there is a tendency to consider JME as part of a broader syndrome of IGE in adolescence.5,86 Myoclonic Jerks Myoclonic jerks occurring after awakening are the most prominent and characteristic seizure type (Figures 10.3 and 10.11). They are shock-like, irregular and arrhythmic, clonic movements of proximal and distal muscles mainly of the upper extremities. They are often inconspicuous, restricted to the fingers, making the patient prone to drop things or look clumsy. They may be violent enough to cause falls. One-fifth of patients describe their jerks as unilateral, but video EEG shows that the jerks affect both sides (Figure 10.3).199,208 Figure 10.11 Violent Myoclonic Jerks of the Hands Associated with Typical EEG Manifestations. From Panayiotopoulos et al. (1994) with the permission of the Editor of Epilepsia. Some patients (< 10%) with mild forms of JME have myoclonic jerks only without developing GTCS.60,209 Typical Absence Seizures One-third of patients have typical absences, which are brief with subtle impairment of consciousness. They are different from the absence seizures of childhood or JAE.24,25,60 Absences appearing before the age of 10 years may be more severe. They become less frequent and less severe with age.24,25,60 One-tenth of patients, do not perceive absences despite GSWD lasting more than 3 s.25,60 However, the GSWD seen on video EEG with breath counting during hyperventilation often manifest with mild impairment of cognition, eyelid flickering or both (Figures 10.12, and 10.13).4 Figure 10.12 Video EEG of two patients with JME. Top: GSWD are not associated with apparent clinical manifestations (but these may have been revealed if breath counting was performed during hyperventilation). Bottom: GSWD are associated with mild (more...) Figure 10.13 Sample from a video EEG of a woman aged 28 years who had suffered from JME since the age of 9 years. This woman was referred for a routine EEG because she had experienced “probable IGE-absences from age 9 until (more...) Generalised Tonic Clonic Seizures GTCS usually follow the onset of myoclonic jerks. Myoclonic jerks, which usually occur in clusters and often with an accelerating frequency and severity, may precede GTCS, a so-called clonic-tonic-clonic generalised seizure.199 Status Epilepticus Myoclonic status epilepticus is probably more common than appreciated.60,210 It almost invariably starts on awakening and is associated with precipitating factors, such as sleep deprivation or missing medication. Consciousness may not be impaired, though in some patients absences are often interspersed with myoclonic jerks (myoclonic-absence status epilepticus) ( Figure 6.10 ). Pure ASE with impairment of consciousness only is exceptional.19 Convulsive generalised tonic clonic status epilepticus is relatively rare. Circadian Distribution Seizures, principally myoclonic jerks, occurring within 30 min to 1 hour of awakening are characteristic of JME. Myoclonic jerks rarely occur at other times unless the patient is tired. GTCS occur mainly on awakening, but may also be purely nocturnal or random. Absence seizures occur during any time of the day while the patient is awake. Seizure-Precipitating Factors Clinical note Sleep deprivation and fatigue, particularly after excessive alcohol intake, are the most powerful precipitants of jerks and GTCS in JME. Sleep deprivation means a late night followed by a brief sleep suddenly interrupted by either compulsory early awakening in order to go to work or on a trip. An unscheduled telephone call early next morning may frequently have disastrous effects. Photosensitivity is confirmed by the EEG in one-third of patients, but probably less than one-tenth of patients with JME have clinical seizures induced by photic stimulation in daily life (Figure 10.13). Other common and prominent precipitants of seizures are mental stress, emotion and, in particular, excitement, concentration, mental and psychological arousal, failed expectations and frustration.60,211,212 Video games may precipitate seizures because of the photic/pattern effect, mental and psychological excitement, or both.60 Reading,213 writing214 and proprioceptive stimuli56 are exceptional precipitants of seizures in patients with typical features of JME. Women often have their seizures premenstrually, particularly if other precipitating factors are also present.60 Hyperventilation is an effective precipitant of EEG generalised discharges, which may often be the only abnormality in a routine EEG.60 Frequently, patients with JME have seizures when these precipitating factors cluster together, such as during exams, trips or vacations. Aetiology JME is genetically determined.61,199,215,216 Between 50% and 60% of families of probands with JME report seizures in first- or second-degree relatives.216,217 Inheritance is probably complex.9,203,218,219 Families with autosomal recessive215 or dominant215,218 Mendelian inheritance have been described. Susceptibility loci for JME have been found in chromosome 6p11–12 (EJM1)9 and 15q14 (EJM2).220,221 Genes C6orf33222 or BRD2 (RING3)223 in the EJM1 region have been recently identified. An association reported between JME and an HLA-DR allele224,225 has not been replicated.226 Genetic heterogeneity of JME is a possible explanation for such discordant observations. Families with phenotypic overlap between JME and idiopathic photosensitive occipital epilepsy have been described.227 Magnetic resonance spectroscopy studies found reductions of N-acetyl aspartate (NAA) in the prefrontal and frontal brain regions of JME patients, but not in other forms of IGE.42 JME patients with reduced frontal NAA concentrations showed poor performance on neuropsychological tests of executive functions, but not JME patients with normal frontal NAA levels. Frontal glutamate plus glutamine concentrations were elevated, which suggests possible increased neuronal excitability.42 Diagnostic Procedures All tests other than the EEG are normal. Using new MRI technologies, abnormalities of cerebral structure in some patients with JME involving mesiofrontal cortical structures have been reported.42,228 Electroencephalography25,32,60,198,229 Interictal EEG Clinical note A normal EEG in a patient suspected of having JME should prompt an EEG on sleep and awakening. The EEG in untreated patients is usually abnormal, with generalised discharges of an irregular mixture of 3–6 Hz spike/polyspikes-slow waves, with intradischarge fragmentations and unstable intradischarge frequency (Figures 2.1, 10.12 and 10.13). One-third of patients have photoparoxysmal responses. Focal abnormalities are recorded in approximately one-third of patients.32 These consist of focal single spikes, spike–wave complexes or focal slow waves. Ictal EEG The typical EEG discharge of a myoclonic jerk is a generalised burst of multiple spikes of 0.5–2 s duration (Figures 10.3, and 10.11). The ictal discharges of absences in JME are distinctly different from those in CAE and JAE.24,25 They consist of single, double, treble or multiple spikes usually preceding or superimposed on a slow wave. Multiple spikes consist of up to 8–10 spikes with a characteristic ‘worm-like’ or “compressed capital W” appearance (‘Ws’). The number and amplitude of the spikes shows considerable inter- and intradischarge variation. The intradischarge frequency of the GSWD varies from 2–10 Hz and is mainly 3–5 Hz. The frequency is often higher in the first second of onset. Fragmentations of the discharge are common and characteristic. Ws and fragmentation of discharges are observed in all patients, but vary quantitatively between patients and between discharges (Figures 2.1, 10.12, and 10.13). Brief GSWD are far more common than long ones and most of them last for 1–4 s. Photoparoxysmal discharges are evoked in 27% of patients (Figure 10.12). Differential Diagnosis JME is a typical example of a frequently misdiagnosed common epileptic syndrome resulting in avoidable morbidity.205,230 Failure to diagnose JME is a serious medical error, because JME defies all aspects of general advice regarding ‘epilepsy’. Diagnosis should improve with heightened medical awareness. Physicians should be ever alert to the possibility of JME, which is common. Important note High rate of misdiagnosis The rate of misdiagnosis of JME is as high as 90%.205,230 Factors responsible include lack of familiarity with JME, failure to elicit a history of myoclonic jerks, misinterpretation of absences as complex focal seizures, misinterpretation of jerks as focal motor seizures, and a high prevalence of focal EEG abnormalities. JME is easy to diagnose because of the characteristic clustering of myoclonic and other generalised seizures of IGEs, the circadian distribution, precipitating factors and EEG manifestations. Patients are otherwise normal and there is no mental or physical deterioration if properly diagnosed and treated early. Of the other IGEs, JAE is more difficult to differentiate, because this syndrome may also manifest with similar clinical and EEG manifestations (Table 10.7). The main differentiating factor is that absences with severe impairment of consciousness, not the myoclonic jerks, are the main seizure type in JAE. Myoclonic jerks, if they occur, are mild and random often lacking the circadian distribution of JME. Another formidable situation is when JME starts with absences in childhood prior to the development of myoclonic jerks. There are no prospective video EEG studies of these patients. Examining the EEG and clinical manifestations of these patients retrospectively, I am of the opinion that their absences are distinct from CAE or JAE in that they are usually shorter, milder and the ictal EEG often contains multiple spike–slow waves. Certainly, this situation is not CAE progressing to JME as some authors have reported:140 it is JME starting with absences prior to the development of myoclonic jerks. Clinical note Diagnostic tips for juvenile myoclonic epilepsy GTCS, usually preceded by myoclonic jerks, are nearly pathognomonic of JME if they occur in the morning after: a party to celebrate a birthday, end of school term or New Year’s eve waking up early in the morning to travel on a vacation, particularly after a late night replacement of valproate with carbamazepine in women wishing to start a family withdrawal of appropriate medication after many seizure-free years. Prognosis All seizures are probably lifelong, though improve after the fourth decade of life. JME may vary in severity from mild myoclonic jerks to frequent and severe falls and GTCS if not appropriately diagnosed and treated. Seizures are generally well controlled with appropriate medication in up to 90% of patients.60,198,199,231 Patients with all three types of seizures are more likely to be resistant to treatment.232 Accepted Practice for the Management of ‘Epilepsy' Is Often Inappropriate in JME “Treatment of epilepsy should not start after a first GTCS.” Withholding treatment in JME is an erroneous medical decision. A JME patient may have his/her first GTCS long after many myoclonic jerks on awakening, absences or both. “Treatment for epilepsy may be withdrawn after 2–3 years freedom of seizures.” Withdrawing treatment in JME is often an erroneous medical decision. More than 80% of patients will relapse. Management Lifestyle and Avoidance of Seizure Precipitants Advice regarding circadian distribution, lifestyle and seizure precipitants may be as important as drug treatment. Avoidance of precipitating factors and adherence to long-term medication is essential to avoid seizures. Some patients experience GTCS or myoclonic jerks only after encountering precipitating factors. Patients may have myoclonic jerks despite treatment only after excessively violating these factors and others may have seizures exclusively after awakening that may not affect their daily duties. Advice on the risk of sleep deprivation and alcohol is mandatory; avoidance of alcohol indulgence and compensating for sleep deprivation is essential. A young person can not probably do without late night entertainment. It is natural and understandable. However, sleeping longer the next morning may well compensate for staying up late at night. A glass of wine is acceptable providing that this is not followed by another five or more. Patients with JME are often intelligent. They understand these simple rules. There are some restrictions, but they do not deprive patients from their rights in life. Sometimes, they experiment against these rules. They find what is right or wrong for them by trial and error. Anti-Epileptic Medication The current state of knowledge of the drug treatment of JME is mainly based on clinical experience of prospective and retrospective studies with little evidence from randomised control trials (RCTs). There are no head-to-head comparisons between old and new AEDs.233 Thus, the assessment of new AED monotherapy or polytherapy in JME and, indeed all IGEs, is notoriously problematic. My evaluation of the AED treatment of JME is based on a thorough review of the literature including abstracts, exchanging views with eminent colleagues in the field and my own clinical research over the last 25 years. The following are the most pragmatic recommendations that I could reach. Briefly these are as follows: Monotherapy: Valproate is unquestionably the most effective AED in JME but is humbled by serious adverse reactions in women. Other options include levetiracetam and lamotrigine. When cost is of concern, phenobarbitone is effective in around 60% of patients. In mild JME with myoclonic jerks only, clonazepam alone may be recommended. Polytherapy: Valproate with small doses of clonazepam or small doses of lamotrigine is the most effective combination. When valproate is undesirable, combining levetiracetam with lamotrigine or lamotrigine with clonazepam (to counteract the pro-myoclonic effect of lamotrigine) may be the best option. Persisting absences are practically never a problem so ethosuximide may not have a place in the polytherapy of JME. Combination regimens with levetiracetam have the advantage that other comedications may be gradually withdrawn without seizure deterioration.234–236 Contraindicated or ineffective AEDs should be excluded and these are: carbamazepine, gabapentin, oxcarbazepine, phenytoin, tiagabine and vigabatrin. Old Anti-Epileptic Drugs in JME The indications and contraindications of old AEDs in JME have been established through numerous prospective and retrospective studies, and clinical experience over many years of use. Valproate has been recognised as the most effective drug with full control of seizures in about 80% of patients with JME.4,237 Valproate monotherapy controls absence seizures in around 75%, GTCS in 70% and myoclonic jerks in 75%.4 The valproate dosage depends on the severity of JME. In a RCT, no difference in control of various types of JME could be demonstrated between 1000 mg and 2000 mg of valproate. 238 The usual dose is 500 mg twice daily. Resistant cases require higher doses of up to 1500 mg twice daily. Mild cases may be well controlled with smaller doses of 300 mg twice daily. The best way to find the most appropriate dose in individual patients is to start with a low dose and increase in small increments until all seizures stop. Similarly, any attempt at withdrawal should be tested with small decrements of approximately 200 mg daily every 3 months. Relapses may be heralded by myoclonic jerks indicating the need to reinstate drug treatment. The major problem is that valproate is undesirable in women because of its teratogenic effects239–244 and its tendency to cause weight gain and polycystic ovary syndrome.245 Women exposed to valproate during pregnancy have a relatively small, but statistically significant increased risk of having a child with a major malformation (spina bifida, cardiac and kidney abnormalities, extra fingers and clubfoot) compared with mothers not exposed to AEDs. This risk with valproate was found to be 5% against a background risk of 1.62%243 or 5.9% versus 2.4%.244 Currently, media, internet and various other campaigns make it nearly impossible to prescribe valproate to women. There are increasing numbers of litigations against physicians and health authorities by parents of children with foetal valproate syndrome (even if the risks were appropriately explained to them) on a scale characterised as ‘bigger than thalidomide’. Phenobarbitone is extensively used as a very effective form of monotherapy in JME by European neurologists.61,201,216 It is effective in controlling GTCS and myoclonic jerks, but may exacerbate absences. It is the AED of choice when cost is of concern. One dose of 100–200 mg of phenobarbitone prior to going to sleep is often sufficient. Some patients may be controlled with 60–90 mg nocte. Teratogenicity and AEDs See references 246,247 The FDA categorization of drug risks to the foetus runs from “Category A” (safest) to “Category X” (known danger—do not use). Valproate is a category D drug that is: there is positive evidence of human foetal risk, but the benefits from use in pregnant women may be acceptable despite the risk (e.g., if the drug is needed in a life-threatening situation or for a serious disease for which safer drugs cannot be used or are ineffective). All newer AEDs are classified as Category C that is: either studies in animals have revealed adverse effects on the foetus (teratogenic or embryocidal or other) and there are no controlled studies in women, or studies in women and animals are not available. Drugs should be given only if the potential benefit justifies the potential risk to the foetus. On preliminary basis from humans, gabapentin,248 lamotrigine249 and levetiracetam250 may be relatively safe but certainly their use in women of child-bearing age should be extremely cautious and their use should be based on the risk-benefit ratio. See also Chapter 14, on pregnancy registries as a new method for assessing the foetal risks from exposures in pregnancy.251,252 Overall, the results are generally encouraging, in that of the total groups of all pregnancies, over 95% of offspring do not have a major congenital malformation. Clonazepam is predominantly used as adjunctive treatment. It is one of the most effective anti-myoclonic drugs, but clonazepam alone may not suppress and may even precipitate GTCS. 253 Furthermore, clonazepam may deprive patients of the warning of an impending GTCS provided by the myoclonic jerks.60,253–255 Clonazepam should be given in small add-on doses (0.5–2 mg at night) when myoclonic jerks persist and are troublesome despite adequate monotherapy with another broad-spectrum AED. Monotherapy with clonazepam may be considered for mild cases of JME with myoclonic jerks only. Clobazam has not been evaluated in JME, but it may be effective in some cases.233,256 Clobazam is a very useful AED in focal epilepsies as I described on pages 432. However, in IGEs, I found clobazam to be far inferior to clonazepam in controlling myoclonic jerks, and to valproate and ethosuximide in controlling absences. Acetazolamide has been used for treating GTCS in cases resistant to conventional treatment, though its use may induce nephrolithiasis.257 New Anti-Epileptic Drugs in JME The adverse effects of valproate and its lack of efficacy in 20% of patients with JME have prompted the search for alternatives. Of the new AEDs, only lamotrigine,4,193,258,259 levetiracetam,234,260–265 topiramate266,267 and possibly zonisamide268 appear to be therapeutic agents in JME. Levetiracetam, because of its efficacy in all seizure types234,260–265,269 and safer adverse reactions profile,270,271 appears to be the most promising substitute for valproate (Table 10.8). Table 10.8 Efficacy and safety of new AEDs in the JME triad of seizures and photosensitivity in comparison with valproate Levetiracetam Levetiracetam fulfils all expectations as probably the best new AED in the treatment of JME and it is the likely candidate to replace valproate in the treatment of this disorder, because of its high and sustained efficacy, fast action, good safety profile and lack of clinically meaningful interactions with other drugs. The results of treatment with levetiracetam in JME are very impressive.234–236,272,273 In three independent studies, 62%,234 67%235 and 63%236 of patients with intractable JME became seizure free with levetiracetam monotherapy or polytherapy. In a USA study, levetiracetam monotherapy was assessed in 24 patients with JME and GTCS.235 Clinical note “Sixteen (66.7%) had been free of GTCS; 3 of them had a single convulsive seizure after either stopping levetiracetam for 24 hours or reducing the dose to 500 mg per day but subsequently have remained free of convulsive seizures. Myoclonic seizures were effectively controlled in 22 of 24 patients.”235 In a UK case-note review of more than 30 patients with resistant JME, 62% became seizure free with levetiracetam.234 In a Dutch study of 16 intractable JME patients, levetiracetam had excellent effects achieving freedom from seizures in 10 (63%) and a greater than 50% seizure reduction in 2 patients (12%). No clinically meaningful change was seen in two patients (12%); one patient experienced an increase in myoclonic jerks (6%).236 A controlled study compared 12-week baseline and levetiracetam treatment periods in 55 patients with idiopathic generalised seizures (myoclonic, GTCS and absence seizures), in whom other anticonvulsants had failed. Three-quarters (76%) of patients had a greater than 50% seizure reduction with levetiracetam therapy and 40% became seizure free; 15% discontinued levetiracetam because of adverse events, mostly sedation.264 The high efficacy of levetiracetam in JME is consistent with its effectiveness against all types of epileptic seizure, epileptic and non-epileptic myoclonus and photosensitivity. Levetiracetam has a potent antimyoclonic effect,254,255,274,275 even in severe myoclonic epilepsies, such as post-anoxic myoclonus and Unverricht-Lundborg disease,276 as well as non-epileptic myoclonus.262,277–279 This may be explained by its structural similarity to piracetam. The prevalence of photosensitivity in IGE is high,280,281 with 30% of patients having EEG photoparoxysmal responses.60 Levetiracetam is the only new AED with well-established efficacy in EEG and clinical photosensitivity.263,282,283 Levetiracetam reduces or eliminates both the photoparoxysmal responses and the myoclonic jerks elicited by IPS.282 Clinical note “Long-term use of levetiracetam in some visually sensitive patients has shown a remarkably good suppressive effect for > 4 years; discontinuation of the drug before becoming pregnant resulted in a return of myoclonic jerks and an increase of IPS sensitivity.”283 Levetiracetam monotherapy was also studied in 18 patients aged 6–22 years with well-documented IGEs.261 The majority of patients (14) had the typical JME seizure triad (absences, myoclonic jerks and GTCS) and most were photosensitive. Eight patients (44.5%), five of whom were photosensitive, became clinically seizure free and the EEG normalised in seven patients. All but one of the other six patients improved with levetiracetam alone or in combination with valproate. One patient with absences and myoclonic jerks worsened with levetiracetam, but responded well to valproate.261 In a case series of highly refractory IGE (absence seizures, myoclonic jerks and GTCS), three patients in whom treatment with at least three AEDs had failed became seizure free when treated with levetiracetam monotherapy.260 Further evidence for the efficacy of levetiracetam in IGEs is that it reduces the density and duration of GSWD, which is documented with continuous EEG monitoring.265 In view of some conflicting evidence in regard to behavioural adverse reactions particularly in children treated with levetiracetam 284–287 it is advisable to apply the rule ‘start slow and go slow’. Thus, the starting dose of levetiracetam should be lower (250 mg) and titration should be slower (250 mg daily per week) than recommended by the manufacturers. Control of seizures is usually achieved at a maintenance dose of 1000–1500 mg in two divided doses daily. Higher doses may be unnecessary. Lamotrigine Lamotrigine is a very useful AED in IGEs, because of its efficacy in controlling GTCS and absence seizures. However, lamotrigine is often a pro-myoclonic AED that exaggerates myoclonic jerks in around 50% of patients. Cases with severe deterioration of JME are well reported.288,289 In view of the fact that JME is predominantly a myoclonic epileptic syndrome, it is unlikely that lamotrigine will be the successor to valproate, because of its weaker efficacy and the deterioration of myoclonic syndromes (Table 10.8). Lamotrigine monotherapy in JME is debatable and probably not recommended.4,290–292 Lamotrigine was the first of the new AEDs that appeared to be effective in JME.4,258,259 However, in most reports, the efficacy of lamotrigine in JME was evaluated not as monotherapy, but in polytherapy with valproate.4,258,259 Withdrawal of valproate often resulted in relapses. Lamotrigine added to valproate is very effective in resistant cases,4,258 because of beneficial pharmacokinetic interactions.4,293 I should emphasise that small doses of lamotrigine (25–50 mg) added to adequate dosage of valproate are sufficient. This is clinically important for two reasons. Firstly, the combination of these drugs has an additive effect on adverse reactions associated with one or both of them.294 Secondly, on anecdotal evidence the beneficial pharmacodynamic interaction with valproate may be lost with increasing dosage of lamotrigine.295 With regard to photosensitivity, lamotrigine also appears to be effective;193 suppression of EEG photoparoxysmal discharges with lamotrigine has been reported in five patients, of whom four were also taking valproate.296 Small doses of lamotrigine added to valproate are very effective in resistant cases,4,258 because of pharmacokinetic interactions.4,293 Withdrawal of valproate often results in relapses even when lamotrigine dosage is increased. In a recent report of 962 patients with various IGE syndromes (including JME), a 1-year period of remission was achieved mainly with valproate monotherapy (52.1%) with lower rates for lamotrigine (16.7%) and topiramate (34.6%).297 The combination of valproate and lamotrigine achieved a remission rate of 15.3%.297 In another report of JME patients who had also received phenytoin and carbamazepine often as initial treatment, poor outcome was more likely with lamotrigine than valproate; worsening of seizures occurred in 6% of patients with lamotrigine.291,292 Topiramate Topiramate266,267,291,298 is another broad-spectrum AED that is effective in primarily GTCS,299–302 but with a weak anti-absence303 and anti-myoclonic action. Irrespective of efficacy, the major problem with topiramate is the high incidence of adverse reactions, some of which are very serious and include significant cognitive disturbances.304,305 Also, topiramate imposes severe drug–drug interactions, including with hormonal contraception.306 In IGE, including JME, topiramate appears to be much less effective than valproate, but more effective than lamotrigine;297 a 1-year period of remission was achieved in 34.6% of patients treated with topiramate compared with 52.1% with valproate.297 In one study, a poor outcome was less likely with valproate than topiramate, which worsened seizures in 5% of patients. Furthermore, tolerability of topiramate was low and worse than valproate.291 Topiramate polytherapy may be a remote option for the few patients in whom treatment with valproate, levetiracetam and lamotrigine treatment alone or in combination fails.291 Topiramate, which is less effective and has a worse safety profile than valproate, is unlikely to be useful as monotherapy in JME. Zonisamide307–309 is also a broad-spectrum AED, but its role in JME is largely unknown and probably weak.310 AEDs Contraindicated in JME Carbamazepine is contraindicated in JME as 68% of patients suffer an increase in seizures,195,196,200,311,312 though it may improve control of GTCS in a few patients.313 Oxcarbazepine like carbamazepine is also contraindicated in JME.314 Phenytoin is often ineffective and may worsen seizures of JME. 312 In the original studies of Janz, it was significantly inferior to phenobarbitone.201 Gabapentin, tiagabine and vigabatrin are detailed on pages 337. Prevention of GTCS and Termination of Myoclonic-Absence Status Epilepticus It is important to remember that patients with JME often experience myoclonic jerks or myoclonic-absence status epilepticus long before terminating to a GTCS, which can be prevented by home administration of an appropriate benzodiazepine preparation. Rectal absorption of liquid diazepam is very rapid, reaches the brain within minutes and has a near-intravenous efficacy. Rectal tubes containing liquid diazepam are the most widely used formulation (Stesolid). Diazepam rectal gel is now available (Diastad). Buccal or nasal application of midazolam is another alternative. Duration of AED Treatment and Withdrawal of Medication in JME Lifelong AED treatment is usually considered necessary in patients with JME. Withdrawal of medication results in relapses, even in patients who have been seizure free for many years with an appropriate AED.60 In mild forms of JME, it may be safe to reduce the dose of medication slowly over months or years, especially after the fourth decade of life. Persistence or recrudescence of myoclonic jerks necessitates continuation of medication. Addendum A just recently completed control study confirmed the efficacy and tolerability of levetiracetam (3000 mg/day) when given as adjunctive treatment in adolescents (□ 12 years) and adults (□ 65 years) suffering from idiopathic generalized epilepsy with myoclonic seizures.395 This was a double-blind, multicenter, randomized, placebo-controlled trial in which 122 patients were randomized. To be eligible, patients had to experience at least 8 days with myoclonic seizures during the 8-week baseline period and had to be treated with one concomitant AED. Patients were uptitrated over 4 weeks and treated at a stable dose over 12 weeks (evaluation period). Seizure activity was recorded by the patients on a seizure diary and evaluated by the investigators using the ILAE classification. Tolerability was assessed using adverse event reporting, ECG, standard clinical examinations and laboratory analyses including plasma concentrations of AEDs and levetiracetam. Entry into a long-term follow-up trial was offered to patients who had benefited from the treatment. The primary efficacy endpoint was the responder rate (□ 50% reduction in days with myoclonic seizures during the treatment period versus baseline). One hundred and twenty one patients were included in the intention to treat analysis (placebo: n=60; levetiracetam: n=61). Responder rate was 58.3% under levetiracetam and 23.3% under placebo (p=0.0002). The corresponding odds ratio [95% CI] was 4.77 [2.12; 10.77]. Thirteen patients on levetiracetam versus 2 patients on placebo were seizure-free during the evaluation period. The most common adverse effect was headache (23.3% in placebo; 21.6% in levetiracetam). One placebo and 2 levetiracetam patients prematurely discontinued due to adverse reactions prior to the end of the evaluation period.395 Based on these results the authors concluded that “Levetiracetam proved to be highly efficacious in the treatment of refractory patients with idiopathic generalized epilepsy experiencing myoclonic seizures. Levetiracetam’s outstanding tolerability profile was also confirmed”.395 Idiopathic Generalised Epilepsy with Generalised Tonic Clonic Seizures Only Clinical note ‘Idiopathic generalised epilepsy with GTCS only’ is a newly proposed IGE syndrome of undetermined definition and boundaries.5 All patients suffer from primarily GTCS occurring at any time in wakefulness, sleep or awakening. Thus, this syndrome is to include ‘epilepsy with GTCS on awakening’ (EGTCSA) which has been extensively studied by Janz. 315–317 Demographic Data Age at onset varies from 6 to 47 years with a peak at 16–17 years; 80% of patients have their first GTCS in the second decade of life. Men (55%) predominate slightly, probably because of differences in alcohol exposure and sleep habits to women. The prevalence of ‘IGE with GTCS only’ is unknown. In my experience, it is very rare if strict criteria are applied (GTCS only). Of 1000 patients with one or more afebrile seizures, 356 patients (35.6%) had various syndromes of IGEs, but only 9 patients (0.9%) had GTCS only, though this was often the reason for referral. The low yield of ‘IGE with GTCS only’ in this sample reflects both the fact that we methodically question patients and witnesses regarding the occurrence of minor seizures and that we make long video EEG recordings, including video EEG during sleep and on awakening. As a result of this approach, for example, we found that 14 patients mainly referred for late onset GTCS had ‘IGE with phantom absences’.58,319 In the experience of other authors, the prevalence of ‘IGE with GTCS only’ varies from 10–15%314,320–322 to as high as 62%14 among IGEs. Of 253 patients with IGEs, 30 (12%) had EGTCSA and 39 (15%) had “a mild form of IGE characterised by infrequent GTCS and generalised interictal EEG discharges of spike wave”.320 Of 1033 patients with IGEs, 138 (13%) had GTCS only with onset between 3 and 18 years of age.321 Of 101 patients with IGE beginning in adolescence, 10 had GTCS only, but neither on awakening nor in the evening period of relaxation.322 The reported prevalence of epilepsy with GTCS on awakening also varies from 0%207 to as high as 17%315 in patients with epileptic seizures. ILAE Definition and Considerations on Classification The 1989 ILAE Commission1 recognised a syndrome of ‘epilepsy with GTCS on awakening’ among IGEs and defined it as follows: “Epilepsy with GTCS on awakening is a syndrome with onset occurring mostly in the second decade of life. The GTCS occur exclusively or predominantly (> 90% of the time) shortly after awakening regardless of the time of day or in a second seizure peak in the evening period of relaxation. If other seizures occur, they are mostly absence or myoclonic, as in JME. Seizures may be precipitated by sleep deprivation and other external factors. Genetic predisposition is relatively frequent. The EEG shows one of the patterns of idiopathic generalised epilepsy. There is a significant correlation with photosensitivity.”1 The new ILAE diagnostic scheme 5,86 broadens this to a syndrome of ‘IGE-GTCS only’ in which ‘epilepsy with GTCS on awakening (EGTCSA)’ is incorporated though there is some evidence that this is genetically different.318 ‘IGE with GTCS only’ has not been defined by the ILAE Task Force.5 Its name implies that it includes only those patients with GTCS alone (i.e. without absences and/or jerks) and that these may occur at any time. However, it is more likely that it is a broader category (rather than a syndrome) of ‘IGE with predominantly GTCS’ (which also includes patients with mild absences, myoclonic jerks or both). If this is so, it is undetermined what proportion of patients also has other generalised seizures (jerks or absences) and there may be significant overlap with other syndromes of IGEs. GTCS commonly feature in IGEs and occur predominantly on awakening.319 GTCS are the most severe form of epileptic seizures, while absences and myoclonic jerks may be mild and sometimes inconspicuous to the patient and imperceptible to the observer.58 They are often detected only by taking a meticulous history or video EEG. A patient with a first GTCS has often suffered from minor seizures (absences, myoclonic jerks or both), sometimes for many years prior to a GTCS. Clinical Manifestations GTCS are the defining clinical manifestations. 63,316,317,319 In EGTCSA, GTCS occur within 1–2 hours of awakening from either nocturnal or diurnal sleep. The seizure may occur while the patient is still in bed, having his breakfast or on arriving at work. Seizures may also occur during relaxation or leisure. However, GTCS in the syndrome of ‘IGE with GTCS only’ may also occur at any other time during sleep, wakefulness or awakening. Overall, GTCS are reported to occur on awakening in 17–53% of patients, diffusely while awake in 23–36%, during sleep in 27–44% or randomly in 13–26%.63 With age, GTCS tend to increase in frequency and become more unpredictable. Janz described patients with EGTCSA as unreliable, unstable and prone to neglect.315,316 Their sleep patterns are particularly unstable and modifiable by external factors (i.e. AEDs), and patients may suffer from chronic sleep deficit.63,316,317 Precipitating Factors Sleep deprivation, fatigue and excessive alcohol consumption are the main precipitants of seizures. Shift work, changes in sleep habits, particularly during holidays and celebrations, predispose to GTCS on awakening. A few patients may be clinically photosensitive. Aetiology There is a high incidence of epileptic disorders in other family members.63,316 Recently, a link to the EJM-1 locus has been reported in EGTCSA while no such link was found in adolescent-onset idiopathic epilepsy with GTCS at any time while awake.318 Microdysgenesis has been reported in pathological specimens of EGTCSA.323 Diagnostic Procedures By definition, all tests apart from the EEG are normal. Electroencephalography The EEG shows generalised discharges of 3–4 Hz spike/multiple spike–wave complexes (GSWD) in 50% of patients with pure EGTCSA (Figure 10.15) and 70% of those with additional absences or myoclonic jerks. Figure 10.15 Asymptomatic GSWD on video EEG of a 19-year-old university student who had two GTCS at 14 and 18 years of age. They both occurred half an hour after awakening from a brief sleep during exam periods. There was no clinical (more...) A normal routine EEG should prompt a video EEG during sleep and on awakening. Myoclonic jerks or, more frequently, brief absences will often be revealed. Focal EEG abnormalities occur in one-third of patients but these are exceptional in the absence of GSWD. Photoparoxysmal responses are reported in 17% of females and 9% of males with EGTCSA.63 Differential Diagnosis The differential diagnosis is mainly from other IGEs that share with EGTCSA the same propensity to seizures after awakening and the same precipitating factors. JME, JAE and eyelid myoclonia with absences are examples of IGE syndromes that may cause diagnostic difficulties (see Chapter 14). Symptomatic and focal epileptic seizures with secondary generalisation may also occur, predominantly on awakening. Prognosis As in all other types of IGEs with onset in the mid-teens, EGTCSA is probably a lifelong disease with a high incidence of relapse (83%) on withdrawal of treatment.316 Characteristically, the intervals between seizures become shorter with time, the precipitating factors less obvious and GTCS may become more random (diurnal and nocturnal), either as a result of the evolution of the disease or drug-induced modifications.63,316,317 Management Patients should be warned of the common seizure precipitants, namely sleep deprivation with early awakening and alcohol consumption, and when possible should avoid occupational night shifts. Patients, after adjusting their lifestyles, may become seizure free. Treatment is with AEDs that control primarily GTCS, such as valproate, phenobarbitone, lamotrigine, levetiracetam and topiramate. In the pure forms of ‘GTCS only’, carbamazepine, oxcarbazepine and phenytoin may be used, but require exclusion of other types of seizures that may be exacerbated by these AEDS. Primarily GTCS in AED RCT Results of RCTs of AEDs in primarily GTCS present significant difficulties in interpretation: secondarily GTCS may contaminate the selected populations for the active AED and the comparator agent (established AED or placebo) patients with primarily GTCS and absences or myoclonic jerks are often included though these seizures respond differently to AEDs inappropriate AEDs for IGEs, such as carbamazepine, have been used in one-quarter of IGE patients as a comparator agent. Important note The results of such studies are of uncertain value and should be interpreted with caution. IGE with Absences of Early Childhood Clinical note Typical absence seizures starting in early childhood (between a few months and 4 years of age)187,324–327 are not a specific expression of a distinct syndrome. This may be the first manifestation of various syndromes of IGEs with absences or more severe forms of generalised epilepsies. By excluding all these conditions, it is realistic to accept that there is a syndrome of IGE that starts in early childhood primarily manifesting with absences, often combined with GTCS and possibly with myoclonic jerks. Doose,325,326 having studied 140 cases with onset of absences in early childhood, rightly concluded that: “this is an heterogeneous subgroup within IGE. There is a distinct overlap with early childhood epilepsy with GTCS and myoclonic-astatic epilepsy on the one side and with childhood absence epilepsy on the other. Thus it should not be regarded as a special syndrome”.325,326 I am in complete agreement with this statement. Age at onset of absence seizures alone can not define an epileptic syndrome. However, with improved diagnostic skills, applying inclusion (e.g. including absences and GTCS) and exclusion criteria (e.g. excluding CAE and symptomatic cases), it appears that there is a rare IGE, which needs a precise definition. This is an IGE syndrome (occurring in otherwise normal children). 325,326 Onset of absences occurs between 1–5 years of age. Absences are markedly different from CAE. Clinically, they are less severe and less frequent. Ictal EEG 2.5–4 Hz GSWD is very irregular and termination is not abrupt, but often fades with spike–wave complexes. GTCS are common (affecting two-thirds of patients) and are often the first seizure type. Boys are more likely to suffer GTCS than girls. Myoclonic jerks and myoclonic-astatic seizures occur in 40% of patients. Absence status epilepticus may lead to cognitive impairment. Background EEG shows a moderate excess of slow waves. Long-term prognosis is worse than in CAE. There is a strong family history of IGE and GSWD in the EEG of unaffected members, particularly mothers. Perioral Myoclonia with Absences Clinical note Typical absences with ictal motor symptoms of perioral myoclonia is a type of seizure.16,26,328–331 However, this is often combined with a clustering of other clinical and EEG features probably constituting an interesting IGE syndrome of perioral myoclonia with absences (PMA).16,18,328,331 Demographic Data Age at onset covers a wide range from 2–13 years (median 10 years). Girls are far more frequently affected than boys. The syndrome is uncommon in children (< 1% with typical absences) but, because it fails to remit, is relatively common in adults (9.3%) with typical absence seizures.16,328,331 Clinical Manifestations Typical absence seizures with perioral myoclonia are the defining symptom. The characteristic feature is perioral myoclonia, which consists of rhythmic contractions of the orbicularis oris muscle that cause protrusion of the lips, contractions of the depressor anguli oris resulting in twitching of the corners of the mouth or, rarely, more widespread involvement, including the muscles of mastication producing jaw jerking (Figure 10.16). Impairment of consciousness varies from severe to mild. Most patients are usually aware of the perioral myoclonia. Duration is usually brief, lasting a mean of 4 s (range 2–9 s). Absences with perioral myoclonia may be very frequent and occur many times a day or 1–2 times a week, or they are rare. Generalised tonic clonic seizures occur in all patients. GTCS often start before or soon after the onset of clinically apparent absences. Exceptionally, GTCS may start many years after the onset of absences. GTCS are usually infrequent (ranging from once in a lifetime to 12/year) and are often heralded by clusters of absences or ASE. Absence status epilepticus is very common in PMA (57%) and frequently ends with GTCS (Figure 10.16). It is more common than in any other syndrome of IGE with typical absences.19 Perioral myoclonia may be more apparent than impairment of consciousness or vice versa. Aetiology Half of patients with PMA have first degree relatives and mainly siblings with IGE and absences.141 Diagnostic Procedures All tests apart from the EEG are normal. Electroencephalography Interictal EEG frequently shows: (a). abortive bursts or brief less than 1 s generalised discharges of 4–7 Hz spikes/multiple spikes-waves, which are usually asymmetrical and may give the impression of a localised focus; and (b). focal abnormalities, including single spikes, spike–wave complexes and theta waves with variable side emphasis. Ictal EEG consists of 3–4 Hz GSWD with frequent irregularities in terms of the number of spikes in the spike–wave complex, the fluctuations in spike amplitude and the occurrence of fragmentations (Figure 10.16). There is no photosensitivity. Differential Diagnosis Patients with PMA are frequently erroneously diagnosed as having focal motor seizures because: (a). the prominent motor features of the absences, which are often reported or sometimes recorded as unilateral; and (b). the presence of interictal focal EEG abnormalities. However, this error is unlikely to happen if EEG is properly recorded and interpreted. Also, patients with focal motor seizures are unlikely to suffer ASE, which is common in PMA. Considerations on Classification Absences with perioral myoclonia despite unequivocal documentation with video EEG16,328,331 have not been recognised by the ILAE as a seizure type.1,5 Perioral myoclonia may occur in typical and atypical absence seizures of idiopathic and symptomatic epilepsies. Therefore, absences with perioral myoclonia alone can not be taken as evidence of any syndrome. However, they often occur together with other symptoms that cluster in a non-fortuitous manner thus constituting the main manifestation of a syndrome within the broad spectrum of IGE, which we proposed to call perioral myoclonia with absences. 16,328 Other manifestations of this syndrome include GTCS, which often start early prior or together with the absences, frequent occurrence of ASE, resistance to treatment and persistence in adult life.16,328,331 The main differential diagnosis is from other syndromes of IGEs such as CAE, JAE and MAE depends on the age at onset. Video EEG invariably reveals perioral myoclonia that sometimes, and particularly in treated patients, may be subtle. Onset of GTCS before or at the same age as typical absences, the relatively brief duration of the absences with the concurrent perioral myoclonia and the frequent occurrence of ASE are useful clinical indicators in favour of PMA and against childhood, juvenile or other forms of IGE. PMA may be difficult to differentiate from MAE, particularly if the latter presents with mild myoclonic jerks localised in the face.26 However, MAE is often symptomatic and is rarely associated with generalised non-convulsive status epilepticus.177 Prognosis Absences and GTCS may be resistant to medication, unremitting and possibly lifelong.16,328,331 Management Treatment is with valproate alone or combined with ethosuximide, small doses of lamotrigine or clonazepam. Levetiracetam may be effective, because of the myoclonic elements of the absences. ASE with perioral myoclonia, of which most patients are aware, should be terminated with immediate self-administered medication of oral midazolam or rectal diazepam. Idiopathic Generalised Epilepsy with Phantom Absences Clinical note The syndrome of IGE with phantom absences58 is characterised by the triad of: phantom absences that are inconspicuous or never appreciated prior to the onset of GTCS GTCS, which are commonly the first overt clinical manifestations, usually start in adulthood and are infrequent ASE, which occurs in 50% of patients. Demographic Data The first overt clinical manifestations of GTCS appear in adult life, though absences may have started much earlier. ASE as the first overt symptom in childhood is rare.77 Men and women are equally affected. The prevalence was estimated to be 15% among IGE with typical absences, 10% of IGE and 3% of 410 consecutive patients over 16 years of age with epileptic seizures.58 Genton et al.338 reported that, among 253 consecutive cases of IGE, 32 (15.4%) patients had rare GTCS with GSWD in the interictal EEG. It is possible that these patients suffered from IGE with GTCS only and/or IGE with phantom absences. Definition of Phantom Absences ‘Phantom absences’ denote typical absence seizures, which are so mild that they are inconspicuous to the patient and imperceptible to the observer.16,58 Why the name phantom absences? Is this synonymous with ‘subclinical or larval absences’? We coined the term ‘phantom absences’ because of their clinically elusive and inconspicuous character.58 Cognitive impairment during ‘subclinical, larval’ GSWD are well documented332,333 and phantom absences may be a good example of them. However, it should be emphasised that patients with phantom absences have, by definition,1,15 active, clinical absence seizures that manifest with mild impairment of cognition, as demonstrated by errors and discontinuation during breath counting on video EEG (Figure 10.17).334 It should also be emphasised that phantom absences in these adults do not represent aborted past childhood or juvenile absences modified by age or medical treatment.58 Furthemore, phantom absences is not synonymous with the EEG pattern of abortive “phantom 4–6 Hz spike and slow wave”, which is a non-specific EEG abnormality of no clinical significance.335–337 Figure 10.17 Video EEG of two patients suffering from IGE with phantom absences. The numbers denote the actual breath counting during hyperventilation. Note that errors, when they occur, are only related to these brief GSWD. Errors consist of hesitation (more...) Considerations on Classification Phantom absences, or mild absence seizures, have not been categorised as such by the ILAE.1,5 The absences are simple, brief (usually 2–4 s) causing only inconspicuous impairment of cognition, which is not clinically disturbing to the patient. Though not classical, they fulfil the criteria of typical absences with more than 2.5 Hz GSWD.58 There is reasonable evidence to suggest that phantom absences are not only discrete seizures, but may also constitute the main symptom of a syndrome within the broad spectrum of IGE. There is non-fortuitous clustering of other symptoms, such as GTCS of usually late onset, frequent occurrence of ASE and persistence in adult life.58 That these patients have IGE is beyond any doubt, as they all are of normal intelligence and physical state, high resolution MRI is normal, the EEG shows GSWD, and the seizures are generalised. The syndrome of IGE with phantom absences has not been recognized by the ILAE.1,5 Accordingly, these cases are probably orphaned or probably categorised among undefined IGE or other syndromes of IGE. Clinical Manifestations Phantom absences are the defining and consistent symptom in these patients.58 Phantom absences manifest with mild, but definite, impairment of cognition documented by video EEG (Figure 10.17). There are no other clinical symptoms except eyelid flickering that consistently occurs in some patients. However, patients are not usually aware, even retrospectively, that absences interfere with their daily life, even when driving or in demanding professions, such as computer programming, civil engineering, major business and administration.58 Patients may retrospectively admit to momentary lack of concentration and forgetfulness, which in their opinion was of no practical significance.58 Rarely patients or witnesses also become aware of some minor motor manifestations during the absence: Patient note occasional, very brief episodes of quick flickering of the eyeballs upwards accompanied by a brief lack of concentration Phantom absences are common in patients with IGEs, but are often unrecognised. These absences are impossible to detect without breath counting and video EEG. Generalised tonic clonic seizures occur in all adult patients that seek medical attention. They are usually the first overt clinical manifestation.58 They are of late onset, infrequent and without consistent circadian distribution or specific precipitating factors. Absence status epilepticus occurs in 50% of patients. This often lasts for many hours alone or prior to GTCS (Figure 10.18).58,76 It manifests with cognitive impairment, which is usually of mild or moderate severity. Patient note At the age of 28, this patient was able to drive for 2–3 hours while in ASE. This was on Christmas Eve with no precipitating factors. For the whole day he was feeling unwell, “tired, unable to compute conversations, clouded mind, missing words, as if sleep waking, slow mind”. Despite this, he had his haircut and then drove alone on a 1-hour trip through busy roads to relatives, offering the wrong Christmas gifts to the wrong persons. He then drove back home for another hour, picked up his wife and drove again towards a friend’s house. His wife noticed that he appeared “slow and very quiet during the drive, but I though he was just tired. I was not alerted because he was driving very well, not making any errors”. She did not notice any abnormal movements, including gestural automatisms or jerking. He became completely silent for 1–2 minutes prior to a GTCS, which occurred while parking; his wife managed to stop the car and avoid an accident. No further seizures occurred in the next 5 years while on valproate 1000 mg daily. Impaired cognition during ASE varies. Most patients often communicate poorly and slowly, feel strange and confused, make errors at work and look depressed, but do not become unresponsive (Figures 10.17, and 10.18). Experiential, mental and sensational symptoms are more common than is usually appreciated.30 Frequently, they have a good recollection of the ictal events and may be able to write down their experiences while in ASE.30,58 Neuropsychological examination under video EEG monitoring during ASE in 2 patients revealed only mild attentional and executive disturbances.339GSWD were associated with selective impairment in the initiation of response and self-generated action, whereas short-term storage of external information was fully preserved.339 Patients with recurrent ASE are often aware of the impeding GTCS and try to find a safe place to have it. Aetiology IGE with phantom absences is probably genetically determined.58 It is difficult to explain the high frequency of ASE in these patients, with such brief and mild absences and infrequent GTCS. It is possible that, under the influence of precipitating factors that are not fully understood, phantom absences might cluster and evolve into ASE facilitated either by lack of or inappropriate treatment. Vuilleumier et al.339 proposed that a predominant involvement of frontomesial thalamocortical circuitry may underlie an ‘inconspicuous’ disorder of consciousness, as seen in phantom absences with selective loss of initiation and goal-oriented behaviour, whereas involvement of more lateral frontal areas in typical absences may also disrupt working memory processes.339 Diagnostic Procedures All tests apart from the EEG are normal. Interictal EEG: The background activity is normal, 50% of patients have EEG focal paroxysmal abnormalities consisting of short transients of localised slow, sharp waves or spikes, or both, occurring either independently or in association with brief GSWD. 58,340 EEG photosensitivity is exceptional. Ictal EEG consists of 3–4 Hz GSWD with occasional fragmentations (Figure 10.17). They are typically brief lasting of no more than 5 s. Mild cognitive impairment manifested with hesitation, discontinuation and errors in breath counting is the only clinical ictal symptom during GSWD. A few may also have mild ictal eyelid fluttering.58,340 Hyperventilation is a major provocative factor. During ASE, the EEG shows continuous, generalised, mainly 3 Hz spike/multiple spike slow wave activity (Figure 10.18). Differential Diagnosis The diagnostic and management errors involved in adult patients with IGE and typical absence seizures have been well reported.56,58 The magnitude of the problem is worse in IGE with phantom absences, in which the absences are very mild, ASE is confused with non-epileptic events or temporal lobe epilepsy, and GTCS are of late onset. This is compounded by frequent EEG focal abnormalities and the current practice of most EEG departments to not test cognition appropriately during brief GSWD. The main problems to consider in IGE with phantom absences are: the first overt unprovoked GTCS appears in adult life absence status epilepticus differentiation from other syndromes of IGE. It is essential to take a careful clinical history and to interpret symptoms correctly, which may be suggestive of typical absences and ASE. A history of altered consciousness preceding GTCS should not be taken as evidence of complex focal seizures, depression or an unspecified seizure prodrome. Other forms of the so-called ‘adult onset IGE’ may be otherwise typical examples of JME, JAE or other IGE syndromes that start or become clinically identifiable after the age of 20 years.202,341–344 Some of the patients described may suffer from IGE with phantom absences. Prognosis IGE with phantom absences may be a lifelong propensity to seizures, which is of undetermined onset and remission. Patients are of normal intelligence, which does not show any signs of deterioration. Further, phantom absences, though frequent, do not appear to affect daily activity. Management There are many unanswered questions as to whether patients with phantom absences need treatment. All patients with IGE with phantom absences had a normal life without medication until their first GTCS, probably many years after the onset of frequent daily mild absence seizures. We do not know how many people there are in the general population with the same problem but without GTCS or conspicuous ASE. If treatment is considered necessary (driving a car is a significant factor), valproate, levetiracetam and lamotrigine are AEDs to consider. A Reminder of “Adult Onset Idiopathic Generalised Epilepsies” Recently, there has been increasing interest in the so-called adult-onset IGE.202,341–344 Seizures consist of GTCS, myoclonic jerks and absences, alone or in combination, which first become clinically detectable after the age of 20 years. I emphasise clinically detectable, because these patients may have had minor seizures (as is the case with phantom absences) long before seeking medical advice, which is usually prompted by a GTCS. Thus, adult onset IGE appears to consist of classical syndromes of IGE, but with delayed onset or delayed identification. Other than age at the time of the first overt seizure, all other clinical and EEG manifestations are typical of JME, JAE or IGE with GTCS only. For example, ‘adult myoclonic epilepsy’ is identical to JME.202 When grouped together on the basis of age at onset, clinical manifestations, EEG and family history of epilepsies, there is a similarity between ‘adult onset IGE’ and ‘IGE of classical late childhood or adolescence onset’.342–344 However, an exemption exists with ‘benign familial adult myoclonic epilepsy’,345 which is a distinct autosomal epileptic syndrome with seizures first appearing in adult life (page 332). Syndromes of autosomal recessive inheritance such as ‘familial infantile myoclonic epilepsy’356,357 have also been described and may be considered in ILAE revisions. Familial (Autosomal Dominant) Generalised Epileptic Syndromes Clinical note There is an increasing number of reports of familial idiopathic generalised epilepsies which have not been recognised by the ILAE yet. These include: Benign adult familial myoclonic epilepsy345–353 Autosomal dominant cortical myoclonus and epilepsy354 Benign Familial Adult Myoclonic Epilepsy Benign familial adult myoclonic epilepsy (FAME)345 is probably the most common of the autosomal epileptic syndromes not yet recognised by the ILAE. FAME has been initially identified in Japanese346,347,352,355 and more recently in European families.351,353 FAME is a relatively benign non-progressive autosomal dominant idiopathic generalised epileptic syndrome with high penetrance. Clinically,FAME is characterised by adult-onset cortical myoclonus and tremor of the fingers; 80% of patients also have infrequent GTCS in periods of worsening myoclonus. Movement and emotional stress intensify the myoclonus. Age at onset varies from 30 to 60 years (mean 40 years). Families with members having concurrent migraine or blindness have been reported. The EEG shows generalised polyspikes and waves, photosensitivity and giant somatosensory evoked potentials. Consistent with cortical myoclonus, long-loop C reflexes are enhanced and there is a preceding wave on jerk-locked back EEG averaging. The genes for FAME have been mapped to 8q24 in Japanese families346,347,352,355 and 2p11.1-q12.2 in European351,353 families. There may be allelism with ‘autosomal dominant cortical myoclonus and epilepsy syndrome’. Autosomal Dominant Cortical Myoclonus and Epilepsy Autosomal dominant cortical myoclonus and epilepsy (ADCME) with complex focal and generalised seizures is based on a study by Guerrini et al354 of a pedigree, in which eight individuals presented with a non-progressive disorder with onset between the ages of 12 and 50 years.354 It was characterised by predominantly distal, semicontinuous rhythmic myoclonus (all patients), GTCS (all patients) and complex focal seizures (three patients). Most patients had suffered infrequent seizures and had a normal cognitive level, but three patients with intractable seizures had mild mental retardation. The pattern of inheritance was autosomal dominant with high penetrance. All patients had frontotemporal as well as generalised interictal EEG abnormalities. Back-averaging analysis and other neurophysiological studies of the myoclonus suggested a cortical origin. The C-reflex at rest was enhanced and somatosensory, and visual evoked potentials were of high amplitude. The resting motor threshold intensity in response to transcranial magnetic stimulation was significantly reduced and the post-motor evoked potential silent period was significantly shortened compared with the controls. These clinical and neurophysiological characteristics suggest diffuse cortical hyperexcitability and a high propensity for intra-hemispheric and inter-hemispheric cortical spread, as well as rhythmic myoclonic activity. The disease has been linked to chromosome 2p11.1-q12.2.354 Treatment of Idiopathic Generalised Epilepsies Consider the following facts: Patient note “A majority (48%) of patients with idiopathic generalised epilepsy initially receive ill-advised AED, which cause IGE to appear intractable.” Benbadis et al.14 Patient note “Carbamazepine is still being inappropriately prescribed to children (47%) with typical absence syndromes to their detriment.” Parker et al.358 Patient note “In a class 1 control AED study, one quarter (27%) of patients with IGE, including JME, were treated with carbamazepine (a contra-indicated drug in IGE) which was the comparator drug on the basis of the physicians’ ‘intention to treat’.” Privitera et al.301 These are striking, recent examples of commonly occurring and disturbing inappropriate medications in patients with IGEs. The most important of the multiple reasons for this continuing error are: diagnostic misclassification of IGEs as focal epilepsies233 sparse or methodologically ambiguous AED RCTs 233 official guidelines and publications paying scant attention to important management aspects of IGEs 359 a ‘one size fits all’ policy with regard to ‘how to treat epilepsy’. RCTs studying ‘epilepsy’ and its treatment in its ‘universe’301,302 are of no benefit and should be discouraged.359 As Faught said, even for IGEs “we can not lump them all together in clinical trials”.233 Formal national formularies are conspicuously void of warning against established pro-epileptic action of certain AEDs in certain types of seizures and syndromes of IGEs. I take some examples from the British National Formulary (BNF), which are more or less the same everywhere: Patient note “Indications for carbamazepine: partial and secondary generalised tonic-clonic seizures and some primary generalised seizures.” Thus, physicians may reasonably prescribe carbamazepine for absences and myoclonic jerks (which are primary generalised seizures as also defined in the BNF) with disastrous effects. Patient note “Indications for tiagabine or vigabatrin: adjunctive treatment for partial seizures with or without secondary generalisation.” There is no warning that these AEDs are deleterious for IGEs, which are often misdiagnosed as partial epilepsies. IGEs demand different treatment strategies to focal epilepsies.4,14,310,360–363 Ignoring this fact results in avoidable morbidity and sometimes mortality. There is an urgent need for clear and unequivocal guidelines in the use of old and new AEDs in IGEs and its various types of seizures. Practising physicians have a colossal task in not only properly diagnosing IGE, but also in deciding which of the many old and new AEDs is the most suitable and which is contraindicated for the seizures and preferably the syndromes of IGEs. The methodology followed in this book regarding treatment recommendations for AEDs is detailed in Chapter 4. The treatment of each individual IGE syndrome has been detailed in the appropriate chapter. Recent reviews and publications on the AED treatment of IGEs are recommended reading.14,233,310,314,360,362,363 Important established documentation to remember is that: Certain AEDs that are beneficial in focal epilepsies are ineffective or even contraindicated in IGE.4,276,292,364,365 Tiagabine and vigabatrin are major pro-absence agents. Carbamazepine, oxcarbazepine and phenytoin exacerbate absences and myoclonic jerks. Gabapentin is ineffective in all types of idiopathic epileptic seizures and may exacerbate some of them. A drug efficacious in one type of generalised seizure may be ineffective or exaggerate another type of generalised seizure. Clonazepam is the best choice of drug for myoclonic jerks, but is ineffective in GTCS. Ethosuximide is highly efficacious only for absence seizures and negative epileptic myoclonus, but is ineffective or may exaggerate GTCS. Carbamazepine and oxcarbazepine are effective in primarily GTCS, but often aggravate myoclonic jerks and absences. Lamotrigine is effective in primarily GTCS and absences, but may exacerbate myoclonic jerks. If a drug is found to be efficacious in ‘generalised’ childhood epileptic encephalopathies, it does not mean that this is also the case in IGEs. Vigabatrin is the drug of first choice in West syndrome, but it is contraindicated in IGE. A drug found to be efficacious in secondarily GTCS, may be ineffective in primarily GTCS or deleterious in IGEs. Gabapentin, an AED licensed for the treatment of focal and secondarily GTCS is ineffective in primarily GTCS and may aggravate other types of IGE seizures. Tiagabine, an AED licensed for the treatment of focal and secondarily GTCS, is a potent pro-absence agent that induces absence seizures and provokes ASE often ending with GTCS in IGE. IGEs are often easily treatable which means that asmall doseof an appropriate AED is as good as alarge dose. 233 No difference in control of various types of JME could be demonstrated between 1000 mg and 2000 mg of valproate. 238 Diagnosis and Treatment of Newly Identified IGEs Diagnosis should first establish that the patient suffers from genuine epileptic seizures and then define: this is IGE and not focal epilepsy the types of seizures that the patient suffers if possible, the IGE syndrome. In choosing the first AED to be recommended from Table 10.8 efficacy and adverse reactions have to be carefully balanced, because treatment is often lifelong. Old Anti-Epileptic Drugs in IGEs Briefly, prior to the introduction of new AEDs, the position was as follows. Valproate has superior efficacy in all seizures and syndromes of IGEs but its use in certain populations and particularly in women of reproductive age is highly problematic and sometimes impossible. Clonazepam, even in small doses of 0.5 –1 mg, is probably the most potent antimyoclonic drug with some anti-absence effect;4 it may deteriorate GTCS or deprive patients from the warning symptoms of an impeding GTCS in JME (see page 317).253–255 Ethosuximide is a potent drug against absences. It may improve myoclonic seizures (particularly negative epileptic myoclonus), but is ineffective and may worsen GTCS.4 Phenytoin is effective in primarily GTCS, but deteriorates absences and possibly myoclonic jerks; it is often ineffective or worsens JME.312 Carbamazepine is effective in primarily GTCS, but aggravates absences and myoclonic jerks. Phenobarbitone was historically the preferred drug in JME and is still used in Europe and developing countries; it worsens absences.61,201 Evolving Treatment of IGE in the Era of New Anti-Epileptic Drugs The adverse effects of valproate and its lack of efficacy in 20% of patients has prompted the search for alternatives. Half of the new AEDs appear to be either ineffective or contraindicated (vigabatrin, tiagabine, oxcarbazepine, gabapentin)366 in IGE (Table 4.1). Of the other four new AEDs, lamotrigine,4,258,259 levetiracetam,234,260–265 topiramate266,267,298 and zonisamide268 appear to be therapeutic agents. Levetiracetam, because of its efficacy in all types of seizure234,260–265 and safer adverse reactions profile,270,271 appears to be the most promising substitute for valproate. Lamotrigine is effective in GTCS and absence seizures, but often aggravates myoclonic jerks. It has important beneficial synergistic interactions with valproate. Topiramate is mainly effective in GTCS, but its adverse effect profile and drug–drug interactions are possibly worse than those of valproate. Zonisamide may have a weak therapeutic effect. Prescribing Errors in IGEs The error of prescribing tiagabine or vigabatrin in patients with absence seizures may be of the same magnitude as prescribing a gluten-rich diet in the treatment of coeliac disease.4Prescribing carbamazepine in patients with JME is a similar situation. New Anti-Epileptic Drugs Useful in IGEs These have been detailed on pages 315–18 in the treatment of JME. Briefly, in order of preference: Levetiracetam,4,367 so far fulfils all expectations as probably the best new AED in the treatment of IGEs and is the likely candidate to replace valproate in at least JME, with its high and sustained efficacy, fast action, excellent safety profile, lack of clinically meaningful interactions with other drugs and no need for laboratory tests. Its role in syndromes with predominant absence seizures (CAE and JAE) is yet unclear though it has a unique mode of action in animal models of absence seizures.368 In a recent report 3 of 4 patients with CAE and 5 of 5 patients with JAE became seizure free with levetiracetam alone or in combination with other AEDs. 183 Lamotrigine has proven efficacy in controlling GTCS and absence seizures in at least 50% of patients, but may exaggerate myoclonic jerks. It is inferior to valproate in terms of efficacy, but superior with respect to adverse reactions despite a high incidence of idiosyncratic reactions that can occasionally be fatal. Efficacy and adverse reactions have to be carefully balanced in these cases, because treatment is often lifelong. Lamotrigine is recommended in the treatment of CAE and JAE either alone170,171 or in combination with other anti-absence AEDs.4,170,171,194,246,247,271 However, lamotrigine monotherapy in JME is probably not recommended but small doses of lamotrigine added to valproate are probably the most effective combination treatment for resistant cases. Topiramate266,298,369 is another broad-spectrum AED that is effective in primary GTCS299–302 with a weak anti-absence303 and anti-myoclonic action.291,297 In IGEs, including JME, it appears to be much less effective than valproate, but more effective than lamotrigine.297 In a well-cited class 1 study, topiramate has been found to be effective in primarily GTCS.299,301,302 Methodological drawbacks of these studies included: (a). patients with symptomatic generalised seizures were enrolled; (b). one fourth (27%) of patients with IGEs were assigned to carbamazepine as a comparator AED; and (c). carbamazepine was statistically decreased in the topiramate group thus favouring the results in this group for at least the ‘subtypes of generalised seizures’, such as absences and myoclonic jerks that deteriorate with carbamazepine. Topiramate is unlikely to achieve monotherapy status in the long-term treatment of IGEs mainly because of its many short and long term adverse effects and drug-drug interactions. Zonisamide307–310 is also a broad-spectrum AED, but its role in IGEs is largely unknown and probably weak. In the long experience of eminent Japanese colleagues that I consulted, zonisamide is effective in the treatment of focal seizures, secondarily GTCS, epileptic encephalopathies (West syndrome responds much better than Lennox-Gastaut syndrome) and progressive myoclonic epilepsies. Zonisamide appears to be much less effective in IGE, primarily GTCS, absences and jerks, though a few patients may have an excellent response. In children, cognitive adverse reactions may be troublesome. Currently in Japan, zonisamide is a second-line AED for focal epilepsies after carbamazepine and for symptomatic generalised epilepsy after valproate, clonazepam and clobazam. Polytherapy Clinical note For patients in whom monotherapy with valproate fails, the combination of valproate with small doses of lamotrigine (25–50 mg) appears to be the most effective.4 In those with persistent myoclonic jerks, clonazepam is the best add-on drug to valproate.4 Ethosuximide should be added only for uncontrolled absence seizures. However, in all these scenarios, valproate is the principal AED. The situation is likely to change dramatically with use of levetiracetam instead of valproate but this needs RCTs. The consensus is that IGEs have a better prognosis with, and a more favourable response to, appropriate AEDs than symptomatic and focal epilepsies. “Most patients with IGE are easily controlled with appropriate medication, refractory patients are rare.”246,247 However, there is no reliable figure regarding the prevalence of intractable IGE, which may be in the order of 10–30%.370 A factor contributing to this uncertainty and probably high incidence of intractable IGEs is that IGEs are often inappropriately treated with AEDs that are either ineffective or contraindicated. Table 4.1 lists AEDs that are indicated and contraindicated in IGE seizures. Management of patients with intractable IGEs, providing that they truly suffer from epileptic seizures, should take the following steps: Based on clinical and EEG evidence, establish the type or types of seizures (absences, myoclonic jerks and GTCS alone or in combination), and make sure that these are primarily and not secondarily generalised. Previous EEGs particularly, in untreated stages, are invaluable. Establish precipitating factors and circadian distribution as well as their effect regarding intractability. List in chronological order, all AEDs used, and in what doses and combinations. Establish which drugs were beneficial and which made the situation worse. Consider thoroughly the current situation regarding: (a). seizures – which are the more predominant and more disturbing; and (b). AEDs – which are definitely or possibly effective, ineffective or contraindicated with respect to seizures and adverse reactions. Consider thoroughly all the above, including compliance, in making a definite plan of which AEDs with adverse effects (seizure efficacy and patient tolerability) should be withdrawn and which of the indicated AEDs should be increased in dosage or added to the scheme. Of the new AEDs, those which are likely to be effective as monotherapy (Table 4.1 and Table 10.8) are also the most likely to be suitable in polytherapy. The order of priority as determined by efficacy, safety, drug–drug interactions and other parameters are levetiracetam, lamotrigine, topiramate and zonisamide. All other new AEDs –tiagabine, oxcarbazepine and gabapentin –are unsuitable. New AEDs Contraindicated in IGEs Gabapentin: all relevant studies have shown that gabapentin is, at least, ineffective in IGEs including primary GTCS, and it may exacerbate absences and myoclonic jerks.194,246,247,271,371–373 Oxcarbazepine appears to have a similar seizure profile to carbamazepine. In IGEs, like carbamazepine, oxcarbazepine mainly aggravates absences and myoclonic jerks; GTCS may also worsen.365 In a case series, all six IGE patients showed significant deterioration in terms of absences and myoclonic jerks related to oxcarbazepine treatment; GTCS also worsened in three patients.365 Vigabatrin and tiagabine are contraindicated. These are pro-absence AEDs, because of their GABAergic action, which explains the high incidence of drug-induced ASE and the appearance of new types of seizures.4 Drug Withdrawal In CAE, treatment may be slowly withdrawn 1–3 years after controlling all absences.4 All other IGE syndromes are probably lifelong and confront the usual textbook advice of withdrawal of medication after 2–3 years from the last seizure. Relapses are probably unavoidable. However, if seizures are mild and infrequent, drug withdrawal may be attempted. This should be done in small decrements, probably over years, warning the patient that re-emergence of even minor seizures, such as absences or myoclonic jerks, mandates continuation of treatment. EEG confirmation of the seizure-free state is needed during the withdrawal period.4 First-Line Drugs Valproate (the most effective of all but often unsuitable for women) Levetiracetam (but needs RCTs) Lamotrigine (may exaggerate myoclonic jerks) Ethosuximide (only for absences and negative myoclonus; may exaggerate GTCS) Clonazepam (only for pure myoclonic syndromes; may exaggerate GTCS) Second-Line or Adjunctive Drugs Topiramate (but with serious adverse reactions and drug–drug interactions) Zonisamide (needs further documentation regarding efficacy and adverse reactions) Phenobarbitone (may be first line AED for those without absence seizures if cost is a major issue) Acetazolamide (only for absences) Contraindicated Drugs Carbamazepine, oxcarbazepine and phenytoin (though they may control primarily GTCS if added to first-line drugs) Gabapentin (ineffective in primarily GTCS and may exacerbate absences and myoclonic jerks) Tiagabine and vigabatrin (pro-absence drugs with a high incidence of induced ASE) Treatment of Status Epilepticus in Idiopathic Generalised Epilepsy Idiopathic generalised ASE with all its variations (generalised non-convulsive status epilepticus) has a high prevalence in IGE, though it is often unrecognised, because symptoms may be mild. Generalised tonic clonic status epilepticus is less common in IGE than symptomatic and probably symptomatic focal or generalised epilepsies. The emergency treatment of convulsive or non-convulsive status epilepticus is the same irrespective of causes, idiopathic or symptomatic, in focal or generalised epilepsies. There are recent major publications and reviews for more extensive reading.363,371,374–379,379–388 Convulsive Status Epilepticus Convulsive status epilepticus (CSE) is a medical emergency and should be managed urgently and properly according to established and well-publicised protocols.371,374–379 Following any of these protocols results in appropriate and rapid management of CSE, which reduces morbidity and mortality.377 The mortality of CSE is around 10%. The commonest cause of CSE is withdrawal of anti-epileptic medication, which should be reintroduced as soon as possible after the onset of CSE. The longer CSE lasts, the harder it is to treat and the greater the morbidity and mortality. The aim of treatment is early termination of CSE in order to prevent neuronal damage caused by systemic and metabolic disturbances and by the direct excitotoxic effect of electrical seizure discharges. Control of overt and electrical seizures is imperative. The risk of brain damage increases progressively if continuous CSE persists for more than 30 minutes and particularly after 1–2 hours. This is because compensatory mechanisms to prevent brain damage are relatively satisfactory during the first 30 minutes. Subsequently, compensatory mechanisms break down with increasing speed if the seizures are not stopped during the first 30 minutes. Neuronal damage leads to transient or permanent neurological, epileptic and cognitive sequelae or even death. Drug Treatment of CSE Can Be Divided into Three Stages In the early stage (first 30 minutes), treatment comprises intravenous administration of a fast-acting benzodiazepine of which diazepam, lorazepam or midazolam are the most effective agents. Diazepam is the traditional drug, then lorazepam came to prominence as the drug of first choice377 and, recently, midazolam infusion has become increasingly popular as an effective and well-tolerated therapeutic agent.389 Most cases are controlled with this approach. If CSE is not controlled at this stage, the patient enters into the second stage of established status epilepticus, which carries an appreciable morbidity. In the stage of established CSE, first-line drug options are intravenous phenytoin or fosphenytoin. If these are ineffective, subanaesthetic doses of phenobarbitone are used. If seizures are not controlled at this stage the patient enters refractory status epilepticus. The last stage of refractory status epilepticus requires general anaesthesia and a continuous infusion of AEDs, such as pentobarbital, midazolam or propofol, and concomitant EEG monitoring of seizure or EEG background suppression.381 Despite these measures, the mortality of refractory CSE is more than 20%.381 Absence (Generalised Non-Convulsive) Status Epilepticus Absence status epilepticus19,66,390 occurs in 10–20% of cases of IGE and in as many as 50% of cases of some syndromes of IGE such as those manifesting with phantom absences or perioral myoclonia. Additionally, nearly all patients are fully aware of this epileptic status and know that it may inevitably lead to a GTCS, though it is avoidable: Patient note It is the same feeling of: “slowing down”, “uncontrollable rush of thoughts”, “losing control of my mind”, “taking me much longer to formulate my response which occasionally is inappropriate and bumbled. Then I know that I will have the fit. If I can, I just go to a private place and wait for it.” This stage is unlikely to be considered as a genuine status epilepticus by the physicians in accident and emergency departments. Therefore, advice to the patient regarding therapeutic options for self-administration of drugs is imperative. Benzodiazepines and, mainly, diazepam, lorazepam or midazolam are the most effective agents. Self-administration: Rectal diazepam (10–20 mg for adults and 0.5 mg/kg for children) as soon as the first symptoms appear may stop ASE and prevent an impending GTCS. Rectal absorption of liquid diazepam is very rapid, reaches the brain within minutes and has a near-intravenous efficacy. Proprietary rectal tubes (Stesolid) containing ready-made liquid diazepam are the most widely used formulation. Suppositories of diazepam are not useful, because of their slow absorption. Diazepam rectal gel is now available in the USA and some other countries.391 However, adult patients, either because of embarrassment or inconvenience, rarely use rectal preparations. An oral bolus dose of valproate (usually twice the daily prophylactic dose) from onset of symptoms is often effective in terminating the ASE and preventing GTCS (most patients prefer this to rectal preparations of drugs). Buccal392,393 or intranasal394 application of midazolam may be the best practical and effective therapeutic option. Midazolam buccal administration has equal efficacy and rapidity of action to rectal diazepam. It is more convenient and less traumatising than rectal preparations of diazepam. Midazolam (0.3 mg/kg for children and 10–20 mg for adults) drawn up from an ampoule is dissolved with peppermint (otherwise it smells and tastes terrible) and should be swirled around the mouth for 4–5 min and then spat out. 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Claassen J, Hirsch LJ, Emerson RG, Mayer SA. Treatment of refractory status epilepticus with pentobarbital, propofol, or midazolam: a systematic review. Epilepsia. 2002;43:146–53. [PubMed: 11903460] 382. Claassen J, Hirsch LJ, Mayer SA. Treatment of status epilepticus: a survey of neurologists. J.Neurol Sci. 2003;211:37–41. [PubMed: 12767495] 383. Gaitanis JN, Drislane FW. Status epilepticus: a review of different syndromes, their current evaluation, and treatment. Neurolog. 2003;9:61–76. [PubMed: 12808369] 384. Manno EM. New management strategies in the treatment of status epilepticus. Mayo Clin.Proc. 2003;78:508–18. [PubMed: 12683704] 385. Ruegg SJ, Dichter MA. Diagnosis and Treatment of Nonconvulsive Status Epilepticus in an Intensive Care Unit Setting. Curr Treat Options Neurol. 2003;5:93–110. [PubMed: 12628059] 386. Sirven JI, Waterhouse E. Management of status epilepticus. Am Fam Physician. 2003;68:469–76. [PubMed: 12924830] 387. Walker MC. Status epilepticus on the intensive care unit. J Neurol. 2003;250:401–6. [PubMed: 12700903] 388. Lowenstein DH. Treatment options for status epilepticus. Curr Opin Pharmacol. 2003;3:6–11. [PubMed: 12550735] 389. De Negri M, Baglietto MG. Treatment of status epilepticus in children. Paediatr Drugs. 2001;3:411–20. [PubMed: 11437186] 390. Walker MC. Diagnosis and treatment of nonconvulsive status epilepticus. CNS Drugs. 2001;15:931–9. [PubMed: 11735613] 391. Fitzgerald BJ, Okos AJ, Miller JW. Treatment of out-of-hospital status epilepticus with diazepam rectal gel. Seizure. 2003;12:52–5. [PubMed: 12495650] 392. Scott RC, Besag FM, Neville BG. Buccal midazolam and rectal diazepam for treatment of prolonged seizures in childhood and adolescence: a randomised trial. Lancet. 1999;353:623–6. [PubMed: 10030327] 393. Kutlu NO, Dogrul M, Yakinci C, Soylu H. Buccal midazolam for treatment of prolonged seizures in children. Brain Dev. 2003;25:275–8. [PubMed: 12767460] 394. Lahat E, Goldman M, Barr J, Bistritzer T, Berkovitch M. Comparison of intranasal midazolam with intravenous diazepam for treating febrile seizures in children: prospective randomised study. BMJ. 2000;321:83–6. [PMC free article: PMC27427] [PubMed: 10884257] 395. Verdru P, Wajgt A, Schiemann Delgado J, Noachtar S. Efficacy and safety of levetiracetam 3000 mg/d as adjunctive treatment in adolescents and adults suffering from idiopathic generalized epilepsy with myoclonic seizures. Epilepsia. 2005;46(Suppl 6):56–7. Go to: Footnotes The term ‘typical’ is used not to characterise them as ‘classical’, but to differentiate them from ‘atypical’ absence seizures. The terminological differences between primarily as opposed to primary and secondarily as opposed to secondary have been detailed in Chapter 1. Non-convulsive status epilepticus is a term that has been rightly discarded in the new diagnostic scheme,5 because it encompasses heterogeneous conditions which may be focal, such as limbic status epilepticus, or generalised, such as ASE.71,72 Convulsive elements and particularly myoclonic jerks are common in generalised non-convulsive status epilepticus as, for example, in eyelid or perioral status epilepticus. Non-convulsive status epilepticus is not synonymous with ASE. If this term is used, the distinction between ‘focal non-convulsive’ and ‘generalised non-convulsive’ should be made for clinical and management purposes. Syndrome in development Astatic is not synonymous with atonic seizures. The term astatic seizures is abandoned in the new ILAE diagnostic scheme, which uses only the term ‘atonic’ or ‘myoclonic-atonic’ seizures.5 I use the eponymic nomenclature ‘Doose syndrome’ only for the pure form of ‘idiopathic epilepsy with myoclonic-astatic seizures’ to exclude cases of symptomatic cause manifesting with myoclonic-astatic seizures. Seizures of Idiopathic Generalised Epilepsies Epileptic Syndromes of Idiopathic Generalised Epilepsies References Footnotes Copyright © 2005, Bladon Medical Publishing, an imprint of Springer Science+Business Media. Bookshelf ID: NBK2608 Contents < PrevNext > Share on Facebook Share on Twitter Views PubReader Print View Cite this Page Disable Glossary Links In this Page Seizures of Idiopathic Generalised Epilepsies Epileptic Syndromes of Idiopathic Generalised Epilepsies References Related Items in Bookshelf All Reference Works All Textbooks Related information PMCPubMed Central citations PubMedLinks to PubMed Recent Activity Clear)Turn Off)Turn On) Idiopathic Generalised Epilepsies - The EpilepsiesIdiopathic Generalised Epilepsies - The Epilepsies Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Panayiotopoulos CP. The Epilepsies: Seizures, Syndromes and Management. 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3364	https://jcadonline.com/review-of-woods-lamp-in-dermatology/	Skip to content Subscribe Revealing The Unseen: A Review of Wood’s Lamp in Dermatology PrevPrevious PostSurgical Outcomes Following Mohs Micrographic Surgery for Basal Cell Carcinoma on the Distal Third of the Nose Next PostInsights on Skin Quality and Clinical Practice Trends in Asia Pacific and a Practical Guide to Good Skin Quality from the Inside OutNext Categories: J Clin Aesthet Dermatol. 2022;15(6):25-30. by Joseph M. Dyer, DO, FAAD and Valerie M. Foy, BS Dr. Dyer is Clinical Assistant Professor at the Philadelphia College of Osteopathic Medicine in Suwanee, Georgia. Ms. Foy is a medical student at the Philadelphia College of Osteopathic Medicine in Philadelphia, Pennsylvania. FUNDING: No funding was provided for this study. DISCLOSURES: The authors report no conflicts of interest relevant to the content of this article. ABSTRACT: In use for over a century, the Wood’s lamp is a time-tested tool to aid in the diagnosis of certain superficial infections, pigmentary disorders, and metabolic diseases. To achieve its high utility, the Wood’s lamp projects ultraviolet light onto the skin which in turn reflects a visible light that a trained eye can use to diagnose and monitor multiple dermatological ailments. Although new alternatives to Wood’s lamp have been considered, it still remains a favored method of diagnosis because it is safe, cost-effective, and reliable. In this review, the authors explore the myriad applications of Wood’s lamp in the field of dermatology. Keywords: Wood’s lamp, ultraviolet light, fluorescence, phosphor, porphyria, dermatophyte In 1903, Robert Williams Wood, a well-known American physicist, developed an instrument that produced an apparent paradox: invisible light.1 This light source, the Wood’s lamp, had a special filter comprised of barium silicate with 9% nickel oxide that blocked much of the visible electromagnetic spectrum and allowed transmission of ultraviolet (UV) light.1-2 Wood primarily used his invention in UV photography. Later, the Wood’s lamp found applications in other scientific fields, including criminal forensics, emergency medicine, ophthalmology, gynecology, and veterinary medicine.3-9 However, Wood’s lamp has arguably gained greatest recognition for its role in dermatology, where it is used to diagnose and monitor an array of fungal and bacterial infections, pigmentary conditions, and metabolic disorders. A brief discussion of electromagnetic radiation is warranted to contextualize the role of the Wood’s lamp. The electromagnetic spectrum comprises a range of energy that radiates, or travels in waves. Arranged within this spectrum are seven energies: radio, microwave, infrared, visible, ultraviolet, X-ray, and gamma. Radio waves have the lowest energy and longest wavelengths, whereas gamma rays have the highest energy and shortest wavelengths. Perceptible by the human retina, visible light has a narrow range from 700nm (red) to 400nm (violet). Slightly more energetic, and invisible, is ultraviolet light. It is in this part of the electromagnetic spectrum that Wood’s lamp emits light, from 320nm to 400nm, with a peak wavelength of 365nm. Physics of fluorescence To understand how Wood’s lamp functions, one must take a closer look at what is occurring at a molecular level. When the relatively energetic and short waves of UV light shine on certain substances, known as phosphors, visible light of lower energy and longer wavelength is produced (Figure 1). This is known as fluorescence. Thus, phosphors essentially convert invisible UV light to visible light. The skin contains natural phosphors, such as collagen and elastin.2 To excite other electrons, photons need to have a certain amount of energy. Elastin contains several phosphors, one of which is a crosslinking tricarboxylic amino acid with a pyridinium ring, and collagen’s main phosphor is tyrosine, an amino acid.10-11 Both of these phosphors contain electrons that can be excited by UV light. Once excited, almost immediately the electrons become unstable and seek their lower energy ground state.12 This electronic transition from high to low energy states occurs through vibrational relaxation.13 This energy decay process releases photons of visible light.13 Alternatives A traditional Wood’s lamp contains Wood’s glass, which is a mixture of barium-sodium-silicate glass and 9% nickel oxide.1-2 This glass coats the inside of tubes through which the UV light is transmitted. However, Wood’s glass is unique in that it blocks much of the visible light passing through the filter. A recent study has suggested that ordinary blacklights are a less expensive and comparable alternative to Wood’s lamp.14 All blacklights, including light emitting diodes, fluorescent light bulbs, and blacklight blue bulbs, emit UV light, similar to Wood’s lamp.15 However, not all blacklights emit the same intensity and peak wavelength of UV light; each blacklight produces slightly different fluorescence.16 Wood’s lamp produces a peak wavelength of 365nm, whereas commercially available blacklight sources can have peak wavelengths of 375nm, 385nm, or 395nm.17 The longer the peak UV wavelength, the more visible light will be produced, which will result in less fluorescence. Blacklights that produce less fluorescence make diagnosing dermatological conditions more difficult. Another difference between a conventional blacklight and Wood’s light is that Wood’s light can magnify objects by 1.5 times.18 Interestingly, a second alternative to Wood’s lamp can be created by simply generating a blue screen background on a smartphone.19,20 Blue light is absorbed well by melanin, however, the luminescence is much harder to see, given the increased levels of visible light.2 Digital screens produce short-wavelength visible light, not UV light.21 A blue screen on a smartphone is a practical option when in a resource-poor setting where the provider may not have access to Wood’s lamp. Clinicians can use an internet search engine to search for “blue image” and download it. Then, the clinician should increase the smartphone screen brightness to the maximum level and turn off the screen timeout setting.19 The rest of the procedure is identical to using a Wood’s lamp; simply darken the room and hold the phone 4 to 5 inches away from the skin. While Wood’s lamp is always the superior option, conventional blacklights and even a blue-lit smartphone screen are acceptable alternatives in resource-poor settings. Performing an examination To conduct a Wood’s lamp examination, the provider will need a light source and a dark room.1 As per the manual, it is ideal to allow the lamp to warm up for about one minute.22 Other sources say only 20 seconds is needed.1 It is recommended to hold the Wood’s lamp 4 to 5 inches away from the surface of the skin.1 Many extraneous artifacts could influence the lamp’s fluorescence. For example, if the patient has showered or bathed recently, it will decrease the fluorescence.2 There are also certain medications, detergents, and fibers that will cause inappropriate fluorescence.2 Many substances produce false positives from Wood’s lamp, especially in pediatric dermatology. These include colored markers (highlighters), dried soap and laundry detergents with optical brighteners, hyperkeratotic scale, invisible ink, lemon juice, lint, semen, serum, saliva, milk, cosmetics and hair dyes, select sunscreens and ointments, and wet ear wax.23 When there is unusual fluorescence observed in atypical locations, it is important to consider exposure to these contaminants. Ophthalmologists have indicated that Wood’s lamp has no harmful effects on the superficial structures of the eye.8 However, in pediatric dermatology, the use of UV light protective goggles is recommended to shield the retina, given that children may impulsively look directly into the light.24 Chronic exposure to Wood’s lamp may cause cataract formation and ocular aging.24 Applications in dermatology Literature on Wood’s lamp use in dermatology is discussed in detail below and summarized in Table 1. Certain pathogenic bacteria of the skin fluoresce with the use of the Wood’s lamp. Pseudomonas aeruginosa infections are commonly associated with burns, onycholysis, and interdigital spaces. Pseudomonas secretes pyoverdine, a phosphor, which will fluoresce green under Wood’s lamp.25 Thus, green fluorescence, in the proper clinical context, insinuates infection with Pseudomonas (Figure 2).26-28 Similarly, erythrasma is a superficial bacterial infection caused by Corynebacterium minutissimum.29 Erythrasma can mimic other intertriginous dermatoses, such as inverse psoriasis or candidiasis, sometimes making diagnosis difficult.30 However, C. minutissimum produces coproporphyrin III and fluoresces a coral-red color, which elucidates its diagnosis.31 Other causes of intertrigo will not fluoresce similarly.31 Trichomycosis axillaris is a superficial bacterial infection in which small concretions form on axillary hair. Trichomycosis is usually associated with Corynebacterium tenuis and other diphtheroid species.32 Instead of a coral-red fluorescence like erythrasma, trichomycosis axillaris displays a pale-yellow fluorescence under Wood’s lamp.32 Although the precise phosphor in the case of trichomycosis axillaris is unknown, it is postulated to be the actual gluelike concretions, whether bacteriogenic or from apocrine sweat.33-34 Lastly, Cutibacterium acnes causes progressive macular hypomelanosis (PMH).35 This hypopigmentation disorder mimics tinea versicolor and post-inflammatory hypopigmentation.30 Under Wood’s lamp, Cutibacterium acnes fluoresces an orange-red color in pilosebaceous follicles, clinching the diagnosis of PMH (Figure 3).35-36 In addition to bacterial infections, Wood’s lamp allows for rapid diagnosis of certain fungal infections, especially when compared to the delay associated with fungal cultures. Of the three genera of dermatophytes—Trichophyton, Epidermophyton, and Microsporum—only Microsporum and a few Trichophyton species will produce fluorescence.37 Dermatophytes in the genus Microsporum will produce a blue-green fluorescence from the porphyrin pteridine; most in genus Trichophyton will not produce any fluorescence, although T. schoenleinii fluoresces dull blue and T. verrucosum fluoresces in cattle (not humans).2,38,39 Historically, most cases of tinea capitis in the United States were caused by M. audouinii, and Wood’s lamp provided a reliable and quick diagnosis.40 Currently, T. tonsurans is the leading cause of tinea capitis in the United States and does not fluoresce.40 Therefore, when tinea capitis is suspected, a negative Wood’s lamp test does not rule out this infection, and a fungal culture should be performed. M. ferrugineum is a prominent cause of juvenile tinea capitis in Russia, Asia, Africa, and Eastern Europe.41 This dermatophyte also creates a green-yellow fluorescence due to tryptophan metabolite accumulation.41 Another common superficial fungal skin infection is pityriasis versicolor, also known as tinea versicolor, caused primarily by Malassezia globosa. Wood’s lamp displays a yellow-orange fluorescence with tinea versicolor infections due to the porphyrin pityrialactone.42,43 Wood’s lamp has limited utility with diagnosing onychomycosis since dermatophytes, yeasts, and nondermatophyte molds are all causative agents, and the most common culprits do not fluoresce.44 Pigmentary disorders are a standard purview for Wood’s lamp diagnosis. Skin devoid of melanin fluoresces brightly. In particular, the depigmentation of vitiligo appears strikingly white and sharply delineated under a Wood’s lamp.2 Nevus depigmentosus, a misleading appellation, is often hypopigmented, and thus shows only a dull, off-white glow with Wood’s lamp.45 Similarly, the hypopigmented macules of tuberous sclerosis complex (TSC) can also be accentuated by Wood’s lamp. These elliptical or lancinate macules, termed ash-leaf spots, are more apparent under a Wood’s lamp.46 Whenever there is a history of TSC in a family, a newborn should receive a complete skin examination with a Wood’s lamp to screen for ash-leaf macules.46 Wood’s lamp effectively magnifies hypopigmentation that the unaided eye can miss otherwise. A Wood’s lamp aids in the evaluation of linear scleroderma and morphea, considering active areas contain loss of pigment.47, 48 Under Wood’s lamp, a new plaque of morphea can be detected before induration is appreciable. Similarly, expanding scleroderma might be detected earlier with Wood’s lamp.49 Apart from diagnosing and delineating disorders of hypopigmentation, Wood’s lamp has further particular use in vitiligo: to assess disease stability, to aid in skin graft harvesting, and to gauge success of surgical grafting itself. A stable disease state is a prerequisite for successful vitiligo surgery and is characterized by sharply demarcated borders.50 Repigmentation after skin grafting procedures may be easily and inexpensively assessed using Wood’s lamp.50 One pilot study utilized Wood’s lamp as an ultraviolet source to improve the speed and quality of suction blister harvesting for use in vitiligo; however, its authors did not enumerate technical parameters, including fluence, of the Wood’s lamp device used. Further, they noted that induction of blisters might have been associated with heat, not UV light, produced by the Wood’s lamp.51 Wood’s lamp assists in diagnosing hyperpigmentation disorders, such as melasma. There is an increase in melanin in the epidermis and/or the dermis with melasma.52 When using Wood’s lamp, epidermal melasma will appear as a sharply circumscribed brown or black patch, and dermal melasma will appear unaccentuated grey-blue.53 Knowing the depth of melanin deposition may help tailor an appropriate treatment regimen. Wood’s lamp has distinctive findings in some cutaneous porphyrias, which arise from enzymatic defects in heme synthesis and subsequent accumulation of photoreactive precursors called porphyrins in tissue, blood, urine, and stool.2 Diagnosis of a specific porphyria in modern medicine may utilize biochemical analysis to identify a specific porphyrin and its ratio in blood, urine, or feces, as well as genetic analysis to identify the specific mutation underlying an enzyme defect. Historically, however, Wood’s lamp highlighted the presence of porphyrins as a classical screening test. Specifically, the porphyrins produced with porphyria cutanea tarda (PCT) fluoresce pink or orange-red in urine and feces under Wood’s lamp (Figure 4). Urine specimens should be shielded from light, and acidification with undiluted acetic acid may enhance fluorescence.2,54-55 Hepatoerythropoietic porphyria, as a biallelic form of PCT,has a more severe phenotype and striking Wood’s lamp findings, including fluorescent red blood cells and erythrodontia.56 Congenital erythropoietic porphyria can present with red or violet wet diapers, intense photosensitivity, and similarly fluorescent erythrodontia; urine, blood, and teeth fluoresce bright coral, due to massive amounts of uroporphyrin I and coproporphyrin I.2, 57-58 Skin, teeth, nails, and urine in patients with erythropoietic protoporphyria fail to fluoresce, though affected erythrocytes will.59 Lastly, Wood’s lamp is utilized in many miscellaneous cutaneous conditions. In particular, the diagnosis of solar urticaria includes exposure to UVA light to produce a reaction, so a Wood’s lamp is commonly used.60 A common condition seen in dermatology is milia; which are small cysts caused by retention of keratin in the dermis.61 The keratin produces a bright yellow fluorescence under Wood’s lamp; thus, milia can be differentiated easily from other minute facial papules.61 Finally, one study asserts that a Wood’s lamp can delineate among porokeratosis, granuloma annulare, and lichen planus annularis. Granuloma annulare fluoresces blue-black at the center and red on the edges; lichen planus annularis fluoresces blue-black without a ring structure; porokeratosis fluoresce blue-brown at the center and white on the edge, resembling a diamond.18 Surgical applications Determination of appropriate surgical margins can be aided by Wood’s lamp.62 To answer this question when removing lentigo maligna melanomas, Wood’s lamp is used to outline the borders and excise the full lesion.62 This method will accentuate the hyperpigmentation in the epidermis of the lesion. When trying to take the narrowest margin, the Wood’s lamp is an excellent additional step. In a study, Wood’s lamp was used to delineate the margins before surgical resection of lentigo maligna. In this particular study, only one of the 16 patients had a recurrence of melanoma eight years after surgery and most lesions were resected with a margin of 0.6 to 1.0 cm.63 Along with using Wood’s lamp to delineate neoplastic lesional margins, there is a new pre-operative technique that allows surgeons to only take damaged tissue and keep as much healthy skin as possible. Photodiagnosis uses methyl aminolevulinate (MAL) cream as a phosphor when applied to patients with basal cell carcinoma (BCC) and then this is fluoresced with Wood’s lamp.64 The bright red fluorescence is the main area of cancerous tissue, but there is also a fainter fluorescence termed the “gray zone.”64 Photodiagnosis has allowed surgeons to achieve radical excision in 90 percent of patients by taking the gray zone, without taking too much tissue.64 Wood’s lamp is valuable in surgical procedures when visualizing cancerous margins. Medical applications Superficial chemical peels are used frequently in dermatology to address lentigines, rhytides, keratoses, comedonal acne, and dyschromia.65 Wood’s lamp can be used after applying a chemical peel to avoid missing or over-treating areas on the skin.65 Salicylic acid and fluorescein sodium both can be added to chemical peels to create fluorescence. These two peels produce two different fluorescence; salicylic acid creates a green fluorescence, and the fluorescein sodium peel is a yellow-orange color.65 Wood’s lamp can also be used before and after applying the peel to measure its effectiveness based on the depth of pigment.65 Similarly, Wood’s lamp is also able to assess the proper application of sunscreen, especially in relatively inaccessible areas such as the back.66,67 Wood’s lamp is advantageous when evaluating patient adherence to tetracycline by examining the toenails for yellow fluorescence.68 Dihydroxyacetone (DHA) is a chemical in sunless tanning cosmetics and is often used to cover vitiligo lesions.69 Wood’s lamp can determine the actual size of vitiligo lesions, as DHA fluoresces a salmon color and vitiligo lesions are bright blue-white.69 Conclusion Wood’s lamp is an inexpensive yet invaluable asset in a dermatologist’s arsenal. It is easy to use, safe, cost effective, and yields rapid results. Its utility in dermatology encompasses superficial infections, pigmentary disorders, and metabolic diseases. It also has many practical applications in cutaneous surgery and cosmetics. Although it has been over a century since its invention, the Wood’s lamp is a valuable tool in dermatology, helping to reveal the unseen. References Sharma S, Sharma A. Robert Williams Wood: pioneer of invisible light. Photodermatol Photoimmunol Photomed. 2016;32(2):60–65. Klatte JL, van der Beek N, Kemperman PM. 100 years of Wood’s lamp revised. J Eur Acad Dermatol Venereol. 2015;29(5):842–847. Pasquetti M, Min ARM, Scacchetti S, et al. Infection by Microsporum canis in Paediatric Patients: A Veterinary Perspective. Vet Sci. 2017;4(3):E46. Santucci KA, Nelson DG, McQuillen KK, Duffy SJ, Linakis JG. Wood’s lamp utility in the identification of semen. Pediatrics. 1999;104(6):1342–1344. Casavant MJ, Shah MN, Battels R. Does fluorescent urine indicate antifreeze ingestion by children. 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Wood’s lamp fluorescence of dihydroxyacetone treated skin. J Eur Acad Dermatol Venereol. 2016;30(11):e12–e126. Share: Recent Articles: Diego Ruiz Dasilva, MD: Nemolizumab Demonstrates Marked Efficacy in Challenging Chronic Pruritus: A Real-world Multicenter Case Read More Letters to the Editor: September 2025 Read More Hidradenitis Suppurativa Patients Experience a Significant Musculoskeletal Symptom Burden: A Quality Improvement Initiative Using the IDEOM MSK-Q Read More Remission of Alopecia Areata Post-colectomy in a Patient with Crohn's Disease Read More What Are Topical Adaptogens? 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Tools & Reference>Rheumatology Raynaud Phenomenon Workup Updated: Jun 24, 2025 Author: Heather Hansen-Dispenza, MD; Chief Editor: Herbert S Diamond, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Raynaud Phenomenon Sections Raynaud Phenomenon Overview Background Pathophysiology Etiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Laboratory Studies Magnetic Resonance Imaging Show All Treatment Approach Considerations Nonpharmacologic Therapy Pharmacologic Therapy Treatment of Critical Digital Ischemia Consultations Diet Show All Guidelines Medication Medication Summary Calcium channel blockers Prostaglandins Nitrates, Angina Phosphodiesterase-5 Enzyme Inhibitors Antidepressants, SSRIs Endothelin Antagonists ARBs Local Anesthetics, Amides Metabolic & Endocrine, Other Neuromuscular Blocker Agents, Botulinum Toxins Anticoagulants, Hematologic Show All Media Gallery;) References;) Workup Approach Considerations Raynaud phenomenon can be diagnosed on clinical grounds. Laboratory testing is necessary to assess for conditions that can mimic Raynaud phenomenon or cause secondary Raynaud phenomenon. The selection of tests should be guided by the clinical findings. Nailfold capillaroscopy is the gold standard for differentiating primary Raynaud phenomenon from secondary Raynaud phenomenon (in particular, for identifying a scleroderma pattern). [1, 34] European guidelines note that capillaroscopy is not often performed in primary care, but recommend its use in secondary care, as abnormal capillary patterns are strong predictors of connective tissue disease. Magnetic resonance imaging has been studied in some forms of secondary Raynaud phenomenon. Next: Laboratory Studies The following laboratory studies may be considered in patients with Raynaud phenomenon: Complete blood cell count - To evaluate for polycythemic disorders, underlying malignancies, or autoimmune disorders Blood urea nitrogen - To evaluate for possible renal impairment or dehydration Creatinine - To evaluate for kidney dysfunction Prothrombin time - To evaluate forr liver dysfunction Activated partial thromboplastin time - To evaluate for antiphospholipid antibody disorder or liver dysfunction Serum glucose - To evaluate for diabetes Thyroid-stimulating hormone - To test for thyroid disorders Optional laboratory tests are as follows: Antinuclear antibody - May be positive in autoimmune disorders and should be obtained in patients with features of these disorders Serum viscosity - Elevated in hyperviscosity syndromes such as paraproteinemias Serum creatine kinase - Elevated in muscle damage such as polymyositis and dermatomyositis Rheumatoid factor - May be elevated in rheumatoid arthritis, other autoimmune disorders, and some forms of cryoglobulinemia (monoclonal proteins in multiple myeloma and Waldenström macroglobulinemia have an increased frequency of rheumatoid factor activity) Hepatitis panel - Positive for hepatitis B or C virus infection in many patients with cryoglobulinemia Cold agglutinins - Present in Mycoplasma infections and lymphomas Heavy metal screen - To asses for neuropathic pain due to poisoning Growth hormone - To evaluate for acromegaly Plasma metanephrine testing or 24-hour urinary collection for catecholamines and metanephrines - To evaluate for pheochromocytoma Leukocyte alkaline phosphatase - To evaluate for leukemias in appropriate patients Antiphospholipid antibodies studies - Including dilute Russell viper venom studies, anticardiolipin antibodies, and anti-beta-1-glycoprotein-2 antibodies Serum protein and urine electrophoresis - To evaluate for paraproteinemias Flow cytometry or acidified serum lysis (Ham) test - To evaluate for paroxysmal nocturnal hemoglobinuria Previous Next: Magnetic Resonance Imaging Smitaman and colleagues report that magnetic resonance imaging (MRI) scans of the feet of patients with Raynaud phenomenon demonstrate a progressive distal-to-proximal pattern of phalangeal bone marrow edema. 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Effects of phosphodiesterase type 5 inhibitors on Raynaud's phenomenon. Rheumatol Int. 2014 Nov. 34 (11):1623-6. [QxMD MEDLINE Link]. Maltez N, Maxwell LJ, Rirash F, et al. Phosphodiesterase 5 inhibitors (PDE5i) for the treatment of Raynaud's phenomenon. Cochrane Database Syst Rev. 2023 Nov 6. 11 (11):CD014089. [QxMD MEDLINE Link].[Full Text]. Dziadzio M, Denton CP, Smith R, et al. Losartan therapy for Raynaud's phenomenon and scleroderma: clinical and biochemical findings in a fifteen-week, randomized, parallel-group, controlled trial. Arthritis Rheum. 1999 Dec. 42(12):2646-55. [QxMD MEDLINE Link]. Dhaliwal K, Griffin M, Denton CP, Butler PEM. The novel use of botulinum toxin A for the treatment of Raynaud's phenomenon in the toes. BMJ Case Rep. 2018 Mar 9. 2018:[QxMD MEDLINE Link]. Nagarajan M, McArthur P. Targeted high concentration botulinum toxin A injections in patients with Raynaud's phenomenon: a retrospective single-centre experience. Rheumatol Int. 2021 May. 41 (5):943-949. [QxMD MEDLINE Link]. Motegi SI, Uehara A, Yamada K, Sekiguchi A, Fujiwara C, Toki S, et al. Efficacy of Botulinum Toxin B Injection for Raynaud's Phenomenon and Digital Ulcers in Patients with Systemic Sclerosis. Acta Derm Venereol. 2017 Jul 6. 97 (7):843-850. [QxMD MEDLINE Link]. Destors JM, Gauthier E, Lelong S, Boissel JP. Failure of a pure anti-platelet drug to decrease the number of attacks more than placebo in patients with Raynaud's phenomenon. Angiology. 1986 Aug. 37 (8):565-9. [QxMD MEDLINE Link]. Gliddon AE, Dore CJ, Black CM, et al. Prevention of vascular damage in scleroderma and autoimmune Raynaud's phenomenon: a multicenter, randomized, double-blind, placebo-controlled trial of the angiotensin-converting enzyme inhibitor quinapril. Arthritis Rheum. 2007 Nov. 56(11):3837-46. [QxMD MEDLINE Link]. Merritt WH. Role and rationale for extended periarterial sympathectomy in the management of severe Raynaud syndrome: techniques and results. Hand Clin. 2015 Feb. 31 (1):101-20. [QxMD MEDLINE Link]. DiGiacomo RA, Kremer JM, Shah DM. Fish-oil dietary supplementation in patients with Raynaud's phenomenon: a double-blind, controlled, prospective study. Am J Med. 1989 Feb. 86(2):158-64. [QxMD MEDLINE Link]. Media Gallery A 9-year-old with Raynaud phenomenon. Notice the discoloration of the fingers. Photo of a patient with Raynaud phenomenon that resulted from working with a jackhammer. Courtesy of the CDC. Raynaud phenomenon showing demarcation of color difference. of 3 Tables Back to List Contributor Information and Disclosures Author Heather Hansen-Dispenza, MD Rheumatology Fellow, University of Arizona College of Medicine Heather Hansen-Dispenza, MD is a member of the following medical societies: American College of Physicians, American College of Rheumatology Disclosure: Nothing to disclose. Coauthor(s) S Anita Narayanan, MD Fellow in Rheumatology, University of Arizona College of Medicine S Anita Narayanan, MD is a member of the following medical societies: American College of Physicians, American College of Rheumatology, American Medical Association Disclosure: Nothing to disclose. Mayra Oberto-Medina, DO Fellow, Department of Rheumatology, University of Arizona Disclosure: Nothing to disclose. Specialty Editor Board Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Received salary from Medscape for employment. for: Medscape. Elliot Goldberg, MD Dean of the Western Pennsylvania Clinical Campus, Professor, Department of Medicine, Lewis Katz School of Medicine at Temple University Elliot Goldberg, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American College of Rheumatology Disclosure: Nothing to disclose. Chief Editor Herbert S Diamond, MD Visiting Professor of Medicine, Division of Rheumatology, State University of New York Downstate Medical Center; Chairman Emeritus, Department of Internal Medicine, Western Pennsylvania Hospital Herbert S Diamond, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American College of Rheumatology, American Medical Association, Phi Beta Kappa Disclosure: Nothing to disclose. Additional Contributors Jeffrey R Lisse, MD, FACP Professor, Department of Internal Medicine, Chief, Section of Rheumatology, University of Arizona School of Medicine Jeffrey R Lisse, MD, FACP is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians-American Society of Internal Medicine, American College of Rheumatology, American Geriatrics Society, Sigma Xi, The Scientific Research Honor Society Disclosure: Received consulting fee from Genentech for consulting; Received consulting fee from Centacor for consulting; Received consulting fee from Novartis for review panel membership. John Varga, MD Professor, Department of Internal Medicine, Division of Rheumatology, Northwestern University John Varga, MD is a member of the following medical societies: American College of Physicians, American College of Rheumatology, Central Society for Clinical and Translational Research, Society for Investigative Dermatology Disclosure: Nothing to disclose. encoded search term (Raynaud Phenomenon) and Raynaud Phenomenon What to Read Next on Medscape
3367	https://terrytao.wordpress.com/2008/01/15/254a-lecture-4-multiple-recurrence/	254A, Lecture 4: Multiple recurrence \| What's new What's new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao Home About Career advice On writing Books Mastodon+ Applets Subscribe to feed 254A, Lecture 4: Multiple recurrence 15 January, 2008 in 254A - ergodic theory, math.CO, math.DS \| Tags: multiple recurrence, van der Waerden's theorem \| by Terence Tao In the previous lecture, we established single recurrence properties for both open sets and for sequences inside a topological dynamical system . In this lecture, we generalise these results to multiple recurrence. More precisely, we shall show Theorem 1. (Multiple recurrence in open covers) Let be a topological dynamical system, and let be an open cover of X. Then there exists such that for every , we have for infinitely many r. Note that this theorem includes Theorem 1 from the previous lecture as the special case . This theorem is also equivalent to the following well-known combinatorial result: Theorem 2. (van der Waerden’s theorem) Suppose the integers are finitely coloured. Then one of the colour classes contains arbitrarily long arithmetic progressions. Exercise 1. Show that Theorem 1 and Theorem 2 are equivalent. Exercise 2. Show that Theorem 2 fails if “arbitrarily long” is replaced by “infinitely long”. Deduce that a similar strengthening of Theorem 1 also fails. Exercise 3. Use Theorem 2 to deduce a finitary version: given any positive integers m and k, there exists an integer N such that whenever is coloured into m colour classes, one of the colour classes contains an arithmetic progression of length k. (Hint: use a “compactness and contradiction” argument, as in my article on hard and soft analysis.) We also have a stronger version of Theorem 1: Theorem 3. (Multiple Birkhoff recurrence theorem) Let be a topological dynamical system. Then for any there exists a point and a sequence of integers such that as for all . These results already have some application to equidistribution of explicit sequences. Here is a simple example (which is also a consequence of Weyl’s equidistribution theorem): Corollary 1. Let be a real number. Then there exists a sequence of integers such that as . Proof. Consider the skew shift system with . By Theorem 3, there exists and a sequence such that and both convege to . If we then use the easily verified identity (1) we obtain the claim. Exercise 4. Use Theorem 1 or Theorem 2 in place of Theorem 3 to give an alternate derivation of Corollary 1. As in the previous lecture, we will give both a traditional topological proof and an ultrafilter-based proof of Theorem 1 and Theorem 3; the reader is invited to see how the various proofs are ultimately equivalent to each other. — Topological proof of van der Waerden — We begin by giving a topological proof of Theorem 1, due to Furstenberg and Weiss, which is secretly a translation of van der Waerden’s original “colour focusing” combinatorial proof of Theorem 2 into the dynamical setting. To prove Theorem 1, it suffices to show the following slightly weaker statement: Theorem 4. Let be a topological dynamical system, and let be an open cover of X. Then for every there exists an open set which contains an arithmetic progression for some and . To see how Theorem 4 implies Theorem 1, first observe from compactness that we can take the open cover to be a finite cover. Then by the infinite pigeonhole principle, it suffices to establish Theorem 1 for each separately. For each such k, Theorem 4 gives a single arithmetic progression inside one of the . By replacing the system with the product system for some large N and replacing the open cover of X with the open cover of , one can make the spacing r in the arithmetic progression larger than any specified integer N. Thus by another application of the infinite pigeonhole principle, one of the contains arithmetic progressions with arbitrarily large step r, and the claim follows. Now we need to prove Theorem 4. By Lemma 1 of Lecture 2 to establish this theorem for minimal dynamical systems. We will need to note that for minimal systems, Theorem 4 automatically implies the following stronger-looking statement: Theorem 5. Let be a minimal topological dynamical system, let U be a non-empty open set in X, and let . Then U contains an arithmetic progression for some and . Indeed, the deduction of Theorem 5 from Theorem 4 is immediate from the following useful fact (cf. Lemma 1 from Lecture 3): Lemma 1. Let be a minimal topological dynamical system, and let U be a non-empty open set in X. Then X can be covered by a finite number of translates of U. Proof. The set is a proper closed invariant subset of X, which must therefore be empty since X is minimal. The claim then follows from the compactness of X. . (Of course, the claim is highly false for non-minimal systems; consider for instance the case when T is the identity. More generally, if X is non-minimal, consider an open set U which is the complement of a proper subsystem of X.) Now we need to prove Theorem 4. We do this by induction on k. The case k=1 is trivial, so suppose and the claim has already been shown for k-1. By the above discussion, we see that Theorem 5 is also true for k-1. Now fix a minimal system and an open cover , which we can take to be finite. We need to show that one of the contains an arithmetic progression of length k. To do this, we first need an auxiliary construction. Lemma 2. (Construction of colour focusing sequence) Let the notation and assumptions be as above. Then for any there exists a sequence of points in X, a sequence of sets in the open cover (not necessarily distinct), and a sequence of positive integers such that for all and . Proof. We induct on J. The case J=0 is trivial. Now suppose inductively that , and that we have already constructed , , and with the required properties. Now let V be a suitably small neighbourhood of (depending on all the above data) to be chosen later. By Theorem 5 for k-1, V contains an arithmetic progression of length k-1. If one sets , and lets be an arbitrary set in the open cover containing , then we observe that (2) for all and . If V is a sufficiently small neighbourhood of , we thus see (from the continuity of the that we verify all the required properties needed to close the induction. We apply the above lemma with J equal to the number of sets in the open cover. By the pigeonhole principle, we can thus find such that . If we then set and we obtain Theorem 4 as required. It is instructive to compare the k=2 case of the above arguments with the proof of Theorem 1 from the previous lecture. (For a comparison of this type of proof with the more classical combinatorial proof, see my Montreal lecture notes.) — Ultrafilter proof of van der Waerden — We now give a translation of the above proof into the language of ultrafilters (or more precisely, the language of Stone-Čech compactifications). This language may look a little strange, but it will be convenient when we study more general colouring theorems in the next lecture. As before, we will prove Theorem 4 instead of Theorem 1 (thus we only need to find one progression, rather than infinitely many). The key proposition is Proposition 1. (Ultrafilter version of van der Waerden) Let be a minimal element of . Then for any there exists such that . (3) Suppose for the moment that this proposition is true. Applying it with some minimal element p of (which must exist, thanks to Exercise 10 of the previous lecture), we obtain obeying (3). If we let for some arbitrary , we thus obtain . (4) If we let be an element of the open cover that contains x, we thus see that for all and all which lie in a sufficiently small neighbourhood of q. Since a LCH space is always dense in its Stone-Čech compactification, the space of all (n,r) with this property is non-empty, and Theorem 4 follows. Proof of Proposition 1. We induct on k. The case k=1 is trivial (one could take e.g. , so suppose and that the claim has already been proven for k-1. Then we can find such that (5) for all . Now consider the expression (6) for any and , where . (7) Applying (5) to the limit in (6), we obtain the recursion for all . Iterating this, we conclude that (8) for all . For i=0, (8) need not hold, but instead we have the easily verified identity . (8′) Now let be a non-principal ultrafilter and define . Observe from (6) that all the lie in the closed set , and so p’ does also. Since p is minimal, there must exist such that p = p” + p’. Expanding this out using (8) or (8′), we conclude that (9) for all . Applying (6), we conclude (10) where and . Now, define to be the limit (11) then we obtain Proposition 1 as desired. Exercise 5. Strengthen Proposition 1 by adding the additional conclusion . Using this stronger version, deduce Theorem 1 directly without using the trick of multiplying X with a cyclic shift system that was used to deduce Theorem 1 from Theorem 4. Theorem 1 can be generalised to multiple commuting shifts: Theorem 6. (Multiple recurrence in open covers) Let be a compact topological space, and let be commuting homeomorphisms. Let be an open cover of X. Then there exists such that for infinitely many r. Exercise 6. By adapting one of the above arguments, prove Theorem 6. Exercise 7. Use Theorem 6 to establish the following the multidimensional van der Waerden theorem (due to Gallai): if a lattice is finitely coloured, and , then one of the colour classes contains a pattern of the form for some and some non-zero r. Exercise 8. Show that Theorem 6 can fail, even for and , if the shift maps are not assumed to commute. (Hint: First show that in the free group on two generators , and any word and non-zero integer r, the three words cannot all begin with the same generator after reduction. This can be used to disprove a non-commutative multidimensional van der Waerden theorem, which can turn be used to disprove a non-commutative version of Theorem 6.) — Proof of multiple Birkhoff — We now use van der Waerden’s theorem and an additional Baire category argument to deduce Theorem 3 from Theorem 1. The key new ingredient is Lemma 3. (Semicontinuous functions are usually continuous) Let be a metric space, and let be semicontinuous. Then the set of points x where F is discontinuous is a set of the first category (i.e. a countable union of nowhere dense sets). In particular, by the Baire category theorem, if X is complete and non-empty, then F is continuous at at least one point. Proof. Without loss of generality we can take F to be upper semicontinuous. Suppose F is discontinuous at some point x. Then, by upper continuity, there exists a rational number q such that . (3) In other words, x lies in the boundary of the closed set . But boundaries of closed sets are always nowhere dense, and the claim follows. Now we prove Theorem 3. Without loss of generality we can take X to be minimal. Let us place a metric d on the space X. Define the function by the formula . (4) It will suffice to show that F(x)=0 for at least one x (notice that if the infimum is actually attained at zero for some n, then x is a periodic point and the claim is obvious). Suppose for contradiction that F is always positive. Observe that F is upper semicontinuous, and so by Lemma 3 there exists a point of continuity of F. In particular there exists a non-empty open set U such that F is bounded away from zero. By uniform continuity of , we see that if F is bounded away from zero on U, it is also bounded away from zero on for any n (though the bound from below depends on n). Applying Lemma 1, we conclude that F is bounded away from zero on all of X, thus there exists such that for all . But this contradicts Theorem 1 (or Theorem 4), using the balls of radius as the open cover. This contradiction completes the proof of Theorem 3. Exercise 9. Generalise Theorem 3 to the case in which T is merely assumed to be continuous, rather than be a homeomorphism. (Hint: let denote the space of all sequences with for all n, with the topology induced from the product space . Use a limiting argument to show that is non-empty. Then turn into a topological dynamical system and apply Theorem 3.) Exercise 10. Generalise Theorem 3 to multiple commuting shifts (analogously to how Theorem 6 generalises Theorem 1). Exercise 11. Combine Exercises 9 and 10 by obtaining a generalisation of Theorem 3 to multiple non-invertible commuting shifts. Exercise 12. Let be a minimal topological dynamical system, and let . Call a point x in X k-fold recurrent if there exists a sequence such that for all . Show that the set of k-fold recurrent points in X is residual (i.e. the complement is of the first category). In particular, the set of k-fold recurrent points is dense. Exercise 13. In the boolean Bernoulli system , show that the set A consisting of all non-zero integers which are divisible by 2 an even number of times is almost periodic. Conclude that there exists a minimal topological dynamical system such that not every point in X is 3-fold recurrent (in the sense of the previous exercise). (Compare this with the arguments in the previous lecture, which imply that every point in X is 2-fold recurrent.) Exercise 14. Suppose that a sequence of continuous functions on a metric space converges pointwise everywhere to another function . Show that f is continuous on a residual set. Exercise 15. Let be a minimal topological dynamical system, and let be a function which is T-invariant, thus Tf = f. Show that if f is continuous at even one point , then it has to be constant. (Hint: is in the orbit closure of every point in X.) [Update, Jan 15: bad link fixed.] [Update, Jan 21: Additional exercise added.] [Update, Jan 26: Another additional exercise added.] [Update, Mar 4: Slight correction to proof of Proposition 1.] Share this: Click to print (Opens in new window)Print Click to email a link to a friend (Opens in new window)Email More Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Click to share on Reddit (Opens in new window)Reddit Click to share on Pinterest (Opens in new window)Pinterest Like Loading... 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Polymath 13 – a success! Non-transitive Dice over Gowers’s Blog Rota’s Basis Conjecture: Polymath 12, post 3 34 comments Comments feed for this article 15 January, 2008 at 7:23 pm Anonymous The first link (to the previous lecture) is broken. 0 0 i Rate This Reply 15 January, 2008 at 7:44 pm Terence Tao Thanks for the correction! 0 0 i Rate This Reply 15 January, 2008 at 8:14 pm richard borcherds How easy is it to make these results about dynamical systems quantitative? For example, in theorem 4 can you give an explicit bound for r in terms of k and and some invariants of the dynamical system and open sets? Presumably any such bounds would have to be rather large, given that its not easy to find reasonable bounds in van der Waerden’s theorem. 1 0 i Rate This Reply 16 January, 2008 at 1:54 am Ed Dean Richard, You might be interested in one of Philipp Gerhardy’s papers (click on my name for a link to the PDF), in which he applies proof-theoretic methods to topological dynamics. Girard extracted bounds for van der Waerden numbers from the Furstenberg/Weiss proof of Multiple Birkhoff which are identical to van der Waerden’s original bounds. Gerhardy proves an effective version of Multiple Birkhoff, again via an analysis of Furstenberg/Weiss, and again obtains the same bounds on vdW numbers. But Gerhardy indicates that the reason behind this is that the topological proof and the combinatorial proof are really the same, just in different languages. He suggests the possibility of carrying out work to obtain a topological proof that, rather than mirroring van der Waerden’s original, corresponds to Shelah’s combinatorial proof, and which would enjoy the same improved bounds on van der Waerden numbers. (I hope this information was useful, I’m new at all of this.) 0 0 i Rate This Reply 16 January, 2008 at 9:10 am Terence Tao Dear Richard, One can indeed make the argument quantitative, yielding the same Ackermann-type bounds as in the classical finitary proof of van der Waerden, as in the papers of Gerhardy and Girard mentioned by Ed above. To answer your specific question, to get Theorem 4 for minimal systems with a designated metric d, the argument above gives quite concrete bounds in terms of (a) the uniform continuity modulus of the shift T, and (b) a quantitative bound for Lemma 1, in particular how many translates are needed for a ball of a given radius to cover all of the space. The only real difficulty is in reducing to the minimal case, using the lemma from the previous lecture: the argument I gave there uses Zorn’s lemma and is thus not immediately finitisable. But one can develop substitutes for that lemma which do have quantitative (albeit Ackermann-type) bounds. It’s easiest to describe this in terms of the systems that are of immediate interest to van der Waerden’s theorem, namely orbit closures of a single word . In this setting, c generates a minimal system if and only if every block in c appears syndetically, thus for every L there exists F(L) such that every block of length L that appears at least once in c, will in fact appear in every F(L)-block of c. The key issue with respect to creating quantitative bounds is to obtain some good bound on F (this is the counterpart of (b) above in this setting). This is not possible if we insist on c being infinite, but we can get somewhere by truncating c to an extremely long but finite length (in particular, much larger than any given function of F(L)). Indeed, by using the finite convergence principle as discussed in my earlier post Soft analysis, hard analysis, and the finite convergence principle one can obtain a quantitative version of Lemma 1 from the previous lecture: given any L and any function G, there exists an N such that in any word c of length N there exists an M and a subword of c of length G(M) such that every block of length L in that subword in fact appears in every block of length M in the subword. (One should think of the parameters being arranged in the order L << M << G(M) << N. The quantity M is playing the role of F(L) here.) Basically, N is going to equal an iterated power of G applied to L, where the number of iterations is equal to the number of possible blocks of length L (i.e. ). The intuition here is that if an L-block fails to appear syndetically, one can pass to a long subword in which that block fails to appear at all. Iterating this at most times, one eventually reaches a long subword in which all surviving L-blocks appear syndetically. (Actually, one needs to apply the above lemma for k different values of L in order to get van der Waerden, because minimality is invoked k times in the proof of Theorem 4 once one unpacks the induction. This is what leads to the Ackermann-type behaviour in k. One has to take some care in arranging the quantifiers (i.e. deciding what parameters are large with respect to what other parameters); this is the price we pay for not using the axiom of infinity, which conveniently offers sets (namely, infinite sets) that are guaranteed to be larger than every finite set one could hope to generate by any conceivable future means, thus removing a lot of the need for epsilon management or quantifier management.) 0 0 i Rate This Reply 16 January, 2008 at 10:28 am ben green Terry, It’s a nice observation that van der Waerden can be deduced without introducing a metric. In preparing my course I’ve been following Furstenberg’s book on this point, and he proceeds by proving Birkhoff Multiple Recurrence, only then pausing to derive combinatorial consequences. One only needs BMR in the case that is a homeomorphism, but then the proof of BMR seems to require that too. As you point out in your exercises one can deduce the result for continuous maps by a clever kind of lifting trick. By the way Benji Weiss’s book “Single Orbit Dynamics” has a discussion of how to get minimality without using Zorn’s Lemma, as does “Ergodic Theory via Joinings”. Furstenberg’s book has a more erudite approach (there is a uniquely ergodic -invariant measure on ) but this seems to be an unnecessarily large sledgehammer. Ben 1 0 i Rate This Reply 16 January, 2008 at 1:08 pm Terence Tao Dear Ben, Thanks for the comments. I think (but haven’t fully checked) that one can also use Proposition 1 directly to imply Theorem 1, at least, for non-invertible T, simply by taking p to be a positive minimal ultrafilter (thus ). Then n+ir+p will be positive too, and we can get Theorem 1 without ever having to apply a negative power of T. The derivation of Theorem 3 from Theorem 1 seems to not require invertibility of T either. 1 0 i Rate This Reply 16 January, 2008 at 11:57 pm Pedro Lauridsen Ribeiro Prof. Green’s comment raises an issue about how “nonconstructive” van der Waerden’s theorem really is. I’ve been once told that van der Waerden’s theorem is actually equivalent to Baire’s theorem, but I don’t recall who actually proved this. Does anyone know? (still on the constructivity issue, Baire’s theorem turns out, on its turn, to be equivalent to the Axiom of Dependent Choices, which is also independent of the remaning axioms of Zermelo-Fraenkel set theory and can be used as a “countable” substitute for the Axiom of Choice. For instance, it’s equivalent to a countable version of Tychonoff’s theorem) 0 0 i Rate This Reply 19 January, 2008 at 4:41 am Nilay Sir I am trying to map quantities used in Theorem 1 to those in Theorem 2. In my understanding — X ==> compactified version of Z. Since X is compact, we can have a finite subcover. We will take a finite subcover of compactified Z. The sets in the subcover will be same as the colour classes. k denotes the length of the arithmetic progression. I think that r denotes the common difference between successive terms of the arithmetic progression but Theorem 2 has nothing to say anything on the common difference. To me it seems that Theorem 2 is weaker than Theorem 1. 0 0 i Rate This Reply 19 January, 2008 at 9:14 am Terence Tao Dear Nilan, One can rephrase Theorem 2 to resemble Theorem 1 more closely by replacing the phrase “contains arbitrarily long arithmetic progressions” by the equivalent phrase “contains infinitely many arithmetic progressions of length k for each k”. (Note that as a long arithmetic progression can always be chopped up into many shorter arithmetic progressions, the two statements imply each other.) To deduce Theorem 2 from Theorem 1, one does not actually use the compactified integers (the colour classes do not cover the whole of unless one takes the closure, in which case they are no longer open) but instead looks at a subsystem of the Bernoulli system , where m is the number of colours used. The equivalence between Theorem 1 and Theorem 2 here is in fact very analogous to the equivalence between Theorem 1 of the previous lecture and the infinite pigeonhole principle, as discussed in Exercise 1 of that lecture. 0 0 i Rate This Reply 21 January, 2008 at 9:13 pm 254A, Lecture 5: Other topological recurrence results « What’s new […] We first prove a significant generalisation of van der Waerden’s theorem (Theorem 2 from the previous lecture): […] 0 0 i Rate This Reply 24 January, 2008 at 7:04 pm 254A, Lecture 6: Isometric systems and isometric extensions « What’s new […] short proof of (a slight strengthening of) the multiple Birkhoff recurrence theorem (Theorem 3 from Lecture 4) as […] 0 0 i Rate This Reply 10 February, 2008 at 1:22 pm 254A, Lecture 10: The Furstenberg correspondence principle « What’s new […] The Furstenberg correspondence principle can be extended to relate several other recurrence theorems to their combinatorial analogues. We give some representative examples here (without proofs). Firstly, there is a multidimensional version of Szemerédi’s theorem (compare with Exercise 7 from Lecture 4): […] 0 0 i Rate This Reply 11 February, 2008 at 9:19 pm 254A, Lecture 11: Compact systems « What’s new […] dynamics, such as Proposition 3 from Lecture 6 or the Birkhoff multiple recurrence theorem from Lecture 4 (though to get the full strength of the results, one needs the syndetic van der Waerden theorem, […] 0 0 i Rate This Reply 28 February, 2008 at 3:09 pm 254A, Lecture 13: Compact extensions « What’s new […] this as a colouring of into colours. Applying van der Waerden’s theorem (Exercise 3 from Lecture 4), we can thus find (if K is sufficiently large depending on ) an arithmetic progression in for […] 0 0 i Rate This Reply 3 March, 2008 at 10:47 pm Nilay Sir In the proof of proposition 1, should not be defined as ? 0 0 i Rate This Reply 4 March, 2008 at 7:46 am Terence Tao Dear Nilay: Thanks for the correction! 0 0 i Rate This Reply 15 January, 2009 at 6:37 pm liuxiaochuan Dear Professor Tao: Sorry I ask questions so radomly. About exercise 9, I don’t see the reason to use a limiting argument to show is non empty. Could I just take a x and then is in ? 0 0 i Rate This Reply 16 January, 2009 at 12:07 am liuxiaochuan I think I figured it out: one can use the diagonal argument to find the limit point and then to prove it is also in , right? 0 0 i Rate This Reply 20 July, 2009 at 4:10 am 陶哲轩遍历论习题解答：第四讲 « Liu Xiaochuan’s Weblog […] (注：遍历论为陶哲轩教授于今年年初的一门课程，我尝试将所有习题做出来，这是第四讲的十五个习题。这里是本讲的链接。关于ultrafilter，我写过一个帖子。) […] 0 0 i Rate This Reply 29 July, 2009 at 4:00 am 陶哲轩遍历论习题解答：第五讲 « Liu Xiaochuan’s Weblog […] (注：遍历论为陶哲轩教授于08年年初的一门课程，我尝试将所有习题做出来，这是第五讲的二十一个习题。这里是这门课程的主页，这里是本讲的链接。有一些题目我花费的时间比较多，我将所需要的知识补充在另外的帖子中。关于ultrafilter, 我写过一个帖子.习题一中涉及到一点数论知识，尤其关于二次剩余的知识。我写过一个帖子是关于Hales-Jewett定理的归纳法证明.习题4的解答比较长，我另写了一个帖子。该过程是参考了上一讲中定理四的证明。注意这个证明方法与前述Hales-Jewett定理的归纳证明的异同。仍有几个问题没有完成，我会慢慢补上。) […] 0 0 i Rate This Reply 5 October, 2009 at 3:49 am Ergodic Ramsey Theory (by Yuri Lima) « Disquisitiones Mathematicae […] 2. Multiple Poincaré’s Recurrence Theorem. […] 0 0 i Rate This Reply 24 December, 2009 at 2:57 pm PDEbeginner Dear Prof. Tao, I have some problems on exercises 10 and 14: Ex 10. According to the hint, we can construct the . However, to apply Theorem 3, we have to first prove Theorem 4. In its proof, we need the restricted (in ) to be invertible. This is not necessarily true for the restricted . But we seems able to re-prove Theorem 4 by the same procedure as in the note, only modifying the definition of by taking $x_J$ as the element $r_J$ step in front of $y$ in the orbit. I was wondering if this would be okay. Ex 14. I spent more than 6 hours on it, I am really silly. I checked the proof given by Xiaochuan Liu in the above link, it seems his proof is not right. I was wondering if you could give a hint on how to prove it. Thanks a lot in advance! 0 0 i Rate This Reply 2 October, 2010 at 1:36 pm Salvatore Stuvard Dear Prof. Tao, I’m sorry to bother you, but I’m working on a graduation thesis about topological dynamical systems, and I have some problems about Lemma 2 of this lecture. I will be grateful if you answer a couple of questions. – First of all, you say about the existence of integers such that for all and , with . Now, if should I expect the existence of two integers or of just one? And in the case , how is it possible to consider , which is undefined? – After having proved it, you apply Lemma 2 to the proof of Theorem 4, to close the induction. But if I set and , by the Lemma I obtain that contains for all , that is an arithmetic progression of lenght and not as I needed to complete the induction step. Where am I getting wrong? Thanks a lot for your helpfulness, and good work. Salvatore Stuvard 0 0 i Rate This Reply 2 October, 2010 at 1:55 pm Terence Tao When J=0 the sequence is empty, and the claim is trivial. If a=b, then is an empty sum and is equal to zero. When i=0, then is trivially contained in , so the length k-1 progression with can be automatically upgraded to a length k progression with . 0 0 i Rate This Reply 4 October, 2010 at 3:18 am Salvatore Stuvard Thank you very much, it was a problem of notation. Now the theorem is very clear. Thank you for the quickness of the answer also… and compliments for your work! Salvatore Stuvard 0 0 i Rate This Reply 11 January, 2013 at 11:02 am cuttheknot Dear Professor, I think in the proof of Proposition 1 we should take such that , as we need to be able to assume when they tend to [Corrected, thanks; -T.] 0 0 i Rate This Reply 19 June, 2013 at 4:07 am Anonymous Shouldn’t in Lemma 1 read ? 0 0 i Rate This Reply 19 June, 2013 at 4:11 am Anonymous Sry, just now realised that is assumed invertible and . 0 0 i Rate This Reply 7 December, 2013 at 4:06 pm Ultraproducts as a Bridge Between Discrete and Continuous Analysis \| What's new […] that Theorem 11 implies Theorem 10. (We will not prove either of these theorems here, but see e.g. this previous blog post for a […] 0 0 i Rate This Reply 26 July, 2018 at 12:13 am domotorp The links (to lectures 2 and 3) around Thm 5 don’t work, because the dates are 01/15 instead of the correct dates. [Corrected, thanks – T.] 0 1 i Rate This Reply 30 November, 2018 at 9:05 am Abhishek Khetan Dear Professor Tao, Thank you for the most interesting post. I have one question. To prove Corollary 1 you define a map as . As you have shown, applying multiple Birkhoff recurrence to this map proves Corollary 1. Can you please say something as to what motivated you to consider the map $T$? Thank you. 1 1 i Rate This Reply 30 November, 2018 at 12:16 pm Terence Tao I am not sure where the connection between Diophantine approximation of polynomials and dynamics of skew shift systems such as this one first arise, but the connection is for instance made in Section 1.7 of Furstenberg’s 1981 book. 2 1 i Rate This Reply 11 June, 2025 at 12:51 pm Anonymous Dear Professor Tao, In the proof of Theorem 3 (Multiple Birkoff Recurrance Theorem), I think there is a typos. After the end of the proof of Lemma 3, in Line 12, I think it should be T^U instead of T^V. Regards, Nazrul [Corrected, thanks – T.] 0 0 i Rate This Reply Leave a comment Cancel reply Δ For commenters To enter in LaTeX in comments, use $latex $ (without the < and > signs, of course; in fact, these signs should be avoided as they can cause formatting errors). Also, backslashes \ need to be doubled as \. See the about page for details and for other commenting policy. « The asymptotic behaviour of large data solutions to NLS On the testability and repair of hereditary hypergraph properties » Blog at WordPress.com.Ben Eastaugh and Chris Sternal-Johnson. Subscribe to feed. Comment Reblog SubscribeSubscribed What's new Join 11,973 other subscribers Sign me up Already have a WordPress.com account? Log in now. What's new SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar %d
3368	https://math.stackexchange.com/questions/4475057/understanding-counting-using-multinomial-coefficients	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 Understanding counting using multinomial coefficients Ask Question Asked Modified 3 years, 3 months ago Viewed 579 times 9 $\begingroup$ I'm studying Chapter 1 of Ross A First Course in Probability Theory (8th Edition) and I'm grappling with multinomial coefficients. All given examples come from this chapter. Specifically $${n \choose n_1 ... n_r}=\frac{n!}{n_1! ... n_r!}$$ gives the number of ways to choose groups of objects of sizes $n_1, ..., n_r$ where $\sum_{i=1}^{i=r} n_i = n$. Then we have the following 3 examples: 10 officers are to be divided as follows. 5 on patrol, 2 at the station and 3 in reserve. In how many ways can this be done? The answer is of course $$\frac{10!}{5!2!3!}$$ 10 kids are to be divided into two teams A and B of size 5 each where each team will play in a separate division. In how many ways can this be done? Again, the answer is similar to what we'd expect $$\frac{10!}{5!5!}$$ 10 kids divide themselves up into two teams of 5 to play basketball at the playground. In how many ways can this be done. This is where my confusion begins. The example says the answer is $$(\frac{10!}{5!5!})/2!$$ because even though this looks like the previous problem, it is different since the order doesn't matter here. Firstly, it looks exactly the same. I do not see why order matters in EITHER of examples #$2$ and #$3$. Secondly, example #$1$ looks exactly like the situation of example #$2$ except with $3$ groups instead of $2$ so, if order mattered in example #$2$, then it should have mattered in example #$1$, no?. So, my question: What am I missing here? Any feedback is much appreciated. probability combinatorics multinomial-coefficients Share edited Jun 18, 2022 at 19:02 RobPratt 51.8k44 gold badges3232 silver badges6969 bronze badges asked Jun 18, 2022 at 2:05 Salazar_3854708Salazar_3854708 81366 silver badges1111 bronze badges $\endgroup$ Add a comment \| 3 Answers 3 Reset to default 9 $\begingroup$ Maybe it's easier to see with a simpler example: 2 kids, Amanda and Ling, are to be divided into two teams A and B of size 1 ... This is counting these combinations separately: Team A: {Amanda}; Team B: {Ling}. Team A: {Ling}; Team B: {Amanda}. Thus we get $$\frac{2!}{1!\ 1!} = 2$$ possibilities. 2 kids, Amanda and Ling, divide themselves up into two teams of 1 to play basketball at the playground This is counting both possibilities above as the same thing. Or in other words, it only counts this possibility: Team: {Amanda}. Team: {Ling}. Thus we get $$\frac{2!}{1!\ 1!} / 2! = 1$$ possibility. Now just replace 2 with 10, and 1 with 5. Share answered Jun 18, 2022 at 11:36 Rebecca J. StonesRebecca J. Stones 27.3k22 gold badges4848 silver badges118118 bronze badges $\endgroup$ 3 $\begingroup$ Hi Rebecca, thank you very much for your response. While it is quite clear, I've seen another question where "12 people are to be divided into 3 groups of sizes 3, 4 and 5" and the answer given is the one that DOES take order into account. But to me, the language doesn't make a distinction between the groups so I think the order shouldn't matter here. Do you agree with my assessment of this? $\endgroup$ Salazar_3854708 – Salazar_3854708 2022-06-18 12:20:23 +00:00 Commented Jun 18, 2022 at 12:20 4 $\begingroup$ I agree, the phrasing isn't fully unambiguous. In practice, you often have to figure out the most likely interpretation of an author's wording (labelled vs. unlabelled, distinguishable vs. indistinguishable). For distinct group sizes (e.g., 3, 4, and 5), it can only mean "distinguishable" since we can distinguish the groups by their size (e.g., we can't swap a team of size 3 with a team of size 4 and consider it the same thing). But for non-distinct group sizes (5 and 5) it could be either depending on the wording of the problem. $\endgroup$ Rebecca J. Stones – Rebecca J. Stones 2022-06-18 12:27:32 +00:00 Commented Jun 18, 2022 at 12:27 $\begingroup$ This is immensely helpful Rebecca. I appreciate the clarification. $\endgroup$ Salazar_3854708 – Salazar_3854708 2022-06-18 13:04:36 +00:00 Commented Jun 18, 2022 at 13:04 Add a comment \| 3 $\begingroup$ The sets patrol, station, reserve in example 1 are clearly distinct. Thus the ordering among these sets matter. The teams A and B in example 2 are clearly distinct, they are given names. It makes a difference whether someone is in Team A or Team B. In example 3, it doesn't matter what you call the teams, so the order doesn't matter. Share answered Jun 18, 2022 at 2:24 kodlukodlu 10.5k22 gold badges1717 silver badges2929 bronze badges $\endgroup$ 0 Add a comment \| 3 $\begingroup$ The distinction is best seen as one between colouring and partitioning a (finite) set (possibly subject to some restrictions). A colouring is simply the association of a colour (presumably from a given palette) to each element. On the other hand a partitioning just defines a notion of "togetherness" or "being on the same team" holding within certain subsets (called classes of the partitioning); technically this is an equivalence relation on the set. There is nothing external to distinguish the classes; there are just its members that can be used to identify a class. So every colouring defines a partitioning (into classes consisting of elements having a common colour), but a partitioning does not determine a single colouring, since going from a partitioning to a colouring we are free to choose a colour to identify each class with (as long as distinct classes get distinct colours). In problems 1 and 2 there are the palettes { patrol, station, reserve } respectively ${A,B}$ that can be considered to define the colours to be used, but in 3. there is no such thing: each team is just defined by its members. It we had say red and blue shirts to identify the teams, then we would have been talking about a colouring, and the answer would be twice as large, but the actual question is about a partitioning (into $2$ classes of size$~5$). Share answered Jun 18, 2022 at 13:38 Marc van LeeuwenMarc van Leeuwen 120k88 gold badges183183 silver badges373373 bronze badges $\endgroup$ 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 probability combinatorics multinomial-coefficients 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 Related Counting Boys and Girls in a Team Multinomial Theorem Example Questions 10 children divide into 2 teams each. 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3369	https://www.youtube.com/watch?v=Hkc-EyHD-M0	"What is the Difference Between Tone and Mood?": A Literary Guide for English Students and Teachers OSU School of Writing, Literature and Film 42200 subscribers 521 likes Description 50831 views Posted: 14 Nov 2022 How do tone and mood differ in literature? What are some examples of mood vs tone in literature? In this short lesson, English Instructor Marcos Norris answers these questions using examples from Dracula and Interview with a Vampire. The video is a collaboration between Dr. Norris and English Instructor Lucia Stone and is designed to help high school and college English students make meaningful arguments about the literature they read. The video is sponsored by the School of Writing, Literature, and Film at Oregon State University. For more discussions of literary topics and essay writing tips, please subscribe to the free SWLF YouTube Channel or visit Spanish subtitles are now available for this video. To access these subtitles, click on the settings icon in the video. Timestamps 0:00 Definitions of Mood and Tone 1:43 Example #1 Dracula 3:55 Example #2 Interview with a Vampire View the full transcript and accompanying lesson here: 32 comments Transcript: Definitions of Mood and Tone Mood and Tone: What’s the Difference? Two ways in which authors communicate with readers is by the use of mood and tone. Although both techniques can elicit particular emotions central to understanding a story, the terms are easily confused. Mood in literature is firmly rooted in the locale or setting of the story that reveals the subject. The physical atmosphere is built scene by scene to create a sense of time, place and reality. Is the world depicted familiar to the reader in its contemporary realism or is it fantastic and reminiscent of the distant past? How does everything look, smell and feel? And, most importantly, what does each scene reveal about the subject at hand? These are some of the questions we can ask to delve deeper into the mood emphasized in each sequence of an unfolding story. Tone, on the other hand, is less sensual play and more the attitude of the characters toward the subject at hand. It is strongly related to the narrator’s point of view, delivered most reliably through choice of words, either explicitly or implicitly. Tone certainly contributes to the mood of a story, but it is less about creating emotional resonance within the readers and more about communicating the narrator’s thoughts or state of mind. Here is another way of understanding the difference between mood and tone: mood shows the subject of the story while tone tells the reader what the characters think of that subject. Example #1 Dracula To illustrate, let’s look at two examples from literature from different eras that share similar themes, DRACULA by Bram Stoker (that’s Dracula, if you didn’t catch it) and Interview with a Vampire by Anne Rice. Vampire literature is a genre in which mood and tone are almost as important as plot and story, so the characters in each novel become conduits for communicating a unique other-worldly atmosphere that can only exist through their perceptions. Dracula is an epistolary novel in which the narration is delivered through a series of journal entries. The mood is set as the scene unfolds with the protagonist Jonathan Harker’s travel from London, England to the Carpathian Mountains in TRANSYLANIA (that’s Transylvania) during the late nineteenth century. The mood first affected is one of disorientation, with the physical contrast drawn sharply between Western and Eastern Europe by the reference to the literal bridges over the Danube leading eastward. Later, the contrast is accentuated as we follow the narrator further into this unfamiliar realm by reading about his first meal, “paprika hendl” (it’s chicken), a dish drawn to be distinct, presumably, from food familiar to the English palette at that time. As in the opening, the mood continues to be shown with sight and taste, with the senses directed toward an unfamiliar scene–and the subject of the novel. To foreshadow the horror to come, the mood is punctuated with the narrator’s attitude about that subject. The tone is one of apprehension and fear as the narrator explicitly tells us about his first night sleeping in a foreign hotel: “I did not sleep well…There was a dog howling all night under my window” and “I had to drink up all the water” but “was still very thirsty”--presumably from the strong, unfamiliar seasoning in the food served the night before. In this case, the protagonist’s tone matches the mood. Example #2 Interview with a Vampire However, sometimes tone and mood are at odds with one another. Interview with a Vampire begins quite literally with the viewpoint of the protagonist, the vampire himself, who languidly opens the novel with, “I see…,” while preparing himself for an interview with a young journalist. His attitude, or tone, is one of quiet ease. His tone matches the mood, which is set by a rather unexotic backdrop of a cityscape through the window of an ordinary hotel room. The dialogue bounces between vampire and journalist, monster and human, while the mood of prosaic reality is revealed in the simple details of a chair, table and recording device. The tonal horror necessary for the tension to unfold is projected by the very different attitude of the journalist toward the scene: the readers are told that the interviewer “shuddered” and “recoiled” with “cold sweat running down the side of his face.” In other words, reason meets emotion in this clash between tone and mood. A sharp contrast is drawn between the attitude of the two protagonists toward the scene, and the audience is sucked right in. (You know, because vampires, they suck blood. THEY SUCK BLOOD!) For the student of literature, such moments of tension are exciting and revelatory. Mood shows the particular scenes that direct us toward the subject of a story, but tone tells what each character actually thinks of that subject. Both are necessary devices to make a world come alive on the page or on the screen.
3370	https://www.history.com/articles/fertile-crescent	Home Shows This Day in History U.S. U.S. History U.S. History All the major chapters in the American story, from Indigenous beginnings to the present day. 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Inventions & Science Inventions & Science Natural Disasters & Environment Natural Disasters & Environment Space Exploration Space Exploration Archaeology Archaeology HISTORY Honors 250 By: HISTORY.com Editors Fertile Crescent HISTORY.com Editors PHAS/Universal Images Group via Getty Images Published: December 20, 2017 Last Updated: February 27, 2025 Print Copy Table of contents Table of contents 1 What Is the Fertile Crescent? 2 Ancient Mesopotamia 3 Sumerians 4 Important Archaeological Sites 5 Fertile Crescent Today 6 SOURCES Table of contents Table of contents The Fertile Crescent is the boomerang-shaped region of the Middle East that was home to some of the earliest human civilizations. Also known as the “Cradle of Civilization,” this area was the birthplace of a number of technological innovations, including writing, the wheel, agriculture and the use of irrigation. The Fertile Crescent includes ancient Mesopotamia. What Is the Fertile Crescent? American archaeologist James Henry Breasted coined the term “Fertile Crescent” in a 1914 high school textbook to describe this archaeologically significant region of the Middle East that contains parts of present day Egypt, Jordan, Lebanon, Palestine, Israel, Syria, Turkey, Iran, Iraq and Cyprus. On a map, the Fertile Crescent looks like a crescent or quarter-moon. It extends from the Nile River on Egypt’s Sinai Peninsula in the south to the southern fringe of Turkey in the north. The Fertile Crescent is bounded on the west by the Mediterranean Sea and on the East by the Persian Gulf. The Tigris and Euphrates rivers flow through the heart of the Fertile Crescent. The region historically contained unusually fertile soil and productive freshwater and brackish wetlands. These produced an abundance of wild edible plant species. It was here that humans began to experiment with the cultivation of grains and cereals around 10,000 B.C. as they transitioned from hunter-gatherer groups to permanent agricultural societies. The Origins of Writing The invention of written language replaced the oral tradition and allowed civilizations to store and share knowledge. Ancient Mesopotamia Mesopotamia is an ancient, historical region that lies between the Tigris and Euphrates rivers in modern-day Iraq and parts of Kuwait, Syria, Turkey and Iran. Part of the Fertile Crescent, Mesopotamia was home to the earliest known human civilizations. Scholars believe the Agricultural Revolution started here. The earliest occupants of Mesopotamia lived in circular dwellings made of mud and brick along the upper reaches of the Tigris and Euphrates river valleys. They began to practice agriculture by domesticating sheep and pigs around 11,000 to 9,000 B.C. Domesticated plants, including flax, wheat, barley and lentils, first appeared around 9,500 B.C. Some of the earliest evidence of farming comes from the archaeological site of Tell Abu Hureyra, a small village located along the Euphrates River in modern Syria. The village was inhabited from roughly 11,500 to 7,000 B.C. Inhabitants initially hunted gazelle and other game before beginning to harvest wild grains around 9,700 BCE. Several large stone tools for grinding grain have been found at the site. One of the oldest known Mesopotamian cities, Nineveh (near Mosul in modern Iraq), may have been settled as early as 6,000 B.C. Sumer civilization arose in the lower Tigris-Euphrates valley around 5,000 B.C. In addition to farming and cities, ancient Mesopotamian societies developed irrigation and aqueducts, temples, pottery, early systems of banking and credit, property ownership and the first codes of law. Sumerians The origins of Sumer civilization are debated, but archaeologists suggest Sumerians had established roughly a dozen city-states by the fourth millennium B.C., including Eridu and Uruk in what is now southern Iraq. Sumer is the earliest known civilization in ancient Mesopotamia and may have been the first human civilization anywhere in the world. They called themselves the Sag-giga, the “black-headed ones.” Ancient Sumerians were among the first to use bronze. They pioneered the use of levees and canals for irrigation. Sumerians invented cuneiform script, one of the earliest forms of writing. They also built large stepped pyramids called ziggurats. Sumerians celebrated art and literature. The 3,000-line poem, the Epic of Gilgamesh, follows the adventures of a Sumer king as he battles a forest monster and quests after the secrets of eternal life. Important Archaeological Sites British and French archaeologists began exploring the Fertile Crescent for the remains of storied Mesopotamian cities such as Assyria and Babylonia as early as the mid-1800s. Some of the most famous Mesopotamian archaeological sites include: Ziggurat of Ur: It’s an enormous temple in southern Iraq and one of the best remaining examples of Sumerian architecture. Archaeologists think it was built around 2100 B.C. Babylon: Founded nearly 5,000 years ago on the Euphrates River in present-day Iraq, this ancient metropolis and Biblical city was the last major power in Mesopotamia to fall under Persian control in 539 B.C. Hattusha: This UNESCO World Heritage site is one of Turkey’s greatest ruins and was once the capital of the Hittite Empire, which reached its peak in the second millennium B.C. Persepolis: An ancient Mesopotamian city in southern Iran, Persepolis ranks among the world’s greatest archaeological sites with a large number of architecturally significant Persian buildings. Fertile Crescent Today Today the Fertile Crescent is not so fertile: Beginning in the 1950s, a series of large-scale irrigation projects diverted water away from the famed Mesopotamian marshes of the Tigris-Euphrates river system, causing them to dry up. In 1991, the government of Saddam Hussein built a series of dikes and dams to further drain the Iraqi marshes and punish dissident Marsh Arabs who made a living cultivating rice and raising water buffalo there. NASA satellite images showed that that by 1992 roughly 90 percent of the marshland had disappeared, turning more than a thousand square miles into desert. More than 200,000 Marsh Arabs lost their homes. Many of the Hussein-era dams have since been removed, though the wetlands remain only about half of their pre-drained level. SOURCES Where is the Fertile Crescent?; Wonderopolis. The World’s First Farmers Were Surprisingly Diverse; Science. The Crimes of Saddam Hussein; PBS Frontline. From Egypt to Greece, explore fascinating documentaries about the ancient world. 7-DAY FREE TRIAL Commercial-free, Cancel anytime Exclusions & terms apply Related Articles Related Articles Canadian T. Rex Is Officially the Biggest Ever The massive dinosaur also lived longer than any other T. rex discovered to date. When Did Humans Start Waging Wars? Organized warfare appears to have started in the Neolithic Age and then ramped up during the Bronze Age. The Mystery of Siberia's 'Ice Maiden' Mummy A 2,500-year-old mummy of a tattooed woman in a silk blouse was found in the remote Altai region of Central Asia. Who was she—and what does she reveal about her ancient people? How Early Humans First Reached the Americas: 3 Theories Did humans first set foot in the Americas after walking—or sailing or paddling by sea? 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3371	https://oeis.org/wiki/Pascal_triangle	This site is supported by donations to The OEIS Foundation. Pascal triangle From OeisWiki This is the latest approved revision, approved on 28 April 2011. The draft has 87 changes awaiting review.(+) Readability: Reviewed Jump to: navigation, search Pascal's triangle is a geometric arrangement of numbers produced recursively which generates the binomial coefficients. It is named after the French mathematician Blaise Pascal (who studied it in the 17th century) in much of the Western world, although other mathematicians studied it centuries before him in Italy, India, Persia, and China. The triangle is thus known by other names, such as Tartaglia's triangle in Italy and much earlier (c. 500 BC) as the Yanghui triangle in China. : : The rectangular version of Pascal's triangle(Figurate Number Triangle) \| = 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| 5 \| 1 \| 5 \| 10 \| 10 \| 5 \| 1 \| \| 6 \| 1 \| 6 \| 15 \| 20 \| 15 \| 6 \| 1 \| \| 7 \| 1 \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 1 \| \| 8 \| 1 \| 8 \| 28 \| 56 \| 70 \| 56 \| 28 \| 8 \| 1 \| \| 9 \| 1 \| 9 \| 36 \| 84 \| 126 \| 126 \| 84 \| 36 \| 9 \| 1 \| \| 10 \| 1 \| 10 \| 45 \| 120 \| 210 \| 252 \| 210 \| 120 \| 45 \| 10 \| 1 \| \| 11 \| 1 \| 11 \| 55 \| 165 \| 330 \| 462 \| 462 \| 330 \| 165 \| 55 \| 11 \| 1 \| \| 12 \| 1 \| 12 \| 66 \| 220 \| 495 \| 792 \| 924 \| 792 \| 495 \| 220 \| 66 \| 12 \| 1 \| \| \| = 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| In the equilateral version of Pascal's triangle, we start with a cell (row 0) initialized to 1 in a staggered array of empty (0) cells. We then recursively evaluate the cells as the sum of the two staggered above. The triangle thus grows into an equilateral triangle. In the rectangular version of Pascal's triangle, we start with a cell (row 0) initialized to 1 in a regular array of empty (0) cells. We then recursively evaluate the cells as the sum of the one above left and the one directly above. The triangle thus grows into a rectangular triangle. The outermost nonzero cells on each rows are therefore set to 1. All the interior cells are necessarily greater than or equal to 2 and the number of cells from rows 0 to which are equal to 1 is (Cf. A005408) and the number of cells from rows 0 to which are greater than or equal to 2 is , the th triangular number. Contents 1 Recursion rule 2 Pascal's triangle and binomial coefficients 3 Pascal's triangle rows 3.1 Pascal's triangle rows sums 3.2 Pascal's triangle rows alternating sign sums 3.3 Pascal's triangle rows and Schläfli's (n-1)-dimensional polytopic formula 3.4 Pascal's triangle rows and the number of (d-1)-dimensional elements of the (n-1)-dimensional simplex 3.5 Pascal's triangle rows and primes (and prime powers?) 4 Pascal's (rectangular) triangle columns (or falling diagonals, due to symmetry) and simplicial polytopic numbers 4.1 Pascal's (rectangular) triangle third column (d = 2) or falling diagonal from the right (f = 2) and square numbers 5 Pascal's (rectangular) triangle rising diagonals and Fibonacci numbers 6 Pascal's triangle central elements 7 Generalizations 8 See also 9 Notes 10 External links Recursion rule Pascal's triangle recursion rule is or equivalently, using binomial coefficient notation Pascal's triangle and binomial coefficients Pascal's triangle is a table of binomial coefficients, i.e. the coefficients of the expanded binomial : which is the generating function for the th row (finite sequence) of Pascal's triangle. Pascal's triangle rows : : \| \| \| --- \| \| = 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| 5 \| 1 \| 5 \| 10 \| 10 \| 5 \| 1 \| \| 6 \| 1 \| 6 \| 15 \| 20 \| 15 \| 6 \| 1 \| \| 7 \| 1 \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 1 \| \| 8 \| 1 \| 8 \| 28 \| 56 \| 70 \| 56 \| 28 \| 8 \| 1 \| \| 9 \| 1 \| 9 \| 36 \| 84 \| 126 \| 126 \| 84 \| 36 \| 9 \| 1 \| \| 10 \| 1 \| 10 \| 45 \| 120 \| 210 \| 252 \| 210 \| 120 \| 45 \| 10 \| 1 \| \| 11 \| 1 \| 11 \| 55 \| 165 \| 330 \| 462 \| 462 \| 330 \| 165 \| 55 \| 11 \| 1 \| \| 12 \| 1 \| 12 \| 66 \| 220 \| 495 \| 792 \| 924 \| 792 \| 495 \| 220 \| 66 \| 12 \| 1 \| \| \| = 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| Pascal's triangle rows give an infinite sequence of finite sequences : {{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}, {1, 5, 10, 10, 5, 1}, {1, 6, 15, 20, 15, 6, 1}, {1, 7, 21, 35, 35, 21, 7, 1}, {1, 8, 28, 56, 70, 56, 28, 8, 1}, {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}, ...} The generating function for the th, , member of the th, , subsequence is The concatenation of the infinite sequence of finite sequences gives the infinite sequence (Cf. A007318) : {1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, ...} The generating function for the th, , member is Pascal's triangle rows sums The sums of the respective finite sequences give the infinite sequence (Cf. A000079) : {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, ...} The sum for the th row gives the powers of 2, which corresponds to evaluated at The generating function is The partial sums across rows of Pascal's triangle, i.e. partial sums of binomial coefficients, give Bernoulli's triangle. Pascal's triangle rows alternating sign sums The alternating sign sums of the respective finite sequences give the infinite sequence (Cf. A000007) : {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...} The alternating sign sum for the th row gives the powers of 0, (which equals 1 for and 0 for which corresponds to evaluated at . The generating function is Pascal's triangle rows and Schläfli's (n-1)-dimensional polytopic formula Schläfli's -dimensional polytopic formula (for convex polytopes of genus 0) is a generalization of the Descartes-Euler polyhedral formula (for convex polyhedrons of genus 0) to dimensions higher than 3. The alternate sum for row of the interior numbers equals the Euler-Poincaré characteristic for convex -dimensional polytopes of genus 0, e.g. which is 0 for even and 2 for odd. If we consider (the one way of choosing the empty vertex set) and (the one way of choosing the full vertex set, which is the polytope itself) then we get thus showing that is simply the result of not counting the one way of choosing the empty vertex set and the one way of choosing the full vertex set. Pascal's triangle rows and the number of (d-1)-dimensional elements of the (n-1)-dimensional simplex The number of -dimensional elements of the -dimensional simplex is where 0-dimensional elements are points, 1-dimensional elements are edges, 2-dimensional elements are faces, ... Pascal's triangle rows and primes (and prime powers?) is prime iff the GCD of all the interior cells of the th row is . Actually, it seems that is a prime power , iff the GCD of all the interior cells of the th row is , else it is 1. (THIS NEED TO BE CONFIRMED...) GCD of all the interior cells of the th row for 2 to 12 : {2, 3, 2, 5, 1, 7, 2, 3, 1, 11, 1, ...} (Cf. A014963) Pascal's (rectangular) triangle columns (or falling diagonals, due to symmetry) and simplicial polytopic numbers : : (Symplicial Polytopic) Figurate Number Triangle \| = 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| 5 \| 1 \| 5 \| 10 \| 10 \| 5 \| 1 \| \| 6 \| 1 \| 6 \| 15 \| 20 \| 15 \| 6 \| 1 \| \| 7 \| 1 \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 1 \| \| 8 \| 1 \| 8 \| 28 \| 56 \| 70 \| 56 \| 28 \| 8 \| 1 \| \| 9 \| 1 \| 9 \| 36 \| 84 \| 126 \| 126 \| 84 \| 36 \| 9 \| 1 \| \| 10 \| 1 \| 10 \| 45 \| 120 \| 210 \| 252 \| 210 \| 120 \| 45 \| 10 \| 1 \| \| 11 \| 1 \| 11 \| 55 \| 165 \| 330 \| 462 \| 462 \| 330 \| 165 \| 55 \| 11 \| 1 \| \| 12 \| 1 \| 12 \| 66 \| 220 \| 495 \| 792 \| 924 \| 792 \| 495 \| 220 \| 66 \| 12 \| 1 \| \| \| = 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| The partial sums of the th column, , build the entries of the th column, thus begetting the -dimensional simplicial polytopic numbers from the -dimensional ones The partial sums of the th falling diagonal from the right, , build the entries of the th falling diagonal, thus begetting the -dimensional simplicial polytopic numbers from the -dimensional ones The th column gives the -dimensional (the th falling diagonal from the right gives the -dimensional) simplicial polytopic numbers, forming simplex polytopes, e.g. : : : \| \| \| \| \| \| --- --- \| d=0, f=0 \| 0-dimensional simplicial numbers \| Point numbers \| (1 (-1)-cells facets) \| (0-simplex) \| \| d=1, f=1 \| 1-dimensional simplicial numbers \| Triangular gnomonic numbers \| (2 0-cells facets) \| (1-simplex) \| \| d=2, f=2 \| 2-dimensional simplicial numbers \| Triangular numbers \| (3 1-cells facets) \| (2-simplex) \| \| d=3, f=3 \| 3-dimensional simplicial numbers \| Tetrahedral numbers \| (4 2-cells facets) \| (3-simplex) \| \| d=4, f=4 \| 4-dimensional simplicial numbers \| Pentachoron numbers \| (5 3-cells facets) \| (4-simplex) \| \| d=5, f=5 \| 5-dimensional simplicial numbers \| Hexateron numbers \| (6 4-cells facets) \| (5-simplex) \| \| d=6, f=6 \| 6-dimensional simplicial numbers \| Heptapeton numbers \| (7 5-cells facets) \| (6-simplex) \| \| d=7, f=7 \| 7-dimensional simplicial numbers \| Octahexon numbers \| (8 6-cells facets) \| (7-simplex) \| \| d=8, f=8 \| 8-dimensional simplicial numbers \| Nonahepton numbers \| (9 7-cells facets) \| (8-simplex) \| where (-1)-cells correspond to the empty set, 0-cells are vertices, 1-cells are edges, 2-cells are faces, and so on... Pascal's (rectangular) triangle third column (d = 2) or falling diagonal from the right (f = 2) and square numbers In the 3rd column (,) the sum of 2 stacked cells gives the square numbers Similarly, in the 3rd falling diagonal from the right (,) the sum of 2 stacked cells gives the square numbers Pascal's (rectangular) triangle rising diagonals and Fibonacci numbers : : \| \| \| --- \| \| = 0 \| 1 \| \| 1 \| 1 \| 1 \| \| 2 \| 1 \| 2 \| 1 \| \| 3 \| 1 \| 3 \| 3 \| 1 \| \| 4 \| 1 \| 4 \| 6 \| 4 \| 1 \| \| 5 \| 1 \| 5 \| 10 \| 10 \| 5 \| 1 \| \| 6 \| 1 \| 6 \| 15 \| 20 \| 15 \| 6 \| 1 \| \| 7 \| 1 \| 7 \| 21 \| 35 \| 35 \| 21 \| 7 \| 1 \| \| 8 \| 1 \| 8 \| 28 \| 56 \| 70 \| 56 \| 28 \| 8 \| 1 \| \| 9 \| 1 \| 9 \| 36 \| 84 \| 126 \| 126 \| 84 \| 36 \| 9 \| 1 \| \| 10 \| 1 \| 10 \| 45 \| 120 \| 210 \| 252 \| 210 \| 120 \| 45 \| 10 \| 1 \| \| 11 \| 1 \| 11 \| 55 \| 165 \| 330 \| 462 \| 462 \| 330 \| 165 \| 55 \| 11 \| 1 \| \| 12 \| 1 \| 12 \| 66 \| 220 \| 495 \| 792 \| 924 \| 792 \| 495 \| 220 \| 66 \| 12 \| 1 \| \| \| = 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| The rising diagonals (starting with the 0th diagonal) give an infinite sequence of finite sequences : {{1}, {1}, {1, 1}, {1, 2}, {1, 3, 1}, {1, 4, 3}, {1, 5, 6, 1}, {1, 6, 10, 4}, {1, 7, 15, 10, 1}, {1, 8, 21, 20, 5}, ... } whose respective sums give : {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, ...}. which are the th Fibonacci numbers (Cf. A000045) The concatenated infinite sequence of finite sequences gives the infinite sequence (cf. A011973) : {1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, ...} Pascal's triangle central elements The central (or almost/quasi central for odd) elements give the sequence (Cf. A001405) : {1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, ...} which is given by the formulae where is the th, , Catalan number (also called Segner number) (Cf. A000108) : {1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, ...} The generating function is where is the generating function of the Catalan numbers Generalizations Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices. See also A007188 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i). A003590 Rows written as a single base 10 number (the first five terms of that sequence match powers of 11; in general we can say that the first b/2 rows written as a single number give the powers of b + 1 in base b. A006046 Total number of odd entries in first n rows of Pascal's triangle. A003015 Numbers that appear five or more times in Pascal's triangle (at this point it's not known whether any terms appear exactly five times.) (1,1)-Pascal triangle or Pascal's triangle (1,2)-Pascal triangle or Lucas triangle (1,k)-Pascal triangle (2,1)-Pascal triangle (k,1)-Pascal triangle (a,b)-Pascal triangle (a(n),b(n))-Pascal triangle Bernoulli's triangle Pascal's pyramid or Pascal's tetrahedron Pascal's simplices Notes ↑ 1.0 1.1 1.2 Eric W. Weisstein, Binomial Coefficient, from MathWorld.. ↑ 2.0 2.1 Eric W. Weisstein, Figurate Number Triangle, from MathWorld.. ↑ www.vaxasoftware.com, Newton's binomial theorem and Pascal-Tartaglia's triangle. ↑ Eric W. Weisstein, Polyhedral Formula, from MathWorld.. ↑ Eric W. Weisstein, Simplex, from MathWorld.. External links Eric W. Weisstein, Pascal's Triangle, from MathWorld.. Pascal, Blaise, Picture of Pascal's triangle, Traité du triangle arithmétique, 1665. Retrieved from " Categories: Pascal triangle (1,1)-Pascal triangle Navigation menu Views Page Discussion View Pending changes View source History Personal tools Log in Request account Navigation OEIS Wiki Main Page Community portal System Status Recent changes Random page Help Tools What links here Related changes Special pages Printable version Permanent link Page information Content is available under The OEIS End-User License Agreement unless otherwise noted. 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3372	https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(Stitz-Zeager)/01%3A_Relations_and_Functions/1.6%3A_Graphs_of_Functions	Published Time: 2022-03-11T23:34:13Z 1.6: Graphs of Functions - Mathematics 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 1: Relations and Functions Precalculus (Stitz-Zeager) { } { "1.01:_Sets_of_Real_Numbers_and_the_Cartesian_Coordinate_Plane" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_Relations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Introduction_to_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.04:_Function_Notation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.05:_Function_Arithmetic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.07:_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.6:_Graphs_of_Functions" : "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:_Relations_and_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Linear_and_Quadratic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Polynomial_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Rational_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Further_Topics_in_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Exponential_and_Logarithmic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Hooked_on_Conics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Systems_of_Equations_and_Matrices" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Sequences_and_the_Binomial_Theorem" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Foundations_of_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Applications_of_Trigonometry" : "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" } Mon, 03 Oct 2022 02:49:47 GMT 1.6: Graphs of Functions 98530 98530 admin { } Anonymous Anonymous 2 false false [ "article:topic" ] [ "article:topic" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Precalculus & Trigonometry 4. Precalculus (Stitz-Zeager) 5. 1: Relations and Functions 6. 1.6: Graphs of Functions Expand/collapse global location Precalculus (Stitz-Zeager) Front Matter 1: Relations and Functions 2: Linear and Quadratic Functions 3: Polynomial Functions 4: Rational Functions 5: Further Topics in Functions 6: Exponential and Logarithmic Functions 7: Hooked on Conics 8: Systems of Equations and Matrices 9: Sequences and the Binomial Theorem 10: Foundations of Trigonometry 11: Applications of Trigonometry Back Matter 1.6: Graphs of Functions Last updated Oct 3, 2022 Save as PDF 1.7: Transformations 2: Linear and Quadratic Functions picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate Page ID 98530 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. The Fundamental Graphing Principle for Functions 2. Example 1.6.1 1. Solution Example 1.6.2 Solution Definition: 1.9 Testing the Graph of a Function for Symmetry Example 1.6.3 Solution Definition: 1.10 Definition: 1.11 Example 1.6.4 Solution Example 1.6.5 Solution Example 1.6.6 Solution 1.6.2 Exercises 1.6.3 Answers Reference In Section 1.3 we defined a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Then in Section 1.4 we described a function as a process and defined the notation necessary to work with functions algebraically. So now it’s time to look at functions graphically again, only this time we’ll do so with the notation defined in Section 1.4. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions The graph of a function f f is the set of points which satisfy the equation y=f(x)y=f(x). That is, the point (x,y)(x,y) is on the graph of f f if and only if y=f(x)y=f(x). Example 1.6.1 Graph f(x)=x 2−x−6 f(x)=x 2−x−6. Solution To graph f f, we graph the equation y=f(x)y=f(x). To this end, we use the techniques outlined in Section 1.2.1. Specifically, we check for intercepts, test for symmetry, and plot additional points as needed. To find the x-intercepts, we set y=0 y=0. Since y=f(x)y=f(x), this means f(x)=0 f(x)=0. f(x)0 0 x−3=0 x=x 2−x−6=x 2−x−6=(x−3)(x+2)factor or x+2=0=−2,3 f(x)=x 2−x−6 0=x 2−x−6 0=(x−3)(x+2)factor x−3=0 or x+2=0 x=−2,3 So we get (−2, 0) and (3, 0) as x-intercepts. To find the y-intercept, we set x=0 x=0. Using function notation, this is the same as finding f(0)f(0) and f(0)=0 2−0−6=−6 f(0)=0 2−0−6=−6. Thus the y-intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.2.1, plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. Graphing piecewise-defined functions is a bit more of a challenge. Example 1.6.2 Graph: f(x)={4−x 2 x−3,if if x<1 x≥1 f(x)={4−x 2 if x<1 x−3,if x≥1 Solution We proceed as before – finding intercepts, testing for symmetry and then plotting additional points as needed. To find the x-intercepts, as before, we set f(x)=0 f(x)=0. The twist is that we have two formulas for f(x)f(x). For x<1 x<1, we use the formula f(x)=4−x 2 f(x)=4−x 2. Setting f(x)=0 f(x)=0 gives 0=4−x 2 0=4−x 2, so that x=±2 x=±2. However, of these two answers, only x=−2 x=−2 fits in the domain x<1 x<1 for this piece. This means the only x-intercept for the x<1 x<1 region of the x-axis is (−2, 0). For x≥1 x≥1, f(x)=x−3 f(x)=x−3. Setting f(x)=0 f(x)=0 gives 0=x−3 0=x−3, or x=3 x=3. Since x=3 x=3 satisfies the inequality x≥1 x≥1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x=0 x=0 falls in the domain x<1 x<1. Thus f(0)=4−0 2=4 f(0)=4−0 2=4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y=4−x 2 y=4−x 2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x<1 x<1. You can also verify y=x−3 y=x−3 possesses none of the symmetries discussed in the Section 1.2.1. When plotting additional points, it is important to keep in mind the restrictions on x x for each piece of the function. The sticking point for this function is x=1 x=1, since this is where the equations change. When x=1 x=1, we use the formula f(x)=x−3 f(x)=x−3, so the point on the graph (1,f(1))(1,f(1)) is (1,−2)(1,−2). However, for all values less than 1, we use the formula f(x)=4−x 2 f(x)=4−x 2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x=1 x=1 on the number line. Thus for the values x=0.9 x=0.9, x=0.99 x=0.99, x=0.999 x=0.999, and so on, we find the corresponding y values using the formula f(x)=4−x 2 f(x)=4−x 2. Making a table as before, we see that as the x x values sneak up to x=1 x=1 in this fashion, the f(x)f(x) values inch closer and closer1 to 4−1 2=3 4−1 2=3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get In the previous two examples, the x-coordinates of the x-intercepts of the graph of y=f(x)y=f(x) were found by solving f(x)=0 f(x)=0. For this reason, they are called the zeros of f f. Definition: 1.9 The zeros of a function f f are the solutions to the equation f(x)=0 f(x)=0. In other words, x x is a zero of f f if and only if (x,0)(x,0) is an x-intercept of the graph of y=f(x)y=f(x). Of the three symmetries discussed in Section 1.2.1, only two are of significance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether the graph of an equation is symmetric about the y-axis by replacing x x with −x−x and checking to see if an equivalent equation results. If we are graphing the equation y=f(x)y=f(x), substituting −x−x for x x results in the equation y=f(−x)y=f(−x). In order for this equation to be equivalent to the original equation y=f(x)y=f(x) we need f(−x)=f(x)f(−x)=f(x). In a similar fashion, we recall that to test an equation’s graph for symmetry about the origin, we replace x and y with −x−x and −y−y, respectively. Doing this substitution in the equation y=f(x)y=f(x) results in −y=f(−x)−y=f(−x). Solving the latter equation for y y gives y=−f(−x)y=−f(−x). In order for this equation to be equivalent to the original equation y=f(x)y=f(x) we need −f(−x)=f(x)−f(−x)=f(x), or, equivalently, f(−x)=−f(x)f(−x)=−f(x). These results are summarized below. Testing the Graph of a Function for Symmetry The graph of a function f f is symmetric about the y-axis if and only if f(−x)=f(x)f(−x)=f(x) for all x x in the domain of f f. about the origin if and only if f(−x)=−f(x)f(−x)=−f(x) for all x x in the domain of f f. For reasons which won’t become clear until we study polynomials, we call a function even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin. Apart from a very specialized family of functions which are both even and odd,3 functions fall into one of three distinct categories: even, odd, or neither even nor odd. Example 1.6.3 Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. f(x)=5 2−x 2 f(x)=5 2−x 2 g(x)=5 x 2−x 2 g(x)=5 x 2−x 2 h(x)=5 x 2−x 3 h(x)=5 x 2−x 3 i(x)=5 x 2 x−x 3 i(x)=5 x 2 x−x 3 j(x)=x 2−x 100−1 j(x)=x 2−x 100−1 p(x)={x+3−x+3,if if x<0 x≥0 p(x)={x+3 if x<0−x+3,if x≥0 Solution The first step in all of these problems is to replace x x with −x−x and simplify. f(x)f(−x)f(−x)f(−x)=5 2−x 2=5 2−(−x)2=5 2−x 2=f(x)f(x)=5 2−x 2 f(−x)=5 2−(−x)2 f(−x)=5 2−x 2 f(−x)=f(x) Hence, f f is even. The graphing calculator furnishes the following. This suggests4 that the graph of f f is symmetric about the y-axis, as expected. g(x)g(−x)g(−x)=5 x 2−x 2=5(−x)2−(−x)2=−5 x 2−x 2 g(x)=5 x 2−x 2 g(−x)=5(−x)2−(−x)2 g(−x)=−5 x 2−x 2 It doesn’t appear that g(−x)g(−x) is equivalent to g(x)g(x). To prove this, we check with an x x value. After some trial and error, we see that g(1)=5 g(1)=5 whereas g(−1)=−5 g(−1)=−5. This proves that g g is not even, but it doesn’t rule out the possibility that g g is odd. (Why not?) To check if g g is odd, we compare g(−x)g(−x) with −g(x)−g(x) −g(x)−g(x)=−5 x 2−x 2=−5 x 2−x 2=g(−x)−g(x)=−5 x 2−x 2=−5 x 2−x 2−g(x)=g(−x) Hence, g g is odd. Graphically, The calculator indicates the graph of g g is symmetric about the origin, as expected. h(x)h(−x)h(−x)=5 x 2−x 3=5(−x)2−(−x)3=−5 x 2+x 3 h(x)=5 x 2−x 3 h(−x)=5(−x)2−(−x)3 h(−x)=−5 x 2+x 3 Once again, h(−x)h(−x) doesn’t appear to be equivalent to h(x)h(x). We check with an x x value, for example, h(1)=5 h(1)=5 but h(−1)=−5 3 h(−1)=−5 3. This proves that h h is not even and it also shows h h is not odd. (Why?) Graphically, The graph of h h appears to be neither symmetric about the y-axis nor the origin. i(x)i(−x)i(−x)=5 x 2 x−x 3=5(−x)2(−x)−(−x)3=−5 x−2 x+x 3 i(x)=5 x 2 x−x 3 i(−x)=5(−x)2(−x)−(−x)3 i(−x)=−5 x−2 x+x 3 The expression i(−x)i(−x) doesn’t appear to be equivalent to i(x)i(x). However, after checking some x values, for example x=1 x=1 yields i(1)=5 i(1)=5 and i(−1)=5 i(−1)=5, it appears that i(−x)i(−x) does, in fact, equal i(x)i(x). However, while this suggests i i is even, it doesn’t prove it. (It does, however, prove i i is not odd.) To prove i(−x)=i(x)i(−x)=i(x), we need to manipulate our expressions for i(x)i(x) and i(−x)i(−x) and show that they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x)i(x), the numerator is 5 x 5 x and in i(−x)i(−x) the numerator is −5 x−5 x. To re-write i(x)i(x) with a numerator of −5 x−5 x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x)=5 x 2 x−x 3=(−1)5 x(−1)(2 x−x 3)=−5 x−2 x+x 3 i(x)=5 x 2 x−x 3=(−1)5 x(−1)(2 x−x 3)=−5 x−2 x+x 3 Hence, i(x)=i(−x)i(x)=i(−x), so i i is even. The calculator supports our conclusion. j(x)j(−x)j(−x)=x 2−x 100−1=(−x)2−−x 100−1=x 2+x 100−1 j(x)=x 2−x 100−1 j(−x)=(−x)2−−x 100−1 j(−x)=x 2+x 100−1 The expression for j(−x)j(−x) doesn’t seem to be equivalent to j(x)j(x), so we check using x=1 x=1 to get j(1)=−1 100 j(1)=−1 100 and j(−1)=1 100 j(−1)=1 100. This rules out j j being even. However, it doesn’t rule out j j being odd. Examining −j(x)−j(x) gives j(x)−j(x)−j(x)=x 2−x 100−1=−(x 2−x 100−1)=−x 2+x 100+1 j(x)=x 2−x 100−1−j(x)=−(x 2−x 100−1)−j(x)=−x 2+x 100+1 The expression −j(x)−j(x) doesn’t seem to match j(−x)j(−x) either. Testing x=2 x=2 gives j(2)=149 50 j(2)=149 50 and j(−2)=151 50 j(−2)=151 50, so j j is not odd, either. The calculator gives: The calculator suggests that the graph of j j is symmetric about the y-axis which would imply that j j is even. However, we have proven that is not the case. Testing the graph of y=p(x)y=p(x) for symmetry is complicated by the fact p(x)p(x) is a piecewise defined function. As always, we handle this by checking the condition for symmetry by checking it on each piece of the domain. We first consider the case when x<0 x<0 and set about finding the correct expression for p(−x)p(−x). Even though p(x)=x+3 p(x)=x+3 for x<0 x<0, p(−x)≠−x+3 p(−x)≠−x+3 here. The reason for this is that since x<0 x<0, −x>0−x>0 which means to find p(−x)p(−x), we need to use the other formula for p(x)p(x), namely p(x)=−x+3 p(x)=−x+3. Hence, for x<0 x<0, p(−x)=−(−x)+3=x+3=p(x)p(−x)=−(−x)+3=x+3=p(x). For x≥0 x≥0, p(x)=−x+3 p(x)=−x+3 and we have two cases. If x>0 x>0, then −x<0−x<0 so p(−x)=(−x)+3=−x+3=p(x)p(−x)=(−x)+3=−x+3=p(x). If x=0 x=0, then p(0)=3=p(−0)p(0)=3=p(−0). Hence, in all cases, p(−x)=p(x)p(−x)=p(x), so p p is even. Since p(0)=3 p(0)=3 but p(−0)=p(0)=3≠−3 p(−0)=p(0)=3≠−3, we also have p p is not odd. While graphing y=p(x)y=p(x) is not onerous to do by hand, it is instructive to see how to enter this into our calculator. By using some of the logical commands,5 we have: The calculator bears shows that the graph appears to be symmetric about the y-axis. There are two lessons to be learned from the last example. The first is that sampling function values at particular x values is not enough to prove that a function is even or odd − despite the fact that j(−1)=−j(1)j(−1)=−j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the Algebra which proves mathematical truths.6 The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them all in great detail. The purpose of this section’s discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing and constant. Before defining the concepts algebraically, it is instructive to first look at them graphically. Consider the graph of the function f f below. Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point (6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we are walking ‘uphill’; then between (−2, 4.5) and (3, −8), we are walking ‘downhill’; and between (3, −8) and (4, −6), we are walking ‘uphill’ once more. From (4, −6) to (5, −6), we ‘level off’, and then resume walking ‘uphill’ from (5, −6) to (6, 5.5). In other words, for the x values between −4 and −2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y=f(x)y=f(x), the y values on the graph are the function values, and we say that the function f f is increasing on the interval [−4, −2]. Analogously, we say that f f is decreasing on the interval [−2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and finally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f f is increasing on [−4, −2] we do not mention the actual y y values that f f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs.7 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run into serious trouble in our previous example if we tried to do so because x=−2 x=−2 is contained in an interval on which f f was increasing and one on which it is decreasing. (There’s more on this issue – and many others – in the Exercises.) We’re now ready for the more formal algebraic definitions of what it means for a function to be increasing, decreasing or constant. Definition: 1.10 Suppose f f is a function defined on an interval I I. We say f f is: increasing on I I if and only if f(a)<f(b)f(a)<f(b) for all real numbers a a, b b in I I with a<b a<b. decreasing on I I if and only if f(a)>f(b)f(a)>f(b) for all real numbers a a, b b in I I with a<b a<b. constant on I I if and only if f(a)=f(b)f(a)=f(b) for all real numbers a a, b b in I I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing and constant as stated in Definition 1.10 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you find an example of a function for which none of the concepts in Definition 1.10 apply? Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5) does not have the largest y y value of all of the points on the graph of f f − indeed that honor goes to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’ between x=−4 x=−4 and x=3 x=3. We say that the function f has a localmaximum8 at the point (−2, 4.5), because the y-coordinate 4.5 is the largest y-value (hence, function value) on the curve ‘near’9x=−2 x=−2. Similarly, we say that the function f has a localminimum10 at the point (3, −8), since the y-coordinate −8 is the smallest function value near x=3 x=3. Although it is tempting to say that local extrema11 occur when the function changes from increasing to decreasing or vice versa, it is not a precise enough way to define the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional definitions and thoroughly vet the pathologies they induce in the Exercises. We have one last observation to make before we proceed to the algebraic definitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see that the largest y y value (the largest function value) is 5.5 at x=6 x=6. In this case, we say the maximum12 of f is 5.5; similarly, the minimum13 of f f is −8. We formalize these concepts in the following definitions. Definition: 1.11 Suppose f f is a function with f(a)=b f(a)=b. We say f f has a local maximum at the point (a,b)(a,b) if and only if there is an open interval I I containing a for which f(a)≥f(x)f(a)≥f(x) for all x x in I I. The value f(a)=b f(a)=b is called ‘a local maximum value of f′f′ in this case. We say f f has a local minimum at the point (a,b)(a,b) if and only if there is an open interval I I containing a a for which f(a)≤f(x)f(a)≤f(x) for all x x in I I. The value f(a)=b f(a)=b is called ‘a local minimum value of f′f′ in this case. The value b b is called the maximum of f f if b≥f(x)b≥f(x) for all x x in the domain of f f. The value b b is called the minimum of f f if b≤f(x)b≤f(x) for all x x in the domain of f f. It’s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function’s graph would have to look like?) We shall see examples of functions in the Exercises which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic definitions above as we explore the wonderful diversity of graphs that functions provide us. Here is the ‘tame’ example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the Exercises. Example 1.6.4 Given the graph of y=f(x)y=f(x) below, answer all of the following questions. Find the domain of f f. Find the range of f f. List the x-intercepts, if any exist. List the y-intercepts, if any exist. Find the zeros of f f. Solve f(x)<0 f(x)<0. Determine f(2)f(2). Solve f(x)=−3 f(x)=−3. Find the number of solutions to f(x)=1 f(x)=1. Does f f appear to be even, odd, or neither? List the intervals on which f f is increasing. List the intervals on which f f is decreasing. List the local maximums, if any exist. List the local minimums, if any exist. Find the maximum, if it exists. Find the minimum, if it exists. Solution To find the domain of f f, we proceed as in Section 1.3. By projecting the graph to the x-axis, we see that the portion of the x-axis which corresponds to a point on the graph is everything from −4 to 4, inclusive. Hence, the domain is [−4, 4]. To find the range, we project the graph to the y-axis. We see that the y y values from −3 to 3, inclusive, constitute the range of f f. Hence, our answer is [−3, 3]. The x-intercepts are the points on the graph with y-coordinate 0, namely (−2, 0) and (2, 0). The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). The zeros of f f are the x-coordinates of the x-intercepts of the graph of y=f(x)y=f(x) which are x = −2, 2. To solve f(x)<0 f(x)<0, we look for the x x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking for where the graph is below the x-axis. This happens for the x x values from −4 to −2 and again from 2 to 4. So our answer is [−4,−2)∪(2,4][−4,−2)∪(2,4]. Since the graph of f is the graph of the equation y=f(x)y=f(x), f(2)f(2) is the y-coordinate of the point which corresponds to x=2 x=2. Since the point (2, 0) is on the graph, we have f(2)=0 f(2)=0. To solve f(x)=−3 f(x)=−3, we look where y=f(x)=−3 y=f(x)=−3. We find two points with a y-coordinate of −3, namely (−4, −3) and (4, −3). Hence, the solutions to f(x)=−3 f(x)=−3 are x=±4 x=±4. As in the previous problem, to solve f(x)=1 f(x)=1, we look for points on the graph where the y-coordinate is 1. Even though these points aren’t specified, we see that the curve has two points with a y y value of 1, as seen in the graph below. That means there are two solutions to f(x)=1 f(x)=1. The graph appears to be symmetric about the y-axis. This suggests14 that f is even. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f f is increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the x-axis.) As we move from left to right, the graph falls from (0, 3) to (4, −3). This means f f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) The function has its only local maximum at (0, 3) so f(0)=3 f(0)=3 is the local minimum value. There are no local minimums. Why don’t (−4, −3) and (4, −3) count? Let’s consider the point (−4, −3) for a moment. Recall that, in the definition of local minimum, there needs to be an open interval I which contains x = −4 such that f(−4)<f(x)f(−4)<f(x) for all x x in I I different from −4. But if we put an open interval around x=−4 x=−4 a portion of that interval will lie outside of the domain of f f. Because we are unable to fulfill the requirements of the definition for a local minimum, we cannot claim that f f has one at (−4, −3). The point (4, −3) fails for the same reason − no open interval around x=4 x=4 stays within the domain of f f. The maximum value of f f is the largest y-coordinate which is 3. The minimum value of f f is the smallest y-coordinate which is −3. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to find the local maximums and local minimums analytically; this is the business of Calculus.15 When we have need to find such beasts, we will resort to the calculator. Most graphing calculators have ‘Minimum’ and ‘Maximum’ features which can be used to approximate these values, as we now demonstrate. Example 1.6.5 Let f(x)=15 x x 2+3 f(x)=15 x x 2+3. Use a graphing calculator to approximate the intervals on which f f is increasing and those on which it is decreasing. Approximate all extrema. Solution Entering this function into the calculator gives Using the Minimum and Maximum features, we get To two decimal places, f f appears to have its only local minimum at (−1.73, −4.33) and its only local maximum at (1.73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn’t be too surprising. The function appears to be increasing on [−1.73, 1.73] and decreasing on (−∞,−1.73]∪[1.73,∞)(−∞,−1.73]∪[1.73,∞). This makes −4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Example 1.6.6 Find the points on the graph of y=(x−3)2 y=(x−3)2 which are closest to the origin. Round your answers to two decimal places. Solution Suppose a point (x,y)(x,y) is on the graph of y=(x−3)2 y=(x−3)2. Its distance to the origin (0, 0) is given by d=(x−0)2+(y−0)2−−−−−−−−−−−−−−−√=x 2+y 2−−−−−−√=x 2+[(x−3)2]2−−−−−−−−−−−−−√Since y=(x−3)2=x 2+(x−3)4−−−−−−−−−−−√d=(x−0)2+(y−0)2=x 2+y 2=x 2+[(x−3)2]2 Since y=(x−3)2=x 2+(x−3)4 Given a value for x x, the formula d=x 2+(x−3)4−−−−−−−−−−−√d=x 2+(x−3)4 is the distance from (0, 0) to the point (x, y) on the curve y=(x−3)2 y=(x−3)2. What we have defined, then, is a function d(x)d(x) which we wish to minimize over all values of x x. To accomplish this task analytically would require Calculus so as we’ve mentioned before, we can use a graphing calculator to find an approximate solution. Using the calculator, we enter the function d(x)d(x) as shown below and graph. Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x=2 x=2. Rounding to two decimal places, we get that the minimum distance occurs when x=2.00 x=2.00. To find the y value on the parabola associated with x=2.00 x=2.00, we substitute 2.00 into the equation to get y=(x−3)2=(2.00−3)2=1.00 y=(x−3)2=(2.00−3)2=1.00. So, our final answer is (2.00, 1.00).16 (What does the y y value listed on the calculator screen mean in this problem?) 1.6.2 Exercises In Exercises 1 - 12, sketch the graph of the given function. State the domain of the function, identify any intercepts and test for symmetry.\ f(x)=2−x f(x)=2−x f(x)=x−2 3 f(x)=x−2 3 f(x)=x 2+1 f(x)=x 2+1 f(x)=4−x 2 f(x)=4−x 2 f(x)=2 f(x)=2 f(x)=x 3 f(x)=x 3 f(x)=x(x−1)(x+2)f(x)=x(x−1)(x+2) f(x)=x−2−−−−−√f(x)=x−2 f(x)=5−x−−−−−√f(x)=5−x f(x)=3−2 x+2−−−−−√f(x)=3−2 x+2 f(x)=x−−√3 f(x)=x 3 f(x)=1 x 2+1 f(x)=1 x 2+1 In Exercises 13 - 20, sketch the graph of the given piecewise-defined function. f(x)={4−x 2 if if x≤3 x>3 f(x)={4−x if x≤3 2 if x>3 f(x)={x 2 2 x if if x≤0 x>0 f(x)={x 2 if x≤0 2 x if x>0 f(x)=⎧⎩⎨⎪⎪−3 2 x−3 3 if if if x<0 0≤x≤3 x>3 f(x)={−3 if x<0 2 x−3 if 0≤x≤3 3 if x>3 f(x)=⎧⎩⎨⎪⎪x 2−4 4−x 2 x 2−4 if if if x≤−2−2<x<2 x≥2 f(x)={x 2−4 if x≤−2 4−x 2 if−2<x<2 x 2−4 if x≥2 f(x)={−2 x−4 3 x if if x<0 x≥0 f(x)={−2 x−4 if x<0 3 x if x≥0 f(x)={x+4−−−−−√x−1−−−−−√if if−4≤x<5 x≥5 f(x)={x+4 if−4≤x<5 x−1 if x≥5 f(x)=⎧⎩⎨⎪⎪x 2 3−x 4 if if if x≤−2−2<x<2 x≥2 f(x)={x 2 if x≤−2 3−x if−2<x<2 4 if x≥2 f(x)=⎧⎩⎨⎪⎪1 x x x−−√if if if−6<x<−1−1<x<1 1<x<9 f(x)={1 x if−6<x<−1 x if−1<x<1 x if 1<x<9 In Exercises 21 - 41, determine analytically if the following functions are even, odd or neither. f(x)=7 x f(x)=7 x f(x)=7 x+2 f(x)=7 x+2 f(x)=7 f(x)=7 f(x)=3 x 2−4 f(x)=3 x 2−4 f(x)=4−x 2 f(x)=4−x 2 f(x)=x 2−x−6 f(x)=x 2−x−6 f(x)=2 x 3−x f(x)=2 x 3−x f(x)=−x 5+2 x 3−x f(x)=−x 5+2 x 3−x f(x)=x 6−x 4+x 2+9 f(x)=x 6−x 4+x 2+9 f(x)=x 3+x 2+x+1 f(x)=x 3+x 2+x+1 f(x)=1−x−−−−−√f(x)=1−x f(x)=1−x 2−−−−−√f(x)=1−x 2 f(x)=0 f(x)=0 f(x)=x−−√3 f(x)=x 3 f(x)=x 2−−√3 f(x)=x 2 3 f(x)=3 x 2 f(x)=3 x 2 f(x)=2 x−1 x+1 f(x)=2 x−1 x+1 f(x)=3 x x 2+1 f(x)=3 x x 2+1 f(x)=x 2−3 x−4 x 3 f(x)=x 2−3 x−4 x 3 f(x)=9 4−x 2√f(x)=9 4−x 2 f(x)=x 3+x√3 5 x f(x)=x 3+x 3 5 x In Exercises 42 - 57, use the graph of y=f(x)y=f(x) given below to answer the question. Find the domain of f f. Find the range of f f. Determine f(−2)f(−2). Solve f(x)=4 f(x)=4. List the x-intercepts, if any exist. List the y-intercepts, if any exist. Find the zeros of f f. Solve f(x)≥0 f(x)≥0. Find the number of solutions to f(x)=1 f(x)=1. Does f f appear to be even, odd, or neither? List the intervals where f f is increasing. List the intervals where f f is decreasing. List the local maximums, if any exist. List the local minimums, if any exist. Find the maximum, if it exists. Find the minimum, if it exists. In Exercises 58 - 73, use the graph of y=f(x)y=f(x) given below to answer the question. Find the domain of f f. Find the range of f f. Determine f(2)f(2). Solve f(x)=−5 f(x)=−5. List the x-intercepts, if any exist. List the y-intercepts, if any exist. Find the zeros of f f. Solve f(x)≤0 f(x)≤0. Find the number of solutions to f(x)=3 f(x)=3. Does f f appear to be even, odd, or neither? List the intervals where f f is increasing. List the intervals where f f is decreasing. List the local maximums, if any exist. List the local minimums, if any exist. Find the maximum, if it exists. Find the minimum, if it exists. In Exercises 74 - 77, use your graphing calculator to approximate the local and absolute extrema of the given function. Approximate the intervals on which the function is increasing and those on which it is decreasing. Round your answers to two decimal places. f(x)=x 4−3 x 3−24 x 2+28 x+48 f(x)=x 4−3 x 3−24 x 2+28 x+48 f(x)=x 2/3(x−4)f(x)=x 2/3(x−4) f(x)=9−x 2−−−−−√f(x)=9−x 2 f(x)=x 9−x 2−−−−−√f(x)=x 9−x 2 In Exercises 78 - 85, use the graphs of y=f(x)y=f(x) and y=g(x)y=g(x) below to find the function value. (f+g)(0)(f+g)(0) (f+g)(1)(f+g)(1) (f−g)(1)(f−g)(1) (g−f)(2)(g−f)(2) (f g)(2)(f g)(2) (f g)(1)(f g)(1) (f g)(4)(f g)(4) (g f)(2)(g f)(2) The graph below represents the height h h of a Sasquatch (in feet) as a function of its age N N in years. Use it to answer the questions in Exercises 86 - 90. Find and interpret h(0)h(0). How tall is the Sasquatch when she is 15 years old? Solve h(N)=6 h(N)=6 and interpret. List the interval over which h h is constant and interpret your answer. List the interval over which h h is decreasing and interpret your answer. For Exercises 91 - 93, let f(x)=⌊x⌋f(x)=⌊x⌋ be the greatest integer function as defined in Exercise 75 in Section 1.4. Graph y=f(x)y=f(x). Be careful to correctly describe the behavior of the graph near the integers. Is f f even, odd, or neither? Explain. Discuss with your classmates which points on the graph are local minimums, local maximums or both. Is f f ever increasing? Decreasing? Constant? In Exercises 94 - 95, use your graphing calculator to show that the given function does not have any extrema, neither local nor absolute. f(x)=x 3+x−12 f(x)=x 3+x−12 f(x)=−5 x+2 f(x)=−5 x+2 In Exercise 71 in Section 1.4, we saw that the population of Sasquatch in Portage County could be modeled by the function P(t)=150 t t+15 P(t)=150 t t+15, where t=0 t=0 represents the year 1803. Use your graphing calculator to analyze the general function behavior of P P. Will there ever be a time when 200 Sasquatch roam Portage County? Suppose f f and g g are both even functions. What can be said about the functions f+g f+g, f−g f−g, f g f g and f g f g? What if f f and g are both odd? What if f is even but g is odd? One of the most important aspects of the Cartesian Coordinate Plane is its ability to put Algebra into geometric terms and Geometry into algebraic terms. We’ve spent most of this chapter looking at this very phenomenon and now you should spend some time with your classmates reviewing what we’ve done. What major results do we have that tie Algebra and Geometry together? What concepts from Geometry have we not yet described algebraically? What topics from Intermediate Algebra have we not yet discussed geometrically? It’s now time to “thoroughly vet the pathologies induced” by the precise definitions of local maximum and local minimum. We’ll do this by providing you and your classmates a series of Exercises to discuss. You will need to refer back to Definition 1.10 (Increasing, Decreasing and Constant) and Definition 1.11 (Maximum and Minimum) during the discussion. Consider the graph of the function f f given below. Show that f f has a local maximum but not a local minimum at the point (−1, 1). Show that f f has a local minimum but not a local maximum at the point (1, 1). Show that f f has a local maximum AND a local minimum at the point (0, 1). Show that f f is constant on the interval [−1, 1] and thus has both a local maximum AND a local minimum at every point (x,f(x))(x,f(x)) where −1<x<1−1<x<1. Using Example 1.6.4 as a guide, show that the function g whose graph is given below does not have a local maximum at (−3, 5) nor does it have a local minimum at (3, −3). Find its extrema, both local and absolute. What’s unique about the point (0, −4) on this graph? Also find the intervals on which g g is increasing and those on which g g is decreasing. We said earlier in the section that it is not good enough to say local extrema exist where a function changes from increasing to decreasing or vice versa. As a previous exercise showed, we could have local extrema when a function is constant so now we need to examine some functions whose graphs do indeed change direction. Consider the functions graphed below. Notice that all four of them change direction at an open circle on the graph. Examine each for local extrema. What is the effect of placing the “dot” on the y-axis above or below the open circle? What could you say if no function value were assigned to x=0 x=0? Function I Function II Function III Function IV 1.6.3 Answers f(x)=2−x f(x)=2−x Domain: (−∞,∞)(−∞,∞) x-intercept: (2, 0) y-intercept: (0, 2) No symmetry f(x)=x−2 3 f(x)=x−2 3 Domain: (−∞,∞)(−∞,∞) x-intercept: (2, 0) y-intercept: (0,−2 3)(0,−2 3) No symmetr f(x)=x 2+1 f(x)=x 2+1 Domain: (−∞,∞)(−∞,∞) x-intercept: None y-intercept: (0, 1) Even f(x)=4−x 2 f(x)=4−x 2 Domain: (−∞,∞)(−∞,∞) x-intercepts: (−2, 0), (2, 0) y-intercept: (0, 4) Even f(x)=2 f(x)=2 Domain: (−∞,∞)(−∞,∞) x-intercept: None y-intercept: (0, 2) Even f(x)=x 3 f(x)=x 3 Domain: (−∞,∞)(−∞,∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd f(x)=x(x−1)(x+2)f(x)=x(x−1)(x+2) Domain: (−∞,∞)(−∞,∞) x-intercepts: (−2, 0), (0, 0), (1, 0) y-intercept: (0, 0) No symmetry f(x)=x−2−−−−−√f(x)=x−2 Domain: [2,∞)[2,∞) x-intercept: (2, 0) y-intercept: None No symmetry f(x)=x−2−−−−−√f(x)=x−2 Domain: [2,∞)[2,∞) x-intercept: (5, 0) y-intercept: (0,5–√)(0,5) No symmetry f(x)=3−2 x+2−−−−−√f(x)=3−2 x+2 Domain: [−2,∞)[−2,∞) x-intercept: (1 4,0)(1 4,0) y-intercept: (0,3−2 2–√)(0,3−2 2) No symmetry f(x)=x−−√3 f(x)=x 3 Domain: (−∞,∞)(−∞,∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd f(x)=1 x 2+1 f(x)=1 x 2+1 Domain: (−∞,∞)(−∞,∞) x-intercept: None y-intercept: (0, 1) Even odd neither even even even neither odd odd even neither neither even even and odd odd even even neither odd odd even even [−5, 3] [−5, 4] f(−2)=2 f(−2)=2 x = −3 (−4, 0), (−1, 0), (1, 0) (0, −1) −4, −1, 1 [−4,−1]∪[1,3][−4,−1]∪[1,3] 4 neither [−5, −3], [0, 2] [−3, 0], [2, 3] f(−3)=4,f(2)=3 f(−3)=4,f(2)=3 f(0)=−1 f(0)=−1 f(−3)=4 f(−3)=4 f(−5)=−5 f(−5)=−5 [−4, 4] [−5, 5) f(2)=3 f(2)=3 x = −2 (−4, 0), (0, 0), (4, 0) (0, 0) −4, 0, 4 [−4,0]∪{4}[−4,0]∪{4} 3 neither [−2, 2) [−4, −2], (2, 4] none f(−2)=−5,f(2)=3 f(−2)=−5,f(2)=3 none f(−2)=−5 f(−2)=−5 No absolute maximum Absolute minimum f(4.55)≈−175.46 f(4.55)≈−175.46 Local minimum at (−2.84, −91.32) Local maximum at (0.54, 55.73) Local minimum at (4.55, −175.46) Increasing on [−2.84, 0.54], [4.55, ∞) Decreasing on (−∞, −2.84], [0.54, 4.55] No absolute maximum No absolute minimum Local maximum at (0, 0) Local minimum at (1.60, −3.28) Increasing on (−∞, 0], [1.60, ∞) Decreasing on [0, 1.60] Absolute maximum f(0)=3 f(0)=3 Absolute minimum f(±3) = 0 Local maximum at (0, 3) No local minimum Increasing on [−3, 0] Decreasing on [0, 3] Absolute maximum f(2.12) ≈ 4.50 Absolute minimum f(−2.12) ≈ −4.50 Local maximum (2.12, 4.50) Local minimum (−2.12, −4.50) Increasing on [−2.12, 2.12] Decreasing on [−3, −2.12], [2.12, 3] (f+g)(0)=4(f+g)(0)=4 (f+g)(1)=5(f+g)(1)=5 (f−g)(1)=−1(f−g)(1)=−1 (g−f)(2)=0(g−f)(2)=0 (f g)(2)=9(f g)(2)=9 (f g)(1)=6(f g)(1)=6 (f g)(4)=0(f g)(4)=0 (g f)(2)=1(g f)(2)=1 h(0)=2 h(0)=2, so the Sasquatch is 2 feet tall at birth. h(15)=6 h(15)=6, so the Saquatch is 6 feet tall when she is 15 years old. h(N)=6 h(N)=6 when N=15 N=15 and N=60 N=60. This means the Sasquatch is 6 feet tall when she is 15 and 60 years old. h h is constant on [30, 45]. This means the Sasquatch’s height is constant (at 8 feet) for these years. h h is decreasing on [45, 60]. This means the Sasquatch is getting shorter from the age of 45 to the age of 60. (Sasquatchteoporosis, perhaps?) Note that f(1.1)=1 f(1.1)=1, but f(−1.1)=−2 f(−1.1)=−2, so f is neither even nor odd. Reference 1 We’ve just stepped into Calculus here! 2 Why are we so dismissive about symmetry about the x-axis for graphs of functions? 3 Any ideas? 4 ‘Suggests’ is about the extent of what it can do. 5 Consult your owner’s manual, instructor, or favorite video site! 6 Or, in other words, don’t rely too heavily on the machine! 7 The notions of how quickly or how slowly a function increases or decreases are explored in Calculus 8 Also called ‘relative maximum’. 9 We will make this more precise in a moment. 10 Also called a ‘relative minimum’. 11 ‘Maxima’ is the plural of ‘maximum’ and ‘mimima’ is the plural of ‘minimum’. ‘Extrema’ is the plural of ‘extremum’ which combines maximum and minimum. 12 Sometimes called the ‘absolute’ or ‘global’ maximum. 13 Again, ‘absolute’ or ‘global’ minimum can be used. 14 but does not prove 15 Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra. 16 It seems silly to list a final answer as (2.00, 1.00). Indeed, Calculus confirms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the authors are overly pedantic, and as such, use the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such. 1.6: Graphs of Functions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 1.7: Transformations 2: Linear and Quadratic Functions Was this article helpful? Yes No Recommended articles 1.1: Sets of Real Numbers and the Cartesian Coordinate PlaneA brief refresher on some basic notions is welcome, if not completely necessary, at this stage. To that end, we present a brief summary of 'set theory... 1.2: RelationsAll of Precalculus can be thought of as studying sets of points in the plane. With the Cartesian Plane now fresh in our memory we can discuss those se... 1.3: Introduction to FunctionsOne of the core concepts in College Algebra is the function. There are many ways to describe a function and we begin by defining a function as a speci... 1.4: Function NotationIf we think of the domain of a function as a set of inputs and the range as a set of outputs, we can think of a function f as a process by which each... 1.5: Function ArithmeticIt would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The following de... Article typeSection or Page Tags This page has no tags. © Copyright 2025 Mathematics 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× contents readability resources tools ☰ 1.7: Transformations 2: Linear and Quadratic Functions Complete your gift to make an impact
3373	https://cdn.aaai.org/ojs/20250/20250-13-24263-1-2-20220628.pdf	End-to-End Learning the Partial Permutation Matrix for Robust 3D Point Cloud Registration Zhiyuan Zhang1, Jiadai Sun1, Yuchao Dai1, Dingfu Zhou2, Xibin Song2, Mingyi He1 1 Northwestern Polytechnical University 2 Baidu Research {zhangzhiyuan,sunjiadai}@mail.nwpu.edu.cn, daiyuchao@gmail.com, {zhoudingfu,songxibin}@baidu.com, myhe@nwpu.edu.cn Abstract Even though considerable progress has been made in deep learning-based 3D point cloud processing, how to obtain accurate correspondences for robust registration remains a major challenge because existing hard assignment methods cannot deal with outliers naturally. Alternatively, the soft matching-based methods have been proposed to learn the matching probability rather than hard assignment. However, in this paper, we prove that these methods have an inherent ambiguity causing many deceptive correspondences. To ad-dress the above challenges, we propose to learn a partial per-mutation matching matrix, which does not assign correspond-ing points to outliers, and implements hard assignment to pre-vent ambiguity. However, this proposal poses two new prob-lems, i.e. existing hard assignment algorithms can only solve a full rank permutation matrix rather than a partial permuta-tion matrix, and this desired matrix is defined in the discrete space, which is non-differentiable. In response, we design a dedicated soft-to-hard (S2H) matching procedure within the registration pipeline consisting of two steps: solving the soft matching matrix (S-step) and projecting this soft matrix to the partial permutation matrix (H-step). Specifically, we aug-ment the profit matrix before the hard assignment to solve an augmented permutation matrix, which is cropped to achieve the final partial permutation matrix. Moreover, to guarantee end-to-end learning, we supervise the learned partial permu-tation matrix but propagate the gradient to the soft matrix in-stead. Our S2H matching procedure can be easily integrated with existing registration frameworks, which has been veri-fied in representative frameworks including DCP, RPMNet, and DGR. Extensive experiments have validated our method, which creates a new state-of-the-art performance. Introduction 3D point cloud registration is a well-known task in 3D vi-sion with wide applications including object pose estimation (Wong et al. 2017), 3D reconstruction (Deschaud 2018), si-multaneous localization and mapping (Shiratori et al. 2015; Ding and Feng 2019), etc. Although the increasingly pros-perous deep learning technique has achieved great success in point cloud registration (Lu et al. 2019), how to obtain accu-rate correspondences for robust registration remains a stub-Yuchao Dai is the corresponding author. Copyright © 2022, Association for the Advancement of Artificial Intelligence (www.aaai.org). All rights reserved. born problem, which can be formulated as solving a match-ing matrix to relate the input two point clouds. And each entry indicates the point pair is a correspondence or not. For ideal consistent point clouds, where the inputs are ex-actly the same except for the pose, i.e. each point can find a corresponding point in the other point cloud, the correspon-dences are built by searching a permutation matching ma-trix, which implements the one-to-one matching principle. However, in practical applications, input point clouds are usually not consistent due to the outliers (i.e. the points with-out corresponding points). To handle the outliers, a widely adopted strategy is to select reliable correspondences after the initial matching (Choy, Dong, and Koltun 2020; Probst et al. 2019; Pais et al. 2020; Deng, Birdal, and Ilic 2018b,a; Bai et al. 2021). Nonetheless, this remedial operation is complicated. Alternatively, we propose to handle the out-liers in the matching stage synchronously. In this case, the desired matching matrix is turned to a partial permutation matrix (notated as PPM) formulated by a binary matrix, where the sum of row or column corresponding to inlier/out-lier is one/zero. PPM embeds two important principles: one-to-one matching and outliers pruning. Unfortunately, exist-ing hard assignment algorithms are not competent to directly solve this PPM since they cannot distinguish inliers and out-liers, and they will solve a full rank permutation matrix as-signing corresponding points to outliers incorrectly. Moreover, PPM is defined in the discrete space, and the hard assignment algorithm is non-differentiable, which is fatal for the deep learning pipeline. To address this issue, soft matching-based methods are proposed. They relax the discrete matching matrix into a continuous one, where each entry is either zero or one, and then the virtual points are achieved by performing weighted average to replace the real corresponding points. However, since the geometric con-straint is ignored, the network does not learn the underly-ing physics, which results in an inherent ambiguity making the distribution of the virtual points degenerate seriously. DCP (Wang and Solomon 2019a) is a typical soft matching method, which suffers from this drawback as shown in Ta-ble 1. RPMNet (Yew and Lee 2020) applies the Augmented-Sinkhorn algorithm replying to outliers in the soft matching process, where the “trash bin” strategy (Sarlin et al. 2020) is employed. However, the degeneration has not been re-mitted. To avoid this ambiguity trap, DeepVCP (Lu et al. The Thirty-Sixth AAAI Conference on Artificial Intelligence (AAAI-22) 3399 Methods DCP (Soft matching) RPMNet (Soft matching) PRNet (Hard matching) Ours (Hard matching) Matrix Constraints mij ∈[0, 1] PNY j mij = 1 mij ∈[0, 1] PNX i mij ∈[0, 1] PNY j mij ∈[0, 1] mij ∈{0, 1} PNY j mij = 1 mij ∈{0, 1} PNX i mij ∈{0, 1} PNY j mij ∈{0, 1} Matching Results Table 1: Comparison of learning-based point cloud registration methods based on “soft” matching and “hard” matching. Points with different colors indicate the source (green), target (blue), and virtual points (red). mij is the entry of matching matrix M ∈RNX ×NY, where NX , NY are size of the source and target. (The matching result of ours is returned by SHMDCP.) 2019) takes an unconvincing assumption, i.e. accurate ini-tial motion parameters are provided as prior. PRNet (Wang and Solomon 2019b) and CorrNet3D (Zeng et al. 2021) ad-vocate turning the learned soft matching matrix to a hard matrix by taking the most similar points as the correspond-ing points of the source points. However, this strategy leads to the drawback of one-to-many matching. In this paper, we first theoretically analyze the inherent ambiguity in these soft matching-based methods, and then devote to achieving the real PPM matrix to handle the out-liers and prevent ambiguity synchronously for robust 3D point cloud registration. However, as mentioned above, two problems block our pace. 1) Although it is well-known that the full rank permutation matrix can be solved by existing hard assignment algorithms, how to solve a PPM has yet not been explored; 2) PPM is defined in the discrete space, which is non-differentiable. To resolve these issues, we de-sign a dedicated soft-to-hard (S2H) matching procedure con-sists of S-step: solving the soft matching matrix, and H-step: projecting this soft matrix to the PPM. Specif-ically, we propose to augment the profit matrix before the hard assignment to solve an augmented permutation matrix, which is cropped to achieve the final real PPM. Moreover, to guarantee end-to-end learning, we supervise the learned final PPM but propagate the gradient to the soft matrix in-stead. Our matching procedure can be easily integrated with existing 3D registration frameworks, which has been veri-fied in DCP, RPMNet, and DGR (Choy, Dong, and Koltun 2020). Extensive experiments on benchmark datasets show that our method achieves state-of-the-art performance. Our main contributions can be summarized as follows: • We theoretically analyze the inherent ambiguity in the soft matching-based methods, which causes serious de-generation of the learned virtual corresponding points. • We propose a novel S2H matching procedure to learn the PPM, which handles the outliers and prevents ambiguity synchronously. This matching procedure not only solves a real PPM, but also guarantees end-to-end learning. • Remarkable performance on benchmark datasets veri-fies the proposed method, which achieves state-of-the-art performance in robust 3D point cloud registration. Related Works Herein, we briefly review the learning-based point cloud reg-istration methods. More detailed summaries have been pro-vided in (Rusinkiewicz and Levoy 2001; Pomerleau, rancis Colas, and Siegwart 2015; Zhang, Dai, and Sun 2020). Correspondences-free methods: These methods estimate the rigid motion by comparing the global representations of input two point clouds, and generally consist of two stages: global feature extraction and rigid motion estimation. Point-NetLK (Aoki et al. 2019) utilizes PointNet (Qi et al. 2017a) to extract global features, and then a modified LK algorithm is applied to solve the rigid motion. From the perspective of reconstruction, Huang et al. (Huang, Mei, and Zhang 2020) utilize an encoder-decoder structure network to learn a more comprehensive global feature. Correspondences-based methods: These methods esti-mate the rigid motion based on correspondences, which gen-erally consists of feature extractor, correspondences build-ing, and rigid motion estimation modules. For feature ex-tractors, various well-designed networks are used, such as set abstraction module (Qi et al. 2017b; Yew and Lee 2018), DGCNN (Wang et al. 2019; Wang and Solomon 2019a), FCGF (Choy, Park, and Koltun 2019; Choy, Dong, and Koltun 2020), globally informed 3D local feature (Deng, Birdal, and Ilic 2018b), KPConv (Thomas et al. 2019; Bai et al. 2020), capsule network (Zhao et al. 2020) and vari-ous rotation invariant features (Deng, Birdal, and Ilic 2018a; Gojcic et al. 2019). With the rise of deep learning, learning-based feature extractors have approached standard compo-nents, which can be integrated easily. For correspondence building, both soft matching-based (Lu et al. 2019; Zhang et al. 2022) and hard matching-based (Wang and Solomon 2019b) methods are representative. (Yew and Lee 2020; Huang et al. 2021) build correspondences on the identified inliers only. Besides, reliable correspondences selection is the widely adopted subsequent step. It is achieved by learn-ing the reliability weight of each initial correspondence (Pais et al. 2020; Choy, Dong, and Koltun 2020; Probst et al. 2019) or selecting consistent correspondences (Deng, Birdal, and Ilic 2018b; Bai et al. 2021). For motion estimation, Pro-crustes (Gower 1975) is the most widely used algorithm in (Wang and Solomon 2019b; Yew and Lee 2020). Recently, regressing the motion parameters directly has become a new hot spot (Pais et al. 2020). Preliminaries Given the source point cloud X = [xi]3×NX and the tar-get point cloud Y = [yj]3×NY, where NX and NY repre-3400 sent the sizes of the two point clouds, 3D point cloud regis-tration aims at solving a rigid motion to best align X with Y. Here, we model the rigid motion by the rotation ma-trix R ∈SO(3) and the translation vector t ∈R3. Since the Procrustes algorithm (Gower 1975) can optimally solve the rigid motion based on the correspondences, point match-ing becomes crucial, which is formulated as searching for a matching matrix M = [mij]NX ×NY to relate the source and target point clouds, and the entry mij = 1 or 0 indicates point xi and point yj are correspondence or not. Ideally, X and Y are consistent, i.e. points in X and Y are exactly one-to-one correspondence. In this case, the match-ing matrix M is a permutation matrix, i.e. mij ∈{0, 1}, PN i=1 mij = 1, ∀j and PN j=1 mij = 1, ∀i, where NX = NY = N. However, in practical applications, outliers al-ways exist without corresponding points. They challenge the matching problem, and turn it into a special assignment problem while the desired matching matrix becomes a PPM for outliers pruning. We reformulate this special PPM M as, {M \| mij ∈{0, 1}; XNX i=1 mij ∈{0, 1}, ∀j; XNY j=1 mij ∈{0, 1}, ∀i}. (1) If the point is an inlier, the sum of the corresponding row/-column equals to 1. Otherwise, the point is an outlier and the sum of the corresponding row/column equals to 0. Ambiguity in Soft Matching-Based Methods As introduced above, different from the hard matching-based methods that build correspondences on the real points, the soft matching-based methods use the virtual correspond-ing points instead. Specifically, these soft matching-based methods generate a “soft map” between the source and tar-get, i.e. xi ∈X is assigned to Y by a probability vector. In this case, the matching matrix M becomes a soft proba-bility matrix P. In DCP (Wang and Solomon 2019a), P is a single stochastic matrix, where pij ∈[0, 1], PNX i=1 pij = 1, ∀j. In RPMNet (Yew and Lee 2020), P is optimized to a partial doubly stochastic matrix (notated as PDSM) by the Augmented-Sinkhorn algorithm, where pij ∈[0, 1], PNX i=1 pij ≤1, ∀j and PNY j=1 pij ≤1, ∀i. Then, the vir-tual corresponding points Y′ are obtained by performing weighted average on Y using P, i.e. Y′ = YPT. However, there is an ambiguity trap here. The geometric constraints and underlying physics are ignored by relaxing the hard matching matrix to a soft probability matrix, hence, the distribution of virtual corresponding points is not unique resulting in serious degeneration. We can conclude as fol-lows. The theoretical proof and more analysis are provided in supplementary materials. Theorem 1 Considering the consistent subset point clouds X, Y with ground truth motion R, t, there exists more than one soft matching matrix P satisfying YPT = RX + t. Since there will be an infinite number of virtual point cloud distributions corresponding to the same rigid motion, this inherent ambiguity will cause serious degeneration of virtual points as shown in Table 1. Essentially, the process of weighted average can also be regarded as a special defor-mation of the point cloud. Although a seemingly good trans-formation estimation is obtained (Yew and Lee 2020), this process violates the rigid motion assumption, which cannot be supported by the Procrustes algorithm (Gower 1975). Soft-to-Hard Matching for Registration In this paper, we devote to solving a PPM to handle the out-liers and prevent the ambiguity synchronously and thus de-sign a meticulous S2H matching procedure. In this section, to clearly present S2H and how to integrate it into existing registration pipeline, we give a pipeline example as shown in Fig. 1. Finally, the proposed loss function is presented. Feature Extractor and Similarity Matrix Solving To achieve robust point matching, distinguished descriptors are crucial. With the rise of deep learning, many standard modules for deep features are proposed. These feature ex-tractors can be easily integrated into our robust point cloud registration pipelines according to different requirements. We denote the point features of the source and target as ΦX ∈RNX ×c, ΦY ∈RNY×c. c is the feature dimension. Then, based on a certain similarity metric e.g. scale dot prod-uct attention (Vaswani et al. 2017), the similarity matrix is returned as S = [sij]NX ×NY, where each entry sij repre-sents the similarity between points xi ∈X and yi ∈Y. S2H Matching A well-known solution to hard matching is to formulate it as a special assignment problem, i.e. zero-one integer program-ming problem. However, two natural but challenging prob-lems exist for learning-based pipeline. 1) Existing integer programming algorithms usually achieve a row or column full rank permutation matrix (rank(M) = min(NX , NY)) rather than a PPM. It means all points of the source or target will be assigned corresponding points without distinguish-ing inliers and outliers. 2) PPM is defined in the discrete space, which is non-differentiable. This characteristic is fatal for the deep learning pipeline. To solve these problems, we propose to augment the profit matrix to solve the PPM and design an S2H matching procedure for end-to-end learning. • Augmenting the profit matrix to solve PPM: Con-ventionally, the matching task is formulated as a zero-one integer programming problem taking S as the profit matrix, M∗= arg max M∈MN < M, S >F , (2) where MN denotes the set of partial permutation matrices. < M, S >F = trace(MTS) denotes the (Frobenius) inner product. Note that the traditional assignment algorithm will achieve a full rank permutation solution. In response to out-liers, we propose to augment the profit matrix by adding ad-ditional rows and columns to solve an augmented permuta-tion matrix. Then the PPM will be returned by cropping this augmented permutation matrix as shown in Fig. 1. Thus, the following two questions should be addressed properly. 1. How many rows and columns should be added? In the Augmented-Sinkhorn algorithm (Yew and Lee 2020), which solves a soft matching matrix, one row and one column 3401 Weighted Procrustes 11 12 13 1 21 22 23 2 31 32 33 3 r r r t r r r t r r r t           S2H Matching Rotation & Translation PPM Feature Extractor Similarity Matrix Point Features Feature Extractor S-step H-step Figure 1: S2H matching procedure in registration pipeline. Given the source and target, the similarity matrix is obtained based on the point features. Then, S2H is applied for the final PPM. Finally, the rigid motion is estimated by the weighted Procrustes. are added as “trash bin”. However, for partial permutation matching, the number of rows and columns augmented to the profit matrix is more crucial since it implies the upper bound of the number of outliers potentially. Thus, we pro-pose to supplement the original NX × NY profit matrix to a (NX + NY) × (NX + NY) matrix, which is a maximum redundant operation replying to the case of all points are out-liers. Specifically, as shown in Fig. 1, left upper is the origi-nal matrix, right upper block is a NX × NX diagonal matrix and left bottom block is a NY × NY diagonal matrix, and right bottom block is a NY × NX zero matrix. 2. What value to set? As aforesaid, the right upper block and the left bottom block are two diagonal matrices, what values should be set to these diagonal positions? Note that these values are roughly used as thresholds to distinguish outliers and inliers potentially. Meanwhile, we observe that in the profit matrix, if the row/column corresponds to an out-lier, the entries in this row/column approximate a uniform distribution with low value, which means the outliers have no similar points in another point cloud. Otherwise, if the row/column corresponds to the inlier, in this row/column, the entries are close to a unimodal distribution, which means the inlier has only one similar point in another point cloud ideally. Thus, we propose to self-adaptively fill the diagonal position with σ according to the corresponding row/column of the input profit matrix: σ=1/ var(v), v is the correspond-ing row/column vector, and var(·) is the variance function. In this case, for outlier, the filled value is large, which en-forces the learned augmented permutation matrix to make the value of this position as one to gain more profit. For in-lier, the filled value is low, the value of this position in the learned augmented permutation matrix is enforced to zero. • S2H matching in end-to-end learning: Since M is defined in the discrete space, and the integer programming algorithm is non-differentiable, we design the S2H match-ing procedure to guarantee end-to-end learning. Specifically, S-step learns a soft matrix and H-step projects this soft matrix to discrete solution space for final PPM. S-step: To deal with the outliers, we use Augmented-Sinkhorn (Yew and Lee 2020) to obtain a soft matrix by adding an additional row and column of ones to the input matrix during the Sinkhorn normalization. In Augmented-Sinkhorn, the additional row and column are regarded as “trash bin”, and the matching weights of outliers are ex-pected to “flow” to these additional row and column to distinguish outliers and inliers. Specifically, given the ob-tained similarity matrix S ∈RNX ×NY, this soft matrix P ∈RNX ×NY is achieved by cropping the output (NX + 1) × (NY + 1) matrix as shown in Fig. 1. P is a PDSM. H-step: The resultant PDSM P is still a soft matrix, and we project it to PPM by applying the proposed profit ma-trix augmenting strategy to P and solving this zero-one inte-ger programming problem. In our implementation, we chose the classical Hungarian algorithm (Kuhn 1955) to solve this assignment problem. After cropping the output augmented matrix, the final PPM M ∈RNX ×NY is obtained. For a clearer understanding, we stress the ingeniousness of the proposed S2H matching structure from two folds: 1. End-to-end learning. We propose a deceptive operation as shown in the left of Fig. 2 to guarantee end-to-end learn-ing. During the forward propagation, the loss is calculated based on the learned PPM M. However, during the back-ward propagation, the gradient is not propagated to the PPM, but directly skipped to the learned PDSM P instead. This in-genious structure guarantees the accuracy of the calculated loss and the backward propagation simultaneously. 2. Hard Matching vs. S2H Matching. To solve a PPM, we give a more straightforward hard matching method in the right of Fig. 2, which can also make sense by learn-ing M considering the augmented similarity matrix as the profit matrix. Nonetheless, there is an obvious local super-vision risk in this case. That is, the gradient will be propa-gated to the input similarity matrix directly if only use the hard matching. And the correlation of entries, which is con-sidered in the integer programming process, is ignored in backward propagation. In other words, the loss calculated from mij can only supervise sij. This will result in only a few sparse points being supervised by the ground truth and the remaining positions will be trained without supervision. For example, assuming mij = 1, the gradient will be prop-agated to sij directly, and enforce the feature extractor to enlarge sij. However, the positions where mij = 0 are not supervised (refer to loss function section for more details). Hence, the corresponding point features will not suppress their similarity. An extreme result is that all entries of the similarity matrix are very large since the point features lose the distinctiveness and each point in X is very similar to all points in Y. This will result in a divergence of training. 3402 PDSM Similarity Matrix PPM S-step S2H Matching Loss H-step H-step only Loss Similarity Matrix PPM Hard Matching Figure 2: Comparison of our S2H matching (Left) and hard matching (Right). Yellow and red lines represent the forward and backward propagation respectively. Our S2H solution effectively avoids the local supervision risk by propagating the gradient to all entries in S-step, i.e. the correlation of all entries is reconsidered. As shown in the left of Fig. 2, during the backward propagation, the sim-ilarity of the correct correspondence will be boosted, mean-while the similarity of the incorrect correspondence will be effectively suppressed. Weighted Procrustes When we get a PPM M, the corresponding point set of X could be achieved as Y′ = YMT. However, the corre-sponding points of outliers in Y′ are obtained as [0, 0, 0]T, which should be filtered out when estimating the transfor-mation. To this end, the weighted Procrustes algorithm is used here. Given X, Y′, the weight vector w is obtained as w = PNY j=1 mij, where w ∈{0, 1}NX . Then, w is normalized to ¯ w. Inspired by DGR (Choy, Dong, and Koltun 2020), the rigid motion is computed as follows: H = Y′KWKX T, where W = diag( ¯ w), K = I − √ ¯ w √ ¯ w T. I is an identity matrix. Then, R = UEVT, t = (Y′ −RX)W1, where UDVT = SVD(H), 1 = (1, ..., 1)T and E = diag(1, ..., 1, det (U) det (V)). Loss Function In this paper, we supervise the matching matrix directly, which is defined as: L1 = − PNX i=1 PNY j=1 mpred ij mgt ij PNX i=1 PNY j=1 mgt ij , (3) where the superscript “pred” and “gt” represent the predic-tion and ground truth respectively. When mgt ij = 1, mpred ij is enforced to 1. But when mgt ij = 0, mpred ij diverges without supervision. Moreover, due to the introduction of augmenta-tion operation, all points tend to be labeled as outliers, which makes the learned PPM close to a full zero matrix. To avoid this degeneracy, we encourage the number of inliers by: L2 = − PNX i=1 PNY j=1(mpred ij ) NX + NY . (4) Besides, we also supervise the final rigid motion, i.e., L3 = ∥RgtTRpred −I3∥2 + ∥tgt −tpred∥2, (5) where I3 is a 3×3 identity matrix. Then, our final loss func-tion is reached as L = λ1L1 + λ2L2 + λ3L3, the trade-off parameters are set to λ1 = λ2 = λ3 = 1 in this paper. Experiments and Evaluation In this section, we evaluate the proposed S2H matching procedure in several representative point cloud registra-tion frameworks including DCP, RPMNet and DGR, on benchmark datasets including ModelNet40 (Wu et al. 2015), 3DMatch (Zeng et al. 2017) and KITTI (Geiger et al. 2013). Implementation details. To validate the proposed matching method can be generally integrated, we evaluate S2H match-ing within three typical frameworks, i.e. DCP, RPMNet, and DGR, notated as SHMDCP, SHMRPMNet, and SHMDGR re-spectively. These three frameworks are typical and repre-sentative, where DCP and RPMNet are soft matching-based methods using different feature extractors, and mainly con-centrate on the synthetic dataset, ModelNet40. DGR is a hard matching-based method focusing on large-scale real datasets, 3DMatch and KITTI. Note that PRNet (Wang and Solomon 2019b) also inherits the framework of DCP, which is compared with SHMDCP herein. The complete loss is used in SHMDCP, SHMRPMNet. Only L3 is used in SHMDGR. Refer to supplementary materials for more details. Evaluation on Synthetic Dataset: ModelNet40 In this section, we validate our proposed S2H matching pro-cedure with SHMDCP, SHMRPMNet on a synthetic dataset, ModelNet40. Following DCP and RPNet, we construct a point cloud by randomly sampling 1024 points, and then ap-ply a rigid transformation to this point cloud, where the ro-tation and translation are uniformly sampled from [0◦, 45◦] and [−0.5, 0.5]3 respectively along each axis. Next, we ran-domly sample 768 points from the original point cloud and the transformed point cloud as the source and target to en-sure the random distribution of the outliers. In addition, following DCP and RPMNet, we test ours on two different dataset settings. 1) Unseen categories (clean): ModelNet40 will be divided into training and test splits based on the object category, i.e. the first 20 categories are selected for training and the rest categories for testing. 2) Noisy data (noisy): For robustness testing, random Gaus-sian noise (i.e., N(0, 0.01)) is added to each point, while the sampled noise out of the range of [−0.05, 0.05] will be clipped. The dataset splitting strategy is the same as clean. Matching. Accurate correspondences estimation is crucial for robust point cloud registration. Here, we evaluate the constructed correspondences for a clear comparison. • Metric: We calculate the discrepancy between the pre-dicted and the ground truth corresponding points. The pre-dicted corresponding points Y′ pred of X can be obtained by two methods. First, based on the predicted matching matrix Mpred, Y′ pred can be obtained by Y′ pred = YMpredT. Second, inspired by the iteration strategy in ICP, Y′ pred can be ob-tained based on the predicted transformation {Rpred, tpred} and nearest neighbor principle, i.e. Y′ pred = NNY(RpredX + tpred), where NNY(·) solves the nearest neighbor point in Y. Given the ground truth corresponding points Y′ gt, the root mean squared error (RMSE) and mean absolute error (MAE) in Euclidean distance between Y′ pred and Y′ gt, notated as RMSE(dis) and MAE(dis) are presented. 3403 Methods RMSE(dis) ↓MAE(dis) ↓RMSE(dis) ↓MAE(dis)↓ clean noisy clean noisy clean noisy clean noisy DCP-v2 0.500 0.466 0.705 0.657 0.088 0.095 0.078 0.095 PRNet 0.311 0.303 0.357 0.351 0.046 0.057 0.023 0.036 SHMDCP 0.106 0.188 0.027 0.082 0.033 0.053 0.008 0.019 RPMNet 0.139 0.137 0.182 0.181 0.019 0.028 0.004 0.013 SHMRPMNet 0.057 0.121 0.009 0.041 0.007 0.027 0.001 0.010 Table 2: Discrepancy based on predicted matching matrix (Left) and predicted transformation, nearest neighbor prin-ciple (Right). We also report the matching recall (%) based on a pro-posed self-adaptive threshold, τi = (1/K)PK j=1 dy′ π(i),yj, where y′ π(i) is the correct corresponding point of xi, yj ∈ KNNY(y′ π(i)), KNNY(·) solves the K-nearest neighbor points (exclude the self-point) in Y with pre-defined param-eter K, dy′ π(i),yj is the distance between y′ π(i) and yj. To sum up, for the i-th point, τi is computed as the average dis-tance of K-nearest points around the correct corresponding point in Y. If the distance between the correct correspond-ing point and the predicted one is less than the threshold, this pair will be confirmed as a correct matching pair. • Evaluation: In Table 2, we report RMSE(dis) and MAE(dis) results in clean and noisy. For the discrepancy based on the predicted matching matrix, our method im-proves the performance with a big margin in both DCP and RPMNet frameworks. These obtained results are reason-able because the DCP-v2, RPMNet are virtual point-based methods, where the corresponding points degenerate seri-ously with a large distance to correct corresponding points. PRNet presents weaker matching performance due to the one-to-many matching. For the discrepancy based on the predicted transformation and nearest neighbor principle, all methods achieve superior performance, whereas SHMDCP, SHMRPMNet remain the overall best performance. Besides, we also draw the matching recall with different thresholds in Fig. 3. For the results based on the predicted matching matrix, the DCP-v2, RPMNet fail to obtain accu-rate correspondences. And ours obtains the best results. For the results based on the predicted transformation and nearest neighbor principle, all methods achieve better performance, and SHMDCP, SHMRPMNet remain the best performance. Registration. In this section, we evaluate the rigid motion estimation performance of SHMDCP, SHMRPMNet. • Metric: Following DCP, RMSE and MAE between the ground truth and prediction in Euler angle and translation vector are used as the evaluation metrics here, notated as RMSE(R), MAE(R), RMSE(t) and MAE(t) respectively. • Evaluation: The evaluation results are provided in Ta-ble 3. Both SHMDCP and SHMRPMNet outperform the hand-craft registration methods, ICP (Besl and McKay 1992), FGR (Zhou, Park, and Koltun 2016). Furthermore, in clean, SHMDCP achieves more accurate performance than DCP-v2 and PRNet in all metrics, SHMRPMNet is more accurate than RPMNet in all metrics. In noisy, we train and test ours and 0 5 10 15 (a-1) 0 50 100 Matching recall (%) 0 5 10 15 (a-2) 0 50 100 0 5 10 15 (b-1) 40 60 80 100 0 5 10 15 (b-2) 40 60 80 100 DCP PRNet RPMNet SHMDCP SHMRPMNet Figure 3: Matching recall with different thresholds based on the predicted matching matrix ((a-1), (a-2)) and the pre-dicted transformation, nearest neighbor principle ((b-1), (b-2)). (a-1), (b-1) are conducted in clean and (a-2), (b-2) in noisy. The X-axis is the number of K-nearest points for threshold, and K = 0 means τi = 0. Methods RMSE(R)↓ MAE(R)↓ RMSE(t)↓ MAE(t)↓ clean noisy clean noisy clean noisy clean noisy ICP 12.545 12.723 5.438 5.298 0.046 0.045 0.024 0.023 FGR 20.042 40.829 7.203 21.065 0.035 0.060 0.019 0.039 DCP-v2 6.265 6.347 3.990 4.294 0.014 0.016 0.011 0.012 PRNet 3.532 4.321 1.760 1.826 0.013 0.013 0.010 0.010 SHMDCP 2.522 3.886 0.833 1.510 0.005 0.006 0.003 0.004 RPMNet 0.886 1.631 0.345 0.565 0.006 0.011 0.003 0.004 SHMRPMNet 0.514 1.456 0.247 0.378 0.004 0.008 0.002 0.003 Table 3: Registration perfomance on clean, noisy. Figure 4: For each pair of point clouds, left upper is the transformed source using the predicted rigid motion and right bottom is the target. The same color represents the cor-respondences, and black indicates the abandoned points. Methods DCP PRNet SHMDCP RPMNet SHMRPMNet Time(ms) 10.66 45.48 98.03 54.75 409.95 Table 4: Inference time of learning-based methods. all baselines with noisy data. And the methods with our S2H matching also achieves the best performance in these two frameworks in all evaluation metrics. These evaluations val-idate the improvement of our S2H matching, which creates a new state-of-the-art results. More intuitively, we provide some qualitative registration performance in Fig. 4. • Time-efficiency: We counted the average inference time of learning-based methods in Table 4. The experiments are conducted in noisy on ModelNet40 using a Xeon E5-2640 v4@2.40GHz CPU and a GTX 1080Ti. S2H matching-based methods are slower because the integer programming algorithm is time-consuming. SHMRPMNet is slower than SHMDCP due to the iteration in RPMNet framework. 3404 Methods TE(cm)↓RE(deg)↓Recall(%)↑Time(s)↓ ICP 18.1 8.25 6.04 0.25 FGR 10.6 4.08 42.7 0.31 Go-ICP 14.7 5.38 22.9 771.0 Super4PCS 14.1 5.25 21.6 4.55 RANSAC 9.16 2.95 70.7 2.32 DCP-v2 21.4 8.42 3.22 0.07 PointNetLK 21.3 8.04 1.61 0.12 DGR w/o safeguard 7.73 2.58 85.2 0.70 DGR 7.34 2.43 91.3 1.21 PointDSC 6.55 2.06 93.28 0.09 SHMDGR 6.41 1.75 91.7 37.2 Table 5: Evaluation on 3DMatch dataset. Evaluation on Real Indoor Data: 3DMatch In this part, we evaluate our S2H matching in SHMDGR on real indoor dataset, 3DMatch (Zeng et al. 2017). SHMDGR takes the S2H matching replacing the original matching of DGR (Choy, Dong, and Koltun 2020) (more details can be seen in supplementary materials). Following DGR, the input point clouds have been voxelized with the voxel size of 5cm, then each of them contains ∼50k points. • Metric: For a fair comparison, we follow the protocols and evaluation metrics of DGR here. The average Rotation Error (RE: arccos((trace(Rpred−1Rgt) −1)/2) 180 π ), aver-age Translation Error (TE: ∥tpred −tgt∥2 2), recall, and time-efficiency are reported. Recall indicates the ratio of success-ful pairwise registrations. Here, the successful pair is con-firmed if the RE and TE are smaller than pre-defined thresh-olds (i.e., RE<15◦, TE<30cm). • Evaluation: From the Table 5, we find that ICP achieves the weak registration result since the dataset contains chal-lenging sequences with large motion while no reliable prior is provided. Super4PCS (Mellado, Aiger, and Mitra 2014), and Go-ICP (Yang et al. 2015), which are sampling-based algorithm and the variant of ICP with branch-and-bound search respectively, present similar performance here. FGR and RANSAC perform better due to the extracted point fea-tures. 3DRegNet (Pais et al. 2020) is also tested here, how-ever, it does not converge, which outputs the error above 30◦ and 1m. DGR is the state-of-the-art learning-based method, which is designed for scene data specifically. However, DGR also takes the most similar points as the corresponding points ignoring the one-to-one matching principle. SHMDGR achieves better registration performance including transfor-mation estimation and recall comparing with the original DGR. PointDSC (Bai et al. 2021) achieves the best recall but the registration performance is weaker than SHMDGR. Evaluation on Real Outdoor Data: KITTI We evaluate SHMDGR on the real outdoor data, KITTI (Geiger et al. 2013). Here, we also follow the protocols of DGR, where the evaluation metrics are identical to the 3DMatch evaluation. The thresholds to confirm the success-ful pair are set to 0.6m and 5◦. From Table 6, the SHMDGR achieves the best performance with respect to rigid transfor-mation estimation and recall, which outperforms the original DGR and the state-of-the-art method, PointDSC. Methods TE(cm)↓RE(deg)↓Recall(%)↑Time(s)↓ FGR 40.7 1.02 0.2 1.42 RANSAC 25.9 1.39 34.2 1.37 FCGF 10.2 0.33 98.2 6.38 DGR 21.7 0.34 96.9 2.29 PointDSC 20.94 0.33 98.20 0.31 SHMDGR 9.32 0.28 99.3 52.4 Table 6: Evaluation on KITTI dataset. Methods RMSE(R)↓ MAE(R)↓ RMSE(t)↓ MAE(t)↓ clean noisy clean noisy clean noisy clean noisy SHM− DCP 3.325 4.570 1.039 1.683 5.056 6.624 3.204 4.316 SHM+ DCP 2.522 3.886 0.833 1.510 4.780 5.758 3.052 3.774 SHM− RPMNet 0.760 1.537 0.270 0.449 4.927 10.120 2.356 3.602 SHM− RPMNet 0.514 1.456 0.247 0.378 3.883 8.421 2.328 3.145 Table 7: Comparison between the end-to-end learning and post-processing setting. “-” indicates the post-processing, and “+” indicates the end-to-end learning. Note: RMSE(t) and MAE(t) results (×1000) are reported. Ablation Studies End-to-end vs. post-processing. In this paper, we advocate learning the one-to-one matching in an end-to-end manner, i.e. achieving the PPM in the matching stage. Oppositely, another natural idea is to solve the PPM in post-processing, i.e. using only the S-step to learn the soft matrix during the training and adding the H-step during the test for the final hard matrix, PPM. The comparison of these two ideas is given in Table 7, where the experiments are conducted on ModelNet40. We can see that the end-to-end learning achieves better results in all metrics because more accurate loss are calculated in this pipeline. Other important ablation studies. Due to the limitation of space, we provide some other import experiments in supple-mentary materials, including the robustness to different out-liers generation strategy, the robustness to different outliers ratio, the influence of different loss function combinations, and more qualitative results of registration, etc. Conclusion In this paper, we tackle the point matching problem in robust 3D point cloud registration. First, we analyze the inherent ambiguity in soft matching-based methods. Second, to re-solve the ambiguity and handle the outliers in the matching stage, we propose to learn the partial permutation matching (PPM) matrix. To address the consequent problem that PPM is defined in non-differentiable space and cannot be solved by existing hard assignment algorithms, we design a soft-to-hard matching method. We have validated the effective-ness by integrating it with various registration frameworks including DCP, RPMNet, and DGR and conducting exten-sive experiments in both synthetic data and real scan data, which created a new state-of-the-art performance for robust 3D point cloud registration. In the future, we plan to extend our framework to non-rigid point cloud registration. 3405 Acknowledgments This research was supported in part by National Key Research and Development Program of China (2018AAA0102803) and National Natural Science Founda-tion of China (61871325, 62001394, 61901387). 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In Pro-ceedings of the IEEE Conference on Computer Vision and Pattern Recognition, 199–208. Zeng, Y.; Qian, Y.; Zhu, Z.; Hou, J.; Yuan, H.; and He, Y. 2021. CorrNet3D: Unsupervised End-to-end Learning of Dense Correspondence for 3D Point Clouds. In Proceed-ings of the IEEE Conference on Computer Vision and Pat-tern Recognition, 6052–6061. Zhang, Z.; Dai, Y.; and Sun, J. 2020. Deep learning based point cloud registration: an overview. Virtual Reality and Intelligent Hardware., 2(3): 222–246. Zhang, Z.; Sun, J.; Dai, Y.; Fan, B.; and He, M. 2022. VR-Net: Learning the Rectified Virtual Corresponding Points for 3D Point Cloud Registration. IEEE Transactions on Circuits and Systems for Video Technology. Zhao, Y.; Birdal, T.; Lenssen, J. E.; Menegatti, E.; Guibas, L. J.; and Tombari, F. 2020. Quaternion Equivariant Cap-sule Networks for 3D Point Clouds. In Proceedings of the European Conference on Computer Vision, 1–19. Zhou, Q.-Y.; Park, J.; and Koltun, V. 2016. Fast global reg-istration. 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3374	https://www.federalregister.gov/documents/2018/02/13/2018-02871/bacillus-calmette-gurin-unresponsive-nonmuscle-invasive-bladder-cancer-developing-drugs-and	Federal Register :: Bacillus Calmette-Guérin-Unresponsive Nonmuscle Invasive Bladder Cancer: Developing Drugs and Biologics for Treatment; Guidance for Industry; Availability Skip to Content Home Sections Money Environment World Science & Technology Business & Industry Health & Public Welfare Browse Agencies Topics (CFR Indexing Terms) Dates Public Inspection Presidential Documents Search Document Search Advanced Document Search Public Inspection Search FR Index Reader Aids Reader Aids Home Office of the Federal Register Announcements Using FederalRegister.Gov Understanding the Federal Register Recent Site Updates Federal Register & CFR Statistics Videos & Tutorials Developer Resources Government Policy and OFR Procedures My FR My Account My Clipboard My Comments My Subscriptions Sign In / Sign Up Site Feedback Search the Federal Register Legal Status This site displays a prototype of a “Web 2.0” version of the daily Federal Register. 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SUMMARY: The Food and Drug Administration (FDA or Agency) is announcing the availability of a guidance for industry entitled “Bacillus Calmette-Guérin (BCG)-Unresponsive Nonmuscle Invasive Bladder Cancer: Developing Drugs and Biologics for Treatment.” This guidance was developed to assist in the development of drugs and biologics for patients with a form of bladder cancer that is not amenable to currently available medical therapy and remains an unmet medical need. This guidance finalizes the draft guidance of the same name issued on November 18, 2016. DATES: The announcement of the guidance is published in the Federal Register on February 13, 2018. ADDRESSES: You may submit either electronic or written comments on Agency guidances at any time as follows: Electronic Submissions Submit electronic comments in the following way: Federal eRulemaking Portal: Follow the instructions for submitting comments. 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3375	https://www.geeksforgeeks.org/electrical-engineering/theory-of-transformer-on-load-and-no-load-operation/	Theory of Transformer on Load and No Load Operation Last Updated : 27 Feb, 2024 Suggest changes 4 Likes In this article, we will study the theory of transformer on load and no load operation. A transformer is a static electrical machine used to increase or decrease the value of voltage and current in an electrical circuit. The transformer operates on the principle of electromagnetic induction and mutual inductance. A transformer typically consists of two copper winding and a magnetic core. The windings are named as primary winding and secondary winding. The input supply is connected to the primary winding and the output electrical supply is taken from the secondary winding. Hence, the secondary winding is one to which the electrical load is connected. Let us understand the operation of a transformer on load and no-load conditions. Transformer on No-Load A transformer is said to be operated in no-load condition if no electrical load is connected across its secondary winding terminals. In other words, when the secondary winding of a transformer remains open-circuited and no current flows through it, then the transformer is said to be in no-load condition. When a transformer operates in no-load condition, no current flows in the secondary winding, but a small current called no-load current (I0) which is around 2% to 10% of the rated current flows in the primary winding. This no-load current I0 magnetizes the core of the transformer and results in some core losses (i.e., iron losses and copper losses). The following figure depicts the circuit diagram of an electrical transformer in no-load condition. In this circuit, we can see that there is an input voltage of V1 volts is applied to the primary which forces a no-load current of I0 ampere through it. The secondary winding is open-circuited, hence the secondary I2 is equal to zero amperes. As we discussed that when a transformer is operated in no-load condition, it draws only a small current called no-load current from the input power supply. This no-load current has two components namely, Magnetizing component Power component The magnetizing component is also called the reactive component and it is responsible for setting up the magnetic flux in the core. This component of the no-load current lags the supply voltage V1 by an angle of 90°, as it is due to the inductive effect. This component is usually denoted by Im. The power component is also called the active component and it is responsible for supplying core losses i.e., iron and copper losses. The active component remains in-phase with the supply voltage. In other words, the angle between the supply voltage V1 and the active component of the no-load current. This component is designated by the symbol Iw. Therefore, we can express the no-load current of the transformer as the phasor sum of the magnetizing component and the power component i.e., I0=Im+Iw Operation of Transformer on No-Load Here is the operation of a transformer in no-load condition: Firstly, the input voltage V1 is applied across the primary winding of the transformer. It magnetizes the core of the transformer by setting up a magnetic flux (ϕ) in it. This magnetic flux induces EMFs in the primary and secondary windings due to electromagnetic induction. These induced EMFs lag the applied voltage by an angle of 90°. Here, we are neglecting the primary winding copper loss and the secondary winding copper loss is zero because I2 = 0. The no-load current I0 lags behind the supply voltage V1 by an angle of ϕ0 which is called the no-load power factor angle. It is also important to note that the EMF E1 induced in the primary winding is equal to the supply voltage V1. There are also some losses in the primary winding (copper loss) and the core (iron losses) of the transformer which are represented by the active component (Iw) of the no-load current. This component is equivalent to the resistive effect and hence remains in-phase with the supply voltage (V1). This is how an electrical transformer operates under no-load conditions. This complete operation of the transformer on no-load can be illustrated with the help of a phasor diagram which is shown in the following figure. Phasor Diagram Explanation The induced EMF (E1 and E2) are out of phase with respect to the supply voltage (V1) and lags the magnetic flux by 90°. The reactive component Im lags the supply voltage V1 by 90°. The active component Iw is in-phase with the supply voltage (V1). The no-load current I0 lags the supply voltage V1 by an angle φ0 which is the no-load power factor angle of the transformer. From this phasor diagram, we can derive the important relations of different electrical parameters of the transformer. They are, (1). The magnitude of no-load current: I0=Im2+Iw2 (2). The magnetizing component of no-load current: Im=I0sinϕ0 (3). The active component of no-load current: Iw=I0cosϕ0 (4). The no-load power factor: cosϕ0=I0Iw (5). The power consumed under no-load: P0=V1I0cosϕ0 Hence, this is all about the theory and operation of an electrical transformer in no-load condition. Let us now study the theory and operation of an electrical transformer under loaded conditions. Transformer on Load A transformer is said to be on load condition if an electrical load is connected to its secondary winding and a current circulates in the secondary winding circuit. The load connected across the secondary winding can be a resistive load or an inductive load or a capacitive load or a combination of the three. Therefore, the magnitude of the secondary winding current also called load current depends on the load impedance and secondary voltage (V2). Also, the phase angle between the secondary voltage and load current depends on the type of the load. For example, if the load is of inductive nature, the load current will lag the secondary voltage. Operation of Transformer on Load Here is the operation of an electrical transformer operation under loaded condition: An electrical input supply voltage V1 is connected across the primary winding. Due to the application of this voltage an electric current I1 will starting flowing in the primary winding and sets up a magnetic flux in the core as shown in the above figure. This magnetic flux follows a path through the core and links to the secondary winding. An EMF E2 is induced in the secondary winding that developers a voltage V2 across its terminals. When a load is connected between the secondary winding terminals, a current will flow in the secondary winding and load circuit which is denoted by I2. The secondary winding current also induces a counter magnetic flux that reduces the main flux in the core. But the main flux must be maintained at a constant value for operation of the transformer. Thus, an additional current is taken by the primary winding from the supply to cancel out the demagnetizing effect of secondary winding current. This is represented by I'1 which is in-phase with the secondary winding current (I2). Thus, the total current flowing in the primary winding under loaded condition of the transformer is I'1. Therefore, if N1 and N2 are the primary and secondary winding turns and I'1 and I2 are the primary and secondary currents, then the EMF balance equation of the transformer on load is given by, N1I1′=N2I2 Hence, the primary winding current will be, I1′=(N1N2)I2=kI2 Also, the total primary winding current (I1) has two main components namely, No-load component - to set up main magnetic flux in the core and supply the losses. Counter balancing component - to overcome the effect of secondary winding current. Thus, the total primary winding current of the transformer on load condition is given by, I1=I0+I1′ The bold facing letter denotes the phasor sum of current components. In practice, the transformers mostly have inductive loads. So, let's assume the load connected across the secondary of the transformer and draw the phasor diagram of it on load condition. Phasor Diagram Explanation The steps given below are to be followed to draw the phasor diagram of the transformer on load: Step (1) : Taking main magnetic flux (ϕ) as the reference axis. Step (2) : This magnetic flux induces EMFs E1 and E2 in primary and secondary windings. Hence, E1 and E2 lags the flux (ϕ) by an angle of 90°. Step (3) : The applied voltage across the primary winding is utilized in three components namely, to induce emf E1, voltage drop in primary winding resistance I1R1, and voltage drop in primary winding reactance I1X1. Hence, the primary winding voltage V1 is the phasor sum of voltage component corresponding to E1 and the voltage drops in primary winding. Step (4) : If ϕ1 is the power factor angle of the primary winding, then the current I1 will lag the voltage V1 by ϕ1. It is the phasor sum of no-load current I0 (lags the supply voltage by 90°) and counter balancing current I'1 (in anti-phase with the secondary voltage V2). Step (5) : The secondary winding EMF E2 is the phasor sum of voltage V2 and the voltage drops in secondary winding resistance and reactance. Step (6) : The magnitude and phase angle of the load current or secondary current I2 depends on the load connected to the transformer. As in this case, we are assuming an inductive load, hence the load current I2 will lag the voltage V2. If the load is capacitive, the load current will lead the voltage V2. Thus, when we operate a transformer on load condition, it takes both no-load current and current required to handle the load connected across its secondary winding. Let us now understand the theory of transformer on load with two other load conditions Transformer on load with purely resistive load Transformer on load with purely inductive load Transformer on load with purely capacitive load Transformer on Load with Purely Resistive Load When a purely resistive load is connected across the secondary winding of a transformer, it draws only active current from the transformer. The resistive load creates a phase difference of 0° between the load current and the secondary winding voltage. The phasor diagram of the transformer on load with purely resistive load is shown in the following figure. Transformer on Load with Purely Inductive Load When a purely inductive load is connected across the secondary winding of the transformer. It cause a phase different of exactly 90° between the secondary voltage and load current. The phasor diagram of a transformer on load with purely inductive load is shown in the following figure. It can be observed that the load current I2 lags the secondary voltage V2 by an angle of 90°. Transformer on Load with Purely Capacitive Load When a purely capacitive load is attached to the secondary winding of the transformer, it draws a leading current from the secondary winding. This causes a phase difference of 90° between the load current and the secondary voltage. Where, the load current I2 leads the secondary voltage V2 by an angle of 90° as shown in the following phasor diagram. Till this section of the article, we have discussed about the operation of a practical transformer on no-load and on-load (with resistive load, inductive load, and capacitive load). Also, we have analyzed a practical transformer on load with winding resistances and leakage reactances. However, there can be three other possible cases, they are: Transformer on Load with No Winding Resistances and Leakage Reactances Transformer on Load with Winding Resistances but No Leakage Reactances Transformer on Load with Leakage Reactances but No Winding Resistances Let us understand these three cases of transformer operation on load in detail. Transformer on Load with No Winding Resistances and Leakage Reactances When we consider a transformer on load with no winding resistances and leakage reactances, then its electrical circuit will look like as shown in the following figure. In this case, there are no voltage drops due to resistances and leakage reactances of the windings. Thus, the terminal voltages and induced emfs are equal i.e., E1 = V1 and E2 = V2 If we consider a practical inductive load connected across the secondary winding, then the load current I2 will lags behind the secondary winding voltage V2 by an angle of ϕ2. Where, ϕ2 is the power factor angle of the load. In this condition, the primary winding current I1 will supply the following two things: It supplies the no-load current I0 to establish magnetic flux in the core and iron losses. It cancel out the demagnetising effect of the secondary winding current. Therefore, the primary winding current is given by, I1=I0+I1′ Here, the component I'1 is the counter balance current of the secondary winding current. If we draw the phasor diagram for this case of the transformer on load, then it will look like as shown below. In this phasor diagram, the secondary winding voltage V2 is equal to the secondary winding emf E2 and they are in-phase. While, the primary winding voltage V1 is equal to the primary winding emf E1 but they are out of phase with each other. The primary winding current is the phasor sum of I0 and I'1. Transformer on Load with Winding Resistances but No Leakage Reactances The circuit diagram of a transformer on load having winding resistances but no leakage reactances is shown below. In this circuit diagram, the resistors R1 and R2 represent the primary and secondary winding resistances. A practical inductive load is connected across the secondary winding terminals. When the input voltage V1 is connected to the primary winding, there will be a voltage drop in the primary winding resistance R1 and the secondary winding resistance R2. Therefore, we get the following equations of the voltages, V1=E1+I1R1 And E2=V2+I2R2 The phasor diagram for this case of transformer on load is depicted in the following figure. Transformer on Load with Leakage Reactances but No Winding Resistances The circuit diagram of a transformer on load having leakage reactances only and not having the winding resistances is shown in the figure. Here, X1 and X2 represents the leakage reactances of the primary and secondary windings respectively. The voltage equations are given by, V1=E1+I1X1 And E2=V2+I2X2 The phasor diagram for this case of transformer on load is depicted below. Where, we have considered a practical inductive load. Conclusion In conclusion, a transformer is a device used to change the levels of voltage and current in an electric circuit. A transformer when operated with open-circuited secondary winding, it is said to be in no-load operation and when it is operated with attaching a load across its secondary, it is said to be in on-load operation. In this detailed article on "Theory of Transformer on Load and No-Load Operation", we have discussed the operation of a transformer under both conditions and explained its behavior using the phasor diagrams. 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3376	https://www.kristakingmath.com/blog/opposite-of-a-number	What are opposite numbers? — Krista King Math \| Online math help About Pricing Login Subscribe Risk Free AboutPricingLogin Step-by-step math courses covering Pre-Algebra through Calculus 3. Subscribe Risk Free What are opposite numbers? What are opposite numbers, and what can we do with them? When we talk about the “opposite of a number,” we’re specifically talking about the positive and negative versions of the same number. Now that’s not a technical definition by any means, so let me show you what I mean. Hi! I'm krista. I create online courses to help you rock your math class. Read more. But before we get into that, let’s take a step back and talk about negative numbers. The easiest way to illustrate a negative number is by using the number line. On this number line, we can see the number ???0??? in the center. The numbers to the right of ???0??? are always positive numbers; the numbers to the left of ???0??? are always negative numbers, and this is always true when we’re talking about a number line like this. The other important thing to realize about this number line is that the numbers ???1??? and ???-1??? are the same distance away from ???0???. We could say they’re both “one unit” away from ???0???, in fact. In the same way, ???4??? and ???-4??? are the same distance from ???0???; they’re both “four units” away from ???0???. And this brings us back to the concept of “opposite” numbers. Opposite numbers are numbers which are the same distance away from ???0???. And that’s why ???1??? and ???-1??? are opposites (because they are the same distance from ???0???), and why ???4??? and ???-4??? are opposites (because they are also the same distance from ???0???). On the other hand, ???2??? and ???4??? are not opposites of each other, because the number ???2??? is two units away from ???0???, whereas the number ???4??? is four units away from ???0???. Two numbers are not opposites of each other if they are different distances away from ???0???. The easiest way to remember how to find the opposite of a number is to add a negative sign if the number is positive, or to take away the negative sign if the number is negative. Opposite numbers graphed on a number line Take the course Want to learn more about Pre-Algebra? I have a step-by-step course for that. :) Learn More Finding the opposite of a positive number Example What is the opposite of ???7???? To find the opposite of the positive number ???7???, we simply add a negative sign to get ???-7???. Both ???7??? and ???-7??? are seven units from ???0??? on the number line, so ???7??? and ???-7??? are opposites of each other. Let’s try an example with a negative number. Opposite numbers are numbers which are the same distance away from 0. Example What is the opposite of ???-10???? To find the opposite of the negative number ???-10???, we simply take away the negative sign to get ???10???. Both ???10??? and ???-10??? are ten units from ???0??? on the number line, so ???10??? and ???-10??? are opposites of each other. How about ???0???? Does ???0??? have an opposite? We know from looking at the number line from earlier that we can find the opposite of a number by measuring out the same distance from ???0???, but just on the opposite side of ???0???. So since ???5??? is five units away from ???0??? to the right of ???0???, then the opposite of ???5??? is the number that’s five units away from ???0??? to the left of ???0???, which is ???-5???. But ???0??? itself is “zero units” away from ???0??? (keep in mind that it’s the ONLY number that’s “zero units” away from ???0???). ???0??? isn’t a positive number or a negative number. So the opposite of ???0??? is actually ???0???. In other words, ???0??? is its own opposite! Get access to the complete Pre-Algebra course Get started Learn mathKrista KingMarch 28, 2021math, learn online, online course, online math, prealgebra, pre-algebra, fundamentals, fundamentals of math, foundations, foundations of math, opposite, opposites, opposite numbers, opposite of a number, number line, positive numbers, negative numbers, positive and negative numbers, same distance from 0 Facebook0 Twitter LinkedIn0 Reddit Tumblr Pinterest0 0 Likes Previous Using unit multipliers to convert between different units Learn mathKrista KingMarch 29, 2021math, learn online, online course, online math, prealgebra, pre-algebra, foundations, foundations of math, fundamentals, fundamentals of math, unit multipliers, converting units, unit conversions, converting. units Next Finding the equations of the normal and osculating planes Learn mathKrista KingMarch 27, 2021math, learn online, online course, online math, calculus iii, calculus 3, calc iii, calc 3, multivariable calc, multivariable calculus, vector calc, vector calculus, normal plane, osculating plane, unit tangent vector, unit normal vector, binormal vector, normal and osculating planes Online math courses Subscribe Risk Free Courses Placement Course Custom Path Pre-Algebra Algebra 1 Geometry Algebra 2 Trigonometry Precalculus Probability & Statistics Calculus 1 Calculus 2 Calculus 3 Differential Equations Linear Algebra Hours BlogContactCoursesFAQ Copyright © 2025 Krista King Math
3377	https://blog.csdn.net/qq_34734303/article/details/80582882	空间统计--二项分布和泊松分布案例代码分析_poisson编码实例-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作空间统计--二项分布和泊松分布案例代码分析最新推荐文章于 2025-07-11 14:51:42 发布原创于 2018-06-05 17:16:08 发布·3.4k 阅读 · 2 · 12· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #python学习#泊松分布#二项分布 Python学习同时被 3 个专栏收录 14 篇文章订阅专栏泊松分布 1 篇文章订阅专栏二项分布 1 篇文章订阅专栏本文通过美国1982年至2012年枪击案的实际数据，介绍了如何利用泊松分布来评估犯罪趋势，并探讨了二项分布及其与泊松分布的关系。空间统计–二项分布和泊松分布案例代码分析一、泊松分布日常生活中，大量的事件是有固定频率的：它们的特点就是，我们可以预估这些事件的总数，但是没法知道具体的发生时间。已知平均每小时出生3个婴儿，请问下一个小时，会出生几个？有可能一下子出生6个，也有可能一个都不出生。这是我们没法知道的。上面就是泊松分布的公式。等号的左边，P 表示概率，N 表示某种函数关系，t 表示时间，n 表示数量，1小时内出生3个婴儿的概率，就表示为 P(N(1)=3) 。等号的右边，λ 表示事件的频率。此处用1982–2012年枪击案的分布来学习泊松分布这个例子给你了美国30年来每年的枪击案发生数目，需要解决的问题是能否从每年发生枪击案的数目判断美国枪击犯罪是否恶化。假设美国枪击案犯罪没有恶化，而是非常稳定，我们可以假设：枪击案的发生为泊松过程，每年平均发生枪击案的数目恒定（条件3），各个年份之间发生枪击案的数目不互相影响（条件2），任一时刻发生枪击案的概率很小（条件1），所以每年发生枪击案的数目服从泊松分布。假设美国枪击案满足“泊松分布”三个条件：（1）以下是，1982–2012年枪击案的分布情况：（2）计算得到，平均每年发生2起枪击案，所以 λ=2。在我们的假设下每年发生枪击案的数目服从泊松分布，那么一年内发生0起枪击案的概率为: 一年内发生1起枪击案的概率为: 依此类推，那么我们可以得到一张分布图：我们假设出的年枪击案数目分布和实际枪击案发生数目的对照表：上图中，蓝色的条形柱是实际的观察值，红色的虚线是理论的预期值。可以看到，观察值与期望值还是相当接近的。可以看到，在频率（2）附近，事件的发生概率最高，然后向两边对称下降，即变得越大和越小都不太可能。平均每年发生2起枪击案，这是最可能的结果，发生的得越多或越少，就越不可能。然后可以得到一张我们假设出的年枪击案数目分布和实际枪击案发生数目的对照表：再由一些统计学的计算方法,计算出我们假设的值与实际观测的值是否接近。如果接近，则说明我们的假设-枪击案发生为泊松过程-是正确的。可以用“卡方检验”如下：更多详细信息可点击更多二、伯努利分布伯努利分布，这个是最简单的分布，就是0-1分布以抛硬币为例，为正面的概率为p，反面的概率为q是一种离散型概率分布，也是很多分布的基础，它是二项分布的特殊情况。三、二项分布简单地说当一个实验满足：（1）每次试验中事件只有两种结果：事件发生或者不发生，如硬币正面或反面，患病或没患病；（2）每次试验中事件发生的概率是相同的，注意不一定是0.5；（3）n次试验的事件相互之间独立。以上三个条件时，即满足了伯努利试验。如抛硬币抛，在一次试验中硬币要么正面朝上，要么反面朝上，每次正面朝上的概率都一样 p=0.5，且每次抛硬币的事件相互独立，即每次正面朝上的概率不受其他试验的影响。若只抛一次则其满足伯努利分布，而二项分布就是重复n次独立的伯努利试验，X~B(n,p), 如实例：某人篮球投篮的命中率是0.3，总共投篮10次，问至少投中2次的概率? 分析：（1）每次投篮有2种结果，投中或没投中；（2）每次投篮的投中概率是相同的，都为0.3；（3）每次投篮可认为是独立事件。因此，符合二项分布。显然，二项分布属于离散型分布。至少2次投中概率即： P(X>=2)=P(X=2)+P(X=3)+P(X=4)+...+P(X=10)P(X>=2)=P(X=2)+P(X=3)+P(X=4)+...+P(X=10) 输出结果：0.85 四、二项分布和泊松分布关系 n 很大，p 很小时泊松分布可以用来近似二项分布,简单说泊松分布可看成是二项分布的极限而得到，记常数 λ=n p。更多详细关系可点击更多备注：博客中泊松分布图python 实现代码为： ``` import numpy as np import scipy.stats as st #引入stats包。其内部包含有许多分布 import matplotlib.pyplot as plt rv = st.poisson(2) # 强度为2的泊松分布 num_years = [4, 10, 7, 5, 4, 0, 0, 1] #测量值 x = range(8) #可看成柱状图数量 plt.bar(np.array(x), num_years, label='Observed instances') #画直方图（横坐标，纵数值，标签） plt.plot(x, sum(num_years)rv.pmf(x), ls='dashed', lw=2, c='r', label='Poisson distribution\n$(\lambda=2.0)$') #画直方图（横坐标，纵数值，虚线，线宽为2，线为红色，标签） print(sum(num_years)rv.pmf(x)) #泊松分布计算值 plt.xlim([-1, 8]) #横轴 plt.ylim([0, 11]) #纵轴标题 plt.title('Mass Shootings in USA 1982-2012') plt.xlabel('Number of mass ahooting in a year') plt.ylabel('Number of years') plt.legend(loc='best') plt.show() #显示图 ``` AI写代码 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 qq_34734303 关注关注 2点赞踩 12 收藏觉得还不错? 一键收藏 2评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录 21、空间数据插值与点分布统计分析 y9z0a1b的博客 08-27 25 本文详细介绍了空间数据处理中的插值方法和点分布统计分析技术。内容涵盖双线性插值、双调和样条插值、立方插值、自然邻域插值和最近邻插值等常用插值方法的特点与应用，并结合MATLAB示例代码说明了不同方法的实现方式和适用场景。同时，还讨论了点分布统计分析的三种主要测试方法：均匀分布测试、随机分布测试和聚类测试，以及它们在考古学、古生物学、地质学、城市规划和生态研究等领域的应用。文章旨在帮助读者更好地理解和应用空间数据处理技术，为相关研究和实践提供支持。 MATLAB基础应用精讲 -【数模应用】负二项回归（附R语言和 python 代码实现） qq_36130719的博客 10-18 1022 为了解某抗癫痫药物的作用，某研究者对58名癫痫患者进行临床试验，每将其随机分为2组，一组服用试验药，一组服用对照药，治疗8周，观察最后两周的发作次数，其中治疗方法包括安慰剂和试验药，请试着分析抗癫痫药的疗效。 2 条评论您还未登录，请先登录后发表或查看评论 SNN图像脉冲化——泊松编码 9-28 该实例中,首先将像素值归一化作为脉冲发放概率,模拟255个时间步,每个像素各个通道产生的泊松编码保存在文件 Poisson.out中。某像素三通道泊松编码片段重建图像reconduct.png能够较准确地还原图像,说明得到的编码较好地保留了原图像信息;reconduct.gif复现了编码过程中不同时间步下的重建图像。 reconduct.gif Poisson Recon:鱼类表面重建资源 9-23 Screened Poisson Surface Reconstruction(Version 6.11)根据官网的步骤执行的结果,并通过meshlab打开执行结果,从多个角度观察,此文档用来分析。格式:pdf 资源大小:9.3MB 页数:38 鱼类:鱼类项目浏览:104 《鱼类:鱼类项目》在IT领域,"鱼类:鱼类项目"可能是一个使用C语言编写的软件开发项目,旨在模拟、研究或者管理... 机器学习储备（12）：二项分布的例子解析算法channel 11-29 5892 请点击上面公众号，免费订阅。交流思想，注重分析，更注重通过实例让您通俗易懂。包含但不限于：经典算法，机器学习，深度学习，LeetCode 题解，Kaggle 实战。期待您的到来！ 01 — 二项分布如果实验满足以下两种条件：在每次试验中只有两种可能的结果，而且两种结果发生与否互相对立；相互独立，与其它各次试验结果无关；事件发生泊松分布函数 10-27 泊松分布【问题描述】泊松分布是一种常用的离散型概率分布，数学期望为m的泊松分布的分布函数定义如下： P(m, k) = mk e - m/k! (k = 0, 1, 2, 3, …) 对于给定的m 和 k (0<m<2000, 0<= k < 2500)，计算其概率，以科学格式输出，保留小数点后6位有效数字。可以使用数学库函数，误差不超过0.000001。【输入形式】输入文件为当前目录下的 poisson.in。文件中包含两个数字，分别为m,k的值。【输出形式】输出文件为当前目录下的 poisson.out。文件中输出泊松分布的值，值以科学格式输出，保留小数点后6位有效数字。【输入样例】 1 0 【输出样例】 3.678794e - 01 【时间限制】 1s 【空间限制】 65536KB 【机器学习】R语言进行机器学习方法及实例 r语言机器 _学习算法 - CSDN... 9-25 glmnet(x, y, family=c("gaussian","binomial","poisson","multinomial","cox","mgaussian"), weights, offset=NULL, alpha = 1, nlambda = 100, lambda.min.ratio = ifelse(nobs<nvars,0.01,0.0001), lambda=NULL, ... Pytorch 学习之十九种损失函数_pytorch损失函数 9-14 torch.nn.Poisson NLLLoss(log_input=True, full=False, eps=1e - 08, reduction='mean') AI运行代码 1 参数: log_input (bool, optional) – 如果设置为 True , loss 将会按照公式 exp(input) - target input 来计算, 如果设置为 False , loss 将会按照 input - target log(input+eps) 计算. ... 【常见分布及其特征(4)】离散型随机变量 - 泊松分布最新发布 qq_37293230的博客 07-11 467 泊松分布（Poisson Distribution）是一种离散型概率分布，用于描述在固定时间或空间内独立随机事件发生次数的概率分布X∼πλ或X∼Poisson λ\boxed{X \sim \pi(\lambda)\quad或\quad X\sim \text{Poisson}(\lambda)}X∼πλ或X∼Poisson λXXX服从参数为λ\lambdaλ的泊松分布; 空间统计分析_（案例）空间分析6.4江西省地级市社会经济统计分析 weixin_39960503的博客 12-16 1167 空间统计分析案例案例 4：江西省地级市社会经济统计分析(数据与视频文件下载见文末)•研究目的：了解地理学第一定律与空间邻接的含义，了解ArcGIS 空间权重矩阵/空间关系的概念化的方法，掌握全局与局部Moran's I、全局与局部Getis - Ord G指标方法的实践操作步骤；能够综合运用这些空间自相关分析方法来解答点、面数据分布特征与模式的相关实际应用问题，并能够将应用方法解决问题的... 机器学习(ML)、深度学习(DL)和图像处理(opencv)专用英语词典_机器学习 m... 9-27 poisson distribution柏松分布 pertinent相关的 PCA,主成分分析 ( Principal Component Analysis , PCA )或者主元分析。是一种掌握事物主要矛盾的统计分析方法,它可以从多元事物中解析出主要影响因素,揭示事物的本质,简化复杂的问题。计算主成分的目的是将高维数据投影到较低维空间。用二元泊松模型预测2022年世界杯结果模型源码_泊松模型足球软件... 9-22 本资源是用二元泊松模型预测2022年世界杯结果的R语言模型源码网上有很多文章用双泊松(Double Poisson)模型来预测世界杯比赛结果。但是双泊松模型有一个严重的缺陷,那就是它假设比赛中两队的比分是条件独立的。而我们都知道,在对抗性比赛中,两队的比分是存在关联的,因为两队都会根据场上的比分形势调整策略。比如足球... 泊松分布采样（Poisson - Disk - Sample）代码及详细注释【OpenCV】热门推荐 ShaderJoy 的兴趣技术杂货铺 09-23 2万+ 算法步骤简述关键代码如下： // 以center为圆心radius为半径的圆环范围内随机产生新的采样点 template void sample_annulus(T radius, const Vec ¢re, unsigned int &seed, Vec<N, T&gt... 二项分布近似泊松分布 Random南荞的博客 07-16 2639 一般来说，当n≥20,p≤0.02的时候，二项分布近似泊松分布。且泊松分布满足可加性。 mgwr:多尺度地理加权回归(MGWR)_mgwr资源 9-5 GWR model calibration via iteratively weighted least squares for Gaussian, Poisson, and binomial probability models. GWR bandwidth selection via golden section search or equal interval search GWR - specific model diagnostics, including a multiple hypothesis test correction and local collinearity ... Poisson 分布随机变量生成与应用 9-7 Poisson 分布随机变量生成与应用本文还有配套的精品资源,点击获取简介:泊松分布是统计学和概率论中的基础概念,其随机变量生成在金融、质量管理、保险等领域有着广泛的应用。泊松分布描述了固定时间间隔或空间单位内,某一事件发生次数的概率分布。本内容详细探讨了泊松分布的概率质量函数及其在C语言和 MATLAB中的实现方法,... Desktop6.zip_matlab 随机曲线_正态分布曲线_ 泊松分布 _ 泊松分布图_ 泊松分布曲线 07-15 泊松分布是一种离散概率分布，常用于描述在一定时间或空间区域内独立事件发生的次数。MATLAB的poissrnd函数可以生成泊松分布的随机数。为了绘制泊松分布图，我们可以使用bar函数创建柱状图，表示每个可能的事件... 二项分布 vs 泊松分布：比较分析与在数据分析中的选择首先介绍了二项分布和泊松分布的基本定义、特性、数学期望和方差，并通过实际案例展示它们在数据分析中的具体应用，如质量控制和市场调研。随后，文章对二项分布与泊松分布进行了深入的比较分析，指出了它们的相似点... 泊松分布与二项分布对比分析：掌握选择的关键文章详细阐述了泊松分布的数学表达、期望、方差以及在事件计数和时间序列分析中的应用，并探讨了二项分布的概率公式、期望、方差以及在质量控制和统计决策中的应用。通过对比两种分布的适用条件、计算方法和统计推断... 二项分布可以由泊松分布近似： 04-25 二项分布可以由泊松分布近似： c语言泊松分布的计算 04-05 【问题描述】泊松分布是一种常用的离散型概率分布，数学期望为m的泊松分布的分布函数定义如下： P(m, k) = mk e - m/k! (k = 0, 1, 2, 3, …) 对于给定的m 和 k (0<m<2000, 0<= k < 2500)，计算其概率，以科学格式输出，保留小数点后6位有效数字。可以使用数学库函数，误差不超过0.000001。【输入形式】输入文件为当前目录下的 poisson.in。文件中包含两个数字，分别为m,k的值。【输出形式】输出文件为当前目录下的 poisson.out。文件中输出泊松分布的值，值以科学格式输出，保留小数点后6位有效数字。【输入样例】 1 0 【输出样例】 3.678794e - 01 Poisson 泊松分布的C++程序 06-21 泊松分布的概率密度函数为： P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!} 泊松分布的参数λ是单位时间(或单位面积)内随机事件的平均发生率。泊松分布适合于描述单位时间内随机事件发生的次数。如某一服务设施在一定时间内到达的人数，电话交换机接到呼叫的次数，汽车站台的候客人数，机器出现的故障数，自然灾害发生的次数等等。泊松算法 08-14 泊松算法, 从点云中拟合三维网凹包. 泊松分布代码实现 weixin_42496466的博客 12-28 973 public class Poisson Distribution : DiscreteDistribution { private double mean; public override double Mean => this.mean; public override double Variance => this.mean; public override double Skewness => 1.0 / ... 二项分布近似成泊松分布、正态分布天天向上的专栏 07-12 1万+ 二项分布可以在一定条件下近似为泊松分布、正态分布。泊松分布近似计算二项分布 m0_54654081的博客 09-25 1万+ 二项分布二项分布是多次伯努利分布实验的概率分布。以抛硬币举例，在抛硬币事件当中，每一次抛硬币的结果是独立的，并且每次抛硬币正面朝上的概率是恒定的，所以单次抛硬币符合伯努利分布。我们假设硬币正面朝上的概率是p，忽略中间朝上的情况，那么反面朝上的概率是q=(1 - p)。泊松分布当一个事件的发生满足以下条件时，可以认为这个事件在某一固定时间段内的发生次数满足柏松分布。事件是独立发生的事件发生的概率在给定的固定时间内不随时间变化总结起来就是，事件的发生是随机且独立的。 ... 【概率统计】用正态分布和泊松分布近似表示二项分布 Sakukou_Bai的博客 07-07 3209 用正态分布和泊松分布近似表示二项分布的证明关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 qq_34734303 博客等级码龄9年 35 原创102 点赞 353 收藏 48 粉丝关注私信 🔥2核2G ARM架构云服务器每月750小时免费试用🆓 灵活弹性又可靠💪早用早省～广告 TA的精选新 Python 无法将表达式转换为float(Can‘t convert expression to float) 22591 阅读新 tensorflow-gpu安装报错：DLL load failed:找不到指定的模块解决办法 3094 阅读热 Python取numpy数组的某几行某几列方法 183924 阅读热 cmd输入JAVA报错Could not create the Java virtual machine，解决方案 138808 阅读热 SQL语言 group by 和 LEFT JOIN 关键字语句使用 36859 阅读查看更多 2021年 1篇 2020年 2篇 2019年 3篇 2018年 29篇大家在看计算机操作系统内存管理——页式管理高性价比之选：4G Cat.1模组在智慧路灯中的应用前景 1615 ✨WPF编程基础【2.2】：布局面板实战构建易受攻击的AWS DevOps环境：CloudGoat场景实践 273 从《深圳市APP保护指引》看技术合规新范式：15条核心条款的开发落地指南分类专栏 tensorflow-gpu1篇 cuda2篇 pytorch gpu安装1篇 DataFrame1篇 groupby1篇 Python学习14篇 iscroll1篇 jsp5篇 js5篇加权移动平均法1篇 EWMA1篇 MySQl3篇数据库2篇爬虫1篇数据库Mysql2篇 Java10篇一元二次方程java实现求根1篇正则表达式3篇 HTML2篇 mailto1篇异常机制1篇 SimpleDateForma1篇 File2篇 FileInputStream1篇 FileOutputStream1篇字符串2篇线程1篇定时器1篇 Timer类1篇 Thread类1篇 Mann-whitney 检验算法1篇空间统计4篇概率统计1篇 CSR1篇时空扫描2篇空间距离计算1篇泊松分布1篇二项分布1篇 plot1篇对数似然比1篇矩阵截取1篇 numpy1篇 SQL语句1篇 GROUP BY 语句1篇 LEFT JOIN语句1篇 Request获取表单1篇判断1篇 eclipse1篇 tomcat1篇字典1篇算法导论1篇算法1篇排序2篇 echarts1篇预测模型检测1篇 Gini1篇基尼系数1篇 list复制1篇列表处理1篇 ast包1篇 json包1篇展开全部收起上一篇：基于经纬度的空间距离计算--空间统计下一篇： Python 实现对数似然比对比曲线及其代码问题分析目录空间统计–二项分布和泊松分布案例代码分析一、泊松分布二、伯努利分布三、二项分布四、二项分布和泊松分布关系展开全部收起目录空间统计–二项分布和泊松分布案例代码分析一、泊松分布二、伯努利分布三、二项分布四、二项分布和泊松分布关系展开全部收起 🔥2核2G ARM架构云服务器每月750小时免费试用🆓 灵活弹性又可靠💪早用早省～广告上一篇：基于经纬度的空间距离计算--空间统计下一篇： Python 实现对数似然比对比曲线及其代码问题分析分类专栏 tensorflow-gpu1篇 cuda2篇 pytorch gpu安装1篇 DataFrame1篇 groupby1篇 Python学习14篇 iscroll1篇 jsp5篇 js5篇加权移动平均法1篇 EWMA1篇 MySQl3篇数据库2篇爬虫1篇数据库Mysql2篇 Java10篇一元二次方程java实现求根1篇正则表达式3篇 HTML2篇 mailto1篇异常机制1篇 SimpleDateForma1篇 File2篇 FileInputStream1篇 FileOutputStream1篇字符串2篇线程1篇定时器1篇 Timer类1篇 Thread类1篇 Mann-whitney 检验算法1篇空间统计4篇概率统计1篇 CSR1篇时空扫描2篇空间距离计算1篇泊松分布1篇二项分布1篇 plot1篇对数似然比1篇矩阵截取1篇 numpy1篇 SQL语句1篇 GROUP BY 语句1篇 LEFT JOIN语句1篇 Request获取表单1篇判断1篇 eclipse1篇 tomcat1篇字典1篇算法导论1篇算法1篇排序2篇 echarts1篇预测模型检测1篇 Gini1篇基尼系数1篇 list复制1篇列表处理1篇 ast包1篇 json包1篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论 2 被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
3378	https://www.msdmanuals.com/professional/pulmonary-disorders/tests-of-pulmonary-function-pft/airflow-lung-volumes-and-flow-volume-loop	honeypot link skip to main content MSD ManualProfessional Version Airflow, Lung Volumes, and Flow-Volume Loop ByKaren L. Wood, MD, Grant Medical Center, Ohio Health Reviewed ByRichard K. Albert, MD, Department of Medicine, University of Colorado Denver - Anschutz Medical Reviewed/Revised Apr 2024 \| Modified Sept 2024 v912529 View Patient Education Patterns of Abnormalities\| Bronchoprovocation Challenge\| Airflow and lung volume measurements can be used to differentiate obstructive from restrictive pulmonary disorders, to characterize severity, and to measure responses to therapy. Tests include spirometry to measure inspiratory and expiratory air flow and lung volumes and sometimes flow-volume loop testing to define specific obstructive and restrictive abnormalities. In patients with obstructive abnormalities, spirometry is repeated after administration of inhaled short-acting bronchodilators to assess reversibility and response to treatment. Measurements are typically reported as absolute flows and volumes and as percentages of predicted values using data derived from large populations of people presumed to have normal lung function. Variables used to predict normal values include age, sex, ethnicity, and height. Clinical Calculators PFT predicted values by 2022 race-neutral GLI equations, ages 3 to 95 It is controversial whether to adjust for race and ethnicity when interpreting pulmonary function tests. There is increasing evidence that race and ethnicity do not accurately account for observed differences in measured pulmonary function, but rather reflect the effects of social and environmental factors, further contributing to health care disparities. Studies have shown that the use of race and ethnicity-based reference equations likely underestimate pulmonary disease severity (and thus result in undertreatment) in non-White individuals (1, 2). The American Thoracic Society (ATS) has recommended replacing previously used ethnicity or race-specific reference equations (3) with those derived from race-neutral reference equations such as those derived from the Global Lung Function Initiative (GLI) average equation (4, 5). The 2022 European Respiratory Society (ERS)/ATS has acknowledged that the use of race-neutral reference sets may result in changes in qualifications for specific treatments (eg, surgery, lung transplantation), which highlights the need for ongoing research to understand the potential impact on clinical decision-making and patient outcomes (6). Airflow Quantitative measures of inspiratory and expiratory flow are obtained by forced spirometry. Nose clips are used to occlude the nares. In inspiratory flow and volume assessments, patients exhale as completely as possible, then forcibly inhale. The peak inspiratory volume is the maximum amount of air inhaled in one deep breath, and the inspiratory flow is the volume inspired per second. In expiratory flow and volume assessments, patients inhale as deeply as possible, seal their lips around a mouthpiece, and exhale as forcefully and completely as possible into an apparatus that records the exhaled volume (forced vital capacity [FVC]) and the volume exhaled in the first second (the forced expiratory volume in 1 second [FEV1]—see figure Normal Spirogram). These maneuvers provide several measures: FVC: Maximal amount of air that the patient can forcibly exhale after taking a maximal inhalation FEV1: Volume exhaled in the first second Peak expiratory flow (PEF): Maximal speed of airflow as the patient exhales FEV1 is the most reproducible flow parameter and is especially useful in diagnosing and monitoring patients with obstructive pulmonary disorders (eg, asthma, COPD [chronic obstructive pulmonary disease]). FEV1 and FVC help differentiate obstructive and restrictive lung disorders. A normal FEV1 makes irreversible obstructive lung disease unlikely. A normal FVC makes restrictive disease unlikely. A decreased ratio of FEV1/FVC indicates obstruction. Repeat measurement of FEV1 and FVC with a short-acting bronchodilator in patients with evidence of obstruction on initial testing is useful in differentiating patients with reversible bronchospasm as occurs in asthma from those with fixed obstruction in COPD. Some people have risk factors for COPD (eg, cigarette smoking, previous infection, occupational exposure, air pollution exposure) but do not demonstrate definite obstruction on pulmonary function testing. These people are said to have pre-COPD (7). Further studies are needed to characterize this population, but following spirometry values over time may be helpful identifying patients who are likely to develop COPD. Normal Spirogram \| \| \| FEF25–75% = forced expiratory flow during expiration of 25 to 75% of the FVC; FEV1 = forced expiratory volume in the first second of forced vital capacity maneuver; FVC = forced vital capacity (the maximum amount of air forcibly expired after maximum inspiration). \| The forced expiratory flow averaged over the time during which 25 to 75% of the FVC is exhaled may be a more sensitive marker of mild, small airway airflow limitation than the FEV1, but the reproducibility of this variable is poor. The peak expiratory flow (PEF) is the peak flow occurring during exhalation. This variable is used primarily for home monitoring of patients with asthma and for determining diurnal variations in airflow. Asthma can be monitored by comparing PEF to one's own personal best. Interpretation of these measures depends on good patient effort, which is often improved by coaching during the actual maneuver. Acceptable spirograms demonstrate Good test initiation (eg, a quick and forceful onset of exhalation) No coughing Smooth curves Absence of early termination of expiration (eg, minimum exhalation time of 6 seconds with no change in volume for the last 1 second) Reproducible efforts agree within 5% or 100 mL with other efforts. Results not meeting these minimum acceptable criteria should be interpreted with caution. Lung volume Lung volumes are measured by determining functional residual capacity (FRC). FRC is the amount of air remaining in the lungs after normal exhalation. The total lung capacity (TLC) is the volume of gas that is contained in the lungs at the end of maximal inspiration. Knowing FRC allows the lungs to be divided into subvolumes that are either measured with spirometry or calculated (see figure Normal Lung Volumes). Normally the FRC represents about 40% of TLC. Normal Lung Volumes \| \| \| ERV = expiratory reserve volume; FRC = functional residual capacity; IC = inspiratory capacity; IRV = inspiratory reserve volume; RV = residual volume; TLC = total lung capacity; VC = vital capacity; VT= tidal volume. FRC = RV + ERV; IC = VT + IRV; VC = VT+ IRV + ERV. \| FRC is measured using gas dilution techniques or a plethysmograph (which is more accurate in patients who have airflow limitation and trapped gas). Gas dilution techniques include Nitrogen washout Helium equilibration With nitrogen washout, the patient exhales to FRC and then breathes from a spirometer containing 100% oxygen. The test ends when the exhaled nitrogen concentration is zero. The collected volume of exhaled nitrogen is equal to 81% of the initial FRC. With helium equilibration, the patient exhales to FRC and then is connected to a closed system containing known volumes of helium and oxygen. Helium concentration is measured until it is the same on inhalation and exhalation, indicating it has equilibrated with the volume of gas in the lungs, which can then be estimated from the change in helium concentration that has occurred. Both of these techniques may underestimate FRC because they measure only the lung volume that communicates with the airways. In patients with severe airflow limitation, a considerable volume of trapped gas may communicate very poorly or not at all. Body plethysmography uses Boyle’s law (P1V1 = P2V2, where P is pressure and V is volume) to measure the compressible gas volume within the thorax. Body plethysmography is more accurate than gas dilution techniques. While sitting in an airtight box, the patient tries to inhale against a closed mouthpiece from FRC. As the chest wall expands, the pressure in the closed box rises. Knowing the pre-inspiratory box volume and the pressure in the box before and after the inspiratory effort allows for calculation of the change in box volume, which must equal the change in lung volume. Clinical Calculators Lung Volume Multicalc Flow-volume loop In contrast to the spirogram, which displays airflow (in L) over time (in seconds), the flow-volume loop displays airflow (in L/second) as it relates to lung volume (in L) during maximal inspiration from complete exhalation (residual volume [RV]) and during maximum expiration from complete inhalation (TLC). The principal advantage of the flow-volume loop is that it can show whether airflow is appropriate for a particular lung volume. For example, airflow is normally slower at low lung volumes because elastic recoil is lower at lower lung volumes. Patients with pulmonary fibrosis have low lung volumes and their airflow appears to be decreased if measured alone. However, when airflow is presented as a function of lung volume, it becomes apparent that airflow is actually higher than normal (as a result of the increased elastic recoil characteristic of fibrotic lungs). Flow-volume loops require that absolute lung volumes be measured. Unfortunately, many laboratories simply plot airflow against the FVC; the flow-FVC loop does not have an inspiratory limb and therefore does not provide as much information. General references Baugh AD, Shiboski S, Hansel NN, et al. Reconsidering the Utility of Race-Specific Lung Function Prediction Equations [published correction appears in Am J Respir Crit Care Med 2022 Jul 15;206(2):230]. Am J Respir Crit Care Med 2022;205(7):819-829. doi:10.1164/rccm.202105-1246OC Ekström M, Mannino D. Research race-specific reference values and lung function impairment, breathlessness and prognosis: Analysis of NHANES 2007-2012 [published correction appears in Respir Res 2023 Feb 3;24(1):41]. Respir Res 2022;23(1):271. Published 2022 Oct 1. doi:10.1186/s12931-022-02194-4 Bhakta NR, Bime C, Kaminsky DA, et al. Race and Ethnicity in Pulmonary Function Test Interpretation: An Official American Thoracic Society Statement. Am J Respir Crit Care Med 2023;207(8):978-995. doi:10.1164/rccm.202302-0310ST Quanjer PH, Stanojevic S, Cole TJ, et al. Multi-ethnic reference values for spirometry for the 3-95-yr age range: the global lung function 2012 equations. Eur Respir J 2012;40(6):1324-1343. doi:10.1183/09031936.00080312 Bowerman C, Bhakta NR, Brazzale D, et al. A Race-neutral Approach to the Interpretation of Lung Function Measurements. Am J Respir Crit Care Med 2023;207(6):768-774. doi:10.1164/rccm.202205-0963OC Stanojevic S, Kaminsky DA, Miller MR, et al. ERS/ATS technical standard on interpretive strategies for routine lung function tests. Eur Respir J 2022;60(1):2101499. Published 2022 Jul 13. doi:10.1183/13993003.01499-2021 Han MK, Agusti A, Celli BR, et al. From GOLD 0 to Pre-COPD. Am J Respir Crit Care Med 2021;203(4):414-423. doi:10.1164/rccm.202008-3328PP Patterns of Abnormalities Most common respiratory disorders can be categorized as obstructive or restrictive on the basis of airflow and lung volumes (see table Characteristic Physiologic Changes Associated With Pulmonary Disorders). Table Characteristic Physiologic Changes Associated With Pulmonary Disorders Table Characteristic Physiologic Changes Associated With Pulmonary Disorders Characteristic Physiologic Changes Associated With Pulmonary Disorders \| Measure \| Obstructive Disorders \| Restrictive Disorders \| Mixed Disorders \| \| FEV1/FVC \| Decreased \| Normal or increased \| Decreased \| \| FEV1 \| Decreased \| Decreased, normal, or increased \| Decreased \| \| FVC \| Decreased or normal \| Decreased \| Decreased or normal \| \| TLC \| Normal or increased \| Decreased \| Decreased, normal, or increased \| \| RV \| Normal or increased \| Decreased \| Decreased, normal, or increased \| \| FEV1 = forced expiratory volume in 1 second; FVC = forced vital capacity; RV = residual volume; TLC = total lung capacity. \| \| \| \| FEV1 = forced expiratory volume in 1 second; FVC = forced vital capacity; RV = residual volume; TLC = total lung capacity. \| Obstructive disorders Obstructive disorders are characterized by a reduction in airflow, particularly the FEV1 and the FEV1 expressed as a percentage of the FVC (FEV1/FVC). The degree of reduction in FEV1 compared with predicted values determines the degree of the obstructive defect. Obstructive defects are caused by Increased resistance to airflow due to abnormalities within the airway lumen (eg, tumors, secretions, mucosal thickening) Changes in the wall of the airway (eg, contraction of smooth muscle, edema) Decreased elastic recoil (eg, the parenchymal destruction that occurs in emphysema) With decreased airflow, expiratory times are longer than usual, and air may become trapped in the lungs due to incomplete emptying, thereby increasing lung volumes (eg, TLC, RV). The most common examples of obstructive disorders are COPD, asthma, and bronchiectasis. The ERS and ATS have updated their guidelines on interpretation of pulmonary function tests in grading severity of obstructive lung disease (see table Severity of Lung Impairment) (1). These guidelines recommend expressing all measurements, including spirometry, lung volumes, and diffusing capacity of the lungs for carbon monoxide (DLCO), as z-scores rather than as percentages of predicted values to grade severity. A z-score less than -1.645 indicates that the value is less than 5th percentile of predicted based on healthy matched controls. In judging response to bronchodilators, the guidelines now recommend the use of percent change relative to an individual's predicted value (instead of baseline value) and they recommend using improvement of FEV1 and/or FVC ≥ 10% as the criterion for airway hyperresponsiveness. Table Severity of Lung Impairment Table Severity of Lung Impairment Severity of Lung Impairment \| Severity \| Z Score \| \| Mild \| -1.65 to -2.5 \| \| Moderate \| -2.51 to -4.0 \| \| Severe \| <-4.1 \| \| Data from Stanojevic S, Kaminsky DA, Miller MR, et al. ERS/ATS technical standard on interpretive strategies for routine lung function tests. Eur Respir J 2022;60(1):2101499. Published 2022 Jul 13. doi:10.1183/13993003.01499-2021 \| \| \| \| Data from Stanojevic S, Kaminsky DA, Miller MR, et al. ERS/ATS technical standard on interpretive strategies for routine lung function tests. Eur Respir J 2022;60(1):2101499. Published 2022 Jul 13. doi:10.1183/13993003.01499-2021 \| Restrictive disorders Restrictive disorders are characterized by a reduction in lung volume, specifically a TLC less than the lower limit of normal (a z-score less than -1.65, corresponding to less than the fifth percentile of predicted based on healthy matched controls). However, in early restrictive disease, the TLC can be normal (as a result of strong inspiratory effort) and the only abnormality might be a reduction in RV. The decrease in TLC determines the severity of restriction. The decrease in lung volumes causes a decrease in airflow (reduced FEV1). However, airflow relative to lung volume is increased, so the FEV1/FVC ratio is normal or increased. Restrictive defects are caused by the following: Loss in lung volume (eg, lobectomy) Abnormalities of structures surrounding the lung (eg, pleural disorder, kyphosis, obesity) Weakness of the inspiratory muscles of respiration (eg, neuromuscular disorders) Abnormalities of the lung parenchyma (eg, pulmonary fibrosis) The feature common to all is a decrease in the compliance of the lungs, the chest wall, or both. Patterns of abnormalities references Stanojevic S, Kaminsky DA, Miller MR, et al. ERS/ATS technical standard on interpretive strategies for routine lung function tests. Eur Respir J 2022;60(1):2101499. Published 2022 Jul 13. doi:10.1183/13993003.01499-2021 Coates AL, Wanger J, Cockcroft DW, et al. ERS technical standard on bronchial challenge testing: general considerations and performance of methacholine challenge tests. . ERS technical standard on bronchial challenge testing: general considerations and performance of methacholine challenge tests.Eur Respir J 2017;49(5):1601526. Published 2017 May 1. doi:10.1183/13993003.01526-2016 Parsons JP, Hallstrand TS, Mastronarde JG, et al. An official American Thoracic Society clinical practice guideline: exercise-induced bronchoconstriction. Am J Respir Crit Care Med 2013;187(9):1016-1027. doi:10.1164/rccm.201303-0437ST Bronchoprovocation Challenge Bronchprovocation challenge is used to diagnose conditions such as asthma, particularly when spirometry is normal but there is a suspicion of airway hyperreactivity. Testing can be done with inhaled methacholine, exercise, or eucapnic voluntary hyperventilation (EVH) using air at ambient or cold temperatures.Bronchprovocation challenge is used to diagnose conditions such as asthma, particularly when spirometry is normal but there is a suspicion of airway hyperreactivity. Testing can be done with inhaled methacholine, exercise, or eucapnic voluntary hyperventilation (EVH) using air at ambient or cold temperatures. Some patients with asthma can have normal pulmonary function and normal spirometric parameters between exacerbations. When suspicion of asthma remains high despite normal spirometry results, bronchprovocation challenge testing with methacholine, a synthetic analog of with methacholine, a synthetic analog ofacetylcholine that is a nonspecific bronchial irritant, is indicated to detect or exclude bronchoconstriction. In a methacholine challenge test, spirometric parameters are measured at baseline and after inhalation of increasing doses of methacholine. The dose of methacholine that causes a 20% drop in FEV1 is called the PD20. Laboratories have different definitions of airway hyperreactivity, but in general, patients showing at least a 20% drop in FEV1 from baseline (PD20) when the delivered dose of inhaled methacholine is < 25 mcg is considered diagnostic of increased bronchial reactivity, whereas a PD20 > 400 mcg excludes the diagnosis. PD20 values between 25 and 400 mcg are inconclusive (1). Exercise testing may be used to detect exercise-induced asthma but is less sensitive than methacholine challenge testing for detecting general airway hyperresponsiveness. The patient does a constant level of work on a treadmill or cycle ergometer for 6 to 8 minutes at an intensity selected to produce a heart rate of 80% of predicted maximum. The FEV1 and FVC are measured before and 5, 15, and 30 minutes after exercise. Exercise-induced bronchospasm reduces FEV1 or FVC may be used to detect exercise-induced asthma but is less sensitive than methacholine challenge testing for detecting general airway hyperresponsiveness. The patient does a constant level of work on a treadmill or cycle ergometer for 6 to 8 minutes at an intensity selected to produce a heart rate of 80% of predicted maximum. The FEV1 and FVC are measured before and 5, 15, and 30 minutes after exercise. Exercise-induced bronchospasm reduces FEV1 or FVC≥ 10 to 15% after exercise (2). EVH may also be used to diagnose exercised-induced asthma. EVH involves hyperventilation of a gas mixture of 5% carbon dioxide and 21% oxygen at 85% of maximum voluntary ventilation for 6 minutes. FEV1 is then measured at specified intervals after the test. As with other bronchial provocation tests, the drop in FEV1 that is diagnostic of exercise-induced bronchospasm varies by laboratory. Cold-induced hyperreactivity can be assessed with a similar test in which the patient hyperventilates for 3 to 6 minutes with the gas mixture cooled to between -10° C and -20° C. This test requires specialized cooling equipment that may not be available in many testing laboratories. Bronchprovocation challenge references Coates AL, Wanger J, Cockcroft DW, et al. ERS technical standard on bronchial challenge testing: general considerations and performance of methacholine challenge tests. . ERS technical standard on bronchial challenge testing: general considerations and performance of methacholine challenge tests.Eur Respir J 2017;49(5):1601526. Published 2017 May 1. doi:10.1183/13993003.01526-2016 Parsons JP, Hallstrand TS, Mastronarde JG, et al. An official American Thoracic Society clinical practice guideline: exercise-induced bronchoconstriction. Am J Respir Crit Care Med 2013;187(9):1016-1027. doi:10.1164/rccm.201303-0437ST Test your KnowledgeTake a Quiz! 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Learn From Your Errors Paul's Online Notes Notes Quick Nav Download × Custom Search Go To Notes Practice Problems Assignment Problems Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections Parametric Equations and Curves Area with Parametric Equations Chapters Applications of Integrals Series & Sequences 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/ Parametric Equations and Polar Coordinates / Tangents with Parametric Equations Prev. SectionNotesPractice ProblemsAssignment ProblemsNext 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 9.2 : Tangents with Parametric Equations In this section we want to find the tangent lines to the parametric equations given by, x=f(t)y=g(t)x=f(t)y=g(t) To do this let’s first recall how to find the tangent line to y=F(x)y=F(x) at x=a x=a. Here the tangent line is given by, y=F(a)+m(x−a),where m=d y d x∣∣∣x=a=F′(a)y=F(a)+m(x−a),where m=d y d x\|x=a=F′(a) Now, notice that if we could figure out how to get the derivative d y d x d y d x from the parametric equations we could simply reuse this formula since we will be able to use the parametric equations to find the x x and y y coordinates of the point. So, just for a second let’s suppose that we were able to eliminate the parameter from the parametric form and write the parametric equations in the form y=F(x)y=F(x). Now, plug the parametric equations in for x x and y y. Yes, it seems silly to eliminate the parameter, then immediately put it back in, but it’s what we need to do in order to get our hands on the derivative. Doing this gives, g(t)=F(f(t))g(t)=F(f(t)) Now, differentiate with respect to t t and notice that we’ll need to use the Chain Rule on the right-hand side. g′(t)=F′(f(t))f′(t)g′(t)=F′(f(t))f′(t) Let’s do another change in notation. We need to be careful with our derivatives here. Derivatives of the lower case function are with respect to t t while derivatives of upper case functions are with respect to x x. So, to make sure that we keep this straight let’s rewrite things as follows. d y d t=F′(x)d x d t d y d t=F′(x)d x d t At this point we should remind ourselves just what we are after. We needed a formula for d y d x d y d x or F′(x)F′(x) that is in terms of the parametric formulas. Notice however that we can get that from the above equation. Derivative for Parametric Equations d y d x=d y d t d x d t,provided d x d t≠0 d y d x=d y d t d x d t,provided d x d t≠0 Notice as well that this will be a function of t t and not x x. As an aside, notice that we could also get the following formula with a similar derivation if we needed to, d x d y=d x d t d y d t,provided d y d t≠0 d x d y=d x d t d y d t,provided d y d t≠0 Why would we want to do this? Well, recall that in the arc length section of the Applications of Integral section we actually needed this derivative on occasion. So, let’s find a tangent line. Example 1 Find the tangent line(s) to the parametric curve given by x=t 5−4 t 3 y=t 2 x=t 5−4 t 3 y=t 2 at (0,4)(0,4). Show Solution Note that there is apparently the potential for more than one tangent line here! We will look into this more after we’re done with the example. The first thing that we should do is find the derivative so we can get the slope of the tangent line. d y d x=d y d t d x d t=2 t 5 t 4−12 t 2=2 5 t 3−12 t d y d x=d y d t d x d t=2 t 5 t 4−12 t 2=2 5 t 3−12 t At this point we’ve got a small problem. The derivative is in terms of t t and all we’ve got is an x-y coordinate pair. The next step then is to determine the value(s) of t t which will give this point. We find these by plugging the x x and y y values into the parametric equations and solving for t t. 0=t 5−4 t 3=t 3(t 2−4)⇒t=0,±2 4=t 2⇒t=±2 0=t 5−4 t 3=t 3(t 2−4)⇒t=0,±2 4=t 2⇒t=±2 Any value of t t which appears in both lists will give the point. So, since there are two values of t t that give the point we will in fact get two tangent lines. That’s definitely not something that happened back in Calculus I and we’re going to need to look into this a little more. However, before we do that let’s actually get the tangent lines. t=−2:t=−2: Since we already know the x x and y y-coordinates of the point all that we need to do is find the slope of the tangent line. m=d y d x∣∣∣t=−2=−1 8 m=d y d x\|t=−2=−1 8 The tangent line (at t=−2 t=−2) is then, y=4−1 8 x y=4−1 8 x t=2:t=2: Again, all we need is the slope. m=d y d x∣∣∣t=2=1 8 m=d y d x\|t=2=1 8 The tangent line (at t=2 t=2) is then, y=4+1 8 x y=4+1 8 x Before we leave this example let’s take a look at just how we could possibly get two tangents lines at a point. This was definitely not possible back in Calculus I where we first ran across tangent lines. A quick graph of the parametric curve will explain what is going on here. So, the parametric curve crosses itself! That explains how there can be more than one tangent line. There is one tangent line for each instance that the curve goes through the point. The next topic that we need to discuss in this section is that of horizontal and vertical tangents. We can easily identify where these will occur (or at least the t t’s that will give them) by looking at the derivative formula. d y d x=d y d t d x d t d y d x=d y d t d x d t Horizontal tangents will occur where the derivative is zero and that means that we’ll get horizontal tangent at values of t t for which we have, Horizontal Tangent for Parametric Equations d y d t=0,provided d x d t≠0 d y d t=0,provided d x d t≠0 Vertical tangents will occur where the derivative is not defined and so we’ll get vertical tangents at values of t t for which we have, Vertical Tangent for Parametric Equations d x d t=0,provided d y d t≠0 d x d t=0,provided d y d t≠0 Let’s take a quick look at an example of this. Example 2 Determine the x x-y y coordinates of the points where the following parametric equations will have horizontal or vertical tangents.x=t 3−3 t y=3 t 2−9 x=t 3−3 t y=3 t 2−9 Show Solution We’ll first need the derivatives of the parametric equations. d x d t=3 t 2−3=3(t 2−1)d y d t=6 t d x d t=3 t 2−3=3(t 2−1)d y d t=6 t Horizontal Tangents We’ll have horizontal tangents where, 6 t=0⇒t=0 6 t=0⇒t=0 Now, this is the value of t t which gives the horizontal tangents and we were asked to find the x-y coordinates of the point. To get these we just need to plug t t into the parametric equations. Therefore, the only horizontal tangent will occur at the point (0,−9)(0,−9). Vertical Tangents In this case we need to solve, 3(t 2−1)=0⇒t=±1 3(t 2−1)=0⇒t=±1 The two vertical tangents will occur at the points (2,−6)(2,−6) and (−2,−6)(−2,−6). For the sake of completeness and at least partial verification here is the sketch of the parametric curve. The final topic that we need to discuss in this section really isn’t related to tangent lines but does fit in nicely with the derivation of the derivative that we needed to get the slope of the tangent line. Before moving into the new topic let’s first remind ourselves of the formula for the first derivative and in the process rewrite it slightly. d y d x=d d x(y)=d d t(y)d x d t d y d x=d d x(y)=d d t(y)d x d t Written in this way we can see that the formula actually tells us how to differentiate a function y y (as a function of t t) with respect to x x (when x x is also a function of t t) when we are using parametric equations. Now let’s move onto the final topic of this section. We would also like to know how to get the second derivative of y y with respect to x x. d 2 y d x 2 d 2 y d x 2 Getting a formula for this is fairly simple if we remember the rewritten formula for the first derivative above. Second Derivative for Parametric Equations d 2 y d x 2=d d x(d y d x)=d d t(d y d x)d x d t d 2 y d x 2=d d x(d y d x)=d d t(d y d x)d x d t It is important to note that, d 2 y d x 2≠d 2 y d t 2 d 2 x d t 2 d 2 y d x 2≠d 2 y d t 2 d 2 x d t 2 Let’s work a quick example. Example 3 Find the second derivative for the following set of parametric equations. x=t 5−4 t 3 y=t 2 x=t 5−4 t 3 y=t 2 Show Solution This is the set of parametric equations that we used in the first example and so we already have the following computations completed. d y d t=2 t d x d t=5 t 4−12 t 2 d y d x=2 5 t 3−12 t d y d t=2 t d x d t=5 t 4−12 t 2 d y d x=2 5 t 3−12 t We will first need the following, d d t(2 5 t 3−12 t)=−2(15 t 2−12)(5 t 3−12 t)2=24−30 t 2(5 t 3−12 t)2 d d t(2 5 t 3−12 t)=−2(15 t 2−12)(5 t 3−12 t)2=24−30 t 2(5 t 3−12 t)2 The second derivative is then, d 2 y d x 2=d d t(d y d x)d x d t=24−30 t 2(5 t 3−12 t)2 5 t 4−12 t 2=24−30 t 2(5 t 4−12 t 2)(5 t 3−12 t)2=24−30 t 2 t(5 t 3−12 t)3 d 2 y d x 2=d d t(d y d x)d x d t=24−30 t 2(5 t 3−12 t)2 5 t 4−12 t 2=24−30 t 2(5 t 4−12 t 2)(5 t 3−12 t)2=24−30 t 2 t(5 t 3−12 t)3 So, why would we want the second derivative? Well, recall from your Calculus I class that with the second derivative we can determine where a curve is concave up and concave down. We could do the same thing with parametric equations if we wanted to. Example 4 Determine the values of t t for which the parametric curve given by the following set of parametric equations is concave up and concave down. x=1−t 2 y=t 7+t 5 x=1−t 2 y=t 7+t 5 Show Solution To compute the second derivative we’ll first need the following. d y d t=7 t 6+5 t 4 d x d t=−2 t d y d x=7 t 6+5 t 4−2 t=−1 2(7 t 5+5 t 3)d y d t=7 t 6+5 t 4 d x d t=−2 t d y d x=7 t 6+5 t 4−2 t=−1 2(7 t 5+5 t 3) Note that we can also use the first derivative above to get some information about the increasing/decreasing nature of the curve as well. In this case it looks like the parametric curve will be increasing if t<0 t<0 and decreasing if t>0 t>0. Now let’s move on to the second derivative. d 2 y d x 2=−1 2(35 t 4+15 t 2)−2 t=1 4(35 t 3+15 t)d 2 y d x 2=−1 2(35 t 4+15 t 2)−2 t=1 4(35 t 3+15 t) It’s clear, hopefully, that the second derivative will only be zero at t=0 t=0. Using this we can see that the second derivative will be negative if t<0 t<0 and positive if t>0 t>0. So the parametric curve will be concave down for t<0 t<0 and concave up for t>0 t>0. Here is a sketch of the curve for completeness sake. [Contact Me][Privacy Statement][Site Help & FAQ][Terms of Use] © 2003 - 2025 Paul Dawkins Page Last Modified : 11/16/2022
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Skip to lesson content Algebra 1 Course: Algebra 1>Unit 9 Lesson 2: Constructing arithmetic sequences Recursive formulas for arithmetic sequences Recursive formulas for arithmetic sequences Recursive formulas for arithmetic sequences Explicit formulas for arithmetic sequences Explicit formulas for arithmetic sequences Explicit formulas for arithmetic sequences Arithmetic sequence problem Converting recursive & explicit forms of arithmetic sequences Converting recursive & explicit forms of arithmetic sequences Convert recursive & explicit forms of arithmetic sequences Arithmetic sequences review Math> Algebra 1> Sequences> Constructing arithmetic sequences © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Arithmetic sequence problem NY.Math: AI‑F.IF.3, AI‑F.LE.1.a, AII‑F.IF.3 Google Classroom Microsoft Teams About About this video Transcript Sal finds the 100th term in the sequence 15, 9, 3, -3...Created by Sal Khan and Monterey Institute for Technology and Education. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted sarra 2 years ago Posted 2 years ago. Direct link to sarra's post “this is clearer approach:...” more this is clearer approach: 15-6(100-1) =15-594 =-579 Answer Button navigates to signup page •1 comment Comment on sarra's post “this is clearer approach:...” (13 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer vedanshiprd 2 years ago Posted 2 years ago. Direct link to vedanshiprd's post “how do you know when to u...” more how do you know when to use a recursive or explicit formula for a math problem? Do they give you a general rule of when to use it or not use it? What I think is that since the previous terms weren't given, Sal couldn't use a recursive formula since he would have to know all the terms. However, I'm not sure that this is correct. Answer Button navigates to signup page •Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Eternity 2 years ago Posted 2 years ago. Direct link to Eternity's post “You use whichever is easi...” more You use whichever is easier with that specific equation. For example, like in the video, finding 99 terms before getting the value you're looking for would be very time consuming and very annoying. So, an explicit formula works better. Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more jabbar 9 years ago Posted 9 years ago. Direct link to jabbar's post “To find the sum for arith...” more To find the sum for arithmetic sequence, sn= n(n+1)/2, it is shown (n+1)/2, can be replaced with the average of nth term and first term. How do we understand that we should not replace the "n" outside the bracket should not be replaced with nth term too. Answer Button navigates to signup page •2 comments Comment on jabbar's post “To find the sum for arith...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Judith Gibson 9 years ago Posted 9 years ago. Direct link to Judith Gibson's post “Confusingly, "n" IS the n...” more Confusingly, "n" IS the nth term in this particular sequence! The ( n + 1 ) represents the sum of the last term (n) and the first term (1). Dividing by 2 gives us their average. Then we multiply that by the number of terms (n). Hope this makes things clearer! 4 comments Comment on Judith Gibson's post “Confusingly, "n" IS the n...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Kwabzin 12 years ago Posted 12 years ago. Direct link to Kwabzin's post “Finding the 100th term (o...” more Finding the 100th term (or any term that's not given) is pretty straightfoward with an explicit(ly defined) equation. But how do you do it with a recursive(ly defined) equation? eg with the recursive equation for this video's example: a(100)=a("subscript" 100-1) - 6 As in, you don't have the 99th term's value so how do you find it so you can then subtract 6 from it and get the 100th term's value? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ArDeeJ 12 years ago Posted 12 years ago. Direct link to ArDeeJ's post “You want to get the analy...” more You want to get the analytic form (= explicitly defined) for your recursive sequence. One, kind of hand-wavey way to do it would be to calculate some amount of the first terms, try to spot the pattern and define the analytic expression. Another way to do it, presuming it's of the appropriate form, would be to use the first-order linear recurrence equation. If you have a recursively defined sequence a_n = ca_(n-1) + d, and you're given the first term a_0, then the sequence explicitly defined is: a_n = a_0 c^n + d (c^n - 1) / (c - 1). Notice that if c = 1, then you have just a regular arithmetic sequence. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Bedcloud 8 years ago Posted 8 years ago. Direct link to Bedcloud's post “The setup equation is 15-...” more The setup equation is 15-(n-1) 6 How did you know to multiply by 6? I would have automatically made the mistake of putting -6, in the set up equation charting the sequences lol. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 8 years ago Posted 8 years ago. Direct link to David Severin's post “technically it would be t...” more technically it would be the same thing, your way would just yield an = 15 + (n-1) -6. So in fact, you would not have made a mistake at all. Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more DAVID 8 years ago Posted 8 years ago. Direct link to DAVID's post “Is 15-(n-1)6 equivalent ...” more Is 15-(n-1)6 equivalent to 15-6(n-1)? Are they the same? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “Yes, those 2 expressions ...” more Yes, those 2 expressions are equal to each other. The commutative property of multiplication lets you move the 6 from right side of the parentheses to the left side. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Habib Mahamat 6 days ago Posted 6 days ago. Direct link to Habib Mahamat's post “Calculate the 100th term” more Calculate the 100th term Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 6 days ago Posted 6 days ago. Direct link to Kim Seidel's post “You haven't given enough ...” more You haven't given enough info. Where is the equations / functions for the sequence? If you have the equation, you can calculate this yourself. To find the 100th term, you would use 100 in your equation for the variable and calculate the 100th term. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Clare Jolley 2 years ago Posted 2 years ago. Direct link to Clare Jolley's post “what is the formula for f...” more what is the formula for finding the nth term Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer sugarnlight 2 years ago Posted 2 years ago. Direct link to sugarnlight's post “The formula is a_n = a_1 ...” more The formula is a_n = a_1 + d(n-1). The underscores signify subscripts. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Diamond Maitland 11 years ago Posted 11 years ago. Direct link to Diamond Maitland's post “how did he change it 2 a ...” more how did he change it 2 a positive 6? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Dhanush 6 years ago Posted 6 years ago. Direct link to Dhanush's post “So what is the nth term o...” more So what is the nth term of the following sequence 1,2,4,7,11,16,22 Answer Button navigates to signup page •2 comments Comment on Dhanush's post “So what is the nth term o...” (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Twitch 4 years ago Posted 4 years ago. Direct link to Twitch's post “You will notice that the ...” more You will notice that the sequence is adding +1,+2,+3 and so on. Can you pick up from here? Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Video transcript [Instructor] We are asked what is the value of the 100th term in this sequence, and the first term is 15, then nine, then three, then negative three. So let's write it like this in a table. So if we have the term, just so we have things straight, and then we have the value, and then we have the value of the term. I'll do a nice little table here. So our first term we saw is 15. Our second term is nine. Our third term is three. I'm just really copying this down, but I'm making sure we associate it with the right term. And then our fourth term, our fourth term is negative three. And they wanna ask, they want us to figure out what the 100th term of this sequence is going to be. So let's see what's happening here, if we can discern some type of pattern. So we went from the first term to the second term, what happened? 15 to nine, looks like we went down by six. It's always good to think about just how much the numbers changed by. That's always the simplest type of pattern. So we went down by six, we subtracted six. Then to go from nine to three, well we subtracted six again. We subtracted six again. And then to go from three to negative three, well we, we subtracted six again. We subtracted six again. So it looks like every term, you subtract six. So the second term is going to be six less than the first term. The third term is going to be 12 minus from the first term, or six subtracted twice. So in the third term, you subtract a six twice. In the fourth term, you subtract six three times. So whatever term you're looking at, you subtract six one less than that many times. Let me write this down just so, notice when your first term, you have 15 and you don't subtract six at all, or you could say you subtract six zero times. So you could say this is 15 minus six times or let me write it better this way, minus zero times six. That's what that first term is right there. What's the second term? This is 15, it's just we just subtracted six once, or you could say minus one times six. Or you could say plus one times negative six, either way, we're subtracting the six once. Now what's happening here? This is 15, this is 15 minus two times negative six, or sorry minus two times six, minus two times six. We're subtracting a six twice. What's the fourth term? This is 15 minus, we're subtracting the six three times from the 15, so minus three times six. So if you see the pattern here, when our term, when we have our fourth term, we have the term minus one right there, the fourth term we have a three, the third term we have a two, the second term we have a one. So if we had the nth term, if we just had the nth term here, what's this going to be? It's going to be 15 minus, you see it's going to be n minus one right here, right when n is four, n minus one is three. When n is three, n minus one is two. When n is two, n minus one is one. When n is one, n minus one is zero. So we're going to have, this term right here is n minus one, so minus n minus one times six. So if you wanna figure out the 100th term of this sequence, I didn't even have to write it in this general term, you could just look at this pattern. It's going to be, and I'll do it in pink, the 100th term in our sequence, I'll continue our table down, is gonna be what? It's going to be 15 minus 100 minus one, which is 99, times six, right? I just followed the pattern. One, you had a zero here, two, you had a one here, three, you had a two here, 100, you're gonna have a 99 here. So let's just calculate what this is. What's 99 times six? So 99 times six, actually you could do this in your head. You could say that's going to be six less than 100 times six, which is 600, and six less is 594. But if you didn't wanna do it that way, you just do it the old-fashioned way. Six times nine is 54, carry the five. Nine times six or six times nine is 54. 54 plus five is 594. So this right here is 594, and then to figure out what 15, so we wanna figure out, we wanna figure out what 15 minus 594 is, and this can sometimes be confusing, but the way I always process this in my head is I say that this is the exact same thing as the negative of 594 minus 15. And if you don't believe me, distribute out this negative sign. Negative one times 594 is negative 594, negative one times negative 15 is positive 15. So these two statements are equivalent. This is much easier for my brain to understand. So what's 594 minus 15? You should do it, we could do this in our head. 594 minus 14 would be 580, and then 580 minus one more would be 579. So that right there is 579, and then we have this negative sign sitting out there. So our the 100th term in our sequence will be negative 579. 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3381	https://www.doubtnut.com/qna/642711204	Simplify the following: 6a-2b-ab-(3a+b-ab)+2ab-b+aSolution in Bengali Step by step video & image solution for Simplify the following: 6a-2b-ab-(3a+b-ab)+2ab-b+a by Maths experts to help you in doubts & scoring excellent marks in Class 7 exams. Topper's Solved these Questions Explore 178 Videos Explore 40 Videos Explore 16 Videos Similar Questions Write the algebraic expression and find their number of terms 4x, 3x+1, 2x+1, 6p-1, 3y+6. Find the value of the following (+2) + (+3). Let's write the approximate values of the following fractions correct up to two, three and four places of decimal. 1317. Simplify: a2b2(a+2b)(3a+b) Simplify: ab(a^(2)-b^(2))+b^(3)(a-2b) If the position vectors of A, B are 2a - 3b, 3a + 2b respectively then the position of vector of C in AB produced such that AC = 2 AB is সরল করো : 6a−2b−ab−(3a+b−ab)+2ab−b+a (a−b)2=(a2−2ab+b2) এর সাহায্যে (3a+2b)2−2(3a+2b)(a+2b)+(a+2b)2 -এর সরল করো \| Simplify: a2b2(a+2b)(3a+b) Consider ∣∣ ∣∣aa+ba+b+c2a3a+2b4a+3b+2c3a6a+3b10a+6b+3c∣∣ ∣∣ ਸਮਾਨ ਪਦਾਂ ਨੂੰ ਇੱਕਠਾ ਕਰਕੇ ਸਰਲ ਕਰੋ :- 3a−2b−ab−(a−b+ab)+3ab+b−a সরল করো : 6a−2b−ab−(3a+b−ab)+2ab−b+a (a−b)2=(a2−2ab+b2) এর সাহায্যে (3a+2b)2−2(3a+2b)(a+2b)+(a+2b)2 -এর সরল করো \| Simplify the following with the help of factor (a^2b-b^2a) /(ab) Simplify in following. 6a - 2b - ab - (3a+ b - ab) + 2ab - b + a NCERT BANGLISH-ALGEBRIC OPERATIONS-Exercise Simplify the following: (x^2+2x-5) + (3x^2-8x+5) Simplify the following: (7x^2-3x+3) - (2x^2-13x-7) Simplify the following: 6a-2b-ab-(3a+b-ab)+2ab-b+a Ramu had (13x^2+x-3). He spent (4x^2-3x-12). Find, how much money Ramu... The length of three sides of a triangle are (x+4)cm, (2x+1)cm, and (4x... How much must be added to -8x^2+8x+1 to get -14x^2+11x-3. Find, what must be subtracted from -11x-7y-9z to get -7x+3y-5z. How much is the sum of (3x^2+4x) and (5x^2-x) more than (3x-5x^2), cal... Subtract the sum (x^2-9x) and (-2x^2+3x+5) from the sum of (5+9x) and ... If x=5 find the value of the algebraic expressions: 6x+11. If x=5 find the value of the algebraic expressions: x/5+2 If x=5 find the value of the algebraic expressions: x^2+2x-1. If x=5 find the value of the algebraic expressions: x^3+8. If x=5 find the value of the algebraic expressions: 10-x. If y=-3, find the value of the algebraic expressions: y+5/4. If y=-3, find the value of the algebraic expressions: 5-y. If y=-3, find the value of the algebraic expressions: y+8. If y=-3, find the value of the algebraic expressions: y^2+2y+3. If y=-3, find the value of the algebraic expressions: y^3-1. Find the value of the 2x+7y, when x=2, y=-1 Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
3382	https://people.revoledu.com/kardi/tutorial/LinearAlgebra/VectorNorm.html	Vector Norm By Kardi Teknomo, PhD. <Next \| Previous \| Index> Vector Norm Based on Pythagorean Theorem, the vector from the origin to the point (3, 4) in 2D Euclidean plane has length of ( \sqrt{3^{2}+4^{2}}=\sqrt{25}=5 ) and the vector from the origin to the point ( (a,b)) has length ( \sqrt{a^{2}+b^{2}} ). The length of a vector with two elements is the square root of the sum of each element squared. The magnitude of a vector is sometimes called the length of a vector, or norm of a vector. Basically, norm of a vector is a measure of distance, symbolized by double vertical bar ( \left \\| \mathbf{a} \right \\| ) The magnitude of a vector can be extended to ( n ) dimensions. A vector a with ( n ) elements has length $$ \left \\| \mathbf{a} \right \\| = \sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}} $$ The vector length is called Euclidean length or Euclidean norm. Mathematician often used term norm instead of length. Vector norm is defined as any function that associated a scalar with a vector and obeys the three rules below Norm of a vector is always positive or zero ( \left \\| \mathbf{a} \right \\| \geqslant 0 ). The norm of a vector is zero if and only if the vector is a zero vector ( \mathbf{a} = \mathbf{0} ). A scalar multiple to a norm is equal to the product of the absolute value of the scalar and the norm( \left \\| k\mathbf{a} \right \\|=\left \| k \right \|\left \\| \mathbf{a} \right \\| ). Norm of a vector obeys triangular inequality that the norm of a sum of two vectors is less than or equal to the sum of the norms ( \left \\| \mathbf{a} + \mathbf{b}\right \\| \leqslant \left \\| \mathbf{a} \right \\| + \left \\| \mathbf{b} \right \\| ). There are many common norms: 1-norm is defined by the sum of absolute value of the vector elements( \left \\| \mathbf{a} \right \\|_{1} = \left \| a_{1} \right \| + \left \| a_{2} \right \| + ... + \left \| a_{n} \right \| ) . 2-norm is the most often used vector norm, sometimes called Euclidean norm. When the subscript index of the vector norm is not specified, you may think that it is a Euclidean norm ( \left \\| \mathbf{a} \right \\| = \left \\| \mathbf{a} \right \\|_{2} = \sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}} ) . p-norm is sometimes called Minskowski norm is defined as ( \left \\| \mathbf{a} \right \\|_{p} = \sqrt[p]{a_{1}^{p}+a_{2}^{p}+...+a_{n}^{p}} ) . p-norm is generalized norm with a parameter ( p) . max-norm is also called Chebyshev norm is the largest absolute element in the vector ( \left \\| \mathbf{a} \right \\|_{\infty} = max[\left \| a_{1} \right \| , \left \| a_{2} \right \| , ... , \left \| a_{n} \right \| ] ) Use the interactive program below to experiment with your own vector input. The program will give you the norm of vector for p=1, 2, 3 and max. The vector input will be redrawn to give you feedback on what you type. Click Random Example button to generate random vector. Properties Some important properties of vector norm are Square of Euclidean norm is equal to the sum of square ( \left \\| \mathbf{a} \right \\|^{2} = a_{1}^{2} + a_{2}^{2} + ... + a_{n}^{2} ). Pythagorean Theorem is hold if and only if the two vectors are orthogonal ( \left \\| \mathbf{a}-\mathbf{b} \right \\|^{2} = \left \\| \mathbf{a} \right \\|^{2}+\left \\| \mathbf{b} \right \\|^{2}, \mathbf{a} \perp \mathbf{b} ). The law of cosine ( \left \\| \mathbf{a}-\mathbf{b} \right \\|^{2} = \left \\| \mathbf{a} \right \\|^{2}+\left \\| \mathbf{b} \right \\|^{2}-2\left \\| \mathbf{a} \right \\|\left \\| \mathbf{b} \right \\|cos\phi ) Norm of the dot product of two vectors is equal to the product of their norms ( \left \\| \mathbf{a}^{T}\mathbf{b} \right \\| = \left \\| \mathbf{a} \right \\|\left \\| \mathbf{b} \right \\| ). Relationship to vector inner products Square of Euclidean norm of a vector is equal to the inner product to itself ( \left \\| \mathbf{a} \right \\|^{2} = \mathbf{a}^{T}\mathbf{a} ) ( \mathbf{a}^{T}\mathbf{b}=\left \\| \mathbf{a} \right \\|\left \\| \mathbf{b} \right \\|cos\phi ) where, ( \phi ) is the angle between the two vectors ( \mathbf{a}^{T}\mathbf{b} = \frac{1}{4} (\left \\| \mathbf{a}+\mathbf{b} \right \\|^{2}-\left \\| \mathbf{a}-\mathbf{b} \right \\|^{2}) ) Norm of addition or subtraction follow the law of cosine ( \left \\| \mathbf{a} \pm \mathbf{b} \right \\|^{2} = \left \\| \mathbf{a} \right \\|^{2} + \left \\| \mathbf{b} \right \\|^{2} \pm 2 \mathbf{a}^{T} \mathbf{b} ) Addition of two square of norm of vectors follow parallelogram law ( \left \\| \mathbf{a} \right \\|^{2} + \left \\| \mathbf{b} \right \\|^{2} =\frac{1}{2}(\left \\| \mathbf{a} + \mathbf{b} \right \\|^{2} + \left \\| \mathbf{a} - \mathbf{b} \right \\|^{2}) ) p-norm is greater than the max-norm but less than ( n^{\frac{1}{p}} ) times the max-norm, that is ( 1\leqslant \frac{\left \\| \mathbf{a} \right \\|_{p}}{\left \\| \mathbf{a} \right \\|_{\infty }}\leq n^{\frac{1}{p}} ) . The norm ratio satisfies the inequality ( 1\leqslant \frac{\left \\| \mathbf{a} \right \\|_{p}}{\left \\| \mathbf{a} \right \\|_{q }}\leq n^{\frac{q-p}{pq}} ) . As ( q ) tends to infinity, the ( \frac{q-p}{pq} ) approaches ( \frac{1}{p} ) and ( \left \\| \mathbf{a} \right \\|_{p} ) approaches ( \left \\| \mathbf{a} \right \\|_{\infty} ). Cauchy-Schwartz inequality stated that the absolute value of vector dot product is always less than or equal to the product of their norms ( \left \| \mathbf{a}^{T}\mathbf{b} \right \|\leq \left \\| \mathbf{a} \right \\|\left \\| \mathbf{b} \right \\| ) . The equality ( \left \| \mathbf{a}^{T}\mathbf{b} \right \| = \left \\| \mathbf{a} \right \\|\left \\| \mathbf{b} \right \\| ) holds if and only if the vectors are linearly dependent. Relationship of norm of cross product and dot product is ( \left \\| \mathbf{a} \times \mathbf{b} \right \\|^{2} + \left \\| \mathbf{a}\cdot \mathbf{b} \right \\|^{2}=\left \\| \mathbf{a} \right \\|^{2}\left \\| \mathbf{b} \right \\|^{2} ) . See Also: Similarity tutorial, Minskowski distance, Chebyshev distance, Euclidean distance, City Block distance, Resources on Linear Algebra <Next \| Previous \| Index> Rate this tutorial or give your comments about this tutorial This tutorial is copyrighted. Preferable reference for this tutorial is Teknomo, Kardi (2011) Linear Algebra tutorial. https:\people.revoledu.com\kardi\tutorial\LinearAlgebra\
3383	https://phys.libretexts.org/Courses/Grand_Rapids_Community_College/PH246_Calculus_Physics_II_(2025)/03%3A_Electric_Potential	3: Electric Potential - Physics 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 PH246 Calculus Physics II (2025) Grand Rapids Community College { "3.01:Prelude_to_Electric_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.02:_Electric_Potential_Energy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.03:_Electric_Potential_and_Potential_Difference" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.04:_Calculations_of_Electric_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.05:_Determining_Field_from_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.06:_Equipotential_Surfaces_and_Conductors" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.07:_Electric_Potential(Exercises)" : "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:_Electric_Charges_and_Fields" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Gauss\'s_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Electric_Potential" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Capacitance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Current_and_Resistance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Direct-Current_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Magnetic_Forces_and_Fields" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Sources_of_Magnetic_Fields" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Electromagnetic_Induction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Alternating-Current_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Electromagnetic_Waves" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Interference" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Diffraction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:__Relativity" : "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" } { "PH246_Calculus_Physics_II_(2025)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Thu, 10 Apr 2025 17:18:05 GMT 3: Electric Potential 107508 107508 Nathan Lindy { } Anonymous Anonymous 2 false false [ "article:topic-guide", "authorname:openstax", "electric potential", "license:ccby", "showtoc:no", "program:openstax", "source-phys-4392", "licenseversion:40", "source@ ] [ "article:topic-guide", "authorname:openstax", "electric potential", "license:ccby", "showtoc:no", "program:openstax", "source-phys-4392", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Grand Rapids Community College 4. PH246 Calculus Physics II (2025) 5. 3: Electric Potential Expand/collapse global location PH246 Calculus Physics II (2025) Front Matter 1: Electric Charges and Fields 2: Gauss's Law 3: Electric Potential 4: Capacitance 5: Current and Resistance 6: Direct-Current Circuits 7: Magnetic Forces and Fields 8: Sources of Magnetic Fields 9: Electromagnetic Induction 10: Alternating-Current Circuits 11: Electromagnetic Waves 12: Interference 13: Diffraction 14: Relativity Back Matter 3: Electric Potential Last updated Apr 10, 2025 Save as PDF 2.6: Gauss's Law (Exercises) 3.1: Prelude to Electric Potential picture_as_pdf Full Book Chapter Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate Page ID 107508 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers In this chapter, we examine the relationship between voltage and electrical energy, and begin to explore some of the many applications of electricity. 3.1: Prelude to Electric PotentialTwo terms commonly used to describe electricity are its energy and voltage, which we show in this chapter is directly related to the potential energy in a system. We know, for example, that great amounts of electrical energy can be stored in batteries, are transmitted cross-country via currents through power lines, and may jump from clouds to explode the sap of trees. In a similar manner, at the molecular level, ions cross cell membranes and transfer information. 3.2: Electric Potential EnergyWhen a free positive charge q is accelerated by an electric field, it is given kinetic energy (Figure). The process is analogous to an object being accelerated by a gravitational field, as if the charge were going down an electrical hill where its electric potential energy is converted into kinetic energy, although of course the sources of the forces are very different. 3.3: Electric Potential and Potential DifferenceElectric potential is potential energy per unit charge. The potential difference between points A and B, V B−V A, that is, the change in potential of a charge q moved from A to B, is equal to the change in potential energy divided by the charge. Potential difference is commonly called voltage, represented by the symbol Δ⁢V. 3.4: Calculations of Electric PotentialPoint charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. 3.5: Determining Field from PotentialIn certain systems, we can calculate the potential by integrating over the electric field. As you may already suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need 𝐸⃗E→\vec{E} to calculate the force in a system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate 𝑉VV and then d 3.5.1: Calculating the Electric Field from the Electric Potential 3.6: Equipotential Surfaces and ConductorsWe can represent electric potentials pictorially, just as we drew pictures to illustrate electric fields. This is not surprising, since the two concepts are related. We use arrows to represent the magnitude and direction of the electric field, and we use green lines to represent places where the electric potential is constant. These are called equipotential surfaces in three dimensions, or equipotential lines in two dimensions. 3.7: Electric Potential (Exercises) This page titled 3: Electric Potential is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. 7: Electric Potential by OpenStax is licensed CC BY 4.0. Original source: Toggle block-level attributions Back to top 2.6: Gauss's Law (Exercises) 3.1: Prelude to Electric Potential Was this article helpful? Yes No Recommended articles 3: Electric PotentialIn this chapter, we examine the relationship between voltage and electrical energy, and begin to explore some of the many applications of electricity. 3.3: Electric Potential and Potential DifferenceElectric potential is potential energy per unit charge. The potential difference between points A and B, V B−V A, that is, the change in... 3.4: Calculations of Electric PotentialPoint charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge on ... 3.5: Determining Field from PotentialIn certain systems, we can calculate the potential by integrating over the electric field. As you may already suspect, this means that we may calculat... 16.4: Calculating Electric Potential of Charge DistributionsThis section describes how to calculate the electric potential for several common distributions of charge including the line segment, ring, disk, and ... Article typeChapterAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow TOCno Tags electric potential source-phys-4392 source@ © Copyright 2025 Physics 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× contents readability resources tools ☰) 2.6: Gauss's Law (Exercises) 3.1: Prelude to Electric Potential Complete your gift to make an impact
3384	https://www.geeksforgeeks.org/physics/inverse-square-law-formula/	Published Time: 2022-05-09 12:00:09 Inverse Square Law Formula - GeeksforGeeks Skip to content Tutorials Python Java Data Structures & Algorithms ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps And Linux Software and Tools School Learning Practice Coding Problems Go Premium Switch to Dark Mode Sign In Physics CBSE Physics Class 8 Notes CBSE Class 9 Physics Notes CBSE Notes for Class 10 Physics CBSE Class 11 Physics Notes CBSE Class 12 Physics Notes CBSE Class 8 Science Notes CBSE Class 9 Science Notes Class 10 Science Notes Sign In ▲ Open In App Next Article: Snell's Law Formula Inverse Square Law Formula Last Updated : 04 Feb, 2024 Comments Improve Suggest changes Like Article Like Report The inverse square law describes the intensity of light in relation to its distance from the source. It states that the intensity of the radiation is inversely proportional to the square of the distance. In other words, the intensity of light to an observer from a source is inversely proportional to the square of the distance between the observer and the source. It is used to calculate the distance or intensity of a particular radiation. As the distance between the source and the observer rises, the intensity of the light from the source decreases. Formula The ratio of the intensity of the light source for two different time intervals is equal to the reciprocal of the squares of their respective distances of the object from the source. The intensity and distance are denoted by the symbol I and d respectively. The standard unit of intensity is candelas or lumens while for distance, it is meters. The dimensional formula of intensity and distance is [M 1 L 0 T-3] and [M 0 L 1 T 0]. I ∝ 1/d2 where, I is the intensity of light, d is the distance between light source and observer. Consider light sources of intensity I 1 and I 2 at distances d 1 and d 2 respectively. For this scenario, the inverse square formula is given by, I1/I2= d22/d12 Sample Problems Problem 1. A light source has an intensity of 20 candelas at a distance of 3 m from the object. Calculate its intensity if it is at a distance of 6 m from the object. Solution: We have, I 1 = 20 d 1 = 3 d 2 = 6 Using the formula we get, I 1/I 2 = d 2 2/d 1 2 => 20/I 2 = 6 2/3 2 => I 2 = 20 (9/36) => I 2 = 5 candelas Problem 2. A light source has the intensity of 50 candelas at a distance of 30 m from the object. Calculate its intensity if it is at a distance of 10 m from the object. Solution: We have, I 1 = 50 d 1 = 30 d 2 = 10 Using the formula we get, => 50/I 2 = 10 2/30 2 => I 2 = 50 (900/100) => I 2 = 50 (9) => I 2 = 450 candelas Problem 3. A light source has the intensity of 100 candelas at a distance of 20 m from the object. Calculate its distance if its intensity is 80 candelas. Solution: We have, I 1 = 100 I 2 = 80 d 1 = 20 Using the formula we get, => 100/80 = d 2 2/20 2 => d 2 2 = (400) (5/4) => d 2 2 = 500 => d 2 = 22.4 m Problem 4. A light source has the intensity of 20 candelas at a distance of 1 m from the object. Calculate its distance if its intensity is 40 candelas. Solution: We have, I 1 = 20 I 2 = 40 d 1 = 1 Using the formula we get, => 20/40 = d 2 2/1 2 => d 2 2 = 1/2 => d 2 = 0.7071 m Problem 5. Calculate the value of d1for d2= 100 m, I1= 200 candelas and I2= 300 candelas. Solution: We have, d 2 = 100 I 1 = 200 I 2 = 300 Using the formula we get, => 200/300 = 100/d 1 2 => d 1 2 = 100 (2/3) => d 1 = 122.48 m Problem 6. Calculate the value of I1for I2= 150 candelas, d1= 5 m and d2= 15 m. Solution: We have, I 2 = 150 d 1 = 5 d 2 = 15 Using the formula we get, => I 1/100 = 15 2/5 2 => I 1 = 32 (100) => I 1 = 900 candelas Problem 7. Calculate the value of I2for I1= 650 candelas, d1= 36 m and d2= 24 m. Solution: We have, I 1 = 650 d 1 = 36 d 2 = 24 Using the formula we get, => 650/I 2 = 24 2/36 2 => I 2 = 650 (1296/576) => I 2 = 1462.5 candelas Comment More info Advertise with us Next Article Snell's Law Formula J jatinxcx Follow Improve Article Tags : School Learning Physics Physics-MAQ Physics-Formulas Similar Reads Inverse Tangent Formula In trigonometry, angles are evaluated with respect to the basic trigonometric functions of trigonometry which are sine, cosine, tangent, cotangent, secant, and cosecant. These trigonometric functions have their own trigonometric ratios under different angles which are used in trigonometric operation 3 min readSquare Root Formula Square root formula is a mathematical expression that calculates the square root of a number. For a non-negative real number xx, the square root y is a number such that y2 = x. A square root is an operation that is used in many formulas and different fields of mathematics. 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3385	https://communities.sas.com/t5/SAS-Programming/Generate-all-valid-combinations-of-4-length-character-strings/td-p/830805	Sign In SAS Programming DATA Step, Macro, Functions and more Turn on suggestions Showing results for Search instead for Did you mean: Home / Programming / Programming / Generate all valid combinations of 4 length character strings Options RSS Feed Mark Topic as New Mark Topic as Read Float this Topic for Current User Bookmark Subscribe Mute Printer Friendly Page BookmarkSubscribeRSS Feed All forum topics Previous Next ☑ This topic is solved. Need further help from the community? Please sign in and ask a new question. DominicCavenagh Fluorite \| Level 6 Go to Solution Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Generate all valid combinations of 4 length character strings Posted 08-28-2022 08:09 PM (7709 views) Hi All, I am trying to compute all valid character strings based on some constraints: Each string is 4 alpha characters long i.e. "ABCD" or "QRTS" A string is declared invalid if there is already a valid string where there are the same two characters next to each other. i.e. If "ABCD" is already a valid string, then "BACD" would be invalid due to AB being next to each other in the first two positions. Consequently "ABCD" would make the following three strings invalid: "BACD", "ACBD", "ABDC". My approach so far has been to generate all 26^4 combinations (CHAR4 in the code). Then for each string, generate the strings that this string makes invalid (T_12, T_23, T_34 in the code). Then loop through each row and merge back into the full list of strings, marking any string that is the same as {T_12, T_23, T_34} as invalid. Then move to the next string and repeat. The resulting dataset should have all the strings, and those that are invalid are flagged i.e. ineligible=1. As you can probably imagine due to the ~500,000 strings and the amount of repeated merging, this process is taking a long time and I havent been able to run it till finish yet. So far I have left it going for 24 hours without any luck. I feel like there is a better way to do this, but cant put my finger on it, does anyone have any advice on a way to optimise my program, or a better way to produce these strings altogether? Thanks in advance See below for my current approach: ``` SET UP POSSIBLE PERMUTATIONS OF THE 4-PART TEXT STRING - 26^4=456,976 combinations ; proc format; value alpha 1="A" 2="B" 3="C" 4="D" 5="E" 6="F" 7="G" 8="H" 9="I" 10="J" 11="K" 12="L" 13="M" 14="N" 15="O" 16="P" 17="Q" 18="R" 19="S" 20="T" 21="U" 22="V" 23="W" 24="X" 25="Y" 26="Z"; run; data alpha; do N1=1 to 26; do N2=1 to 26; do N3=1 to 26; do N4=1 to 26; output; end; end; end; end; format N1 N2 N3 N4 alpha.; run; data new1; set alpha; length C1 C2 C3 C4 $1.; c1=compress(vvalue(N1)); c2=compress(vvalue(N2)); c3=compress(vvalue(N3)); c4=compress(vvalue(N4)); CHAR4=C1\|\|C2\|\|C3\|\|C4; by default: ineligible=0; INELIGIBLE=0; this variable will be used for 1-to-many merging; ALL=1; set up the combinations with potential transcription errors; T_12=c2\|\|c1\|\|c3\|\|c4; T_23=c1\|\|c3\|\|c2\|\|c4; T_34=c1\|\|c2\|\|c4\|\|c3; run; this is the dataset to be overwritten with each loop ; data new2; set new1; run; proc sort; by all; run; %macro loopy (start, stop); %do i=&start. %to &stop.; data select1; set new2; if N=&i. and ineligible=0; keep the combinations with potential transcription errors; TRANSPOSE12=T_12; TRANSPOSE23=T_23; TRANSPOSE34=T_34; keep ALL TRANSPOSE12 TRANSPOSE23 TRANSPOSE34; run; data new2; merge NEW2 (in=in1) SELECT1; by all; if in1; if CHAR4 is the same as any of the transposed combinations, then indicate as INELIGIBLE=1; if CHAR4=TRANSPOSE12 or CHAR4=TRANSPOSE23 or CHAR4=TRANSPOSE34 then INELIGIBLE=1; drop TRANSPOSE12 TRANSPOSE23 TRANSPOSE34; run; %end; %mend; ``` 0 Likes Reply 1 ACCEPTED SOLUTION Accepted Solutions mkeintz PROC Star Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 06:13 PM (7527 views) \| In reply to DominicCavenagh Editted note: After this solution was accepted, a much more optimal solution was shown by @FreelanceReinh. Take look at it in this topic thread. Your task is effectively to examine all 358,800 four-character strings taken from the 26 letter alphabet, where no letter is present in the string more than once. You need to consider the string as 3 overlapping letter-pairs, starting in positions 1, 2, and 3 respectively. For a string to be declared valid (i.e. VALID='Y'), none of its 3 letter-pairs can be a reverse of the corresponding letter-pair in other valid strings. This suggests that, as you determine a string to be valid, you can record its three letter-pairs (reversed) in a hash (three hashes actually: _RP1 for reversed position 1 pairs, _RP2 for reversed position 2 pairs, and _RP3). Then any subsequent string that has a pair starting in position 1 that is found in hash _RP1, must not be valid. This task is doable in one data step, but I put it in two steps below. It takes less than 1 second: It produces 24,173 with VALID='Y', and 334,627 with VALID='N'. ``` data need (drop=i j k l p) ; array LTR {26} $1 temporary ('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'); length strng $4 ; array pr {3} $2; /Each 4-letter string has 3 pairs / array rp {3} $2; /Reversed Pairs / n=0; do i=1 to 26; do j=1 to 26; if j=i then continue / See discussion /; do k=1 to 26; if k=i or k=j then continue / See discussion /; do l=1 to 26; if l=i or l=j or l=k then continue / See discussion /; strng=cats(ltr{i},ltr{j},ltr{k},ltr{l}); do p=1 to length(strng)-1; pr{p}=substr(strng,p,2); rp{p}=reverse(pr{p}); end; n=n+1; output; end; end; end; end; run; data want (drop=_:); set need ; label link_rp1='Earliest N generating this rp1 value' link_rp2='Earliest N generating this rp2 value' link_rp3='Earliest N generating this rp3 value' ; if n=1 then do; declare hash _rp1 (); / reverse pair1 values encountered so far / _rp1.definekey('rp1'); _rp1.definedata('rp1','link_rp1'); /Link_rp1, / _rp1.definedone(); declare hash _rp2 (); _rp2.definekey('rp2'); _rp2.definedata('rp2','link_rp2'); _rp2.definedone(); declare hash _rp3 (); _rp3.definekey('rp3'); _rp3.definedata('rp3','link_rp3'); _rp3.definedone(); end; call missing (of link_:); _rc=_rp1.find(key:pr1); /Get prior LINK_RP1, if any, for current PR1/ _rc=_rp2.find(key:pr2); _rc=_rp3.find(key:pr3); length valid $1; / will be either "Y" or "N" / if N(of link_:)>0 then valid='N'; / At least 1 current pair found in a prior reverse pair/ else valid='Y'; if valid='Y' then do; /If current is valid, then add new reverse pairs for lookup / if _rp1.check(key:rp1) then _rp1.add(key:rp1,data:rp1,data:n); if _rp2.check(key:rp2) then _rp2.add(key:rp2,data:rp2,data:n); if _rp3.check(key:rp3) then _rp3.add(key:rp3,data:rp3,data:n); end; run; proc freq data=want; tables valid;run; ``` Note this generates candidate strings in lexicographic order (i.e. ABCD is valid, so BACD cannot be valid). You could take away this lexicographic "bias" by randomly re-ordering the values in the LTR array. Editted annotation: Regarding the three statements with / See discussion / : Remove these statements to allow a letter to appear in the strings more than once. The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for Allow PROC SORT to output multiple datasets View solution in original post 3 Likes Reply 13 REPLIES 13 ballardw Super User Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-28-2022 10:09 PM (7670 views) \| In reply to DominicCavenagh I am not sure if you are clear on the difference between "combination" and "permutation". Permutation, which you have created, considers order of the values. Combination will just have one value with any specific set of values, so if done properly and have "ABCD", then combinations will not include ABDC, BACD, DABC and so on. So you can start by creating COMBINATIONS. Which is moderately easy after some time playing with the different combinatorial functions: data start; array x(26) $ 1 ('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'); n=dim(x); k=4; ncomb=comb(n, k); do j=1 to ncomb; rc=allcomb(j, k, of x[]); string=cats(x1,x2,x3,x4); output; end; keep string; run; Not the function COMB will give you the number of N items taken K at a time. In this case 14,590. So I suspect starting there might reduce the scope of this exercise quite a bit as if ABCD is generated won't have a BACD in the output (or other permutations of the specific 4 characters. I personally might be more concerned with a list of "valid" values. Once I have those it is extremely easy to mark anything else as invalid and typically the combinations of "valid" are generally shorter starting something like code values. Perhaps there is a data set with the valid values somewhere? I am wondering about why you have used the comment set up the combinations with potential transcription errors; T_12=c2\|\|c1\|\|c3\|\|c4; Transcription tends to make me think that you have some data that you want compared and perhaps this is a poor start to begin with and should be looking at actual "transcribed values". This may be a case of considering one piece of the problem in detail is hiding the actual need and it may help to describe the whole problem a bit more. 2 Likes Reply DominicCavenagh Fluorite \| Level 6 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-28-2022 11:00 PM (7663 views) \| In reply to ballardw Hi, thanks for your reply. The combinations idea is very interesting and has given me some food for thought. In terms of the "valid" values, these are generated in the same loop. So they are highly dependent on which is your first choice. i.e. Generate all string permutations. Choose one string as "valid" value (call it "str") Mark all strings that "str" makes invalid as invalid Choose another string that has not yet been marked invalid. 5.repeat steps 2-4, until you run out of valid strings to choose. Obviously from above, if the first choice is ABCD, then BACD will not be valid. However if your first choice is BACD then ABCD will not be valid. In my original program I was just going through the list of permutations as sorted alphabetically, however this may not be the best approach, and it may be better to just pick at random. In terms of the transcription comment, this is domain specific. The purpose of producing these ID's is that they need to be handwritten on samples and we don't want people accidently writing BACD when they read ABCD. There was some corncern this would happen, so we sought to produce IDs that if this did occur, it would not be a valid ID and we would know something wrong happened, rather than the sample being assigned to the wrong ID. While the combination method you describe guarantees success for our criteria, it is only producing a subset of the valid list of ID's. I think this is what you meant when saying this is a good place to start. If the combination method chooses ABCD, then there can be no other string with all these characters. However, based on our constraints, DBCA, CBAD and ADCB would be valid, as the repeated (/swapped) letters are not next to each other. Producing these extras is relatively easy to add in to the method, so I will have a go at that. Thanks again for your help. 0 Likes Reply ballardw Super User Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 10:33 AM (7589 views) \| In reply to DominicCavenagh You could define specific "allowed" 2 letter combinations instead of 4 single letters for creating permutations or combinations. You do need to define a "start" rule in any case. And what about same letter twice such as "AA" together? Unfortunately a very small example often does not completely describe the rejection rule. I humbly suggest that if the "entry codes" have not been defined yet that you consider exactly how many are actually needed. That might reduce the scope of the problem. If that is not possible then you may well be in an environment when the next thing is , "Now we need 5 characters" and have a lot more exclusion rules to impose. 1 Like Reply Ksharp Super User Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 08:05 AM (7612 views) \| In reply to DominicCavenagh ``` proc format; value alpha 1="A" 2="B" 3="C" 4="D" 5="E" 6="F" 7="G" 8="H" 9="I" 10="J" 11="K" 12="L" 13="M" 14="N" 15="O" 16="P" 17="Q" 18="R" 19="S" 20="T" 21="U" 22="V" 23="W" 24="X" 25="Y" 26="Z"; run; %let total=%eval(264); proc plan seed=123 ; factors id=&total. ordered x=4 of 26 comb/noprint; output out=temp; quit; data temp2; set temp; alpha=put(x,alpha.); run; proc transpose data=temp2 out=temp3(drop=_:); by id; var alpha; run; data want; set temp3; length want $ 20; want=cats(of col:); keep id want; run; ``` 0 Likes Reply FreelanceReinh Jade \| Level 19 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 10:45 AM (7580 views) \| In reply to DominicCavenagh Hi @DominicCavenagh and welcome to the SAS Support Communities! I think this task (as I understand it) can be viewed as a problem in graph theory (which I don't know much about, unfortunately): The four-letter combinations can be thought of as the vertices of a graph and an edge connects two vertices if (and only if) the combinations differ by exactly one permutation of two neighboring letters. While the entire graph with its 264=456976 vertices is huge, it can be decomposed into small disjoint components ("subgraphs") comprising only 24 or fewer vertices. This is because switching two letters doesn't change the choice of the letters. The components are comb(26,4) = 14950 combinations of four different letters (with 24 possible permutations each) comb(26,3)3 = 7800 combinations of three different letters one of which occurs twice (with 12 possible permutations each) comb(26,2)2 = 650 combinations of two different letters one of which occurs three times (with 4 possible permutations each) comb(26,2) = 325 combinations of two different letters both of which occur twice (with 6 possible permutations each) 26 combinations consisting of four equal letters. (1495024 + 780012 + 6504 + 3256 + 26 = 456976) Each of the above five classes of components can be investigated separately, based on an arbitrarily selected single element of the class (because a one-to-one mapping of the letters doesn't change the structure of a combination). The connection rule implies that a vertex of this particular graph can have at most three edges. Let's start with the easiest case: class 5. Obviously, all of the 26 constituent combinations AAAA, BBBB, ..., ZZZZ can be selected as "valid" because they are isolated points in the graph. Class 4: The components of this class can be represented by a subgraph like this: ABBA / \ AABB──ABAB BABA──BBAA \ / BAAB Now you see that some selections are more favorable than others: For example, if you selected ABAB and BABA, you would end up with only these two elements -- the other four elements would then be invalid. However, the selection AABB, ABBA, BAAB, BBAA yields four valid strings -- there is no direct connection (edge) between any two of these four permutations in the graph. It's easy to see that you can't get more than four valid elements out of those six. So, if you want to obtain as many valid four-letter strings as possible, you need to find "optimum" selections like this, i.e., with the maximum number of valid elements, for each class. [Continuation #1] Class 3: The representatives of this class have a simple chain structure: ABBB──BABB──BBAB──BBBA Obviously, there are three possible "optimum" selections: {ABBB, BBAB}, {ABBB, BBBA} and {BABB, BBBA}. Class 2: Using the letter choice A, B, C, C, the representative of this class has the below structure: ACBC──ABCC──BACC──BCAC / \ / \ ACCB CABC────────CBAC BCCA \ / \ / CACB──CCAB──CCBA──CBCA The subset of the six permutations highlighted in bold constitutes one of the "optimum" selections (one of two, as it turns out). While this statement seems plausible, it's not obvious (to me). So, let's try to prove it (using SAS): A "better" selection than the above must contain seven or more out of the twelve permutations. If we can show that a selection of seven would always contain an invalid element, there couldn't be more than seven a fortiori. %let n=12; %let k=7; data test; array x[&n] $4 ('ABCC' 'ACBC' 'ACCB' 'BACC' 'BCAC' 'BCCA' 'CABC' 'CACB' 'CBAC' 'CBCA' 'CCAB' 'CCBA'); do i=1 to comb(&n, &k); rc=allcomb(i, &k, of x[]); f=0; do j=1 to &k-1 until(f); do k=j+1 to &k until(f); if compged(x[j],x[k])=20 then f=1; end; end; if ~f then output; end; run; / --> 0 observations / Apparently, each of the comb(12, 7)=792 subsets of 7 out of the 12 permutations in the array creates one or more pairs with generalized edit distance 20 (which in this case is equivalent to a "SWAP", i.e., a transposition of two neighboring letters). The same code with k=6 yields two observations: the two optimum selections. (The second one being the complement of the first, i.e., comprising the six non-bold permutations in the graph above.) [Continuation #2] Class 1: Using the letter choice A, B, C, D, the representative of this class has the below structure: ____________________ / \ ABDC ACBD──ACDB ADBC / \ / │ │ \ / \ BADC ABCD │ │ ADCB DABC / \ / │ │ \ / \ │ BACD CABD──CADB DACB │ │ │ / \ │ │ │ │ CBAD CDAB │ │ │ │ / \ / \ │ │ │ BCAD CBDA──CDBA DCAB │ │ \ / \ / │ │ BCDA DCBA │ │ \ / │ \ BDCA──DBCA / \ │ │ / \ │ │ / \ │ │ / \──────BDAC──DBAC─────/ Actually it's rotational symmetric ... Each vertex is linked to exactly three others (as it should, with four different letters, considering the connection rule). We can use almost the same code as above to create the optimum selections (two again) -- each comprising 12 out of the 24 permutations -- and to prove that selecting more than 12 valid elements is impossible. One of the optimum selections is shown above in boldface. The other one is just the complement of it, i.e., the non-bold permutations. %let n=24; %let k=12; data test2; array c $1 ('A' 'B' 'C' 'D'); array x[&n] $4; do p=1 to &n; rc=allperm(p, of c[]); x[p]=cat(of c[]); end; do i=1 to comb(&n, &k); rc=allcomb(i, &k, of x[]); f=0; do j=1 to &k-1 until(f); do k=j+1 to &k until(f); if compged(x[j],x[k])=20 then f=1; end; end; if ~f then output; end; run; / --> 2 observations / Again, with k=13 this step results in 0 observations. Either way the run time was only a few seconds on my workstation. Given the above results, it should be feasible to select a subset of the original 264 four-letter combinations with the maximum possible number of valid elements. And this number should be 1495012 + 78006 + 6502 + 3254 + 26 = 228826. Just translate one of the optimum selections given for each of the five classes to each of the 14950, 7800, 650, 325 and 26 letter combinations, respectively (e.g., by replacing A, B, C, D with K, R, W, Y, etc.). I can help you with that tomorrow (CEST) if needed. (And if a graph theory adept in the forum recognizes this as something for which a standard solution is known, please chime in.) 3 Likes Reply DominicCavenagh Fluorite \| Level 6 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 09:51 PM (7502 views) \| In reply to FreelanceReinh Thanks for this very thorough solution to the problem. It really opened my eyes to the differences in number of solutions that can be produced given the initial selection. I love the graph theory approach. Is it your feeling that all the strings could be produced by starting with one of the optional strings you mention and then just iterating through the letters? Then do this for each case?I would be interested to see your approach to this. If i could have, I would have really liked to mark this as an acceptable solution as well. Many thanks! 0 Likes Reply FreelanceReinh Jade \| Level 19 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-30-2022 11:46 AM (7434 views) \| In reply to DominicCavenagh Here is the implementation of what I outlined in my previous post: ``` / Create a maximum-size set of permutations of four capital letters so that any two permutations from the set differ by more than just a transposition of two neighboring letters / data want(keep=string); call streaminit(27182818); length string $4; / resulting four-letter strings / array a $1; / the alphabet: 'A', 'B', ..., 'Z' / array c $1; / selected letters for a combination / array x $4; / systematic permutations of four letters / array d temporary; / indices for selecting from four characters / array v[47,4] temporary; / combinations of indices 1, 2, 3, 4 for "optimum" selections / array m[5,4] temporary (1 2 3 4 1 1 2 3 1 1 1 2 1 1 2 2 1 1 1 1); / see array D / array t temporary (24 12 4 6 1); / number of permutations per selection for class 1, ..., 5 / array u temporary (12 6 2 4 1); / maximum number of valid strings per selection for class 1, ..., 5 / array s temporary ( 2 2 3 1 1); / number of different "optimum" solutions for class 1, ..., 5 / / Phase 1: Prepare "optimum" selections of combinations / do g=1 to dim(t); / g=1, ..., 5 is the "class" number / do i=1 to dim(d); d[i]=m[g,i]; end; do i=1 to t[g]; call lexperm(i, of d[]); x[i]=cat(of d[]); end; do i=1 to comb(t[g], u[g]); / search through maximum-size selections of permutations of four digits / select(g); when(1) rc=allcomb(i, u[g], of x[]); when(2) rc=allcomb(i, u[g], of x1-x12); when(3) rc=allcomb(i, u[g], of x1-x4); when(4) rc=allcomb(i, u[g], of x1-x6); otherwise; end; f=0; / flag variable to indicate the detection of an invalid selection / do j=1 to u[g]-1 until(f); / check validity of pairs of selected permutations / do k=j+1 to u[g] until(f); if compged(x[j],x[k])=20 then f=1; / here, 20 means one "SWAP" (transposition of neighboring digits) / end; end; if ~f then do; do j=1 to u[g]; / write indices for valid selections to array V / y+1; do k=1 to dim(d); v[y,k]=input(char(x[j],k),1.); end; end; end; end; end; / Phase 2: Create strings based on the above results / do g=1 to dim(t); / again, g=1, ..., 5 is the "class" number / link init; / reset the "alphabet" array A for a clean start / select(g); when(1) do i=1 to comb(26, 4); rc=allcomb(i, 4, of a[]); / In class 1, a selection of four letters (e.g., A, B, C, D) / do j=1 to dim(c); / leads to one combination: ABCD. / c[j]=a[j]; end; link write; end; when(2) do i=1 to comb(26, 3); / In class 2, a selection of three letters (e.g., A, B, C) / rc=allcomb(i, 3, of a[]); / gives rise to three different combinations: / c1=a1; c2=a2; c3=a3; link write; / AABC, / c1=a2; c2=a1; c3=a3; link write; / BBAC, / c1=a3; c2=a1; c3=a2; link write; / CCAB. / end; when(3) do i=1 to comb(26, 2); / In class 3, a selection of two letters (e.g., A, B) / rc=allcomb(i, 2, of a[]); / gives rise to two different combinations: / c1=a1; c2=a2; link write; / AAAB, / c1=a2; c2=a1; link write; / BBBA. / end; when(4) do i=1 to comb(26, 2); / In class 4, a selection of two letters (e.g., A, B) / rc=allcomb(i, 2, of a[]); / leads to one combination: AABB. / c1=a1; c2=a2; link write; end; when(5) do i=1 to 26; / In class 5, a selected letter (e.g., A) / do j=1 to dim(c); / leads to one combination: AAAA. / c[j]=a[i]; end; link write; end; otherwise; end; e+u[g]s[g]; / offset for dimension 1 of array V / end; return; init: do i=1 to 26; a[i]=byte(i+rank('A')-1); / write letters 'A', ..., 'Z' to array A / end; return; write: r=rand('integer',s[g]); / randomly select from several possible "optimum" solutions (if any) / do j=e+ug+1 to e+u[g]r; string=cat(c[v[j,1]],c[v[j,2]],c[v[j,3]],c[v[j,4]]); / combine selected letters according to the / output; / patterns stored in "solution" array V / end; return; run; / 1495012 + 78006 + 6502 + 3254 + 26 = 228826 obs. (from class 1 through 5) / ``` As predicted, 228,826 "valid" four-letter strings were found. The typical run time of 4 - 5 seconds includes the time needed to find optimum solutions (see "phase 1" of the code). In cases where more than one selection of maximum size exists, one of the possible solutions is chosen randomly on a case-by-case basis. 2 Likes Reply DominicCavenagh Fluorite \| Level 6 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-30-2022 07:37 PM (7410 views) \| In reply to FreelanceReinh Wow, thanks so much for this - this is a beautiful solution to the problem! Thanks so much for your effort to implement this! I wish I could accept multiple solutions! 0 Likes Reply Quentin Super User Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 10:51 AM (7573 views) \| In reply to DominicCavenagh Maybe a hash approach could help? So start with your data set of ~500,000 candidate 4-character strings. In a data step, create an empty hash table with just a key value, which is a 2-character string. Then read through your dataset of candidate strings. Each candidate string has three pairs of values you want to check to see if they would introduce risk of transposition errors. For each candidate string, check all three pairs of values to see if they are already in the hash table. So if the candidate string is 'ABCD', check if 'AB' 'BC' or 'CD' is already in the hash table. If any are, then this string is INELIGIBLE. If none of them are in the table, then this string is ELIGIBLE, and you add six values to the hash table 'AB' 'BA' 'BC' 'CB' 'CD' 'DC'. EDIT: In re-reading your question, you only want to make a string ineligible if the 2-character combination exists in the same position as another string. So I think you would want a hash table with two keys: Position and Substring. Position would be 1, 2, or 3. So in my example, you would add six values to the hash table: position:1 substr:'AB' position:1 substr:'BA' position:2 substr:'BC' position:2 substr:'CB' position:3 substr:'CD' position:3 substr:'DC' 1 Like Reply mkeintz PROC Star Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 06:13 PM (7528 views) \| In reply to DominicCavenagh Editted note: After this solution was accepted, a much more optimal solution was shown by @FreelanceReinh. Take look at it in this topic thread. Your task is effectively to examine all 358,800 four-character strings taken from the 26 letter alphabet, where no letter is present in the string more than once. You need to consider the string as 3 overlapping letter-pairs, starting in positions 1, 2, and 3 respectively. For a string to be declared valid (i.e. VALID='Y'), none of its 3 letter-pairs can be a reverse of the corresponding letter-pair in other valid strings. This suggests that, as you determine a string to be valid, you can record its three letter-pairs (reversed) in a hash (three hashes actually: _RP1 for reversed position 1 pairs, _RP2 for reversed position 2 pairs, and _RP3). Then any subsequent string that has a pair starting in position 1 that is found in hash _RP1, must not be valid. This task is doable in one data step, but I put it in two steps below. It takes less than 1 second: It produces 24,173 with VALID='Y', and 334,627 with VALID='N'. ``` data need (drop=i j k l p) ; array LTR {26} $1 temporary ('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'); length strng $4 ; array pr {3} $2; /Each 4-letter string has 3 pairs / array rp {3} $2; /Reversed Pairs / n=0; do i=1 to 26; do j=1 to 26; if j=i then continue / See discussion /; do k=1 to 26; if k=i or k=j then continue / See discussion /; do l=1 to 26; if l=i or l=j or l=k then continue / See discussion /; strng=cats(ltr{i},ltr{j},ltr{k},ltr{l}); do p=1 to length(strng)-1; pr{p}=substr(strng,p,2); rp{p}=reverse(pr{p}); end; n=n+1; output; end; end; end; end; run; data want (drop=_:); set need ; label link_rp1='Earliest N generating this rp1 value' link_rp2='Earliest N generating this rp2 value' link_rp3='Earliest N generating this rp3 value' ; if n=1 then do; declare hash _rp1 (); / reverse pair1 values encountered so far / _rp1.definekey('rp1'); _rp1.definedata('rp1','link_rp1'); /Link_rp1, / _rp1.definedone(); declare hash _rp2 (); _rp2.definekey('rp2'); _rp2.definedata('rp2','link_rp2'); _rp2.definedone(); declare hash _rp3 (); _rp3.definekey('rp3'); _rp3.definedata('rp3','link_rp3'); _rp3.definedone(); end; call missing (of link_:); _rc=_rp1.find(key:pr1); /Get prior LINK_RP1, if any, for current PR1/ _rc=_rp2.find(key:pr2); _rc=_rp3.find(key:pr3); length valid $1; / will be either "Y" or "N" / if N(of link_:)>0 then valid='N'; / At least 1 current pair found in a prior reverse pair/ else valid='Y'; if valid='Y' then do; /If current is valid, then add new reverse pairs for lookup / if _rp1.check(key:rp1) then _rp1.add(key:rp1,data:rp1,data:n); if _rp2.check(key:rp2) then _rp2.add(key:rp2,data:rp2,data:n); if _rp3.check(key:rp3) then _rp3.add(key:rp3,data:rp3,data:n); end; run; proc freq data=want; tables valid;run; ``` Note this generates candidate strings in lexicographic order (i.e. ABCD is valid, so BACD cannot be valid). You could take away this lexicographic "bias" by randomly re-ordering the values in the LTR array. Editted annotation: Regarding the three statements with / See discussion / : Remove these statements to allow a letter to appear in the strings more than once. The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for Allow PROC SORT to output multiple datasets 3 Likes Reply DominicCavenagh Fluorite \| Level 6 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 09:55 PM (7500 views) \| In reply to mkeintz Thanks for this solution, it is very elegant. I have not come across hashes in SAS before so these are new to me and I will need to do a bit of work to fully understand your implementation. Your solution answers how the initial question was posed but does not allow for multiples of the same letter. Technically this is allowed by the constraints only if they do not violate the other constraints. Your code was very easy to edit to allow this however (just commented out the equality tests in the initial loop). As mentioned by @FreelanceReinhard in his solution, there appears to be many possible solutions and some will produce more valid strings than others. I will be interested to have a play around with the selection order of the strings to see how we can change the number of solutions.It is possible to see this by randomly sorting the first dataset produced in your code before running the second part. Anyways, I really appreciate you taking time out to help. Many thanks! 0 Likes Reply mkeintz PROC Star Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 11:13 PM (7479 views) \| In reply to DominicCavenagh @DominicCavenagh wrote: Thanks for this solution, it is very elegant. I have not come across hashes in SAS before so these are new to me and I will need to do a bit of work to fully understand your implementation. Your solution answers how the initial question was posed but does not allow for multiples of the same letter. Technically this is allowed by the constraints only if they do not violate the other constraints. Your code was very easy to edit to allow this however (just commented out the equality tests in the initial loop). As mentioned by @FreelanceReinhard in his solution, there appears to be many possible solutions and some will produce more valid strings than others. I will be interested to have a play around with the selection order of the strings to see how we can change the number of solutions.It is possible to see this by randomly sorting the first dataset produced in your code before running the second part. Anyways, I really appreciate you taking time out to help. Many thanks! Your comments on the order of the strings sent to the second data step are interesting. I reordered of data set NEED by changing ``` n=n+1; output; ``` to ``` n=n+1; call streaminit(1295086); rn=rand('uniform'); output; ``` and then sorted by RN. After permitting letters to appears multiple times, the number of VALID='Y' jumped from 25,750 (using original order) to 53,040 with the above random sort. Do you know of some formula that would provide the size of the largest possible set of strings satisfying your criteria? The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for Allow PROC SORT to output multiple datasets 0 Likes Reply DominicCavenagh Fluorite \| Level 6 Mark as New Bookmark Subscribe Mute RSS Feed Permalink Print Report Inappropriate Content Re: Generate all valid combinations of 4 length character strings Posted 08-29-2022 11:51 PM (7469 views) \| In reply to mkeintz Not really. After looking at @FreelanceReinhard s graph theory solution, it seems like this would take a little bit more thought, and someone much smarter than me to figure out! 0 Likes Reply Join the 2025 SAS Hackathon! Calling all data scientists and open-source enthusiasts! Want to solve real problems that impact your company or the world? Register to hack by August 31st! 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3386	https://pressbooks-dev.oer.hawaii.edu/collegephysics/chapter/5-3-elasticity-stress-and-strain/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 5 Further Applications of Newton’s Laws: Friction, Drag, and Elasticity 33 5.3 Elasticity: Stress and Strain Summary State Hooke’s law. Explain Hooke’s law using graphical representation between deformation and applied force. Discuss the three types of deformations such as changes in length, sideways shear and changes in volume. Describe with examples the young’s modulus, shear modulus and bulk modulus. Determine the change in length given mass, length and radius. We now move from consideration of forces that affect the motion of an object (such as friction and drag) to those that affect an object’s shape. If a bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the force—that is, for small deformations, Hooke’s law is obeyed. In equation form, Hooke’s law is given by whereis the amount of deformation (the change in length, for example) produced by the forceandis a proportionality constant that depends on the shape and composition of the object and the direction of the force. Note that this force is a function of the deformation—it is not constant as a kinetic friction force is. Rearranging this to makes it clear that the deformation is proportional to the applied force. Figure 1 shows the Hooke’s law relationship between the extensionof a spring or of a human bone. For metals or springs, the straight line region in which Hooke’s law pertains is much larger. Bones are brittle and the elastic region is small and the fracture abrupt. Eventually a large enough stress to the material will cause it to break or fracture. Tensile strength is the breaking stress that will cause permanent deformation or fracture of a material. HOOKE’S LAW whereis the amount of deformation (the change in length, for example) produced by the forceandis a proportionality constant that depends on the shape and composition of the object and the direction of the force. The proportionality constantdepends upon a number of factors for the material. For example, a guitar string made of nylon stretches when it is tightened, and the elongationis proportional to the force applied (at least for small deformations). Thicker nylon strings and ones made of steel stretch less for the same applied force, implying they have a larger(see Figure 2). Finally, all three strings return to their normal lengths when the force is removed, provided the deformation is small. Most materials will behave in this manner if the deformation is less than about 0.1% or about 1 part in Figure 2. The same force, in this case a weight (w), applied to three different guitar strings of identical length produces the three different deformations shown as shaded segments. The string on the left is thin nylon, the one in the middle is thicker nylon, and the one on the right is steel. STRETCH YOURSELF A LITTLE How would you go about measuring the proportionality constantof a rubber band? If a rubber band stretched 3 cm when a 100-g mass was attached to it, then how much would it stretch if two similar rubber bands were attached to the same mass—even if put together in parallel or alternatively if tied together in series? We now consider three specific types of deformations: changes in length (tension and compression), sideways shear (stress), and changes in volume. All deformations are assumed to be small unless otherwise stated. Changes in Length—Tension and Compression: Elastic Modulus A change in lengthis produced when a force is applied to a wire or rod parallel to its lengtheither stretching it (a tension) or compressing it. (See Figure 3.) Figure 3. (a) Tension. The rod is stretched a length ΔL when a force is applied parallel to its length. (b) Compression. The same rod is compressed by forces with the same magnitude in the opposite direction. For very small deformations and uniform materials, ΔL is approximately the same for the same magnitude of tension or compression. For larger deformations, the cross-sectional area changes as the rod is compressed or stretched. Experiments have shown that the change in length () depends on only a few variables. As already noted,is proportional to the forceand depends on the substance from which the object is made. Additionally, the change in length is proportional to the original lengthand inversely proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these factors into one equation for whereis the change in length,the applied force,is a factor, called the elastic modulus or Young’s modulus, that depends on the substance,is the cross-sectional area, andis the original length. Table 3 lists values offor several materials—those with a largeare said to have a large tensile stiffness because they deform less for a given tension or compression. \| Material \| Young’s modulus (tension–compression)Y \| Shear modulus S \| Bulk modulus B \| --- --- \| \| Aluminum \| 70 \| 25 \| 75 \| \| Bone – tension \| 16 \| 80 \| 8 \| \| Bone – compression \| 9 \| \| \| \| Brass \| 90 \| 35 \| 75 \| \| Brick \| 15 \| \| \| \| Concrete \| 20 \| \| \| \| Glass \| 70 \| 20 \| 30 \| \| Granite \| 45 \| 20 \| 45 \| \| Hair (human) \| 10 \| \| \| \| Hardwood \| 15 \| 10 \| \| \| Iron, cast \| 100 \| 40 \| 90 \| \| Lead \| 16 \| 5 \| 50 \| \| Marble \| 60 \| 20 \| 70 \| \| Nylon \| 5 \| \| \| \| Polystyrene \| 3 \| \| \| \| Silk \| 6 \| \| \| \| Spider thread \| 3 \| \| \| \| Steel \| 210 \| 80 \| 130 \| \| Tendon \| 1 \| \| \| \| Acetone \| \| \| 0.7 \| \| Ethanol \| \| \| 0.9 \| \| Glycerin \| \| \| 4.5 \| \| Mercury \| \| \| 25 \| \| Water \| \| \| 2.2 \| \| Table 3. Elastic Moduli1. \| \| \| \| Young’s moduli are not listed for liquids and gases in Table 3 because they cannot be stretched or compressed in only one direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitudeacting in opposite directions. For example, the strings in Figure 3 are being pulled down by a force of magnitudeand held up by the ceiling, which also exerts a force of magnitude Example 1: The Stretch of a Long Cable Suspension cables are used to carry gondolas at ski resorts. (See Figure 4) Consider a suspension cable that includes an unsupported span of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is Figure 4. Gondolas travel along suspension cables at the Gala Yuzawa ski resort in Japan. (credit: Rudy Herman, Flickr) Strategy The force is equal to the maximum tension, orThe cross-sectional area isThe equationcan be used to find the change in length. Solution All quantities are known. Thus, Discussion This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments. Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behavior of bones under tension and compression is important because it determines the load the bones can carry. Bones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus the bone in the top of the femur is arranged in thin sheets separated by marrow while in other places the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in bone joints and tendons. Another biological example of Hooke’s law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone) must stretch easily at first when a force is applied, but offer a much greater restoring force for a greater strain. Figure 5 shows a stress-strain relationship for a human tendon. Some tendons have a high collagen content so there is relatively little strain, or length change; others, like support tendons (as in the leg) can change length up to 10%. Note that this stress-strain curve is nonlinear, since the slope of the line changes in different regions. In the first part of the stretch called the toe region, the fibers in the tendon begin to align in the direction of the stress—this is called uncrimping. In the linear region, the fibrils will be stretched, and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this chapter. Ligaments (tissue connecting bone to bone) behave in a similar way. Figure 5. Typical stress-strain curve for mammalian tendon. Three regions are shown: (1) toe region (2) linear region, and (3) failure region. Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The elastic properties of the arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch when the blood is pumped out of the heart. When the aortic valve shuts, the pressure in the arteries drops and the arterial walls relax to maintain the blood flow. When you feel your pulse, you are feeling exactly this—the elastic behavior of the arteries as the blood gushes through with each pump of the heart. If the arteries were rigid, you would not feel a pulse. The heart is also an organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely and elastically when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg with no visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in elasticity starts in the early 20s. Example 2: Calculating Deformation: How Much Does Your Leg Shorten When You Stand on It? Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius. Strategy The force is equal to the weight supported, or and the cross-sectional area isThe equationcan be used to find the change in length. Solution All quantities exceptare known. Note that the compression value for Young’s modulus for bone must be used here. Thus, Discussion This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 3 have larger values of Young’s modulusIn other words, they are more rigid. The equation for change in length is traditionally rearranged and written in the following form: The ratio of force to area,is defined as stress (measured in), and the ratio of the change in length to length,is defined as strain (a unitless quantity). In other words, In this form, the equation is analogous to Hooke’s law, with stress analogous to force and strain analogous to deformation. If we again rearrange this equation to the form we see that it is the same as Hooke’s law with a proportionality constant This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in length, sideways bending, and changes in volume. STRESS The ratio of force to area,is defined as stress measured in N/m2. STRAIN The ratio of the change in length to length,is defined as strain (a unitless quantity). In other words, Sideways Stress: Shear Modulus Figure 6 illustrates what is meant by a sideways stress or a shearing force. Here the deformation is calledand it is perpendicular torather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with similar equations. The expression for shear deformation is whereis the shear modulus (see Table 3) andis the force applied perpendicular toand parallel to the cross-sectional areaAgain, to keep the object from accelerating, there are actually two equal and opposite forcesapplied across opposite faces, as illustrated in Figure 6. The equation is logical—for example, it is easier to bend a long thin pencil (small) than a short thick one, and both are more easily bent than similar steel rods (large). SHEAR DEFORMATION whereis the shear modulus andis the force applied perpendicular toand parallel to the cross-sectional area Figure 6. Shearing forces are applied perpendicular to the length L0 and parallel to the area A, producing a deformation Δx. Vertical forces are not shown, but it should be kept in mind that in addition to the two shearing forces, F, there must be supporting forces to keep the object from rotating. The distorting effects of these supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually negligible compared with forces large enough to cause significant deformations. Examination of the shear moduli in Table 3 reveals some telling patterns. For example, shear moduli are less than Young’s moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young’s modulus, but it is as large as that of steel. This is why bones are so rigid. The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some of both. Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge shaped disc below the last vertebrae) is particularly at risk because of its location. The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces. Example 3: Calculating Force Required to Deform: That Nail Does Not Bend Much Under a Load Find the mass of the picture hanging from a steel nail as shown in Figure 7, given that the nail bends only(Assume the shear modulus is known to two significant figures.) Figure 7. Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing effect of the supported weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied across opposite cross sections of the nail. See Example 3 for a calculation of the mass of the picture. Strategy The forceon the nail (neglecting the nail’s own weight) is the weight of the pictureIf we can findthen the mass of the picture is justThe equationcan be solved for Solution Solving the equationforwe see that all other quantities can be found: S is found in Table 3 and isThe radiusis 0.750 mm (as seen in the figure), so the cross-sectional area is The value foris also shown in the figure. Thus, This 51 N force is the weightof the picture, so the picture’s mass is Discussion This is a fairly massive picture, and it is impressive that the nail flexes only—an amount undetectable to the unaided eye. Changes in Volume: Bulk Modulus An object will be compressed in all directions if inward forces are applied evenly on all its surfaces as in Figure 8. It is relatively easy to compress gases and extremely difficult to compress liquids and solids. For example, air in a wine bottle is compressed when it is corked. But if you try corking a brim-full bottle, you cannot compress the wine—some must be removed if the cork is to be inserted. The reason for these different compressibilities is that atoms and molecules are separated by large empty spaces in gases but packed close together in liquids and solids. To compress a gas, you must force its atoms and molecules closer together. To compress liquids and solids, you must actually compress their atoms and molecules, and very strong electromagnetic forces in them oppose this compression. Figure 8. An inward force on all surfaces compresses this cube. Its change in volume is proportional to the force per unit area and its original volume, and is related to the compressibility of the substance. We can describe the compression or volume deformation of an object with an equation. First, we note that a force “applied evenly” is defined to have the same stress, or ratio of force to areaon all surfaces. The deformation produced is a change in volumewhich is found to behave very similarly to the shear, tension, and compression previously discussed. (This is not surprising, since a compression of the entire object is equivalent to compressing each of its three dimensions.) The relationship of the change in volume to other physical quantities is given by whereis the bulk modulus (see Table 3),is the original volume, andis the force per unit area applied uniformly inward on all surfaces. Note that no bulk moduli are given for gases. What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example illustrates. Example 4: Calculating Change in Volume with Deformation: How Much Is Water Compressed at Great Ocean Depths? Calculate the fractional decrease in volume () for seawater at 5.00 km depth, where the force per unit area is Strategy Equationis the correct physical relationship. All quantities in the equation exceptare known. Solution Solving for the unknowngives Substituting known values with the value for the bulk modulusfrom Table 3, Discussion Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress. Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way. Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here. Section Summary Hooke’s law is given by whereis the amount of deformation (the change in length),is the applied force, andis a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as whereis Young’s modulus, which depends on the substance,is the cross-sectional area, andis the original length. The ratio of force to area,is defined as stress, measured in N/m2. The ratio of the change in length to length,is defined as strain (a unitless quantity). In other words, The expression for shear deformation is whereis the shear modulus andis the force applied perpendicular toand parallel to the cross-sectional area The relationship of the change in volume to other physical quantities is given by whereis the bulk modulus,is the original volume, and is the force per unit area applied uniformly inward on all surfaces. Conceptual Questions 1: The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood (pulsating or continuous). 2: What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference? 3: Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? What differences will dry and wet conditions make for these surfaces? 4: Would you expect your height to be different depending upon the time of day? Why or why not? 5: Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 6: Explain why pregnant women often suffer from back strain late in their pregnancy. 7: An old carpenter’s trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail firmly with pliers. Why does this help? 8: When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more with temperature than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a pocket of air above the vinegar would prevent the break. (This is the function of the air above liquids in glass containers.) Problems & Exercises 1: During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. Her mass is 60.0 kg. 2: During a wrestling match, a 150 kg wrestler briefly stands on one hand during a maneuver designed to perplex his already moribund adversary. By how much does the upper arm bone shorten in length? The bone can be represented by a uniform rod 38.0 cm in length and 2.10 cm in radius. 3: (a) The “lead” in pencils is a graphite composition with a Young’s modulus of aboutCalculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils? 4: TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius? 5: (a) By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (b) Does the answer seem to be consistent with what you have observed for nylon ropes? Would it make sense if the rope were actually a bungee cord? 6: A 20.0-m tall hollow aluminum flagpole is equivalent in stiffness to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 900 N exerted at the top would. How far to the side does the top of the pole flex? 7: As an oil well is drilled, each new section of drill pipe supports its own weight and that of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in stiffness to a solid cylinder 5.00 cm in diameter. 8: Calculate the force a piano tuner applies to stretch a steel piano wire 8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long. 9: A vertebra is subjected to a shearing force of 500 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 cm in diameter. 10: A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus ofThe disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter. 11: When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle ofto the horizontal. (a) By how much does the wood flex perpendicular to its length? (b) How much is it compressed lengthwise? 12: To consider the effect of wires hung on poles, we take data from Chapter 4.7 Example 2, in which tensions in wires supporting a traffic light were calculated. The left wire made an anglebelow the horizontal with the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow aluminum pole is equivalent in stiffness to a 4.50 cm diameter solid cylinder. (a) How far is it bent to the side? (b) By how much is it compressed? 13: A farmer making grape juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.2% (that is,) relative to the space available. Calculate the magnitude of the normal force exerted by the juice per square centimeter if its bulk modulus isassuming the bottle does not break. In view of your answer, do you think the bottle will survive? 14: (a) When water freezes, its volume increases by 9.05% (that is,). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.) (b) Is it surprising that such forces can fracture engine blocks, boulders, and the like? 15: This problem returns to the tightrope walker studied in Chapter 4.5 Example 2, who created a tension ofin a wire making an anglebelow the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter. 16: The pole in Figure 9 is at abend in a power line and is therefore subjected to more shear force than poles in straight parts of the line. The tension in each line isat the angles shown. The pole is 15.0 m tall, has an 18.0 cm diameter, and can be considered to have half the stiffness of hardwood. (a) Calculate the compression of the pole. (b) Find how much it bends and in what direction. (c) Find the tension in a guy wire used to keep the pole straight if it is attached to the top of the pole at an angle ofwith the vertical. (Clearly, the guy wire must be in the opposite direction of the bend.) Figure 9. This telephone pole is at a 900 bend in a power line. A guy wire is attached to the top of the pole at an angle of 300 with the vertical. Footnotes 1 Approximate and average values. Young’s modulifor tension and compression sometimes differ but are averaged here. Bone has significantly different Young’s moduli for tension and compression. Glossary deformation : change in shape due to the application of force Hooke’s law : proportional relationship between the forceon a material and the deformationit causes, tensile strength : the breaking stress that will cause permanent deformation or fraction of a material stress : ratio of force to area strain : ratio of change in length to original length shear deformation : deformation perpendicular to the original length of an object Solutions Problems & Exercises (a)1 mm (b) This does seem reasonable, since the lead does seem to shrink a little when you push on it. (a)9 cm (b)This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much. 8.59 mm (a) (b) This is about 36 atm, greater than a typical jar can withstand. 1.4 cm
3387	https://en.wikipedia.org/wiki/Barycenter_(astronomy)	Jump to content Barycenter (astronomy) العربية Asturianu বাংলা Чӑвашла Čeština Dansk Deutsch Español فارسی Galego 한국어 Bahasa Indonesia Latviešu Lëtzebuergesch Bahasa Melayu Монгол Norsk nynorsk Русский Simple English Slovenčina Taqbaylit Татарча / tatarça ไทย Türkçe Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Sorry to interrupt, but our fundraiser won't last long. This Wednesday, we ask you to join the 2% of readers who give. If everyone reading this right now gave just $2.75, we'd hit our goal quickly. $2.75 is all we ask. September 3: Knowledge is human. We're sorry we've asked you a few times recently, but it's Wednesday, September 3—please don't wait until tomorrow to help. We're happy you consult Wikipedia often. If just 2% of our most loyal readers gave $2.75 today, we'd reach our goal quickly. Most readers donate because Wikipedia is useful, others because they realize knowledge needs humans. If you agree, please give. Any contribution helps, whether it's $2.75 one time or monthly. 25 years ago Wikipedia was a dream. A dream built piece by piece by people, not machines. Now, with 65 million articles and 260,000 volunteers across the world, Wikipedia is proof that knowledge is human—a place of free, collaborative, and accessible knowledge. Your donation isn't just supporting a website; it's investing in the world's largest collaborative project of human intelligence—crafted by humans, for humans. Please join the 2% of readers who give what they can to help keep Wikipedia strong and growing. Thank you. Center of mass of multiple bodies orbiting each other "Barycenter" redirects here. For the general concept, see Barycenter (physics). Animation of barycenters Two bodies with similar mass, like the 90 Antiope asteroid system Two bodies with slightly different masses, like Pluto and Charon Two bodies with significant difference in masses, like Earth and the Moon Two bodies with an extreme difference in mass, like the Sun and Earth Two bodies with the same mass with eccentric elliptic orbits, common for binary stars \| \| \| Part of a series on \| \| Astrodynamics \| \| \| \| Orbital mechanics \| \| Orbital elements Apsis Argument of periapsis Eccentricity Inclination Mean anomaly Orbital nodes Semi-major axis True anomaly \| \| Types of two-body orbits by eccentricity Circular orbit Elliptic orbit Transfer orbit (Hohmann transfer orbit Bi-elliptic transfer orbit) Parabolic orbit Hyperbolic orbit Radial orbit Decaying orbit \| \| Equations Dynamical friction Escape velocity Kepler's equation Kepler's laws of planetary motion Orbital period Orbital velocity Surface gravity Specific orbital energy Vis-viva equation \| \| Celestial mechanics \| \| Gravitational influences Barycenter Hill sphere Perturbations Sphere of influence \| \| N-body orbits Lagrangian points (Halo orbits) Lissajous orbits Lyapunov orbits \| \| Engineering and efficiency \| \| Preflight engineering Mass ratio Payload fraction Propellant mass fraction Tsiolkovsky rocket equation \| \| Efficiency measures Gravity assist Oberth effect \| \| Propulsive maneuvers Orbital maneuver Orbit insertion \| \| v t e \| In astronomy, the barycenter (or barycentre; from Ancient Greek βαρύς (barús) 'heavy' and κέντρον (kéntron) 'center') is the center of mass of two or more bodies that orbit one another and is the point about which the bodies orbit. A barycenter is a dynamical point, not a physical object. It is an important concept in fields such as astronomy and astrophysics. The distance from a body's center of mass to the barycenter can be calculated as a two-body problem. If one of the two orbiting bodies is much more massive than the other and the bodies are relatively close to one another, the barycenter will typically be located within the more massive object. In this case, rather than the two bodies appearing to orbit a point between them, the less massive body will appear to orbit about the more massive body, while the more massive body might be observed to wobble slightly. This is the case for the Earth–Moon system, whose barycenter is located on average 4,671 km (2,902 mi) from Earth's center, which is 74% of Earth's radius of 6,378 km (3,963 mi). When the two bodies are of similar masses, the barycenter will generally be located between them and both bodies will orbit around it. This is the case for Pluto and Charon, one of Pluto's natural satellites, as well as for many binary asteroids and binary stars. When the less massive object is far away, the barycenter can be located outside the more massive object. This is the case for Jupiter and the Sun; despite the Sun being a thousandfold more massive than Jupiter, their barycenter is slightly outside the Sun due to the relatively large distance between them. In astronomy, barycentric coordinates are non-rotating coordinates with the origin at the barycenter of two or more bodies. The International Celestial Reference System (ICRS) is a barycentric coordinate system centered on the Solar System's barycenter. Two-body problem [edit] Main article: Two-body problem The barycenter is one of the foci of the elliptical orbit of each body. This is an important concept in the fields of astronomy and astrophysics. In a simple two-body case, the distance from the center of the primary to the barycenter, r1, is given by: where : r1 is the distance from body 1's center to the barycenter a is the distance between the centers of the two bodies m1 and m2 are the masses of the two bodies. The semi-major axis of the secondary's orbit, r2, is given by r2 = a − r1. When the barycenter is located within the more massive body, that body will appear to "wobble" rather than to follow a discernible orbit. Primary–secondary examples [edit] The following table sets out some examples from the Solar System. Figures are given rounded to three significant figures. The terms "primary" and "secondary" are used to distinguish between involved participants, with the larger being the primary and the smaller being the secondary. m1 is the mass of the primary in Earth masses (M🜨) m2 is the mass of the secondary in Earth masses (M🜨) a (km) is the average orbital distance between the centers of the two bodies r1 (km) is the distance from the center of the primary to the barycenter R1 (km) is the radius of the primary ⁠r1/R1⁠ a value less than one means the barycenter lies inside the primary Primary–secondary examples \| Primary \| m1 (M🜨) \| Secondary \| m2 (M🜨) \| a (km) \| r1 (km) \| R1 (km) \| ⁠r1/R1⁠ \| \| Earth \| 1 \| Moon \| 0.0123 \| 384,400 \| 4,671 \| 6,371 \| 0.733[a] \| \| Pluto \| 0.0021 \| Charon \| 0.000254 (0.121 M♇) \| 19,600 \| 2,110 \| 1,188.3 \| 1.78[b] \| \| Sun \| 333,000 \| Earth \| 1 \| 150,000,000 (1 AU) \| 449 \| 695,700 \| 0.000645[c] \| \| Sun \| 333,000 \| Jupiter \| 318 (0.000955 M☉) \| 778,000,000 (5.20 AU) \| 742,370 \| 695,700 \| 1.07[d] \| \| Sun \| 333,000 \| Saturn \| 95.2 \| 1,433,530,000 (9.58 AU) \| 409,700 \| 695,700 \| 0.59 \| ^ The Earth has a perceptible "wobble". Also see tides. ^ Pluto and Charon are sometimes considered a binary system because their barycenter does not lie within either body. ^ The Sun's wobble is barely perceptible. ^ The Sun orbits a barycenter just above its surface. Example with the Sun [edit] If m1 ≫ m2—which is true for the Sun and any planet—then the ratio ⁠r1/R1⁠ approximates to: Hence, the barycenter of the Sun–planet system will lie outside the Sun only if: —that is, where the planet is massive and far from the Sun. If Jupiter had Mercury's orbit (57,900,000 km, 0.387 AU), the Sun–Jupiter barycenter would be approximately 55,000 km from the center of the Sun (⁠r1/R1⁠ ≈ 0.08). But even if the Earth had Eris's orbit (1.02×1010 km, 68 AU), the Sun–Earth barycenter would still be within the Sun (just over 30,000 km from the center). To calculate the actual motion of the Sun, only the motions of the four giant planets (Jupiter, Saturn, Uranus, Neptune) need to be considered. The contributions of all other planets, dwarf planets, etc. are negligible. If the four giant planets were on a straight line on the same side of the Sun, the combined center of mass would lie at about 1.17 solar radii, or just over 810,000 km, above the Sun's surface. The calculations above are based on the mean distance between the bodies and yield the mean value r1. But all celestial orbits are elliptical, and the distance between the bodies varies between the apses, depending on the eccentricity, e. Hence, the position of the barycenter varies too, and it is possible in some systems for the barycenter to be sometimes inside and sometimes outside the more massive body. This occurs where: The Sun–Jupiter system, with eJupiter = 0.0484, just fails to qualify: 1.05 < 1.07 > 0.954. Relativistic corrections [edit] In classical mechanics (Newtonian gravitation), this definition simplifies calculations and introduces no known problems. In general relativity (Einsteinian gravitation), complications arise because, while it is possible, within reasonable approximations, to define the barycenter, we find that the associated coordinate system does not fully reflect the inequality of clock rates at different locations. Brumberg explains how to set up barycentric coordinates in general relativity. The coordinate systems involve a world-time, i.e. a global time coordinate that could be set up by telemetry. Individual clocks of similar construction will not agree with this standard, because they are subject to differing gravitational potentials or move at various velocities, so the world-time must be synchronized with some ideal clock that is assumed to be very far from the whole self-gravitating system. This time standard is called Barycentric Coordinate Time (TCB). Selected barycentric orbital elements [edit] Barycentric osculating orbital elements for some objects in the Solar System are as follows: \| Object \| Semi-major axis (in AU) \| Apoapsis (in AU) \| Orbital period (in years) \| --- --- \| \| C/2006 P1 (McNaught) \| 2,050 \| 4,100 \| 92,600 \| \| C/1996 B2 (Hyakutake) \| 1,700 \| 3,410 \| 70,000 \| \| C/2006 M4 (SWAN) \| 1,300 \| 2,600 \| 47,000 \| \| (308933) 2006 SQ372 \| 799 \| 1,570 \| 22,600 \| \| (87269) 2000 OO67 \| 549 \| 1,078 \| 12,800 \| \| 90377 Sedna \| 506 \| 937 \| 11,400 \| \| 2007 TG422 \| 501 \| 967 \| 11,200 \| For objects at such high eccentricity, barycentric coordinates are more stable than heliocentric coordinates for a given epoch because the barycentric osculating orbit is not as greatly affected by where Jupiter is on its 11.8 year orbit. See also [edit] Barycentric Dynamical Time Centers of gravity in non-uniform fields Center of mass Lagrange point Mass point geometry Roll center Weight distribution References [edit] ^ "barycentre". Oxford English Dictionary (2nd ed.). Oxford University Press. 1989. ^ MacDougal, Douglas W. (December 2012). Newton's Gravity: An Introductory Guide to the Mechanics of the Universe. Berlin: Springer Science & Business Media. p. 199. ISBN 978-1-4614-5444-1. ^ Moore, P. (2005). "SOLAR SYSTEM \| Moon". Encyclopedia of Geology. pp. 264–272. doi:10.1016/B0-12-369396-9/00077-0. ISBN 978-0-12-369396-9. barycentre lies 1700 km below the Earth's surface (6370km–1700km) ^ Olkin, C. B.; Young, L. A.; Borncamp, D.; et al. (January 2015). "Evidence that Pluto's atmosphere does not collapse from occultations including the 2013 May 04 event". Icarus. 246: 220–225. Bibcode:2015Icar..246..220O. doi:10.1016/j.icarus.2014.03.026. hdl:10261/167246. ^ "If You Think Jupiter Orbits the Sun, You're Mistaken". HowStuffWorks. 9 August 2016. The Sol-Jupiter barycenter sits 1.07 times the radius of the sun ^ "What's a Barycenter?". Space Place @ NASA. 8 September 2005. Archived from the original on 23 December 2010. Retrieved 20 January 2011. ^ Meeus, Jean (1997), Mathematical Astronomy Morsels, Richmond, Virginia: Willmann-Bell, pp. 165–168, ISBN 0-943396-51-4 ^ Brumberg, Victor A. (1991). Essential Relativistic Celestial Mechanics. London: Adam Hilger. ISBN 0-7503-0062-0. ^ Horizons output (30 January 2011). "Barycentric Osculating Orbital Elements for 2007 TG422". Archived from the original on 28 March 2014. Retrieved 31 January 2011. (Select Ephemeris Type:Elements and Center:@0) ^ Kaib, Nathan A.; Becker, Andrew C.; Jones, R. Lynne; Puckett, Andrew W.; Bizyaev, Dmitry; Dilday, Benjamin; Frieman, Joshua A.; Oravetz, Daniel J.; Pan, Kaike; Quinn, Thomas; Schneider, Donald P.; Watters, Shannon (2009). "2006 SQ372: A Likely Long-Period Comet from the Inner Oort Cloud". The Astrophysical Journal. 695 (1): 268–275. arXiv:0901.1690. Bibcode:2009ApJ...695..268K. doi:10.1088/0004-637X/695/1/268. S2CID 16987581. Retrieved from " Categories: Astronomical coordinate systems Celestial mechanics Mass Geometric centers Hidden categories: Articles with short description Short description is different from Wikidata Use American English from December 2020 All Wikipedia articles written in American English Use dmy dates from December 2020 Pages using multiple image with auto scaled images
3388	https://www.youtube.com/watch?v=6E7RYcJXVPc	Comparing Fractions With Cross Multiplication Mr. W Reading and Math 3210 subscribers 4 likes Description 260 views Posted: 19 Feb 2021 This lesson shows how to compare two fractions using cross multiplication. Transcript: our topic for today is fractions part 10 comparing fractions with cross multiplication comparing means that we're going to put two fractions side by side here on the left side we have two thirds and on the right side we have seven twelfths a lot of times when you compare fractions if you have models you can look at them and get a sense of which one is larger and which one is smaller but there are times when you're not sure so cross multiplication is when you would take the two-thirds you would multiply the three times the seven and then write it up there and then the 12 times the two and then when you get yourself your product you would compare them and it has that cross and that's what we mean here so let's try it three times 7 is 21. 12 times 2 is 24 and when we compare 24 and 21 i see that 24 is greater and i'm going to put my greater than so this would read two thirds is greater than 7 12. comparing fractions mean that we place the fractions side-by-side in this case 6 8 is on the left side and seven tenths and we see the model for six eighths and the model for seven tenths and they're pretty close when you have models you can tell i can tell that this one's a little bit bigger but if we did not have these models it's a little bit hard to tell and that's where the cross multiplication comes in we have three outcomes equal greater than and less than remember when we read this we read this from the left side across to the right okay let's try our cross multiplication i'm going to go 8 times 7 that would be 56 and 10 times six would be sixty now i'm going to compare sixty and fifty six i can see that sixty is greater so this would read six eights is greater than seven tenths compare the fractions write greater than less than or equal to four twelfths and five eighths twelve times five is sixty and eight times four 32 when i compare so we would read this as 4 12 is less than 5 8 4 8 and one half 8 times one is eight two times four is eight those are equal four eighths is equal to one-half three-fourths compared to five-eighths four times five is twenty eight times three is 24 and when i compare 24 is greater so this would read 3 4 is greater than 5 8. three-fifths compared to four-tenths five times four is twenty and ten times three is thirty when i compare 30 and 20 that is greater so three-fifths is greater than four-tenths in review our lesson today is comparing fractions using cross multiplication and cross multiplication is just like it sounds we just would multiply the bottom of the first fraction times the top then do the same thing going the other way compare the products and then put in either greater than less than or equal to okay thanks everybody
3389	https://www.entelechyjournal.com/defense_mechanisms_in_ep.htm	Defense Mechanisms in EP Defense Mechanisms in EP?: Nope by James Brody A lady of uncertain decisions but sparkling prose remarked that she loved The Adapted Mind's (Barkow et al., 1992) chapter on "defense mechanisms" (Nesse and Lloyd, 1992). I remembered my antipathy to that same chapter but neither its content nor my reasoning and promised to give it another look. "Defense mechanisms" grew from Freud’s scribblings, first Sigmund, then Anna (Kaplan & Sadock, 1998): that is, puns and slips of the tongue, inconsistency between what was said and what was done, differences between infant and adult behavior, and forgetting of very large, nasty experiences reflected the interference of filters and lenses, erasers and amplifiers, called projection, incorporation, denial, displacement, suppression, repression, reaction formation, sublimation, or rationalization. (I'm missing several.) That is, memory is equal and accurate for all experiences; variable recall demands services from a team of special agents. Defense mechanisms intervene between our instinctive but shameful impulses and our conscience. Psychoanalysis collapses primitive mechanisms such as denial or projection and replaces them with new ones such as repression, symbolization, or sublimation. (Of course, defense mechanisms reproduced with each generation of practitioners. Vaillant, (1977, cited in Kaplan & Sadock, 1998)¹ for example, listed 27 of them in four clusters.) Neither time nor money was an object so long as you had plenty of both. Symptoms (e.g., a phobia) represented psychic conflict between nonverbal instincts and verbal rules about good or bad: curing the phobia but not the conflict was expected to produce "symptom substitution," that is, more phobias. Also, emotions drove thoughts and should be drained like dirty engine oil. Problem: not one instance of symptom substitution could ever be found in controlled research. Another problem: fear of snakes could be managed simply by thinking of holding a snake until you relax and then actually holding one. Dreams and memories of early childhood became irrelevan t (Franks, 1969) and we now believe that thoughts drive emotions as often as the reverse is true (Beck & Emery, 1985), and that "draining" emotions may actually make emotional outbursts more likely instead of less. Psychoanalysis collapsed like one more primitive defense mechanism while first behaviorism and then cognitive-behavioral therapy demystified treatment for most of the fears and rituals in human experience. Psychiatry turned to medication management, defense mechanisms were adopted by clinical social workers, and behavior therapy is now done by nearly anyone, including computer programs. Unfortunately, theoretical inertia characterizes some practitioners as much as it characterizes Rhesus potato cleaners. That is, a young Japanese macaque (Imo) discovered how to wash sand off of potatoes and quickly taught her siblings and peers: the old males, however, never did learn the new skill but still picked the sand from their potatoes and dry cereal, one grain at a time. Likewise, a small group of psychiatrists retains their concepts (e. g., defense mechanisms or personality disorders) and justifies them with evolutionary stories. (See Beck, 1998; Beck & Freeman, 1990; Millon, 1990). The old ways might have died with their adherents where they were had not Bob Trivers reasoned through his hairline and invented "self deception." That is, we lie and we also detect lying in each other. An evolutionary contest should produce not only better liars but also liars who are better at concealing their lies, so good, in fact as to be unaware, themselves, that they were lying. Bob gave traditionalists a bridge from their original island to a new one called "evolutionary psychology," a refuge that might nurture their old beliefs. (Trivers also inspired some interesting studies, whatever you think of their explanations. See Krebs & Denton, 1997) Egad. First, there are no measurements of "defense mechanisms." They are like Jesus, you either believe in them or you don't. And like the man's face on Mars, you will either see them or you won't. And if you believe in them, no one will talk you out of your position. (I think that a little imprinting is involved here or perhaps a genetic loading of some kind.) I suggest, for those who will believe, that modern neurology and network theory (1) describe patterns found widely from electrical networks to cellular biochemistry to neural organizations to the structure of our language (Barabasi, 2002; Strogatz, 2003) and (2) handily account for the outcomes that we traditionally ascribe to defense mechanisms but do so in a domain-specific way, open to measurement and without special agents. Second, I'm puzzled that the Adapted Mind rails against "domain general," problem-solving algorithms but smuggles a chapter into its closing sections on the evolution of exactly those processes. Thus, we are to have modules for the specific tasks of mating and fighting but we also have erasers and amplifiers and editors that roam across our modules. (I suspect that friendships between Randy, Leda, Jerry, and John had influence.) I'm reminded of a talk by Krebs (1998, 1999) on morality: he found it to consist of three or four stages, to be reversible but not domain general: morals shift with contexts and Kohlberg, who wanted to do for morals what Piaget tried to do for thinking, was wrong even before he committed suicide. So was Piaget about thinking.) I'm puzzled that my Jewish Venus can be enamored of The Blank Slate (Pinker, 2002) but adore defense mechanisms that are a dozen gremlins and each one a ghost in our machine. I reject a model that says that we have a thought or impulse but hide it from our awareness. I also dismiss its adherents. According to "whatever gods may be" as well as to Max Planck and Lynn Margulis (1995), they do eventually die. g Jim is an evolutionist and a clinical psychologist, practicing both faiths in a small but very traditional Pennsylvania community. He earned his PhD from the University of Pittsburgh in physiological-experimental psychology and has two decades' experience working in large institutions and a like amount of time in independent practice. Jim has published more than a dozen academic journal articles, as well as over 700 essays forBehavior OnLine's Evolutionary Psychology/Clinical Sociobiology Forum, which he founded. He organized four courses on human evolution and clinical practice for the Cape Cod Institute, recruiting Robert Wright and other notables to help. He also taught evolution in church basements and has given talks to many professional organizations. Jim is self-taught in evolutionary-developmental biology, behavior genetics, and network physics and applies all three to his understanding of both psychology and human evolution. Jim is also recognized by some as the "founder of clinical sociobiology." ¹ Kaplan & Sadock, 1998, devote only 4 pages of 1401 to defense mechanisms, plus another 6 pages scattered through their Synopsis of Psychiatry. References Barabasi, A-L (2002) Linked: The New Science of Networks. NY: Perseus. Barkow, J., Cosmides, L., & Tooby, J. (Eds.)(1992)The Adapted Mind: Evolutionary Psychology and the Generation of Culture. NY: Oxford. Beck, A. 1998. Cognitive aspects of personality disorders and their relation to syndromal disorders: A psychoevolutionary approach. In R. Cloninger (Ed.), Personality and Psychopathology. American Psychiatric Association, Washington, D. C., pp. 411-429. Beck, A. & Emery, G. (1985) Anxiety Disorders & Phobias: A Cognitive Perspective. NY: Basic. Beck, A., & Freeman, A. (1990) Cognitive Therapy of Personality Disorders. NY: Guilford. Franks, C. (1969) Behavior Therapy: Appraisal and Status. NY: McGraw-Hill. Kaplan, H. & Sadock, B. (1998) Synopsis of Psychiatry (8th Ed.). Baltimore: Williams & Wilkins. Krebs, D. (1998) Evolution of moral behaviors. In Crawford, C., & Krebs, D. Handbook of Evolutionary Psychology. Mahwah, NJ: Erlbaum, pp. 337-368. Krebs, D. (1999) Evolution of moral dispositions in the human species. Presentation at Hunter School of Social Work, Manhattan, NY, May 5. Krebs, D., & Denton, K. (1997) Social illusions and self-deception: The evolution of biases in person perception. In Simpson, J. & Kendrick, D. (Eds.) Evolutionary Social Psychology. Mahwah, NJ: Erlbaum, 21-48. Margulis, L. (1995) Gaia is a tough bitch. In Brockman, J. (Ed). The Third Culture: Beyond the Scientific Revolution. New York: Simon & Schuster, 130-140. Millon, T. (1990) Toward a New Personology: An Evolutionary Model. NY: Wiley. Nesse, R. & Lloyd, A. (1992) Evolution of psychodynamic mechanisms. In J. Barkow, L. Cosmides, & J. Tooby (Eds.) The Adapted Mind: Evolutionary Psychology and the Generation of Culture. NY: Oxford, pp. 601-626. Pinker, S. (2002) The Blank Slate: The Modern Denial of Human Nature. NY: Viking. Strogatz, S. (2003) Sync: The Emerging Science of Spontaneous Order. NY: Hyperion. Copyright by James Brody, 2003. All rights reserved. Copyright © 2003 Entelechy: Mind & Culture. All rights reserved.
3390	https://www.youtube.com/watch?v=o8xULufjEf4	Desmos Graphing Calculator Tips for Algebra 1 (Part 1) Steve Boring 224 subscribers 252 likes Description 27733 views Posted: 7 May 2019 Video on how to use the Desmos Graphing calculator on Algebra 1 type problems. This video specifically covers problems students may see on the Virginia State Standards of Learning (SOL) Test for Algebra 1. Part 1 covers: -graphing equations & inequalities -substitution -systems of equations & inequalities -zeros / roots / solutions / x-intercepts Part 2 covers: -Linear & Quadratic Regression Transcript: Intro in this video I'm going to show you some desmos graphing calculator tips specifically I'm gonna look at algebra 1 type tips for that aligned with the Virginia State Standards of Learning so the stuff I'm going to talk about here is I'm going to talk about graphing equations than equality's substitution systems of equations inequalities zeroes roots solutions X intercepts which is where y equals 0 f of X equals 0 and in part two of my video because I can only upload 15-minute videos I'm going to do regression as a short video separately so that's why it's separated here so let's go ahead and look at some old sol-type problems so graphing equations Graphing Equations is actually really easy in the desmos calculator it's actually easier than the graphing calculator let's see I graphing calculators because the TI graphing calculators only graph functions so you would have to put this in function form which means solve it for y and you could do that and if you're good at that it's no big deal but the nice thing about the desmos calculator is the new graph an equation in basically any form so we can just graph it in this form which is standard form so let's go to the desmos graphing calculator first let me show you how to get there you're going to go Getting to Desmos to if you're one of my students are going to go to LCPs go and login I think mine's a little different because mine's obviously the teacher view you're gonna go to the math folder here see a few things in here I don't think you're gonna see exactly the same stuff but you're obviously not gonna have this the scientific calculators for math 6 math 7 math 8 so we're not gonna worry about that you could use that this video is not really going to cover that the thing you're going to be using is the desmos graphing calculator so that's what you want to use if you can't get to it that way if you're somebody else watching this video you could always Google desmos test testing graphing calculator and then whatever state you're in the reason I'm doing testing graphing calculator is because these are the functions that are available on the desmos calculator when you're taking some kind of standardized test or what you're going to be available to you on the test it's different for different states in a North Carolina for instance it has its own but I'm just gonna work on Virginia I'm gonna go in through the app which is right here it doesn't Graphing matter which one you use so it looks like that here's our problem so I want to graph that to graph that you're going to go ahead and just type this in so that's basically how you do so it's 4x plus 5y equals negative 20 so 4x plus 5y equals negative 20 in the graphing calculators there's a negative sign and a separate subtraction sign for desmos it's the same thing which is nice if you'll notice that it graphs it which is convenient this is how you turn the graph on and off so if you just want this equation sitting here and you don't really want to see it right now you can just click this toggle switch which is cool let's see it gives you a couple points if I click on this it'll keep the label on it so the labels are negative 5 0 and 0 negative 4 so all I have to do really is find which one has the same points which would be answer choice a super easy inequalities aren't that much Inequalities harder the only difference is we have a funny symbol here so we have less than or equal to here so it's a slightly different there's only one extra step so 2 7 X minus 2 so let's try that here type Y and the problem here is and my screencast isn't doing the whole screen so I'm gonna make this smaller problem here is uh where is the key for less than or equal to well if you go to this thing which is the keypad you'll see this right here so here's all your symbols and we're gonna use this one and then it was 2/7 X minus 2 to type in a fraction you just type 2 divided by that divided by symbol is where the question mark key is 7 and then it's gonna keep us in the denominator that's why the it's blinking right here in the denominator you hit the right key on your keyboard to get out of there and I want to put in X minus 2 if you put this X down here in the denominator it's actually gonna do 2/7 X which isn't the same as 2/7 X okay anyway so a graph that you'll see it has a couple nice points it looks like there's a nice point there at 7 0 a nice point here at 0 negative - and it's shaded down so we're just gonna look for the graph that has these two points zero negative two seven zero and shaded down well we know it's shaded down so instantly we can get rid of these two so now we have a 50-50 shot negative two zeroes here and seven zeroes here so that one definitely not right I think it looks like they tried to do it backwards negative 7 and positive 2 so the answer here is gonna be C pretty simple stuff substitution Substitution can be a little bit tricky so make sure you're you're following along with me here this is probably the most tricky one on here besides your regression so please just stay with me if you lose me feel free to pause the video or rewind it and watch it again but we're gonna try this equation and put it in the Des Moines ago to desmos slide this over I'm just going to type this in so I'm gonna click X on this to get rid of that negative two well click x and then we need a cube root that's gonna be in this functions button right here this is um trig stuff so if you're an algebra 2 tree or geometry you cover this a little bit this would be good for you stats you don't really need to worry about we're not gonna do any of this stuff in algebra class anyway and I don't think using geometry either this is combinations permutations factorials standard deviation standard deviation for population I assume that's variants quartiles median and Max mean might be helpful but I wouldn't worry about any of this we went in here this is Lisa called multiple greatest common denominator modulus division not sure what all these do I'm not sure at the round it does but the one we want is we're gonna want this one which is the nth root we want to change that to cube root so right here we want to put a 3 to make it a cube root so whatever root it is put 3 when I click right to go under the radical and put a in the radicand click write to get out of that radical and out of the right of can we return the 2 plus B squared and you can either hit this to square it or you can hit shift 6 and that will put you in the exponent click - now I'm stuck up there in the exponent technically you should probably get yourself out by clicking the right era now you'll notice some stuff popped up here it says add slider you can do this or you can just type a equals B equals let's just add the sliders so the cool thing about sliders is I can just do this and it will just tell me what the value of this algebraic expression is whenever a is whatever number I'm clicking here which is kind of cool I can do the same thing with B I can change both moment doesn't matter notice it only goes up to 10 but you can change it if you click on that number I'm not gonna mess with that too much so what I would do is I would just type in 64 and I would type in negative 5 and if you go up here gives you the output for that expression super easy so the answer here is going to be 17 done Tricky Parts all right so here's tricky parts because you're probably wondering why was that tricky well the problem with desmos is make this a little bigger let me get rid of that okay for all with desmos is that if you let's say instead of a and B the formula is set x and y so i'm going to do the same formula but instead of a it's gonna be X instead of B it's gonna be why it gives you an error here when you do that so try adding an equal sign to turn this into an equation it wants an equation because it thinks you want to graph it using x and y coordinates cuz this is x axis y axis so that's a problem so watch what happens if i do x equals 64 and i do y equals negative 5 it's not out putting that thing up here so that's gonna give me an issue now there's two options you can do you can either know to change this to some other variable like a and B for instance and that's what I would use or you can make this X sub 1 so just put if you go by beside this and click and then put a 1 beside it it's X sub 1 which means like a specific x-coordinate and change this to Y sub 1 by putting a 1 beside it now it kind of looks weird y sub 1 this means Y sub 1 squares that looks a little kooky because the two directly above the one that's a little weird but anyway you can do the same thing down here and just change your sliders to x1 and y1 and then it instantly works so that's fine so just know that x and y are going to give you errors and that's kind of the area you're gonna get another type of Function Form substitution that could give you some issue or at least another thing I'll show you is function form so we're gonna put this in I'm gonna do F of negative 8 so let's go ahead and put that in here let me make this smaller again I'm gonna get rid of these guys clicking X I'm just gonna retype so f of X and I just use parentheses for this so that's the shift 9 and shift 0 equals 11 parentheses X minus 24 divided by 2 now that didn't work out too well because I just typed it straight like that that doesn't look the same as this so I'm gonna back up I'm gonna close my parentheses and now if I click / - you see how it puts it and it makes it exactly like this so you got to be careful that they match up correctly if you run into trouble just put parentheses around what you want the numerator to be and then parentheses around what you want the denominator to be until it works nice thing about this is the easy way to type this in because I remember I told you X usually doesn't work for this but we do have an equation so it should work we're gonna type negative 8 so this says f of negative 8 which means instead of f of X they want f of negative 8 which means the function when you input negative 8 for X right here like that so to do that in desmos all you really have to do is type f of negative 8 and it gives you the output of negative 176 easy okay Multiple Outputs that's your answer now there's gonna be some problems that they ask for multiple outputs so there's some problems that are gonna say like the domain I'll say the domain is I don't know let's pick some numbers let's just pick some easy ones say negative 1 0 and 1 and they're gonna say find the range ok so we're just gonna use the same equation use this domain I'm gonna go back to here and I'm gonna get rid of this the cool thing you can do is if you're typing a lot of stuff you can just enter a table and here's how you put in table you hit the plus sign and table and it says x1 y1 this is gonna be separate from this thing up here but I just leave it there so I can see it so I want domain or X values of negative 1 0 & 1 so I'm gonna type negative 1 enter 0 enter 1 okay now we want to know what the output is and it doesn't automatically give it to you I'm not sure if you can type in f of X I haven't actually tried this let's see I think that works okay so make this bigger so can I'll see it here okay now this specifically says the input is X 1 this is saying put X 1 into this formula up here so it's outputting all the correct stuff so for instance if I put negative 8 it should say negative 176 again and it does so that worked the other thing you can do is if that's not working for you you can just retype this equation so you can just say think you can just put in 11 times the quantity X and has to be X 1 because this says X 1 minus 24 parenthesis divided by 2 and it'll give you the same output so if you get stuck there you can always just retype it in just be careful with the X ones you got to put an X sub 1 or it's not gonna work because it wants a specific X for X sub 1 okay so that's a really good feature of desmos you just got to remember these X sub ones and Y sub ones and when they work if not it usually pops up with some kind of error until you do that let's see systems of Systems of Equations equations are actually super easy I'm just going to cover them really fast it's basically just graphing two equations so get rid of these 2x plus 4y equals 22 7x plus y equals 12 okay if you know anything about systems of equations what you should already know about if your watch this video you know that to find the solution graphically it's where the lines intersect that's right here at what point there so the answer is 1/5 super easy now on the SOL I think they're gonna try and make it harder by giving you a word problem and you would have to come up with these equations and then it's easy to solve so I think that's what they're going to try and do in the future I don't actually know that that's just my guess last one is finding roots roots are the same thing as zeros solutions or x-intercepts which is when y 0 or f of X equals 0 so to do that also pretty simple get rid of these we're just gonna type in that formula here this equation 3x squared get the exponent by clicking right plus 11x equals 20 okay so there are going to be the X intercepts which if you look it kind of highlights them for you now this has a really low vertex so I'm not gonna try and find it here so let me just zoom in here's one at negative 5 0 and here's one at one point three repeating zero which is the same as positive 4/3 so that would be answer choice D easy all right so there's some things you can do in desmos I'm going to post another video I think I'll just call it part 2 for regression because I'm gonna run out of time here in a second so go ahead and look at Part two and I'm going to show you some linear regression and quadratic regression and you're going to need to know which one to use win if you want to go back and look at some of this feel free to pause it and rewind Thanks
3391	https://www.quora.com/How-do-I-complete-the-square-of-x-2-2x-3	Something went wrong. Wait a moment and try again. Solving Quadratic Equatio... Algebraic Manipulation Mathematics Homework Ques... Solving Equations Mathematical Problems Algebra Class Quadratic Equations Algebraic Techniques 5 How do I complete the square of x^2-2x=3? Durga's Curiosity Answered by Ramani Mohan Das · Author has 1.7K answers and 242.3K answer views · Oct 16, 2022 Or, X^2–2X+1=3+1 Or, X-1=+_ 2 Either, X=3 Or X=—2+1=—1 Ans, 3, —1 Kurt Mager Enjoys solving math problems · Author has 17.6K answers and 7.4M answer views · Jul 21 x2−2x=3 For simplicity, Let y=−1 Assistant Bot · 1y To complete the square for the equation , follow these steps: Move the constant to the other side : Rearrange the equation : Find the value to complete the square : Take the coefficient of (which is ), divide it by 2, and square it: 4. Add and subtract this value on the left side : This simplifies to: Move the constant back to the right side : Take the square root of both sides : Solve for : To complete the square for the equation , follow these steps: Move the constant to the other side : Rearrange the equation : Find the value to complete the square : Take the coefficient of (which is ), divide it by 2, and square it: 4. Add and subtract this value on the left side : This simplifies to: Move the constant back to the right side : Take the square root of both sides : Solve for : So, the solutions to the equation are and . Related questions Why does 2x = 2^x? What is [x-2/3] 3 square? If (3^x – 1) (2^2x) = 48, what is x? If 2^(x-1) +2^(x-2) +2^(x-3) =1/16, what is 2x? How do you complete square root x^2+9 square roots x =3? Durga's Curiosity Answered by Ravi Sharma · Author has 15.1K answers and 3.8M answer views · Oct 16, 2022 X²-2X= 3 X²-2X+1= 3+1 (X-1)²= 4= 2² X-1= ±2 X-1= 2 OR X= 3 X-1= —2 OR X= —1 Audley Willacey B.Sc. in Mechanical Engineering, University of the West Indies, St. Augustine (Graduated 1976) · Author has 5K answers and 1.5M answer views · Updated 1y x^2 - 2x = 3 We complete the square by adding “the square of half the coefficient of x”. And whatever we add to one side of an equ'n,we have to add to the other side. x^2 - 2x + (-2/2)^2 = 3 + (-2/2)^2 (x - 1)^2 = 3 + 1 = 4 Taking squareroot of both sides : (x - 1) = +/- 2 x = 1 +/- 2 x = 1+2 ; x = 1 - 2 x = 3; x = -1. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Michael Lamar PhD in Applied Mathematics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3.7K answers and 17.5M answer views · Feb 26 Related How do you complete the square of x^2-2x? Just for fun, here’s an unusual way that works: [math]x^2-2x=x(x-2)=\left((x-1)+1\right)\left((x-1)-1\right)[/math] The last step looks unhelpful until we notice that these two terms are the factoring of the difference of squares! [math]x^2-2x=(x-1)^2-1^2[/math] And just like that, we’ve completed the square. Notice that this approach is general even if it doesn’t appear to be the formulaic approach you are taught in school. [math]x^2+ax+b=x(x+a)+b[/math] [math]x^2+ax+b=\left(\left(x+\frac a2\right)-\frac a2\right)\left(\left(x+\frac a2\right)+\frac a2\right)+b[/math] [math]x^2+ax+b=\left(x+\frac a2\right)^2+\left(b-\left(\frac a2\right)^2 \right)[/math] And just li Just for fun, here’s an unusual way that works: [math]x^2-2x=x(x-2)=\left((x-1)+1\right)\left((x-1)-1\right)[/math] The last step looks unhelpful until we notice that these two terms are the factoring of the difference of squares! [math]x^2-2x=(x-1)^2-1^2[/math] And just like that, we’ve completed the square. Notice that this approach is general even if it doesn’t appear to be the formulaic approach you are taught in school. [math]x^2+ax+b=x(x+a)+b[/math] [math]x^2+ax+b=\left(\left(x+\frac a2\right)-\frac a2\right)\left(\left(x+\frac a2\right)+\frac a2\right)+b[/math] [math]x^2+ax+b=\left(x+\frac a2\right)^2+\left(b-\left(\frac a2\right)^2 \right)[/math] And just like that, we have completed the square. Related questions What is [math]\int x\sqrt {3-2x-x^2} dx[/math] ? How do you complete the square and solve for x: [math]x^2+x-1=0[/math] ? If math^2=(2x-3)^2[/math] , what is [math]x[/math] ? If [math]2^x=3[/math] , what's [math]3^x[/math] ? What is the square of (3x - 4)? Neal Schermerhorn Lives in Massachusetts (1967–present) · Author has 7K answers and 1.8M answer views · Feb 26 Related How do you complete the square of x^2-2x? Let’s assume this equals 0. x^2 - 2x = 0 To solve for x by completing the square, make sure the coefficient of x^2 is 1, and the constant is on the opposite side of the equation. Our x^2 coefficient is 1, and the constant is 0, so we’re all set. Divide the coefficient of x by 2 and then square it. Here, half -2 is -1, and squaring it gives us 1. Add 1 to both sides. x^2 - 2x + 1 = 1 The left side is factorable as (x - 1)^2. (x - 1)^2 = 1 Square root both sides, being sure to use both positive and negative on the right. x - 1 = +/- 1 We now have two linear equations we can solve. (1) x - 1 = 1 ==> x = 2 ( Let’s assume this equals 0. x^2 - 2x = 0 To solve for x by completing the square, make sure the coefficient of x^2 is 1, and the constant is on the opposite side of the equation. Our x^2 coefficient is 1, and the constant is 0, so we’re all set. Divide the coefficient of x by 2 and then square it. Here, half -2 is -1, and squaring it gives us 1. Add 1 to both sides. x^2 - 2x + 1 = 1 The left side is factorable as (x - 1)^2. (x - 1)^2 = 1 Square root both sides, being sure to use both positive and negative on the right. x - 1 = +/- 1 We now have two linear equations we can solve. (1) x - 1 = 1 ==> x = 2 (2) x - 1 = -1 ==> x = 0 However,, this one can be solved in a better way. Note that the constant is 0. If the constant is 0, then (x) is a factor and 0 is a root. So we would start by factoring out the x. (x)(x - 2) = 0 x = 0 x - 2 = 0 ==> x = 2 Sponsored by CDW Corporation How well do your distributed teams work together? Enable seamless and secure communication for everyone with Cisco collaboration solutions from CDW. Jack Blake Honours Degree in Pure Physics, University of Leeds (Graduated 1968) · Author has 712 answers and 728.8K answer views · Updated Jun 8 Related How do you solve for x by completing a square: 2x²-3x=5? . 2x²-3x=5? 1. 2X² - 3X = 5? The following generic quadratic equations are well worth the effort of becoming familiar with them. 1. aX^2 + bX + c = 0 2. ( X + p ) ( X + q ) = 0 Importantly, (pq) are not directly related to (abc) in #1. To complete the Square, #2 is simpler to use. Most “books” teach as follows. Given that equations, more often than not, look like #1. ( or an extract from #1. ) 1. The drawback is that, if the coefficient, a, is not 1, it calculations and comparisons are made more difficult. 2. By comparison, #2. seems more simple. Complexity may be added later, if essential. Complet . 2x²-3x=5? 1. 2X² - 3X = 5? The following generic quadratic equations are well worth the effort of becoming familiar with them. 1. aX^2 + bX + c = 0 2. ( X + p ) ( X + q ) = 0 Importantly, (pq) are not directly related to (abc) in #1. To complete the Square, #2 is simpler to use. Most “books” teach as follows. Given that equations, more often than not, look like #1. ( or an extract from #1. ) 1. The drawback is that, if the coefficient, a, is not 1, it calculations and comparisons are made more difficult. 2. By comparison, #2. seems more simple. Complexity may be added later, if essential. Complete the Square to solve 1. 2X² - 3X = 5. Rearrange Divide thru by 2. to make the coefficient of X^2 eual to 1. 2. 2X^2 - 3X - 5 = 0 3. X^2 - ( 3/2).X - ( 5/2) == 0 Rewrite the equation. x^2 + 12x +10 = 0 x^2 + 12x +10 [ +2 -2 ] = 0 x^2 + 12x +12 = 2 Factorise (x + 12) • (x + 1) = 2 1. x^2 + 12x +10 = 0 by completing the square The following is helpful to be aware of, to know where to find it, even be able to recall on demand. The Sum of the Roots is the negative ratio of the coefficient of x and x^2. In this case we have :- (a, b) of ( ax, bx^2 ) ==> (1 : 12) The Product of the roots to the ratio of the constant term c and the x^2 coefficient. By Inspection (a,b) are in the negative ratio of (1 : 12). From #1. ERROR ? We see that to complete the square we require “ dc”to be 12, and not 10. Rewrite the equation. as ••••••• We see that to complete the square we require “ dc”to be 12, and not 10. Rewrite the equation. Query. The square seems to mean. . a = 1, b = 12, c = 12. (b & c) are matched. x^2 + 12x +10 = 0 x^2 + 12x +10 [ +2 -2 ] = 0 x^2 + 12x +12 = 2 Factorise (x 12) • (x + 1) = 2 Then what ?? Elizabeth Jean Stapel tutor (1989 to present), instructor (1991+), author (1998+) · Author has 5.9K answers and 3.3M answer views · 9mo Related How do you solve for x by completing a square: 2x²-3x=5? How do you solve for x by completing a square: 2x²-3x=5? To solve by completing the square, follow the standard steps: [math]\qquad 2x^2 - 3x = 5[/math] Make sure that the variable terms are on one side of the “equals” sign, with any loose numbers on the other side. (In your case, this step is already done.) If anything other that [math]1[/math] is multiplied on the squared term, factor this out of both variable terms: [math]\qqu[/math] How do you solve for x by completing a square: 2x²-3x=5? To solve by completing the square, follow the standard steps: [math]\qquad 2x^2 - 3x = 5[/math] Make sure that the variable terms are on one side of the “equals” sign, with any loose numbers on the other side. (In your case, this step is already done.) If anything other that [math]1[/math] is multiplied on the squared term, factor this out of both variable terms: [math]\qquad 2\left(x^2 - \frac{3}{2}x\right) = 5[/math] Find half of the linear term’s coefficient (and note its sign): [math]\qquad \frac{1}{2}\left(-\frac{3}{2}\right) = -\frac{3}{4}[/math] Square this value: [math]\qquad \left(-\frac{3}{4}\right)^2 = \frac{9}{16}[/math] Add this value to both sides, remembering to include any factored-out value on the numerical side: [math]\qquad 2\left(x^2 - \frac{3}{2}x + \frac{9}{16}\right) = 5 + 2\left(\frac{9}{16}\right)[/math] Convert the variable side to completed square form, remembering the sign on the halved value: [math]\qquad 2\left(x - \frac{3}{4}\right)^2 = 5 + 2\left(\frac{9}{16}\right)[/math] Simplify the numerical side: [math]\qquad 2\left(x - \frac{3}{4}\right)^2 = 5 + \frac{18}{16} = \frac{98}{16}[/math] Divide through by the multiplier on the variable side: [math]\qquad \left(x - \frac{3}{4}\right)^2 = \frac{49}{16}[/math] Take the square root of each side, remembering the “plus-minus” on the numerical side: [math]\qqu... 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Complete the square for the equation that remains, just as you would any other quadratic wherein the leading coefficient is 1. You can safely disregard the factor -3 altogether, because it has no effect on the roots of the equation. It’s just a scaling factor that reflects the parabola about the x-axis, and stretches the parabola vertically by a factor of 3. (See the graph below for confirmation.) Move the constant term +2 to the right; then factor out -3 from the entire equation immediately. Complete the square for the equation that remains, just as you would any other quadratic wherein the leading coefficient is 1. You can safely disregard the factor -3 altogether, because it has no effect on the roots of the equation. It’s just a scaling factor that reflects the parabola about the x-axis, and stretches the parabola vertically by a factor of 3. (See the graph below for confirmation.) Elizabeth Jean Stapel tutor (1989 to present), instructor (1991+), author (1998+) · Author has 5.9K answers and 3.3M answer views · 7y Related I have (2×x^2)-x, how can I complete this square? There are many lessons online which explain the process. completing the square quadratic In your case, the first step would be to put the quadratic in a more standard form: [math]2x^2\, -\, x[/math] Then factor out the 2: [math]2\left(x^2\, -\, \frac{1}{2}x\right)[/math] Take the coefficient of the x term (not the x-squared term), divide it by 2, square the result, and add this new value inside the parenthetical: [math]2\left(x^2\, -\, \frac{1}{2}x\, +\, \frac{1}{16}\right)[/math] Then write the parenthetical as a squared binomial. Rajeev Raj B.sc-Hons ♡-->Mathematics from Magadh University (Graduated 2021) · Author has 8.4K answers and 2M answer views · 9mo Related How do you solve for x by completing a square: 2x²-3x=5? solⁿ :→ Given 2x² - 3x =5 or x² - 3x/2 =5/2 or (x - 3/4)² - (3/4)² =5/2 or (x - 3/4)² - 9/16=5/2 or (x - 3/4)² =5/2 + 9/16 or (x - 3/4)² =49/16 or x - 3/4 =± 7/4 or x =3/4 ± 7/4 or x =(3 ± 7)/4 Saba Hemati Knows Persian · Author has 1.8K answers and 1.6M answer views · 2y x^2 - 2x = 3 x^2 - 2x + (2/2)^2 - (2/2)^2 = 3 (x - (2/2))^2 - (2/2)^2 = 3 (x - 1)^2 = 4 Related questions Why does 2x = 2^x? What is [x-2/3] 3 square? If (3^x – 1) (2^2x) = 48, what is x? If 2^(x-1) +2^(x-2) +2^(x-3) =1/16, what is 2x? How do you complete square root x^2+9 square roots x =3? What is ? How do you complete the square and solve for x: ? If , what is ? If , what's ? What is the square of (3x - 4)? How do I complete the square of 4x^2-32=-28? How do I complete the square: ? How do you solve 2x^3+4x+3/x+2? How do you complete the square of -x^2-5x-3? Can you please outline the steps one by one? How do you simplify the expression: (2x^3 + 3x^2 - 4x + 5) / (x ^2 + 2x - 3)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
3392	https://www.cuemath.com/algebra/factoring-polynomials/	LearnPracticeDownload Factoring Polynomials Factoring Polynomials means decomposing the given polynomial into a product of two or more polynomials using prime factorization. Factoring polynomials help in simplifying the polynomials easily. The first step is to write each term of the larger expression as a product of its factors. As a second step, the common factors across the terms are taken out in common to create the required factors. Let's discuss the methods of factoring polynomials, and some of the important concepts related to factoring polynomials: remainder theorem, factor theorem, GCF, long division. \| \| \| --- \| \| 1. \| What is Factoring Polynomials? \| \| 2. \| Process of Factoring Polynomials \| \| 3. \| Methods of Factoring Polynomials \| \| 4. \| Concepts Relating to Factoring Polynomials \| \| 5. \| FAQs on Factoring Polynomials \| What is Factoring of Polynomials? The process of factoring polynomials involves expressing the polynomial as the product of its factors. Factoring polynomials help in finding the values of the variables of the given expression or to find the zeros of the polynomial expression. A polynomial is of the form axn + bxn - 1 + cxn - 2+ .........px + q, which can be factorized using numerous methods: grouping, using identities and substituting. Here in this polynomial, the exponent of x is n and it has n factors. The number of factors is equal to the degree of the variable in the polynomial expression. Higher degree polynomials are reduced to a simpler lower degree, linear or quadratic expressions to obtain the required factors. Factoring polynomials can be understood with the help of a simple example. The quadratic polynomial x2 + x(a + b) + ab can be factorized as (x + a)(x + b). Process of Factoring Polynomials The following steps help for the process of factoring polynomials. Follow the below sequence of steps to factorize a polynomial. Factor out if there is a factor common to all the terms of the polynomial. Identify the appropriate method for factoring polynomials. You can use regrouping or algebraic identities to find the factors of the polynomial. Write polynomial as the product of its factors. Methods of Factoring Polynomials There are numerous methods of factoring polynomials, based on the expression. The method of factorization depends on the degree of the polynomial and the number of variables included in the expression. The four important methods of factoring polynomials are as follows. Method of Common Factors Grouping Method Factoring by splitting terms Factoring Using Algebraic Identities Let us discuss each of the methods of factoring polynomials. Method of Common Factors This is the simplest method of factoring an algebraic expression by taking common factors of each of the terms of the given expression. As a first step, the factors of each of the terms of the algebraic expression are written. Further, the common factors across the terms are taken to obtain the possible factors. This is equivalent to using the distributive property in reverse. Let us understand this better with the help of an example. Consider a simple example: 3x+9 By factoring each term we get, 3 x + 3 . 3 By distributive law, 3x+9= 3.x + 3.3 = 3(x+3) Factoring Polynomials By Grouping The method of grouping for factoring polynomials is a further step to the method of finding common factors. Here we aim at finding groups from the common factors, to obtain the factors of the given polynomial expression. The number of terms of the polynomial expression is reduced to a lesser number of groups. First, we split each term of the given expression into its factors and further aim at taking common terms to find the group of factors. Let us try to understand grouping for factorizing with the help of the following example. Let us solve an example problem to more clearly understand the process of factoring polynomials. Consider a polynomial: 8ab+8b+28a+28. Notice that 4 is a single factor common to all the terms of this polynomial. So, we can write 8ab+8b+28a+28 =4(2ab+2b+7a+7) Let us group 2ab+2b and 7a+7 in the factor form separately. 2ab + 2b = 2b(a + 1), and 7a + 7 = 7(a + 1) Now we have 8ab+8b+28a+28 = 4(2ab+2b+7a+7)= 4 (2b(a + 1) + 7(a + 1)) = 4(2b + 7)(a + 1) Thus the factoring polynomials is done by grouping. 8ab + 8b + 28a + 28 = 4(2b + 7)(a + 1) Factoring Polynomials by Splitting Terms The process of factoring polynomials is often used for quadratic equations. While factoring polynomials we often reduce the higher degree polynomial into a quadratic expression. Further, the quadratic equation has to be factorized to obtain the factors needed for the higher degree polynomial. The general form of a quadratic equation is x2 + x(a + b) + ab = 0, which can be split into two factors (x + a)(x + b) = 0. Consider the quadratic polynomial of the form x2 + x(a + b) + ab. =x.x + ax + bx + ab =x(x + a) + b(x + a) =(x + a)(x + b) Here in the above polynomial, the middle term is split as the sum of two factors, and the constant term is expressed as the product of these two factors. Thus the given quadratic polynomial is expressed as the product of two expressions. Let us understand this better, by factoring a quadratic polynomial x2 + 7x + 12. x2 + 7x + 12 = x.x + 3x + 4x + 3.4 = x(x + 3) + 4(x + 3) x2 + 7x + 12 = (x + 3)(x + 4) Thus factoring polynomials is done using splitting the middle terms as in a quadratic polynomial. Factoring Polynomials Using Algebraic Identities The process of factoring polynomials can be easily performed using algebraic identities. The given polynomial expressions represent one of the algebraic identities. Also sometimes the given expression has to be modified so as to match with the expression of the algebraic identities. A few of the algebraic identities are helpful in factoring polynomials. a2 - b2 = (a + b)(a - b) a3 - b3 = (a - b)(a2 + ab + b2) a3 + b3 = (a + b)(a2 - ab + b2) a4 - b4 = (a2 + b2)(a + b)(a - b) Let's factorize the polynomial 4z2-12z+9 Observe that 4z2=(2z)2, 12z=2 × 3 × 2z, and 9 = 32 So, we can write 4z2-12z+9 = (2x)2 + 2(2x)(3) + 32 = (2z - 3)2 Concepts Relating to Factoring Polynomials The following concepts are helpful in factoring polynomials. Remainder Theorem The remainder theorem is helpful to find the remainder on dividing an algebraic expression with another expression, without actually performing the division. The remainder obtained when the algebraic expression f(x) is divided by ( x - a) is f(a). If f(a) = 0, then (x - a) is a factor of f(x). For a polynomial expression f(x) = 12x3 - 9x2 + 5x + 17, the remainder obtained on dividing it with (x - 2) is f(2) = 12(2)3 - 9(2)2 + 5(2) + 17 = 12(8) -9(4) + 10 + 17 = 96 - 36 + 27 = 87. Factor Theorem The factor theorem helps in connecting the factors and zeros of polynomials. If f(x) is a polynomial of degree n , a is a real number such that (x - a) is a factor of f(x), then f(a) = 0. Also if f(a) = 0 then (x - a) is a factor of f(x). The factor theorem is helpful to find if a given expression is a factor of a higher degree polynomial expression without actually performing the division. Greatest Common Factors The process of obtaining the greatest common factor for two or more terms includes two simple steps. First, split each of the terms into its prime factors, and then take as many common factors as possible from the given terms. Let us understand this by taking a simple expression of two terms 12x2 + 9x. Here we split the terms into its prime factors 12x2 + 9x = 2.2.3.x.x + 3.3.x. Among these two terms, we can take the maximum common terms to obtain the greatest common factors. Here we have the maximum common factor as 3x, and hence 12x2 + 9x = 2.2.3.x.x + 3.3.x = 3x(4x + 3). Long Division The process of long division involving polynomials is similar to the process of long division of natural numbers. Long division of polynomials is greatly helpful to find the factors of the given algebraic expression. The division resulting in a remainder of zero has the divisor as a factor of the polynomial expression. Divisions resulting in a remainder of zero can be written as Dividend = Divisor × Quotient. Thus the given polynomial expression gets divided into two factors. Further, the division of the below polynomial expression can be written as 4x2 - 5x - 21 = (x - 3)(4x + 7). ☛ Also Check: Factorization of quadratic equations Equations Linear Equations Long Division Formula Factoring Polynomials Examples Example 1: Carlos finds that the cost of a notebook is twice more than $4 for a pen. Represent this above information using a polynomial. Can you help him in factoring polynomial? Solution: Let's assume the cost of a pen = $x According to the given information, the cost of the notebook can be expressed as (2x + 4) 2 is a common factor in the polynomial (2x + 4) Answer: Therefore on factoring polynomials, the factors of (2x + 4) are 2 and (x + 2) 2. Example 2: Factorize the polynomial 6xy-4y+6-9x using the method of regrouping of factoring a polynomial. Solution: For factoring polynomials we observe that we have no common factor among all the terms in the expression 6xy-4y+6-9x. Let's try regrouping them as (6xy-4y) and (6-9x). (6xy - 4y) + (6 - 9x) = 2y(3x - 2)- 3(3x - 2) = (3x - 2)(2y - 3) Answer: Therefore on factoring polynomial 6xy - 4y + 6 - 9x, we get (2y - 3) and (3x - 2) as the factors. 3. Example 3: Use the factoring polynomials techniques and factor x3 + 5x2 + 6x. Solution: Before factoring polynomial, let us reduce the degree of the polynomial from 3 to 2.Notice that x is a common factor in x3 + 5x2 + 6x. So, x3 + 5x2 + 6x = x(x2 + 5x + 6) We can now split x2 +5x+6 as x2 + 3x + 2x + 6 x2 + 5x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2) Thus, on factoring the cubic polynomial x3 + 5x2 + 6x we get x(x + 2)(x + 3) as its factors. Answer: Therefore x3 + 5x2 + 6x = x(x+2)(x+3). View Answer > go to slidego to slidego to slide Great learning in high school using simple cues Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes. Book a Free Trial Class Practice Questions on Factoring Polynomials Check Answer > go to slidego to slide FAQs on Factoring Polynomials What is Factoring Polynomials? The process of factoring polynomials is to split the given expression and write it as a product of these expressions. For example, to factorize x2 + 2x, we split it into two factors x and (x + 2), and write it as a product of these two factors x2 + 2x = x(x + 2). Here the process of factoring polynomials involves polynomials of higher degrees and involves concepts of the greatest common factor, factor theorem, long division. How Do you Find the Factors of a Polynomial? To write a polynomial in factored form, it must be expressed as a product of terms in their simplest form. The terms could be constant or linear equation or any polynomial expression, and which cannot be further factorized. How to Factorize Polynomials in Two Variables? For factoring polynomials in two variables we factorize using a factoring method or by using a formula. A polynomial in two variables is of the form x2 + (x(a + b) + ab = 0, and can be factorized as x2 + (x(a + b) + ab = (x + a)(x + b) . Also, the factoring polynomials in two variables is needed for further factoring polynomials of high degree. How Do you Do the Prime Factorization of Polynomials? The following methods mentioned below can be used for factoring polynomials into their prime factors: Method of Common Factors Method of Grouping Method Using Algebraic Identities How to Factorize Polynomials in 3 Degree? The process of factorization of polynomials of 3 degrees involves three simple steps. First for the given n degree polynomial f(x), substitute a value 'a' such that f(a) = 0, and (x - a) is a factor. As a second step divide f(x) by (x - a) to obtain a quadratic equation. Finally, factorize the quadratic equation to obtain its two factors and hence we can obtain all the three factors of the 3-degree polynomial. How Is Factor Theorem Useful in Factoring Polynomials? The factor theorem is used to find the factors of an n-degree polynomial without actual division. If a value x = a satisfies a n-degree polynomial f(x), and f(a) =0, then (x - a) is a factor of the polynomial expression. Further, we can find a few factors using the factor theorem and the remaining can be found using the factorization of a quadratic equation. What is the Meaning of Factoring Polynomials by Grouping? Factoring polynomials by grouping means factoring the polynomial by the method of grouping that allows us to rearrange the terms of the expression, to easily identify and find factors of the polynomial expression. How Do you Find the Factors a Polynomial With 5 Terms? The process of factoring polynomials with 5 terms is as follows. Write the polynomial in the standard form. Take the greatest common factor out if it exists. Try to find at least 3 roots of the polynomial. If (\alpha) is a root of the polynomial, then (x-\alpha) is a factor of the polynomial. After finding the 3 linear factors, we are left with a quadratic polynomial. Find the product of the leading coefficient and the constant term. Determine the factors of the product found in step 3 and check which factor pair will result in the coefficient of x. After choosing the appropriate factor pair, keep the sign in each number such that while operating them we get the result as the coefficient of (x), and on finding their product the number is equal to the number found in step 3. Now, you have 4 terms in the expression and so we use the method of regrouping to factorize. What Are the Four Methods of Factoring Polynomials? The four methods of factoring polynomials are: Method of Common Factors Method of Grouping Method Using Algebraic Identities Method of Finding Roots Q1: Factor the trinomial $$x2 + 8x + 16.$$ Q2: Factor the polynomial $$8x2+12x$$ by looking for a common factor. Q3: If Β $$is a root of a polynomial, then which of the following is a factor of the polynomial? Q4: What are the factors of the polynomial $$2x2 + 5x - 3$$? Q5: Which polynomial has the factor $$(x - 2)$$? 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3393	https://impactful.ninja/impactful-synonyms-for-burden/	Top 10 Positive Synonyms for “Burden” (With Meanings & Examples) \| Impactful Ninja Skip to content DONATE Menu News Newsletter Podcast Blog Shop About Founder’s Journey Team of Experts Our Partners Contact DONATE Impactful Ninja Menu News Newsletter Podcast Blog Shop About Founder’s Journey Team of Experts Our Partners Contact DONATE Impactful Words Top 10 Positive & Impactful Synonyms for “Burden” (With Meanings & Examples) by Alexis Ingram Top 10 Positive & Impactful Synonyms for “Burden” (With Meanings & Examples) By Alexis Ingram Read Time:7 Minutes CLICK TO SUBSCRIBE ListenNoticeDonateShare Impactful Ninja is reader-supported. When you buy through links on our site, we may earn an affiliate commission. Learn moreLearn more. Affiliate Disclosure Hey fellow impactful ninja ? You may have noticed that Impactful Ninja is all about providing helpful information to make a positive impact on the world and society. 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It is a passion project of mine and I love to share helpful information with you to make a positive impact on the world and society. However, it's a project in that I invest a lot of time and also quite some money. Eventually, my dream is to one day turn this passion project into my full-time job and provide even more helpful information. But that's still a long time to go. Stay impactful, Share on EmailShare on X (Twitter)Share on PinterestShare on LinkedInShare on FacebookShare on Reddit Responsibility, calling, and trust—positive and impactful synonyms for “burden” enhance your vocabulary and help you foster a mindset geared toward making a positive impact. So, we had to ask: What are the top ten positive & impactful synonyms for “burden”? The top 10 positive & impactful synonyms for “burden” are responsibility, commitment, charge, obligation, duty, mission, calling, stewardship, custody, and trust. Using these synonyms helps you enhance both your communication and psychological resilience in several meaningful ways. In the table below, you can see all these top ten synonyms including their descriptions, why they are positive and impactful synonyms for “burden,” and example sentences that highlight how you can use each of these. We’ll then also share ten benefits of why you should use these synonyms, ten interesting facts about the word “burden,” and a brief history of the development of our alphabet. Related: Are you looking for even more positive & impactful words? Then you might also want to explore those words that start with all the other letters of the alphabet: A \| B \| C \| D \| E \| F \| G \| H \| I \| J \| K \| L \| M \| N \| ‍O \| P \| Q \| R \| S \| T \| U \| V \| W \| X \| Y \| Z Here Are the Top 10 Positive & Impactful Synonyms for “Burden” Our list of positive & impactful synonyms for “burden” help you expand your vocabulary and enhance both your communication and psychological resilience in several meaningful ways (you can read more about it in the next section). That’s why it’s so important to focus on synonyms that can be used in a positive and impactful way. Burden:a load, typically a heavy one \| the main theme or gist of a speech, book, or argument \| load heavily Oxford Dictionary Our top ten synonyms for “burden” exemplify the beauty of our language—their meaning is not just fixed but can be shaped by the context they are used in. SynonymDescriptionExample Sentence ResponsibilityA duty or task that one is required or expected to fulfill, similar to ‘burden’ but emphasizes honor and trust.“Taking care of the family business became his responsibility after his father retired.” CommitmentAn engagement or obligation that restricts freedom of action, akin to ‘burden’ but highlights dedication and loyalty.“Her commitment to environmental causes inspired many to join her in activism.” ChargeA task or duty assigned to someone, comparable to ‘burden’ but implies authority and trust bestowed upon someone.“He accepted the charge of leading the project with enthusiasm.” ObligationA moral or legal duty to do something, similar to ‘burden’ but conveys a sense of duty and necessity.“She felt a deep obligation to help those in need in her community.” DutyA task or action that someone is required to perform, akin to ‘burden’ but emphasizes respect and moral imperatives.“Serving his country was a duty he performed with pride.” MissionA specific task or goal that someone is assigned to accomplish, parallel to ‘burden’ but with a sense of purpose and importance.“Their mission to provide clean water to the village brought them together.” CallingA strong urge or vocation towards a particular way of life or career, akin to ‘burden’ but imbued with personal fulfillment and destiny.“He found his calling in teaching, despite the challenges it presented.” StewardshipThe responsible overseeing and protection of something worth caring for, similar to ‘burden’ but implies caretaking and preservation.“Her stewardship of the land ensured its preservation for future generations.” CustodyThe protective care or guardianship of something, akin to ‘burden’ but focuses on the responsibility of protection and care.“Taking custody of the historical artifacts was an honor for the museum.” TrustThe obligation or responsibility imposed on someone who is trusted, parallel to ‘burden’ but centers on confidence and faith placed in a person.“The trust placed in him by his mentor was a significant motivator in his career.” 10 Benefits of Using More Positive & Impactful Synonyms Our positive & impactful synonyms for “burden” help you expand your vocabulary and enhance both your communication and psychological resilience in several meaningful ways: Encouraging Positive Framing: Using positive synonyms allows for a more optimistic and affirmative way of expressing thoughts. This can influence not only the speaker’s or writer’s mindset but also positively impact the audience’s perception and reaction. Improving Emotional Intelligence: Learning different positive synonyms helps in accurately expressing emotions. This aids in emotional intelligence, as one can more precisely convey feelings and understand the emotions of others. Enhancing Persuasive Communication: In persuasive writing and speaking, using positive synonyms can be more effective in convincing an audience, as people generally respond better to positive language. Broadening Emotional Vocabulary: A range of positive synonyms enriches your emotional vocabulary. It’s one thing to say you’re “happy” and another to express that you’re “elated,” “joyful,” or “content.” Each word carries a unique emotional hue. Creating a Positive Atmosphere: The use of positive language can create a more constructive and encouraging atmosphere in both personal and professional settings. This can lead to better teamwork, more effective communication, and improved interpersonal relationships. Enhancing Creative Writing: For those engaged in creative writing, a repertoire of positive synonyms can help in vividly depicting scenes, characters, and emotions, making the narrative more engaging and lively. Improving Mental Health and Well-being: Regularly using and thinking in terms of positive words can influence one’s mental state and outlook on life. Positive language has been linked to greater well-being and a more optimistic outlook. Improving Cognitive Flexibility: Expanding your vocabulary with positive synonyms enhances your cognitive flexibility. This means you become more adept at thinking creatively and adapting your language use to different situations. The mental exercise involved in learning and using a variety of positive words can also contribute to overall cognitive health, keeping your mind sharp and responsive. Building Social Skills and Empathy: When you have a variety of positive words at your disposal, you’re better equipped to offer compliments, encouragement, and empathetic responses in social interactions. Facilitating Conflict Resolution: In situations of conflict, the use of positive language can help de-escalate tension. Having a range of positive synonyms allows for more constructive and diplomatic communication. Overall, your use of positive synonyms not only broadens your vocabulary but also positively influences your thought processes, emotional expression, and interpersonal interactions. 10 Interesting Facts About the Word “Burden” Let’s take a step back and have a look at some interesting facts about the word “burden”. Etymology: The word “burden” originates from the Old English “byrðen,” which meant a load, weight, or charge. It is related to the Old English word “beran,” meaning to carry or to bear, highlighting its roots in the concept of carrying weight. Historical Use: Historically, “burden” referred not only to physical loads but also to metaphorical ones, such as responsibilities or obligations, illustrating its dual usage from early on. Nautical Term: In nautical contexts, “burden” referred to the carrying capacity of a ship, indicating the volume or weight it could safely transport, showcasing the term’s application in maritime trade. Legal Contexts: In legal terminology, “burden” can refer to the burden of proof, indicating the obligation to prove one’s assertions in court, underscoring its significance in judicial processes. Psychological Concept: Psychologically, “burden” is used to describe the emotional or mental load one carries, such as stress or grief, reflecting its broad application to human experiences. Literary Motif: The concept of a burden is a common motif in literature, symbolizing challenges, responsibilities, or guilt that characters must bear, offering deep insights into human nature and societal expectations. Economic Implications: Economically, “burden” can refer to financial obligations or the strain of economic conditions on individuals and communities, highlighting its relevance in discussions of poverty and wealth distribution. Musical Usage: In music, particularly in older folk songs, the “burden” refers to the chorus or refrain of a song, a usage that metaphorically suggests something that is carried throughout the piece. Medical Use: Medically, “burden” can describe the impact of a disease on a patient, in terms of both physical symptoms and the psychological toll, emphasizing the comprehensive effects of illness. Social and Cultural Dimensions: The concept of a “burden” is also used in discussions about social and cultural responsibilities, such as the care for elders or the expectations placed on specific demographic groups, reflecting its significance in societal roles and relationships. A Brief History of Our Alphabet The story ofour alphabet has a rich and compelling history, beginning with ancient civilizations and carrying forward into the present day. The history of our modern alphabet is a fascinating journey that spans several millennia and cultures. It’s commonly referred to as the Latin or Roman alphabet, and here’s a brief overview of its evolution: Phoenician Alphabet (circa 1050 BCE): The story begins with the Phoenician alphabet, one of the oldest writing systems known to use a one-to-one correspondence between sounds and symbols. This Semitic alphabet had about 22 consonants, but no vowels, and was primarily used for trade. Greek Alphabet (circa 800 BCE): The Greeks borrowed and adapted the Phoenician script. Crucially, they introduced vowels, making it one of the first true alphabets where each symbol represented a distinct sound (both vowel and consonant). The Greek alphabet had a significant influence on the development of other alphabets. Etruscan Alphabet (circa 700 BCE): The Etruscan civilization in Italy adapted the Greek alphabet to their own language. While Etruscan was largely replaced by Latin, their version of the alphabet was a key predecessor to the Roman one. Latin Alphabet (circa 700 BCE – Present): The Latin alphabet emerged from the adaptation of the Etruscan script. Ancient Rome used this alphabet, and it spread across Europe as the Roman Empire expanded. The original Latin alphabet did not contain the letters J, U, and W. These were added much later along with other modifications to suit different languages and phonetic needs. Modern Variations: Today, the Latin alphabet is the most widely used alphabetic writing system in the world. It has undergone various changes to accommodate different languages and sounds. For instance, English—among other languages—added letters like ‘J’, ‘U’, and ‘W’, while other languages incorporate additional characters like ‘Ñ’ in Spanish or ‘Ç’ in French. This evolution reflects not just linguistic changes but also cultural and historical shifts, as the alphabet was adapted by different societies across centuries. Related: Are you looking for even more positive & impactful words? Then you might also want to explore those words that start with all the other letters of the alphabet: A \| B \| C \| D \| E \| F \| G \| H \| I \| J \| K \| L \| M \| N \| ‍O \| P \| Q \| R \| S \| T \| U \| V \| W \| X \| Y \| Z Final Thoughts Expanding your vocabulary is akin to broadening your intellectual horizons and enhancing your capacity to express your thoughts and emotions with precision. By embracing additional synonyms for “burden,” you’re not just learning new terms, but you’re also gaining nuanced ways to communicate positivity and impact. The more words you have at your disposal, the more accurately and vividly you can paint your thoughts into speech and writing. So, by growing your vocabulary, especially with positive and impactful words, you’re empowering yourself to engage more effectively and inspiringly with the world around you. Stay impactful, Sources Society for Personality and Social Psychology: Why a Simple Act of Kindness Is Not as Simple as It Seems: Underestimating the Positive Impact of Our Compliments on Others Journal of Personality: Psychological Resilience and Positive Emotional Granularity: Examining the Benefits of Positive Emotions on Coping and Health David Sacks: Letter Perfect: The Marvelous History of Our Alphabet From A to Z Impactful Ninja: Positive & Impactful Words Starting With A Impactful Ninja: Positive & Impactful Words Starting With B Impactful Ninja: Positive & Impactful Words Starting With C Impactful Ninja: Positive & Impactful Words Starting With D Impactful Ninja: Positive & Impactful Words Starting With E Impactful Ninja: Positive & Impactful Words Starting With F Impactful Ninja: Positive & Impactful Words Starting With G Impactful Ninja: Positive & Impactful Words Starting With H Impactful Ninja: Positive & Impactful Words Starting With I Impactful Ninja: Positive & Impactful Words Starting With J Impactful Ninja: Positive & Impactful Words Starting With K Impactful Ninja: Positive & Impactful Words Starting With L Impactful Ninja: Positive & Impactful Words Starting With M Impactful Ninja: Positive & Impactful Words Starting With N Impactful Ninja: Positive & Impactful Words Starting With O Impactful Ninja: Positive & Impactful Words Starting With P Impactful Ninja: Positive & Impactful Words Starting With Q Impactful Ninja: Positive & Impactful Words Starting With R Impactful Ninja: Positive & Impactful Words Starting With S Impactful Ninja: Positive & Impactful Words Starting With T Impactful Ninja: Positive & Impactful Words Starting With U Impactful Ninja: Positive & Impactful Words Starting With V Impactful Ninja: Positive & Impactful Words Starting With W Impactful Ninja: Positive & Impactful Words Starting With X Impactful Ninja: Positive & Impactful Words Starting With Y Impactful Ninja: Positive & Impactful Words Starting With Z Show sources ↧ Alexis Ingram Alexis is passionate about providing creative and insightful information about the nuances of our everyday language and vocabulary. They have the unique advantage of being trained on a vast lexical database that helps provide unparalleled depth with their extensive linguistic analysis. Articles authored by Alexis are supplemented by the use of a multimodal large language model, helping us to share a scientific perspective on how we use our language and vocabulary. Was this article helpful? Yay - we are happy that you found this article helpful :) Become more impactful and share the information! Share on EmailShare on X (Twitter)Share on PinterestShare on LinkedInShare on FacebookShare on Reddit Feedback Form Oh no - we are sorry that you didn’t find this article helpful :( Become more impactful and help us improve it! Feeback Name Email Share your feedback Follow this site to get new posts directly to your inbox: Follow This Site Cancel Did you like this article? Get the 5-minute newsletter that makes reading impactful news enjoyable—packed with actionable insights to make a positive impact in your daily life. 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3394	https://www.cs.princeton.edu/~chazelle/pubs/FClowerbounds.pdf	Journal of Computer and System Sciences 68 (2004) 269–284 Lower bounds for intersection searching and fractional cascading in higher dimension$ Bernard Chazellea,b and Ding Liub, aNEC Research Institute, Princeton University, Princeton, NJ, USA bDepartment of Computer Science, Princeton University, Princeton, NJ, USA Received 24 November 2001; revised 12 December 2002 Abstract Given an n-edge convex subdivision of the plane, is it possible to report its k intersections with a query line segment in Oðk þ polylogðnÞÞ time, using subquadratic storage? If the query is a plane and the input is a polytope with n vertices, can one achieve Oðk þ polylogðnÞÞ time with subcubic storage? Does any convex polytope have a boundary dominant Dobkin–Kirkpatrick hierarchy? Can fractional cascading be generalized to planar maps instead of linear lists? We prove that the answer to all of these questions is no, and we derive near-optimal solutions to these classical problems. r 2003 Elsevier Inc. All rights reserved. 1. Introduction Fractional cascading is a general technique for speeding up lookup queries in catalogs associated with the nodes of a graph . Speciﬁcally, suppose that G is a bounded-degree graph, where each node v is associated with a catalog Cv (which is just a sorted list of numbers). The successor of x in Cv is deﬁned as minfyACv,fNg j yXxg: Given a subset of k nodes in G whose induced subgraph is connected, it is easy, by repeated binary search, to compute the successors of any query x in the k catalogs in time Oðk log nÞ; where n is the combined size of all the catalogs. Fractional cascading reduces the query time to Oðk þ log nÞ while increasing the storage by only a constant factor. The technique has found numerous applications in computational geometry (from which it originated), but also in constraint databases , IP routing , packet classiﬁcation ARTICLE IN PRESS $A preliminary version of this paper appeared in Proceedings of the 33rd Annual ACM Symposium Theory of Computing (2001), 322–329. Corresponding author. E-mail address: dingliu@CS.Princeton.EDU (D. Liu). 0022-0000/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.jcss.2003.07.003 , data mining , geographical information systems , etc. This versatility has motivated the design of all sorts of variants: dynamic, probabilistic, parallel, external-memory fractional cascading [4,3,16,18,23,27,26]. An outstanding open problem has been the two-dimensional generalization of fractional cascading: what if we replaced the catalogs at the nodes by planar subdivisions? The question is motivated by more than mere curiosity. Indeed, the wide applicability of the technique stems from the fact that many data structures for multidimensional searching [5,22] consist of a skeleton graph whose nodes are themselves repositories of auxiliary (often recursively deﬁned) data structures. As long as linear lists reside at the bottom of the hierarchy, fractional cascading can be called into action. In some cases, however, the bottom layer consists of planar maps (i.e., 2D catalogs) and the current technology fails. 2D fractional cascading would immediately lead to improved algorithms for nearest-neighbor searching in E3; intersection search for query segments in planar line arrangements and convex subdivisions, intersection search for query planes and polygons in polytopes, ﬁxed-directional ray shooting in 3D, not to mention numerous instances of range searching. This paper dashes all such hopes. We show that not only generalizing FC to 2D catalogs is impossible but that no pointer machine solutions can provide the sort of logarithmic speed-up associated with fractional cascading. Furthermore, the counterexamples are hardly pathological: in fact, catalogs consisting of simple parallel strips are enough to break the whole fractional cascading scheme apart. This can seem rather surprising in view of previous results suggesting that such generalizations might, indeed, be possible. For example, navigation among geodesic triangles as described in [13,19] can be naturally viewed as an instance of 2D fractional cascading. So, it appears that the catalog graphs of geodesically triangulated polygons are more exceptional than previously thought. We also resolve a question that had been open since the mid-1980s. The Dobkin–Kirkpatrick hierarchy is a simplicial complex subdivision of an n-vertex convex polytope with the property that no line can intersect more than Oðlog nÞ simplices. It is an essential tool for all sorts of polyhedral operations, mesh simpliﬁcation, etc. If the DK hierarchy is boundary dominant, meaning that, up to a logarithmic additive term, the number of simplices intersected by any given plane is at most proportional to the boundary size of the intersection, then it can also be used for intersection searching or silhouette computation. For example, we could use such a hierarchy to compute, in optimal time, the shadow of a polytope cast by a single light source. Constructing a DK hierarchy is highly nondeterministic: Among the (usually) exponentially many possible realizations, could it be that at least one of them is boundary dominant? We prove that the answer is no. Of anecdotal interest, we should mention that this purely combinatorial question is resolved by using algorithmically inspired arguments. Here is a summary of our results. All complexity results hold in the standard pointer machine model . 1. There is a graph with 2D catalogs attached to its nodes such that to perform the same point location query at the k nodes of a connected subgraph in time Oðk þ polylogðnÞÞ requires storage Oðn2Þ; where n is the combined size of all the catalogs.1 ARTICLE IN PRESS 1The tilde notation hides a polylogarithmic factor. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 270 2. For any n large enough, there is a convex planar subdivision with n vertices such that to compute all k edges intersected by a query line in Oðk þ polylogðnÞÞ time requires storage Oðn2Þ: On the other hand, Oðn2Þ storage is sufﬁcient to achieve Oðk þ log nÞ query time (even if the query is a line segment). The same lower bound holds for intersecting a query line with a simple polygon with n vertices. 3. For any n large enough, there is a (nonconvex) polytope P in R3 with n vertices such that, given a query plane p; to compute its intersection P-p in time Oðk þ polylogðnÞÞ; where k is the number of vertices in P-p; requires storage Oðn3eÞ; where e is an arbitrarily small positive constant. On the other hand, Oðn3Þ storage is sufﬁcient to achieve Oðk þ log nÞ query time. 4. For any n large enough, there exists a convex polytope with n vertices that admits of no boundary dominant DK hierarchy. Remarks. In all cases, the lower bounds remain true, up to a factor of ne; if we replace the polylog term in the query time by a function of the form nd; for small enough d; e40: Our third result also holds for convex polytopes but the lower bound then drops to Oðn2Þ: (We have been unable to ﬁnd a matching upper bound for that restricted version of the problem.) How do our results compare to previous work? The upper bounds are mostly applications of known techniques; in particular, duality, ﬁltering search, navigation in arrangements. The lower bounds, on the other hand, should come as more of a surprise. Although not particularly difﬁcult technically, they are rather counter-intuitive. Pointer machine lower bounds exist for all sorts of problems, including union-ﬁnd [24,28] and range searching [10,15]; see surveys [1,21]. However, we are not aware of previous lower bounds for intersection searching or 2D fractional cascading. Our proofs rely on a general volume argument developed in and a Heilbronn-type result proven in . The rest is new and self-contained. 2. Preliminaries Intersection searching refers to the following type of problem: Given a set S of geometric ‘‘objects’’ (e.g., edges of a planar subdivision, facets of a polytope), we wish to preprocess S into a data structure so that, given any query q from a predeﬁned class (line segment, hyperplane), the set of all objects in S intersected by q can be reported efﬁciently. It is understood that the set S is ﬁxed once for all, while queries are presented and answered on-line. Note that this deﬁnition need not make any reference to geometry; in fact, it is often useful to view intersection searching abstractly by simply assuming a relation between each q and a subset of S (the ‘‘intersected’’ objects). In the pointer machine framework , a data structure for intersection searching is modeled as a directed graph G with bounded outdegree. The graph G is referred to as a search structure for S. Some of the nodes are assigned objects of S (not necessarily injectively). Given a query q; the algorithm navigates through the graph G; beginning at some start node, until the subgraph traversed contains at least one node assigned to each object of S intersected by q: For lower bound purposes, it sufﬁces to count the number of nodes visited as a conservative estimate of the query time. (This also covers randomized query-answering algorithms as well.) ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 271 As in [10,15], we ﬁrst give a general lemma on the size of any pointer machine data structure for efﬁcient intersection searching; then we establish the desired storage lower bound by exhibiting a set of ‘‘hard’’ queries. Deﬁnition 2.1. A search structure G for set S is (a; o)-effective, with a a positive constant and o an additive overhead, if for any query q; we have jGðqÞjpaðk þ oÞ: Here GðqÞ is the set of nodes visited in G while answering query q; and k is the output size, i.e., the number of objects in S intersected by q: This deﬁnition expresses our focus on search structures that answer queries in time linear in the output size, aside from an additive overhead of o ¼ oðjSjÞ: Of course, we can also handle arbitrary multiplicative overheads by setting a ¼ aðjSjÞ: Deﬁnition 2.2. A collection of queries Q ¼ fqig is ðm; oÞ-favorable for S; with m41; if Q satisﬁes the following relevance and independence conditions. 1. Relevance: jS-qijXo; for any query qiAQ: 2. Independence: jS-qi1-?-qimj ¼ Oð1Þ; for all possible i1o?oim: Here S-qi denotes the set of objects in S intersected by qi; and S-qi1-?-qim has a similar meaning. The relevance condition means that each query intersects enough objects so make the output size dominate the additive overhead. The independence condition gets to the heart of the matter: It is a Zarankiewicz-type condition stating that the bipartite graph induced by objects and queries should be free of large complete subgraphs. The motivation for these deﬁnitions comes from a general volume argument [10,15] about pointer machines. Adapted to our purposes it states that: Lemma 2.3. Given a search structure G for S and a set Q of queries, assume that G is ða; oÞ-effective for some constant a; and Q is ðm; oÞ-favorable for S. For any o large enough, the size of G is OðjQjo=mÞ: Lemma 2.3 is the main vehicle for proving lower bounds on the storage required by a search structure with a given query time. Complexity questions are in this way reduced to purely combinatorial ones, involving the existence of ‘‘favorable’’ sets of queries. The rest of this paper is organized as follows. In Section 3 we establish lower bounds on the complexity of intersecting a planar subdivision (resp. convex polytope) with a query line segment (resp. query plane). These results give us the stepping stone from which we can conclude that 2D fractional cascading is impossible (Section 4), and that not all convex polytopes admit of a boundary dominant Dobkin–Kirkpatrick hierarchy (Section 5). We establish the Oðn3eÞ lower bound for (nonconvex) polytopes in Section 6, and also provide a nearly matching upper bound. ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 272 3. Planar subdivisions and convex polytopes Theorem 3.1. Given a planar subdivision with n vertices, to compute all k edges intersected by a query line in Oðk þ oÞ time requires Oðn2=o2Þ storage. Theorem 3.2. Given an n-vertex convex polytope in R3; to compute all k edges intersected by a query plane in Oðk þ oÞ time requires Oðn2=o2Þ storage. The lower bound for planar subdivisions is essentially optimal: an Oðk þ log nÞ query time solution using Oðn2Þ storage is described in . Note that the lower bound also holds for the problem of intersecting a query line with a simple polygon, since any n-vertex planar subdivision can be easily transformed into a simple OðnÞ-gon (by essentially duplicating each edge and creating a polygon of very small area). For convex polytopes the complexity gap is still large. The best solution with Oðk þ polylogðnÞÞ query time requires Oðn3Þ storage and is obtained by applying the solution for the nonconvex case described in Section 6. Both theorems use the same basic starting construction. Take Jon congruent vertical segments evenly distributed horizontally, labeled 1; y; Jon; from right to left (Fig. 1). Next, decompose each segment into t ¼ Jn=on subsegments of the same length, bringing the total number of edges to YðnÞ: Let ai (resp. bi) denote the midpoint of the ith subsegment on segment 1 (resp. 2), counting top down. Of the t2 lines passing through pairs ðai; bjÞ; any one that intersects segment Jon is called hitting. For example, line l1 in Fig. 1 is such a hitting line. A few simple observations: Fact 3.3. Every hitting line intersects all vertical segments 1 to Jon; and every intersection point is the midpoint of some subsegment. To bound the number of hitting lines, observe that any point on segment 1 can join, via a hitting line, at least Jn=on=ðJon 1Þ points on segment 2; and hence, Fact 3.4. The number of hitting lines is at least n2=o3: Let the canonical subset of a hitting line refer to the set of subsegments that intersect it. By Fact 3.3, intersections take place at midpoints, and so two lines that intersect the same subsegments pass through the same two points and therefore are identical. ARTICLE IN PRESS n ω 1 2 ω 2 l 2 a l1 b3 Fig. 1. Towards a hard convex subdivision. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 273 Fact 3.5. Each canonical subset is of size Jon and the intersection of any two of them is of size at most 1. Now let S (resp. Q) be the set of subsegments (resp. hitting lines). Fact 3.5 implies that Q is ð2; oÞ-favorable for S: We are now ready to build a convex subdivision around S: In this subdivision, as well as all the others in this paper, we shall ensure that the subdivision is in ‘‘general position’’, meaning that no two adjacent edges are collinear. Thicken the vertical segments in S; turn them into thin ladder-like concave strips, and add horizontal rungs to ensure the convexity of their decompositions (Fig. 2). Finally, cap the top and bottom parts of the ladders with two suitable convex pieces. We redeﬁne S to be the left subsegments of the ladders, and check that, by keeping the ladders thin enough, Q still is ð2; oÞ-favorable for S: we then are ready to apply Lemma 2.3 to derive a lower bound. The only problem is that the intersection search problem deﬁned by Q and S is not, properly speaking, a ‘‘line-intersects-subdivision’’ problem since the output contains edges outside of S: However, by keeping the ladders thin enough, we see that each set of intersected edges in S is a subset at least half the size of the actual set of intersected edges. And so the lower bound for the ðS; QÞ problem also applies to the ‘‘line-intersects-subdivision.’’ Theorem 3.1 follows from the fact that the number n of vertices in the resulting subdivision is OðjSjÞ: & The reader might wonder why we did not choose to augment S by including in it all the edges of the subdivision, and then check again that the relevance and independence conditions still hold. The reason is that although augmenting S works in this case, it does not in the more complex conﬁgurations discussed below. So, we prefer to use a reduction argument, where the actual geometric problem is reduced to an abstract intersection searching problem speciﬁed by a map Q/2S: It is easy to modify the construction used in Theorem 3.1 to derive Theorem 3.2. The modiﬁcations are sufﬁciently straightforward to dispense with a formal explanation. The idea is to raise the planar subdivision to form the ‘‘roof ’’ of a shed-like polytope in R3: Each hitting line is replaced by a plane passing through that line and perpendicular to the ground, then the set of planes is favorable for the box. A brief sketch follows: We begin with the polytope of Fig. 3, whose xy-projection reproduces the conﬁguration of segments in Fig. 1. Next, we replace each rectangle on the roof by a convex polygon with 2ðJn=on þ 1Þ edges, as shown in Fig. 4. (The bounding box is drawn only for illustration purposes.) ARTICLE IN PRESS δ l1 Fig. 2. Completing the convex subdivision. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 274 We patch the gaps between consecutive polygons on the roof by taking the convex hull (Fig. 5). Note that the polygonal chains p1; p2; y and q1; q2; y both lie in a plane but are not coplanar. Our construction is a lifted version of the hard convex subdivision of Fig. 2. The ﬂoor and vertical walls of the shed add only a constant number of edges to the intersection with any hitting plane, and so the same lower bound argument applies, leading to Theorem 3.2. 4. Fractional cascading in higher dimension We use the results of the previous section to show that 2D fractional cascading is impossible. Recall that fractional cascading is a general technique for speeding up lookup queries in catalog graphs . For our purposes, a catalog graph G is a connected bounded-degree graph, where each node v is associated with a catalog Cv (which is just a sorted list of numbers). The successor of x in Cv is deﬁned as minfyACv,fNg j yXxg: A query is speciﬁed by a key x and a connected subgraph H of G: Its answer is the list of successors of x in the catalogs associated with the nodes of H: Fractional cascading is a scheme that allows a query to be answered in time Oðk þ log nÞ; using OðnÞ storage, where k is the number of nodes in H and n is the combined size of all the catalogs. This improves upon the naive Oðk log nÞ query time solution. The construction of Fig. 1 tells the whole story: The graph G is a simple path whose nodes are associated with the Jon vertical segments. The catalog at a node consists of the lines dual to the ARTICLE IN PRESS ω Fig. 3. A ‘‘house’’ and its projection. p p p y y y q q q δ 1 2 3 1 2 3 1 2 3 2 Fig. 4. Carving out the details. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 275 midpoints of the Jn=on subsegments. We use the duality: ða; bÞ/y ¼ ax þ b and y ¼ ax þ b/ða; bÞ because it respects above/below relationships. In this way, a catalog appears as a collection of parallel strips, i.e., a planar convex subdivision (Fig. 6). A hitting query line dualizes to a query point, and the line/subsegment intersections identify which of the strips contain the dual point. Thus, 2D fractional cascading is seen to suffer from exactly the same lower bound as intersection searching for line/planar subdivision (Theorem 3.1). & Theorem 4.1. There is a graph with planar subdivisions attached to its nodes such that to perform the same point location query at the k nodes of a connected subgraph in time Oðk þ polylogðnÞÞ requires storage Oðn2Þ; where n is the combined size of all the catalogs. 5. The DK hierarchy Let P be an n-vertex convex polytope in R3 whose vertex set is VðPÞ: A sequence of convex polytopes, HðPÞ ¼ P0; y; Pk; is called a (polyhedral) hierarchy if: (i) P0 ¼ P and Pk is a simplex; (ii) Piþ1CPi and VðPiþ1ÞCVðPiÞ for 0piok; (iii) VðPiÞ\VðPiþ1Þ forms an independent set with respect to the facial graph of @Pi: The size of HðPÞ is deﬁned as Pk i¼0jVðPiÞj; its height is k; and its degree is max 0pipk # edges of Pi incident to a vertex of VðPiÞ\VðPiþ1Þ: ARTICLE IN PRESS gaps fixed 1 p 1 q2 p2 q 1 p q1 2 q p2 Fig. 5. Gaps are ﬁlled by taking the convex hull. v1 v2 G Fig. 6. A catalog graph in which 2D fractional cascading is impossible. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 276 The hierarchy is called Dobkin–Kirkpatrick (or DK for short) if its size is OðnÞ; its degree is Oð1Þ; and its height is Oðlog nÞ: It is well known that any polytope P admits of a DK hierarchy. In fact, a polytope will typically have an exponential number of distinct DK hierarchies. Of particular interest are the boundary dominant hierarchies: These have the additional property that, given any plane p; X k i¼0 jPi-pj ¼ OðjP-pj þ log nÞ; where jpolygonj denotes the number of vertices in the polygon. Intuitively, it means that any planar cross-section should essentially be like the 2D equivalent of a DK hierarchy: a collection of concentric layers with a fraction of the complexity on the outermost layer. It is folk knowledge that the existence of a boundary dominant DK hierarchy would carry with it all sorts of algorithmic beneﬁts: For example, we could use such a hierarchy to compute, in optimal time, the shadow of a polytope cast by a single light source. We dash all such hopes by exhibiting polytopes with no boundary dominant DK hierarchies. Of anecdotal interest, we should mention that this purely combinatorial question is resolved by using algorithmically inspired arguments. A remarkable feature of a DK hierarchy is that it lends itself naturally to navigation along piecewise linear curves and surfaces. Speciﬁcally, the pockets formed by Pi\Piþ1 can be triangulated so as to turn the whole polytope P into a simplicial cell complex C (with the standard glueing properties one expects of a cell complex). Given any plane p; one can ﬁnd, in Oðlog nÞ time, a starting point in P-p; and then explore every simplex of C that intersects p in a breadth ﬁrst search traversal that takes constant time per simplex . Suppose now that the DK hierarchy is boundary dominant. Because the pockets are each of constant size, the number of intersected simplices would be OðjP-pjÞ: So, we could compute the intersection between a query plane and a convex polytope in Oðk þ log nÞ time, using only OðnÞ storage. This would stand in contradiction with Theorem 3.2. & Theorem 5.1. For any n large enough, there exists a convex polytope with n vertices that admits of no boundary dominant DK hierarchy. 6. Nonconvex polytopes We consider the problem of intersecting a (possibly nonconvex) n-vertex polytope in R3 with a query plane. We prove a lower bound ﬁrst; then we give a nearly matching upper bound for the case where o ¼ Oðlog nÞ: Theorem 6.1. Given an n-vertex polytope in R3; to compute all k edges intersected by a query plane in Oðk þ oÞ time requires Oðn3e=o3Þ storage, where e is an arbitrarily small positive constant. Note that this implies our earlier claim, the Oðn3eÞ lower bound when oðnÞ is polylogarithmic or even of the form nd; for any constant d40 with e ¼ eðdÞ: From now on, e denotes an arbitrarily ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 277 small positive constant. Since the storage is at least linear in n; we may assume with no loss of generality that o=nð2eÞ=3 is small enough. Our proof relies on the construction of a ‘‘hard’’ polytope P and a set Q of query planes that is ðYðlog nÞ; oÞ-favorable for some designated edges of P that constitute the set S: For notational convenience, we use n as a parameter that differs from the number of vertices of P by at most a constant factor. As usual, we ensure that of all the edges intersected by any query plane of Q; at least a ﬁxed fraction of them are designated. To achieve a query time of OðjP-pj þ oÞ; we know from Lemma 2.3 that the size of the data structure should be OðjQjo=log nÞ: The input polytope P is carved out of the unit cube C ¼ ½0; 13 in several ‘‘carving’’ steps. Let e040 be a small enough absolute constant (i.e., with no dependence on e). To begin with, we choose Jon random points in the subsquare j2y 1jpe0; j2z 1jpe0 of the face x ¼ 0 of C; independently and uniformly, and join them to the face x ¼ 1 by x-parallel segments. Next, decompose each such unit-length segment s into J6n=on congruent subsegments. For technical reasons, we need to perform a random shift of this decomposition: For each s; pick a random real a uniformly between 1 2J6n=on1 and 1 2J6n=on1 and move each of the J6n=on 1 interior endpoints along s (not including the two endpoints at each end of s) by a (Fig. 7). Note that each interior endpoint in a given s is shifted by the same amount, but that the Jon random shifts are independent. The next step is to turn this conﬁguration of edges into a bona ﬁde polytope. First, make each horizontal segment into a cylinder with a tiny square base, say, of side length 1=n2: Next, attach these cylinders to the face x ¼ 0 but not to the face x ¼ 1; let it come very near the latter but not touch it. What happens to the (randomly shifted) decompositions? They naturally partition the boundary of each cylinder into rectangles. The polytope P consists of the unit cube with the protrusions through x ¼ 0 formed by the cylinders (Fig. 8). Each shifted endpoint gives rise to four vertices of P: all the x-parallel edges that are incident upon any of them are called designated and form the set S: Note that many faces are coplanar in this construction but it is routine to perturb the polytope P to make it simplicial and in general position. Trivally, Fact 6.2. The number of vertices of P and the number of designated edges are both in YðnÞ: Furthermore, no designated edge is of length larger than o=2n: Next, we deﬁne a large set Q of query planes that satisfy both the relevance and independence conditions (Deﬁnition 2.2). Choose t random points q1; y; qt uniformly in the square ARTICLE IN PRESS x z o y shift random Fig. 7. Carving P out of a cube. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 278 1 e0px; yp1 on the face z ¼ 1; where t ¼ Jn2e=o3n: Next, along each segment Oqi; add the points ð3jo=nÞqi; for all integers n=6oojon=5o: This deﬁnes a set fqg of Yðtn=oÞ points,2 which in turn speciﬁes the set Q of query planes, each deﬁned by the equation pq : /p; qS ¼ jjqjj2 2 (Fig. 9). In other words, pq is the plane passing through q and normal to Oq: Let Sq be the slab consisting of all points at distance at most o=n from the plane pq: By using a Heilbronn-type argument , we can prove along the lines of : Lemma 6.3. With probability 1 oð1Þ the query set Q satisﬁes the following property: for some ﬁxed c large enough, any subset of at least c log n planes in Q contains three planes pˆ q1; pˆ q2; pˆ q3 such that the volume of the polyhedron Sˆ q1-Sˆ q2-Sˆ q3 is Oð1=n1þeÞ: Note that the probability given in the lemma is over the initial random choices of the set Q of query planes, and the statement is over all large enough subsets of Q: Proof. We may allow ourselves a subset S of size c log t planes, since log t ¼ Oðlog nÞ: Recall that each plane is normal to some segment Oqi where qi is a point on the face z ¼ 1: Slabs normal to the same Oqi do not intersect, and so we may assume without loss of generality that the planes of S are normal to distinct segments Oq1; Oq2; y; OqjSj: Because the qi’s are chosen randomly in an e0-by-e0 square at z ¼ 1; they have the following Heilbronn-type property (proven in ). For c large enough, with probability 1 oð1Þ; the convex hull of fq1; y; qjSjg is a polygon of area Oðe2 0ðlog tÞ=tÞ: A triangulation of this convex hull produces at least one triangle of area Oðe2 0=ðctÞÞ; which we relabel q1q2q3: We now show that the corresponding planes pˆ q1; pˆ q2; pˆ q3 satisfy the ARTICLE IN PRESS Fig. 8. Turning transversals into protrusions. o q o n ω Fig. 9. The query set Q: 2By abuse of terminology we also use q to denote the vector Oq: B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 279 desired condition (note that ˆ qi is a point that lies on segment Oqi; for i ¼ 1; 2; 3): vol \ 3 i¼1 Sˆ qi ¼ Oð1=n1þeÞ: ð1Þ The polyhedron Sˆ q1-Sˆ q2-Sˆ q3 is spanned by w1; w2; w3 where, /wj; ˆ qiS ¼ ð2o=nÞjjˆ qijj2 if i ¼ j; and 0 otherwise: In other words, Sˆ q1-Sˆ q2-Sˆ q3 is the set of linear combinations x0 þ P i aiwi; for 0paip1; where x0 is the unique point satisfying /x0; ˆ qiS ¼ jjˆ qijj2 o n jjˆ qijj2 for 1pip3 (Fig. 10). Denote by ½w the 3-by-3 matrix ðw1; w2; w3Þ and deﬁne the matrices ½q and ½ˆ q similarly. Finally, let L be the diagonal matrix with Lii ¼ jjˆ qijj2: It follows from ½wT½ˆ q ¼ ð2o=nÞL that det½w det½ˆ q ¼ 2o n 3 Y 3 i¼1 jjˆ qijj2: Recall that qi (i ¼ 1; 2; 3) is a point in the small square in the plane z ¼ 1; and ˆ qi lies on Oqi: It then follows Q i jjˆ qijj2ðdet½qÞ ¼ Q i jjqijj2ðdet½ˆ qÞ; and each jjqijj2 is Yð1Þ; det½wdet½q ¼ 2o n 3 Y 3 i¼1 jjqijj2 ¼ Y o3 n3 : The determinant of ½q is, in absolute value, at least proportional to the area of triangle q1q2q3; which is Oð1=tÞ: By our choice of t ¼ Jn2e=o3n; jdet½wj ¼ Oðto3=n3Þ ¼ Oð1=n1þeÞ: Note that vol T i Sˆ qi ¼ jdet½wj; which proves (1). & Lemma 6.3 shows that, for any subset KDQ of size at least c log n; the volume of TfSq j pqAKg is itself in Oð1=n1þeÞ: We now verify that with high enough probability the relevance and independence conditions hold. Lemma 6.4. Any plane in Q intersects at least Jon designated edges. ARTICLE IN PRESS w1 w2 q2 q1 0 x Fig. 10. Intersection parallelotope. B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 280 Proof. It sufﬁces to show that, given any y; z with j2y 1jpe0 and j2z 1jpe0; the points p0 ¼ ð0; y; zÞ and p1 ¼ ð1; y; zÞ lie cleanly on opposite sides of any plane pˆ q of Q: (The adjective ‘‘cleanly’’ refers to the fact that this must remain true after turning segments into thin cylinders and perturbing P slightly.) Recall that ˆ q ¼ ðbu; bv; bÞ; where 1 e0pu; vp1 and 1 2obo3 5: We easily verify that for some small enough c040: /p0; ˆ qS jjˆ qjj2 2 ¼ bvy þ bz b2ðu2 þ v2 þ 1Þ ¼ b 2 þ Oðe0Þo c0: /p1; ˆ qS jjˆ qjj2 2 ¼ bu þ bvy þ bz b2ðu2 þ v2 þ 1Þ ¼ b 5 Oðe0Þ4c0: Lemma 6.5. With probability 1 oð1Þ; given any set of at least c log n planes in Q, at most a constant number of designated edges intersect all of them. Proof. Let K be a set of at least c log n planes in Q and let K be the (interior of the) polyhedron C- TfSq j pqAKg: Recall that to build P we choose random points on the face x ¼ 0 of C: Let ð0; y; zÞ be one of them and let c ¼ cðy; zÞ be the line passing through it parallel to the x-axis. The construction of P proceeds by choosing points on c at regular intervals of length l ¼ J6n=on1 and shifting the J6n=on 1 endpoints by a ﬁxed, random amount between l=2 and l=2: At the end of this process, let NðK; cÞ denote the number of endpoints in K-c: A simple calculation shows that the expected value of NðK; cÞ (over the random shift) is at most jK-cj=l; where jK-cj designates the length of the corresponding segment (we say ‘‘at most’’ and not ‘‘equal to’’ for the rare case where K intersects c very near the boundary of C). It follows that vol K ¼ Z ½0;12 jK-cðy; zÞj dy dz X l Z ½0;12 E NðK; cðy; zÞÞ dy dz: Restricting the integration domain to the e0-by-e0 squares to which each c is actually adjacent and identifying E NðK; cðy; zÞÞ with the conditional expectation of NðK; cÞ given c ¼ cðy; zÞ; we ﬁnd that vol KX l Z ½ð1e0Þ=2;ð1þe0Þ=22 E NðK; cðy; zÞÞ dy dz ¼ Oðe2 0lÞ Eðy;zÞ E ½NðK; cÞ j c ¼ cðy; zÞ ¼ Oðe2 0lÞ E NðK; cÞ; where c is a random x-parallel segment of the type used in the construction (i.e., with endpoints in the tiny squares at x ¼ 0; 1). It follows that Prob½NðK; cÞ40 ¼ O vol K e2 0l : ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 281 On the other hand, a simple geometric argument shows that jSq-cj ¼ ð2o=nÞ=cos y; where y is the angle between Oq and the x-axis. cos y ¼ qx=jjqjj2 ¼ Yð1Þ; and so jSq-cj ¼ YðlÞ and, therefore, any given c can contribute only Oð1Þ such endpoints. By Lemma 6.3, it follows that a given c contributes either no endpoint to K; or if it does, the number of endpoints is Oð1Þ and this event happens with probability Oðvol KÞ=e2 0l ¼ Oðe2 0 Þ=one: The choice of c is repeated Jon times independently, so the expected number is OðneÞ: By a Chernoff-type estimate ([2, Theorem A.1.12]), the probability that the number of endpoints in K exceeds neD is at most Oð1=DÞOðneDÞ: Setting D ¼ bne; for some large enough constant b ¼ bðeÞ; we ﬁnd that Oð1Þ endpoints lie in K with probability at most n10: The size of Q is Oðn3Þ; which bounds by Oðn9Þ the number of triplets of slabs that can be formed in Lemma 6.3. We can thus ensure that with high probability no K contains more than a constant number of endpoints. By Fact 6.2, we know that the designated edges of P are all of length at most o=2n and so at most Oð1Þ of them can intersect any set of query planes of size at least c log n: (The 2 in the denominator is an overly generous slack factor to account for the distortions resulting from turning c into a cylinder.) We observed earlier that, because of Lemma 2.3, to achieve a query time of OðjP-pj þ oÞ requires OðjQjo=log nÞ storage. By Lemmas 6.4 and 6.5, we now have a polytope P and a set Q of query planes that is ðYðlog nÞ; oÞ-favorable for the YðnÞ designated edges of P: The storage requirement is OðjQjo=log nÞ ¼ Oðn3e=o3 log nÞ; which, after suitably readjusting e; proves Theorem 6.1. & We prove a nearly matching upper bound for the case where o ¼ Oðlog nÞ: Theorem 6.6. Given an n-vertex polytope in R3; there is a data structure of size Oðn3Þ that allows us to compute all k edges intersected by a query plane in Oðk þ log nÞ time. Proof. Actually, our solution is more general than that: the input can be any set of line segments in R3: We begin with an Oðn4Þ solution, which we improve to Oðn3Þ in a second stage. We dualize the problem by transforming the endpoints of the input segments into 2n planes, using the polarity ax þ by þ cz ¼ 1: A line segment is dualized into a double wedge. We can preprocess the arrangement of 2n planes for fast point location so that, given a point (here, the dual of the query plane), the convex cell that contains it can be found in Oðlog nÞ time. If each cell keeps a list of the double wedges that enclose it, then an intersection query can be answered in time Oðk þ log nÞ; using Oðn4Þ storage. We use ﬁltering search to reduce the storage to Oðn3Þ: Think of the cells as forming the nodes of a graph, with an edge joining two nodes whose corresponding cells share a facet. By doubling each edge, we can form an Eulerian tour that visits all the nodes. Now observe that the wedge lists in the Oðn4Þ solution have a great deal of ‘‘coherence.’’ Indeed, ﬁx a double wedge and observe the ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 282 nodes of the tour in whose lists it appears: these nodes form intervals along the tour, whose endpoints correspond to an entry into or an exit from the wedge in question. Thus, there are only Oðn3Þ such intervals. So the problem is reduced to this: given m intervals on a line, where m ¼ Oðn3Þ; build a data structure of size OðmÞ; such that, given a query x; all k intervals that contain x can be reported in Oðk þ log mÞ time. The window list of does just that. & References P.K. Agarwal, J. Erickson, Geometric range searching and its relatives, in: B. Chazelle, J.E. Goodman, R. Pollack (Eds.), Advances in Discrete and Computational Geometry, Contemporary Mathematics, 223, Amer. Math. Soc., 1999, pp. 1–56. N. Alon, J.H. Spencer, The Probabilistic Method, 2nd Edition, Wiley-Interscience, New York, 2000. L. Arge, D.E. Vengroff, J.S. Vitter, External-memory algorithms for processing line segments in geographic information systems, Proceedings of the 3rd Annual European Symposium on Algorithm 1995, pp. 295–310. M.J. Atallah, R. Cole, M.T. Goodrich, Cascading divide-and-conquer: a technique for designing parallel algorithms, SIAM J. Comput. 18 (1989) 499–532. M. de Berg, M. van Kreveld, M. Overmars, O. Schwarzkopf, Computational Geometry, Springer, Berlin, 1997. E. Bertino, B. Catania, B. Shidlovsky, Towards optimal indexing for segment databases, Technical report, University of Milano, Italy, 1998. M.M. Buddhikot, S. Suri, M. Waldvogel, Fast layer-4 packet classiﬁcation using space decomposition techniques, Proceedings of the Protocols for High Speed Networks, Salem, MA, 1999. B. Chazelle, Filtering search: a new approach to query-answering, SIAM J. Comput. 15 (1986) 703–724. B. Chazelle, Lower bounds on the complexity of polytope range searching, J. Amer. Math. Soc. 2 (1989) 637–666. B. Chazelle, Lower bounds for orthogonal range searching: I. The reporting case, J. ACM 37 (1990) 200–212. B. Chazelle, An optimal algorithm for intersecting three-dimensional convex polyhedra, SIAM J. Comput. 21 (1992) 671–696. B. Chazelle, The Discrepancy Method: Randomness and Complexity, Cambridge University Press, Cambridge, 2000. B. Chazelle, H. Edelsbrunner, M. Grigni, L.J. Guibas, J. Hershberger, M. Sharir, J. Snoeyink, Ray shooting in polygons using geodesic triangulations, Algorithmica 12 (1994) 54–68. B. Chazelle, L.J. Guibas, Fractional cascading: I. A data structuring technique, II. Applications, Algorithmica 1 (1986) 133–191. B. Chazelle, B. Rosenberg, Simplex range reporting on a pointer machine, Comput. Geom.: Theory and Appl. 5 (1996) 237–247. F. Dehne, A. Ferreira, A. Rau-Chaplin, Parallel fractional cascading on hypercube multiprocessors, Comput. Geom.: Theory and Appl. 2 (1992) 141–167. D.P. Dobkin, D.G. Kirkpatrick, A linear algorithm for determining the separation of convex polyhedra, J. Algorithms 6 (1985) 381–392. M.T. Goodrich, Efﬁcient parallel techniques for computational geometry, Ph.D. Thesis, Department of Comput. Science, Purdue University, West Lafayette, IN, 1987. J. Hershberger, S. Suri, A pedestrian approach to ray shooting: shoot a ray, take a walk, J. Algorithms 18 (1995) 403–431. T.V. Lakshman, D. Stiliadis, High-speed policy-based packet forwarding using efﬁcient multi-dimensional range matching, Proc. ACM SIGCOMM (1998), 191–202. J. Matous ˇ ek, Geometric range searching, ACM Comput. Surv. 26 (1994) 421–461. K. Mehlhorn, Data Structures and Algorithms 3: Multidimensional Searching and Computational Geometry, Springer, Berlin, 1984. K. Mehlhorn, S. Na ¨ her, Dynamic fractional cascading, Algorithmica 5 (1990) 215–241. ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 283 K. Mehlhorn, S. Na ¨ her, H. Alt, A lower bound on the complexity of the Union-Split-Find problem, SIAM J. Comput. 17 (1988) 1093–1102. Y. Morimoto, T. Fukuda, S. Morishita, T. Tokuyama, Implementation and evaluation of decision trees with range and region splitting, Constraint 2 (1997) 163–189. S. Sen, Fractional cascading revisited, J. Algorithms 19 (1995) 161–172. R. Tamassia, J.S. Vitter, Optimal cooperative search in fractional cascaded data structures, Algorithmica 15 (1996) 154–171. R.E. Tarjan, A class of algorithms which require nonlinear time to maintain disjoint sets, J. Comput. System Sci. 18 (1979) 110–127. ARTICLE IN PRESS B. Chazelle, D. Liu / Journal of Computer and System Sciences 68 (2004) 269–284 284
3395	https://codegolf.stackexchange.com/questions/73143/knights-and-knaves	code golf - Knights and Knaves - Code Golf Stack Exchange Join Code Golf 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 Knights and Knaves Ask Question Asked 9 years, 7 months ago Modified9 years, 7 months ago Viewed 2k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. This is code-golf. In this challenge, we will be writing programs/functions that solve "Knights and Knaves" puzzles. Background You find yourself on an island ... etc. ... every person on the island except for you is either a knight or a knave. Knights can only make true statements. Knaves can only make false statements. I don't want to rigorously define "statement," but we will say a statement is anything which is either "true" or "false." Note that this excludes paradoxical sentences. For the purposes of this challenge, you will be coming across groups of islanders; they will make statements to you. Your task is to determine who is a Knight and who is a Knave. Input: You will be given (in any reasonable format) the following information: A list of the people present. They will be named with uppercase alphabet characters "A-Z". The limit on the number of people imposed by this will not be exceeded. The statements that each person makes. See below for important details about this. Output You will then output (in any reasonable format) what each person is. For example, if there were players A B C D and A is a knight, but the rest are knaves, you could output A: 1 B: 0 C: 0 D: 0 Important details: Uppercase alphabet characters A-Z refer to islanders. The characters 0 (zero) and 1 (one) refer to a "Knave" and a "Knight", respectively. (You can any other two non A-Z characters, as long as you specify) Each islander present may make any natural number of statements, or may choose to say nothing. The normal logical operators can be used in statements (IS, AND, OR, NOT). On top of this, De Morgan's Laws and Conditionals may be used. The following are examples of how they might be presented in a spoken puzzle followed by how they might be input into your program. ( on a more technical note. The "IS" operator is really used as containment (which isn't a logical operator). When I say "A is a Knight", I really mean "A is a member of the set of Knights". The true operator used would be 'ϵ'. For simplicity's sake, we will instead be using '='.) I use the following (you may use whatever, as long as it is reasonable and consistent): ^ AND v OR = IS ~ NOT => IMPLIES X: Person X claims that... Person Z could make any combination of any of the following types of statements: Person Z says that... Person A is a Knight. Z: A = 1 Person Q is a Knave. Z: Q = 0 I am a Knight. Z: Z = 1 Person A is a Knight OR Person B is a Knight. Z: ( A = 1 ) v ( B = 1) Person C is a Knight AND I am a Knave. Z: ( C = 1 ) ^ ( Z = 0 ) Person R is NOT a Knight. Z: ~( R = 1 ) On top of this, input may also use De Morgan's Laws It is NOT True that both person A and Person B are Knaves Z: ~( ( A = 0 ) ^ ( B = 0 ) ) It is False that either person A or person B is a Knight Z: ~( ( A = 1 ) v ( B = 1) ) Finally, conditionals and their negations may be used If I am a Knight, then person B is a Knave Z: ( Z = 1 ) => ( B = 0 ) It is NOT True that If person B is a Knight, Then Person C is a Knave. Z: ~( ( B = 1 ) => ( C = 0 ) ) Notes on conditionals Check out wikipedia for more info. A conditional statement takes the form p =>q, where p and q are themselves statements. p is the "antecedent " and q is the "consequent". Here is some useful info The negation of a condition looks like this: ~( p => q ) is equivalent to p ^ ~q A false premise implies anything. That is: if p is false, then any statement p =>q is true, regardless of what q is. For example: "if 2+2=5 then I am Spiderman" is a true statement. Some simple test cases These cases are given in the following fashion (1) how we would pose the problem in speech (2) how we might pose it to the computer (3) the correct output that the computer might give. Person A and Person B approach you on the road and make the following statements: A: B is a knave or I am a knight. B: A is a knight. Answer: B is a Knight and A is a Knight. Input: A B # Cast of Characters A: ( B = 0 ) v ( A = 1) B: A = 1 Output: A = 1 B = 1 Persons A, B, and F approach you on the road and make the following statements: A: If I am a Knight, then B is a Knave. B: If that is true, then F is a Knave too. Answer: A is a Knight, B is a Knave, F is a Knight. Input A B F A: ( A = 1 ) => ( B = 0 ) B: ( A = 1 ) => ( F = 0 ) Output: A = 1 B = 0 F = 1 Q, X, and W approach you on the road and make the following statements: W: It is not true that both Q and X are Knights. Q: That is true. X: If what W says is true, then what Q says is false. Answer: W and Q are Knights. X is a Knave. Input Q X W W: ~( ( Q = 1 ) ^ ( X = 1 ) ) Q: W = 1 X: ( W = 1 ) => ( Q = 0 ) Output W = 1 Q = 1 X = 0 There is a similar challenge from 3 years ago that focuses on parsing and does not contain conditionals or De Morgan's. And therefore, I would argue, is different enough in focus and implementation to avoid this being a dupe. This challenge was briefly closed as a dupe. It has since been reopened. I claim that this challenge is, first off, different in focus. The other challenge focuses on English parsing, this does not. Second it uses only AND and OR whereas this uses conditionals and allows for the solving of many more puzzles. At the end of the day, the question is whether or not answers from that challenge can be trivially substituted to this one, and I believe that the inclusion of conditionals and conditional negations adds sufficient complexity that more robust changes would need to be made in order to fit this challenge. code-golf Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:39 CommunityBot 1 asked Feb 15, 2016 at 4:32 LiamLiam 3,233 2 2 gold badges 24 24 silver badges 44 44 bronze badges 12 What can we conclude if a Knave says (B=1)=>(C=0)? ~((B=1)=>(C=0)) or (B=1)=>(C=1) or something else?njpipeorgan –njpipeorgan 2016-02-15 10:39:15 +00:00 Commented Feb 15, 2016 at 10:39 This is impossible to do in less than 5 minutes. This problem is known as SAT, and is exponential in complexity. Thus for n=26 in the general case (not 2 SAT), it is impossible to solve on a computer in a reasonable time.Labo –Labo 2016-02-15 10:50:33 +00:00 Commented Feb 15, 2016 at 10:50 Your first test case have 2 possible output. As you're using logical OR, it could be A:1 B:1 or A:1 B:0 because B's B=1 could be false while A would still be true.Katenkyo –Katenkyo 2016-02-15 12:35:23 +00:00 Commented Feb 15, 2016 at 12:35 @njpipeorgan If the Knave is B, he cannot say that. A false premise implies anything and therefore that statement would be true. If the Knave what a different character, you would take the negation, which is (B=1)^(C=1) as per how conditionals are normally dealt with Liam –Liam 2016-02-15 17:48:01 +00:00 Commented Feb 15, 2016 at 17:48 1 For those wondering, the real issue was because I was looking at the input query and he was looking at the worded puzzle. That has been fixed Cameron Aavik –Cameron Aavik 2016-02-16 05:31:34 +00:00 Commented Feb 16, 2016 at 5:31 \|Show 7 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Python 3, ~~450~~~~342~~ 307 bytes Edit: it turns out I forgot an import... My first solution takes advantage of having flexible naming for queries from functools import def g(c,r):c=c.split();l,y=len(c),range;d=[dict((c[i],n>>i&1)for i in y(l))for n in y(2l)];return[nfor n in]+'('+s[2:].replace('->','<1or')for s in r)+')')),d[i]]for i in y(len(d))]if n] You can call the above one with g('Q X W', ['W: not( ( Q == 1 ) and ( X == 1 ) )','Q: W == 1', 'X: ( W == 1 ) -> ( Q == 0 )']) The next one here uses the same format of queries as seen in the OP, it also doesn't have some of the modifications I made to the first one. It is 417 bytes because it converts between the two formats. from functools import def g(c,r):c=c.split();l,y=len(c),range;d=[{dict((c[i],n>>i&1)for i in y(l)),{'v':'or','^':'and','=':'==','~':'not'}}for n in y(2l)];f=lambda r,c:reduce(lambda x,y:x.replace(y,str(c[y])),c,('(0<1'+''.join([')^ '+['~',''][c[t]]+'('+tfor t in[s.split(":")for s in r]])+')').replace('=>','<1or'));return[dict((o,j) for o,j in n.items() if o in c) for n in And it can be called by: g('Q X W', ['W: ~( ( Q = 1 ) ^ ( X = 1 ) )','Q: W = 1', 'X: ( W = 1 ) => ( Q = 0 )']) They both return [{'X': 0, 'W': 1, 'Q': 1}] Ungolfed Explanation: from functools import def knight_and_knaves(cast,rules): # turns 'A B C' into ['A','B','C'] cast = cast.split() # gets all numbers that can fit in len(cast) bits bitmasks = range(2 len(cast)) # for every bitmask, apply the value for a bit to the boolean value for each cast member. # This returns a dictionary of all possible outcomes. d=[dict((cast[i], n>>i & 1) for i in range(len(cast))) for n in bitmasks] # Split rules at colon rules = [s.split(":")for s in rules] # Turns list of rules into one python expression, joins each rule with ')and ', maybe a 'not' depending on if the hypothesis has the rule as a lie, and '('. # Also replaces '->' with '<1or' which is equivalent to it. Also starts with '(True' and ends with ')' to resolve missing parentheses transform_rules = lambda d, rules: ('(True' + ''.join([')and ' + ['not', ''][d[rule]] + '(' + rule.replace('->','<1or') for rule in rules]) + ')') # Applys transform_rules on each outcome and evaluates the result, storing it into a list of lists where each element is [outcome, value] outcomes=[[d[i],eval(reduce(lambda x,y:x.replace(y,str(d[i][y])),d[i],transform_rules(d[i], rules)))] for i in range(len(d))] # Filters outcomes if value is True return[nfor n in outcomes if n] Also, the second solution needs 3.5 (not 3.4) due to the use of PEP 448 Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 17, 2020 at 9:04 CommunityBot 1 answered Feb 16, 2016 at 4:51 Cameron AavikCameron Aavik 773 4 4 silver badges 8 8 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Mathematica, 80 bytes F[c_,s_]:=Select[Thread[c->#]&/@{True,False}~Tuples~Length@c,And@@Equal@@@s/.#&] Explanation The function F takes two arguments, c is a list of all characters' names, s is a list of statements, each of which contains two parts - who says what. For example, there are three characters, Q, X and W. characters={q,x,w}; And they say, statements= {{w, !((q==True)&&(x==True)) }, {q, w==True }, {x, Implies[w==True,q==False] }}; where True and False means Knights and Knaves respectively. Then F[characters, statements] will give {{q->True, x->False, w->True}}, which means there is only one solution that Q and W are Knights while X is a Knave. If there are more than one solution, the output will look like {{...},{...},...} The algorithm is very simple. {True,False}~Tuples~Length@c gives all possible combinations of Knights and Knaves among the characters. Then Thread[c->#]&/@ construct an array of rules based on these combinations. In the case of two characters A and B, the array will be {{a->True, b->True }, {a->True, b->False}, {a->False,b->True }, {a->False,b->False}} Substituting the statements with one row of these rules, we will get an array looks like {{True,True},{True,False},{False,False}} The first column of this array is the identities of the speakers, and the second column indicates whether their statements are true or false. A valid solution requires the accordance between speakers' identities and their statements. The array above means that this combination is not a solution, since the second speaker, a Knight, makes an incorrect statement. Select[...,And@@Equal@@@s/.#&] does the substitutions and select those combinations that satisfy the condition. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Feb 16, 2016 at 6:33 answered Feb 16, 2016 at 3:37 njpipeorgannjpipeorgan 3,112 15 15 silver badges 15 15 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Ruby, 128 This is the same algorithm as everyone else, try all possible combinations of knaves and knights and see which sticks. I have another that I'm working on, but I think it'll be longer (though more interesting). The statement inputs must be: & AND \| OR == IS ! NOT > IMPLIES X: Person X claims that... I also require each statement and sub-statement to be in parentheses. The only problem with this version is that it goes through at most 2^26 iterations, and if they're not all knaves, at least 2^(26-n) iterations! To put that in perspective, that means that if there are two people, and at least one is not a knave, it will take a minimum of 16,777,216 iterations! To limit that, I submit the following at 168 bytes. Sub in 26 for #{o.size} to cut it back to 161. ruby ->s{o=s[/.?$/].split i=0 eval h=o.zip(("%0#{o.size}b"%i+=1).chars).map{\|k\|k?=}?;until h&&o.all?{\|t\|!s[/#{t}:(.)$/]\|\|eval("(#{t}<1)^(#{$1.gsub(?>,'!=true\|\|')})")} h} But if I can instead use an array of people and a map of statements e.g.: ruby c[[?A, ?B], { ?A=> "( B == 0 ) \| ( A == 1)", ?B=> "A == 1" } ] Then I get it down to 128. ruby ->o,s{i=0 eval h=o.zip(("%026b"%i+=1).chars).map{\|k\|k?=}?;until h&&s.all?{\|t,k\|eval("(#{t}<1)^(#{k.gsub(?>,'!=true\|\|')})")} h} Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Feb 24, 2016 at 14:38 answered Feb 24, 2016 at 5:57 Not that CharlesNot that Charles 1,965 12 12 silver badges 26 26 bronze badges Add a comment\| Your Answer If this is an answer to a challenge… …Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. 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3396	https://stackoverflow.com/questions/2807771/find-the-intersection-of-two-parallel-collinear-line-segments	geometry - Find the intersection of two parallel (collinear) line segments - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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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 Find the intersection of two parallel (collinear) line segments [closed] Ask Question Asked 15 years, 4 months ago Modified4 months ago Viewed 7k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Closed. This question is not about programming or software development. It is not currently accepting answers. This question does not appear to be about a specific programming problem, a software algorithm, or software tools primarily used by programmers. If you believe the question would be on-topic on another Stack Exchange site, you can leave a comment to explain where the question may be able to be answered. Closed 7 months ago. The community reviewed whether to reopen this question 5 months ago and left it closed: Original close reason(s) were not resolved Improve this question I understand there are many algorithms to determine whether two line segments intersect. However, many of these algorithms simply return "No" when the segments are parallel, without checking whether the line segments are collinear. If they are collinear, there can be overlap or shared endpoints. I know I can compute the distance between the two segments—if it's zero, I know the segments are collinear and can then check whether their endpoints overlap. But this approach requires numerous if-else conditions with logical operators (&&, \|\|). While this isn't difficult, my question is: Is there a algorithmic trick or formula to handle this special case of parallel line segments more efficiently? geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Apr 22 at 7:54 rasputin 395 2 2 silver badges 14 14 bronze badges asked May 11, 2010 at 2:49 JudarknessJudarkness 274 4 4 silver badges 10 10 bronze badges 5 3 What exactly are you trying to find? By definition, parallel lines never intersect.Tyler –Tyler 2010-05-11 02:53:41 +00:00 Commented May 11, 2010 at 2:53 2 Parallel lines never intersect unless the distance is 0. But since their distance is 0, they are overlaped. However, my question is about the line segments. The stretched lines are overlapped, but the line segments are remain unknown. So I try to ask a solution to reveal the unkown.Judarkness –Judarkness 2010-05-11 03:08:04 +00:00 Commented May 11, 2010 at 3:08 1 Judarkness you probably need to refine your question a little bit. What does it matter if they are line segments or not if we are checking for parallelism?WhirlWind –WhirlWind 2010-05-11 03:23:05 +00:00 Commented May 11, 2010 at 3:23 1 Parallel line segments can only overlap if they are collinear, meaning that all 4 endpoints lie on the same straight line.rasputin –rasputin 2025-03-10 15:37:52 +00:00 Commented Mar 10 at 15:37 5 This question is being discussed on Meta.F1Krazy –F1Krazy 2025-05-06 15:47:02 +00:00 Commented May 6 at 15:47 Add a comment\| 7 Answers 7 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. I can't say there is a "trick" but there is indeed an efficient way of knowing if two collinear line segment intersect. Let's assume we have two collinear line segments : [AB] and [CD] Translating coordinates to 1D First of all, we need to translate the coordinates of the points into a 1D environnement, because it will be way easier to compare the positions of the points. The process is rather straight forward : we will shave off from our coordinates the y and z values. It works because there is no two points on a line which have the same x value (as long as the line segment does not belong to a plane orthogonal to the x axis). But if the points all have the same x value, then we will instead take the y coordinate. And if the points all have the same y coordinates, we will account for the z coordinate. And if we arrived at that point, the points cannot all have the same z coordinate. The method used to know if all points have the same x/y coordinate is to calculate the vector AB. If AB.x equals 0, it means that we don't have to move on the x axis to get from point A to point B, thus the points A and B have the same x value, and so is it for the points C and D. I am mostly talking about 3D here, but the point is the same for 2D, 4D, 5D... The code would look like if AB.x == 0 if AB.y == 0 A = A.z B = B.z C = C.z D = D.z else A = A.y B = B.y C = C.y D = D.y else A = A.x B = B.x C = C.x D = D.x Detecting the collision Detecting the collision is also quite straight forward in the end. There is exactly 52 different relative positions the points A, B, C and D can have (including 2 points having the exact same coordinate). Among these 52 positions, there are 44 that indicate there is a collision and only 8 that indicate the contrary. Thus the obvious solution is to only look for those 8 positions. But we can reduce this number to 2. Indeed, if we consider that the coordinate of A and B, and C and D are in ascending order, there are only 13 different relative positions the four points could have (see Notes section). And among these 13 positions, only 2 indicate that there isn't any collision. Those 2 positions are when D has a lower x value than A (C-D A-B) or when C has a greater x value than B (A-B C-D). So after sorting the points A and B, and C and D, we would return false if D < A or B < C and true otherwise, which gives us return not( D < A or B < C ) or in a more simplified way : return A <= D and C <= B This final part of the algorithm should look like : if A < B if C < D return A <= D and C <= B else return A <= C and D <= B else if C < D return B <= D and C <= A else return B <= C and D <= A Notes The 13 different positions A, B, C and D can have are : ┏━━━━━━━━━━━━┳━━━━━━━━━━━━┳━━━━━━━━━━━━┓ ┃ C D ┃ C D ┃ C D ┃ ┃ AB ┃ A B ┃ A B ┃ ┃━━━━━━━━━━━━╋━━━━━━━━━━━━╋━━━━━━━━━━━━┫ ┃ C D ┃ C D ┃ C D ┃ ┃ A B ┃ A B ┃ A B ┃ ┣━━━━━━━━━━━━╋━━━━━━━━━━━━╋━━━━━━━━━━━━┫ ┃ C D ┃ C D ┃ C D ┃ ┃ A B ┃ A B ┃ AB ┃ ┣━━━━━━━━━━━━╋━━━━━━━━━━━━╋━━━━━━━━━━━━┫ ┃ C D ┃ C D ┃ C D ┃ ┃ A B ┃ A B ┃ A B ┃ ┣━━━━━━━━━━━━╋━━━━━━━━━━━━╋━━━━━━━━━━━━┫ ┃ C D ┃ ┃ ┃ ┃ AB ┃ ┃ ┃ ┗━━━━━━━━━━━━┻━━━━━━━━━━━━┻━━━━━━━━━━━━┛ Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Jun 12, 2023 at 14:43 LygenLygen 76 3 3 bronze badges Comments Add a comment This answer is useful 2 Save this answer. Show activity on this post. Yes, given the formulas for both of the lines, test whether their slopes are equal. If they are, the lines are parallel and never intersect. If you have points on each of the lines, you can use the slope formula. If both are perpendicular to the x-axis, they will both have infinite slopes, but they will be parallel. All points on each line will have equal x coordinates. To deal with line segments, calculate the point of intersection, then determine if that point of intersection exists for both of the segments. 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 11, 2010 at 3:10 answered May 11, 2010 at 2:52 WhirlWindWhirlWind 14.1k 3 3 gold badges 44 44 silver badges 42 42 bronze badges 7 Comments Add a comment Chris Cooper Chris CooperOver a year ago @Tyler: But I would add what my original answer says: I think determining if a given point is on both lines should be an easier way to distinguish the two cases than finding the distance bewteen two lines. 2010-05-11T02:57:50.227Z+00:00 0 Reply Copy link Judarkness JudarknessOver a year ago Make it simple, my line segments are lengthed lines, but not infinite straight lines. Excep the intersection or not, I have all other info like slope, distance, vector. That's for the other calcualtions :) 2010-05-11T03:13:02.923Z+00:00 0 Reply Copy link WhirlWind WhirlWindOver a year ago Your question is really confusing then. If two line segments are parallel, they don't intersect. Isn't that what you asked? That's what I thought, since you wrote: "Is there a trick( or mathematics) method to calculate this special parallel case?" 2010-05-11T03:20:12.953Z+00:00 0 Reply Copy link David Gelhar David GelharOver a year ago @whirlwind he's talking line segments, not lines. The question is, for 2 (colinear) parallel line segments, how to determine if there is overlap? 2010-05-11T03:33:00.757Z+00:00 3 Reply Copy link WhirlWind WhirlWindOver a year ago I didn't see the picture... or the word "colinear." His edit resolves that. 2010-05-11T13:12:25.393Z+00:00 0 Reply Copy link Add a comment\|Show 2 more comments This answer is useful 0 Save this answer. Show activity on this post. I assume the case you're interested in is where the two line segments are parallel (as determined by checking the slope, as Whirlwind says), and you're trying to determine whether the two segments overlap. Rather than worrying about the distance between the lines, I would think the easiest way to do that would to if either endpoint of one segment lies within the other: if (segment_contains_point(segment_A, segment_B.start) \|\| segment_contains_point(segment_A, segment_B.end)) { // lines overlap } Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered May 11, 2010 at 3:05 David GelharDavid Gelhar 27.9k 3 3 gold badges 69 69 silver badges 87 87 bronze badges 1 Comment Add a comment Judarkness JudarknessOver a year ago This is just what I want, I have my own segment_contains_points()(in other name) But It's way too inefficient. And I have to call it three times because segment B might be larger and contains the whole segment A. And I want to know if there is a better solution? 2010-05-11T03:21:06.98Z+00:00 2 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. I just got the same problem: The easiest way I have come up with just to check whether the lines overlap: Assuming the segments are colinear (parallel and have the same intersection with the x axis). Take one point A from the longer Segment (A,B) as starting point. Now find the point among the other three points that has the minimal distance to point A (squared Distance is better, even manhattan-length might work too) measuring the distance in the direction of B. If the closest point to A is B, the lines do not intersect. If it belongs to the other segment they do. Perhaps you have to check for special cases like zero length lines or identical lines but this should be easy. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 23, 2012 at 16:59 answered Jun 11, 2012 at 16:16 MartinMartin 4,890 4 4 gold badges 31 31 silver badges 60 60 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Parallel line segments can only overlap if they are collinear, meaning that all 4 endpoints lie on the same straight line. If the line segments are parallel but not collinear, the 4 endpoints will not lie on a straight line, but will instead form a trapezoid. Checking whether parallel line segments are collinear If one segment is defined by points a and b, and the other segment by points c and d, you can check whether the segments are collinear by calculating the area of the triangle formed by any 3 of the points. Area = 0.5 abs(a(b-c) + b(c-a) + c(a-b)) If the area is 0, then the 3 points are collinear. (And, since the segments are parallel, the fourth point is as well.) If the area is non-zero, then the segments are parallel, but not collinear, and the segments do not intersect. Since we don't care what the area of the triangle is (we only want to know whether it's 0), we can simplify the formula a bit more: if a(b-c) + b(c-a) + c(a-b) != 0: return False # segments are not collinear and do not intersect Check for overlap between collinear line segments Now that we know our line segments are collinear, we can determine whether they overlap at all. ``` sort lexicographically. this will sort the points along the line. aligned = sorted([a, b, c, d]) check if the collinear line segments share an endpoint if aligned == aligned: return aligned check if the collinear line segments don't overlap if set(aligned[:2]) == {a, b} or set(aligned[:2]) == {c, d}: return False return the middle 2 points. the lines overlap between these points. return aligned[1:3] ``` There are 4 steps here: Sort the points along the collinear line. Since the points are collinear, we can sort them lexicographically. This just means we sort by x first. If x is identical for all points (i.e. the collinear line is vertical), then we sort by y. Compare the middle 2 points to determine if the segments share an endpoint. If they do, return that point. Test if the segments overlap by comparing the first two sorted points to the two segments. If the first two points is identical to one of the segments, then that means the segments don't overlap and we return False. If the segments do overlap, we return the middle two points. These points correspond to the overlapping portion of the segments. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Mar 11 at 18:45 answered Mar 7 at 21:18 rasputinrasputin 395 2 2 silver badges 14 14 bronze badges 1 Comment Add a comment rasputin rasputinMar 11 at 15:45 See my complete function for finding the intersection of line segements here 2025-03-11T15:45:50.607Z+00:00 0 Reply Copy link This answer is useful -1 Save this answer. Show activity on this post. Let's assume that you have two lines described by formulas a.x + b.y + c = 0 and d.x + e.y + f = 0. The two lines are parallel when a = 0 and d = 0 or b/a = e/d. Perhaps instead of doing the division just make sure that b.d = a.e. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered May 11, 2010 at 3:13 Ricardo MarimonRicardo Marimon 10.8k 9 9 gold badges 55 55 silver badges 62 62 bronze badges Comments Add a comment This answer is useful -1 Save this answer. Show activity on this post. I found this (modified a little by me to suit), it will return the intersection x,y else if no intersection found it will return -1,-1 ``` Public Function intersection(ByVal ax As Integer, ByVal ay As Integer, ByVal bx As Integer, ByVal by As Integer, ByVal cx As Integer, ByVal cy As Integer, ByVal dx As Integer, ByVal dy As Integer) As Point '// Determines the intersection point of the line segment defined by points A and B '// with the line segment defined by points C and D. '// '// Returns YES if the intersection point was found, and stores that point in X,Y. '// Returns NO if there is no determinable intersection point, in which case X,Y will '// be unmodified. Dim distAB, theCos, theSin, newX, ABpos As Double '// Fail if either line segment is zero-length. If ax = bx And ay = by Or cx = dx And cy = dy Then Return New Point(-1, -1) '// Fail if the segments share an end-point. If ax = cx And ay = cy Or bx = cx And by = cy Or ax = dx And ay = dy Or bx = dx And by = dy Then Return New Point(-1, -1) '// (1) Translate the system so that point A is on the origin. bx -= ax by -= ay cx -= ax cy -= ay dx -= ax dy -= ay '// Discover the length of segment A-B. distAB = Math.Sqrt(bx bx + by by) '// (2) Rotate the system so that point B is on the positive X axis. theCos = bx / distAB theSin = by / distAB newX = cx theCos + cy theSin cy = cy theCos - cx theSin cx = newX newX = dx theCos + dy theSin dy = dy theCos - dx theSin dx = newX '// Fail if segment C-D doesn't cross line A-B. If cy < 0 And dy < 0 Or cy >= 0 And dy >= 0 Then Return New Point(-1, -1) '// (3) Discover the position of the intersection point along line A-B. ABpos = dx + (cx - dx) dy / (dy - cy) '// Fail if segment C-D crosses line A-B outside of segment A-B. If ABpos < 0 Or ABpos > distAB Then Return New Point(-1, -1) '// (4) Apply the discovered position to line A-B in the original coordinate system. 'X=Ax+ABpostheCos 'Y=Ay+ABpostheSin '// Success. Return New Point(ax + ABpos theCos, ay + ABpos theSin) End Function ``` Origin 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 6 at 16:12 BDL 22.3k 33 33 gold badges 57 57 silver badges 64 64 bronze badges answered Jul 31, 2010 at 10:36 RobertRobert 89 1 1 silver badge 4 4 bronze badges 1 Comment Add a comment Wil Shipley Wil ShipleyOver a year ago This doesn't solve the question of determining if colinear line segments overlap—in fact in the comments to this code it specifically says it'll fail for colinear (and parallel) lines: // Fail if segment C-D doesn't cross line A-B 2013-08-03T10:06:08.903Z+00:00 0 Reply Copy link Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry See similar questions with these tags. 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3397	https://brainly.com/question/15338615	[FREE] The formula for any arithmetic sequence is a_n = a_1 + d(n - 1), where: - a_n represents the value of the - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +74,1k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +14,8k Ace exams faster, with practice that adapts to you Practice Worksheets +8,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified The formula for any arithmetic sequence is a n=a 1+d(n−1), where: a n represents the value of the nth term, a 1 represents the value of the first term, d represents the common difference, and n represents the term number. What is the formula for the sequence 10, 8, 6, 4, ...? A. a n=10+2(n−1) B. a n=2+10(n−1) C. a n=−2+10(n−1) D. a n=10+(−2)(n−1) 2 See answers Explain with Learning Companion NEW Asked by angelwolfrum18 • 03/26/2020 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 57861 people 57K 4.9 21 Upload your school material for a more relevant answer Answer: an = 10 + (-2)(n-1) Step-by-step explanation: We are given from the sequence first term, a1 = 10 common difference, d = a2 - a1 Where; a2 = second term = 8 d = 8 - 10 = -2 From the formula, an = a1 + d(n - 1), We substitute the value of a1 and d therefore, an = 10 + (-2)(n - 1) Answered by preciousamadasun35 •147 answers•57.9K people helped Thanks 21 4.9 (19 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 60962503 people 60M 4.9 12 College Physics 1e - OpenStax Mechanics - Benjamin Crowell Introductory Physics - Building Models to Describe Our World - Ryan D. Martin, Emma Neary, Joshua Rinaldo, Olivia Woodman Upload your school material for a more relevant answer The formula for the arithmetic sequence 10, 8, 6, 4, ... is a n=10+(−2)(n−1). This shows that the first term is 10 and the common difference is -2. Therefore, option D is the correct choice. Explanation To find the formula for the arithmetic sequence 10, 8, 6, 4, ..., we start by identifying the key components of the sequence: First term (a 1): This is the first number in the sequence, which is 10. Common difference (d): To find this, we subtract the second term from the first term. The second term is 8, so: d=a 2−a 1=8−10=−2 Term number (n): This is the position of the term in the sequence. For example, the first term is when n=1, the second term when n=2, and so on. Now we can substitute these values into the formula for the nth term of an arithmetic sequence, which is given by: a n=a 1+d(n−1) Substituting in our values: a n=10+(−2)(n−1) Which simplifies to: a n=10−2(n−1) Thus, among the options provided, the correct answer for the formula of the sequence is: D. a n=10+(−2)(n−1) Examples & Evidence For example, to find the 3rd term of the sequence, substitute n=3 into the formula: a 3=10+(−2)(3−1)=10−4=6 which is the third term in the sequence. The formula for an arithmetic sequence is well-established in mathematics, showing a constant difference between consecutive terms. In this case, checking the common difference and first term confirms the formula is derived correctly. Thanks 12 4.9 (13 votes) Advertisement Community Answer This answer helped 60962503 people 60M 4.9 12 Answer: an = 10 + (-2)( n - 1) Step-by-step explanation: In an arithmetic progression, the consecutive terms differ by a common difference. The formula for determining the nth term of an arithmetic sequence is expressed as an = a + (n - 1)d Where a represents the first term of the sequence. d represents the common difference. n represents the number of terms in the sequence. Looking at the given sequence, a = 10 d = 8 - 10 = 6 - 8 = - 2 Therefore, the formula for the sequence is an = 10 + (n - 1)- 2 an = 10 + (-2)( n - 1) Answered by Favouredlyf •6.4K answers•61M people helped Thanks 12 4.9 (13 votes) Advertisement ### Free Mathematics solutions and answers Community Answer The formula for any arithmetic sequence is an = a1 + d(n - 1), where a represents the value of the nth term, a1 represents the value of the first term, d represents the common difference, and n represents the term number. What is the formula for the sequence -15, -11, -7, ...? an = -15 + (-4)( n - 1) an = -15 + ( n - 1) an = 4 + (-15)( n - 1) an = -15 + 4( n - 1) Community Answer The formula for any geometric sequence is an = a1 · rn - 1, where an represents the value of the nth term, a1 represents the value of the first term, r represents the common ratio, and n represents the term number. What is the formula for the geometric sequence 1, -2, 4, -8, ...? an = 1 · (-2)n - 1 an = -2 · 1n - 1 an = -8 · (-2)n - 1 an = 1 · 2n - 1 Community Answer The formula for any geometric sequence is an = a1 · rn - 1, where an represents the value of the nth term, a1 represents the value of the first term, r represents the common ratio, and n represents the term number. What is the formula for the sequence 2, -6, 18, -54, ...? an = 2 · 3 n - 1 an = 2 · (-3) n - 1 an = -3 · 2 n - 1 an = 3 · 2 n - 1 Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Using the Pythagorean theorem, find the length of a leg of a right triangle if the other leg is 8 feet long and the hypotenuse is 10 feet long. A. 41 f t B. 12.81 ft. C. 6 ft. D. 36 ft. Hence, for 0⩽u⩽360 and 0⩽v⩽36 0∘, solve the simultaneous equations 2 sin u+cos v=2 sin u−cos v=−2 1. [2 2 x−3]={3 x−2}. According to the Rational Root Theorem, the following are potential roots of f(x)=2 x 2+2 x−24:−4,−3,2,3,4 Which are actual roots of f(x)?A. -4 and 3B. −4,2, and 3C. −3,2, and 4D. -3 and 4 According to the Rational Root Theorem, −8 7 is a potential rational root of which function? A. f(x)=24 x 7+3 x 6+4 x 3−x−28 B. f(x)=28 x 7+3 x 6+4 x 3−x−24 C. f(x)=30 x 7+3 x 6+4 x 3−x−56 D. f(x)=56 x 7+3 x 6+4 x 3−x−30 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
3398	https://digitalcommons.library.umaine.edu/fac_monographs/76/	"Plant Systematics: A Phylogenetic Approach " by Walter S. Judd, Christopher S. Campbell et al. Menu Home Search Browse Collections My Account About Digital Commons Network™ Skip to main content Home About FAQ My Account Contacts <Previous Next> Home>Research Centers and Institutes>FAC_MONOGRAPHS>76 Faculty and Staff Monograph Publications Plant Systematics: A Phylogenetic Approach Authors Walter S. Judd Christopher S. CampbellFollow Peter F. Stevens Michael J. Donoghue Files Description A comprehensive introduction to vascular plant phylogeny, the Third Edition of Plant Systematics reflects changes in the circumscription of many orders and families to represent monophyletic groups, following the most recent classification of the Angiosperm Phylogeny Group. The taxonomic evidence described includes data from morphology, anatomy, embryology, chromosomes, palynology, secondary plant compounds, proteins, and DNA. Molecular taxonomic methods are fully presented, as are the results of many recent studies, both molecular and morphological. A chapter on the history of plant classification puts current systematic methods into historical context. Issues relating to variation in plant populations and species, including speciation and species concepts, polyploidy, hybridization, breeding systems, and introgression are carefully considered. Appendices cover botanical nomenclature as well as field and herbarium methodology. ISBN 9780878934072 Publication Date 2008 Publisher Sinauer Associates City Sunderland, MA Keywords Plants, Classification Disciplines Biology \| Botany Comments Third edition Recommended Citation Judd, Walter S.; Campbell, Christopher S.; Stevens, Peter F.; and Donoghue, Michael J., "Plant Systematics: A Phylogenetic Approach" (2008). Faculty and Staff Monograph Publications. 76. Buy this Book Find @ the Library DOWNLOADS Since January 17, 2014 PlumX Metrics Usage Abstract Views: 2507 see details Share FacebookLinkedInWhatsAppEmailShare Search Enter search terms: Select context to search: Advanced Search Notify me via email or RSS Browse Collections Disciplines Authors Contacts Contact the Repository Author Corner Author FAQ Elsevier - Digital Commons Home \| My Account \| Accessibility Statement \| Send Feedback PrivacyCopyrightDigitalCommons@UMaine ISSN: 2476-2547 ✓ Thanks for sharing! AddToAny More… We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookie Details List‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Targeting Cookies [x] Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. If you do not allow these cookies, you will experience less targeted advertising. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices
3399	https://math.stackexchange.com/questions/3421170/which-satisfy-the-system-of-inequalities-below	algebra precalculus - Which satisfy the system of inequalities below: - 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 Which satisfy the system of inequalities below: Ask Question Asked 5 years, 11 months ago Modified5 years, 8 months ago Viewed 155 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. For each nonnegative integer n n, calculate the number of triples (a,b,c)(a,b,c) of nonnegative integers which satisfy the system of inequalities below: ⎧⎩⎨a+b≤2 n a+c≤2 n c+b≤2 n{a+b≤2 n a+c≤2 n c+b≤2 n What I thought: We can solve this by plotting the inequalities with the bounds x,y,z≥0 x,y,z≥0 and getting that all such (a,b,c)(a,b,c) are lattice points bounded by the axis and x+y+z=2 n x+y+z=2 n. algebra-precalculus inequality integer-lattices Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 8, 2020 at 2:30 RobPratt 51.8k 4 4 gold badges 32 32 silver badges 69 69 bronze badges asked Nov 4, 2019 at 6:14 Lambert macuseLambert macuse 4,264 10 10 silver badges 22 22 bronze badges 4 2 Unfortunately your approach will miss some triples, such as (n,n,n)(n,n,n).Greg Martin –Greg Martin 2019-11-04 06:40:07 +00:00 Commented Nov 4, 2019 at 6:40 2 If you fix a a, can you find a formula for the number of legal pairs b,c b,c? Can you then sum this formula over all a a?Greg Martin –Greg Martin 2019-11-04 06:41:32 +00:00 Commented Nov 4, 2019 at 6:41 The shape produced is a right tetrahedron.Lambert macuse –Lambert macuse 2019-11-04 06:45:52 +00:00 Commented Nov 4, 2019 at 6:45 1 You must mean non-negative integers, not non-integer.Macavity –Macavity 2019-11-04 09:41:45 +00:00 Commented Nov 4, 2019 at 9:41 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The constraints define an n n-fold dilation of the 3-dimensional polytope with vertices (0,0,0)(0,0,0), (1,1,0)(1,1,0), (1,0,1)(1,0,1), and (0,1,1)(0,1,1). The number of lattice points is hence a cubic Ehrhart polynomial. By inspection, the counts are 1,11,42,106 1,11,42,106, for n=0,1,2,3 n=0,1,2,3, respectively. The resulting polynomial is hence 2 n 3+9 n 2 2+7 n 2+1.2 n 3+9 n 2 2+7 n 2+1. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 8, 2020 at 2:26 RobPrattRobPratt 51.8k 4 4 gold badges 32 32 silver badges 69 69 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Consider the triples (a,b,c)(a,b,c) of nonnegative integers that satisfy the following relationships: ⎧⎩⎨a+b=x a+c≤2 n b+c≤2 n{a+b=x a+c≤2 n b+c≤2 n Let's divide the count of these triples into two steps. In the first we consider x=2 k x=2 k, while in the second we consider x=2 k+1 x=2 k+1. Thus, the (a,b)(a,b) pairs that satisfy a+b=2 k a+b=2 k are (2 k,0),...,(k,k),...,(0.2 k)(2 k,0),...,(k,k),...,(0.2 k). For each of these pairs, c c will have to obey c≤2 n−M c≤2 n−M where M=max(a,b)M=max(a,b), which gives us 2 n−M+1 2 n−M+1 solutions. Assuming that among the (a,b)(a,b) pairs that satisfy the equation, the value of M M ranges from k+1 k+1 to 2 k 2 k twice and then goes to k k, so the number of solutions for this case it will be: (2∑M=k+1 2 k(2 n−M+1))+2 n−k+1(2∑M=k+1 2 k(2 n−M+1))+2 n−k+1 =−3 k 2+4 n k+2 n+1=−3 k 2+4 n k+2 n+1 For the second case, be n=2 k+1 n=2 k+1. The difference is that we will not have the extra solution in which the components of pair (a,b)(a,b) are equal. Like this: (2∑M=k+1 2 k+1(2 k+1−M+1))(2∑M=k+1 2 k+1(2 k+1−M+1)) =−3 k 2+(4 n−3)k+4 n=−3 k 2+(4 n−3)k+4 n So just calculate the sum of all cases where a+b=0 a+b=0, a+b=1 a+b=1,… up to a+b=2 n a+b=2 n and just get the sums for when 2 k+1=1,3,5,…,2 n−1 2 k+1=1,3,5,…,2 n−1 and sum with the sums for when 2 k=0,2,4,…,2 n 2 k=0,2,4,…,2 n: ∑k=0 n 3 k 2+4 n k+2 n+1+∑k=0 n−1−3 k 2+(4 n−3)k+4 n∑k=0 n 3 k 2+4 n k+2 n+1+∑k=0 n−1−3 k 2+(4 n−3)k+4 n =2 n 3+9 n 2 2+11 n 2+1=2 n 3+9 n 2 2+11 n 2+1 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 8, 2020 at 1:47 Lambert macuseLambert macuse 4,264 10 10 silver badges 22 22 bronze badges 1 The count for n=1 n=1 is 11. Looks like your coefficient of n n is wrong.RobPratt –RobPratt 2020-01-08 02:27:40 +00:00 Commented Jan 8, 2020 at 2:27 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 inequality integer-lattices 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 0Inequalities (System) (solving it) 0Solving system of inequalities with floor function 6Algorithm to solve a system of linear inequalities over the natural numbers 0Find all possible ordered triples for the system 5Solve the system of linear inequalities with parameters 4Determining the number of solutions of a system of linear inequalities. 2How many integer solutions satisfy x 2+y 2≤20 x 2+y 2≤20? 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